[contemporary mathematics] chapel hill ergodic theory workshops volume 356 ||

CONTEMPORARY MATHEMATICS

356

Chapel Hill Ergodic Theory Workshops

June 8-9, 2002 and February 14-16, 2003

University of North Carolina Chapel Hill , NC

ldris Assani Editor

Licensed to George Mason Univ. Prepared on Sun Mar 17 12:07:53 EDT 2013 for download from IP 129.174.21.5.

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CoNTEMPORARY MATHEMATICS

356


June 8-9, 2002 and February 14-16,2003

University of North Carolina Chapel Hill, NC

ldris Assani Editor

American Mathematical Society Providence, Rhode Island



Editorial Board Dennis DeTurck, managing editor

Andreas Blass Andy R. Magid Michael Vogelius

2000 Mathematics Subject Classification. Primary 11K55, 28D05, 37 A30, 37B20, 42Al6, 47 A35, 60Fl5.

Library of Congress Cataloging-in-Publication Data Chapel Hill ergodic theory workshops: June 8-9, 2002 and February 14-16, 2003, University

of North Carolina, Chapel Hill, NC / ldris Assani, editor. p. em.- (Contemporary mathematics, ISSN 0271-4132; 356)

Includes bibliographical references. ISBN 0-8218-3313-8 (alk. paper) 1. Ergodic theory-Congresses. I. Assani, ldris. II. University of North Carolina at Chapel

Hill. III. Contemporary mathematics (American Mathematical Society); v. 356.

QA313.C48 2004 515'.48-dc22 2004046326

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10 9 8 7 6 5 4 3 2 1 09 08 07 06 05 04



I would like to dedicate this volume to my mother who passed away on December 2, 2003.



Contents

Preface ix

Why is the 3X + 1 problem hard? 1 ETHAN AKIN

Lectures on Cantor and Mycielski sets for dynamical systems 21 ETHAN AKIN

Duality and the one sided ergodic Hilbert transform 81 I. ASSANI

Rigidity conditions in topological dynamics related to a theorem of George ~ll 91

JOSEPH AUSLANDER AND KENNETH BERG

On strong laws of large numbers with rates 101 GUY COHEN, ROGER L. JONES, AND M. LIN

Besicovitch weights and the necessity of duality restriction in the weighted ergodic theorem 127

CIPRIAN DEMETER AND ROGER L. JONES

Strong sweeping out for lacunary sequences 137 ROGER L. JONES

Some old and new Rokhlin towers 145 ISAAC KORNFELD

vii



Preface

In the summer of 2002 the editor of this volume began organizing Ergodic theory workshops at the University of North Carolina at Chapel Hill. The second workshop was held on February 2003. A third workshop was organized in February 2004. The present volume includes contributions from some of the participants at the 2002 and 2003 workshops. The list of participants outside Chapel Hill to both workshops is the following: Lina Avramidou*, Joe Auslander, Ken Berg, Guy Cohen*, Ciprian Demeter*, Hillel Furstenberg, Geoffrey Goodson, Anish Gosh*, Roger Jones, Jonathan King, Isaac Kornfeld, Bryna Kra, Michael Lin, Igor Mezic, Kimberly Presser, and Arkady Tempelman. The names asterisks were graduate students at the time of these workshops. Some of them seized on this opportunity and together with more senior researchers submitted joint papers to this volume.

The goals of these workshops are mainly the following: 1) to allow interested young researchers (graduate students, Post Doctoral stu-

dents and Assistant professors) to be introduced to active research areas.

2) to energize the graduate program at UNC Chapel Hill. To this effect the speakers were asked to make the first 15 minutes of their talks understandable to advanced graduate students.

3) to promote the interaction between established researchers and younger re-searchers.

4) to target minority and underrepresented groups and facilitate their possible insertion into the graduate program at UNC Chapel Hill.

The success of these two workshops has allowed us to make these February Er-godic theory workshops an annual event. The most recent one was held in February 2004. We are welcoming the support of the MSRI for the forthcoming February 2005 and 2006 workshops.

It is a pleasure to acknowledge the institutions that made these events possible through their support. First we thank the National Science Foundation for its continued support. Thanks also to the Department of Mathematics, the Deans of the College of Arts and Sciences and the Vice Chancellor for research at UNC Chapel Hill. We would like also to acknowledge the energetic support of the staff and some of the interested graduate students in our Department of Mathematics.

Finally we appreciate the support of the American Mathematical Society and its Contemporary Mathematics Staff. Special thanks to Christine Thivierge for her smooth handling of the publication process.

More information about the workshops can be found at the following URL:

ix



X PREFACE

www.math. unc.eduiFaculty I assaniiMiniConferencel ergodicminiconf.html

and at www.math. unc.eduiFaculty I assaniiErgodic Workshop lindex.html

Idris Assani



Contemporary Mathematics Volume 356, 2004

WHY IS THE 3X + 1 PROBLEM HARD?

BY ETHAN AKIN

"It works over the two-adics." - D. Sullivan

1. Introduction

Start with an odd natural number x. Multiply by 3 and add 1. From the resulting even number, divide away the highest power of 2 to get a new odd number T(x). If you keep repeating this operation do you eventually hit 1, no matter what odd number you began with?

Simple to state, this problem, called the 3X + 1 conjecture, remains unsolved. The operation T is easy to program on even a hand cal-culator. The numbers dance about in a tantalizing fashion, at times appearing to take off towards infinity but finally dropping down to 1.

One evening after dinner, Dennis Sullivan and I nibbled on this old chestnut. After the remark quoted above, he added "It illustrates the difficulty of describing particular orbits in an ergodic system." At the time I didn't see what he meant and the conversation meandered off to other subjects.

Over the next two weeks I finally saw the ergodic theory perspective which Dennis had been pointing along. This viewpoint, worked out by him and David Ruelle over lunch one day, does not solve the problem. Instead, it suggests why it is hard to solve.

In what follows I hope I can inspire you to share the delight I found in this peculiar reinterpretation of such an apparently simple system.

2. The Two-adic Integers

Because of the divisions by 2 in the operation it is easiest to deal with the natural numbers by writing them in base 2. So we think of a

Date: November 2002. Mathematics Department, The City College, 137 Street and Convent A venue

New York City, NY 10031 AMS subject classification Primary 37B20 Secondary 11K55.

© 2004 American Mathematical Society

http://dx.doi.org/10.1090/conm/356/06495



2 ETHAN AKIN

natural number as a finite sequence a0a1a2 ... with each ai = 0 or 1. To add:

(2.1) aoa1a2 .. .

+ bob1b2 .. .

first add a0 + b0 by 0 + 0 = 0; 0 + 1 = 1 + 0 = 1; and 1 + 1 = 0, carry the 1 to the next place. Then add a1 + b1 and the carry if any. Notice that I am writing the base 2 digits, i.e. the bits, in the reverse of the standard order so that you carry to the right.

For multiplication we can use the fact that each bi is 0 or 1 to write: aoa1a2 .. .

x bob1b2 .. .

bo · a0 a1a2 ...

(2.2) bl . aoal ...

b2. ao ...

+ where we just cross out row i if bi = 0 and include it if bi = 1. Notice now that these operations will work just fine even if the se-

quences are of infinite length. In multiplication, for example, because the rows are receding to the right, the ith place of the answer will in-volve adding at most 2i 1 's ( i + 1 rows and less than i carries). The infinite sequences of O's and 1 's with this notion of + and x are called two-adic integers and the uncountable set of them is denoted Z2 •

Since we generalized from the set of natural numbers, it is perhaps surprising that subtraction is always defined. Subtraction is again by the grade school rule but the trick is that you are allowed to "borrow" 1 from infinity if necessary. Alternatively, notice that

1111.. .. (2.3) + 1000 .. ..

0000 .. ..

and so -1 = 1111... Now for any sequence a = a0a1 a2 ... define a by ai = 0 if ai = 1 and vice-versa. Clearly, a + a = 1111... = -1. So a+ (a+ 1) = 0, or

(2.4) -a a + 1.

From our original conception it is clear that a natural number, 0, 1, 2, ... is a 2-adic integer whose expansion terminates in a string of O's. From (1) it follows that the negative integers -1, -2, ... can be identified



WHY IS THE 3x + 1 PROBLEM HARD? 3

with the 2-adic integers whose expansions terminate in a string of 1's. We will reserve for the word integer its usual meaning, i.e. the natural numbers and their negatives, and refer to typical elements of 2 2 as 2-adics.

As you might expect from the peculiar folding together of positive and negative integers, the order relation on the integers does not extend to the 2-adics.

The distinction between even and odd does extend. Call a 2-adic a even if a0 = 0 and odd if a0 = 1. Since multiplication by 2 = 0100 ... introduces a zero at the left end, i.e.

(2.5) it is clear that a number is even iff it is divisible by 2. You cannot divide and odd 2-adic by 2.

However, you can divide by any odd 2-adic: bo

(2.6) - boclc2···

dld2···

where c1c2 ... = b0 x a1a2 ... Then continue the usual long division rou-tine. Alternatively, we can construct the reciprocal 1/ a for a odd by reversing the multiplication algorithm. For example:

(2.7)

1 1100000 .. . 1 110000 .. . 0 00000 .. . 1 0 1

1100 .. . 000 .. .

11...

+ 1000000 ...

shows that for 3 = 11000 ... , 1/3 = 11010101... Also, 3 x (101010 ... ) = 111... so that -1/3 = 101010 ...

We can also build the set of 2-adics by an inverse limit construction using congruence mod 2k. This bit of abstract algebra provides a useful complement to the previous algorithmic approach.

Recall that two integers are congruence mod 2k if their difference is divisible by 2k. Because + and x preserve congruence, these operations



4 ETHAN AKIN

can be defined on the mod 2k equivalence classes yielding the ring of integers mod 2k, denoted Z/2k. Two natural numbers a and b are congruent mod 2k, written a b mod 2k, precisely when the base 2 expansions of a and b agree in the first k places. The mod 2k congruence classes can be represented by the 2k possible initial strings of k O's and 1 's. To add or multiply mod 2k we just proceed as usual and ignore the results after the kth bit.

Because congruence mod 2k implies congruence mod 2k-1 there is an obvious restriction map Pk : Z/2k --+ Z/2k-1 which forgets the last bit. The map preserves + and x, i.e. it is a ring homomorphism.

This business of ignoring everything after the kth bit works for infinite sequences as well and so defines a map Z2 --+ Z/2k which preserves + and x. For a E Z2 we define [a]k to be the mod 2k congruence class of a:

So [a]k = [b]k iff ai = bi fori = 0, 1, ... , k -1. Notice that the restriction map corresponds to set inclusion: (2.9)

Pk[a]k = [b]k-1 {=::::} a b mod 2k-1 {=::::} [a]k C [b]k-1.

Z2 is the inverse limit of the sequence of rings Z/2k and the con-necting homomorphisms Pk· This means, first, that each a E Z2

is uniquely described by the coherent sequence of congruence classes {[a]k: k = 1, 2, ... }where a sequence {ak E Z/2k: k = 1, 2, ... }is called coherent when pkak = ak_1. This much is true of the natural numbers as well. But for Z2 the correspondence to coherent sequences of con-gruence classes is onto as well as one-to-one. That is, every coherent sequence { ak} determines a 2-adic. As we run along the coherent se-quence the new information provided by ak given ak-1 is precisely the kth bit in the expension. It tells us whether a 0 or a 1 goes in the 2k-1 place.

This identification between the elements of Z2 and coherent sequences makes it easy to check what we have so far merely presumed implic-itly: the usual arithmetic rules are true for Z2 . The commutative, associative and distributive laws are inherited from the rings Z/2k.

Finally, we will call a 2-adic x rational if it can written x = afb for some integers a, b with b odd. We conclude this section with a series of exercises which shows that a 2-adic x is rational iff it is terminally periodic, that is, there exists a positive integer K such that Xi+K = Xi for all sufficiently large i. If K is the smallest such integer then K is called the period of x. If xi+K = xi for all i then x is called periodic.




We have already seen that the integers are precisely those 2-adics which are terminally periodic of period 1, i.e. which are either eventually 0 or eventually 1. 0 = 000 ... and -1 = 111 ... are the two 2-adics which are periodic of period 1. Exercise 1. (a) If x is periodic of period K then x = -a/(2K -1)

for some natural number a (Hint: Compute x- 2Kx). (b) Compute the expansion of -1/(2K- 1). (c) Assume that x and y are terminally periodic and that n is an

integer. Prove that x + y, -x and n · x are terminally periodic. (d) Prove: If x is terminally periodic then x is rational (Hint: Write

x = n + 2Ly with n a nonnegative integer andy periodic). (e) If n is an odd integer then n divides 2K - 1 for some K (Hint:

Use the Euler ¢ function. Let K = cp(n) and use Fermat's Theorem from elementary number theory.)

(f) Prove: If x is rational then x is terminally periodic.

3. The Two-adic Shift Map

Before wrestling with the operation T itself let us simplify (grossly) by omitting the multiplication by 3. Starting with an odd natural number x, divide away the highest power of 2 in x + 1 to get the odd number S(x). Clearly, S(1) = 1 while S(x) < x if x > 1. So the sequence of iterates of S, {Sn(x)} decreases monotonically to 1.

Instead of performing all of the divisions by 2 at once we can do them one at a time by defining:

(3.1) s(x) { x/2 (x + 1)/2

if xis even if xis odd.

Suppose we are given a set X and a function q : X --+ X. We can define a dynamical system on X by iterating q. We imagine that the points of X evolve according to the rule Xt+I = q(xt) so that with each tick of the clock each point moves to its image under the mapping q. For x EX the q-orbit of xis the sequence in X: {x, q(x), (q o q)(x), ... } = {qn(x): n = 0, 1, 2, ... }.

Starting with an odd natural number x, look at the s-orbit of x. The S-orbit of x is precisely the subsequence of odd numbers in the s-orbit. If s(x), ... , sk-1(x) are all even and sk(x) is odd then k is called the first return time to the set of odds and S(x) = sk(x).

Because even 2-adics can be uniquely divided by 2 definition (2.1) works for the 2-adics as well, defining a function s : z2 --+ z2. It will be more convenient to conjugate by -1 and define a-(x) = -s( -x). So



6

a : z2 ---t z2 is defined by:

(3.2) a(x)

ETHAN AKIN

{ x/2 (x- 1)/2

if x is even if x is odd.

a is called the shift map on Z2 because you get the base 2 expansion of a( x) by deleting x0 and shifting the remaining bits left one place, I.e.

(3.3) fori = 0, 1, ...

So for the a-orbit of x: x = a0(x), a 1 (x), a 2 (x), ... it is clear that

(3.4) Xi ai(x)0 fori = 0, 1, .. .

which we can restate as:

(3.5) {0 if ai(x) is even 1 if a2 (x) is odd.

Thus, the base 2 expansion of x can be thought of as a coded tape describing the successive parities along the a-orbit of x.

While the map a is clearly onto, it is not one-to-one. In fact, for any y E Z2 the equation a(x) = y has precisely two solutions correponding to the possible values of the initial digit of x deleted by a. We can describe these by defining:

(3.6) &o(Y) 2y and 2y+ 1.

So we see that &E(y) is just y shifted right one place withE inserted in the now vacant 2° place. Thus, starting from any point in Z2 there are two different ways of moving backwards one step: an even way, &0 and an odd way &1.

Let us look at some special a-orbits. xis a nonnegative integer iff ai(x) = 000 ... = 0 fori sufficiently large.

On the other hand, x is a negative integer iff ai ( x) = 111 ... = -1 for i sufficiently large. 0 and -1 are the only fixed points of a, the only solutions of a(x) = x. The integers are those 2-adics whose orbits eventually arrive at one of the fixed points.

a has one cycle of period 2. Recall that -1/3 = 101010 ... and so -2/3 = 010101... Hence, a(-1/3) = -2/3 and a( -2/3) = -1/3.

In general, x is periodic iff x is a fixed point for some iterate of a, i.e. aK ( x) = x for some positive integer K. The orbit of x returns to x after K iterates and thereafter repeats the cycle. That is, the base 2 expansion of x is periodic iff the a-orbit of x is periodic. The results of Exercise 1 say that x is rational iff its a-orbit eventually reaches such a periodic point and then enters a cycle.



WHY IS THE 3x + 1 PROBLE!\1 HARD'? 7

4. The Two-adic 3X + 1 Map

Instead of looking at T directly it will be convenient to introduce the single step map T which is related to T as s, defined by ( 2.1), was to S.

( 4.1) T(x) {x/2 if xis even (3:r + 1)/2 if xis odd.

As before, the T-orbit of x is the subsequence of odd numbers in the T-orbit of x. So the original problem is equivalent to the conjecture that Ti(x) = 1 for some i whenever xis a positive integer. Notice, though, that 1 is not a fixed point forT. Instead, T(1) = 2 and T(2) = 1. So the T-orbit of 1 is a cycle of period 2.

Definition ( 3.1) extends as before to define a map T : z2 -----+ z2. Again we look for the sequence encoding the successive parities along the T-

orbit. For the shift map the base 2 expansion of x gave the coding. For T it is no longer true that the expansion of :r bears any simple relation to the coding. Instead we define the function Q : Z2 -----+ Z2 by:

(4.2)

or, equivalently:

( 4.3)

for i = 0, 1, ...

{0 if T~(x) is even 1 if T 2 (x) is odd.

Again T maps Z2 onto itself and the equation T(x) = y two solutions described by the two functions:

( 4.4) 7o(Y) 2y and 71 (y) (2y- 1)/3,

moving backwards along T-orbits either the even way or the odd way. Exactly because we have lost the simple relationship between the

expansion and the map, it is worthwhile pausing here to see how the 7/s are used.

Inductively, for a 0 ... ak_ 1 a list of O's and 1's of length k, we define:

(4.5) 7a0 ... ak_ 1 (Y) =def 7a0 (7a1 ... a~,_ 1 (y)).

In other words, we start with y and move backwards the ak_ 1 way, then the ak-2 way, .. . and finally the ao way, arriving at x = 7a0 ... ak-l (y). Now if we start at x and move forward via T fork steps we are back at y and along the way the parities are given by a0 , a 1 , .... Thus, Tk ( x) = y and

(4.6) a· ~ for i = 0, 1, ... , k - 1.



8 ETHAN AKIN

We now show that Q defines a conjugacy between T and the shift map a. This means that Q provides a recoding or change of coordinates on Z2 and under this change of coordinates T is transformed into a.

Theorem 1. The function Q is a one-to-one map of Z2 onto itself. It defines a conjugacy between T and a. That is, we have following equation of composed maps:

(4.7) QoT (J 0 Q.

Proof: The conjugacy relation (3. 7) is easy because for all x and i

(4.8) Q(T(x))i = Ti(T(x))o = Ti+l(x)i+l = a(Q(x))i

by definitions (2.4) and (3.2). The hard part is to prove that Q is one-to-one and onto. We use the

correspondece between the 2-adics and the coherent sequences of mod 2k congruence classes. We need a bit of algebraic spadework.

Notice first that for k ;::: 1: (4.9) x- x' mod 2k I

Xo = Xo and a(x) _ a(x') mod 2k-l.

This just says that the first k bits agree iff the initial bits do and then the next k - 1 bits agree as well.

The analogous result is true for T as well. That is, for k ;::: 1: (4.10) x- x' mod 2k {:::=::} x0 = x~ and T(x) _ T(x') mod 2k-l.

This time we have to use a bit of algebra. Clearly, x x' mod 2k implies x0 = x~. If this common value is 0 then T(x) = x/2 = a(x) and T(x') = x' /2 = a(x'). So in that case (3.10) follows from (3.9). Now suppose that x0 = x~ = 1. If we let T(x) = y and T(x') = y' then we are going backwards the odd way and sox= 1\(y) = (2y- 1)/3 and x' = 1\(y') = (2y'- 1)/3. Thus, x- x' = 2(y- y')/3. Now y - y' mod 2k-l, i.e. y- y' is divisible by 2k-l iff 2(y- y') is divisible by 2k. Multiplying by 1/3 we see that this is true iff 2(y- y')/3 is divisible by 2k, i.e. iff x- x' mod 2k.

From these equivalences we derive: Proposition 2. For any y E Z2 , Q-1 ([Y]k) = {x E Z2 : Q(x)- y mod 2k} is a single mod 2k congruence class. In particular, for x, x' E Z2 .

(4.11) x- x' mod 2k Q(x)- Q(x') mod 2k.

Proof: If wE z2 and X= TYO···Yk-1 (w) then by (3.6) Q(x)- y mod 2k. Thus, the set Q-1 ([Y]k) is nonempty. It suffices to demonstrate (3.11) in order to complete the proof.




We prove (3.11) by induction on k. For the initial step, k = 0, observe that x0 = Q(x)0 and x~ = Q(x')0. For the inductive step we use the conjugacy equation (3. 7) which we have already proved.

First assume x _ x' mod 2k. Apply (3.10) to get x0 = x~ and 7(x) 7(x') mod 2k-I. By inductive hypothesis, Q(7(x))- h(7(x')) mod 2k-I_ The conjugacy equation (3.7) allows us to rewrite this as a(Q(x)) _ a(Q(x')) mod 2k-I_ Meanwhile, Q(x)0 = x0 = x~ = Q(x')0. So (3.9) implies Q(x) - Q(x') mod 2k. Furthermore, the reasoning we have just used is completely reversible to prove the implication the other way. This completes the proof of Proposition 2. D

We use Proposition 2 to complete the proof of Theorem 1 by showing that given y E Z2 there exists a unique x E .Z2 such that Q(x) = y. Proposition 2 implies that {Q-1 ([Y]k) : k = 0, 1, ... } is a sequence of mod 2k congruence classes. Recall from (1.9) that coherence of the sequence just says that the sequence of sets is monotonically decreas-ing. Because this is true for the sequence {[y]k} it is clearly true for { Q-1([y]k)}. The coherent sequence { Q-1([y]k)} corresponds to the unique x E .Z2 defined by {x} = nkQ-1([Y]k)· Because Q(x) _ y mod 2k for all k, Q(x) = y. D

The conjugacy h transforms 7-orbits to a-orbits. To see this, do an easy induction on (3.7) to get

(4.12) fori= 0, 1, 2, ...

So if we apply Q to the 7-orbit: x, 7(x), 72 (x), ... we get the a-orbit of Q(x): Q(x), a(Q(x)), a 2(Q(x)), ... We use this to see that for any positive integer K, Q(7K(x)) is exactly Q(x) truncated by the removal of the initial K bits.

In order to understand how 7 behaves on a subset N of .Z2 , we need only compute Q(N). In particular, when N is the set of positive integers our problem becomes transformed as follows. Proposition 3. If x is a positive integer then 7i(x) = 1 for some i iff the 2-adic -3Q(x) is a positive integer, in which case, as a positive integer it is relatively prime to 3.

Proof: For 2 and 1/3 the 7-orbits are, respectively: 2, 1, 2, 1, ... and 1/3, 1, 2, 1, ... Consequently:

(4.13) Q(1) Q(2) Q(1/3)

101010 ... 010101...

110101...

-1/3 -2/3

1/3



10 ETHAN AKIN

Now suppose that x is an integer greater than 2 and ri(x) = 1 for some i. Let k be the smallest such i and let y = Q(x). Because ak(Q(x)) = Q(rk(x)) = Q(1), we have Q(x) = YoY1···Yk-d010 ... and

(4.14) Q(rk-1(x)) = ak- 1 (Q(x)) = Yk-11010 ...

Were Yk- 1 = 1 this would mean that Q(rk-1(x)) would be 1/3 and so rk-1(x) = 1/3 which is impossible for integral x. Hence, Yk- 1 = 0. Similarly, Yk- 2 = 0 for if not then Q(rk-2(x)) = 1010 ... = Q(1) and k was defined to be the smallest i such that ri ( x) = 1.

Thus, if we let n be the nonnegative integer with binary expansion YoY1···Yk- 3000 ... then Q(x) = n+2k·(-1/3) and so -3Q(x) = 2k-3n.

If -3Q(x) were divisible by 3 as a whole number then Q(x) would be a negative integer whose expansion terminates in a string of 1 's. But -1 is a fixed point for rand so Q( -1) = 111... = -1. As -1 cannot lie on the r-orbit of a positive integer like x, it follows that Q(x) cannot terminate in a string of 1 's.

Finally, it is easy to check that if Q(x) = -a/3 with a a positive integer prime to 3 then the expansion of Q ( x) terminates in the cycle 1010 ... Since Q is one-to-one the r-orbit of x then terminates in the cycle 1, 2, 1, 2, .... D

While suggestive, these results are not as helpful as they might ap-pear to be. Unfortunately, the only way to compute Q(x) appears to be to use the definition (3.3) which requires knowledge of the entire r-orbit of x.

One idea is to compute Q(x) mod 2k. By (3.11) we can define the bijection [Q]k : Z/2k--+ Z/2k by [Q]k([x]k) = [Q(x)]k. Perhaps we can discern a pattern from these finite approximations.

But probably not. It is time to raise the difficulty which underlies this whole approach to the original problem: The method is too general. It will work just as well if instead of using the map T we define, for a any odd 2-adic, the map Ta : Z2 --+ Z2 by

(4.15) Ta(x) { x/2 if x is even (ax+ 1)/2 if xis odd.

We can then define Qa : z2 --+ z2 by replacing T by Ta in (3.2) and (3.3), i.e.

(4.16) Qa(x)i r~(x)o fori = 0, 1, ... The analogue of Proposition 2 with Q repaced by Qa is still true

with the same proof (replace multiplication by 1/3 with multiplication




by 1/a). Just as before Qa is a one-to-one onto map with

(4.17) QaoTa aoQa.

With a= 5, for example, 13, 33, 83,208, 104,52, 26, 13 and 1, 3, 8, 4, 2, 1 are disjoint cycles. The original problem appears to depend delicately upon the choice of a = 3. However, this approach might be useful for a portion of the problem. Proposition 4. Let a be an odd 2-adic. For X E z2 theTa-orbit even-tually enters a cycle iff Qa(x) is a rational2-adic.

Proof: If K, L are positive integers and r~+K ( x) = T~ ( x) for all i 2: L then by (3.16), Qa(x)i+K = Qa(x)i for all i 2: L. Thus, Q a ( x) is terminally periodic and so is rational by Exercise 1. Con-versely, suppose that Qa(x) is terminally periodic so for some K, L, Qa(x)i+K = Qa(x)i for all i 2: L. Conjugacy with the shift map implies that Qa(T~+K(x)) = Qa(r~(x)) for all i 2: L. Since Qa is one-to-one, it follows that r~+K ( x) = T~ ( x) for all i 2: L. Thus, the Ta-orbit of x is eventually cyclic. 0

Theorem 5. Let a be an odd, rational2-adic. If for X E z2, Qa(x) is rational then x is rational.

Proof: By Proposition 4, the Ta-orbit of x is eventually periodic. That is, there exist positive integers K, L such that T~+K(x) = r~(x) for all i 2: L. Let y = rf(x) = rf+K (x) so that T~+K (y) = T~(y) for all nonegative integers i.

Following (3.4) define

(4.18) Tao(Y) 2y and Ta1(Y) (2y- 1)/a.

Notice that the coefficients are rational because a is rational. Let h0 ... hK_1 be the first K bits of Qa(y). The periodicity of Ta-orbit

of y implies that

(4.19) y

This in turn says that y is the solution of an equation:

(4.20) y 2Kc1y + c2

with c1 and c2 rational. Hence, y = -c2/ (2K c1 - 1) is rational. Finally, let k0 ... k L-1 be the first L bits of Q a ( x). Just as before we

have

(4.21) X



12 ETHAN AKIN

Hence,

( 4.22) X

with d1 and d2 rational. Thus, x is rational as well. 0

We now consider a possible converse to Theorem 5. Rationality Conjecture 6. Let a be an odd, rational 2-adic integer. If for X E z2, X is rational then Qa(x) is rational. That is, the map Qa : z2 ---+ z2 preserves rationality.

This conjecture is too much to hope for. Its truth depends on the value of a, and can be decided in many cases. Theorem 7. ( 1) The Rationality Conjecture is true for a = 1 or -1.

(2) The Rationality Conjecture is false when a is an odd, rational 2-adic that is not an ordinary integer.

Proof: ( 1) If a is an ordinary integer, then the denominators of the iterates of any rational x under Ta never increase. If a = 1 or a = -1 then IT(x)l ~ lx~ 1 . This means iterates monotonely decrease in ab-solute value until they enter the interval [-1, 1], and all subsequent iterates remain in this interval. But given a bound B on the denom-inator, there are only finitely many rationals in this interval having denominator below B. Thus the iterates of x are eventually periodic, and Proposition 4 shows that Qa(x) is rational.

(2) Let a= E in lowest terms, with p and q odd, and q > 1. Proposi-q

tion 4 says that Q a ( x) is rational if and only if x is eventually periodic under Ta, so it suffices to exhibit an odd rational number x which is not eventually periodic under iteration. We show that x = a = ~ has this property. This time under iteration by Ta, the denominators of the iterates of x are nondecreasing, and strictly increase every time some iterate is an odd 2-adic integer, then adding an extra power of q. Since an infinite number of odd iterates occur, the denominators increase indefinitely and hence the orbit of x is not eventually periodic. D.

The remaining cases where the Rationality Conjecture is not settled are those where a is an odd integer with lal 2: 3. This includes the case a = 3 corresponding to the 3X + 1 problem. For integer a, the only way an orbit of a rational x cannot be eventually periodic is that its iterates tend to infinity (in absolute value when viewed in IR), which we call a divergent orbit. The truth of the Rationality Conjecture rules out such divergent orbits for Ta· In particular, if the Rationality Conjecture were




true for a = 3 then every integer x 2: 1 would be eventually periodic, and the only the way the 3x + 1 Conjecture could fail is that there exists some as yet undiscovered cycle of positive integers disjoint from the cycle of period 2 which contains 1. In the final section we give a heuristic argument indicating that the Rationality Conjecture should be true for a = 3, and be false for all odd integers a 2: 5.

The following exercises extend some of the observations of this sec-tion. Exercise 2. For any positive integer k, compute the smallest positive number x such that the T-orbit of x begins with k odd numbers. That is, compute the unique x < 2k such that Q(x)- -1 mod 2k. Exercise 3. For all X E z2 prove Q(2x) = 2Q(x). Exercise 4. For a, b odd 2-adics define the functions /1b, Ta,b : z2 ---+ z2 by f.1b(x) = b · x and

(4.23) { x/2 if x is even

(ax+ b)/2 if x is odd.

Prove that /1b is one-to-one and onto and prove the conjugacy

( 4.24) Ta,b 0 /1b·

Exercise 5. Define the real-valued function v on Z2 by v(O) = 0 and for x =/= 0, v(x) = 2-k where 2k is the highest power of2 dividing x, i.e. k is the number ofO 's which precede the first 1 in the expansion of x. (v is called the 2-adic valuation function.) Let d(x, x') =def v(x- x'). Prove that d : z2 X z2 ---+ lR satisfies the conditions (for all X' x'' x" E Z2):

(1) d(x,x') = 0 iffx = x'. (2) d(x, x') = d(x', x). (3) d(x,x')::; max(d(x,x"),d(x",x')).

Show that condition 3 (called the ultrametric inequality) implies the triangle inequality d( x, x') ::; d( x, x") + d( x", x'). Consequently, we can define distance on Z2 by using the metric d. Show that if r is a real number between 2-k and 2-k+l then for y E Z2

(4.25) {x: d(y,x) < r} {x: d(y,x)::; r} [Y]k· (The resulting topology is the same as the product topology obtained by regarding Z2 as the countable product of copies of { 0, 1}. Thus, the set of 2-adic integers has the structure of a compact topological ring.)

Exercise 6. Using the metric d defined in the previous exercise, prove that Ta,b defined in Exercise 4 satisfies

(4.26) d(Ta,b(x),Ta,b(x')) < 2d(x,x').



14 ETHAN AKIN

Conclude that a (a = 1 and b = -1) and T (a = 3 and b = 1) are continuous. Prove that Q and /-Lb are isometries, e.g. d(Q(x), Q(x')) = d(x,x') (Use (3.11)).

5. Ergodic Theory Viewpoint

Think of y = y0 y1y2 ... in Z2 as the typical outcome of an infinite sequence of independent flips of a so-called "fair coin", labeled 0 on one side and 1 on the other. This tactic introduces probability theory into our study of the 2-adics. We assume that the two outcomes of each flip are equally likely and that the outcomes of the separate flips are independent of one another. The probability of a 0 or a 1 on each flip is thus ~. From the independence assumption all 2k possible outcomes YoYI .. ·Yk-1 are equally likely. For any subset A c Z2 we will write P R(A) for the probability that y lies in A. Technically, this is defined only for certain measurable subsets but these will include all that we will consider. So for y E Z2

(5.1) Probability that x- y mod 2k PR([y]k) 2-k.

A map H : Z2 ~ Z2 is said to preserve probability if for any mea-surable subset A of Z2

(5.2) Probability that H(x) E A PR(H-1(A)) PR(A).

In order to check that ( 4.2) holds, it in fact suffices to check the equation for sets A of the form [y] k.

For example, the shift map a preserves probability since a(x) E [Y]k says that the bits x 1 ... xk are specified by the list y0 ... yk-I and this event has probability 2-k.

The study of such probability preserving mappings is the domain of ergodic theory. For a nice introduction see Billingsley (1965).

In addition to being probability preserving, a is mixing. This prop-erty says that information about the initial state x is gradually lost as we move along the a-orbit of x. For example, if we know that x E [y]k then the first k bits of x are specified by y, but we know nothing about ak(x). The initial data puts no constraint whatever on ai(x) fori~ k. In fact, fori~ k the events x E [Y]k and ai(x) E [y']k' are independent in the same sense that two different flips of the coin are independent.

Contrast this with the translation map a : Z2 ~ Z2 given by a(x) =def x + 1. a preserves probability because

(5.3) a-1 ([Y]k) [y- 1]k,




which has probability 2-k. On the other hand, ai(x) x mod 2k whenever 2k divides i. So the condition x E [Y]k implies ai(x) E [y]k whenever 2k divides i and so for infinitely many i. Thus, a is not mixing.

Like a the map T preserves probability and is mixing. This in turn follows from the fact that the conjugacy map Q preserves measure. To prove the Q result notice that Proposition 2 says that Q-1 ([Y]k) is a mod 2k congruence class and so has probability 2-k.

As described in Billingsley (1965) an important consequence of the mixing property is ergodicity.

Suppose that f : Z2 ---+ lR is a real-valued (and measurable) function. For simplicity we will suppose that f takes on only finitely many values: /I, ... , fn· Define Pi to be the probability that f(x) = fi fori= 1, ... , n. The space average or mean or expected value of f on Z2 is given by

n

(5.4) E(f) =def L fiPi· i=l

The name comes from imagining that we compute the average value of f by choosing randomly a large number of points in Z2: x 1, ... , xN. For approximately N Pi of these points it will be true that f ( x) = k So the statistical average will satisfy

(5.5) E(f).

On the other hand, suppose that H : Z2 ---+ Z2 is a probability pre-serving mapping. Starting from an initial point x E Z2 the associated time average ](x) is obtained by averaging the values along the H-orbit of x.

(5.6) ](x)

provided that the limit exists. If H preserves probability and is mixing then the Birkhoff Ergodic

Theorem says that for almost every initial state x the time average ](x) exists and equals the space average E(f). The remaining points, those at which the limit of ( 4.6) either fails to exist or exists and is unequal to E(f), together form a set of measure zero. {x: ](x) = E(f)} might not be the whole space z2 but it does have probability 1.



16 ETHAN AKIN

Let us apply this to the characteristic function of the congruence class [y]k which is given by

(5.7) { 1 if x _ y mod 2k 0 otherwise.

Clearly, the space average of 1[Y]k is the probability of [Y]k which is 2-k.

On the other hand, the sum

N-1

(5.8) L 1[Y]k(Hi(x)) i=O

is just the number of occurrences of [y]k among the first N elements of the H -orbit of x. The Ergodic Theorem says that for a typical element x the congruence Hi ( x) _ y mod 2k occurs approximately once in every run of 2k elements along the orbit.

We will call a point H -generic if this typical behavior occurs for every congruence class. That is, x is generic if for every congruence class [y]k

(5.9)

We call a point exceptional for H if it is not H -generic. While Z2 is uncountable, there are only countably many congruence

classes and so ( 4.6) imposes only countably many conditions. It follows that the set of exceptional points has measure zero. It is important to notice that any countable set has measure zero and so the exceptional set can be infinite.

Now apply this with H = T. If X is a T generic point in z2 then for every pair of positive integers y and k, ri(x) _ y mod 2k infinitely often. On the other hand, the original conjecture says that for any positive integer x eventually ri ( x) is either 1 or 2. Such a point could not be r-generic. In fact, the r-orbit of a r-generic point cannot enter any cycle. Thus, the Rationality Conjecture (for a= 3) would say that every rational x is r-exceptional.

This at last answers the question of our title. The problem is hard because we are looking at a particular countable subset of the un-countable set Z2 and on it we are trying to demonstrate that a kind of behavior occurs which we know to be completely atypical. This is what Sullivan meant by his comment quoted in the introduction.




6. Concluding Remarks

We mentioned earlier that a proof of the Rationality Conjecture for a = 3 will have to provide an explanation of the essential role of the number 3 (as opposed to odd integers a 2:: 5) in the definition of 7a· From probability theory we can derive a heuristic expanation- possibly misleading - of why 3 is special.

Recall that our original operation T(x) is the first return of the 7-orbit of an odd number x to the set of odds. We let v(x) be the first return time to the set of odds:

(6.1) v(x) =def min { i 2:: 1 : 7i(x)o = 1 }.

If x = 0 or -1/3 then 7(x) = 0 and v(x) = oo. Otherwise, 7(x) is nonzero and v(x) is one more than the number of initial O's in the expansion of 7(x). Recall that when y is even, 7(y) = (J'(y). Alterna-tively, recall that Q(x) is sequence of parities along the 7-orbit of x. Thus, the successive return times are the number of iterations between successive 1's. So if xis an odd number then (6.2)

v(x) > k ~ 7(x) = 0 mod 2k ~ Q(x)- 1 mod 2k+1.

Now if xis an odd number then we define the multiplier M(x) =def

3 · 2-v(x). For an odd integer x, the multiplier M ( x) is approximately the ratio T(x)jx. The initial odd step multiplies by approximately 3/2 and each successive even step multiplies by 1/2. In particular, the multiplier is less than 1 and T(x) < x unless v(x) = 1, i.e. 7(x) is odd.

Proposition 8. Regarding M as a real-valued function on the odd 2-adics, its mean or expected value E(M) is 1.

Proof: When we restrict to the subset of odd numbers we are con-sidering the conditional probability of an event assuming oddness. If A is a subset of the odd numbers then this conditional probability, P Ro(A), is exactly 2 · P R(A) since the probability of the set of odds is ~· From (5.2) we have

(6.3)

and so

(6.4)

PRo({x: Xo = 1 and v(x) > k}) 2 · PR({x: Q(x) = 1 mod 2k+1})

PRo({x: v(x) = k}) PR0 ({x: v(x) > k -1})- PR0 ({x: v(x) > k})



18 ETHAN AKIN

Hence, PRa({x: M(x) = 3 · 2-k}) = 2-k and so

00

(6.5) E(M) 1,

as asserted. 0

On the other hand, suppose we proceed analogously for Ta with a an odd integer larger than 3. Define va(x) as the first return time to the odds for the Ta-orbit of x and define for odd x the multiplier l'v1a(x) = a · 2va(x). The computation in Proposition 8 then yields E(l'v1a) = a/3 > 1.

Thus, 3 is special in that, on average, T(x) is roughly the same size as x. However, when we replace 3 by the odd integer a the size of the odd numbers along the Ta-orbit appear to be increasing geometrically with ratio aj3. This suggests that for such a most orbits should be divergent.

Wait a minute here! The computations of E(Ma) are fine but the ratio interpretations we have given them may be just elaborate flim-flam. The heuristics only apply to integer values of x. As we have seen, not only are the integers of measure zero but they are probably completely atypical points for the systems we are examining. As we seem to be in danger of wandering into nonsense perhaps we should stop here.

At this point I would like to acknowledge the help of the reviewer, who contributed Theorem 7 and provided some further remarks and references which I would like to pass along to the reader.

First, the 2-adic, ergodic theory approach to the 3X + 1 problem has a continuing history which began before the lunchtime discussion be-tween Sullivan and Ruelle. This viewpoint was taken in Matthews and Watts (1984, 1985) and Lagarias (1985), and recent work extending it appears in Venturini (1992), Bernstein (1994), Bernstein and Lagarias (1996), Wirsching (1998) and Monks and Yazinski (2002).

Second, the peculiar expected value computations given above sup-port the idea that divergent trajectories exist for the aX + 1 map on the integers when a is an odd integer greater than 3. This would mean that the Rationality Conjecture is false for Ta with a 2': 5. In fact, various authors have conjectured that such divergent trajectories may in fact be generic points for Ta· For a = 3 divergent trajectories may




or may not exist, but it is a consequence of recent work of Monks and Yazinski (2002) that rational points can never be generic points forT,

Theorem 9. Every rational point of Z2 is an exceptional point for the map T associated with a= 3.

Proof If x = E with q odd then the T-iterates of x all lie in .!z and q q

so either enter a periodic orbit or diverge. If the periodic orbit has period d then the iterates cannot be equidistributed mod 2d+1 since some residue classes are omitted. In the divergent case, Monks and Yazinski (2002) have shown (their Theorem 2.7b) that (6.6)

1 lim infk__,ooy;/Q(x)o + ... + Q(x)k_I) > (log2)/(log3) > 0.63.

However, for a generic point x this limit would exist and equal 0.5. D.

For readers who wish to go further on the 3X + 1 problem, the surveys of Lagarias (1985), Muller (1991) and Wirsching (1998) are good places to start. For more information on p-adic numbers see Gouvea (1993).

Bibliography

D.J. Bernstein (1994), A non-iterative 2-adic statement of the 3x + 1 conjecture, Proc. Amer. Math. Soc. 121: 405-408. D.J. Bernstein and J.C. Lagarias (1996), The 3x + 1 conjugacy map, Canad. J. Math. 48: 1154-1169. P. Billingsley (1965), Ergodic theory and information, John Wiley and Sons, New York. F.P. Gouvea (1993), p-adic numbers, Springer-Verlag, Berlin. J.C. Lagarias (1985), The 3x+1 problem and its generalizations, Amer. Math. Monthly 92:3-21. J. C. Lag arias ( 1990), The set of rational cycles for the 3x + 1 problem, Acta Arith. 56:33-53. G.T. Leavens and M. Vermeulen (1992), 3x+ 1 search programs, Computers and Mathematics with Applications 24: 79-99. K.R. Matthews and A.M. Watts (1984), A generalization of Hasse's generalization of the Syracuse algorithm, Acta Arith-metica 43: 167-175. K.R. Matthews and A.M. Watts (1985), A Markov approach to the generalized Syracuse algorithm, Acta Arithmetica 45: 29-42.



20 ETHAN AKIN

K. Monks and J. Yazinski (2002), The autoconjugacy of the 3x + 1 function, to appear. H. Muller (1991), Das 3n + 1 problem, Mitteilungen der Math. Ges. Hamburg 12:231-251. G. Venturini (1992), Iterates of number theoretic functions with periodic rational coefficients (generalization of the 3x + 1 prob-lem), Studies in Applied Math. 86: 185-218. S. Wagon (1985), The Collatz problem, Math. Intelligencer 7:72-76. G.J. Wirsching (1998), The dynamical system generated by the 3n+l function, Lect. Notes in Math. No. 1681, Springer-Verlag, Berlin.



Contemporary rviathematics Volume 356, 2004

Lectures on Cantor and Mycielski Sets for Dynamical Systems

Ethan Akin

Mathematics Department The City College

137 Street and Convent Avenue New York City, NY 10031

Abstract: Given a residual subset P of the product space X x X we would like to find large subsets A C X such that A x A C P. Often the nature of P precludes the possibility that a residual subset A can be found. However, if X is a perfect Polish space then Mycielski sets A do always exist satisfying the required condition. This is a dense subset of X which is a countable union of Cantor sets. We describe the associated theory and provide some applications to topological dynamics.

AMS 2000 Classification Numbers: 37B20, 37B05, 54H20, 54H05.

Keywords and Phrases: Cantor set, Mycielski set, independent sets, topo-logical transitivity, weak mixing system, Li-Yorke chaos.

1 Introduction

These results were originally inspired for me by a lovely argument of Huang and Ye. Suppose that R is a dense GJ subset of X x X where X is a Polish space, ie. a separable space which admits a complete metric. Assume that

(x,y) E R (y,x),(x,x) E R. (1.1)

@ 2004 American Mathematical Society

21




22 ETHAN AKIN

Huang and Ye (2002)prove there exists a dense, uncountable A C X such that A x A C R. Glasner pointed out to me that by using Baire category arguments one could obtain sets A with more structure. With Glasner's observation as a hint, the results that follow are an elaborate series of application of the Baire Category Theorem. We will see, in particular, that the subset A can be taken to be a countable union of Cantor sets.

The Baire Category Theorem is used so often that we abbreviate it by BCT. To provide some contrast, let us recall a few classical results.

Lemma 1.1 Let H : Z --7 W be a continuous, open surjection with Z and W Polish spaces. IfF is a closed, nowhere dense subset of Z then

ND(F) =def {wE W: H- 1 (w) n F is nowhere dense in H- 1 (w)} (1.2)

is a dense G8 subset of W.

Proof For U a nonempty open subset of Z let

Fu =def H(U) \ H(U \F)

{wE W: U n H-1(w) c F and U n H- 1(w) =1- 0}. (1.3)

Clearly, W \ ND(F) is the union of the sets Fu as U varies over a countable basis for Z. Since His open, Fu is a closed subset of the open set H(U) and so is a countable union of closed subsets of W. Hence, N D(F) is a G8 set. If N D(F) is not dense, i.e. if Uu{Fu} has a nonempty interior, then by the BCT some Fu contains a nonempty open set V. Hence, U n H-1(V) is a nonempty open subset of Z which is entirely contained in F. This contradicts the assumption that F is nowhere dense.

Theorem 1.2 Let H : Z --7 W be a continuous open surjection with Z and W Polish spaces. If R is a dense G8 subset of Z then

H#(R) =def {wE W: H- 1(w) n R is dense in H- 1 (w)} (1.4)

is a dense G8 subset of W.

Proof Assume Z\ G is the union of the sequence of nowhere dense closed subsets F 1 , F2 , ... of Z. By the BCT, H- 1 (w)nRis dense in H- 1(w) iff each H-1 (w)nFi is nowhere dense in H- 1(w). That is, H#(R) is the intersection of the sequence N D(Fl), N D(F2), ... So the result follows from another application of the BCT.

0

From this result we obtain the topological analogue, due to Kuratowski and Ulam, of the Fubini Theorem (see Oxtoby (1980) Chapter 15).



LECTURES ON CANTOR AND MYCIELSKI SETS 23

Corollary 1.3 Let Y and Z be Polish spaces. For w E W let iw : Y ----> Y x W be the inclusion map defined by iw(Y) = (w,y). If R is a dense G/S subset of Y x W, then

1r#(R) =def {wE W: i";;/(R) is dense in Y} (1.5)

is a dense G& subset of W.

Proof For 1r : Y x W ----> W the open, continuous, projection map the set labeled 1r#(R) in (1.4) agrees with the set so labeled in (1.5). Hence, the result follows from Theorem 1.2.

Now let us return to the case of R C X x X, a dense G&, satisfying (1.1). Let us denote by R(x) the subset i;1(R) C X. Corollary 1.3 says that for all x in a dense G& set D C X, the set R(x) is a dense GJ. A set A satisfies Ax A C R exactly when A C R( x) for all x E A. One might hope that we can find a dense G/S in this class of sets. However, this need not be the case.

For example, let f be a homeomorphism on X. We denote by FIX(!) the fixed point set of f. That is,

FIX(!)

If FIX(!)= 0 then

{x: f(x) = x }. (1.6)

(1.7)

satisfies condition (1.1). Notice that we follow the set theorist convention of regarding the function f as a subset of X x X.

For A c X

A X A c St Anj(A) = 0. (1.8)

From the BCT it follows that such a set cannot be a dense G 8 subset of X. In light of this it is not surprising that the general results we achieve produce subsets of first category.

In Section 2 we review results about Cantor sets in Polish spaces. A Polish space is a separable space which admits a complete metric. A Cantor space is a nonempty, compact, metizable, zero-dimensional space. A Cantor set is a subset of a Polish space which is a Cantor space with respect to the relative topology. We make this distinction because we want to extend the classical uniqueness results for Cantor spaces to a newly labeled class of spaces.

We define a Mycielski space to be a second countable, zero-dimensional, locally uncountable, a-compact space which is locally noncompact, ie. every compact subset has empty interior. A Mycielski subset of a Polish space is a a-Cantor subset which is dense and which has an empty interior.The main result of Section 2 is Theorem 2.10 which proves in part:



24 ETHAN AKIN

• A Mycielski subset of a Polish space is a Mycielski space with respect to the relative topology.

• If a dense subset of a locally compact, Polish space is a Mycielski space with respect to the relative topology then it is a Mycielski subset.

• A Mycielski space is homeomorphic to a Mycielski subset of Cantor space.

In Section 3 we show that every Mycielski space is homeomorphic to a My-cielski subset of the real line, JR, and that if A, B are Mycielski subsets of lR then there exists an order preserving homeomorphism f of lR such that f(A) =B. It follows that any two Mycielski spaces are homeomorphic and so any Mycielski space is homeomorphic to Q x C where Q is the set of rationals and C is a Can-tor set. For another representation, consider X = { -, 0, + }N. Let the subset A consist of sequences in which the symbol 0 occurs only finitely often. It is easy to see that A is a Mycielski subset of the Cantor space X.

In Section 4 we consider a complete metric space X and review the basic facts about the space of closed subsets of X equipped with the Hausdorff metric, denoted 2x. It is a complete metric space and the subspaces C(X) of compacta and C'(X) of nonempty compacta are closed. Furthermore, the topology on C(X) depends just on the topology of X. Thus, associated with a Polish space X, the space C(X) is a Polish space well-defined via the choice of complete metric on X. The subset FIN(X) of finite subsets is dense in C(X). When X is perfect, ie. has no isolated points, then C'(X) is perfect and CANTOR(X) the set of Cantor sets in X is a dense, G0 subset of C'(X). We also review results about the sequence space, XN, with the product topology and the space of continuous maps from a compact space, C(K, X), with the topology of uniform convergence. These are both Polish spaces. When K is a Cantor space, the subset Cfin(K, X) of locally constant maps, each of which has a finite range in X, is dense in C(K,X). The subset DENSE of sequences with a dense range in X is a dense, Go subset of xN.

In Section 5 we introduce our main definitions. Starting with a Polish space X, we examine two related structures.

First, we consider Q be a subset of C(X). We say that Q is a hereditary (or finitely hereditary) if A E Q implies that Q contains every compact subset of A (resp. every finite subset of A). Q is called finitely determined when A E Q iff all finite subsets of A lie in Q.

Alternatively, we consider a sequence { Rn : n = 1, 2, ... } with Rn a subset of xn(= X x ... x X with n copies). We call such a sequence a coherent list for (or on) X when for n, m = 1, 2, ... :

For (x1, ... ,xn) E Xn and (yl,···,Ym) E Xm (x1, ... ,xn) ERn and {yl,···,Ym} C {x1, ... ,xn}

=* (yl, ... , Ym) E Rm.

A coherent list is called a residual list when each Rn is a dense G0 .




For example, if R C X x X satisfies (1.1) then we can define a coherent list {Rn} by

(xi, Xj) E R for i, j = 1, ... , n.

We call this coherent list the extension of R. Clearly, (1.1) implies R = R2. If R is a dense Go subset of X x X then the extension of R is a residual list.

A finitely hereditary subset Q determines a coherent list by letting R:;_ c xn be the set of n-tuples (xi. ... , Xn) E Xn such that {x1, .... , Xn} E Q. If Q is a dense, G0 subset then {R:;,} is a residual list.

On the other hand, for a coherent list { Rn} a subset B of X is called an {Rn} set when

for n = 1,2, ....

The set of compact { Rn} subsets is a finitely determined subset of C (X) denoted C( {Rn} ). If {Rn} is a residual list then C( {Rn}) is a dense, G0 subset of C(X).

For a finitely hereditary subset Q we have Q C C( {R:;,}) with equality iff Q is finitely determined.

Our main result, Theorem 5.10, extends to this context results of Mycielski and Kuratowski. If we begin with a G0 , hereditary subset Q of C(X) for a perfect, Polish space X and let {Rn} be the coherent list associated with Q then we obtain a list of equivalent conditions which include the following:

• {Rn} is a residual list.

• Q is a dense subset of C(X).

• CANTOR(X) n Q is a dense, Go subset of C'(X).

• There exists a Mycielski subset B of X such that Bn C Rn for n = 1, 2, ....

• There exists a dense sequence {Ai} in CANTOR(X) such that

for n = 1, 2, ...

Notice that if {Rn} is arbitrary G0 coherent list then we can apply this result tc C( {Rn} ).

Historically, these results grew out of the hunt for large sets which are in-dependent in some algebraic sense, e.g. linearly independent or algebraically independent. Marczewski provided a very general set theoretic context for this search (summarized in Marczewski (1961)). Mycielski (1964) introduced topo-logical conditions and used BCT arguments to obtain what we are calling My-cielski sets of independent elements. In effect, he was looking at what we are calling coherent lists. Since independence consists of avoiding certain sets, the relations considered by Mycielski are the complements of ours and so are small, Fa sets instead of large G0s. Mycielski used direct constructions to build his



26 ETHAN AKIN

Cantor sets. Kuratowski (1973) introduced the "functional analysis" approach which we adopt here, by considering the Cantor sets in X as points in the space C(X). His methods yielded large collections of Cantor sets each with the required independence property.

In Section 6 we follow Iwanik (1989) in applying these results to various hereditary subsets associated with a dynamical system (X, f), consisting of a Polish space X and a continuous map f on X. Letting wf(x) denote the set of limit points of the orbit of x E X, ie. {f(x), JZ(x), ... }, then we define TRANS(!) = {x E X : 'wf(x) = X}. The system is called transitive when TRANS(!) -j. 0 and minimal when TRANS(!)= X. For a transitive system TRANS(!) is a dem;e G(i. The system is called weak mixing when (X x X,f x f) is transitive in which case it follows that the n-fold product (Xn, fXn) is tran:sitive for n = 1, 2, .... Letting TRANS(J)n = TRANS(fxn) we obtain a residual list for a weak mixing system. Similarly, then-fold proximal points and the n-fold recurrent points yield coherent lists.

In particular, we obtain descriptions for various concepts of chaos. For ex-ample, for a system on a complete metric space X which exhibits sensitive dependence on initial conditions we show that there exist a positive E and a Mycielski subset B C X such that any two distinct points in B have orbits which are frequently at least E apart.

In Section 7 we refine the results for weak mixing systems by considering the filter T1 generated by the return time sets for nonempty open subsets of X.

Rather than summarize the results of Sections 6 and 7 we will conclude this introduction with a simple dynamical application of the Kuratowski-Mycielski Theorem of a slightly different type. While the argument is typical of those of used by Mycielski, Kuratowski and lwanik the result appears to be new.

Theorem 1.4 Let G be a countable group of homeomorphisms on a perfect Polish space X. Assume that for each g E G, the fixed point set FIX (g) is nowhere dense. There then exists a Mycielski subset A C X such that for all g,h E G

g(A) n h(A) I- 0 g =h. (1.9)

Proof: First define PER(G) = U9 E0 FIX(g). By hypothesis and the BCT X\PER(G) is a dense G8 . Furthermore, since X is perfect, it follows that each g E G is a nowhere dense closed subset of X 2 = X x X.

Define

Sc =def (X2 \ U g) n (X\ PER(G)) 2 (1.10) gEG

By the BCT again this is a dense, G!i subset of X 2 . It is easy to check that Sc satisfies condition (1.1). By applying the Kuratoski-Mycielski Theorem to the residual list which is the extension of Sc we see that there exists a Mycielski




subset A C X such that

Ax A c Sc. (1.11)

It is easy to check that this condition implies (1.9). <)

For example, let G be the group generated by a transitive homeomorphism f on an infinite, compact metric space X. By tranl')itivity of the system X is perfect and the dense set TRANS (f) is disjoint from the set of periodic points. By Theorem 1.4 there exists a Mycielski subset A C X which is wandering. That is, for integers n, m

n=m. ( 1.12)

On the other hand, for any nonempty open U C X there exists a positive integer n such that U n r(U) # 0. Thus, a set which satisfies (1.12) must have empty interior. That is, every nonempty open subset of X is nonwandering but we can obtain large wandering sets, in fact, Mycielski wandering sets.

2 Cantor Sets and Mycielski Sets in Polish Spaces

All our spaces arc assumed to be Hausdorff. We will call a topological space Cr I if it is a regular space which admits a

countable base. The Urysohn Mctrization Theorem says that a space is Cu iff it is separable and metrizable. A Polish Space is a separable space which admits a complete metric. Any metric compatible with a compact topology is complete and so a compact, metrizablc space is Polish. Any G0 subset of a Polish space is Polish.

A Hausdorff space is called perfect when it contains no isolated points and so every opene subset is infinite, where we use the neologism opene for "open and noncmpty". A space is zero-dimensional when it admits a basis of clopcn sets. A Cantor space is a nonempty, compact, zero-dimensional, perfect metrizable space. A Cantor set in a Cu space is a subset of the space which is a Cantor space with respect to the relative topology.

P!'oposition 2.1 (a) Any nonempty, perfect, complete, metric space contains a Cantor set.

(b) Any two Cantor spaces are homeomorphic.

Proof We review a well-known proof of these two results as we will later usc the construction on which it is based. Let X be a nonempty, perfect, complete, metric space.

Let ko = 1 and Ao = {A0 , A1 } be a pair of opene subsets of X with disjoint closures. Because each openc subset is infinite, one can inductively construct



28 ETHAN AKIN

for n = 1, 2, ... : kn, a positive integer greater than kn-b and An, a collection of opene subsets with pairwise disjoint closures, indexed by the set wkn of words with length kn on the two symbols 0, 1. For each w E Wkn the open set Aw in An has diameter at most 1/n. Furthermore, if a is the initial segment of length kn-1 in the word w then the closure, Aw is contained in Aa.

With N = {0, 1, ... } let C be the Cantor space {0, 1 F11 regarded as the set of infinite words. Because X is complete, the Cantor Intersection Theorem implies that for each x E C there is a unique point f(x) in the intersections

00 00

(2.1) n=l n=l

where xlkn is the initial length kn segment of x. This defines a topological embedding f : C ---+ X. '

If X is a Cantor space, then we can choose a (necessarily complete) metric and sharpen the above construction so that X= A0 UA1 and for each a E Wkn-l

the sets Aw form a partition of Aa as w E Wkn varies over those words with initial segment a. Thus, each An is a clopen partition of X. In that case, f : C ---. X is onto and so is a homeomorphism.

Recall the two maps (3, T : {0, 1 }N ---.I where I= [0, 1] is the unit interval.

(2.2)

The map T is a homeomorphism of {0, 1 }N onto the classical "middle thirds" Cantor set and (3 is a continuous surjection.

We call a space X locally uncountable if every opene subset of X is uncount-able. Clearly, a locally uncountable space is perfect.

For a Cn space X we define the countable part Xco and the uncountable part XuN by

Xco =def U{G: G is a countable, open subset of X}

XuN =def X\ Xco

Lemma 2.2 Let X be a Cn space.

{a) AssumeD is a dense subset of Z. D is perfect iff X is perfect.

(2.3)

{b) The countable part, Xco, is a countable, open subset of X which contains every countable open subset. Every isolated point of X lies in X co.

(c) The uncountable part, Xu N, is a locally uncountable, closed subset of X which contains every locally uncountable subset of X.

{d) If X is Polish and perfect then every opene subset of X has cardinality c and contains a Cantor set. In particular, X is locally uncountable.




(e) If X is Polish then XuN is Polish and perfect and contains every perfect subset of X.

(f) If X is Polish and uncountable then it contains a Cantor set.

Proof (a): If x is an isolated point of D then it is an isolated point of X. Conversely, if D is perfect and U is an opene subset of X then U nD is infinite and so X is perfect.

(b): As the union of open sets, Xco is open and it clearly contains every countable open set. If x is an isolated point then { x} is a countable open set and sox E Xco. We can obtain Xco as the union of the countable members of a countable basis for X. Hence, Xco is countable ..

(c): Clearly, XuN is closed and contains every locally uncountable subset of X. If an open subset U meets XuN then U is uncountable and so Xco countable implies that

UnXuN U\Xco (2.4).

is uncountable. (d): An opene subset of a Polish space is Polish and nonempty. An open

subset of a perfect space is perfect. Proposition 2.1(a) thus implies that an opene subset of a Polish space contains a Cantor set and so has cardinality at least c. Since no two points of a Hausdorff space lie in exactly the same set of elements of a basis, it follows that any Cn space has cardinality at most c.

(e): XuN is perfect because it is locally uncountable. It is Polish because it is closed. If A is a nonempty, perfect subset of X then its closure is Polish and is perfect by (a). By (d), A is locally uncountable and so is contained in XuN by (c).

(f): Since Xco is countable, its complement XuN is a nonempty, perfect Polish space and so it contains a Cantor set by (d).

0 Remark The implicit assumption of completeness in (f) is necessary. There

exists a subset B of the unit interval I which meets every Cantor set in I but which contains none of them. Such a subset is necessarily perfect, and of cardinality c.

Proposition 2.3 Let X be a Polish space. If B is a locally uncountable, Fu subset of X and x E B then there exists a Cantor set C with x E C and C C B.

Proof Assume that B is the union of the sequence of closed subsets { Ak}. Since B is perfect we can choose a sequence {xn} of distinct points in B with limit x. Choose {Un} a sequence of open sets with pairwise disjoint closures such that Xn E Un and the diameter of Un is less than 1/n with respect to some



30 ETHAN AKIN

complete metric on X. Since Un n B is uncountable, there exists kn such that Un n Akn is uncountable. As this intersection is Polish it contains some Cantor set by Proposition 2.2(f). That is, for each n = 1, 2, .. there exists a Cantor set Cn c Un n B. The set

c (2.5) n

is clearly, compact, perfect and zero-dimensional. Thus, it is the required Cantor set.

We will repeatedly use the observation that a subset B of a space X has empty interior iff its complement is dense in X. That is,

IntB 0 X \ B is dense. (2.6)

Lemma 2.4 Let {An} be a sequence of Fa subsets of a complete metric space X with union B. The subset B has empty interior iff each An has empty interior.

Proof If B has empty interior then so does every An. The converse follows from the BCT since X is complete. (Recall that we use the abbreviation BCT for the ubiquitous Baire Category Theorem).

We will call a space X locally noncompact if every compact subset of X has empty interior. A subset is called locally noncompact when it is a locally noncompact space with the relative topology. Clearly, a locally noncompact space has no isolated points and so is perfect.

Proposition 2.5 Let B be a subset of X.

(a) If B is dense in X with I ntB = 0 then B is locally noncom pact.

(b) If B is locally non compact and X is locally compact then I ntB = 0.

Proof (a): If K is a compact subset of B and IntBK f. 0 then there exists U opene in X with U n B c K. Since B is dense in X, U n B is dense in U. Since K is compact, it is closed in X. Hence,

U c UnB c K c B. (2.7)

Thus, B has a nonempty interior. (b): If IntB f. 0 then since X is locally compact there exists U opene in X

with K = U compact and contained in IntB. Hence,

(2.8)




Thus, K is a compact subset of B with nonempty interior. <>

A space,or subset of a space, is called a- Cantor if it is a countable union of Cantor sets. As it is the empty union of Cantor sets, the empty set is a-Cantor.

Proposition 2.6 (a) A a- Cantor space is a- compact, locally uncountable and zero-dimensional.

(b) A a- Cantor subset of a space is an FrY subset.

(c) A countable union of a- Cantor subsets is a a- Cantor subset.

{d) Let B be a dense a-Cantor subset of a Polish space X and let A be a countable subset of X. The union B U A is a a- Cantor subset of X.

Proof (a): a-compactness is obvious. Since each Cantor space is locally un-countable, the same is true of an arbitrary union of Cantor sets. The Sum The-orem of dimension theory says that a CII space which is the countable union of closed zero-dimensional subsets is itself zero-dimensional. See Hurewicz and Wallman (1941) Theorem II.2.

(b) and (c) are obvious. (d): Since B is dense and locally uncountable, BUA is a locally uncountable,

FrY subset of the Polish space X. By Proposition 2.3 there exists for each a E A a Cantor set Ca C BUA with a E Ca. The union ofB and all theCa's is BUA. So (c) implies that B U A is a-Cantor.

<>

Proposition 2. 7 Let X be a Polish space. If a subset B of X is dense, a- compact, locally uncountable and zero-dimensional, then B is a- Cantor.

Proof Let B be the union of the sequence {Kn} of compact sets. Let A be the union of the sequence of countable sets {(Kn)co} so that A is a countable subset of X. For each n let Cn be the uncountable part (Kn)UN· By Lemma 2.2(c) Cn is a closed, locally uncountable subset of Kn. Hence, Cn is compact and perfect. It is zero-dimensional since B is. So each Cn is either a Cantor set or is empty. Thus, the union of the sequence {Cn}, which we denote by B, is a-Cantor and B = B U A.

If U is an opene subset of X then since B is dense, UnB is an opene subset of B. Since B is locally uncountable, UnB is uncountable while UnA is countable. Hence, u n B =I 0. Thus, B is dense in X. It follows from Proposition 2.6(d) that B = B U A is a-Cantor.

<>



32 ETHAN AKIN

Proposition 2.8 A space is homeomorphic to a dense subset of a Cantor space iff it is a nonempty, perfect, zero-dimensional Cu space.

Proof Any subset of a Cantor space is a zero-dimensional Cu space. If it is dense then it is perfect by Lemma 2.2(a).

Conversely, if Z is a C I I zero-dimensional space then we can choose a se-quence gn : Z ----+ { 0, 1} the indicator functions of a countable basis of clopen sub-sets of Z. Concatenate them to define a topological embedding g: Z----+ {0, 1V'. Let X be the closure of the image of g. As a closed subset of a Cantor set, X is compact and zero-dimensional. The embedding g is a homeomorphism of Z onto a dense subset of X. If Z is nonempty and perfect then so is X by Lemma 2.2(a) again. That is, X itself is then a Cantor space.

0 Remark: (a): If Y is a nonempty, perfect, compact metric space we can

choose Z a countable dense subset of Y. The metric d takes on only countably many values on Z x Z we can obtain a countable basis {Un} for Y consisting of balls centered at points of Z and with radii not in d(Y x Y). Perform the above construction with the associated basis for Z obtained by intersection. The homeomorphism g- 1 : g(Z) ----+ Z extends to a continuous map k: X ----+ Y which is necessarily surjective by compactness. For x E X C { 0, 1 V', the point k(x) is the unique point in the intersection of {Un: Xn = 1}.

(b): If Y is any nonempty, compact Cu space, we can choose a metric and a Cantor space K. Then Y x K is a nonempty, perfect, compact metric space and so as above we can obtain a continuous map from a Cantor space onto Y x K. Compose with the projection to Y to obtain a continuous map of a Cantor space onto Y. This reproves the classic result that any compact, metrizable space is the continuous image of a Cantor space.

Definition 2.9 (a) A Mycielski space is a nonempty, u- compact, zero-dimensional, locally uncountable, locally noncom pact C II space.

(b) A subset of a Cu space X is called a Mycielski subset of X if it is a dense, u- Cantor subset with empty interior.

The following results show how the two pieces of the definition fit together.

Theorem 2.10 (a) A Cu space which contains a Mycielski subset is locally uncountable.

(b) A perfect Polish space contains Mycielski subsets.

(c) A Mycielski subset of a nonempty Cu space is a Mycielski space with respect to the relative topology.




(d) A Mycielski space is a u- Cantor set.

(e) If a dense subset of a locally compact Cu space is a Mycielski space with respect to the relative topology then it is a Mycielski subset.

(f) A Mycielski space is homeomorphic to a Mycielski subset of a Cantor space.

Proof (a): If a dense subset is locally uncountable then the space is locally uncountable.

(b): By Lemma 2.2(d) a perfect Polish space X is locally uncountable. Let D be a countable dense set in X. If U is an opene subset of X then U \ D is a locally uncountable Polish space and so it contains a Cantor set Cu by Proposition l.l(a). Since Cu is disjoint from D it is nowhere dense. Taking the union of the Cu 's as U varies over a countable basis we obtain a Mycielski subset by Lemma 2.4.

(c): Let B be a Mycielski subset of a nonempty Cu space X. Since B is u-Cantor, Proposition 2.6(a) implies that it is u-compact, zero-dimensional and locally uncountable. B is nonempty because X is nonempty and B is dense. Finally, Proposition 1.5(a) implies that B is locally noncompact. Thus, B is a Mycielski space.

(d), (e) and (f): Let Z be a Mycielski space. Because Z is zero-dimensional and locally uncountable, Lemma 2.8 implies that Z is homeomorphic to a dense subset B of a Cantor space C. By Proposition 2.7, B is u-Cantor. Since Z is homeomorphic to B, Z is u-Cantor. This completes the proof of part (d). Once we have proved (e) it will follow that B is a Mycielski subset of C, completing the proof of (f).

It remains to prove (e): Assume that the Mycielski space Z is a dense subset of a locally compact Cu space X. Having proved (d) we know that Z is u-Cantor. Finally, because Z is locally noncompact, Proposition 2.5(b) implies that IntZ = 0. Thus, Z is a Mycielski subset of X, proving (e).

3 Applications of Order

The main result of this section is a uniqueness theorem for Mycielski spaces analogous to the uniqueness for Cantor space given by Proposition 2.1(b). Our proof uses order structures on Cantor spaces. We use some results from, and follow the notation of, Akin (1999).

An ordered set (X,:::;) is a set X equipped with an order :::;, ie. a reflexive, anti-symmetric, transitive relation on X which is total, ie. :::; U ~ X x X. An order space(X, :::;) is a compact, metrizable space X equipped with a closed order. That is, the subset :::; of X x X is closed. We use the usual interval notation and write m, M for the minimum and maximum points of X,



34 ETHAN AKIN

respectively. The topology on X is the order topology. That is, the intervals (a, MJ and [rn, a) are open in X and together they form a subbase for the topology (see, eg. Akin (1999)Proposition 2.1).

An order map g between ordered sets is a map which preserves order. That is,

a :S b ===? g(a) :S g(b). (3.1)

An order isomorphism is a bijective order map. The inverse of an order isomorphism is an order isomorphism. An order isomorphism is clearly a home-omorphism with respect to the order topologies.

A pair a, b in an ordered set is called an endpoint pair when a < b and the open interval (a, b) is empty. We then call a the left endpoint and b the right endpoint of the pair. If an ordered set has no endpoint pairs then it is called order dense. This means that for all a < b there exists c such that a < c < b.

The terms "left" and "right" come from the following result whose proof we omit.

Proposition 3.1 Let C be a bounded subset of the real line ~ with supremum M and infimum rn. C is a Cantor set iff it is a perfect, closed, nowhere dense subset of ~ in which case the endpoint pairs are exactly the endpoints of the complementary intervals in (rn, M), ie. the components of [rn, .iH] \C. For such a Cantor set the set of left endpoints and the set of right endpoints are disjoint, countable dense subsets of C.

On the product space {0, 1 }~'~ we usc the lexicographic ordering. That is, distinct points x, y E {0, 1 }~'~ satisfy x < y if Xi < Yi when i = rnin{j : Xj =f. Yj }. The map T of (2.2) is an order isomorphism from {0, 1}~'~ with the lexicographic ordering to the "middle thirds" Cantor set in I. The map f3 of (2.2) is a surjective order map to the unit interval. Furthermore, if x < y then

f3(x) = f3(y) (x,y) = 0. (3.2)

That is, endpoint pairs are identified by (3. Otherwise, distinct points in {0, 1 }~'~ are mapped to distinct points in I. It easily follows that f3 restricts to a home-omorphism from the complement in {0, 1 }~'~ of the countable, dense set of end-points onto the complement in I of the countable, dense set of dyadic rationals in (0, 1).

Since any Cantor space is homeomorphic to a Cantor set in I we see that every Cantor space admits a closed order. We can in fact prescribe the sets of endpoints.

Proposition 3.2 Let X be a Cantor space, with distinct points rn, A1 E X and countable dense subsets L, R of X such that { rn, M}, L and R are pairwise disjoint. There exists a closed total order :S on X such that the minimum point




ism, the maximum point isM, L is the set of left endpoints and R is the set of right endpoints. Furthermore, with this order X is order isomorphic with {0, 1 }N ordered lexicographically.

Proof We adapt the proof of Proposition 2.1 imposing extra conditions on the sequence of partitions An and ordering each partition. We want the orders to be compatible. That is, suppose Au, Aw E An and Aa, Ab E Am with m < n, Au C Aa and Aw CAb. Then

(3.3)

Let r0 = m and let { r 1 , r 2 , ... } be a counting of R. Call these the red points. Similarly, let l0 = M and let {h, h, ... }be a counting of L. Call these the blue points.

Let Ao = { A 0 , Ad be a clopen partition of X with ro E Ao and lo E A1. Order A 0 by A0 < A1 . Let Ro = {ro} and La = {lo}. We will define an increasing sequence of finite sets of red points { Rn} and an increasing sequence of finite sets of blue points { Ln} such that

and (3.4)

and such that each member of An contains exactly one point of Rn and one point of Ln.

Assuming that An, Rn and Ln have been constructed. Let r be the first point of R not in Rn and similarly define f. Construct the refining clopen partition An+1 subject to the additional condition that each member of An+l contains at most one point of Rn U Ln U {r, Z}. Throw in additional red and blue points to define Rn+1 ~ Rn U {i} and Ln+1 ~ Ln U {Z} so that each member of An+1 contains exactly one point from Rn+ 1 and one from Ln+ 1.

Consider A E An containing the points r E Rn and l E Ln. Order members of An+l contained in A so that the one containing r comes first and the one containing l comes last. Extend the ordering by condition (3.3).

For two distinct points x, y E X we define x < y when Aa < Ab for x E Aa, y E Ab and Aa, Ab E An· By (3.3) the order is well-defined. It is easy to check that it is a closed, total order satisfying the requirements of the proposition.

Finally, a bit of care with the labeling by words ensures that the home-omorphism constructed in the proof of Proposition 2.1 is the required order isomorphism. If A E An is labeled by the length kn word w, then we label the members of An+1 contained in A so that the chosen ordering agrees with the lexocographic order on the length kn+l words which begin with w.

\;>

Corollary 3.3 If B is a subset of a Cantor space X with I ntB = 0 then there exists an order on X with respect to which neither the maximum, nor minimum



36 ETHAN AKIN

nor any endpoint lies in B and with respect to which X is order isomorphic with {0, l}N ordered lexicographically.

Proof Because X is a perfect Cn space and X \ B is dense we can choose points m, M E X and countable, dense subsets R, L of X such that B, R, L and { m, M} are pairwise disjoint. The required order exists by Proposition 3.2.

<:; Remark It is in fact true that any two ordered Cantor spaces are order

isomorphic, see eg. Akin (1999) Corollary 2.13.

Theorem 3.4 (a) A nonempty, perfect, zero-dimensional, locally noncom-pact Cn space is homeomorphic to a dense subset of the real line ~ with empty interior.

(b) A Mycielski space is homeomorphic to a Mycielski subset of the real line R

Proof Let B be a nonempty, perfect, zero-dimensional, locally noncompact Cn space. By Lemma 2.8 we can define a homeomorphism g of B onto a dense subset of a Cantor space C. Since g(B) is locally noncompact, it has empty interior by Proposition 2.5(b). We can apply Corollary 3.3 to choose an order on C such that neither maximum, minimum nor any endpoint lies in g(B) and an order isomorphism q from C onto {0, 1}N. Because the map f3 of (2.2) is a homeomorphism on the complement of the set of endpoints, the composition f3 o q o g is a homeomorphism from B onto a subset B of the the complement in (0, 1) of the set of dyadic rationals. Since g(B) is dense in C, B is dense in (0, 1). Finally, follow by a homeomorphism of (0, 1) onto R This completes the proof of (a). If B is a Mycielski space then it is O"-Cantor by Theorem 2.10(d) and so the image is a Mycielski subset of ~ as (b) requires.

We omit the inductive construction which proves the following classical re-sult, see, eg. Akin (1999) Lemma 2.3.

Proposition 3.5 If an ordered set is countable, order dense and has neither maximum nor minimum then it is order isomorphic to the set Q of rationals in the real line R

<:;

The uniqueness results we want exist in the order category for subsets of R If g : ~ ---+ ~ is an order isomorphism then it is a homeomorphism. Conversely, connectedness implies that any homeomorphism of lR either reverses order or preserves order. The latter is the order isomorphism case.




Lemma 3.6 If B1 and B2 are dense subsets of~' then an order isomorphism 9o : B1 --+ B2 extends uniquely to an order isomorphism g : ~ --+ ~.

Proof For x E ~ define

g(x) sup{g0 (b): for all bE B 1 such that b < x} = inf{go(b): for all bE B1 such that b > x}.

(3.5)

Because B 1 and B2 are dense, the two definitions agree, and g is a surjective, order preserving map. Uniqueness is clear from density of B 1 and continuity of an order isomorphism.

Proposition 3. 7 If B 1 and B2 are countable, dense subsets of ~ then there exists an order isomorphism g: ~--+ ~ such that g(B1) = g(B2).

Proof By Propositon 3.5 there exists an order isomorphism from B 1 to B2. By Lemma 3.6 it extends uniquely to R

Corollary 3.8 Any nonempty, perfect, countable, metrizable space is homeo-morphic to the set Q of rationals in ~.

Proof The space is of course separable and a separable metric space is C II. Since a metric takes on only countably many values, the balls with radii in the remaining set of positive reals form a clopen basis. By Theorem 3.4(a) the space is homeomorphic to a countable dense subset of ~ which by Proposition 3. 7 is order isomorphic to Q.

Proposition 3.9 If B 1 and B 2 are Cantor subsets of ~ then there exists an order isomorphism g: ~--+ ~ such that g(Bl) = g(B2).

Proof By using preliminary linear maps with positive slope, we can assume that B 1, B 2 C I with common minimum 0 and common maximum 1.

Let B1 denote the set of complementary intervals for B 1, that is, the set of components of I \ B 1 . With the induced ordering B1 is a countable, order dense ordered set with neither maximum nor minimum. Similarly, define 8 2 .

By Proposition 3.5 there exists an order isomorphism G : B1 --+ B2. On each



38 ETHAN AKIN

interval J E 8 1 , let y0 : J-+ G(J) be the unique affine bijection with positive slope. Let g0 be the identity on lR \ I. This defines an order isomorphism y0 : lR \ B 1 -+ lR \ B 2 . These are dense subsets of lR because the Cantor sets have empty interior. The unique extension g given by Lemma 3.6 is the required order isomorphism of JR.

Theorem 3.10 If B 1 and B 2 are Mycielski subsets of lR then there exists an order isomorphism y: lR-+ lR such that y(B1 ) = y(B2 ).

Proof Replace lR by its homeomorphic image (0, 1) and adjoint 0, 1 to each of B1 and B2 . We will prove that if A and B are Mycielski subsets of the unit interval I, each of which contains 0 and 1 then there exists an order isomorphism y: I-+ I such that y(A) =B. We first show that we can write A as the union of an increasing sequence of Cantor sets { Ak : k = 0, 1, ... } with 0, 1 E A 0 and which satisfies the following condition for all k:

For every component J of I\ Ak,

the intersection ] n Ak+1 is a Cantor set. (3.6)

We let A_ 1 = {0, 1} and construct Ak+l inductively. Since A is a Mycielski subset it is a union of some sequence of Cantor sets {Ck : k = 0, 1, ... }. Each component J of I\Ak is an open subinterval of (0, 1) with endpoints an endpoint pair in Ak. Apply Proposition 1.3 to the dense Fer subset An] in the space]. We can obtain a Cantor set AJ which is contained in An] and which contains the endpoints of J. If Ck+l meets the open interval J then Ck+1 n] is a Cantor set as well. If not then Ck+ 1 n] is contained in the set of endpoints of]. It follows that ( Ck+1 u AJ) n] is always a Cantor set. Define

Ak u ck+l u (U{AJ} ), (3.7) J

where J varies over the components of I\ Ak· Ak+l is closed and perfect. It is a subset of A and so has empty interior. Thus, it is a Cantor set in I which satisfies (3.6).

We write B as a the union of a similar increasing sequence { Bk : k = 0, 1, ... } of Cantor sets.

First, use Proposition 3.9 to define an order isomorphism: y0 from the Cantor set A 0 to the Cantor set B 0 . Inductively, we define an order isomorphism Yk+l : Ak+l -+ Bk+l which extends the previously defined order isomorphism Yk· The endpoint pairs are defined by the order structure of the set. Hence, if J = (a, b) is a complementary interval for Ak, then a, b E Ak and L =def (Yk (a), Yk (b)) is a complementary interval for Bk. We can use Proposition 3.9 again to extend Yk defining gk+l on Ak+l n] as an order isomorphism to Bk+l n I. By condition (3.6) we are just comparing Cantor sets on each interval.




Taking the union of the sequence of functions {gk : k = 0, 1, ... } we obtain an order isomorphism from A to B. By Lemma 3.6 it extends uniquely to an order isomorphism on I.

Notice that continuity comes along for free because the map is order pre-serving.

Corollary 3.11 Any Mycielski space is homeomorphic to the product Q x C where Q is the set of rationals in the real line and C is a Cantor set.

Proof By Theorem 3.4(b) any Mycielski space is homeomorphic to a Mycielski subset of the real line and by Theorem 3.10 all Mycielski subsets of the real line are homeomorphic. Thus, any two Mycielski spaces are homeomorphic. If A is any countable metric space then the product A x C is zero-dimensional, a-Cantor and locally uncountable. When A= Q the product is nonempty and locally noncompact. That is, Q x C is a Mycielski space and so is homeomorphic to any other Mycielski space.

4 The Space of Compacta in a Polish Space

In this section we deal with metric spaces. That is, each is equipped with a particular metric. On a finite product of metric spaces we use the sup metric. Following Kuratowski (1966) Section 17 we describe the associated metric space of closed subsets.

We will assume that our metrics, all denoted d, are bounded so that every set B in a metric space X has a bounded diameter:

diamB =def sup{d(a,b): a,b E B}. ( 4.1)

In particular, diam0 = 0. The distance from a point x to a set B is given by

d(x,B) =def inf{d(x,b): bE B}. (4.2)

By convention,

d(x, 0) = diamX. ( 4.3)

Clearly, the closure of B is given by

{x: d(x, B)= 0}. ( 4.4)



40 ETHAN AKIN

We can regard any subset R of X x Y as a relation from X to Y (or a relation on X when X= Y). Define

R(x) =def {y: (x, y) E R},

R(B) =def {y: (b, y) E R for some bE B} U R(b). (4.5) bEB

In particular, if we define for E 2': 0

11;, =def {(x, y) EX X X: d(x, y) < E} ""' =def {(x, y) EX X X: d(x, y) S E}

(4.6)

then Ve(x) (and Ve(x) ) is the open (resp. closed ) ball of radius E centered at x, and

1/;,(A) {x: d(x, A)< E}. (4.7)

Observe that V0 is the diagonal in X x X, ie. it is the identity map lx. Two subsets A, B of X are disjoint iff A x B is disjoint from lx. The gap

between A and B is

d.(A,B) =def inf {d(a,b): a E A,b E B} inf{d(a, B): a E A} sup{E: An Ve(B) = 0}

(4.8)

with the convention that d.( A, B) = diamX if A orB is empty. It is easy to check that

d.(A, B) > d.(lx, Ax B) > d.(A, B)/2. (4.9)

The Hausdorff distance between two subsets A and B is given by

d(A, B) =def inf { E: A C 1/;,(B) and B C 1/;,(A)} max(sup{d(a,B): a E A},sup{d(b,A): bE B}).

(4.10)

It is easy to check that this is a pseudo-metric and from (4.4) we see that

d(A, B) 0 A B. ( 4.11)

By convention we assume that d(0, A) is diamX,the diameter of X, whenever A is nonempty. Thus, 0 is an isolated point with respect the the Hausdorff pseudo-metric.




For a metric space X we define the following collections of subsets of X:

2x =def {A: A is closed}. 2x -1 =def 2x \ {0}. C(X) =def {A: A is compact}. C'(X) =def C(X) \ {0}.

FIN(X) =def

FIN'(X) =def

INFTY(X)

PERF(X)

ZERODIM(X) =def

CANTOR(X) =def

{A : cardinality A is finite}. F I N(X) \ {0}.

=def 2x \ FIN(X).

=def {A : A is compact and perfect}. {A : A is compact and zero-dimensional}.

{A: A is a Cantor set}.

( 4.12)

All of these are metric spaces with the Hausdorff metric induced from the metric on X.

We can regard gas a function g: 2x x 2x --+ JR;. and we define for n = 1, 2, ...

in : xn --+ 2x - 1 by in(xl, ... , Xn) = {x1, ... xn}·

Vn: (2X- 1)n--+ 2X- 1 by Vn (A1, ... ,An) = A1 U ... U An. (4.13) X xn

Xn: 2 --+ 2 by Xn (A)= An.

If f : Y --+ X is a continuous map of metric spaces then f. : 2Y --+ 2x is defined by

(4.14)

The restriction f. : C(Y)--+ C(X) is given by f,(A) = f(A).

Lemma 4.1 (a) The maps d_, in, Vn and Xn are continuous and Vn is an open surjection. The restriction Vn : (C'(X))n--+ C'(X) is also an open surjection.

(b) If f : Y --+ X is a uniformly continuous map of metric spaces then j. : 2Y --+ 2x is uniformly continuous.

(c) Iff : Y --+ X is continuous, then f. is continuous at every element of C(Y). In particular, the restriction f, : C(Y) --+ C(X) is continuous.

Proof (a): It is easy to check that each map has Lipschitz constant 1 and V n is obviously surjective. Now assume that Vn(A1, ... ,An)= A and E > 0. Assume that d(A, B) < E. Choose 6 so that d(A, B) < 6 < E. Let Bi = B n Vs(Ai) for i = 1, ... , n. It is easy to check that d(Ai, Bi) < f and clearly, Vn(Bl, ... , Bn) = B. Hence, V n is an open map. If the Ai 's and B are compact then so are the Bi's.



42 ETHAN AKIN

(b),(c): If f. > 0 and f is uniformly continuous or A is compact, then there exists o > 0 such that fCVo(x)) C Ve(f(x)) for all x EA. Clearly, d(A, B) < o implies d(f*(A), f*(B)) <f..

Proposition 4.2 Assume that the metric on X is complete.

(a) The Hausdorff metric on 2x is complete.

(b) C(X) is a closed subset. In fact,

C(X) FIN(X) (4.15)

(c) If X is separable, then C(X) is separable. In fact, if D is a dense subset of X then FIN(D) is dense in C(X).

(d) If X is compact then 2x = C(X) is compact.

Proof (a): For a sequence {An} in 2x let Aoo equal the

limsup{An} =def n U Ak (4.16) n k?_n

Clearly, the limsup of any subsequence is contained in A00 •

If {An} is a Cauchy sequence, then for every f. > 0 there exists a positive integer n such that for all m ;::::: n

(4.17)

It easily follows that if the limit exists, it contains A00 •

We show that if X is complete then a Cauchy sequence {An} converges to Aoo. Since the sequence is Cauchy it suffices to show that some subsequence converges to its limsup. By going to a subsequence, we can assume that for n = 0, 1, ...

d(A A ) < 1/2n+l. n, n+l (4.18)

Given f. > 0, let m be any positive integer with 1/2m < f. and let x be any point of Am. It will suffice to find a point y E Aoo such that d(x, y) < f..

Let Xm = x and for n > m use (4.18) to inductively choose Xn E An so that for alln ;::::: m, d(xn,Xn+l) < 1/2n+l. Then {xn} is a Cauchy sequence. By completeness it has a limit X 00 which clearly lies in Aoo and which satisfies d(xm, X 00 ) < 1/2m. Thus, we can let y = X 00 •




(b): Recall that a subset A of a metric space X is compact when it is complete and totally bounded, where A is totally bounded when for every E > 0 there exists a finite set F C A which is E-dense in A. That is, A C V,(F), or, equivalently, the Hausdorff distance d(F, A) is less than E. Notice that if G E F I N(X) satisfies d( G, A) < E then there is a finite subset F of A such that d( G, F) < E and so d(F, A) < 2E. It clearly follows that the closure ofF I N(X) in 2x consists of the totally bounded, closed subsets of X. If X is complete this is C(X).

(c): It is obvious that if Dis dense in X then FIN(D) is dense in FIN(X). So the result follows from (4.15).

(d): It follows from (a) and (4.15) that it suffices to show that X totally bounded implies F I N(X) is totally bounded. For this, note that F E-dense in X implies 2F = FIN(F) is t:-dense in FIN(X).

Examples:

(a) On IR we use the metric d(x,y) = min(lx- yi, 1) so that on the subset N the zero/one metric is induced. On the uncountable subset 2N of 2JR the Hausdorff metric is the zero/one metric. It follows that 2JR is not separable.

(b) If h : X -> X is a homeomorphism and d is a complete metric on X then h*d is a topologically equivalent metric on X where h*d(x, y) = d(h(x), h(y)). If {xn} is an h*d Cauchy sequence then h(xn) is d Cauchy and so converges to some pointy EX. Thus, h*d is complete. Apply this with the map h: IR-> IR given by h(x) = x3 . Let An= {k + (1/n) : k E N} C IR for n = 3, 4, .... Notice that

(n, k) -::J (m,p) ===;. l(k + (1/n))- (p + (1/m))l > 1/mn

and so l(k + (1/n))3 - (p + (1/m))3 1 > kp/(mn) 3 . ( 4·19)

While it is clear that d(An, N) < 1/n, ( 4.19) implies that if n -::J m, then the Hausdorff distance with respect to h*d between An and Am is equal to 1.

It follows that we obtain different topologies on 2JR from the two complete Hausdorff metrics.

(c) The open unit interval (0, 1) is totally bounded and so it follows that the closure ofF IN( (0, 1)) is all of 2(o,l), not just C( (0, 1)). Using homeomor-phism from IR to (0, 1) we obtain a totally bounded metric h*d on IR and a third distinct topology on 2JR with FIN (IR) dense.

It follows that when we replace the metric on X with a uniformly equivalent metric then the corresponding Hausdorff metrics on 2x are uniformly equivalent but that we may obtain different topologies from merely topologically equivalent



44 ETHAN AKIN

metrics on X. On the other hand, Proposition 4.1(c) implies that the topology on C(X) is independent of the choice of metric. In particular, if X is a Polish space then C(X) is a well-defined Polish space.

Proposition 4.3 Let X be a Polish space, with C(X) the Polish space of com-pacta in X. Let n be a positive integer.

(a) The set DIS =def {(A, B) E C(X) x C(X): An B = 0} is open.

(b) Let B, K, U be subsets of X with B closed, K compact and U open. C(B), {A E C(X) : K c A} and {A E C(X) :An B -=1- 0} are closed subsets of C(X). {A : An U -=1- 0} and C(U) are open subsets of C(X).

(c) FIN(X) is an Fa subset of C(X). INFTY(X) n C(X), PERF(X), ZERODIM(X), and CANTOR(X) are G0 subsets ofC(X).

(d) X is perfect iff INFTY(X) is dense in C'(X). If X is perfect then C'(X) is perfect and CANTOR(X) is dense in C'(X).

(e) Let B be a subset of X. If B c X is Go then Bn c xn and C(B) c C(X) are G0 sets. If B is dense in X then Bn is dense in xn and FIN(B) c C(B) is dense in C(X).

(f) If X is perfect and B is a dense, G0 subset of X then C'(B)nCANTOR(X) is a dense, G0 subset of C'(X).

(g) Let Q be a subset of C'(X). If Q is G0 then v;;:1(Q) = {(A1, ... ,An) : A1 U ... U An E Q} C (C'(X))n is G0 and if Q is dense in C'(X) then v;;: 1 (Q) is dense in (C'(X))n.

(h) If R c xn is G0 (or closed) then {A : An c R} c C(X) is G0 (resp. closed).

Proof (a): Using the continuous restriction of d_ to C(X) x C(X), DIS = (d_)- 1 ((0, oo)).

(b): A c B {'} V2 (A,B) = B, K c A{'} V2 (A,K) =A and An B-=/:-0 {'} (A, B) (j. DIS. So the first three sets are closed. If B = X\ U then An U = 0 {'}A c B, and A c U {'}An B = 0. So the latter two sets are open.

(c): If the cardinality of A is greater than n then there exist disjoint open sets Uo, .. , Un such that An Ui -=/:- 0 for i = 0, ... , n. By (b) these are open conditions. Hence, {A : cardinality of A ::::; n} is a closed subset of C(X) and FIN(X) is an Fa. Its complement in C(X), INFTY(X) n C(X)is a G0 .

By letting U0 , .. , Un vary over families of disjoint open subsets contained in a fixed open subset U we see the

{A E C(X): The cardinality of An U is greater than n} (4.20)



LECTURES ON CANTOR AND 1\JYCIELSKI SETS 45

is an open subset of C(X). It follows that for every pair of open sets U, V with VcU

SING(U, V) =def {A E C(X): An U =An V

and this intersection is a singleton} ( 4.21)

is a closed subset of C(X) because this set consists of those A which meet V and for which the cardinality of A n U is at most 1.

Letting U, V vary in a countable base we see that {A : A contains an isolated point} is an Fa subset of C(X) and so its complement P ERF(X)

is G15. If U1 , .. .Un are open subsets of X then by (b) {A : A c U1 u ... u Un}

is an open subset of C(X). Now take the union as (U1 , ... , Un) varies over pairwise disjoint lists of open sets of diameter at most E and then intersect over E > 0. We see that ZERODIM(X) is a G6 . Finally, CANTOR(X) = PERF(X) n ZERODIM(X) \ {0} and so is a G6 .

(d): If x is an isolated point then { x} is an isolated point of C' (X) and is bounded away from IN FTY(X) n C(X). If X is perfect, E > 0, U c X open and B = {xl, ... , Xn} c u then by Lemma 2.2(d), we can choose a Cantor set ci such that Xi E ci c V<(xi) n u and obtain c = cl u ... u Cn a Cantor set which is E close to B. Since FIN'(X) is dense in C'(X), so is CANTOR(X). Finally, if A E C' (X) choose B1 E FIN' (A), E close to A. Since X is perfect we can choose BE FIN(X), E close to B1 but disjoint from B1 . Let U =X\ B1 . As above, we can construct a Cantor set E close to B but contained in U. Hence, C is 3E close to A but distinct from A. It follows that C' (X) has no isolated points.

(e): If B = n{oi} then Bn = n{Of} and C(B) = n{C(Oi)} and so these are G<5's. If B is dense then each Bn is dense in xn and so U{i(Bn) : n = 1, 2, ... } is dense in FIN'(X) and so in C'(X). A fortiori, C(B) is dense in C(X).

(f): If B is a dense, G15 subset of X then C(B) is a dense, G6 subset of C(X) and by (d) CANTOR(X) is a dense, G15 subset of C'(X). The intersection is dense by the BCT.

(g) The continuous preimage of a G15 set is G15 and the open map preimage of a dense set is dense.

(h) The set (xn)-1 (C(R)) is G15 by (e) (or closed by (b)) and continuity of

We will want the following sharpening of part (d). Recall that a Polish space admits a Mycielski subset iff it is nonempty and perfect (see Theorem 2.10).

Proposition 4.4 If B is a Mycielski subset of a Polish space X, then CANTOR( B) is a dense subset of C'(X).

Proof For A E C'(X) and E > 0, choose {U1 , ... , Un} a finite cover of A by open subsets of X with diameter less than E and such that U; n A "I 0 fori = 1, ... , n.



46 ETHAN AKIN

Because B is dense, each B n Ui is a nonempty, locally uncountable F17 set. By Proposition 2.3, there exists a Cantor set Ci c B n Ui. Then C = C 1 U ... U Cn is a Cantor set contained in B which is f close to A.

In addition to the space C(X) of compact subsets, we will use two other constructions. The first is the space of sequences X 111 equipped with the product topology. If d is a bounded metric on X then we use on X 111 the metric given by:

d(x,y) (4.22)

It is easy to check that a sequence in X 111 is Cauchy iff each coordinate sequence is Cauchy in X. Hence the product is complete when X is. If B is a countable dense subset of X and b is any particular point of B, then it is easy to check that

{ x E B 111 : Xi = b for sufficiently large i E N } (4.23)

is a countable dense subset of X 111 • Thus, if X is a Polish space, then X 111 is a Polish space.

Proposition 4.5 Let X be a Polish space, with X 111 the Polish space of se-quences in X. Let B be a subset of X.

(a) The set DENSE =def {x : {xi : i E N} is dense in X} is a dense, G0 subset of X 111 • If B is a dense subset of X then DENSE n B 111 is a dense subset of X 111 •

(b) If B is a Go subset of X then B 111 is a G 0 subset of X 111 •

Proof (a): Let 1ri : X 111 ---+X be the ith coordinate projection map, which is an open surjection. Letting U vary over the members of a countable base we see that DENSE is a dense, G0 because:

DENSE n ( u (rri)-l(U) ) (4.24) U iEN

and it is easy to see that for each nonempty U the union is dense as well as open.

A dense set contains every isolated point. So if x E X 111 and E > 0 then we define a point y E B 111 with d(x, y) :S E by choosing Yi =Xi if Xi is isolated and by choosing Yi E B with d(xi, Yi) < Ej2i+1 otherwise. Since

0, (4.25)

the sequence y is dense if x is.




(b): This is clear from

n (ni)~ 1 (B). ( 4.26) iEI\!

Now we consider, for X, K Cu spaces with K compact, the space C(K, X) of continuous maps equipped with the compact-open topology. If d is a metric on X then we use on C(K, X) the metric given by:

d(k, r) =def sup d(k(.1:), r(.r)). ( 4.27) xEK

Since uniform convergence preserves continuity, C(K, X) is complete when X is. We will see below that C(K, X) is always separble. Thus, C(K, X) is a Polish space when X is.

If H : X __, Y and g : L __, K are continuous maps with K and L compact then we define maps H. : C(K, X) __, C(K, Y) and g* : C(K, X) __, C(L, X) by

H.(k) =def H o k and g*(k) =def k 0 g. ( 4.28)

It is easy to check that these are continuous maps. If g is surjective then g* is an isometric injection.

For x E X we denote by Cx the constant map in C(K, X) which maps all of K to the point x. This defines a map c: X __, C(K, X) which is clearly an isometric injection.

For K a Cantor space and X any CI I space we define:

IM: C(K,X) __, C'(X) by IM(k) =def k(K). Cin(K,X) =def {k E C(K,X): k is injective}.

Cfin(K, X) =def {k E C(K, X): k(K) E FIN(X)} ( 4.29)

{ k E C ( K, X) : k is locally constant}.

Proposition 4.6 Let K be a Cantor space, X be a CI I space and B C X.

(a) Cfin(K, X) is dense in C(K, X). If B is dense in X then Cfin(K, B) zs dense in C(K,X).

(b) Cin(K, X) is a Go subset of C(K, X). It is dense if X is a perfect Polish space. Conversely, if X is Polish and Cin (K, X) is dense then X is perfect.

(c) I lv! is a continuous, open .mrjection.

(d) If A c X is a Cantor set then Cin(K,X) n IM~ 1 (A) = {k E C(K,X): k restricts to a homeomorphism of K onto A } is a dense, Go in IA1~ 1 (A).

(e) If B is a G 0 subset of X then C(K, B) is a G 0 subset of C(K, X).



48 ETHAN AKIN

Proof Choose bounded metrics d on X and K. Use the induced metrics on C(X) and C(K,X). Note first that IM has Lipschitz constant 1 and so is continuous. Recall that nonempty compact metric space is the continuous image of a Cantor space (see, eg. the remarks after Proposition 2.8). It follows that I M is surjective.

For any E > 0, (k x k)(K2 \ 11;,) c (X2 \ 1x) is an open condition on k. So Cin(K,X) is a G0 .

Given k E C(K, X) choose a clopen partition K 1 , .... , KN of K such that diamk(Ki) < E/2 fori = 1, ... ,Nand choose Xi E k(Ki) fori= 1, ... , N. If B is dense in X we instead choose xi E B, E/2 close to a point of k(Ki)· Let k1 E Cfin(K, X) map Ki to Xi for each i. If A E C'(X) with distance from k(K) less than E/2 then Ai = An V.:(xi) defines a cover A1 , ... ,AN of A by nonempty closed sets. Let k2 E C(K, X) map Ki onto Ai for each i. If A is a Cantor set then we can construct a clopen partition B 1 , ... , BN of A with each Bi c 1/;,(xi)· Let k3 E Cin(K, X) map Ki homeomorphically onto Bi for each i. Each of these maps is E close to k. So cfin(K, X) is dense, Cjin(K, B) is dense when B is, IM is open and IM- 1 (A) nCin(K,X) is dense in IM- 1 (A) when A is a Cantor Set. This proves parts (a), (c), (d) and the first part of (b).

If X is a perfect Polish space then CANTOR(X) is dense in C'(X) by Proposition 4.3(d). Since IM is an open map IM- 1(CANTOR(X)) is dense in C(K, X). By (d) Cin(K, X) is dense in IM- 1(CANTOR(X)). Conversely, if Cin(K, X) is dense then IM(Cin(K, X)) =CANTOR( X) is dense in C'(X) since I M is a continuous surjection. Hence, X is perfect by Lemma 1.3b.

If B is an open subset of X then by compactness of K, C(K, B) is an open subset of C(K, X) (in fact, this follows from the definition of the compact open topology). By intersecting over a countable collection of open sets we see that C(K, B) is a G0 when B is.

0 Remark: Any clopen subset of a compact Cn space is a finite union of

members of any countable base. Hence, there are only countably many clopen subsets. If B is a countable, dense subset of X it follows from part (a) that Cfin(K, B) is a countable dense subset of C(K, X) when K is a Cantor space. For any compact Cn space K there exists a continuous surjection g: K' ~ K with K' a Cantor space (see the Remarks following Proposition 2.8). Hence, g* is an isometric injection of C ( K, X) into the separable metric space C ( K', X). It follows that C(K, X) is separable.

5 Sets of Compacta and Coherent Lists

Throughout this section X will be a Polish space. We will choose any of its bounded, complete metrics whenever we need to measure distance. The space of compact subsets C(X) is a Polish space with respect to the associated Hausdorff metrics.




A subset of xn is called symmetric if it is invariant with respect to all permutations of the coordinates. For example, the sets

tln =def {(xl, ... ,xn): For all1 :S i,j :S n X;= :r·j},

FAT.tln =def {(x1, ... ,xn): There exist 1 :S i < j :S n X;= Xj} (5.1)

are symmetric. Notice that tl 1 = X but FAT tl 1 = 0. Also, tl2 = FAT tl2 = 1x.

Definition 5.1 Let X be a Polish space and Q be a subset of C(X). We say that Q is a hereditary if it satisfies the condition

AEQ C(A) c Q. (5.2)

It is called finitely hereditary if it satisfies the condition

AEQ FIN(A) c Q. (5.3)

Q is called finitely determined if it satisfies the condition.

AEQ A E C(X) and FIN(A) c Q. (5.4)

Clearly, we have for Q C C(X) that

finitely determined ==? hereditary ==? finitely hereditary. (5.5)

Assume that Q is a finitely hereditary subset of C(X). For n = 1, 2, ... we de-fine Rr; c xn be the set of n-tuples (xl, ... , Xn) E xn such that {xl, .... , Xn} E Q. That is, ·

(5.6)

We call { Rr;} the coherent list associated with Q . In general,

Definition 5.2 Let X be a Polish space. A coherent list for (or on) X is a sequence { Rn : n = 1, 2, ... } with Rn a subset of xn such that the following condition holds for n, m = 1, 2, ... :

For (xl, ... , Xn) E xn and (yl, ... , Ym) E xm

(xl, ... ,xn)ERn and {yl,···,Ym}C{xl,···,xn} (5.7) ==? (yl,···,Ym) E Rm·

A coherent list { Rn} is called a G 0 coherent list when each Rn is a G 0 subset of xn. It is called a closed coherent list when each Rn is closed.

We will call a coherent list { Rn} trivial when Rn = 0 for all n. By ( 5. 7) Rn =f. 0 for all n when the list is nontrivial.

It is clear that the members of a coherent list are symmetric subsets. Fur-thermore, if n > k and 1r : xn -+ Xk is one of the nCk coordinate projections then a coherent list satisfies 1r(Rn) = Rk·



50 ETHAN AKIN

Definition 5.3 Let X, Y be a Polish spaces, { Rn} be a coherent list on X and f: Y----> X be a c_ontinuous map.

(a) A subset B of X is called an {Rn} set when

for n = 1,2, .... (5.8)

The set of compact {Rn} subsets is denoted C({Rn}) so that

C({Rn}) =def {AEC(X):AncRn forn=1,2, ... }. (5.9)

Let C'( {Rn}) = C( {Rn}) \ {0}.

(b) A sequence X E xN is called an {Rn} sequence when

(xo, ... , Xn-d ERn for n = 1, 2, ...

We denote by Roo the subset of {Rn} sequences in XN.

(c) The f preimage of {Rn} is

J*{Rn} =def {(fxn)- 1(Rn)},

(5.10)

(5.11)

where fxn : yn ____, xn is the product map. When Y is a closed subset of X and f is the inclusion map, we write {Rn} A Y for f*{Rn}.

Proposition 5.4 Let { Rn} be a coherent list on a Polish space X and let Q be a finitely hereditary subset of C(X). Let f : Y----> X be a continuous map with Y Polish.

00

FIN(X) n C( {Rn}) (5.12)

Furthermore, this set is dense in C ( { Rn}).

(b) B is an { Rn} set iff every finite subset of B is an { Rn} set, ie. iff FIN(B) C C( {Rn} ). If B is an {Rn} set then so is every subset of Band so C(B) c C({Rn}).

(c) C({Rn}) is a finitely determined subset ofC(X).

(d) If {Rn} is a G0 coherent list (or a closed coherent list) then C({Rn}) is a G0 (resp. a closed) subset of C(X).

(e) f*{Rn} is a coherent list on Y. A subset D of Y is an f*{Rn} set iff f(D) is an {Rn} set. In particular,

C(f* {Rn})) (5.13)




(f) A sequence x E Xl\1 is an {Rn} sequence iff the set {xn} is an {Rn} set.

(g) { R~} is a coherent list.

(h) B is an {R~} subset iff FIN(B) C Q. In particular,

Q c C( {R~}) (5.14)

Furthermore,

FIN(X) nQ FIN(X) n C({R~}) ( 5.15)

and this set is dense in Q as well as C ({ R~}). In particular, Q is dense in C({R~} ).

(i) If Q is a Gr, (or a closed) subset of C(X) then { R~} is a Gr, coherent list (resp. a closed coherent list).

(j) Rn is the coherent list associated with the s11bset C( { Rn}) of C(X).

(k) Q = C ( { R~}) iff Q is finitely determined.

(l) The coherent list associated with the subset (J*)-I(Q) is J*{R~}.

Proof (a): Since (xi, ... , Xn) E {XI, ... , Xn}n, it is clear that {XI, ... , Xn}n C Rn implies (xi, ... , Xn) E Rn· The converse follows from condition (5.7). If A E C ( { Rn}) and E > 0 then by compactness of A there is a finite subset F of A which is E close A. Since F C A, F E C ( { Rn}) and so the finite sets are dense inC( {Rn})

(b) and hence (c) are obvious from the definitions. (d): Proposition 4.3(h). (e) is obvious. (f) follows from (a) and the fact that C({Rn}) is finitely

determined. (g) merely requires reflection upon the definitions. (h): B is an {R~} subset iff every n-tuple from B lies in R~ and so, by

definition of R~, iff every finite subset of B is in Q. If A E Q then, because Q is finitely hereditary, FIN(A) C Q and so A E C({R~}), proving (5.14). On the other hand, if a finite set is an { Rn} set then it is in Q, proving ( 5.15). By (a) the closure of this collection of finites sets contains C ( { R~}) and so it certainly contains Q. It follows that Q is dense in C ( { R~}).

(i): This follows from continuity of the maps in : xn --. C(X). (j): This follows from part (a). (k): Since C( { Rn}) is finitely determined, Q = C( { Rn}) implies Q is finitely

determined. On the other hand, if A E C({Rn}) then by (h) FIN(A) C Q. When Q is finitely determined this implies A E Q. Together with (5.14) this implies that Q = C ( { Rn}) when Q is finitely determined.

(1): Again a simple consequence of the definitions.



52 ETHAN AKIN

Remark: Since being an { Rn} set is a property of finite type by (h) above, it follows from Zorn's Lemma that every { Rn} set is contained in a maximal {Rn} set.

Proposition 5.5 Let P be a symmetric subset of X 2 . Assume that P satisfies

(x,y) E P (x,x) E P. (5.16)

Define for n = 1, 2, ... the set Pn c xn by

(5.17)

(a) P2 = P.

(b) { Pn} is a coherent list, called the coherent list generated by P.

(c) {Pn} is a Gs coherent list (or a closed coherent list) if P is a Gs subset of X x X (resp. if P is closed).

(d) B C X is a {Pn} subset iff B x B C P.

Proof (a) and (b) are obvious. If 'Trij : xn ----+ X X X is the projection map using the i and j coordinates

then

n(1rij)-l(P) i,j

and so (c) follows from continuity of the maps 'Trij.

(d) is obvious from the relevant definitions.

(5.18)

Remark: The dimension 1 analogue of this result says that for D C X, { Dn} is a coherent list which is G8 or closed if Dis. The { Dn} sets are just the subsets of D.

Definition 5.6 Let X be a Polish space and {Rn} be a coherent list on X. We say that the Perturbation Condition on the list { Rn} holds when

For n = 1, 2, ... , Rn \ F AT~n is dense in Rn· (5.19)

Remark: Clearly, if X x X\ 1x is dense in X x X then X is perfect. Con-versely, if X is perfect then X" \ FAT ~n is dense in X" for all n (recall that FAT ~ 1 = 0). So X is perfect iff the coherent list { xn} satisfies the Perturba-tion Condition.




Theorem 5.7 Let Q be a G0 , hereditary subset ofC(X) for a Polish space X. Let { Rn} be the coherent list on X associated with Q . The following conditions are equivalent.

(i) {Rn} satisfies the Perturbation Condition (5.19}.

(ii) P ERF(X) n Q is a dense subset of Q.

(iii} CANTOR( X) n Q is a dense, G0 subset of Q \0.

(iv) PERF(X)nC({Rn}) is a dense subset ofC({Rn}).

(v) CANTOR(X)nC({Rn}) is a dense, Go subset ofC'({Rn}).

Proof By Proposition 4.3(c) PERF(X) and CANTOR(X) are Go subsets of C(X).

(ii)¢:? (iii): If A E PERF(X) n Q then by Proposition 4.3(f) CANTOR(A) is dense in C'(A) which is contained in Q because Q is hereditary. Hence, A is in the closure of CANTOR(X) n Q. Since CANTOR(X) c PERF(X) the reverse implication is obvious.

(ii) => (i): If (xll .. , Xn) E Rn and E > 0 then by (5.6) {x1, .. , Xn} E Q and by (ii) there exists A E P ERF(X) n Q which is E close to {XI, ... , Xn}· Since A is perfect, we can choose distinct points Yll .. :, Yn E A such that d(xi, Yi) < E

fori= 1, ... , n. So (y1, ... , Yn) E Rn \ F AT~n and E close to (x1, ... , Xn)· Since {y1, ... , Yn} C A and A is a member of the hereditary set Q, {y1, ... , Yn} E Q and so (y1, ... ,yn) ERn by (5.6) again.

(i) => (ii): We will assume that Z =def Q' \ PERF(X) n Q is nonempty, where Q' =def Q \ {0} and show that the Perturbation Condition fails. Notice that Z is an open subset of the Go set Q' and so Z is a Polish Space. From the proof of Proposition 4.3( c), Z is the countable union of closed subsets of the form Z n SING(U, V) defined by (4.21). By the BCT at least one of these sets must have a nonempty interior in Z and so in Q. That is, there exists A E Z such that Z n S(U, V) is a neighborhood of A in Q. By Proposition 5.4(h) the finite sets are dense in Q and so we can perturb A if necessary so that A E FIN(X). Then we can write A= {xll ... , Xn} and we can choose the order so that Xn E V. Since A E Q the point (xb···Xn, Xn) E Rn+l· Let N be a neighborhood in Rn+l of (x1, .... , Xn, Xn)· For (y1, ... , Yn, Yn+1) E N the set A= {y1, ... , Yn, Yn+d E Q is close to A and both Yn and Yn+l are close to Xn. So we can choose N small enough so that (y1, ... , Yn+l) E N implies {Yll ... , Yn+l} C S(U, V) and Yn, Yn+l E V. The definition of S(U, V) implies that Yn = Yn+l· Hence, with this choice of the neighborhood N we have N n Rn+ 1 C FAT ~n+ 1· This shows that the Perturbation Condition does not hold.

By Proposition 5.4(c),(d) C({Rn}) is a finitely determined, G0 subset of C(X). By Proposition 5.4(j) C({Rn}) yields the same coherent list {Rn} as Q. Hence we can apply the equivalence of (i),(ii) and (iii) to C( { Rn}) and obtain the equivalence of (i),(iv) and (v).



54 ETHAN AKIN

Remark: If we begin with any G0 coherent list {Rn} then by Proposition 5.4(j) we can apply the theorem to Q = C({Rn}).

Definition 5.8 A residual list { Rn} on a Polish space X is a coherent list such that Rn is a dense, Go subset of xn for n = 1, 2, ....

Lemma 5.9 (a) If P is a dense, G0 subset of X x X which satisfies (5.16} then the coherent list generated by P is residual.

(b) If { Rn} and { Sn} are coherent (or residual) lists on X then { Rn n Sn} is a coherent (resp. residual) list on X.

(c) Assume that {Rn} and {Sn} are coherent lists on X with Rn n Sn dense in Rn for n = 1, 2, .... If {Rn} satisfies the Perturbation Condition (5.19} then so does {Rn n Sn} and C( {Rn n Sn}) is dense inC( {Rn} ).

(d) If {Rn} is a residual list on X and D is a dense, G0 subset of X then { Rn n Dn} is a residual list.

Proof (a): When n ~ 2 we can restrict the intersection in (5.18) to pairs i,j with i =1- j 0 When i =1- j the projection maps from xn to X X X are continuous open surjections. Hence the preimages of the dense, G0 set P are dense. This intersection is dense as well by the BCT. For n = 1 we have, by (5.16),

1r1 (P), (5.20)

where 1r1 : X x X -->X is the first coordinate projection. So P1 is dense in X. (b): Coherence is obvious. For residual lists, use the BCT again. (c): Rn \ FATtln is an open subset of Rn· By assumption both it and

Rn n Sn are dense. Hence, the intersection is dense. By continuity of in, U{in(Rn n Sn): n = 1, 2, ... }is dense in U{in(Rn) : n = 1, 2, ... }which is dense inC( { Rn}) by Proposition 5.4(a). A fortiori C( { Rn nSn}) is dense inC( { Rn} ).

(d): Apply (b) with {Sn = Dn} the residual list generated by D. ()

Our main result collects a list of equivalent conditions obtained by combining and extending results in Mycielski (1964) and Kuratowski (1973). We will refer to the result as The Kuratowski-Mycielski Theorem.

Theorem 5.10 Let Q be a G0 , hereditary subset of C(X) for a perfect, Polish space X. Let { Rn} be the coherent list on X associated with Q. The following conditions are equivalent.

(i) { Rn} is a residual list, ie. Rn is dense in xn for n = 1, 2, ....




(ii) Rn \ F AT~n i8 den8e in xn for n = 1, 2, ....

(iii) The 8et Roo of {Rn} 8equence8 i8 a den8e, Go 8Ub8et of xN. (iv} The 8et Roo n DENSE of den8e {Rn} 8equence8 i8 a den8e, G0 8Ub8et of

xf\1. (v) There exi8t8 a den8e 8Ub8et A of X 8uch that An C Rn for n = 1, 2, ....

(vi) FIN(X) n C( {Rn}) i8 den8e in C(X).

(vii} C({Rn}) i8 den8e in C(X).

(viii) Q i8 a den8e 8Ub8et of C(X).

(ix} CANTOR(X) n Q i8 a den8e, G0 8Ub8et of C'(X).

(x) There exi8t8 a Myciel8ki 8Ub8et B of X 8UCh that sn c Rn for n = 1, 2, ....

(xi) There exi8t8 a den8e 8equence {Ai} in CANTOR(X) 8uch that

n-1

U Ai E Q for n = 1, 2, ... (5.21) i=O

In particular, a re8idual li8t 8ati8jie8 the Perturbation Condition ( 5.19). If D i8 a countable den8e { Rn} 8Ub8et of X then F I N(D) i8 a countable 8Ub8et of C({Rn}) which i8 den8e in C(X).

Proof(i) ¢:} (ii): Since X is perfect, xn \FAT ~n is a dense open subset of xn. Hence, Rn \ FAT~n is dense if {Rn} is. The converse is obvious. Clearly, (iii) implies (5.19).

(i) ==? (iii): Let 1fn : xN --+ xn be the projection onto the first n coor-dinates. This is an open, continuous surjection and so (ii) implies that the G0 , (7rn)- 1(Rn) C XN, is dense as well. intersecting over n = 1, 2, ... we see that the set Roo of { Rn} sequences is dense by the BCT applied to the Polish space xN.

(iii) ==? (iv): By Proposition 4.5(a) the set DENSE of sequences dense in X is a dense, G0 subset of XN. By assumption (iii) and the BCT, the intersection DENSE n Roo is dense in XN.

(iv) ==? (v): If xis a sequence in DENSE n Roo then by Proposition 5.4(£) the dense set {xi} is an {Rn} set.

If Dis countable, then FIN(D) is countable and if Dis dense in X then FIN(D) is dense in FIN(X) and hence in C(X) by Propositon 4.3(e). If Dis an {Rn} subset then FIN(D) C C({Rn}). In particular, (v)=?(vi).

(vi) ==? (vii): Obvious. (vi) ==? (viii): Obvious from (5.15).



56 ETHAN AKIN

(viii) {o} (ix): Since X is perfect, CANTOR(X) is a dense, G 0 subset of C'(X) by Proposition 4.3(d). So (vii) =?(ix) follows from the BCT. The converse is obvious.

(ix) =? (i) : If (x1 , •.• , xk) E Xk and E > 0 then by (ix) we can choose A E C({Rn}) which is E close to {x 1 , ... ,xk} and so we can choosey; E A with d(x;, y;) < E for i = 1, ... , k. Then (y1 , ... , Yk) is a point in Rk which is E close to (xi, ... ,xk)·

We have completed the proof that (i)-(ix) are equivalent. It is also obvious that (x) =? (v). We finish by showing that (viii) implies (xi) and (xi) implies (x).

(viii)=? (xi):For n = 1, 2, ... let

R(Q)n =def (Vn)- 1(Q) = {(AI, ... , An) E (C'(X))n: A1 U ... U An E Q} (5.22)

Because Q is assumed hereditary, R(Q)n satisfies (5.7) and so is a coherent list on C(X). By Lemma 4.1(d) the map Vn is an open surjection. So assump-tion (viii) implies that R(Q)n is a dense, G0 subset of C'(X)n. Thus, {R(Q)n} is a residual list on C'(X). By Proposition 4.3(d) CANTOR(X) is a dense, G0

subset of C'(X). Lemma 5.9(d) then implies that {R(Q)n n CANTOR(X)n} is also a residual list. We now apply (i) =? (iv) to this residual list and conclude that there is a dense sequence {A;} in CANTOR(X) which is an {R(Q)n} sequence. Then (5.22) implies (5.21).

(xii) =? (xi): If {A;} a dense sequence in CANTOR(X) then, since X is perfect, D = U; A; is a dense, a-Cantor subset of X. The only trouble is that some of the A; 's might have non empty interior. Choose B; a Mycielski subset of the Cantor space A; and let B = U; B;. B is still a dense, a-Cantor subset of X. Furthermore, each B; is an Fu with an empty interior. So B has empty interior by the Lemma 2.4 version of the BCT. By assumption (5.21) every finite subset of B lies in Q and hence in C( { Rn} ). By Proposition 5.4(b) it follows that B is an {Rn} set.

Corollary 5.11 Let Q be a G 0 , hereditary subset ofC(X) for a Polish space X. Let { Rn} be the coherent list on X associated with Q. The following conditions are equivalent.

(i) There exists a Cantor set A E Q.

(ii) There exists a Cantor set A C X which is an {Rn} set, ie. A E C({Rn}).

(iii) There exists an uncountable {Rn} set.

(iv) There exists a nonempty { Rn} set with no isolated points.

(v) There exists a nonempty, closed, perfect Y C X such that yn n Rn is dense in yn for all n.




Proof That (i) implies (ii) is obvious from the inclusion (5.14). By Lemma 2.2(b),(c) an uncountable, Cu space contains an uncountable, perfect subset. Hence, (ii)=>(iii)=>(iv). For (iv)=>(v) let Y = D with D a nonempty, perfect {Rn} set. Y is perfect by Lemma 2.2(a). nn c yn n Rn is dense in yn_

Now assume (v). Y is a perfect, Polish space and assumption (v) says that the coherent list {Rn} A Yon Y is in fact a residual list on Y. By Proposition 5.4(e) {Rn} A Y is the coherent list associated with Q n C(Y). By Theorem 5.10 CANTOR(Y) n Q is dense in Q n C'(Y). Since the latter is nonempty, (i) follows.

As remarked after Theorem 5.7, we can apply these results to any Gli coher-ent list {Rn} by using Q = C( {Rn} ).

Corollary 5.12 Let X be a Polish space. If P is a symmetric Gli subset of X x X which satisfies condition (5.16} then the following conditions on P are equivalent:

(i} There exists a Cantor set A C X such that A x A C P.

(ii} There exists an uncountable set A C X such that Ax A C P.

(iii) There exists a nonempty set A C X with no isolated points such that A X A c P.

(iv) There exists a nonempty, closed, perfect Y C X such that Y x Y n P is dense in Y x Y.

If, in addition, P is dense in X x X and X is perfect then there exists a Mycielski subset B C X such that B x B C P.

Proof By Proposition 5.5, P generates a Gli coherent list {Pn} with P2 = P. The equivalences then follow from Corollary 5.11 . By Lemma 5.9(a) {Pn} is a residual list when Pis dense X x X. The final result then follows from the Kuratowski-Mycielski Theorem 5.10.

6 Applications to Dynamical Systems

Throughout this section a dynamical system (X, f) is a complete metric space X and a continuous map f : X ----> X . Thus, a complete, bounded metric d is given with X. All of the constructions will be invariant with respect to replacement of d by a uniformly equivalent metric but only some of them will remain unchanged when we replace d by a metric which is only topologically equivalent.



G8 ETHAN AKIN

When X is compact, we call (X, f) a compact dynamical system. In that case, of course, all metrics topologically equivalent to d arc uniformly equivalent to d. If the system is compact then for all x E X the limit point set of the orbit sequence {fn(x): n = 1, 2, ... },denoted wf(x), is nonempty.

Recall that on finite product spaces we use the sup metric. We denote by (Xn, rn) then-fold product of copies of (X, f) (or by (X X X, f X f) when n = 2). More generally for any compact metric space A, there is an induced dynamical system, (C(A, X), f.), on the space of continuous maps.

We call the system (X, f) r-eversible iff is a homeomorphism, in which case, (X, f- 1 ) is the reverse system.

A subset A C X is called + invariant (or invariant ) if f(A) C A (resp. f(A) =A).

Let A be a compact subset of X. We call A a uniformly proximal subset if

0. (6.1)

We let Q(P ROX, f) C C(X) denote the set of uniformly proximal subsets of X. That is,

Q(PROX,f) =def {AEC(X): Forevery E>Othereexistsa

positive integer k such that d(.fk(x), fk(y)) < E for all x, yEA}. (6.2)

Since fk(A) x fk(A) c VE is an open condition condition on A E C(X) it follows that Q(PROX, f) is a G0. The singleton sets are always uniformly proximal and so i 1 (X) C Q(P ROX, f).

Example: On R define the homeomorphism h by h(x) = .r113 .The trans-lation map f ( x) = x + 1 is an isometry of the usual metric d and so the only nonempty, uniformly proximal subsets are the singleton sets. On the other hand, with respect to the complete metric h*d every compact subset is uniformly prox-imal because for any fixed K > 0, Limx_,00 (x + K) 113 - x 113 = 0. Thus, if we use h*d, we have Q(PROX, f)= C(R).

Clearly, Q(P ROX, f) is a hereditary subset of C(X), ie. A E Q(PROX) im-plies C(A) C Q(PROX). We denote by {PROX(f)n} the associated coherent list on X so that

For every E > 0 there exists a

positive integer k such that d(.fk(xi), fk(xj)) < E for all i,j = 1, 2, ... (6.3)

We will write PRO X (f) for the pairwise proximality relation PRO X (.fh C X xX.

The {PRO X (.f)n} sets, and in particular the elements of C ( {PRO X (.f)n}), are called proximal sets. So A is a proximal set if x 1 , ... , Xn E A implies (x1 , ... , Xn) E PROX(.f)n. A sequence x E PROX(.f) 00 is called a proximal




sequence. Sox EX~'~ is a proximal sequence when (x1 , ... , xn) E P ROX(f)n for n = 1,2, ...

If (X, f) is a compact dynamical system, then for A E Q(P ROX, f) we can first choose a subsequence fk' so that Lim;___, 00 diam(fn, (A)) = 0 and then by going to a further subsequence we can assume that {fk' (x)} converges to a point in X for some, and hence for all, x E A. Thus, in the compact case, A E Q(PROX, f) iff some subsequence of {fkiA} converges uniformly to a constant function, where fk lA : A -+ X is the restriction of fn. That is, some constant map Cx E C(A, X) is a limit point of the sequence {fk lA} in C(A, X).

In particular, in the compact case, PROX(f) is the usual proximality rela-tion.

Theorem 6.1 Let (X, f) be a dynamical system with X a perfect, complete metric space.

(a) P ROX (f) is a dense subset of X x X iff there exists a Mycielski subset B c X such that B x B c PROX(f).

(b) The following conditions are equivalent.

(i) There exists a Cantor set A C X such that Ax A C PROX(f).

(ii) There exists an uncountable set A c X such that Ax A C PRO X (f).

(iii) There exists a nonempty set A C X with no isolated points such that Ax A c PROX(f).

(iv) There exists a nonempty, closed, perfect Y C X such that Y x Y n PROX(f) is dense in Y x Y.

(c) The following conditions are equivalent.

(i) There exists an uncountable set A C X which is a proximal set, ze. An C PROX(f)n for n = 1, 2, .... ,

(ii) There exists a nonempty proximal set A C X with no isolated points.

(iii) There exists a nonempty, closed, perfect Y c X such that yn n PROX(f)n is dense in yn for n = 1, 2, ....

(iv) There exists a uniformly proximal Cantor set in X.

(d) The following conditions are equivalent.

(i) Q(PROX,J) is a dense subset ofC(X).

(ii) {PROX(f)n} is a residual list, ie. PROX(f)n is dense in xn for n = 1,2, ....

(iii) The set PROX(f)= n DENSE of dense, proximal sequences is a dense, G8 subset of X~'~.

(iv) There exists a dense proximal subset of X.

(v) The set of finite proximal sets,F I N(X) n C( { P ROX(f)n}), is dense in C(X).



60 ETHAN AKIN

(vi) The set of uniformly proximal Cantor sets, CANTOR(X)nQ(PROX, f), is a dense, G0 subset of C'(X).

(vii) There exists a Mycielski subset of X which is a proximal subset, ie. Bn c PROX(f)n for n = 1, 2, ....

(viii) There exists a dense sequence {Ai} in CANTOR(X) such that each U;:01 Ai is uniformly proximal, ie.

n-l

U Ai E Q(PROX,f) for n = 1,2, ... (6.4) i=O

Proof We apply the results of the previous section to Q(P ROX, f). (a),(b): Corollary 5.12. (c): Corollary 5.11. (d): Theorem 5.10.

Recall that a point x E X is called a recurrent point for a dynamical system (X, f) when xis a limit point of the orbit sequence {fk(x)}, ie. x E wf(x). We will write RECU R(f) for the set of recurrent points in X.

For a dynamical system (X, f) we define for subsets A, B of X the hitting time set

N(A,B) =def {k 2:0: Anf-k(B) i- 0}. (6.5)

Recall that

(6.6)

A subset A C X is called wandering when N(A, A) = {0} or, equivalently, when A is disjoint from U:1 fi(A). By (6.6) this is equivalent to the condition An (U:1 f-i(A)) = 0. On the other hand, if for some k > m, f-m(A) n f-k(B) i- 0 then the inclusion

(6.7)

implies, with A = B that A is nonwandering. A point is called nonwandering when every neighborhood of the point is a

nonwandering set. It is easy to see that a recurrent point is nonwandering and the set of nonwandering points is closed. The closure of the set of recurrent points is called the Birkhoff center. If the recurrent points are dense then we call the system central. It then follows that every point is nonwandering. Conversely, if every point is nonwandering, or equivalently, if every opene subset is nonwandering, then the recurrent points are dense. To see. this, let D be a countable dense subset of X. It is easy to check that

00

RECUR(!) n u (V-(x) n (u f-i(V-(x))) (6.8) e>OxED i=l




and of course we can intersect over positive rational E and get the same result. For every opene u c V.(x), u n (U:l ri(U)) is nonempty by hypothesis and so the union over D in ( 6.8) is dense. It follows that RECU R(f) is dense by the BCT.

We call the system (X, f) product central if (Xn, fxn) is central for n = 1, 2, ...

Let A be a compact subset of X. We call A a uniformly rigid subset if some subsequence of {fkiA} converges uniformly to the inclusion map iA : A~ X, that is, iA E C(A, X) is a recurrent point for the dynamical system (C(A, X), f*).

We let Q(RECU R, f) C C(X) denote the set of uniformly rigid subsets of X. That is,

Q(RECU R, f) =def {A E C(X) : For every E > 0 there exists

a positive integer k such that d(fk ( x), x) < E for all x E A}. (6.9)

For g : A~ X, with A a compact subset of X, let !(g) : A~ X x X be defined by 1(g)(x) = (x,g(x)). Since 1Uk)(A) C v;, is an open condition on A E C(X) it follows that Q(RECU R, f) is a G0 .

Clearly, Q(RECUR,f) is a hereditary subset of C(X). We denote by {RECUR(f)n} the associated coherent list on X so that

(x1, ... , Xn) E RECU R(f)n ¢=:::} For every E > 0 there exists

a positive integer k such that d(fk(xi), xi) < E for all i = 1,2, .. ,n .. (6.10)

Clearly, the set of recurrent points, RECU R(f), is equal to RECU R(fh. More generally, RECUR(f)n is the set of recurrent points of (Xn,jXn).

The { RECU R(f)n} sets, and in particular the elements of C( { RECU R(f)n} ), are called weakly rigid subsets. So A is a weakly rigid subset if x 1 , ... , Xn E A implies (x1 , ... , Xn) is a recurrent point for (Xn, fxn). A sequence x E RECU R(f)00 is called a weakly rigid sequence. So x E XN is a weakly rigid sequence when (x1, ... , Xn) E RECU R(f)n for n = 1, 2, ....

The name comes from the fact that a compact set A is a weakly rigid set iff iA is a limit point of the sequence {riA} in XA equipped with the product topology.

Lemma 6.2 Assume A C X is a weakly rigid subset.

(a) The restriction fk lA is injective for k = 1, ... If A is compact, ie. if A E C({RECURn}), then each fkiA: A~ fk(A) is a homeomorphism.

(b) The union U:o Ji(A) is a weakly rigid subset. If A is compact, then U7=o fi(A) E C( { RECU Rn}) for n = 1, ... If the system is reversible then U:-oo Ji(A) is a weakly rigid subset.

(c) If A is a uniformly rigid compact subset, ie. A E Q( RECUR), then U7=o fi(A) E Q(RECU R) forn = 1, ... In the reversible case, U~-n Ji(A) E Q(RECUR).



62 ETHAN AKIN

Proof A weakly rigid means that for every finite subset F C A there exists a sequence { kj} of positive integers tending to infinity, such that {fk' ( x)} ap-proaches x for all x E F. Then {fk1(fi(x)) = fi(fkJ(x))} approaches Ji(x) for all x E F. It follows that U:o Ji(A) is weakly rigid. If A is compact then each u~=O p (A) is compact as well. This proves (b). If F = { Xl' x2} and fk(xl) = fk(x 2 ) for some k then for all kj ;?: k, we have fkJ (x1 ) = fkJ (x2 ) and so the common limit is x 1 = x 2 . Thus, fk is injective, proving (a).

If A is a uniformly rigid subset and E > 0 we can choose <5 between 0 and E

as an E modulus of uniform continuity for f. If for some i d(fi(x),x) < <5 for all x E A then d(Ji(x), x) < E for all x E AU f(A). Part (c) then follows by induction.

In the reversible case, the integer i can be negative in the above arguments. ()


(a) RECU R(f)2 is a dense subset of X x X iff there exists a Mycielski subset B c X such that B x B C RECUR(f)2.

(b) The following conditions are equivalent.

(i) There exists a Cantor set A C X such that Ax A C RECU R(f)2.

(ii) There exists an uncountable set A C X such that Ax A C RECU R(f)2.

(iii) There exists a nonempty set A C X with no isolated points such that A x A c RECU R(f)2.

(iv) There exists a nonempty, closed, perfect Y C X such that Y x Y n RECU R(f)2 is dense in Y x Y.


(i) There exists a uniformly rigid Cantor set A C X.

(ii) There exists an uncountable set A C X which is a weakly rigid set, ie. An C RECUR(f)n for n = 1, 2, .....

(iii) There exists a nonempty weakly rigid set A C X with no isolated points.

(iv) There exists a nonempty, closed, perfect, + invariant subset Y of X such that the subsystem (Y, flY) is product central. If the original system (X, f) is either compact or reversible then Y can be chosen invariant.

(d) The following conditions are equivalent.

(i) Q(RECUR,f) is a dense subset ofC'(X).

(ii) The system (X, f) is product central, ie. RECU R(f)n is dense in xn for n = 1, 2, ....




(iii) The set RECUR(f)oo nDENSE of dense, weakly rigid sequences is a dense, Go subset of X~'~.

(iv) There exists a dense weakly rigid subset of X.

(v) The set of finite weakly rigid sets,FIN(X) n C({RECUR(f)n}), is dense in C(X).

(vi) The set of uniformly rigid Cantor sets, CANTOR(X)nQ(RECUR, f), is a dense, G0 subset of C'(X).

(vii) There exists a Mycielski subset of X which is a weakly rigid subset, ie. Bn C RECU R(f)n for n = 1, 2, .... Furthermore, B can be chosen to be + invariant, ie. f(B) C B. If the system is reversible then B can be chosen invariant, ie. f(B) =B.

(viii) There exists a dense sequence {Ai} in CANTOR(X) such that each U~j~ 0 jJ (Ai) is uniformly rigid, ie.

n-1 U f 1(Ai) E Q(RECUR,f) for n = 1, 2, ... i,j=O

Proof We apply the results of the previous section to Q(RECU R, f). (a),(b): Corollary 5.12.

(6.11)

(c): Begin with Corollary 5.11. To prove the sharpened version of (c)(iv), use (c)(i) to choose a weakly rigid Cantor set A, By and let Y be the closure in X of B = U:o fi(A) which is weakly rigid by Lemma 6.2(b). Since B is weakly recurrent and dense in Y, the subsystem is product central. If f is a homeomorphism we can use the union from -oo to oo.

(d): Theorem 5.10. The extended union in (d)(viii) is obtained from Lemma 6.2(c).

To complete the proof of (d)( vii), apply Theorem 5.10(d) to find B = U:o Bi a weakly recurrent Mycielski subset with each Bi a Cantor set. Choose Ai a Mycielski subset of the Cantor space Bi· By Lemma 6.2(a) each f1(Ai) is a Mycielski subset of the Cantor space jJ (Bi)· It follows that U~=o jJ (Ai) is a a-Cantor, dense subset of X. Each jJ(Ai) has empty interior in f1(Bi) and hence in X. It follows from the Lemma 2.4 version of the BCT that U~=o Jl(Ai) has empty interior and so is a Mycielski subset of X. If the system is reversible then we can take the union with j varying from -oo to +oo.

A dynamical system (X, f) is topologically transitive, or just transitive, if for every pair of opene subsets U, V of X, N(U, V) =f. 0. By letting U and V be disjoint open subsets of X with V C X\ f(X) we see that for a transitive system the image of f is dense in X, and so in the compact case f is surjective.

A point x EX is called a transitive point if X= wf(x). We let TRANS(!) denote the set of transitive points of X. Clearly, if there exists a transitive



64 ETHAN AKIN

point, ie. TRANS(!) =f. 0, then N(U, V) is infinite for every pair of opene subsets U, V. On the other hand,

TRANS(!) n u f-i(V) (6.12) V,k i?_k

where V varies over a countable basis of opene sets and k varies over the set of positive integers. Note that V opene implies f-k(V) is opene because f is continuous with a dense image. So for a transitive system each union in (6.12) meets each opene set U and so is dense as well as open. So by the BCT, TRANS(!) is a dense Gli when the system is transitive.

Since the orbit of a transitive point is dense, a transitive system either con-sists of a single finite, periodic orbit, or else the space X is perfect.

The system is called minimal if TRANS(!) =X. It is easy to check that (X, f) is minimal iff X is the only nonempty, closed, +invariant subset.

A system (X, f) is called weak mixing when (X x X, f x f) is transitive. This clearly is equivalent to the condition that for every four opene subsets U1, V1, U2, V2 of X

(6.13)

is nonempty. The special power of this assumption comes from the following beautiful

observation. Following Akin (1997) we will call this result The Furstenberg Intersection Lemma.

Lemma 6.4 Let (X, f) be a dynamical system. Assume

{a) Either for every pair of opene subsets U, V of X

N(U, V) n N(U, U) =f. 0,

{b) Or for every pair of opene subsets U, V of X

N(U, V) n N(V, V) =f. 0.

(6.14)

(6.15)

The system (X, f) is then weak mixing and for every four opene subsets U1, Vi, U2 , V2 of X there exist opene subsets U3 , V3 of X such that

(6.16)

Proof Either assumption implies that the system is transitive. Since f has a dense image, f-k(U) is opene for every positive integer k and every opene set u.

Let U1, V1, U2, V2 be four opene subsets of X.




Assume (a): Since N ( U1, V1) f. 0 there exists k1 ~ 0 such that U1 n f-k 1 (VI) f. 0 and so is opene. So there similarly exists k2 such that U = U1 n f-k 1 (Vl) n f-k 2 (U2) is opene. By assumption (a),

0 f. N(U,U)nN(U,rk 1 -k2 (V2))

C N(U1, f-k 2 (U2)) n N(f-k 1 (V1), f-k 1 (f-k2 (V2))) c N(Ul> f-k 2 (U2)) n N(V1, f-k2 (V2)).

(6.17)

Now choose k3 E N(Ub f-k 2 (U2)) n N(V1, f-k 2 (V2)) and let m = k3 + k2. Define U3 = U1 n f-m(U2) and V3 = V1 n f-m(V2).

N(U3, V3) c N(Ub V1) n N(f-m(U2), f-m(V2)) c N(U1, V1) n N(f-m(U2), f-m(V2)).

(6.18)

Now assume (b) instead. As before we can choose kb k2 ~ 0 such that V = V1 n f-k 2 (V2) n f-k 1 -k2 (U2) is opene. Then, as in (6.17)

0 f. N(f-k 1 (Ul), V) n N(V, V)

c N(Ul> f-k 2 (U2)) n N(V1, f-k 2 (V2)). (6.19)

Complete the proof as before.

Corollary 6.5 If (X, f) is a weak mixing dynamical system, then for opene sets U1, .... , Un, V1, ... , Vn c X the intersection

(6.20)

for n = 1, 2, ... In particular, (Xn, fxn) is then weak mixing for all n.

Proof From (6.16) we have, by induction, that there exists opene U, V c X such that

N(U, V) c N(Ul, VI) n ... n N(Un, Vn) = N(Ul X ... X Un, vl X ... X Vn), (6.21)

Where the latter is the return time Set for ( xn, f X n). It follOWS that the interSeC-tion is nonempty and so (Xn, fxn) is transitive. Since (X2n, fx 2n) is transitive, (Xn, fxn) is weak mixing.

Remark: While a transitive system need not be weak mixing, it is true that every transitive system is product central, i.e. for (Xn, fxn) the recurrent points are dense (see, e.g. Akin (1997) Proposition 4.8(d)).



66 ETHAN AKIN

Define

Q(PROX#, f) =def {A E C(X): For every E > 0, and p EX there exists a positive integer k such that d(fk ( x), p) < E for all x E A}.

(6.22)

Q(TRANS,f) =def {AEC(X): Forevery E>O,n>O,

pairwise disjoint closed A1, ... ,An C A and P1, ... ,pn EX, there exists a

positive integer k such that d(fk(x),pi) < E for all x E Ai for i = 1, ... ,n}. (6.23)

We call the sets of Q(T RANS, f) the uniformly mixed sets for (X, f). Clearly, both Q(PROX#, f) and Q(TRANS, f) are hereditary subsets of

C(X).We denote by {PROX#(J)n} and by {TRANS(f)n} the coherent lists on X associated with Q(PROX#,f) and Q(TRANS,f), respectively.

Lemma 6.6 (a) Q(PROX#, f) and Q(TRANS, f) are G8 subsets of C(X)

(b) Q(TRANS,f) c Q(PROX#,f) c Q(PROX,f) and so TRANS(f)n c PROX#(J)n c PROX(f)n for n = 1, 2, ...

(c) If(X, f) is a compact minimal system then Q(PROX#, f)= Q(PROX, f) and so PROX#(J)n = PROX(f)n forn = 1,2, ...

(d) Forn=1,2, ...

TRANS(f)n \ F AT,6.n (6.24)

In particular, TRANS(fh =TRANS(!).

(e) Q(TRANS, f)= {0} unless (X, f) is transitive.

(f) If (X, f) is transitive, but not weak mixing, then Q(T RAN S, f) = {0} U i 1 (T RANS(f)).

Proof (a): Let B be a countable basis for X which is closed under finite union and let D be a countable dense subset of X.

We first prove that for rational E > 0, U1 , ... , Un E B with pairwise disjoint closures and q1 , ... , Qn E D the set

0 = {A E C(X): there exists k such that

d(fk(x),qi) < E for all x E AnUi for i = 1, ... ,n} (6.25)

is open. An VJ(Ui) is compact and decreases to An Ui as 8 decreases to 0. So

if A E 0 we can choose positive k, 8 and E1 < E so that d(Jk(x), Qi) < E1




for all x E An \;J(Ui) for i = 1, ... , n. By shrinking 15 if necessary we can assume that it is an E - ~: 1 modulus of continuity for all x E A. That is, for x E A,y E X,d(x,y) < 15 implies d(Jk(x),fk(y)) < E- ~: 1 .

If d(B, A) < 15 andy E B n Ui then there exists x E A such that d(x, y) < 15. Hence, x E AnVJ(Ui)· It follows that d(Jk(x),pi) < ~: 1 and d(fk(y), qi) < ~:-~: 1 . By the triangle inequality, B E 0.

Intersecting the countable collection of open sets 0 we obtain a G8 which we denote Q. Clearly, Q(TRANS, f) C Q. We show, conversely, that if A E Q then A E Q(TRANS,f). Given E > 0, pairwise disjoint, closed A 1, ... ,An C A and Pl, ... ,pn E X, choose q1, ... , qn E D such that d(pi, qi) < t:/2. Since the Ai 's are compact and !3 is a basis closed under finite unions, we can choose U1 , ... , U n E !3 with pairwise disjoint closures such that Ai C Ui for i = 1, ... , n. Since A E Q, there exists k so that d(Jk(x), qi) < t:/2 for all X E An ui for i = 1, ... ,n. It follows that A E Q(TRANS,f).

For Q(P ROX#, f) the argument is similar but much easier. In the definition (6.22), we can restrict E to the rationals and p to lie in D.

(b),(c): It is clear that Q(TRANS, f) c Q(PROX#, f) c Q(PROX, f) which implies the corresponding inclusions for the associated coherent lists. The reverse inclusion for A E Q(PROX, f) can fail in two ways. First, it can happen that for x E A the limit point set wf(x) is empty and so the points do not have any common limit. This occurs in the translation example which was given after (6.2). Compactness fixes this problem. If the system is compact then for A E Q(PROX,f) some subsequence {fkiiA} converges uniformly to some constant function cp- That is, the constant function cp lies in the closed invariant set wf*(iA)· It is easy to see that the isometry c : X ----> C(A, X) is an action map and so c-1(wf*(iA)) is a closed invariant subset of X. To say that A E Q(P ROX#, f) is exactly to say that wf*(iA) contains all the constant functions, ie. that X = c 1 (wf*(iA)). This is true when A E Q(PROX, f) provided that (X, f) is compact and minimal.

(d) is obvious and it implies (e) and (f).

Because of the Furstenberg Intersection Lemma and its corollary the ana-logue of Theorems 6.1 and 6.3 are much more succinct for weak mixing.

Theorem 6. 7 Let (X, f) be a dynamical system with X a perfect, complete metric space. The following conditions are equivalent.

(i} The system (X, f) is weak mixing, i.e. TRANS(f)2 \ 1x -1-0 (ii} PROX#(J)2 is dense in X x X.

(iii) The coherent list {T RAN S(f)n} is residual, ie. T RAN S(f)n is dense in xn for n = 1, 2, ....

(iv) The coherent list {PRO X# (f)n} is residual.

(v) The set TRANS(f)oo n DENSE is a dense, G8 subset of XN.



68 ETHAN AKIN

(vi) The set FIN(X) n C( {TRANS(f)n} ), is dense in C(X).

(vii) Q(PROX#, f) is a dense subset of C'(X).

(viii) Q(TRANS,f) is a dense subset ofC'(X).

(ix) CANTOR(X) n Q(TRANS, f) is a dense, G0 subset of C'(X).

(x) There exists a Mycielski subset B of X such that En C TRANS(f)n for n = 1,2, ....

(xi) There exists a dense sequence {Ai} in CANTOR( X) such that each U::01 Ai is nniformly mixed, ie.

n-1

U Ai E Q(TRANS, f) for n = 1, 2, ... (6.26)

Proof First observe that the equivalence of (iii),(v),(vi),(viii),(ix),(x) and (xi) follow as in Theorem 6.1 and 6.2 from Theorem 5.10 applied to Q(TRANS, f). Similarly, the equivalence of (vii) and (iv) follows from Theorem 5.10 applied to Q(PROX#, f).

Next observe that (iii) =? (iv) and (iv) =? (ii) are obvious. (ii) =? (i): We verify condition (6.15). Given opene U, V C X there exists

(x, y) E (U x V) n P ROX#(J)2 and so with p = y there exists a positive integer k such that fk(x), fk(y) E Y. So k E N(U, V) n N(V, V). By the Furstenberg Intersection Lemma (X, f) is weak mixing, proving (i).

(i) =? (iii): By Corollary 6.5, (i) implies that (Xn, fXn) is transitive and so TRANS(f)n \ F AT~n TRANS(fxn) is dense in xn.

Remark: The implication (i) =? (x) is the original application of the Kuratowski-Mycielski results to dynamical systems given in Iwanik (1989). No-tice that if B is a {TRANS(f)n} set, which Iwanik calls independence, then it is a wandering set. Suppose instead that for some positive integer k there is a point x such that x, fk(x) E B. If x = fk(x) then xis a periodic point and so is not in TRANS (1)1. If x and fk ( x) are distinct then the pair ( x, fk ( x)) lies in the proper, closed invariant subset fk c X x X and so it is not in TRANS(f)2. Thus, B cannot be a {TRANS(f)n} set.

The definition of a uniformly mixed set is given in (6.23) to make these results work. The meaning, initially obscure, is revealed by the following theorem.

Theorem 6.8 Let (X, f) be a weak mixing dynamical system with X a perfect, complete metric space. Let K be a Cantor space.

(a) If A is a uniformly mixed Cantor set in X, ie. A E CANTOR(X) n Q(T RANS, f), then {fklA: k = 0, 1, 2, ... } is dense in C(A, X).




(b) gEC(K,X) isatransitivepointfor(C(K,X),f*), ie. gETRANS(f*), iff g is injective and g( K) is uniformly mixed, that is, iff g is a homeo-morphism of K onto an element of Q(T RAN S, f).

(c) The system (C(K, X), f*) is weak mixing.

Proof (a): Let h E Cfin(K, X) with h(K) consisting of n distinct points P1, ... , Pn. Let Ai = h - 1 (Pi) so that A 1, ... , An is a list of pairwise disjoint clopen sets in A. Since A E Q(TRANS, f), some subsequence of {fkiA: k = 0, 1, 2, ... } converges to h. It follows that the closure of {JkiA: k = 0, 1,2, ... }in C(K,X) contains Ctin(K, X). By Proposition 4.6(a) Cfin(K, X) is dense in C(K, X). So the closure of {fkiA: k = 0, 1, 2, ... }is all of C(K, X).

(b): If g(K) =A and g is injective, so that g: K-+ A is a homeomorphism, then g* : C(A, X) -+ C(K, X) is an isometric isomorphism from (C(A, X), f*) to (C(K, X), f*) mapping the inclusion iA to g. So g is a transitive point iff iA is. By (a) iA is a transitive point if A E Q(TRANS,f). Conversely, suppose that iA is a transitive point. If B 1 , .... , Bn are pairwise disjoint closed sets in A then we can choose pairwise disjoint clopen subsets A1, ... , An with Bi C Ai for i = 1, ... , n. Given points P1, ... ,pn E X we can define h E Cfin(K, X) so that h maps each Ai to Pi· Since iA is a transitive point, some subsequence of {fkiA: k = 0, 1, 2, ... }converges to h. It follows that A E Q(TRANS, f) when iA is a transitive point.

Finally, if g(x1) = g(x2) then h(x1) = h(x2) for every h in the closure of {fk o g : k = 0, 1, 2, ... }. If x 1 f. x2 then there exists h E C(K, X) such that h(x1) f. h(x2). Thus, if g is a transitive point x1 = x2. So any transitive point in C(K, X) is injective.

(c): From (b) it follows that (C(K, X), f*) admits transitive points and so is a transitive system. On the other hand, it is clear that

(6.27)

Since (X, f) is weak mixing, (X x X, f x f) is transitive. It follows that ( C ( K, X x X),(! x f)*) is transitive. From (6.27) it follows that (C(K, X), f*) is weak mixing.

A Cantor set in Q(TRANS,f) is called a Kronecker set for (X, f). The name comes from the case where X is the unit circle in the complex plane and f(z) = z2 (although any integral power greater than 1 will do). The system is clearly weak mixing. In fact, for every opene U C X we have fk(U) =X fork sufficiently large. It then follows from Theorem 6. 7 that there exists a Cantor set A C X such that the restrictions of the power functions fk(z) = z2k to A is a sequence of functions dense in C(A, X). Such a Cantor set is classically called a Kronecker subset of the circle. The proof of the existence of Kronecker subsets for weak mixing dynamical systems, given as Theorem 4.2 of Akin (1997), was



70 ETHAN AKIN

directly inspired by the functional analysis proof given in Katznelson (1968) for the existence of classical Kronecker sets. Katznelson in turn refers to Kaufman (1967).

We conclude this section by considering various notions of chaos. We say that a dynamical system (X, f) exhibits sensitive dependence on

initial conditions if there exists a positive E such that in any opene subset U C X there exist points x, y E U such that

supj=O,l, ... d(JJ(x),fj(y)) > E (6.28)

in which case we call E a modulus of sensitivity. Notice that a singleton set U cannot satisfy this condition. Hence, a system with sensitive dependence has a perfect state space. This condition was used as the definition of chaos in Auslander and Yorke (1980).

Theorem 6.9 Let (X, f) be a dynamical system on a complete, perfect metric space which exhibits sensitive dependence on initial conditions. There exists E > 0 and a Mycielski subset B C X such that for every pair x, y of distinct points in B

(6.29)

Proof Choose E > 0 so that 2E is a modulus of sensitivity. Consider for n = 0, 1, ... the closed subset of X x X:

00

An =def n (fxf)-J(V"_). (6.30) j=n

Recall that V. is the closed subset of X x X consisting of pairs of points at most E apart.

We observe first that An is nowhere dense. If instead it contained the product U x V of opene subsets of X then by the triangle inequality we have for all x,y E U we have

(6.31)

and so by shrinking U we can obtain an opene subset U1 such that for all x, y E u1

(6.32)

which violates the condition that 2E is a modulus of sensitivity. It follows from the BCT that

00

p =def 1x u n (X X X) \An (6.33) n=O




is a symmetric, dense, Gli subset of X x X which satisfies (5.16) since 1x C P. The result then follows from Corollary 5.12.

For a dynamical system (X, f) on a metric space, we denote by ASY M(f) C X x X the set of asymptotic pairs so that

ASYM(f) =def {(x,y) EX x X: LimJ_,oo d(Jl(x),jl(y))

The Li- Yorke set and the strong Li- Yorke set are subsets of X x X:

LI-YORKE(f) =def (PROX(J)\ASYM(J))U1x

sLI- YORKE(!) =def PROX(f) n RECUR(f)2

0}. (6.34)

(6.35)

Li and Yorke (1975) call a set B C X scrambled if every pair of distinct points in B is proximal but not asymptotic. They call the system chaotic when there exists an uncountable scrambled set. We will call such a system Li- Yorke chaotic.

Blanchard, Glasner, Kolyada and Maass (2002) considered the set of Li-Yorke pairs as defined above. They observed that if an off-diagonal pair (x, y) is recurrent then it is certainly not asymptotic. This motivates the definition of the set, sLi - Yorke whose elements we call strong Li- Yorke pairs. We call B C X strongly scrambled if every pair of points in B is proximal and recurrent. Thus, we have

B B is scrambled

is strongly scrambled B x B c LI- YORKE(!).

B x B c sLI- YORKE(!). (6.36)

We will call a system strongly Li- Yorke chaotic when there is an uncountable strongly scrambled set.

The results of Blanchard et al (2002) on Li-Yorke pairs are extended to strong Li-Yorke pairs in Akin, Auslander and Glasner (2004). See also Akin and Kolyada (2003).

Because sLI- YORKE(!), in contrast with LI- YORKE(!), is a Gli we can apply the Kuratowski-Mycielski Theorem to it.


(a) sLI- YORK E(f) is a dense subset of X x X iff there exists a strongly scrambled Mycielski subset B C X.

(b) If (X, f) is transitive then the following are equivalent:

(i} sLI- YORKE(!) is dense in X x X.

(ii} LI- YORK E(f) is dense in X x X.



72 ETHAN AKIN

(iii) PROX(f) is dense in X x X.


(i) There exists a strongly scrambled Cantor set A C X . (ii) The system is strongly Li- Yorke chaotic, i.e. there exists a strongly

scrambled uncountable set A C X.

(iii) There exists a nonempty strongly scrambled set A C X with no iso-lated points.

(iv) There exists a nonempty, closed, perfect Y C X such that Y x Y n sLI- YORKE(!) is dense in Y x Y.

Proof We obtain (a) and (c) by applying the results of the previous section to Q(PROX, f) n Q(RECUR, f) exactly as in Theorem 6.1 and Theorem 6.3.

In (b) the implications ( i) =? ( ii) =? (iii) are obvious. As remarked after Corollary 6.5 any transitive system is product central and so RECU R(f)2 is a dense G0 for such a system. Since PROX(f) is a G0 , (iii) =? (i) follows from the BCT.

0 Remark: If (X, f) is weak mixing and nontrivial then the conditions in

part (b) certainly hold. In fact,

TRANS(! x f) c sLI- YORKE(!). (6.37)

I do not know whether the analogue of part (c) for Li-Yorke chaos is true. If (x,y) E sLI- YORKE(!)\ 1x then it is a transitive point for the sub-

system of (X x X, f x f) obtained by restricting to w(f x f)(x, y). The entire set of transitive points of this subsystem is contained in sLI - Y 0 RK E \ 1x and this set is uncountable. Thus, if sLI - YORKE\ 1x is nonempty then it is uncountable. I do not know whether such a system is necessarily strongly Li-Yorke chaotic or even Li-Yorke chaotic. Recently, examples of systems (X, f) have been constructed with LI- YORK E(f) \ 1x countably infinite (see Blan-chard, Durand and Maass). For such a system sLI- YORKE(!) C 1x

7 Family Matters

We can sharpen the results of the previous section by using families of subsets of N = {0, 1, ... }. This idea, introduced in Gottschalk and Hedlund (1955), was used with great effect by Furstenberg (1981). We follow the notation of Akin (1997) Chapter 2.

A family :F is a collection of subsets of N which is upwards hereditary that is,

(7.1)




A family is proper if it is a proper subset of P the power set of N. From (7.1) it follows that a family F is proper iff N E F and 0 tJ_ F.

If F1 and F 2 are families then

(7.2)

is a family which contains the family F 1 U F 2 . We say that families F 1 and F 2 meet when F 1 · F 2 is proper.

If A C P then the family generated by A is

[A] =def {F: A C F for some A E A}. (7.3)

Clearly, [A] is proper iff 0 tJ_ A and A "I 0. We write [A] for [{A}]. A family F is called countably generated if there is a countable subset A c P

such that [A] = F. For a family F the dual family is

k:F =def { F : F n F1 # 0 for all F1 E F} = {F: N\F tJ_ F}

(7.4)

It is easy to check that k:F is proper iff F is and that kk:F = F. Also the operator k reverses inclusions, ie. F 1 C F 2 implies k:F2 C k:F1 .

A filter F is a proper family such that

F·F=F (7.5)

or, equivalently,

(7.6)

The dual of a filter is called a filterdual. It is a proper family which satisfies what Furstenberg calls the Ramsey Property dual to (7.6)

(7. 7)

Notice that

This implies,

(7.9)

In particular, F is a filter iff

F·kF=kF (7.10)

and so for a filter

F c k:F (7.11)



74 ETHAN AKIN

A C P generates a filter when it is contained in some filter, or, equivalently, when the intersections of finite subcollections of A are nonempty. In that case, the family generated by those intersections is the filter generated by A, that is, the smallest filter containing A. For example, if A1, A 2 have a nonempty intersection then [{A1,A2}] = [A1] U [A2] is the family of all sets which contain either A1 or A2. The filter generated by the pair is [A1 n A2].

Of special importance is E the family of infinite subsets of N. This is a filterdual with dual kE the family generated by the set of tails,{[k, oo): kEN} where [k,oo) = {i EN: i 2 k}.

We denote byTE the family of thick sets. A set F C N is thick when

(7.12)

where gk : N----> N by gk(i) = k + i. So F is thick iff for every n there occur runs in F of length greater than n.

The dual, kTE, is the family of syndetic sets. F is syndetic if there exists N > 0 so that every run inN of length at least N meets F.

In general, for a proper family F we define the family

TF =def {F:Fng-k1 (F)n ... ng-kn(F)EFforall k1, ... knEN}. (7.13)

For a dynamical system (X, f) we define for F C Nand A C X

fF(A) =def U fk(A) and rF(A) =def U rk(A), (7.14) kEF kEF

writing, as usual, j±F(x) for j±F({x}). For a family F and a point x E X we define the set of F limit points

w;:j(x) = n jF(x) FEk:F (7.15)

{y EX: N(x, U) E F for every neighborhood U of y}.

To see that the two expressions are equivalent notice that y E f F ( x) iff for every neighborhood U of y, U n jF(x) =I= 0 and so iff F n N(x, U) =I= 0.

For a dynamical system (X, f) we let Tt be the family generated by the return time sets of opene subsets of X. That is,

Tt =def [{N(U, V): opene U, V C X}]

Lemma 7.1 (a) Tt is a proper family iff (X, f) is transitive.

(b) If X is infinite then kE C TJ.

Proof (a) is clear.

(7.16)




Given a positive integer k and x EX, the closed set {fi(x) : i = 0, ... , k} is a proper subset of X and so we can choose an open neighborhood U of x small enough that there is an opene V disjoint from U7=o fi(U). Hence, N(U, V) is contained in the tail (k, oo) of N. This proves (b).

The crucial result for us is the family reinterpretation of the Furstenberg Intersection Lemma.

Proposition 7.2 For a dynamical system (X, f) the following are equivalent.

(a) (X, f) is weak mixing.

(b) 7f genemtes a filter.

(c) 7f is a filter.

(d) For any pair of opene sets U, V C X, N(U, V) is a thick set, ie. 7f C TB.

(e) (X, f) is tmnsitive and for every opene U C X, N(U, U) is a thick set.

Proof(c) ::::} (b) is obvious. (b) ::::} (a) is clear from (6.14). (a) ::::} (c) is the Furstenberg Intersection Lemma 6.4.

(b) ::::} (d) : For every positive integer k and subsets A, B C X it is clear that

N(A,f-k(B)) = g-k(N(A,B)) = N(fk(A),B). (7.17)

Since 0 is not thick, (b) implies that the system is transitive and so f(X) is dense in X. So for opene sets U, V the preimages f-k;(V) are opene. From (7.17)

N(U, V) ng-k1 (N(U, V)) n ... ng-kn(N(U, V))

= N(U, V) n N(U, f-k 1 (V)) n ... n N(U, f-kn(V)) (7.18)

and the latter is nonempty by assumption (b). (d) ::::} (e): Obvious. Notice that a thick set is nonempty. (e)::::} (a): Given opene sets U, V C X, we can choose a positive integer k so

that W = Vnf-k(U) is opene because (X, f) is transitive. Because N(W, W) is thick, N(W, W) n g-k(N(W, W)) =f. 0. Notice that N(W, W) c N(V, V). Since f(W) C U, (7.17) implies that N(f(W), W) = g-k(N(W, W)). Hence,

N(V, V) n N(U, V) :J N(W, W) n g-k(N(W, W)) (7.19)

and the latter is nonempty since N(W, W) is thick. This verifies condition (6.16) and so by the Furstenberg Intersection Lemma 6.4, the system is weak mixing.

0



76 ETHAN AKIN

Assume that Q is a countably generated family contained in kTj for a weak mixing system (X, f). Define

Q(TRANS, f, Q) =def {A E C(X): For every E > 0, n > 0, G E Q

pairwise disjoint closed A 1 , ... , An C A and p 1 , ... , Pn E X, there exists

kEG such that d(fk(x),pi) < E for all x E Ai for i = 1, ... , n }. (7.20)

Clearly, Q(TRANS, f, Q) is a hereditary subset of C(X).We denote by {TRANS(!, Q)n} the coherent list on X associated with Q(TRANS, f, Q)and by T RANS(f, Q) 00

the set of {TRANS(f,Q)n} sequences in XN.

Theorem 7.3 Let (X, f) be a nontrivial, weak mixing dynamical system. For any countably generated family Q which is contained in kTj the following hold.

(i} Q(TRANS,f,Q) is a dense G0 subset ofC(X).

(ii} The coherent list {TRANS(!, Q)n} is residual, ie. TRANS(!, Q)n is a dense Go in xn for n = 1, 2, ....

(iii} The set TRANS(!, Q)oo n DENSE is a dense, Go subset of XN.

(iv} The set FIN(X) n C( {TRANS(!, Q)n} ), is dense in C(X).

(v) CANTOR(X) n Q(TRANS,J, Q) is a dense, G0 subset of C'(X).

(vi} There exists a Mycielski subset B of X such that Bn C TRANS(!, Q)n for n = 1, 2, ....

(vii} There exists a dense sequence {Ai} in CANTOR(X) such that

n-1

U Ai E Q(TRANS,f,Q) for n = 1, 2, ... (7.21)

(viii} If A E CANTOR( X) n Q(TRANS, f, Q), then {fkiA: k E G} is dense in C(A,X) for every G E Q.

Proof A finite periodic orbit with period greater than one is not weak mixing. As the system is nontrivial, X must be infinite and so is perfect because the system is transitive.

For any fixed G E P the condition descibed in (7.20) defines a G0 subset of C(X). This is proved exactly as in Lemma 6.6(a) which was precisely the case of G = N. We intersect over a countable family generating Q to see that Q(TRANS, f, Q) is a G0 .

Because G E Q meets every N(U, V) it follows that f- 0 (V) is a dense, open subset of X for every such G and every opene subset V of X. Intersecting over




G from a countable family generating g and V from a countable base for X, we see that TRANS(f,9h is a dense Gli subset of X by the BCT.

By (6.20) and Lemma 6.4, Yj = Tpn and so by the argument of the previous paragraph, TRANS(!, 9)n is a dense Gli subset of xn. This proves (ii). As in Theorem 6.7, (ii) implies (i) and (iii)-(vii).

(viii) is proved just as is Theorem 6.8(a).

Remark Since Tt is a countably generated filter, (7.11) implies we can replace g by Tt · g and still have a countably generated family contained in kYj. In addition, we then have Tt · g = Q. By Lemma 7.1(b) the family g then satisfies

kB c Yj c 'Tt · g g c kYj. (7.22)

Lemma 7.4 Assume that (X, f) is a compact, weak mixing dynamical system. For every x EX, the limit set WkTtf(x) is a nonempty, closed+ invariant subset of X. If, in addition, (X, f) is minimal then WkTtf(x) =X for all x EX.

Proof Because (X, f) is weak mixing, Tt is a filter by Proposition 7.2. By (7.15) and the finite intersection property, wgf(x) is nonempty in a compact space for any filterdual g.

Now if y E WkTtf(x), U, V C X are opene and 0 is any neighborhood of f(y) then

0 f. N(x, r 1(0)) n N(U, r 1 (V)) = g- 1(N(x, 0) n N(U, V)) (7.23)

where g : N --> N is the translation map given by g(t) = t + 1. It follows that f(y) E WkTtf(x). Thus, the limit set is+ invariant.

If the system is minimal then the only nonempty, closed + invariant subset of X is X itself.

The major result of this section combines work of Xiong and Yang (1990),of Akin and Kolyada (2003), and of Shao and Ye (2003). It provides a sharpening of Theorem 6.10 in the weak mixing case.

Theorem 7.5 Let (X, f) be a nontrivial, compact, weak mixing dynamical sys-tem. For every x E X there exist y E X and a Mycielski subset B C X such that for every z E B there is an increasing sequence of positive integers { ki} which satisfies:

y (7.24)



78 ETHAN AKIN

is a sequence dense in X. (7.25)

In particular, (x, z) ELI- YORKE(!) for every z E B. If, in addition, (X, f) is minimal then for each x EX we can choosey= x

in the above condition. In that case, ( x, z) E sLI-Y 0 RK E (f) for every z E B.

Proof Let y be any point in Wkr1 f(x). For every E > 0, N(x, V,(y)) E k7f. Hence,

Q =def kB · [{N(x, V,(y)): E > 0}] (7.26)

is a countably generated family contained in the family of tails kTt. By Theorem 7.3 there exists a Mycielski subset B of X which is a union of

a sequence of Cantor sets {Aj E Q(TRANS,f,Q)}. Any z E B is a point of TRANS(!, Q)l. So for any G E Q the set {Jk(z): kEG} is dense in X.

Choose k1 < ... < km, in N(x, V1 (y)) so that {Jk, (z), ... , fkm, (z)} is 1 dense, ie. has Hausdorff distance less 1 from X. Inductively, choose kmn+l < ... < kmn+l in N(x, Vrjn+l (y) )n(kmn, 00) SO that {Jkmn+ 1 (z), ... , fkmn+l (z)} is 1/n+ 1 dense.

The resulting sequence {ki} clearly satisfies (7.24) and (7.25). Some subse-quence of {Jk'(z)} converges toy and so (x, z) E PROX(f). Some subsequence of {Jki(z)} converges to z as well. So both (y,y) and (y,z) are limit points of {(fki(x),jk'(z))}. Hence, (x,z) ELI- YORKE(!).

When the system is minimal then wkr1 f(x) = X and so we can choose y = x. So both (x,x) and (x,z) are limit points of {(Jk;(x),fk'(z))}. Hence, (x, z) E sLI- YORKE(!).

Bibliography

E. Akin (1997) Recurrence in topological dynamics: Furstenberg families and Ellis actions Plenum Publishing Co., New York.

E. Akin (1999) Measures on Cantor space Topology Proceedings 24:1-34.

E. Akin and S. Kolyada (2003) Li- Yorke sensitivity Nonlinearity 16:1421-1433.

E. Akin, J. Auslander and E. Glasner (2004) The topological dynamics of Ellis actions (to appear). Manuscript available from the authors.

J. Auslander and J.A. Yorke (1980) Interval maps, factors of maps and chaos Tohoku Math. J. 32:177-188.

F. Blanchard, E. Glasner, S. Kolyada, and A. Maass (2002) On Li- Yorke pairs J. fUr die Reine und Angewandte Mathematik 547:51-68.




F. Blanchard, F. Durand and A. Maass (2004) Constant-length substitu-tions and countable scrambled sets Nonlinearity (to appear).

H. Furstenberg (1967) Disjointness in ergodic theory, minimal sets and a problem in diophantine approximation Math. Systems Th. 1:1-55.

H. Furstenberg (1981) Recurrence in ergodic theory and combina-torial number theory, Princeton U. Press, Princeton.

E. Glasner and D. Maon (1989) Rigidity in topological dynamics Ergodic Th. & Dynam. Sys. 9:309-320.

W. Huang and X. Ye (2002) Devaney's chaos or 2-scattering implies Li-Yorke 's chaos Topology Appl. 117:259-272.

W. Hurewicz and H. Wallman (1941) Dimension theory, Princeton U. Press, Princeton.

A. Iwanik (1989) Independent sets of transitive points Dynamical sys-tems and ergodic theory Banach Center Publ. # 23, pp. 277-282.

Y. Katznelson (1968) An introduction to harmonic analysis, John Wiley and Sons, New York.

R. Kaufman (1967) A functional method for linear sets Israel J. Math. 5:185-187.

K. Kuratowski (1966) Topology, Academic Press, New York.

K. Kuratowski (1973) Applications of the Baire-category method to the problem of independent sets Fundamenta Mathematicae 81:65-72.

T.Y. Li and J.A. Yorke (1975) Period three implies chaos Amer. Math. Monthly 82:985-992.

E. Marczewski (1961) Independence and homomorphisms in abstract alge-bra Fundamenta Mathematicae 50:45-61.

J. Mycielski (1964) Independent sets in topological algebras Fundamenta Mathematicae 55:139-147.

J. Oxtoby (1980) Measure and category (2nd Ed.) Springer-Verlag, Berlin.

S. Shao and X. Ye (2003) F-mixing and weak disjointness Topology Appl. (to appear)

J. Xiong and Z. Yang (1990) Chaos caused by a topologically mixing map, Advanced Series in Dynamical Systems, World Scientific, 9:550-572.




Duality and the one-sided ergodic Hilbert transform

by I. Assani

ABSTRACT. Let Xn be an iid sequence of random variables with finite p-moment, 1 < p < oo and zero expectation defined on a probability measure space (0, A, P) We prove that we can find a set of full measure O' such that for w E O' , for each 1 < r ::; oo, the random series

f: Xn(w)g(Sny)

n=l n

converges a.e v for all dynamical system (Y, 9, v, S) and all g E Lr(v). We also characterize the functions f E £ 1 for which the averages

f: f(Tnx)Xn(w)

n=l n

converge a.e.(with respect tow).

1. Introduction

In [A1] we proved that if Xn is an iid sequence of symmetric random variables with finite p-moment , for some 1 < p < oo then for P a.e. (w) the averages

1 N N L Xn(w)g(Sny)

n=l

converge a.e v for every dynamical system (Y, A, v, S) for all g E U ( v), 1 < r :::; oo. We noticed later in [A2] that the symmetry assumption could be removed.

One of the novelties of the result in [A1] was the "break of duality". We mean by this the fact that the sequence Xn providing the random weight Xn(w) and the function g belong to spaces £P and Lr that go beyond the duality normally imposed by Holder's inequality. (we can have the pointwise convergence even if 1 < r < pS).

Here we look at the corresponding problem for the one-sided ergodic Hilbert transform given by the series

f= Xn(w):(Sny)

n=l

Department of Mathematics, UNC Chapel Hill, NC 27599 AMS subject classification 60F15 37 A30 email: assani@email. unc.edu.


81




82 I. ASSANI

This is motivated in part by a question raised in [CJR]. They asked if one could still have the break of duality for this one sided ergodic Hilbert transform. We will show in theorem 1 that the situation in this case is different from the Cesaro averages: we do not have a.e. convergence for p = 1 and r = oo.

THEOREM 1. There exist symmetric L 1 random variables for which one can not find a set of w with full measure for which the following series

converges a.e. v for all dynamical systems (Y,Q,v,S) and all g E L 00 •

Our second result is the following theorem that deals with the remaining values of p and r.

THEOREM 2. Let Xn be an iid sequence of random variables with finite p-moment, 1 < p < oo and zero expectation defined on a probability measure space (0, A, P). We can find a set of full measure 0' such that for w E 0' , for each 1 < r :::; oo, the random series

converges a.e v for all dynamical system (Y, Q, v, S) and all g E Lr(v), 1 < r:::; oo.

We will give a proof of this result in the following section. In the final section we . . ~ f(Tnx)

consider the "dual case" of the convergence of the senes L..J Xn(w) where n=l

n

f E L1 (JL) and Xn is a sequence of L1 iid random variables with zero expectation. We will show that in this case we do not have convergence for all function f E L1.

oo f(Tn ) However the almost everywhere convergence of the series L n x Xn(w) holds

n=l as soon as X 1 E LP for some 1 < p:::; oo.

2. Proof of theorem 1

If theorem 1 was false then by applying such a result to the family of rotations (11', B('ll'), m, Sa) on the one dimensional torus 1!' where Sa(Y) = y +a and to the function e21riy we would have the a.e. convergence for all t of the random Fourier series

. By Billard result [Ka] we would have the uniform convergence a. e. of this random Fourier series. But such a conclusion has been shown to be false in [T] as soon as



DUALITY AND THE ONE-SIDED ERGODIC HILBERT TRANSFORM 83

3. Proof of theorem 2

We list in the following proposition and the next theorem some of the elements we will use.

PROPOSITION 3. Let Xn be a iid sequence of random variables with mean zero such that E(IX1 IP) < oo for some p, 1 < p < oo. Define the variables X~ Xnl{IXni:Sn} -lE[Xnl{IXni:Sn}] Then for (3 large enough we have the following:

1 (N+1) 13

(1) For P a.e the sequence Nf3 I: IX~(w)llgi(Sny) converges to zero for n=N/3

all dynamical systems (Y, Q, v, S) and all g E U(v), 1 < r::::; oo. 00 00

(2) For a function g E U(v) we have I: I: ivAi < oo where Ai = {y: N=l _1_

i=[Nr-1] i-1<gS:i}.

PROOF. In [A1] by using lemma 5 we proved that for P a.e. the sequence 1 (N+l)/3

Nf3 I: IXn(w)llgi(Sny) converges to zero for all dynamical systems (Y,Q,v,S) n=N/3

and all g E U(v), 1 < r::::; oo. The details are given on page 148 of the paper.

As IXnl{IXni:Sn}l S: IXnl we only need to show that

1 (N+l)/3

Nf3 I: llE[Xnl{IXni:Sn}JIIg(Sny)l Nil

converges v a.e. to zero. The variables Xn having zero mean we have

IJE[Xnl{IXni:Sn}JI = IJE[Xnl{IXnl>n}]l

. By Holder and Chebyshev inequalities the last absolute value is less than 1 p_ 1

IIXnllp~IIXniiJ = IIXI!I~~ nq nq

Therefore we can bound the L 1 (v) norm of

1 (N+l)/3

Nf3 I: llE[Xnl{IXni:Sn}JIIg(Sny)l Nil

by

Gil II - 1 - 1- N{3-l = c~ g 1 Nf3 Nf3l'. f3l'.+l q n q

Part (1) follows now from the summability of the series.

Part(2) can be obtained by partial summation. It is done on page 144 of [A1]. D

One of the main ingredients for the proof of theorem 1 is a consequence of the proof of theorem 4.3.1 in Salem-Zygmund [SZ]



84 I. ASSANI

THEOREM 4. There exists a finite constant C such that for each positive integer T for all Yn independent sequence, mean zero L 2 random variables we have

T T

(1) II sup I LYne21rintlll2 S C t n=1

(L E(IYnl 2)ylogT n=1

A proof of this theorem was shown to us by M. Weber and is indicated in [A1]. We proceed now with the proof of theorem 2.

We first introduce some notations; We fix 1 < p < oo, 1 < r :::; 2 (it is enough to prove the theorem for r in this range) and consider a sequence Xn of iid, mean zero random variables with E(IX1 IP) = 1. We introduce the variables

M I

I I "'"" X k 2 "kt Xn = Xnl{IXni:'On}- E[Xnl{IXn\:'On}] and the partial sums SM(t) = L ke 1r•

k=1 Mx

and S M ( t) = L T e21rikt. We fix (3 large enough. k=1

LEMMA 1. There exists a finite constant C such that for all positive integer N we have

[(N+1)~'J x' c II s~p I L nne21rintlll1 S N~Y+l/ 2 ~

n=[N~']+l

(2)

PROOF. We can apply (1) to the variables Yn defined as Yn = f for [N~] < n:::; [(N + 1)~] and Yn = 0 for 1:::; n:::; [N~]. We have

[(N+1)~']

II s~p I L Yne21rint Ill S C n=[Ni3] 1

. As

we have

II [(N+1)~'J xl II [(NH)~'J 1

S~PI L nne21rintl sC( L nP)1/2~ n=[Ni3]+1 1 n=[Ni3]

The right hand side of the previous inequality being less than

the lemma follows.

c /3-1 ~ NP~/ 2 N-2- v log N

LEMMA 2. For ~(p-; 1 )+1 > "! + 1 we have for P a. e. w

00

(3) L N' sup 1Bi(N+1)~'] (t) - s;Ni3] (t) I < 00 N=1 t

0




PROOF. Using lemma 1 we have

oo Ill l<N+lJilJ xl Ill oo c "' sup I "' _!!:_e2rrint < "' N"~ y'log N ~ t ~ n - ~ Nf39-+1/2 N=1 n=[Nil]+1 1 N=1

The assumption f3(p-21)+1 > "( + 1 makes the right hand side summable. 0

We consider now a dynamical system (Y, 9, v, S) and a fixed nonnegative func-tion g E U(v). We write g as g = g}v + g'J.r where g}v = g 1\ N r!_l. We denote by O"g}._, the spectral measure of the function g}v.

We have for w fixed in the set of full measure given by lemma 2 and for 'Y = 2-r

2(1-r) · 00 [(N+l)il] I

L II L Xn~w) g}v(Sny)ll N=1 n=[Nil]+l 1

00 [(N+l)il] I

~ L IINr!._l L xn~w) g}v(Sny)ll N=1 n=[Nil]+1 2

00 [(N+l)il] I 1/2 = L (/ INr!._l L xn~w) e2rrintld0"g}._,(t))

N=1 n=[Nil]+1 00 [(N+l)il] I

~ L s~p I L xn~w) e2rrintiii9Jvll2 N=1 n=[Nil]+1

00 [(N+1)il] I

= L s~p I L xn~w) e2rrintl(/ lg}v(y)l2dv(y))1/2 N=1 n=[Nil]+1

00 2-r [(N+l)il] X 1 W

~ LN2(1-r)s~p~ L n~ )e2rrintlll9ll~/2 N=1 n=[Nil]+1

<oo We can also estimate the same partial sums with the sequence g'J.r Using part (2) of Proposition 3, we obtain:

<oo



86 I. ASSANI

From these two series of estimates we can conclude that the sequence [M~'~] I

H[lw/3] = L xn~w) g(Sny) n=l

converges v a.e. .

To establish the a.e. convergence of the series

we just need to show that

lim sup IHn - H[N~'~] I = 0 N [N~'~J+l:'On<[(N+l)r;J

v a.e .. But using part 1 of Proposition 2 we have

lim sup IHn - H[N~'~] I N [N~'~]+l:'On<[(N+l)/3]

<

<

[(N+lJilJ x' lim """" I n(w)llg(Sny)l N L..., n

n=[Nil]+l

. 1 hm-N Nf3

0

[(N+1) 13 ]

L IX~(w)lg(Snyl n=[N~'~]+l

Therefore we have shown that for a.e w the series

converges v a.e for all dynamical systems (Y, 9, v, S) and all g E Lr(v).

To conclude it remains then to prove that the series

converges v a. e.. It will be enough to show that each one of the following series converges v a.e ..

(4) f Xn(w)l~IXnl>n} g(Sny)

n=l

and

(5) f ( -lE[Xn~IXnl>n]) g(Sny)

n=l 00

As the series LP({IXnl > n}) converges because 1E[IX1 1] < oo, P a.e. w belongs n=l

then only to finitely many of the sets {IXnl > n}. Hence the series in (4) converges v a.e. as we will only have finitely non zero terms in the series.




The series in (5) is absolutely summable because it is dominated in absolute value by the series

C~ \g(Sny\ ~ HE n=l n q

(we showed in the proof of part 1 of proposition 2 that \lE[Xnl{IXnl>n}\::; 1~1!: ). n q

This ends the proof of theorem 1.

Remarks (1) One can also obtain the pointwise convergence for the fractional one sided

ergodic Hilbert transform

where 0 < c(p, r) < o ::; 1 and c(p, r) is a constant depending only on r andp.

(2) In summary we have shown that we do not have pointwise convergence for p = 1 and for r = oo. On the other hand theorem 1 shows that we have pointwise convergence if 1 < p::; oo and 1 < r ::; oo. At the present time we do not know if we have convergence for 1 < p ::; oo and r = 1.

(3) G. Cohen pointed out to us that he obtained with similar arguments a proof of theorem 2.

4. The dual case

In this section we look at the convergence of the series f f(Tnx)Xn(Y). n=l n

In this case one can obtain a more complete characterization of those val-ues of p and r for which the convergence holds as the next theorem shows.

THEOREM 5. Let (X, B, p,, T) be an ergodic measure preserving system and 1 ::; p ::; 00

(1) The Case p = 1

There exists f E L 1 (p,) for which one can not find a set of x with

. . ~ f(Tnx)Xn(w) .. full measure on whzch the serzes ~ converges for all zzd

n n=l sequence Xn of L 1 random variables with zero expectation.

(2) The case 1 < p ::; oo For all f E LP(p,) , for p, a.e x, for all Xn E LP iid random variables

. . . ~ f(Tnx)Xn(w) wzth zero expectatzon, the serzes ~ converges for a. e w

n=l n

PROOF. The case p = 1. The proof of theorem 8 in [A3] actually shows that the following are equivalent

for a nonnegative sequence Cn such that limn c.;: = 0. {#k;9s. ~ .l}

(1) sup k n < oo n n



88 I. ASSANI

(2) for any iid sequence Xn of L1 random variables the sequence cnX;:(w)

converges a.e. to zero.

The equivalence is stated for any dynamical system (Y, 9, v, S) but a close look at the proof shows that we used shifted random variables of X 1 (given at the bottom of page 243), thus L1 iid random variables.

In [ABM] we showed that we could find in any ergodic dynamical system { #k· f(Tkx) > .!. }

(X, A, J-L, T) a function f E L1 for which sup ' k - n is not finite a.e .. n n

By the remark above this means that for each x in a set of positive measure X f we can find a sequence Y: of L1 iid random variables such that

n

does not converge a.e. ( w) to zero. As a consequence of this and the a. e. convergence to zero of the sequence f(:nx) we can conclude that the sequence

n

does not converge a.e. (w) to zero, where X~ = Y: -JE(Y:). This implies that the senes f= f(Tnx~X~(w)

n=l

does not converge a.e., as its generic term does not converge to zero.

The case 1 < p :::; oo. We will use for this the three series theorem. By this 00 f(Tnx)Xn(w) 1.f theorem we will have the pointwise convergence of the series 2.:.: :...-:... _ _:.__.:.:..~

n

for each c > 0 the following series converge

(1) f= P[l f(Tnx)Xn(w) I > c] n

n=l

(2) f= 1 Xn(w). f(Tnx) dP(w) n=l {IXn(w)< lt<~;;xll} n

(3) f= { c.n [Xn(w). f(:nx)]2dP(w) n=l }{IXn(w)< lf(Tnx)l}

n=l

We can observe that if some of the terms f(Tnx) are zero then we would have less terms to consider in these series. So without loss of generality we can assume that lf(Tkx)l > 0 for each k.

We will also use the result in [J] which says that for each f E L 1 (J-L) for each 1 < s < oo, for a.e. x, the series

(6)

converges.




For the first series (1) we have the following estimate

f= P[IXn(w)ll f(:nx) I >c) n=1

00

< ~ P[IXn(w)l > lf(~:x)l]

< f= [f(:nx)piiXIil~ < oo n=1

by using Chebyshev 's inequality in LP and (6).

We can write the second series as

f= jxn(w/(:nx)- f= { c.n Xn(w)_f(:nx) dP(w). n=1 n=1 }{IXn(w)?:_ lf(Tnx)j}

As the first term converges because the variables Xn have zero expectation, we can focus only on the second term. We have for any 1 < r < oo, (r* = r.':_ 1 .)

f= 1 f(Tnx)Xn(w) dP(w) n=1 {1Xn(w)J2: lf(~i{x)l} n

= f= f(Tnx) 1 Xn(w)dP(w) n=1 n {IXn(w)l?:. lf<~;;xll}

::;(f=lf(:nx)lr*)1/r*(fl r c.n Xn(w)dP(wW)1/r n=1 n=1 }{1Xn(w)J2: )f(Tnx)j}

by using Holder's inequality. Again by (6) we just need to focus our attention on

(~I ~IXn(w)J2:1f(~;;x)l} Xn(w)dP(wW)1/r.

By using this time Holder's inequality for integrals and Chebyshev's inequality in LP we have

= (f (lf(~:x)lrcp-1J)1/riiX111~ n=1

which is finite if we choose r large enough so that r(p - 1) > 1. This choice of r is consistent with the implicit assumption made on r* namely that n > 1.

Finally for the third series we have

f 1 [Xn(w). f(Tnx)]2dP(w) n=1 {IXn(w)< lf<~;;xll} n



90 I. ASSANI

= f (f(:nx))2 r c.n IXniPIXnl2-pdP(w) n=l }{IXn(w)< lf(Tnx)l}

< ~ (f(:nx) ) 2 I/( 2;::~~~~P J IXniPdP(w) =~I J(:nx) IP C 2 -piiX1II~ < 00 This ends the proof of this theorem. 0

References

[A1] I. Assani: "A weighted pointwise ergodic theorem" Ann. Inst. Henri Poincare, vol 34, 1, 1998, 139-150.

[A2] I. Assani: "Wiener Wintner Dynamical Systems," ?reprint, 1998, to appear in Erg Ther and Dyn. Syst.

[A3] I. Assani: "Strong laws for weighted sums of iid random variables" Duke Math. J., vol 88, 2, 1997, 217-246.

[ABM] I. Assani, Z. Buczolich and D. Mauldin: "An £ 1 counting problem in ergodic theory" preprint, 2003.

[CJL] G. Cohen, R. Jones and M. Lin: "On strong Laws of large numbers with rates" ?reprint, 2002.

[J] R. Jones:"lnequalities for the ergodic maximal function" Studia Mathematica,60, 1977, 111-129.

[Ka] J.P. Kahane: "Some Random Series of Functions" Cambridge, second edition [SZ] R. Salem and A. Zygmund: "Some properties of trigonometric series whose terms have

random signs" Acta Math., 91, 1954, 245-301. [T] M. Talagrand: "A Borderline Random Fourier Series" The Annals of Probability, voL 23,

2, 1995, 776-785.




Rigidity Conditions in Topological Dynamics Related to a Theorem of George Sell

Joseph Auslander and Kenneth Berg

ABSTRACT. A rigidity condition, tightness, is introduced in topological dy-namics and related to a theorm of George Sell concerning the omega limit set of a flow arising in autonomous differential equations

Historically, the study of abstract dynamical systems (flows) has had a strong interaction with the theory of (especially autonomous) differential equations. For example, see [5]. A precise definition of the type of flow we have in mind follows: 1

DEFINITION 1. A real metric flow (rmf) (X, cp) consists of the real numbers R, a metric space X with a metric d, and a jointly continuous mapping cp: (t, x) r-+ 'Pt(x) from R x X--+ X satisfying 'Po(x) = x and 'Pt+s(x) = 'Pt('Ps(x)) for all s, t, x. The flow is called a compact real metric flow if the space X is compact.

Suppose that X is an open subset of Rn and f: X--+ Rn. Under reasonable assumptions on J, for example that f is locally Lipschitz, the initial value problem x' = f(x), x(O) = x0 has a unique solution t r-+ 'Pt(x0 ). If necessary the function f can be rescaled so that solutions are defined for all time. Assuming still that f is locally Lipschitz leads to the conclusion that ( t, x) r-+ 'Pt ( x) defines an rmf.

We will examine a theorem of Sell as stated in Jane Cronin's text. This the-orem has a strong connection with topological dynamics (as Sell was aware). The viewpoint from topological dynamics will provide insight into the proofs and allow us to broaden the results.

First, we need some standard terminology.

DEFINITION 2. Suppose that (X, cp) is an rmf and x EX. (1) O(x) = { y I y = 'Pt(x) for some t E R} is the orbit of x. (2) o+(x) = { y I y = 'Pt(x) for some t E R+} is the positive semi-orbit of x. (3) O(x) = { y I for every E > 0 there is at E R such that d(y, 'Pt(x)) < E}

is the orbit closure of x. (4) O+(x) = { y I for every E > 0 there is at::=: 0 such that d(y, 'Pt(x)) < E}

is the positive orbit closure of x.

1991 Mathematics Subject Classification. Primary 37B05; Secondary 37B20. Key words and phrases. Toplogical dynamics, differential equations. 1The authors appreciate the efforts of a very careful referee. Several vague arguments have,

we hope, been made clearer in response to his comments.


91




92 JOSEPH AUSLANDER AND KENNETH BERG

(5) O(x) = { y I for every E > 0, T > 0 there is at E R such that t > T and d(y, <pt(x)) < E} is the omega limit set of x.

DEFINITION 3. Suppose that (X, <p) is an rmf. A subset Y of X is called invariant if <pt ( x) E Y for every x E Y and t E R.

DEFINITION 4. Suppose that (X, <p) is an rmf. A non-empty subset M of X is called minimal if O(x) = M for every x EM.

It's clear that a minimal subset is invariant and closed and that any two minimal sets are disjoint. It is simple to show, using Zorn's lemma, that every non-empty compact invariant subset of an rmf contains at least one minimal set. If M is a compact minimal set then O(x) = O(x) = M for every x EM.

DEFINITION 5. Suppose that (X, <p) is an rmf and x, y EX. We say x andy are (positively) asymptotic if lim d( <pt ( x), 'Pt (y)) = 0.

t->oo

The next two definitions are less standard.

DEFINITION 6. Suppose that (X,<p) is an rmf and x, y EX. We say x andy are shift asymptotic if there is a real numbers such that lim d( 'Ps+t(x), 'Pt(Y)) = 0.

t->oo That is, some shift of x is asymptotic to y.

DEFINITION 7. Suppose that (X, <p) is an rmf and x EX. We say that x, or that the orbit of x, is shift attracting if there is a 15 > 0 such that if d( 'Pu ( x), 'Pv (y)) < 15 for some real numbers u and v then x and y are shift asymptotic.

The following lemma follows readily from the definitions.

LEMMA 8. Let (X, <p) be an rmf.

(1) Asymptoticity is an invariant equivalence relation. (That is, if x andy are asymptotic, so are <p8 (x) and <p 8 (y), for s E R)

(2) If x andy are shift asymptotic, so are 'Pa(x) and 'Pb(Y) for all a, bE R. (3) Shift asymptoticity is an invariant equivalence relation.

LEMMA 9. Suppose that (X, <p) is an rmf and that x E X is shift attracting . Every point in O(x) is shift asymptotic to x and any two points of O(x) are shift asymptotic to each other. Every point of 0( x) is shift attracting .

PROOF. Choose 15 > 0 so that if y E X and d( <pu ( x), 'Pv (y)) < 15 for some real numbers u and v then x andy are shift asymptotic. Let x E O(x). Choose t E R so that d( 'Pt ( x), x) < 15. Since x is shift attracting , x and x are shift asymptotic. Since shift asymptoticity is an equivalence relation, any two points are shift asymptotic.

Suppose that z E X, u, v E R and d(<pu(x),<pv(z)) < 15. Set E = 15-d( 'Pu(x), 'Pv(z)) and choose t so that d( 'Pt (x ), 'Pu (x)) < E. Since d( 'Pt(X), 'Pv(z)) < 15, the points z and x are shift asymptotic. Since shift asymptoticity is an equivalence relation, z and x are shift asymptotic.

0

THEOREM 10. Suppose that (X, <p) is an rmf, that x E X is shift attracting , and that O(x) is compact. Then O(x) is a periodic orbit.

Before we give the proof, we remark that this is the essence of a theorem of Sell [7], see also [2] (page 246). The Sell theorem as stated in these references includes an extra condition of stability of O(x) which carries over to O(x), but this can easily be dealt with separately (see Lemma 18).



RIGIDITY IN DYNAMICS 93

PROOF. First we show that O(x) is minimal. Since O(x) is compact and in-variant, it contains a minimal set M. The point xis shift asymptotic to every point of M and consequently it is asymptotic to some point y E M. It is clear that any two asymptotic points must have the same omega limit set. Since O(y) = M, we conclude O(x) = M.

We give two proofs that the minimal set O(x) is a periodic orbit.

Proof I, using invariant probability measures: It is well-known that there is at least one probability measure p, on the compact invariant M such that p,( 'Pt (E)) = p,( E) for all t and all Borel sets E. We will exploit the existence of this invariant measure on M.

Let An= { y EM I x and 'f!t(Y) are asymptotic for some t E [n, n+ 1)] }. Since any pointy E M is shift asymptotic to x, M is the union of the sets An. The sets An are Borel sets, and '{!1 (An) = An_1 . We let p, be an invariant measure as described above, so p,(An) = An_ 1 . Since the sets An all have the same measure and exhaust M, they cannot be disjoint and in fact A0 intersects An for some n =1- 0. Therefore some y E M is asymptotic to 'f!T(Y) for some T =1- 0. It follows that 'Pr(fi) = y for some y EM and then, by minimality, that the periodic orbit through y is all of M.

Proof II, using equicontinuous factors. Recall that an rmf (X, 'f!) is said to be equicontinuous if given E > 0 there is a

c5 > 0 such that if d(x, x) < c5 then d('f!t(x), 'Pt(x)) < E for all t E R. An rmf (X, <p) is said to be a factor of (X, 'f!) ifthere is a surjective mapping II: X ---t X such that Il('f!t(x)) = 'Pt(x) for all x and t. Given any compact rmfthere is a (possibly trivial) maximal equicontinuous factor, meaning that all other equicontinuous factors are factors of the maximal one (see [1]).

We first consider a discrete dynamical system (X, T) (Tis a homeomorphism of the compact metric space X). Just as in the case of a real action, we can define asymptoticity and shift asymptoticity of points x and y. (x and y are asymptotic if lim d(Tn(x), Tn(y)) = 0 and x and y are shift asymptotic if x and Tk(y) are

n->oo asymptotic, for some integer k.) The system (X, T) is shift asymptotic if every pair x and y are shift asymptotic.

We show that a minimal shift asymptotic system (X, T) consists of a single periodic orbit. We first note that an equicontinuous minimal set which is shift asymptotic consists of a single periodic orbit. At the other extreme, a weakly mixing shift asymptotic minimal set (X, T) is trivial (consists of a single point). For X x X contains a point (x, y) with a dense orbit (a "transitive point"). For every integer n, (x, Tn(y))is transitive, so it can't be asymptotic, unless X is a one point space.

Now suppose (X, T) is shift asymptotic and minimal. Since a factor of a shift asymptotic system is shift asymptotic, the maximal equicontinuous factor (Xeq, T) is shift asymptotic, and, as noted above, consists of a single periodic orbit which we write as Xeq = {0, 1, ... ,n -1} with T(j) = j + 1 (mod n). Let 1f: X ---t Xeq, let X 0 = 1r-1 (0), and let T0 = Tn. It is easy to see that (X0 ,T0 ) is minimal and shift asymptotic. In fact, it is also the case that (X0 , To) is weak mixing. For, if this were not the case, (X0 , T0 ) would have a non-trivial equicontinuous factor, and it then follow that (X, T) would have an equicontinuous factor which is an extension of { 0, 1, ... , n}. Thus X 0 is trivial as are all the X;, and X consists of a periodic orbit.




Now we return to the case of a real action. Suppose that (X, t.p) is a com-pact minimal shift asymptotic rmf. The maximal equicontinuous factor (X eq, cp) is clearly also a minimal shift asymptotic rmf. For such a flow, distinct points are never asymptotic, so two shift asymptotic points are on the same orbit. This shows that Xeq consists of a single orbit. It follows (this is not entirely obvious, see [1] p. 9) that the action is periodic. Assume for convenience that the period is 1. The space Xeq is a circle, which we write as Xeq = [0, 1). Let Xo = 1r-1(0) and T0 = t.p1, the time one map. As in the discrete case discussed above, (X0 , T0 ) is minimal, shift asymptotic, and weak mixing. It follows that X 0 , and therefore all1r-1(s), for s E [0, 1) are singletons, and so the map 1r: X-> Xeq is one to one.

0

We now define a rigidity condition, which we call tightness.

DEFINITION 11. Let (X, t.p) be an rmf. A closed invariant subset A ~ X is called tight if for every sequence tn -> oo either { 'Ptn (y)} converges for every y E A or { 'Ptn (y)} converges for no y E A. We say that the rmf is tight if X is tight.

We will give three sufficient conditions for an rmf to be tight (Lemmas 12, 14, and 18) as well as of tight minimal rmfs (Theorem 19).

LEMMA 12. If (X, t.p) is an rmf such that x EX is shift attracting then O(x) is tight.

PROOF. If y E O(x) then x andy are shift asymptotic so, if tn -> oo, { 'Ptn (y)} converges if and only if { 'PtJx)} converges.

0

DEFINITION 13. Let (X, t.p) be an rmf. An endomorphism of X is a continuous surjective function 1/J : X -> X such that i/J(t.pt(x)) = 'Pt(i/J(x)) for all t E Rand x E X. If 1/J is also injective we say it is an automorphism.

The following lemma is obvious.

LEMMA 14. If (X, t.p) be an rmf such that for every x, y E X there is an endomorphism 1/J: X-> X with 1/J(x) = y then X is tight.

Our plan (stated more formally below) is to show that for a minimal compact rmf, the converse of the above lemma holds. It will then follow from a theorem of Gottschalk that (X, t.p) is equicontinuous.

For this proof, we will make extensive use of the enveloping semigroup of an rmf, which we now define. Recall, for a metric space Y, that yY is the set of all not necessarily continuous maps of; Y into itself. Given two elements (, 77 of yY let (77 E yY be defined by (ry(y) = ((ry(y)). If ( E yY andy E Y we write (y for ((y) . Recall that if Y is compact then yY is compact in the topology of pointwise convergence.

DEFINITION 15. Let (X,t.p) be an rmf. (1) Let Y be a compact invariant subset of the rmf (X, t.p) . Let E = E(Y) =

{ ( E yY I VE > 0 and (y1, y2, ... , Yn) E yn there is at E R such that d(t.pt(Yi), ((Yi)) < E fori= 1, 2, ... , n }.




(2) Let Y be a compact positively invariant subset of the rmf (X, 'f?). Let E+ = E+(Y) = { ( E yY I VE > 0, T > 0, and (y1, Y2, ... , Yn) E yn there is at > T such that d('Pt(Yi), ((yi)) < E fori= 1, 2, ... , n }.

The set E is called the enveloping semigroup (it is easily seen that it is a semigroup). It can be, and usually is, alternatively defined as the closure of { 'Pt I t E R} (where the mappings are restricted toY), in yY, with respect to the topology of pointwise convergence. The set E+ is called the positive asymptotic enveloping semigroup and can be alternatively defined as the intersection over all T of the closure in yY of { 'Pt It > T }.

The following proposition is both easy to prove and well known.

PROPOSITION 16. Suppose that (X, 'P) is an rmf, that Y is a compact subset of X, and that Y is either invariant or positively invariant as appropriate.

(1) The spaces E and E+ are both closed (and thus compact) in YY. (2) BothE and E+ are semigroups under composition. (3) lft E Rand ( E E ( resp. E+) then 'Pt( = ('Pt and 'Pt( E E (resp. E+ ). (4) If x E Y and x E D(x) then there is a ( E E+ such that (x = x. (5) IfY is invariant the rmf(Y, 'P) is minimal if and only if for every x, x E Y

there is a ( E E+ with (x = x. (6) Let ry E E+, and let x 1, x2, ... , Xn E Y. Then there is a sequence tk ____, oo

such that 'Ptk (xi) ____, ryxi fori = 1, ... , n.

The spaces E and E+ are typically non-metrizable semigroups such that ( f---+ (()

is continuous but() f---+ (()is generally not. Despite this, they have proved very useful. In the case of tightness E and E+ will be found to be very tame. Before we get to that, we discuss the stability condition referred to previously that appears in Sell's theorem.

The following definition for autonomous differential equations appears in [2], page 186. We have changed notation slightly to conform to our rmf viewpoint.

DEFINITION 17. Let (X, 'P) be an rmf. We say a point x (or the function t f---+ 'Pt(x)) is uniformly stable if there exists a constant K such that E > 0 implies there is a positive 8 (E) so that if d( 'Pt 1 (y), 'Pt2 ( x)) < 8 for some t1, t2 > K then d( 'Pt+t1 (y), 'Pt+t2 ( x)) < E for all t 2': 0.

The definition of uniform stability, as given above, could be extended to non-autonomous differential equations. In the autonomous case (this is the situation considered by Cronin, Sell, and ourselves) this definition can be cast in the simpler form: The point x is uniformly stable if for every E > 0 there is a 8 (E) > 0 such that if y EX, to 2': 0, and d(y, 'f?t0 (x)) < 8 then d('Pt(Y), 'Pt+to(x)) < E for all t 2': 0.

LEMMA 18. Suppose that X is uniformly stable and o+ (X) is compact. Let E+ = E+(o+(x))

(1) Every x E D(x) is uniformly stable. (2) Every ( E E+ is continuous at all points of D(x) and maps D(x) onto

itself. (3) Every minimal subset of (D(x), 'P) is tight.

PROOF. Let x, O+(x) and D(x) be as given.




(1) Let E > 0. Choose 8 > 0 such that if y EX, t 1 2: 0 and d(y, i.pt 1 (x)) < 8 then d(~.pt(y), IPt+h (x)) < ~ for all t 2: 0.

Let x E n(x). Suppose that y E X, t2 2: 0, and d(y, i.pt 2 (x)) < ~. Since 1Pt2 (x) E n(x) there is some t 1 2: 0 such that d( 1Pt 1 (x), 1Pt 2 (x)) <

~· Since d(i.pt 1 (x),y) < 8, it follows that d(IPh+t(x),i.pt(Y)) < ~ and d(IPt+t 1 (x), 1Pt+t2 (x)) < ~ for all t 2: 0. This shows d(~.pt(y), IPt+t2 (x)) < E for all t 2: 0, as required.

(2) Let E > 0 and again choose 8 so that if y EX, h 2: 0 and d(y, i.pt1 (x)) < 8 then d(i.pt(Y), 1Pt+t 1 (x)) < ~ for all t 2: 0. Suppose that x1 E n(x) and

---- 0 0 X2 E o+(x) and d(xl, x2) < 2• Choose h 2: 0 so that d(xl, IPtl (x)) < 2' and so d(x2, i.pt 1 (x)) < 8. It follows that d(i.pt(xl), 1Pt(x2)) < E for all t 2: 0. If ( E £+of (n,~.p) then d((x1,(x2):::; E. This shows that every ( E E+ is continuous on n(x).

To show that each ( is surjective observe that the image of the com-pact set n(x) under the continuous map ( is compact, so we must only show that this image is also dense in n(x).

Let x E n(x). Reasoning as before, using the uniform stability of (x, there is a h 2:0 such that d(~.pt((x),i.pt+t 1 (x)),e for all t 2:0. For some t > 0 we have d(IPt+dx),x) < e, and so d(~.pt((x),x) < 2e. Since IPt((x) = (i.pt(x), we have found a point in the image of ( that is within 2E of x. This shows that the image under (of n(x) is n(x).

(3) Since uniform stability at X is passed on to all points in n(x), it suffices to show that if (X, ~.p) is a compact minimal rmf and if one (and therefore all) of the points in X is uniformly stable then (X, ~.p) is tight. This follows immediately since given two points x 1 , x2 E X there is (by minimality) a ( E £+ with (x1 = x 2 and ( is continuous by part 2 of this proof. The conclusion now follows from Lemma 14.

0

We now state our main result for tight flows.

THEOREM 19. Let (X, i.p) be a minimal compact rmf. The following are equiv-alent:

(1) (X, i.p) is tight (2) For every x, y EX there is an automorphism 'ljJ of X with 1/J(x) = y. (3) (X, ~.p) is equicontinuous

PROOF. (3) implies (1) : Since the space X is complete it is sufficient to show that if { IPtn (x)} is a Cauchy sequence for some x then it is a Cauchy sequence for every x. Suppose that { IPtn ( x)} is a Cauchy sequence and x E X. Let E > 0. Choose 8 > 0 so that if d(x1 ,x2 ) < 8 then d(i.pt(x1 ),i.pt(x2 )) < E for all t E R. Choose N so that n > m > N implies d(IPtn(x),i.pt,.(x)) < 8. Then d(IPt+tn(x),i.pt+t,.(x)) < E for all t E R. Since (X, ~.p) is minimal, xis in the orbit closure of x, so d(i.ptn (x), IPt,. (x)):::; E. Therefore { IPtn (x)} is Cauchy.

(2) implies (3) : A theorem of Gottschalk, [4], says that if the automorphism group of a compact metric minimal flow (with an acting abelian group) is transitive then the flow is equicontinuous. (For generalizations see Auslander [1] and Glasner [3]).




We sketch the proof of Gottschalk's theorem found in [1]. To establish equicon-tinuity of the flow it suffices to show that the collection of mappings { 'Pt} is totally bounded in the uniform norm ([6] p. 276). This is equivalent to showing that every sequence { 'Ptn} has a uniformly convergent subsequence. Select some x E X. Passing to a subsequence, we can assume that { 'Ptn ( x)} converges. Given y E X, there is an automorphism 'ljJ with 'lj;(x) = y so 'Ptn(Y) = 'Ptn('lj;(x)) = 'lj;('Ptn(x)) converges. Thus { 'Ptn} converges pointwise.

The main ingredient now is a lemma which says that at each point of a residual subset of X the convergence is "uniform at the point". We omit the formal definition of uniform convergence at a point, but when combined with the transitivity of the group of automorphisms, it implies that { 'Ptn} converge uniformly as required.

(1) implies (2) : This is the heart of the matter. The proof will be given as a sequence of lemmas.

LEMMA 20. Suppose (X, 'P) is a compact tight rmf. Then:

(1) If( E E+ and {tk} is a sequence such that 'Ptk(x)--+ (x for some x EX then 'Ptk (x) --+ (x for all x E X.

(2) If tk --+ oo and {'Ptk(x)} converges for some x E X, then there is an ( E E+ such that 'Ptkx--+ (x for all x EX.

(3) If x EX andy E D(x) then there is a unique ( E E+ such that (x = y. (4) Suppose {(n}is a sequence in E+ and {(nx} converges for some x EX.

Let ( be the unique element of E+ such that (nx--+ (x. Then (nx--+ (x for all x EX.

PROOF. Suppose (X, 'P) is compact and tight. (1) Suppose that 'Ptk(x) --+ (x, and let x EX. Since (X,i.p) is tight, the

sequence { 'Ptk (x)} converges to something. Choose a sequence Sk such that 'Psk(x) --+ (x and 'Psk(x) --+ (x. Choose the sequence {uk} such that tk = u2k and Sk = u2k+l· By tightness, { 'Puk (x)} converges. The subsequence { 'Psk (x)} converges to (x so the subsequence { 'Ptk (x)} also converges to (x as required.

(2) Choose (so that (x = lim'Ptk(x) and apply (1). (3) The existence of (follows from (2) . Suppose that (x = ryx and let x EX.

Choose a sequence tk --+ oo such that 'Pt2 k (x) --+ (x, 'Pt2 k+ 1 (x) --+ ryx. By (1), 'Ptk(x)--+ (x = ryx By tightness, {'Ptk(x)} converges, so (x = ryx.

(4) Let x EX and choose a subsequence {nk} such that {(nkx} converges. Using the definition of E+ we can find tnk --+ oo such that lim (nx =

n->oo lim 'Ptn ( x) and lim (nx = lim 'Ptn (x). Then lim 'Ptn (x) = (x since

k---+oo k n---+oo k-+oo k k---'+-oo k

lim 'Ptn (x) = (x. It follows that lim (nkX = (x whenever the limit k->oo k k->oo exists. Since X is compact, lim (nx = (x.

n->oo

0

The essence of the following lemma, central to our results, was brought to our attention by Eli Glasner.

LEMMA 21. Suppose (X, 'P) is a compact tight rmf. Then:

( 1) Every ( E E+ is continuous at a residual set of points in X. (2) Let(, ryE E+ and suppose that ( is continuous at ryx. Then (ryx = ry(x.




PROOF. Suppose (X, 'P) is compact and tight. (1) From Lemma 20, parts (1) and (2), it follows that every ( E E+ is a se-

quential limit of continuous functions mapping the complete metric space X into itself. It is well-known that the limit function must be continuous on a residual set.

(2) Note that all 'Pt commute with allry E E+. Let tk -+ oo such that 'Ptk -+ 1].

Then (ryx = (lim'Ptk(x) = lim('Ptk(x) = lim'Ptk((x) = ry(x.

LEMMA 22. Suppose (X, 'P) is a compact tight minimal rmf. Then:

(1) If ( E E+ and (x = x for some x EX, then ( is the identity map. (2) E+ =E. In particular, all of the maps 'Pt are in E+.

0

(3) The semigroup E+ is a commutative group. Every ( E E+ is a homeo-morphism of X onto itself.

PROOF. (1) Let x E X and choose e E E+ so that ex= x. Choosey to be a point of continuity of both ( and e. Choose 1] E E+ so that 1]X = y Then (y = (ryx = ry(x = ryx = y. The same argument shows that ey = y, and uniqueness implies (=e. That is, (x =ex= x.

(2) We have just shown that 'Po E E+. It follows that 'Pt = 'Pt o 'Po E E+ for all t E R. Thus E+ contains E, the closure of the collection of all maps 'Pt· The reverse inclusion is obvious so E+ = E.

(3) We have seen that 'Po, the identity mapping on X, is in E+. We will show that every ( E E+ has a left and right inverse mapping in E+, which will simultaneously show that the elements of E+ are bijections and that E+ is a group.

Given ( E E+' choose X E X and choose e E E+ so that e(x = x. Then e( is the identity. This shows that ( is injective.

Now set y = (x and choose 17 so that 1JY = x. Since (ryy = y, (17 is the identity. This shows that ( is surjective.

We now show that E+ is commutative. Let e, ( E E+. Choosey so that e is continUOUS at y and choose X SO that (X = y.

Observe ey = e(x = (ex = (e(- 1y so e = (e(-1 , giving e( = (e as required.

Continuity will now follow from commutativity. Suppose that Xn -+ x E X, and let (nx = Xn, (n E E+. Since the identity mapping is in E+, we see from (1) that (nx-+ x for all x E X. Now let 17 E E+. Then 1JXn = 1](nX = (n1]X -+ 1]X.

0

We have seen, for a compact tight minimal rmf, that E+ is transitive on X, that every element of E+ is a bijection commuting with the flow, and we have just seen that every element of E+ is continuous. This shows that (1) implies (2) and completes the proof of Theorem 19

0

CoROLLARY 23. Let (X, 'P) be a compact tight minimal rmf. Then, topologi-cally, X is the projective limit of tori.

PROOF. Since the flow on X is minimal and equicontinuous, the space X admits a compact group structure (X,+) with Rasa dense subgroup. Standard arguments




in duality show that the dual X of (X,+) is an additive subgroup of R with the discrete topology. Since X is metric, X is countable. Write X1 s:;; X2 · · · / X where each Xk is a torsion free finitely generated abelian group. Let fh : X ---+ Xk be the mapping of X to Xk, the dual of Xk, be defined by (Bk(x))(x) = x(x) for x E Xk. The image of each Bk is a torus, and X is the projective limit of this system.

0

We remark that metrizability of the space X is essential for the implication (2) implies (3) in Theorem 19. A counterexample is provided by the enveloping semigroup (regarded as a space) of a distal non-equicontinuous minimal flow. Let (Y, ¢) be such a flow, and let X= E(Y), the enveloping semigroup of (Y, ¢). Now X is a compact Hausdorff space, and there is an action of R on X, which we still denote by¢. (In fact, this is the case for any flow, since there is an induced action on yY and the enveloping semigroup is an invariant set). Now, since (Y, ¢)is distal, the enveloping semigroup is a group, and right translations define a transitive group of automorphisms. If (X,¢) were equicontinuous, (Y, ¢)which is a factor of (X,¢) would also be equicontinuous.

Tightness is quite a rigid condition, so the following observation may prove useful. If there is a continuous time change so that the new flow becomes tight then the orbit closures and the minimal sets remain unchanged. In particular, such a flow must be on the projective limit of a torus.

The authors are now investigating the structure of (not necessarily minimal) tight rmfs. It is not clear to us whether tightness is preserved under factors. That this holds in the minimal case is a consequence of Theorem 19.

References

[1] J. Auslander, Minimal flows and their extensions, North-Holland, Amsterdam, 1988. [2] J. Cronin, Differential equations and qualitative theory, Marcel Dekker, New York, 1994. [3] S. Glasner, Regular PI metric flows are equicontinuous, Proc. Amer. Math. Soc.114 (1992),

269-277. [4] W. H. Gottschalk, Transitivity and equicontinuity, Bull. Amer. Math. Soc. 54 (1948), 982-

984. [5] V. V. Nemytskii and V. V. Stepanov, Qualitative theory of differential equations Princeton

University Press, Princeton, 1960. [6] J. Munkres, Topology, a first course, Prentice-Hall, Englewood Cliffs, 1975 [7] G. Sell, Periodic solutions and asymptotic stability, J. Diff. Eq. 2 (1966), 143-157.

DEPT. OF MATHEMATICS UNIVERSITY OF MARYLAND, COLLEGE PARK, MD, 20742 E-mail address: jnaiDmath.umd.edu

DEPT. OF MATHEMATICS UNIVERSITY OF MARYLAND, COLLEGE PARK, MD, 20742 E-mail address: krb~math. umd. edu




On strong laws of large numbers with rates

Guy Cohen, Roger L. Jones, and Michael Lin

ABSTRACT. Let {/n} C Lp(J.L), 1 < p < oo, be a sequence of functions with supn 11/niiP < oo. We prove that if for some 0 < (3 :::; 1 we have

1 n 1 1 n sup II 1 _ 13 L fk II < oo, then for 8 < p- (3 the sequence { 1 _ 8 L fk}

n n k=1 P p n k=1 has a.e. bounded p-variation, hence converges, and the p-variation norm func-tion is in Lp(f.L). If we replace supn 11/niiP < oo by supn 11/nll= < oo, then the a.e. convergence holds for 8 < pfrf3. Furthermore, in each case we also have

a.e. convergence of the series f {: 8 for the corresponding values of 8, and k=1 k

in the first case we even have that the sequence of partial sums has bounded p-variation.

Some applications are given. In particular, we show that if {gn} are centered independent (not necessarily identically distributed) random variables with supn IIYnllq < oo for some q 2': 2, then almost every realization an= Yn(Y) has the property that for every Dunford-Schwartz operator T on a probability

. ~ akTkf space (0, J.L) and f E Lp(f.L), p > ~ the senes ~ --k- converges a.e. The

k=1 same result holds for 1 < q < 2 if in addition the random varaibles {gn} are all symmetric. When the {gn} are i.i.d. the symmetry is not needed, and a.e. convergence of the above series holds also for f E L ...!L (f.L).

q-1

1. INTRODUCTION

It is known that there is no general speed of convergence in the pointwise ergodic theorem for ergodic measure preserving transformations; Krengel [Krl] has shown that for every measure preserving transformation () of the unit circle with Lebesgue measure and for every sequence {an} of positive numbers converging to 0 there exists a continuous function f with integral 0 such that lim supn I~ 2:::~= 1 f o ()k I/ an = oo a.e. For further discussion see pp. 14-15 of [Kr2].

1991 Mathematics Subject Classification. Primary 47 A35, 28D05; Secondary 42A16, 60F15. Key words and phrases. strong laws of large numbers, ergodic theorems, speed of conver-

gence, random Fourier series. Roger Jones was partially supported by a research leave granted by the Research Council of

DePaul University.


101




102 GUY COHEN, ROGER L. JONES, AND MICHAEL LIN

Derriennic and Lin [DL] have used a rate of convergence in the mean to obtain pointwise rates of convergence: Let T be a Dunford-Schwartz operator on L1 (!1) of a probability space, and let f E Lp for some (fixed) p > 1. Assume that for some 0 < (3 :::; 1 we have

(1)

(i) If (3 > 1 - 1/p, then the series 2:::~= 1 Tk f jk11P converges a. e. and thus (1jn 11P) 2:::~= 1 Tk f----+ 0 a. e.

(ii) If (3 :::; 1- 1/p, then for every"' > 1- (3 the series 2:::%:1 yk f / k"~ converges a.e. and (1/n"~) 2:::~= 1 Tkf----+ 0 a.e.

Condition (1) had been previously used by Loeve [Lo] (see [Do], p. 492) forT unitary on £ 2 to obtain the strong law of large numbers. Rates of convergence in this case were obtained by Gaposhkin [G].

For T induced by an ergodic probability preserving transformation on (D, /1) and f E £ 1(11) orthogonal to the eigenfunctions ofT, the Wiener-Wintner the-orem [WW] yields that for a.e. x we have limn~ 2:::~= 1 )..kyk f(x) = 0 for ev-ery >.. on the unit circle; in fact, the convergence (for fixed x) is uniform in >.. (see [Al] for f E £ 2 , and [CL] for the extension to f E LI). This yields [CL] llmaxl>-1=1 I~ 2:::~= 1 >..kTk flllP ----+ 0 when f E Lp, p > 1. Independently of [DL], Assani [A3] studied the rate of convergence in the Wiener-Wintner theorem, and considered functions f E £ 2 which for some (3 > 0 satisfy

He showed the existence of such functions for K-automorphisms and other inter-esting systems, and proved that for x in a set of full measure the Fourier series 2:::~= 1 )..kyk f(x)jk converges for every >.. on the unit circle. When f E Lp with p 2 2 and (3 > *' Assani and Nicolaou [AN] strenghtened the result, proving the uniform convergence of 2:::~= 1 >..kTk f(x)jk"~ for any"'> 1- (~- 2~).

A different method of measuring the speed of convergence of a numerical se-quence Xn ----+ x is to check whether 2:::~= 1 lxn - xiP < oo (i.e., { Xn - x} E Cp) for some p 2 1. Note that if for E > 0 we define the f.-deviation of the convergent sequence by D({xn},E) := l{n: lxn- xi> E}l, we obtain

The condition { Xn - x} E CP is obviously very strong, and implies

[ 00

] 1/p sup Llxnk+ 1 -XnkiP ::;2ll{xn-x}llp<oo.

{nk}/ k=1

A sequence { Xn} of complex numbers is said to have bounded p-variation if it

satisfies ll{xn}llvp := sup{nk}/ [ 'L~= 1 Ixnk+ 1 -xnkiPr!P < oo. For fixedp 21 the



ON STRONG LAWS OF LARGE NUMBERS WITH RATES 103

sequences of bounded p-variation are a vector space, with ll{xn}llvv a semi-norm. Since lxnk+ 1 - Xnkl S I:;;,~~- 1 1xi+l- Xjl, we have ll{xn}llv1 = I:~ 1 1xi+l- Xjl·

LEMMA. Every complex sequence of bounded p-variation converges.

PROOF. For p = 1 this is immediate, since Xn = x1 + I:~:i(xk+l- Xk). Fix j > 1, and take n1 = 1, n2 = j, and nk = k + j for k > 2. Then

lxil S lxj- x1l + lx1l S ll{xn}llvv + lxll· Hence {xn} is bounded. Assume {xn} has two different limit points a and b. Then we can find an increasing subsequence { nk} with a = lim Xn 2k and b = lim Xn 2k+l, so lxn2 k+l - Xn2k I 2: lb- al/2 > 0 for large k, contradicting the convergence of the series of p-powers. 0

The Lemma (which should be well-known) shows that ll{xn}llvv is a norm (the p-variation norm) on the space BVP0 of all sequences of bounded p-variation converging to 0, which contains eP'

DEFINITION. The f-jump of a sequence {xk} is defined for f > 0 by

J(f) = max{n: 3 s1 < t1 S s2 < t2 · · · S Sn < tn with lxti- Xsi I> f, 1 S j S n}.

Note that J(f) = J({xk},f) is finite for every f > 0 if (and only if) {xk} converges; it counts the number of jumps of size f that are observed along the sequence {xk}· It is easy to check that D({xn},f/2) 2: J({xn},f)/2.

Let {xn} have bounded p-variation. If J( {xn}, f) =nand the jumps occur at then pairs Sj < tj, 1 S j S n, as in the definition, then

({ } ) Ln (lxti-Xsii)P 1ll{ }liP J Xn , f S S - Xn V: • f fP P j=l

Bourgain [B) showed that for a probability preserving transformation 0 on (O,J.L) and f E L2 the sequence of ergodic averages Anf(x) := ~ E~=l f(Okx) satisfies IIIIAnf(x)llvpll2 S c(p)ll/112 for every p > 2. This was generalized to Lp, 1 < p < oo, by Jones, Kaufman, Rosenblatt, and Wierdl [JKRW), who proved for p > 2 the weak (1,1) inequality

J.L {X: IIAnf(x)llvP > f} S c(p) II/III· f

For further discussion and additional references, see [CJRW).

2. STRONG LAWS OF LARGE NUMBERS WITH RATES

Our main results give more precise information on the SLLN with rate obtained in Cohen and Lin [CL). Throughout this section we assume that(n, J.L) is a a-finite measure space. We start with a rather simple result.

THEOREM 1. Let 1 < p < oo. Let Un} C Lp(J.L), and assume that for some ~ < /3 S 1 we have

(2)




Then for 0::::; J < (3- * we have {nL8 2::~= 1 fk(x)} E Rp a. e. Moreover, the series

~ !k(x) { ~ !k(x)} II~ fk(x) II . . L......- k1_8 converges a.e., L......- k1_8 E Rp a.e., and L......- k1_8 zs m Lp(Jl} k=1 k=n k=n fp

PROOF. Denote Sn = 2::~= 1 fk· Then

Hence 2::~= 1 / n~". 8 /P < oo a.e. Denote~~ = 1- J. For 1 ::::; n < m, Abel's summation by parts (with s0 = 0)

yields

(4) f ~~ = f Sk ~~k-1 = :~ - S~~1 + I:1 (k\ - (k: 1)1) Sk. k=n k=n k=n

The a.e. convergence off ~k 1 ~} is proved as in Theorem 1 of [CL], where the k=1

boundedness of {llfnllp} is not used for the a.e. convergence of the series on the right hand side of (4), so letting m ___, oo in (4) we obtain

(5) ~ fk Sn-1 ~ ( 1 1 ) ~ kl = -----:;;;:~ + ~ kl - (k + 1)1 Sk.

By the first part, for a.e. x the sequence { 8 ';,~x)} is in Rp. Since 2:%"=n k!l\-,

O(n8-!3), and p((3- J) > 1 by assumption, Minkowski's inequality yields

~~ ~~ (:, ~ (k~ I)') '"" 1M~~~ (o ~ kP~o lp1P''I)' dM

~ ~o' (~ k"~' IIPI"''IIJ ~ o•BP; nP<~-'' < oo.

00 I 00 ( 1 1 ) IP { 00

fk(x)} Hence ~ ~ kl - (k + 1)' Sk < oo a.e., so by (5) ~ p-ii E Rp for

a.e. x, and

lit. ~,(x) II, ~ C(p,~,b)B. 0 P Lv(JL)

REMARKS. 1. Unlike the result of [CL], Theorem 1 does not require that supn llfniiP be finite. This is due to the restriction on (3 and the small range for J.

2. For J = (3 - * the above result is no longer valid. Fix 1 < p < oo and * < (3::::; 1. Put fk = k1-!3- (k- 1)1-!3, so (2) is satisfied, but for J = (3- * we have { nL8 2::~= 1 fk} = { n'\v} which is not in Rp·




DEFINITION. The E -deviation function of a sequence of functions {gn} is defined forE> 0 by D({gn},E)(x) = D({gn(x)},E), i.e., for each point x we look at the E-deviation of the sequence of values {gn ( x)}.

COROLLARY. Under the hypothesis of Theorem 1 we have

PROOF. For every point x we have (see the introduction)

and the result follows by integrating and applying (3).

THEOREM 2. Let 1 < p < oo. Let Un} C Lp such that supn llfnllp < oo, and assume that (2) holds for some 0 < (3:::; 1. For fixed 0:::; 8 < fJ(p- 1)/p, define the "averages"

A (1~6) ·- _1_ Ln j n .- 1 6 k· n ~

k=1

Then for a. e x the sequence { A~ 1 ~ 6 ) ( x)} has bounded p-variation and converges

to 0. Moreover, the p-variation norm of { A~ 1 ~ 6 ) ( x)} is in Lp, and satisfies the p-variational inequality

IA (1~6)1 L and thus supn n E p·

PROOF. In view of Theorem 1 and (3), we have to prove the theorem only when either (3 :::; ~, or (3 > ~ and 8 ~ (3 - ~, which will be assumed henceforth.

The measurability of the variation norm that occurs in the left hand side of the p-variational inequality above is handled by first restricting the supremum to all finite increasing sequences of length N (and then the series are summed for k :::; N); this supremum is clearly measurable. These restricted suprema are monotone increasing in N, and hence the limit will also be measurable.

Throughout the arguments, c and C will denote constants that may depend on a, (3, 8 and p, but will not depend on x, nor even on {fn}· The values of these constants may vary from one occurance to the next. We put q = pj(p- 1), the dual index of p.

8 1 Fix 8 < (3 (p -1) j p; this is equivalent to p(fJ _ 8) < q, so for E > 0 small enough

(1 + E)8 1 l+< 1 we have p(fJ _ 8) < -q· For .such E > 0 fixed, put a= p(f3-li), so a8 < q· Note that

if (3 :::; ~ then a > P~ ~ 1, and if (3 > ~ and 8 ~ (3 - ~, then p(fJ - 8) :::; 1; thus in any case a > 1. Let mk = [k"'] + 1, which is strictly increasing since a > 1. We




first prove that 2:%':1 ~A~~ 8 )(x)IP converges a.e. to an integrable function. Since mk :;:: ka, we have

which yields

(6) Jf ~A~~8l(x)IP dJL = f IIA~~8)11: ~ BPf klc+' < CBP. k=1 k=1 k=1

Hence the series 2:%': 1 I A~~ 8 ) ( x) I P converges a.e.

As is now standard in such arguments, we break the variation along any given strictly increasing sequence { nj} into two parts, the "long variation" and the "short variation", described below. For the "long variation" we will later use the variation at times from the above sequence {mk}. First note that for each x we have

(7)

In order to handle the short variation, for each k we put Ik = [mk, mk+1]. For the given subsequence {nj}, let Jk denote the set of j such that [nj,nj+1] C h, and let L be the set of j such that for some i we have nj < mi < nj+l· Of course, Jk and L depend on {nj}· In the series 2:;: 1 ~A~~- 8 )(x)- A~~~~)(x)IP' the long

variation is the sum over the indices in L, and the short variation is the sum over the indices in J := Uk;:: 1 Jk. In order to estimate the short variation, define

1

Sk(x) := (z= ~A~~- 8 l(x)- A~~~~l(x)IP);:; JEJk

Clearly, Sk depends on {nj }. Using the inequality ia+b+ciP ~ 3P- 1 (IaiP+IW+IciP), we obtain

SP- """"'IA(1-8)_A(1-8)1p = """"'1-1-~f---1-~f-lp k - L...t nj nJ+l L...t 1-8 L...t ' 1-8 L...t t

jEJk ]Eh nj i=1 nj+1 i=1 p




where

and p

Using the fact that II · I ltv :::; II · lit,, we obtain

< m1-(31A(1-(3)(x)l (-1-- _1_) < IA(1-f3)(x)l (m~~~- m~-6) m1-(3 - k mk m 1-6 1-6 - mk m 2-26 k ·

k mk+l k

Since 1 + ta. :::; (1 + t)a. for t ~ 0 and a ~ 1, the definition of mk yields

m~~~- m~-6:::; ((k + 2)a.)1-6- (ka.)1-6:::; cka.(1-6)-l,

and we obtain m1-6 _ m1-6 ka.(1-6)-1 c c

k+l k m1-(3 < c ka.(1-(3) < < mk2-26 k - ka.(2-26) - ka.(f3-6)+1 - k · (8)

Hence Uk(x):::; f ~A~;:-f3)(x)l·

Using again the fact that II · II tv :::; II · lit,, we see that




For the third term in Sf(x), we use II· llcp :::; II· lie, and Holder's inequality, and obtain

For fixed k define the following functions (which do not depend on { nj}):

Fk(x) := :P IA};;~t3)(x)IP ,

G,(x) '~ k('+La')' (];:, IJ;(x)l)' , and

ffik+l

Hk(x) := ka(Lc5)p (mk+l- mk)p/q L lfi(x)IP. i=mk+1

We have shown that Sf(x) :::; c1Fk(x) + c2Gk(x) + Hk(x). Putting F = L;~ 1 Fk, G = L:;:'=1 Gk, and H = L;;:'=1 Hk, we conclude that

The "short p-variation" relative to any increasing sequence { nj} satisfies

(9) L IA~~-c5)(x)- A~~~~l(x)IP:::; c1F(x) + c2G(x) + H(x). jEJ

In order to finally estimate the p-variation of a given sequence { nj}, fix j E L, and let i1 = i1 (j) be the smallest i with nj < mi, and let i2 = i2(j) be the largest i with mi < nj+l· We then have mi,-1:::; nj < mi, :::; mi2 < nj+l, and obtain

(10)

< 3p-1(IA(l-c5)- A(1:-c5)1p + IA(l:-c5)- A(1:-c5)1p + IA(1-c5)- A(l-c5)1P). - n1 m'~- 1 mt 1 m'~- 2 m'~- 2 n1 +t

We now define a new increasing sequence of integers { nj} which is the refinement of { nj} by joining all the integers { mi, (j), mi2 (j) : j E L} (if i1 (j) = i2 (j) we add only mi,(j)). Similarly to the definition of J and L for the original sequence { nj}, we define J~ := {j: [nj,nj+l] C h}, J' := UJ~, and L' := {j: nj < mi < nj+1 for some i}. Let j E Jk; we have nj = nj, for some j', and the definition of Jk yields that j' E JL hence {nj : j E J} C {nj : j E J'}. When j E L, there is no element of {mk} between nj and mi,(j), while nj+l > mi,(j) and mi,(j)- 1 :::; nj, so if nj = nj,, then [nj,, nj,+l] C Ji,(j)- 1, so j' E J'. All this means that the short variation of { nj} contains all the variation of the original { nj}. Furthermore, for j E L we always have mi2 (j) E {nj, : j' E J'}; if i2(j) = i1(j) + 1, then also mi,(j) E {nj, : j' E J'}; when i2(j) > i1(j) + 1, then mi,(j) is in {nj, : j' E L'}, so




{nj,: j' E L'} = {mit(j): j E L, i1(j) + 1 < i2(j)}. Using (10), and then applying (9) to the short variation of {nj} and (7) to the long one, we have

00

:::; 3p-l [c1F(x) + c2G(x) + H(x)] + 3p-l2p L IA~~ 8 \x)IP. k=l

Since the estimate does not depend on the sequence { nj}, we have

00 00

(11) sup L IA~~- 8 )- A~~~~)lp:::; 3P-1 (ctF + c2C +H)+ 3p-l2p L IA~~ 8 )IP. {nj}/ j=l k=l

In order to prove the claimed p-variational inequality, we have to show the inte-grability of the right-hand side of (11), with an appropriate estimate. For the last term we use (6). For the integrals ofF, G, and H we look at their summands.

With K := supn llfniiP and using Minkowski's inequality, we obtain

k(a-l)p cKP < cKP ----,----=--- k(l+a-a8)p kP(2-a8) ·

Thus, G = L:k Gk will be integrable, with the desired estimate, if p(2 - aJ) > 1. This is equivalent to 1- aJ > ~ - 1, or aJ < 1 + ~' which certainly holds, since aJ < ~ by the definition of a.

Using pjq = p- 1 and the estimate mk+l - mk :::; cka-l, we obtain

H x d < m - m pjq+l KP < J 1 CKP k( ) f.L - ka(l-8)p ( k+l k) - ka(l-8)p-(a-l)p ·

Thus, H = L:k Hk is integrable, with the desired estimate, since p(1 - aJ) > 1, which is equivalent to aJ < ~, holds.

We therefore have the required p-variational inequality, by (11), which implies the a.e convergence of {A~- 8 )}, and since (2) yields norm convergence to 0, the limit in the a.e. convergence is 0. The inequality supj lxj I :::; II{ xn}llvv + lxtl proved in the Lemma yields that supn{IA~ 1 - 8 )1} is in Lp· 0

DEFINITION. The E-jump function of a sequence of functions {gn} is defined for E > 0 by J( {gn}, t:)(x) = J( {gn(x)}, t:), i.e., for each point x we look at the E-jump of the sequence of values {gn(x)}.





PROOF. For every point x we have (see the introduction) 1

J({A~l-o)(x)},E)t:::; II{A~l-o)(x)llvv = ~ (sup f ~A~lk-ol(x)- A~lk~~l(x)IP) "P E E (nk)/' k=l

So the result follows by taking the Lp-norm of each side and applying Theorem 2.

THEOREM 3. Let 1 < p < oo. Let Un} C Lp such that supn llfnllp = K < oo, and assume that (2) holds for some 0 < f3:::; 1. Then for fixed 0:::; J < f3(p- 1)/p,

the sequence of finite sums {t ~1 ~}} has a. e. bounded p-variation, hence the k=l

series converges. Moreover, we have

PROOF. As before, we use the notations Sn := I:Z=l fk and A~l-o) := n11_ 8 sn, and put"(:= 1- J. For every increasing sequence {nj} we use (4) with n = 1 and m = nj, and after subtracting we obtain

Together with Minkowski's inequality in .eP, this yields

(~ I (A~:~~) - A~:-·)) + ·:r u, -(k: 1)>) ,. I') l 1

< (~ I(A~:~~)- A};,-'>) I')'+ (~ 1 ·:r u, -(k: 1)>) ,, ') • Hence

(12)




sup (f j(A~ 1 ~~)- A~~-S))\P) ~ + sup (f n1f 1 (k1

1 - k \ ,) Sk P) ~ {n;}/ j=1 {n,}/ j=1 k=n1 ( + ) with first term on the right in Lp(P,), with an appropriate estimate of the norm, by Theorem 2. It remains to check the last term. For this put S( { n1}) :=

1

( 2::;':1 \I:~::+~J- 1 ( k\ - (k;1)1') Sk n p. Then the norm inequality II . llep ::::; II . lie, and obvious estimations yield

= n;+,-1 lskl 1 = lskl 1 ::::: c 2::: 2::: p-!3 k!3+, = c 2::: k1-{3 k!3+, ·

j=1 k=n1 k=1 Since the right hand side does not depend on { n1}, and (3 + 'I > 1, we obtain

which shows that also the last term in (12) is in Lp(P,) with the desired estimate of the norm, and the theorem is proved. 0


REMARKS. 1. The a.e. convergence obtained in Theorems 2 and 3 was first proved in[CL].

2. The results of Theorems 2 and 3 (in fact, even the a.e. convergence proved in [CL]) cannot be improved in general, as the following example shows.

EXAMPLE 1. Under the assumptions of Theorem 2, the a. e. convergence of

{nLs 2::~= 1 h} can fail if 6 2 (J(p -1)/p. We will work on [0, 1) with Lebesgue measure, thought of as the unit circle.

Fix p > 1 and (3 < 1. Let nk = [k<>] with a= ~- For each k, let h be a half open interval of length i;, such that h+1 is adjacent to, and to the right of h, mod 1 (i.e., h corresponds to a half open arc). h is the whole space, and for k > 1 the intervals (arcs) h and h+1 are clearly disjoint. Also note that each x E [0, 1) will be in infinitely many of the h.

Let fj(x) = k11PXh(x) if nk < j::::; nk+l· Note that llf1IIP = k11P(1/k)11P = 1 where nk < j ::::; nk+ 1 . Also note that

p




Since a > 1 we have { nk+l - nk} increasing. Define fJ ( x) = jj ( x) - fnk ( x) if nk < j:::; nk + (nk- nk-d and fJ = Jj(x) when nk + (nk- nk-1) < j :::; nk+l·

The idea is that for the first few terms of the k-th block, we both put positive mass on the interval Ik and put negative mass on the interval h-1. We stop putting negative mass on h-1 after we have cancelled all the previous positive masses on it, but continue to put mass on h until we reach nk+1·

Thus II fJ liP :::; 2 for each j, and by the definitions nk+l nk+l L fJ = L Jj = (nk+l- nk)k11PXIk(x). j=1 j=nk+1

Using our choice of nk, we see that

(13) p

For any n, let nk:::; n < nk+1· Since II!JIIP:::; 2, (13) yields

p

1 nk < -~!· - 1-{3 L...J J

nk j=1 p

1 p-a.{3

Since we selected a = ~, we have 1 - a/3 = 0, so (2) is satisfied. However, on I k the height of the "average" is

1 nk+l ka.-1k1fp 1 __ ~ . _ nk+l - nk k1!P ~ _ 1-o L...J !1 - 1-o ~ ka.(1-o) - k1-a.o-1/p

nk+1 j=1 nk+1

Hence on Ik we will have height greater than some fixed positive constant provided 1-8//3 -1/p = 1- a8 -1/p:::; 0, which is 8 2:: j3(p -1)/p. Since every x E [0, 1) is in infinitely many Ik, we obtain limsupk --f=o L::;~t 1 fJ(x) > 0 for every x. Since

nk+l L::;~r IJ(x) = 0 for x ~ Ik, and each x is outside infinitely many Ik, we have liminfk nL8 L:;~r IJ(x) = 0 for every x. Hence {nL 8 L::;=1 /j(x)} is everywhere

k+l divergent.

THEOREM 4. Let 1 :::; p < oo and 1 < q < oo. Let Un} C Lp(J.L) n Lq(J.L) such that supn 11/nllq < oo, and assume that (2} holds for some 0 < /3 :::; 1. Then for

0 :::; 8 < max{/3 - ~, q~(q~~)P} the sequence { nl-8 L::~= 1 fk} converges to 0 a. e.,

. ~ !k(x) and the serzes L...J k1_8 converges a.e.

k=1

PROOF. If j3 > ~, we can apply Theorem 1. We first check when, in this case, the assertion of the theorem enlarges the interval for 8; it turns out that j3 - l > (q- 1 pf3 is equivalent to pq/3 > q + (q- 1)p. Put r := q+(q-1)P. we have p q+ q-1)p pq{3 ' to deal only with the case r 2:: 1 (which is obviously satisfied also when j3 :::; ~).

Fix 8 E [0, q~{q 1 )~fP). We first prove that nL8 L::~= 1 fk(x)- 0 a.e., by modify-ing the proof of Proposition 1 of [CL] (which treats the case q = p). The assumption on 8 yields




(i) (/3- 6)rp > 1 and (ii) (1 - r6)q > 1 since we have equality for the above value of r when 6 = q~(q~~fP.

Define nm = [mr] + 1 (which is strictly increasing since r 2:: 1). Then (i) yields

so I:::=1 I n~-s I:;~;-: 1 !kiP converges a.e., which implies n~-s I:;~;-: 1 fk(x) ~ 0 a.e. For nm ~ n < nm+l we have [CL]

(15)

With C := supn llfnllq we obtain, as in [CL],

[ 1 n,+, lq ( )q ( +2)(r-1)q (2r)q ~ 1-8 2..: ll!kllq ~ cq nm+~:, nm ~ cq ~ (1-r8)q·

nm k=n,+1 nm m m

Since (1- r6)q > 1 by (ii), we have a convergent series, which proves that

1

1 n 1 n, lq max 1""=8 L fk - 1_ 8 L fk ---+ 0 a.s.

nTTt <n<nm+l n n m----+oo - k=1 k=1

Since I nLs I:;~;-: 1 fkl ~ I n~-6 I:;~;-: 1 !kl ~ 0 a.e., we have I nLs I:;~= 1 fkl ~ 0 a.e.

The a. e. convergence of the series f ~k 1 ~} is proved, using ( 4) (with n = 1), k=1

as in Theorem 1 of [CL]; see the proof of our Theorem 1. 0

REMARKS. 1. Note that we may have 1 < q < p, so when JL is finite no convergence follows from Theorem 2.

2. When JL is finite and q > p, we obviously have also supn llfnllp < oo, the assumption of Theorem 2; however, Theorem 4 yields a larger interval for 6. In any CElBe, for fixed p, the larger q is, the larger the interval for 6 is.

3. When JL is finite, we can also prove (as in [CL]) that sup It {:81 is in n>O k= 1 k

Lmin{p,q} ·

When JL is finite and supn llfnlloo < oo, we can apply the previous theorem, and let q ~ oo to obtain the interval for 6, given in the case ak = 1 of the next theorem. However, when JL is not finite this cannot be done. For example, on [0, oo) with Lebesgue's measure let An:= [0, n) and fn = ( -1)nXAni then sup llfnllq = oo for any 1 < q < oo, while for 1 < p < oo (2) is satisfied with f3 = 1 - 1/p.




DEFINITION. Let { ak} be a sequence of (complex) numbers, and let 1 :S t < oo; we say that {ak} E Wt ifsupn>O * 2::~= 1 /ak/t < oo. If {ak} is bounded we say that {ak} E Woo.

THEOREM 5. Let 1 :S p < oo, and let {fn} C Lp(J.L) s11ch that supn llfnlloo < oo. Let 1 < t :S oo with dual index s := tj(t- 1), and let {ak} E Wt. If for some 0 < (3 :S 1 we have

sup 11 ~ t akfk 11 < oo, n>O n k=1 P

then for 0 :S 0 < max{p~s(J, (3- ~} the sequence { n,l_" 2::~= 1 akfk} converges to 0

. ~ akfk(x) . . a. e., and the serzes ~ k1-li converges a. e. When J.L zs fimte, for 0 as above we

k=1

1 n ~~ akfk(x) I also have supn>O I n1 _ 6 Lk= 1 akfk/ E Lp and sup ~ k1_8 E Lp. n>O k=1

PROOF. We want to check when the value of the upper limit for o is (3 - ~. This requires first that (3 > ~ (in which case Theorem 1 applies to { akfk}). The inequality (3 - ~ > p}s (3 is equivalent to ps(J / (p + s) > 1. We therefore have to prove the theorem only when r := $ ~ 1. Then for fixed 0 with 0 :S 0 < p}sf3 we have (since for 0 = p}sf3 = -J:s equality holds)

(i) rp((J- o) > 1 and (ii) 1 - rso > 0.

Let nm = [mr] + 1, which is strictly increasing since r ~ 1. Replacing fk in 1 n,

(14) by akfk we obtain by (i), as in the previous proof, that 1_/i L akfk ----> 0 a. e. nm k=1

Put K1 := supn llfnlloo· Let K2 := supn(* 2:::~= 1 /ak/t) 1!t if t < oo, and K2 := supn /an/ if t = oo. For nm :S n < nm+1 we obtain, using (15) with fk replaced by akfk> and then Holder's inequality in case t < oo (i.e., s > 1),

1"=8 Lakfk- 1"=8 Lakfk :S 1"=8 L /akfk/ :S K1l"=J L /ak/ 1

1 n 1 n, I 1 n-m+1 1 nm+1

n k=1 n k=1 nm k=nrn +1 nm k=n, +1

"f 1 1 ( nm+1 ) i 1 s> t 1

:S K1l"=J L /ak/ (nm+1- nmF nm k=nm+1

1 1 1

<K Knli (nm+1)' (nm+1-nm)8 <(2r+1)iK K (nm+1-nm)8 - 1 2 m - 1 2 1-s8 nm nm nm

(If t = oo we take s = 1, and skip the middle line above). We now use r ~ 1 and the definiton of nm to obtain, as in [CL] (see proof of Theorem 4)

nm+l- nm 2r(m + 2t-1 (m + 2)r-1 -----,--;--- < = 2r -- --c--eo

n~-sli - mr-1m1-rsli m m1-rs8 · 1

Since 1- rso > 0 by (ii), we conclude (with K := K 1K 2(2r + 1)11t(2r)11s) that

(16)




Th f h . ~ akfk(x) . d . [CL] h e a.e. convergence o t e senes 6 p-li IS prove as m ; see t e k=!

proof of our Theorem 1. When fl is finite, the constant functions are in Lp(fl); using (14) with {fk}

replaced by { akfk}, we obtain from (16)

When fl is finite, sup It a~~~) I E Lp is proved as in Theorem 1 of [CL]. 0 n>O k=!

COROLLARY. Let 1 :::; p < oo. Let Un} C Lp(fl) such that (2) holds for some 0 < (3 ::=; 1. In addition, assume thatsupn llfnlloo < oo. ThenforO ::=; 8 < (3pj(p+1)

the sequence { nLs 2::~=! fk} converges to 0 a. e., and the series f ~\~} converges k=1

a. e. When fl is finite, for 8 as above we also have supn>O I n/-s 2::~= 1 fkl E Lp and

I ~ fk(x) I sup 6 k1-li E Lp. n>O k=1

PROOF. Note that P~ 1 (3 > (3- ~'and apply Theorem 5 with ak = 1. 0

REMARKS. 1. The proof of the a.e. convergence in the corollary does not require that {llfnllp} be bounded, but when fl is finite this follows from the bound-edness of the £ 00-norms.

2. Note that Theorem 4 and the previous corollary hold also for p = 1, while in general for p = 1 condition (2) does not imply a.e. convergence of ~ 2::~= 1 fk -see Example 1 in [CL] (the condition 0:::; 8 < (3(p- 1)/p cannot be satisfied when p = 1, so Theorems 1 and 2 are meaningless for p = 1).

3. The speed of convergence obtained in Theorem 2, namely the bounded p-variation of {n11_ 6 2::~= 1 fk(x)}, may fail in the corollary when 8 ;:::: (3(p- 1)/p (although the sequence converges), as shown by the following simple example: let fl be finite and p > 1, and let fk = (-1)k+1 be constant functions. Then (2) is satisfied with (3 = 1, but for 8;:::: (p- 1)/p we have

~ IA(1-Ii) - A(1-li) lp ~ 1 -~ n n+l 2: ~ (n + 1)P(1-Ii) - oo.

EXAMPLE 2. Under the assumptions of the corollary, the a. e. convergence of

{ n'1-s 2::~= 1 !k} can fail if 8 2: (3pj(p + 1). We modify Example 1. We still work on [0, 1) and define the same sets {h},

but we now take nk = [k<"] with a= (p+ 1)/pf3. Put ]1 = Xr. when nk < j :::; nk+1,

and define {fJ} as before: fJ = Jj- fnk when nk < j ::=; nk + (nk- nk_I), and fJ = Jj when nk + (nk- nk_I) < j ::=; nk+1· Thus II!JIIoo = 1 for every j, and by the definitions

nk+l nk+I

L fJ = L f1 = (nk+1- nk)Xh· j=1 j=nk+1




Since IIXh llv = k- 1/P, the definition of nk yields

(17) 1

p

For any n, let nk::;; n < nk+l· Since llhllv::;; IIXhllv + IIXh-tllv < 2(k -1)-1/P when nk < j::;; nk+1 , (17) yields

1 n

n1-f3 Lfi j=1

p

By our choice of n: we have 1 - n:{J + 1/p = 0, so (2) is satisfied. However, on Ik the height of the "average" is

Hence on Ik we will have height greater than some fixed positive constant provided 1-n:8::;; 0, which is 8 2: {Jpj(p+1). Since every x E [0, 1) is in infinitely many h, we obtain limsupk -h :L7~f fi(x) > 0 for every x. Since 2:::7~! 1 fi(x) = 0 for x ~ Ik,

nk+I

and each xis outside infinitely many Ik, we have liminfk -b., 2:::7~! 1 fi(x) = 0 for nk+l

every x. Hence {nL& 2:::7=1 fi(x)} is everywhere divergent.

3. APPLICATIONS

In this section we apply our previous results, especially Theorems 4 and 5, to obtain additional information in some special cases of the results of [CL].

PROPOSITION 6. Let {nk} be a non-decreasing sequence of positive integers, and let { ak} be a sequence of complex numbers such that for some 0 < {3 ::;; 1 we have

(18)

{i) If { ak} is bounded, then for every Dunford-Schwartz operator T on L1 (J-L) of a probability space and every f E Lp (J-L), 2 < p < oo, the series 2::::=1 a~;=~ 1

converges a.e. for any 0 ::;; 8 < ;~::::~{3. Iff E Loc, then the convergence holds for

0::;; 8 < ~{3, and also supn I 2:::~= 1 a~;=~! IE L2(J-L). {ii) If { ak} E Wt for 1 < t < oo with dual index s, then for f E L00 the series

~oo akTnkf f 0 £ { 2 {J {J 1} L..k=1 ~ converges a.e. Jor every ::;; u <max 2+8 , - 2 . {iii) If { ak} E Wt and ( 18) holds for nk = k, then for any f E Ls (J-L) the series

oo Tk f 1 n ""~k converges a. e., and thus-"" akTk f ---+ 0 a. e. ~ n~ n-+oo k=1 k=1




PROOF. As in [CL], we note that (18) implies (by applying the spectral theo-rem for unitary operators and the unitary dilation theorem for contractions) that for any contraction T on a Hilbert space we have

(19)

(i): Putting fk = akTnk J, the sequence Un} is in L2(J.L) and satisfies (2). We now apply Theorem 4 (with q replaced by p), noting that for p > 2 and f3 ::; 1 we always have ~~::::~/3 > f3- !· For f E L00 apply the corollary to Theorem 5 (with p = 2).

(ii) follows from applying Theorem 5 with p = 2 to fk = Tnk f. (iii): By (ii) we have the a.e. convergence of~ L::~=l akTkf for bounded func-

tions, which are dense in Ls(J.L). For any f E Ls(J.L), Holder's inequality yields

1 n I 1 n { 1 n ! 1 n !} s~p n {; akTk f ::; s~p n {; lakTk !I ::; s~p (n: {; laklt) t (n: {; ITk !Is) 8 •

But ITI, the linear modulus ofT, satisfies ITk !Is S (ITiki/W S ITik(lfls) (e.g., p. 65 of [Kr2]). Since {ak} E Wt, the pointwise ergodic theorem for ITI applied to 1/ls E Ll(J.L) yields supn I~ L::~=l akTk fl < oo a.e.; now the Banach principle yields ~ L~=l akTk f ----+ 0 a.e. for every f E Ls(J.L).

n-+oo For f E Ls(J.L), put Snf = L~=l akTk f. Abel's summation by parts yields

n a Tk J S J n-1 1 L T = ~ + L k2 Skf· We have shown that Snf /n--+ 0 a.e., so it remains k=l k=l to check the series. When s ~ 2 (i.e., 1 < t ::; 2), we have f E L2(J.L), and IISnfll2 S Kn1-,BIIJII2 by (19). Since J.L is a probability, we obtain

! ~ ISkfld = ~ IISkflll < ~ IISk/112 < Kll/11 ~ _1_ < L..t k2 J.L L..t k2 - L..t k2 - 2 L..t ki+.B oo, k=l k=l k=l k=l

showing that L::;::1 IS:fl converges a.e., which proves (iii) when s ~ 2. Assume now 1 < s < 2. The operator Sn = L::~=l akTk maps L2 into itself

with norm IIBnll2 S Kn1-.B by (19), and it maps L1(J.L) into itself with norm IIBnll1 S L~= 1 lakl· Since 1 < s < 2, the Riesz-Thorin theorem ([Z], vol. II p. 95) yields that Sn maps Ls(J.L) into itself with norm IISnlls S IISnii~IISnll~-<>, where 0 < a < 1 is defined by ~ = a· ! + (1- a)· 1. Holder's inequality yields IIBnll1 S (l:::~= 1 laknl/tnl/s. Hence

IISnlls S K<>n(l-,B)a(t laklt) 1-;"' n 1 ~"' S K<>n(l-,B)an 1 -;"' (~ t laklt) 1

-;"' n 1 ~"'. k=l k=l

Since {ak} E Wt and i + ~ = 1, we obtain

I IBn lis S C · n(l-,B)an(l-a)(t+~) = C · nl-<>.8.

· · ~~ ISkfl ~ IISkflls ~ 1 Th1s y1elds L..t ~dJ.L S L..t k2 dJ.L S GII/I Is L..t kl+<>.B < oo · k=l k=l k=l

Now the

previous arguments yield (iii) also in the case s < 2. 0




REMARKS. 1. Proposition 6(i) complements Proposition 2(ii) of [CL], which deals with f E Lp for 1 < p ~ 2.

2. Since J.L is assumed finite, f E Lp(J.L) with p > 2 is in £ 2 , and Proposition 2(ii) of [CL] can be applied; however, we obtain here a larger interval for 8 than that given in [CL] for £2 functions (which is the interval for which Theorems 2 and 3 hold).

3. Proposition 2 of [CL] gives additional results under the assumption (18). These can be improved by applying Theorems 1 or 3, according to the value of 8. We omit the statements of these improvements.

4. Examples of sequences {an} satisfying (18) for nk = k were given in [CL]. Another example (not mentioned there) is an = exp[27rin(logn)'] with 'Y > 0; by [I] the series 2::~= 1 p; 2 U~g k) 6 >..k converges uniformly on the unit circle for large enough 8, so (18) is satisfied with any (3 < 1/2.

5. For { ak} bounded satisfying (18), Proposition 2(ii) of [CL] applies also when J.L is not finite. It yields, for 1 < p ~ 2, the estimate of the Lp-norm of

the operators II~ 2::~=1 akTnk liP = O(n!3P) with (3p = 2f3P;1. For f E Loon Lp,

we can now apply the corollary to Theorem 5, with fk = akTnk f, to obtain the a.e. convergence of the series 2::~= 1 a~;:~t when 0 ~ 8 < p!hf3P = ~2(3. For bounded Lp functions, this improves the interval 8 < P;1 (3 obtained in Proposition 2(ii) of [CL].

THEOREM 7. Fix 1 < q < oo, and let {gn} be i.i.d. on a probability space (Y,m), with jjg1Jiq < oo and I g1drn = 0. Then for a.e. y E Y the sequence ak := gk(Y) has the following property:

For every Dunford-Schwartz operator T on L1 (J.L) of a probability space and . oo akTk f

f E L 0 (J.L), the senes L -k- converges a. e. k=1

PROOF. We first note that by the strong law of large numbers, ~ I:~= 1 jgkjq converges a.s. to I Jg1Jqdrn. Hence for a.e. y E Y the sequence {ak} is in Wq.

If q > 2 then also I jg1j2dm < oo, so putting q1 := min{2, q} we have {gn} centerd i.i.d. with finite absolute moment of order q1 ~ 2. Let a E (q1 1 , 1), so a E (~, 1), and 1 < 1/a < q1 yields

E(Jg1j11"'(log+ Jg1J)i--He) < oo for every E > 0.

By the result of Cuzick and Lai [CuLa] we now have that for a.e. y E Y the

series ~ gk(Y) >..k converges uniformly in J>..J = 1. For such y, put ak = gk(y). A L..t k"' k=1

variant of Kronecker's lemma (a Banach space version, in the space of continuous functions) yields that n1"' 2::~= 1 ak>..k converges uniformly to 0, so { ak} satisfies (18) with nk = k and (3 = 1- a (note that (3 < ~). The theorem now follows from Proposition 6(iii). D

REMARKS. 1. The convergence of ~ 2::~= 1 akTk f under the assumptions of the theorem follows from the "return times theorem" (Appendix of [B], see also [Ru]; for the passage from measure preserving transformations to Dunford-Schwartz




operators see [QLO]). Our result improves this convergence (in the particular i.i.d. case).

2. Assani [A4] showed that Theorem 7 fails for q = 1, although the "return times theorem" holds.

3. For an i.i.d. sequence as in the theorem, with the additional assumption that g1 is symmetric, Assani [A2] obtained the a.e convergence of~ ~~= 1 akTk f for every f E Lp (J.L) with p > 1 (even if p < ~). We do not know if in this case

. oo akTk f also the senes L -k- converges a.e. for every Dunford-Schwartz operator and

k=1 every f E Lp(J.L) when 1 < p < ~·

THEOREM 8. Let (0, J.L) be a probability space, and let {fn} C Lp(J.L), 1 :::; p < oo, such that supn llfnllq < oo for some 1 < q < oo. Let {nk} be a sequence of integers such that for some 0 < (3 :::; 1 we have

supllmaxl Lr>tfk).nkll =K<oo. n 1>-1=1 n ,_,

k=1 p

(20)

If q~(q~~fP :2: (3- ~ (e.g., (3 :::; ~ or q :2: p), then there exists a set O' C 0 with

p(O') = 0 such that for x ~ 0' and every 0:::; c5 < q~(q~~fP the series f= ~1 ~} ).nk k=l

converges uniformly in 1>.1 = 1.

PROOF. The proof is sfimi:r to thlat 1of ~~eorem ~~ with the same notations.

Instead of (14) we obtain L max 1_ 6 L fk>.nk < oo, and instead of (15) m=1 1>-1=1 nm k=1

we have

From these we deduce max I~ t fk(x)>.nk I-+ 0 for a.e. x. For the proof of 1>-1=1 n - k=1

the uniform convergence of the series, see the proof of Theorem 9 of [CL]. 0

REMARKS. 1. Since J.L is finite, for q = oo (i.e., when sup llfnlloo < oo), we have the above result for c5 < p!hf3, by using finite q tending to oo.

2. Theorem 8 extends Corollary 6 of [CL]. Theorem 9 there could be similarly extended.

COROLLARY. Let (0, J.L) be a probability space, and let T be a power-bounded operator on Lq(J.L), 1 < q < oo. Iff E Lq(J.L) satisfies, for some (3 > 0,

sup II max I Lr> t >.kTk Jill = K < oo, n 1>-1=1 n ,_, k=1 1

then there exists a set 0' C 0 with p(O') = 0 such that for x ~ 0' and every oo Tk f(x)>.k

"( E (1 - <~;~~, 1] the series L k"~ converges uniformly in 1>.1 = 1. k=1




PROOF. Apply Theorem 8 with fn := Tn f and p = 1.

REMARKS. 1. For q 2: 2 and T induced on Lq(J-L) by a probability preserving transformation, the corollary was proved in [AN]. Since~- 21q < 13 i~=i), our result yields the convergence for a wider range of"(. However, iff is bounded, the limit as q---. oo in the corollary yields the same range as in Theorem 5 of [AN]. Existence functions satisfying the assumption of the corollary was shown in [A3] and [AN].

2. ForT a positively dominated contraction on Lq, 1 < q < oo, the a.e. uniform oo Tk f(x).>..k

convergence of the random Fourier series L k under the assumption of k=1

the corollary was proved in Theorem 8 of [CL] by a different method. 3. For T a positive contraction of L 1 (J-L) with T1 = 1 and f E L 1 satisfying

the hypothesis of the corollary, the a. e. uniform convergence of the random Fourier oo Tk f(x).>..k

series L k was proved in Theorem 8 of [CL]; this does not follow from k=1

our Theorem 8.

THEOREM 9. Let (0, J-L) be a probability space and 2 :::; p :::; oo. Let {in} C Lp(J-L) be independent, with J fndf-L = 0 and supn llfniiP < oo. Then

(21) sup lim~ I ;;4 tfk.>..[Vk]lll < oo n>O 1>-l-1 n k=1 2

and for a. e. x E 0 and b < t;:_14 the series f ~k 1 ~} .>..[Vk] converges uniformly in k=l

1.>..1 = 1.

PROOF. We first prove (21). The assumption yields supn llfnll2 = K < oo. Put Sn = L~=l .>..[Vk] fk. Then

n-1 (j+1) 2 -1 2

ISn2 -1l 2 = L )..J L fk j=l k=j2

n-1 (j+l) 2 -1 (m+1) 2 -1 .>..j-m L L !de

j,m=1 k=j2

n-1 (j+1) 2 -1 (j+1) 2 -l n-1 (j+1) 2 -1 (m+1) 2 -1 = 2: 2: 2: !de+ 2: .>..j-m 2: 2: !de .

j=1 k=j2 e=j2 j,m=1 jf.m

Denote the last two summands by Gn and Hn. and satisfies

n-1 (j+1) 2 -1 (j+l)2 -1 n-1

Then Gn does not depend on .>..,

IIGnll1 :::; L L L llfdelh :::; K 2 L(2j + 1)(2j + 1):::; 4K2(n + 1)3 /3. j=1 k=j2 j=l



ON STRONG LAWS OF LARGE NUMBERS WITH RATES

Since Hn does depend on>., we have

n-1 J max IHnl df-L:::; J "'"' 1>..1=1 .~ J,m=1

j=f.m

:::; {! [,~, rim

(j+I) 2 -1 (m+1) 2 -1 I: I: fkfp dJ-L k=P P=m2

121

1

n {1 (,~, 0 :~.-'(·:~.-' lhl'lftl' + ,~, (J;~q:.~~'MM. )d"} 2

j=f.m j=f.m (k,P)#(r,s)

The restriction j =f. rn puts k and r in one block of integers, while £ and s are in another one; thus when (k, £) =f. (r, s) the independence yields J fkfdrfsdf-L = 0. Hence the independence of l!kl 2 and lhl 2 yields

We conclude that

lim!~ I n2 ~ 1 Sn2 -111[ :::; (n2 ~ 1)2 (11Gnll1 +II fll!~ IHnlll1) :::; ~ Now let n satisfy rn2 :::; n < ( rn + 1 )2 . Then the previous inequality yields

+~II max It >.[Vklfklll rn 1>..1=1 2 k=m2 2

2m+ 1 C' C' < + K< <--- rn2 - y'rn + 1 - n 114 '

which proves inequality (21). The claimed a.e. convergence assertion now follows from Theorem 8, with

(3 = ~, p replaced by 2, and q replaced by p. 0

REMARK. The method of [CL], based on the deep results of Marcus and Pisier [MPl], cannot be applied here since the terms in {[Yk]} are not distinct; regrouping terms according to powers of >. and then following the method of [ CL] yields a worse estimate (i.e., a smaller value of (3).




PROPOSITION 10. Let (0, JL) be a probability space and let Un} C Lp(JL), 1 < p :::; oo, be independent with supn llfniiP < oo. Then for 1 :S: t < p we have

1 n sup- L lhlt < oo a. e. (i.e., for a. e. x E 0 the sequence {fk(x)} is in Wt)· n>O n k= 1

PROOF. We first prove that the assumptions imply supn ~ 2::~= 1 lfkl < oo a.e. (the case t = 1). It is clearly sufficient to prove for {fk} non-negative, and we may certainly assume in this part that 1 < p :::; 2. We then have E(fn) = llfnlh :S: llfniiP, and the centering 9n = fn -E(fn) satisfies ll9nllp :S: 2llfnllp· Hence 00 L E(lgniP)jnP < oo. By the Marcinkiewicz-Zygmund theorem ([MaZ], Theorem

n=1 00

5'; see also [S], Theorem 2.12.2), the series L gn converges a.e., so by Kronecker's n n=l

lemma ~ I:~=l gk ---+ 0 a.e. The claim now follows from

1 n 11 n I 1 n 11 n I * ~ {;h :S: ~ {;gk + ~ {;E(fk) :S: ~ {;gk +s~pllf1llp·

We now prove the proposition. The functions hn = lfnlt E Lp/t are indepen-dent, with supn llhnllpft < oo. Since pjt > 1, we can apply the first part of the proof to {hn} C Lp;t(JL), and obtain

1 n 1 n

sup- L I !kit= sup- L hn < oo a.e. 0 n>O n k=l n>O n k=l

REMARK. Note that {h(x)} need not be in WP. Let {An} be independent sets in non-atomic (0, JL) with JL(An) = nl~gn and fn := (nlogn) 11PXAn· By Borel-Cantelli a.e. xis in infinitely many An, and for x E Anj we have ,;j l:Z~ 1 Ifk(x)IP ::0: log n1.

THEOREM 11. Let { nk} be a strictly increasing sequence of integers with nk :::; cF for some r ::0: 1, let (Y, m) be a probability space, and let {gn} C Lq (Y, m), 2:::; q < oo, be independent with sup ll9nllq < oo and J gndm = 0. Then for a.e. y E Y the sequence ak := gk(Y) has the following property:

For every Dunford-Schwartz operator T on L1 (JL) of a probability space and

. ~ akTnk f 2q-l f E Loo (JL), the senes L...t k"~ converges a. e. for 1 E ( 3q_2 , 1].

k=l

PROOF. Since q ::0: 2, we have supn ll9nll2 < oo. It follows from Theorem 12 of [CL] (by a variant of Kronecker's lemma) that for a.e. y E Y the sequence {ak} satisfies (18) for any f3 < ~· By Proposition 10 {ak} E Wt for 1:::; t < q. We can now apply Proposition 6(ii) (letting t---+ q and f3---+ 1/2). 0

THEOREM 12. Let { nk} be a strictly increasing sequence of integers with nk :::; ckr for some r ::0: 1, let (Y, m) be a probability space, and let {gn} C L00 (Y, m) be independent with sup ll9nlloo < oo and J gndm = 0. Then for a.e. y E Y the sequence ak := gk(Y) has the following property:




For every Dunford-Schwartz operator T on L 1 (JL) of a probability space and

. 00 akTnk f 2 -1

f E Lp(JL), 2 S p < oo, the serzes L k"~ converges a. e. for"( E (~, 1]. k=1

PROOF. As before, {ak} satisfies (18) for any {3 < ~· For p = 2 we apply Proposition 2(i) of [CL], and for p > 2 we apply Proposition 6(i). 0

REMARKS. 1. When f E L:xo and supn JJgnJioo < oo, the lower limit for"( is 2/3, either by letting q----+ oo in Theorem 11 or by letting p----+ oo in Theorem 12.

2. Theorem 12 complements Theorem 14 of [CL], which gives the result for p = 2, with 'Y > 3/4, and uses it also when f E Lp(JL) with p > 2. Theorem 12 gives a better lower bound for 'Y·

THEOREM 13. Let (Y, m) be a probability space, and let {gn} C Lq (Y, m), 2 S q < oo, be independent with sup JJgnJJq < oo and J gndm = 0. Then for a. e. y E Y the sequence ak := gk(y) has the following property:

For every Dunford-Schwartz operator T on L 1 (JL) of a probability space and

. ~ akTkJ 1 n k f E Lp(JL), p > q~ 1 , the serzes L..t --k- converges a.e. and r;: Lk=1 akT f----+ 0

k=1 a. e.

PROOF. As in the proof of Theorem 11, { ak} E Wt for t < q, and { ak} satisfies (18), with nk = k, for any {3 < ~· For a given p, if p > q~ 1 then its dual index tis less than q, and we apply Proposition 6(iii) (with s = p). 0

REMARKS. 1. When q = 2 we obtain the convergence for all f E Lp, p > 2. When q > 2 we obtain convergence for all f E L2 .

2. If the sequence {gn} in Theorem 13 is i.i.d., then Theorem 7 gives the convergence of the series also for p = i!:T, since the SLLN can be used instead of Proposition 10. Moreover, for {gn} i.i.d. Theorem 7 does not require a finite second moment.

In order to extend the previous theorem to the case q < 2, we need the following theorem, which complements Theorem 12 of [CL]. Note that we have an additional assumption of symmetry.

THEOREM 14. Let (0, JL) be a probability space. Let 1 < p < 2, and {fn} C Lp(JL) be symmetric and independent with J fndfL = 0, and supn llfnllP < oo. Let { nk} be a strictly increasing sequence with nk S ckr for some r 2: 1. Then for a. e.

x, the series f :k 1 ~} )..nk converges uniformly in>.., for any 0 S 6 < ~· k=l

PROOF. We will use Theorem B(i) of [MP2], with the group G the unit circle, G the compact neighborhood, the set of characters A := { nk : k 2: 1 }, and the independent random variables ~nk = fk·

By linearity of the model we may and do assume that supn llfnllP S 1; this clearly implies that P(lfn I > c) S cP for every n and c > 0, the assumption in [MP2], p. 247. Fix 0 < 6 < (p-1)/p, and put a= P( 1 ~;)- 1 , so 0 <a< (p-1)/p.

Define { aj} on A by ank = J-s (the sequence need not be defined outside A, but we put aj = 0 for j ~ A). It will be convenient to identify the unit circle with




the interval [0, 21r], with addition modulo 21r. Let t 1 , t2 E [0, 21r] and define the corresponding translation invariant pseudo-metric d(h, t2 ) = a(h - t2 ) (which is uniformly convergent), where

with"(:= p- p8- m > p- p8- p(1- 8) + 1 = 1. Denote by m the Lebesgue measure on [0, 21r]. Then the "distribution" of a

satisfies ma(E) := m{t E [0, 27r]: a(t) < E} 2 c;:f> Ef, ;

hence the 'inverse' function defined on [0, 21r] (which is the non-decreasing re-arrangement of a), satisfies

a(s) := sup{t > 0: ma(t) < s}:::; Cas':.

In order to apply Theorem B(i) of [MP2] (in the form described in the discus-sion beginning at the end of p. 248 there), we estimate

I ( ) ·-127!" a(s)ds C 12

7!" ds P a .- b( ) < a ) o s(log ----f- )liP - o s1-': (log b~ )liP

where b(p) > 21r is a constant depending only on p (see p. 290 of [MP2]). The finiteness of Ip(a) follows from the integrability of 1 ~-"- for a> 0. Now the claimed

s p

convergence follows from [MP2]. 0

REMARKS. 1. The theorem applies to sequences {[kr]: k 2 1} with r 2 1. 2. The integers in the sequence { nk} must be distinct (in addition to the growth

condition), to make it an enumeration of the set of characters A; hence the proof of the theorem does not apply to the sequence { [ Vk]}.

THEOREM 15. Let (Y, m) be a probability space, and let {gn} C Lq(Y, m), 1 < q < 2, be independent and symmetric with sup llgnllq < oo and J gndm = 0. Then for a.e. y E Y the sequence ak := gk(Y) has the following property:

For every Dunford-Schwartz operator T on L 1 (J.L) of a probability space and . 00 akTk f n

f E Lp(J.L), p > q~ 1 , the senes L -k- converges a. e. and~ Ek=1 akTk f-+ 0

k=1 a. e.

PROOF. The proof is similar to that of Theorem 13, but uses Theorem 14 instead of Theorem 12 of [CL]: {ak} E Wt for 1:::; t < q, and by Theorem 14 (and a variant of Kronecker's lemma) { ak} satisfies (18) with nk = k for any 0 < {3 < q~ 1 . For p > ~ the dual index t is less than q and we apply Proposition 6(iii) (with s = p). 0




REMARK. Note that in the i.i.d. case (Theorem 7) symmetry is not required, and the convergence holds also for f E L ...!L •

q-1

THEOREM 16. Let (Y, m) be a probability space, and let {gn} C L 9 (Y, m), 2 S:: q < oo, be independent with sup ll9nll 9 < oo and I 9ndm = 0. Then for a. e. y E Y the sequence ak := 9k(Y) has the following property:

For every Dunford-Schwartz operator T on L 1 (J.L) of a probability space and 00 akT[Vk] f ( ) 2

f E L 00 (J.L), the series L k"Y converges a. e. for 'Y E ( 1 - 399-_12 , 1]. k=l

PROOF. By Theorem 9 (and a variant of Kronecker's lemma), {ak} satisfies (18), with nk = [v'k], for any (3 < 699-_~. By Proposition 10 { ak} E Wt for any t < q. We now apply Proposition 6(ii) with (3---+ 699-_14 and t---+ q. D

THEOREM 17. Let (Y, m) be a probability space, and let {gn} C L 00 (Y, m) be independent, with sup ll9nlloo < oo and I 9ndm = 0. Then for a.e. y E Y the sequence ak := 9k (y) has the following property:

For every Dunford-Schwartz operator T on L 1 (J.L) of a probability space and 00 akT[Vk] f

f E Lp(J.L), 2 S:: p < oo, the series L k"Y converges a. e. for 'Y E ( ~~::::~, 1]. k=l

PROOF. As before, Theorem 9 implies that the sequence { ak} satisfies (18) for nk = [ v'k], this time for any (3 < fi (by letting p ---+ oo in the result). Since { ak} is bounded, we apply Proposition 6(i), letting (3---+ fi. D

ACKNOWLEDGEMENTS

The authors are grateful to Christophe Cuny for many helpful discussions; in particular, his suggestions led to Theorem 4. The authors are also grateful to Idris Assani for his helpful comments, and for sending them the preprint [A4].

References [A1] I. Assani, A Wiener- Wintner property for the helical transform, Ergodic Th. & Dyn.

Syst. 12 (1992), 185-194. [A2] I. Assani, A weighted pointwise ergodic theorem, Ann. Inst. Poincare Proba. Stat. 34

(1998), 139-150. [A3] I. Assani, Wiener Wintner dynamical systems, ?reprint 1998, Ergodic Th. & Dyn.

Syst., to appear. [A4] I. Assani, Duality and the one-sided ergodic Hilbert transform, preprint. [AN] I. Assani and K. Nicolaou, Properties of Wiener Wintner dynamical systems, Bull. Soc.

Math. France 129 (2001), 361-377. [B] J. Bourgain, Pointwise ergodic theorems for arithmetic sets, Pub!. Math. IHES 69

(1989), 5-45. [CJRW] J. Campbell, R. Jones, K. Reinhold, and M. Wierdl, Oscillation and variation in singular

integrals in higher dimensions, Trans. Amer. Math. Soc., to appear. [CL] G. Cohen and M. Lin, Laws of large numbers with rates and the one-sided ergodic Hilbert

transform, Illinois J. Math., to appear. [QLO] D. Qomez, M. Lin, and J. Olsen, Weighted ergodic theorems for mean ergodic Lt con-

tractions, Trans. Amer. Math. Soc. 350 (1998), 101-117.



126

[CuLa]

[DL]

[Do] [G]

[I] [JKRW]

[Kr1]

[Kr2] [Lo] [MaZ]

[MP1]

[MP2]

[Ru]

[S] [WW]

GUY COHEN, ROGER L. JONES, AND MICHAEL LIN

J. Cuzick and T.L. Lai, On random Fourier series, Trans. Amer. Math. Soc. 261 (1980), 53-80. Y. Derriennic and M. Lin, Fractional Poisson equations and ergodic theorems for frac-tional coboundaries, Israel J. Math. 123 (2001), 93-130. J. Doob, Stochastic processes, John Wiley, New York, 1953. V. Gaposhkin, On the dependence of the convergence rate in the SLLN for stationary processes on the rate of decay of the correlation function, Theory of probability and its applications 26 (1981), 706-720. A. E. Ingham, Note on a certain power series, Ann. Math. 31 (1930), 241-245. R. Jones, R. Kaufman, J. Rosenblatt, and M. Wierdl, Oscillation in ergodic theory, Ergodic Th. & Dyn. Syst. 18 (1998), 889--935. U. Krengel, On the speed of convergence in the ergodic theorem, Monatshef. Math. 86 (1978), 3-6. U. Krengel, Ergodic Theorems, De Gruyter, Berlin, 1985. M. Loeve, Surles fonctions aleatoires de second ordre, Rev. Sci. 83 (1945), 297-303. J. Marcinkiewicz and A. Zygmund, Sur les fonctions independentes, Fun. Math. 29 (1937), 60-90. M. B. Marcus and G. Pisier, Random Fourier series with applications to harmonic analysis, Princeton University Press, Princeton, 1981. M. B. Marcus and G. Pisier, Characterizations of almost surely continuous p-stable random Fourier series and strongly stationary processes, Acta Math. 152 (1984), 245-301. D. Rudolph, A joinings proof of Bourgain's return time theorem, Ergodic Th. & Dyn. Syst. 14 (1994), 197-203. W. Stout, Almost sure convergence, Academic Press, New York, 1974. N. Wiener and A, Wintner, Harmonic analysis and ergodic theory, American J. Math. 63 (1941), 415-426.

[Z] A. Zygmund, Trigonometric Series, vol. I-II, corrected second edition, Cambridge Uni-versity Press, Cambridge, 1968.

DEPARTMENT OF ELECTRICAL AND COMPUTER ENGINEERING, BEN-GURION UNIVERSITY OF THE NEGEV, BEER-SHEVA, ISRAEL

E-mail address: guycohenl!lee. bgu. ac. il

DEPARTMENT OF MATHEMATICS, DE PAUL UNIVERSITY, 2320 N. KENMORE, CHICAGO, IL 60614, USA

E-mail address: rj onesl!lcondor. depaul. edu

DEPARTMENT OF MATHEMATICS, BEN-GURION UNIVERSITY OF THE NEGEV, BEER-SHEVA,

IsRAEL E-mail address: linl!lmath. bgu. ac. il




Besicovitch weights and the necessity of duality restrictions in the weighted ergodic theorem

Ciprian Demeter and Roger L. Jones

ABSTRACT. In an earlier paper the first author proved that in order for the weighted ergodic theorem to hold in £P, for all weights in the Besicovitch class B(r), the restriction r 2': q = p~l is necessary. We give a more simplified proof here and describe a possible approach, based on this argument, to the conjecture that duality is also necessary in order for the return times theorem to hold.

1. Duality restrictions in the weighted ergodic theorem

Let T and CJ be measure preserving transformations on the probability spaces (X,L:,m) and (Y,F,p,). We will call (X,L:,m,T) and (Y,F,fJ,CJ) dynamical sys-tems.

DEFINITION 1.1. An operator T acting on a probability space (X, L:, m) is said to be a Dunford-Schwartz operator, if it is a contraction in both the L 1 and £= norm. Note that the operator need not be positive.

It is well known that if T is a Dunford-Schwartz operator, and Ang(x) = ~ 2::~= 1 Tkg(x) then the averages Anf converge a.e. for all f E £P for each p, 1 ::; p < oo. See in particular, [10], page 675.

Further, if T is a Dunford-Schwartz operator, then >..T is also a Dunford-Schwartz operator, where ).., is a complex number of modulus 1. Hence if P(k) = l:f=1 aj>..j then for each j, the averages ~ 2::~= 1 >..jTkg(x) converge a.e., and con-sequently the averages

- L P(k)Tkg(x) =- L L aj-XjTkg(x) = L aj - L -XjTkg(x) 1 n 1 n J J (1 n )

n n n k=1 k=1 j=1 j=l k=1

converge a. e. for all g E £P, 1 ::; p ::; oo.

REMARK 1.2. The observation that the a.e. convergence of~ 2::~= 1 P(k)Tkg(x) follows from the fact that for T a Dunford-Schwartz operator, ,\Tis also a Dunford-Schwartz operator appears to be due to James Olsen [18].

The work of R. Jones was partially supported by a research leave from DePaul University and by a grant from the DePaul University Liberal Art and Science research program.


127




128 C. DEMETER AND R. JONES

DEFINITION 1.3. Let 1 :::; p < oo. A function W : N -. C is said to belong to the Besicovitch class B(p), if for each f > 0 there is a trigonometric polynomial P, such that

THEOREM 1.4. Let T be a Dunford-Schwartz operator. If 1 :::; p < oo and WE B(r) with r 2 q = p~l then the averages

converge a. e. for all g E LP (X).

REMARK 1.5. The above Theorem is contained in [16], although the proof below is slightly different. If WE B(q) n £00 see [7]. Also see [6, 11, 15, 17, 18].

PROOF. First note that if WE B(r) with r 2 q then WE B(q). Thus in what follows, we can assume without loss of generality that WE B(q).

Fix a non-negative sequence (Ej) which converges to zero. Since WE B(q), we know that we can find a sequence ( Pj) of trigonometric polynomials such that for each j we have

Note that by the remarks above we have for a.e. x, the limit

1 n lim - L Pj(k)Tkg(x)

n---+oo n k=l

exists. Denote this limit by L1(x). We claim that for a.e. x the sequence (Lj(x)) is Cauchy. To see this, note that

ILj(x)- Lm(x)l = ~}~ .. ~ ( ~ ~ Pj(k)Tkg(x)- ~ ~ Pm(k)Tkg(x)) I

:::; IJ!.~ ( ~ ~ (Pj - wk) (k)Tkg(x)- ~ ~ (Pm(k)- wk) Tkg(x)) I



DUALITY RESTRICTIONS 129

We now need to argue that

( 1 n ) ~ li:,U_:'~P ~ ~ ITkg(x)IP < oo.

First note that we can dominate the expression~ 2:.:~= 1 1Tkg(x)IP by first replacing T by a positive Dunford-Schwartz operator P that dominates it. Now we use the fact that (Pklfi)P :::; pk(IJIP) (see [14], page 65, Lemma 7.4) to dominate ~ 2:.:~= 1 (Pklg(x)i)P by ~ 2:.:~= 1 pk(lg(x)IP). From this and the fact that g = IJIP E £ 1 (X) implies a. e. convergence of these averages, and hence the desired conclusion.

Since (Lj(x)) is a Cauchy sequence, there is a limit, L(x). We now need to show that

1 n lim - ""'wkTkg(x) = L(x).

n->oo n L..J k=l

For E > 0, except for a set of measure zero, we can select j (depending on x) 1

such that ILj(x)- L(x)l < E/3 and Ej x limsupn->oo (~ 2:.:~= 1 ITkg(x)IP):;; < E/3. Further, we can find N such that n > N implies

11 n I (1 n )~ - ""'P·(k)Tkg(x)- L ·(x) < _: and E · -""' ITkg(x)IP < .:. nL..J 3 3 3 3 nL..J 3

k=l k=l We now have for all n > N

11 n I 11 n ~ ~ wkTkg(x)- L(x) = ~ ~ (wk- Pj(k)) Tkg(x)

+ ( ~ t Pi(k)Tkg(x)- Li(x)) + (Lj(x)- L(x))l

(1 n )~ (1 n )~ :::; ~ ~ iwk- Pj(kW ~ ~ jTkg(x)jP

+ ~~ tPj(k)Tkg(x)- Lj(x)l + ILi(x)- L(x)i.

Each of these terms was selected to be less than E/3 so the sum is less than E. Since E > 0 was arbitrary, we are done. 0

The restriction above, that WE B(q) implies W is a good weight for g E £P, seems to come from the proof. Hence it is natural to ask if the restriction is just an artifact of the proof, or is a real restriction. The same kind of issue appears in the following return times theorem [4], where the use of Holder's inequality apparently restricts p and q to be conjugate exponents.

THEOREM 1.6. For each dynamical system (X, E, m, r), 1 :::; q :::; oo and f E Lq(X), there is a set X 0 C X of full measure, such that for any other dynamical system (Y,F,J.L,a), g E £P(Y)- where 1/p+ 1/q = 1- and x E Xo

1 n lim - L f(rkx)g(aky)

n----?CXJ n k=l




exists for f..L a. e. y.

In [9] it was proved that in fact the restriction in Theorem 1.4 is real. Here we give a proof of this result based on a slightly different type of argument.

THEOREM 1. 7. Let (Y, F, f..L, a) be an ergodic dynamical system. For each 1 :::; p < oo and each r < q, where q is the conjugate exponent of p, there exist a function g E £P(Y) and a sequence W E B(r) of positive numbers such that the averages ~ 2:~= 1 wkg(akx) diverge on a set of positive measure.

REMARK 1.8. The case p = 1, q = oo was first established in [5].

PROOF. We need to construct a sequence W = (wk) such that WE B(r), but such that divergence occurs for the averages Ang(x) = ~ 2:~= 1 wkg(akx) for some g E £P(X). For each Ewe will use the trigonometric polynomial that is identically zero. Thus we will find W such lim supn->oo ~ 2:~= 1 wr; = 0.

Let A*g(x) = supn IAng(x)l. Based on Sawyer's principle, it is enough to prove the failure of the weak maximal inequality for A*.

First, we need a definition.

DEFINITION 1.9. A sequence (n8) is said to be evenly distributed (mod 1) if there is a constant d > 0 such that for any positive integer M, if [0, 1) is partitioned into [M/d] equal intervals, then each of these intervals will have at least one element from the set {e, 28, ... , Me}, (mod 1).

REMARK 1.10. There are e such that (n8) is evenly distributed (mod 1). See [12] or [13].

Fix an irrational e which has the property that (n8) is evenly distributed (mod 1). For each k set ak = [k log k] + ok with 0:::; ok < [log k], with ok chosen as described below.

For each n we will work in the block [22n, 22n+l ). In the interval [22n, 2 x 22n) we will select ok such that B([k log k] + ok) is as close as possible to 0 + ~ 2 ~, subject to the restriction that 0 :::; Ok < [log k]. In the interval [2 x 22n, 22 x 22n) select Ok so that B([k log k] + ok) is as close as possible to 21n + ~ 2~. In general, in the interval [2J x 22" , 2J +1 x 22") we select o k such that e ( [ k log k] + 0 k) is as close as possible t j_ + l_l_ c · - 0 1 2n - 1 0 2n 2 2n !Of J - ' ) • • • ) .

Define Wak = (log k) t and let Wj = 0 if j of. ak for all k. Fix n a positive integer and define 9n = X[o, 2d/2n). If x E [1 - ;jn, 1 - i;J}) then

for k E [2J22n, 2J+122n) we see that the sequence (ak) was constructed in such a way that x + ake E [0, 2d/2n). Consequently if N = 22n2i+ 1 - 1 then

1 aN

AaN9n(x) 2: - L W£9n(X + £.8) aN £=1

1 N !

2: ( N + 1) log N L (log 22n) t

£=N/2

1 N , > X - X (2n)' - 2NlogN 2

> ~2n(t-1l. - 8




Thus for each x E [0, 1) we have A*gn(x) ~ ~2n<t- 1 ). To have a weak (p,p) inequality, we would have to have

1 cP 2d m{x: A*gn(x) > -82n(1/t-1)}:::; X -

(~2n(l/t-1J)P 2n ·

From this we see that

This implies 2n2np(1/t-1) :::; CJ1 X 2d X 8P.

Since t < q we have ~ < t or 1/t- 1 = 1/q- 1 + E where E > 0. Using this, the above becomes

2n2np(1/q-l+<) :::; CJ1 X 2d X 8P.

Since 1/q- 1 = -1/p this can be rewritten as 2np< :::; CJ1 X 2d X 8P

which cannot hold if n is large enough. Hence there is no weak type inequality, and convergence cannot occur. To complete the proof of Theorem 1. 7 we note that

1a"' 1m 1 - L wl. =- L(log kt!t:::; m(logmt!t = (logmt/t-1 ---. o, am k=1 am k=1 m log m

since r < t, so the average converges to 0. The above arguments show that all the required properties are satisfied for this sequence. D

REMARK 1.11. One of the differences between the argument given here and the one from [9] is that the latter uses a single transformation on the torus T(x) = x+O in order to deny the maximal inequality, while the former makes repeated use of all the transformations T(x) = x + f;r. Also, in the argument from [9], the Besicovitch sequence W is supported on a subset of integers which is an infinite union of blocks, each block containing integers which are in the same residue class. This suggested to us a similar approach to the return times theorem, that is described in the next section.

In both of the proofs, the Besicovitch sequence W is supported on a subset of integers A= {ak: k ~ 1}, where (ak) grows roughly like klogk. In other words, if one introduces the counting function Cn = #{k : ak :::; n}, then Cn :::; ,~;n for some constant c independent of n. This growth turns out to be optimal in the sense provided by the following:

THEOREM 1.12. Let 1 < r, s < oo be conjugate exponents, A= (ak) a sequence of integers with Cn :::; (log~•+• for some positive E. Then for every positive sequence

WE B(r) supported on A and every g E L 1(Y)

1 n - L wkg(uky) = 0 n

k=1

for a.e. y E Y.




PROOF. It suffices to show that 00 2n 00 2n

II L 2~ L wkl9l(aky)ll1 = L 2~ L Wk < oo. n=1 k=1 n=1 k=1

But that easily follows since

t, w, ,; (#A,.) 'I• (t, wk) 'I• ,; c, 2"/" (#A,.) 'I• ,; :::::.

by Holder's inequality and using the fact that WE B(r). D

REMARK 1.13. When r gets big, s gets closer and closer to 1, which forces the support sequence A to have a maximal admisible growth closer and closer to that of k log k, in order for the negative result of Theorem 1. 7 to hold. Also, the result above shows that for any p, q and r like in the statement of Theorem 1. 7, there are sequences WE B(r) \ B(q) which are good weights for the ergodic theorem in LP(Y). Indeed, it suffices to choose as above a support sequence A = (ak) with Cn :::; (log';:)•+' for some positive f. Define a sequence W supported on A, Wak = Ck,

such that both of the following are satisfied:

"n r lim L..ti=1 ci = 0

n--+oo an

"n q l . L...i-1 ci liD - = 00.

n-+oo an

2. Duality restrictions in the return times theorem

In [3] it was recently proved that the result of Theorem 1.6 is false when p = q = 1. This still leaves open the question on what happens for the other pairs of indices p and q such that 1/p + 1/q > 1.

Our strategy is to assume that for a certain ergodic dynamical system (X, :E, m, r) the duality can be broken. To be more specific, assume that for some fixed q < oo the following is true:

for each f E Lq(X) there is a set X 0 C X of full measure, such that for any {S) other dynamical system (Y, F, p,, a), g E £ 1 (Y), x E X 0 and p, a. e. y the

following limit exists

Using Sawyer's principle [21] and the fact that the above is known to hold when q = oo, it follows that the statement (S) is equivalent to the following maximal inequality:

for each f E Lq(X) there is a set Xo C X of full measure and finite constants




C J,x for each x E Xo, such that for any other dynamical system (Y, F, J.L, a), {I) g E L 1 (Y), x E Xo and >. > 0

J.L{Y E y: sup I L~=1 f(rkx)g(aky) I > >.}:::; CJ,xllgll1. n n >.

Note that by Conze's principle (8], the constants CJ,x (not necessarily the best constants) can be taken to be independent of the dynamical system (Y, F, J.L, a). They will depend on (X, E, m, r), but it can be shown using the same Conze's principle that if CJ,x < oo for every f E Lq(X) and for all x in a set offull measure depending on f, then the same thing should hold for any other dynamical system. In other words, disproving inequality (I) for some f in some dynamical system, will immediately disprove (S).

Let Y = (0, 1] be equipped with addition modulo 1. Define aN(Y) = y + tt and gN = X[o,1;N)· Note that for each 1:::; i:::; Nand y E [*, ~ ), g(rky) equals 1 if k = N- i (mod N) and 0 otherwise. Hence, for this particular choice of the system and function, inequality (I) immediately gives

{ Lm-1 f( i+kN ) }

sup >.# 1 :::; i :::; N : sup Nk-(O 7 ) _x > >. :::; C J,x A>O m~1 m- 1 + z

for each f 2:: 0, f E Lq(X). Since

Lm,:1 f(ri+kNx) {f(rix) 1 L~ f(ri+kNx)} sup k 0 >max -- -sup =k=1'-'::-::...._-----'-m~1 N(m- 1) + i - i ' 2 m~1 Nm '

in order to disprove (S) it will suffice to prove that either of the following two operators are a.e. infinite for some f E U(X).

Ttf(x) = limsupsup>.# {1:::; i:::; N: f(~ix) >>-}=sup# {i: f(~ix) > >-} N->oo A>O Z A>O Z

. { . Lm- f(ri+kN x) } T2f(x) = hmsupsup>.# 1:::; z:::; N: sup k-1 N > >. .

N->oo A>O m~1 m

The first operator was introduced and studied by Assani in [1] and (2] where it was proved to be finite a. e. in Lq, q > 1. Hence it cannot be used to disprove the return times theorem in this case.

On the other hand it was proved in [3] that Ttf(x) = oo for some f E Ll, which immmediately disproves the return times theorem for p = q = 1.

This paper seems to be the first place where T2 is defined. To better understand it, define

"N 1 "m f( i+kN ) ,., f( ) 1' L...i=1 SUPm>1 m L..k=1 T X .L3 x = 1msup N

N->oo If one denotes aN,i = tt supm~ 1 ~ 2:~ 1 f(ri+kN x), then obviously T3f(x) =

limsupN->oo llaN,ill11 while T2f(x) = limsupN->oo llaN,ill! 1 ,oo, where llll11 ,oo is the weak-it norm of the sequence { aN,i}~ 1 . This shows T2 and T3 are closely related and moreover, T3f(x) 2:: T2f(x). The behavior of T3 is easy to handle in L1(X) as the following shows:




THEOREM 2.1. In every ergodic dynamical system (X, I:, m, T) there exists an f E L1 (X) such that

form a.e. x.

PROOF. By Conze's principle it suffices to prove the result for some particular ergodic dynamical system.

Take T to be the Bernoulli shift on X= [0, 1]z and let f E L 1 (X)\ L log L(X) be such that the variables xk = Tk f are independent. It follows from Orstein's result [19] that X* = supm ,$, 2::=~= 1 Xk is not integrable. Denote

1 m x: N(x) =sup- L xkN+i(x).

, m m k=1

Then for each N, (X~N ){:1 are pairwise independent and have the same distribution as X*, so

lim JID (]:__ ~ X7N >A)= 1 N-+oo N L..,. '

i=1

for any positive ,\, by the law of large numbers [20]. It will follow that

1 N lim sup- L:x: N = oo

N-+oo N i=1 '

form a.e. x, which ends the proof. D

REMARK 2.2. Our hope is that the use of i.i.d. random variables could prove in a similar way that T2 f(x) = oo for a.e x, for some f E L 1 . That would give an alternative proof to the afore-mentioned result from [3].

REMARK 2.3. The behavior of T2 in the other Lq-spaccs is not known. We suspect that here things are as bad as in L1 . Note that while T1 behaves well in Lq, T2 remains a plausible tool to disprove the return times theorem for q > 1.

References

[1] I. Assani, Strong laws for weighted sums of iid random variables. Duke Math. J. 88 (1997) 217-246.

[2] I. Assani, Convergence of the p-Series for Stationary Sequences. New York J. Math. 3A (1997) 15-30.

[3] I. Assani, Z. Buczolich, R. Daniel Mauldin, An L 1 counting problem in ergodic theory, preprint.

[4] J. Bourgain, H. Furstenberg, Y. Katznelson and D. Orstein, Return times of dynamical systems( appendix to Bourgain's pointwise ergodic theorems for arithmetic sets). IHES Pub!. Math. 69 (1989) 47-50.

[5] J. Baxter, R. Jones, M. Lin and J. Olsen, SLLN for weighted independently identically dis-tributed random variables, ( submitted for publication preprint, 20 pages).

[6] J. Baxter, J. Olsen, Weighted and subsequential ergodic theorems, Can. J. Math. 35 (1983) 145-166.

[7] A. Bellow and V. Losert, The weighted pointwise ergodic theorem and the individual ergodic theorem along subsequences, Thans. Amer. Math. Soc., 288 (1985) 307-345.

[8] J. P. Conze, Convergence des moyennes ergodiques pour des sous-suites, Bull. Soc. Math. France 35 (1973) 7-15.




[9] C. Demeter, The best constants associated with some weak maximal inequalities in ergodic theory, (to appear in Can. J. Math.)

[10] N. Dunford and J. T. Schwartz, Linear operators, Part 1, General Theory, John Wiley and Sons, NY, 1988.

[11] R. Jones and J. Olsen, Multiparameter weighted ergodic theorems, Can. J. Math. 46 (1994) no. 2 343-356.

[12] R. Graham and J. van Lint, On the distribution of (n(!), Canadian J. Math. 20 (1968) 1020-1024.

[13] D. Knuth, The art of computer programming, Vol 3, Sorting and Searching, Addison Wesley, 1969, page 543, problem 9.

[14] U. Krengel, Ergodic Theorems, de Gruyter, Berlin, 1985. [15] M. Lin and J. Olsen, Besicovitch functions and weighted ergodic theorems for LCA group

actions, Convergence in ergodic theory and probability (Columbus, OH, 1993), Ohio State Univ. Math. Res. Inst. Pub!., 5, de Gruyter, Berlin, 1996, 277-289.

[16] M. Lin, J. Olsen and A. Tempelman, On modulated ergodic theorems for Dunford-Schwartz operators, Illinois J. Math. 43 (1999) 542-567.

[17] J. Olsen, Calculation of the limit in the return times theorem for Dunford-Schwartz operaotrs, Proceedings of the Conference on Ergodic Theory and its Connections with Harmonic Anal-ysis, Alexandria, Egypt, Cambridge University Press, 1994, 359-367.

[18] J. Olsen, The individual weighted ergodic theorem for bounded Besicovitch sequences, Can. Math. Bull., 25 (1982) 468-471.

[19] D.S. Orstein, A remark on the Birkhoff ergodic theorem, Illinois J. Math. 15 (1971) 77-79. [20] V. V. Petroir, Sums of Independent Random Variables Springer-Verlag, New York Heidelberg

Berlin 1975. [21] S. Sawyer, Maximal inequalities of weak type, Ann. Math. 84 (1966), 157-174.

(C. Demeter) DEPARTMENT OF MATHEMATICS, UNIVERSITY OF ILLINOIS AT URBANA, URBANA, IL 61801

E-mail address: demeter@math. uiuc. edu

(R. Jones) DEPARTMENT OF MATHEMATICS, DEPAUL UNIVERSITY, 2219 N. KENMORE, CHICAGO IL 60614

E-mail address: rj ones@condor. depaul. edu




Strong sweeping out for lacunary sequences

Roger L. Jones

ABSTRACT. We give a new (more transparent) proof of the fact that lacunary sequences have the strong sweeping out property. The proof also shows the same it true for finite unions of lacunary sequences.

1. Introduction

Throughout the paper (X,~, J.L) will denote a complete non-atomic probability space, and T: X-+ X will be an invertible measurable ergodic measure preserving point transformation from X to itself. A sequence (nk), of positive integers, is said to be lacunary if there is a number p > 1 such that n~k 1 ~ p for all k. We will refer to p as the constant of lacunarity.

DEFINITION 1.1. A sequence of positive £1 - L 00 contractions, (Tk), is said to be strong sweeping out if given f. > 0 there is a set E such that J.L( E) < f. but such that

limsupTkXE(x) = 1 a.e. and liminfTkXE(x) = 0 a.e .. k k

The strong sweeping out property was introduced by Bellow, and subsequently studied by several authors. Showing that a sequence of operators is strong sweeping out implies that the operators diverge in the worst possible way. In [I J it was shown that many subsequence averages, such as ~ LJ=l f( r 23 x) have this property. In fact it is shown that the sequence (2j) can be replaced by any lacunary sequence, and we still have strong sweeping out. To do this the C(a) condition was introduced.

DEFINITION 1.2. Let 0 < a < ~- A sequence of real numbers, (nk), is said to satisfy the C(a) condition if given any finite sequence of real numbers, x 1,x2, ... ,x£, there is a real number() so that Onk E Xk + (a,1- a)+ Z for k = 1, 2, ... ,L.

In [I] it is shown that any lacunary sequence, after possibly neglecting the first few terms, satisfies the C (a) condition. A modification of what is given there shows that finite unions of lacunary sequences will also satisfy the C(a) condition. See

2000 Mathematics Subject Classification. Primary 37 A05; Secondary 28D05. Key words and phrases. lacunary sequences, strong sweeping out, ergodic averages. R. Jones was partially supported by a research leave granted by DePaul University's Research

Council, and by a research grant from the College of Liberal Arts and Sciences.


137




138 R. JONES

[8] where the details of this modification are given. In [1] it was also shown that a sequence of averages associated with a sequence that has the C(a) property will have the strong sweeping out property.

While the C (a) condition is not difficult to verify if there is a high degree of lacunarity, it is not easy to verify for sequences where the lacunarity constant is close to 1, and even more difficult in the case of finite unions of lacunary sequences. We give a new proof, not using the C(a) condition, which may be more transparent than the original proof. The first part of the argument given below uses a technique which has been used before. See [6] where Furstenberg used the technique to show that lacunary sequences are not a set of recurrence and [3] where Bellow used the technique to show that any lacunary sequence is "bad universal" for the pointwise ergodic theorem with f E L 1 .

If ( Vn) denotes a sequence of measures on Z, and T is a measurable, measure pre-serving point transformation of a probability space onto itself, we will also use (vn) to denote the associated sequence of operators given by vnf(x) = l:k vn(k)f(Tkx). A sequence of measures on Z, (vn) is said to be dissipative if given any point p E Z we have limn_, 00 vn(P) = 0. Below we will prove the following theorem.

THEOREM 1.3. Let W = (nk) be a finite union of lacunary sequences. Let (vn) be a dissipative sequence of probability measures on Z with support contained in the range of W, then the sequence of induced operators (vn) has the strong sweeping out property.

2. Some tools

We will need to be able to move results from one dynamical system to an-other. This is accomplished using a standard argument involving Roblin towers. For completeness we include the argument in the form that will be most useful here.

Let (X, I:, f..L, T) denote a dynamical system with (X, I:, m) a complete non-atomic probability space and T a measurable measure preserving invertible ergodic transformation from X to itself. Let (X, 't, jl,T) denote a second system with the same properties.

Let {vs : s E S} denote a collection of probability measures on Z. Then each 1!8 induces an operator both on L 1 (X) and L 1 (X), by Vsf(x) = :EkEZ Vs(k)f(Tkx),

and vsf(x) = :EkEZ v8 (k)](ikx) respectively.

PROPOSITION 2.1. LetS be a finite set and let {vs : s E S} denote a collection of probability measures on Z, each with finite support. If for some E > 0 and some A> 0 there is a set E E I: with f..L(E) < E such that f..L{X: supsES VsXe(x) > .\} > 1- E then there is a set E E 't with jl(E) < E such that jl{x : supsES VsXE;(x) > .\} > 1- 2t.

PROOF. Since we have only a finite number of measures, and each has finite support, we can find an N such that all measures have support in [-N, N]. For both dynamical systems, construct a Roblin tower of height L, where L is chosen such that the union of the top N steps have measure less than E/3, and the error set also has measure less than te/3. Remove a bit from the base of the tower with the larger base so that the measure of the base is the same for both towers. Let B denote the base of the tower in X and B the base of the tower in X. If we start with x E B, we can assign an "E-name" to x by looking at the sequence of L O's and 1's that are observed looking at xe(Tkx), k = 0, 1, ... , L- 1, as we move



STRONG SWEEPING OUT 139

from the base up to the top of the tower. These E-names partition B into a finite number of measurable pieces. Find a measure preserving set map ¢ from B to B such that each of the elements of this partition of B is mapped to a set of the same measure in B. Now define E C X so that if Pis one of the partition elements of B and Pis the image of P under¢ then theE-name of each point in Pis the same as the E name of each point in P.

Let R denote the union of the top N levels, the bottom N levels and the error set of the tower in X, and let R denote the corresponding set in X. Note that J,L( R) = [L( R) and that

J.L{X EX\ R: supvsXE(x) > >.} = [L{x EX\ R: supvsXE;(x) > >.}. sES sES

Since

[L{x EX: supvsxE;(x) > >.}:::: [L{x EX\ R: supvsxE;(x) > >.}

the result follows.

sES sES :::: J.L{X EX\ R: SUPVsXE(x) > >.}

sES :::: J.L{X EX: SUPVsXE(x) > >.}- J.L(R)

sES > 1 - f- f = 1 - 2E,

D

We will now prove a simple result, and then see how the ideas in this simple case can be used to prove Theorem 1.3. The idea is that given a degree of lacunarity, R, and any positive integer N, we can find an open set 8 such that for any o: E 8 we can place the points n 1o:, n2o:, ... , nNo: any place we want with an error of at most~-

PROPOSITION 2.2. Fix positive integers N and R. Assume n 1, n 2 , ... , nN sat-isfy n~: 1 :::: R fork= 1, 2, ... , N- 1. Then given any {'Yl, '/'2, ... , 'YN} C [0, 1), we can find an open set 8 ~ (0, 1) such that for any o: E 8 and each k, 1 ~ k ~ N we have nko: E ('Yk- ~' 'Yk + ~) + Z.

PROOF. Let ¢k(O) = nk(} (mod 1). Then cPk is periodic with period .":k, and consists of nk line segments, each segment with slope nk and with the lower left end point of the jth segment at (..!;, 0), j = 0, 1, ... , nk- 1.

Define Gk = {0 E (0, 1): cPk(O) E ('Yk- ~,'Yk + ~) + Z}.

Thus Gk ~ (0, 1) is a union of equally spaced open intervals of length R~k. Now consider ¢k+1 (0) = nk+1(}. This will have period - 1-. If the length of an

nk+l interval in Gk is at least twice the length of a period of ¢k+1 then each interval of Gk must contain a full period cPk+l and hence a full interval of Gk+I· This occurs if -R2 > 2-1 - which is equivalent to~ > R, and this was true by our assumption.

nk - nk+l nk -

Let G = nf:=1Gk. By the above argument, since we only have a finite intersec-tion, G is non-empty and open. D

REMARK 2.3. In the following arguments, to simplify some notation, we will sometimes use x EEl y to denote x + y mod 1.



140 R. JONES

COROLLARY 2.4. Let W = ( nk) denote a lacunary sequence with integer la-cunary constant at least R. Let {vn : n = 1, 2, ... , R} denote a collection of R probability measures on IZ, each with finite support contained in the range of the sequence W. Assume the supports are disjoint. Then there is an open set 8 of (0, 1) such that for any a E 8, if we let T(x) = x EB a and E = (0, ~) then SUP1$n$RllnXE(x) = 1 a.e.

PROOF. Let Jn denote the support of !In- Let N = sup{k: nk E ut;=1Jn}· Let In= [n.R 1, ~) and let rk = R-~+ 2 if nk E Jn. (If nk rf_ ut;=1Jn then we

can leave rk undefined.) We are now in the situation of Proposition 2.2, and we can find the open set e given by that Proposition. Therefore X E In, nk E Jn and a E 8, implies

X+ nk() E [n.R1' ~) + ( R-~+2- ~' R-~+2 + ~) + !Z C (0, ~) + !Z.

Therefore if a E e and we let TX = xEBa then X E In and nk E Jn implies Tnkx E E. Thus llnXE(x) = 1 for all X E In = [n.R 1' ~). Since u;;=1In = [0, 1) we have

sup1$n$R VnXE(x) = 1 for all X E [0, 1). D

To obtain strong sweeping out results, we would like to have a high degree of lacunarity in order to have the set E that occurs in Corollary 2.4 have very small measure. The idea is to break up our given lacunary sequence into a finite number, say ·r, lacunary subsequences (how many depending on how small we would like the set to be) so that each subsequence has a very high degree of lacunarity. We will then work on the r-torus, with irrational rotation, using the ith lacunary subsequence to control what happens on the ith coordinate of the torus. (The same idea has previously been used by Katznelson [7], in his work on coloring graphs. Indeed the idea for the present proof was obtained by reading his argument.) Once we have a certain behavior on the r-torus we can transfer this behavior to our given transformation using Proposition 2.1, preserving the behavior. The details are the content of the results below.

We will use the following notation. Let r and R be positive integers. Let I= [0, 1) let Eo= (0, ~). Define Em,r = Im-1 x Eo x F-m. Let E = U~= 1 Em,r· We see that lEI :::; riEol = ~. Let In= [n.R1, ~)for n = 1, 2, ... , R.

Let Gr = {1, 2, ... , Ry. For each g = (g1,g2, ... , gr) E Gr let Cg = I91 X I92 X

· · · x I 9r. Thus each Cg is a cube in F with side length~' and UgEGrC§ =F.

PROPOSITION 2.5. Assume that { J9 : g E Gr} are Rr disjoint finite subsets of z+. Let 81, 82, ... Sr be a partition of J = UgEGrJ§ into disjoint subsets such that for each C = 1, 2, ... , r the elements of Be form a lacunary sequence with lacunarity constant at least R. Then there is an open set 8 contained in [0, 1 t such that for any i5. E 8 and any g E Gn if x E Cg and n E Jg then x EB ni5. E E.

PROOF. Fix C and note that since the elements of Se form a lacunary sequence with lacunarity constant at least R, we can find an open set 8e such that if n E J9 n Se and x E I 9£ (the projection of Cg onto the Cth coordinate) then for any a E 8e we have X EB na E Eo. Let e = 81 X 82 X ... X er. Fix any i5. E e. If x E Cg and n E Jg then n E J9 n Se for some C E {1, 2, ... , r} and for that C we have xe EB nae E Eo, and thus x EB ni5. E E.

D




The following is immediate, and is essentially a restatement of Proposition 2.5 in terms of averaging operators on the r-torus.

COROLLARY 2.6. Assume that {Jg : § E Gr} are Rr disjoint subsets of z+' and let 81, 82, ... Sr be a partition of J = UgEGrJ§ into disjoint subsets such that for each£= I, 2, ... , r the elements of St form a lacunary sequence with lacunarity constant at least R. If each J9 is the support of a probability measure v9 then there is an open set 8 contained in [0, It such that for any a E 8 if we let TX = x EB a then sup§EG V§Xc(x) =I for all x E [0, It.

This result, together with Proposition 2.I results in the following corollary. We just need to assume that the coordinates of a above are irrational and rationally independent. Since we can select a from the open set 8, there is no problem finding such an a.

COROLLARY 2.7. Assume that {J9: § E Gr} are Rr disjoint subsets ofZ+, and let s1' s2' ... Sr be a partition of J = UgEGr J § into disjoint subsets such that for each£= I, 2, ... , r the elements of Se form a lacunary sequence with lacunarity constant at least R. If each J9 is the support of a probability measure v9 then given any E > 0 and any invertible measurable measure preserving ergodic transformation T on a complete non-atomic probability space (X,~' m) there exists a set E with m(E) :S ~ and m{x: SUP§EG V§XE(x) =I}> I- E.

As is typical with strong sweeping out results, we will use the following result of del Junco and Rosenblatt: [5, Theorem 1.3].

THEOREM 2.8 (del Junco and Rosenblatt). Let (Y, 9, v) denote a probability space, and L+(X) the cone of positive equivalence classes in L1 (X). Let (Tn), Tn: g-+ L+(X) and assume

(I) ForE, FE 9, if E c F then TnD:::; TnF. (2) The Tn are continuous in measure. (3) TnY = I for all n. ( 4) For all E > 0 and M ~ I, there exists A E g with v(A) < E such that

p,{supn~M TnA >I-E}~ I- E.

Then there is a dense G li subset R C g such that if A E R then lim supn->oo TnA = I a.e. and liminfn->oo TnA = 0 a. e., that is we have strong sweeping out.

3. Proof of the main result

We will apply Theorem 2.8. To adapt their notation to ours, we will let TnE = VnXE· It is clear that for any sequence (vn) of probability measures on Z the associated operators on the dynamical system satisfy properties I,2, and 3 of Theorem 2.8. Thus we only need to worry about property 4.

First assume we have a single lacunary sequence (nk) with lacunarity constant p > I, and that we have a sequence of probability measures (vn), each with support in the range of (nk)· For each n let Jn denote the support of vn. For now also assume the sets (Jn) are disjoint and finite.

Let E > 0 and a positive integer M be given. Let [pr] denote the greatest integer in pr, and select r so that [~~] < E. Let R = [pr]. We can ignore the first M measures, and relabel them so that it will be enough to show we can find a set E with m(E) < E and such that m{x : supn VnXe(x) > I - E} > I -E.



142 R. JONES

For € = 1, 2 ... , r let Se = {ne, nf+n nf+2n ... }. These sets clearly partition the supports of the measures (vn) as required by Corollary 2.7. Since the hypothesis of Corollary 2.7 are satisfied, we have the conclusion. We selected Rand r so that m(E) < f. Thus property 4 of Theorem 2.8 is satisfied, and we have the strong sweeping out conclusion.

If the supports are not finite and disjoint we argue as follows. First, for each n we can find a finite set Kn such that vn(Kn) > 1- f/2. Since the sequence (vn) is dissipative, we know vn(S) ---. 0 as n ---. oo for any finite set S C Z. We will now show we can select a subsequence of measures, (vmk), by induction in such a way that the new sequence will have the property required to apply the above argument. Let m1 = 1 so that Vm 1 = v1. Let J1 = K1. If Vm1 , ••• , Vmk have been selected, we select mk+l such that Vmk+l (UjskKm3) < ~. Then let

Jmk+l = Kmk+l \ UjSkKm3 •

Thus f f

Vmk+l ( Jmk+l) ~ Vmk+l (Kmk+l) - Vmk+l (UiSkKm,) ~ 1 - 2 - 2 = 1 - f.

Relabel, and we see that for each n we can find a finite set of integers, Jn such that vn(Jn) > 1- f, and by construction the sets (Jn) are disjoint. Now let

- 1 Vn(s) = Vn(Jn) Vn(S)XJn (s).

For this sequence we have by Corollary 2. 7 and our choice of r and R that there is a set E with m(E) < f such that m{x: supniinXE(x) = 1} > 1- f. Since VnXE(x) ~ Vn(Jn)iinXE(x) ~ (1- f)iinXE(x), we see that

m{x: supVnXE(x) > 1- f} ~ m{x: supiinXE(x) = 1} > 1- f, n n

we are done. To handle the case of a union of K lacunary sequences, simply apply the above

argument, but instead of breaking up the single lacunary sequence into r subse-quences, break up the sequence into K r subsequences, by first breaking up the initial sequence into K lacunary sequences, each with a lacunarity constant of at least p, and then break up each of these K sequences as above. The only modifica-tion is that we will now need to take r such that (;:] < f.

4. ~oving averages

The arguments used to prove the results in the previous sections can be used to obtain strong sweeping out for divergent sequences of moving averages. In particular we have the following results obtained by a more complex argument in [2].

THEOREM 4.1. Let (nk) be a finite union of lacunary sequences. Let (fk) be a non-decreasing sequence of positive integers such that given a positive integer L we have lim infk-+oo ek+L/nk = 0. Then the moving averages

fk-l

Akf(x) = €1 L f(rnk+Jx) k j=O

has the strong sweeping out property.




PROOF. The argument is much like the proofs of the earlier results, so we only sketch the modifications required. For a single lacunary sequence let p be the constant of lacunarity. Like before, given E > 0 let r be a positive integer such that [~~] < E and let R = [pr]. Split the sequence into r subsequences so that each has lacunarity constant at least R. Then as before for any one of these subsequences we can find an open set 8 such that for any o: E 8 we can place nk() where ever we want with an error of at most -k. However if let L = r Rr and we start the construction with ko where ko is selected SO that £ko+L/nk0 < -k, then we see that o: can be taken to be less than 1/ nko. Hence for such an o: we have i(nk + j)o:- nko:i :::; jo: :::; Cko+L/nko < 1/ R provided 0 :::; j < Cko+L· This will occur for all the j that occur in the averages Ak0 , Ako+l> ... , Ako+L· Hence for these averages we can place all the points involved in the averages where ever we want with an error of at most 2/ R. The result now follows as before. For a finite union of lacunary sequences the same argument works, we just break up the initial sequence into more sequences as before, and require [;:] < E where K is the number of lacunary sequences involved. 0

The following two corollaries are immediate.

COROLLARY 4.2. Let nk = [ak] and £k = [bk] where 1 < b < a < oo. Then the moving averages associated with the sequence ( nk, Ck) have the strong sweeping out property.

COROLLARY 4.3. Fix a, 1 <a< oo and let nk = [ka]. If £k = o(nk) then the moving averages associated with the sequence ( nk, Ck) have the strong sweeping out property.

Indeed with more work we can obtain the result from [4]. To state this result we need the following notation.

Let n be an infinite collection of lattice points with positive second coordinate. Define

Da = {(z,s): lz- Yi < o:(s- r) for some (y,r) E !1, (z,s) a lattice point}.

The cross section of Da at height s > 0 is denoted by n"' ( s) and is defined by Da ( s) = { k : ( k, s) E Da. We will say that n satisfies the cone condition if there is a constant c (which can depend on o:) such that for every,\> 0 we have IDa(.A)I :::; c.A. In [4] it was shown that the moving averages Akf(x) = £~ '2:7~~~k j(TJx) converge a. e. if and only if the set n = { ( nk, Ck)} satisfies the cone condition. The following theorem is contained in [4]. However, by selecting an appropriate subsequence, the proof follows by the above methods. See [2] to see how to select an appropriate subsequence. Thus the above methods can be used to establish the following result.

THEOREM 4.4. Assume that the set n = { (nk, Ck)} fails to satisfy the cone con-

dition. Then the moving averages Akf(x) = L '2:7~~:k j(TJx) are strong sweeping out.

References

[1] M. Akcoglu, A. Bellow, R. L. Jones, V. Losert, K. Reinhold-Larsson and M. Wierdl, The strong sweeping out property for lacunary sequences, Riemann sums, convolution powers, and related matters, Ergodic Theory and Dynamical Systems, 16 (1996) 207-253.



144 R. JONES

[2] M. Akcoglu and R. Jones, Strong sweeping out for block sequences and related ergodic aver-ages, Illinois J. Math., 43 (1999) 447-456.

[3] A. Bellow, On "bad universal" sequences in ergodic theory II, Lecture Notes in Mathematics, vol. 1033, Springer-Verlag, 1983.

[4] A. Bellow, R. Jones and J. Rosenblatt, Convergence for moving averages, Ergodic Theory and Dyn. Sys., 10 (1990) 45-62.

[5] A. del Junco and J. Rosenblatt, Counter examples in ergodic theory and number theory, Math. Ann., 245 (1979) 185-197.

[6] H. Furstenberg, Poincar recurrence and number theory, Bull. Amer. Math. Soc. (N.S.) 5 (1981), no. 3, 211-234.

[7] Katznelson, Y., Chromatic numbers of Cayley graphs on Z and recurrence, Paul Erds and his mathematics (Budapest, 1999), Combinatorica 21 (2001), no. 2, 211-219.

[8] Losert, V. A remark on the strong sweeping out property, Convergence in Ergodic Theory and Probability, (Bergelson, March and Rosenblatt, ed.), de Gruyter, Berlin, 1996. Pages 291-294.

[9] Rosenblatt, J., Universally bad sequences in ergodic theory, Almost Everywhere Convergence II, (Bellow and Jones ed.) Academic Press, San Diago, CA., 1991. Pages 227-245.

(R. Jones) DEPARTMENT OF MATHEMATICS, DEPAUL UNIVERSITY, 2330 N. KENMORE, CHICAGO IL 60614

E-mail address: rj onestacondor. depaul. edu




SOME OLD AND NEW ROKHLIN TOWERS

ISAAC KORNFELD

Abstract. We discuss an important and powerful technique in ergodic theory,

which originated from the famous Rokhlin Lemma. We illustrate how the Rokhlin

towers can be used to obtain "negative results" (counterexamples), mostly related

to the behavior of ergodic averages and of measurable cocycles. We also briefly

discuss the role of Rokhlin towers and, especially, nested sequences of such towers

in measurable and topological orbit theory.

0. Introduction.

One of the very basic tools of abstract ergodic theory is a simple fact known as Rokhlin Lemma, or Kakutani-Rokhlin Lemma, or Rokhlin-Halmos Lemma. It belongs to the group of mathematical statements (such as Zorn Lemma in set theory, Schwartz Lemma in complex analysis, and some others) which are traditionally called lemmas, despite the fact that their roles in the respective fields are fundamental.

In the present notes we are using the term Rokhlin Lemma (RL) which seems to be commonly used in the recent literature (see [Rl], [W] for some history of this result). Several variations on the theme of Rokhlin Lemma and some of its applications will be discussed. By no means these notes can be considered a complete survey of the results related to RL, which would have to contain a huge number of topics from different parts of ergodic theory. We include, first, some relatively simple old constructions, which illustrate the flavor of the Rokhlin Tower technique. Second, we briefly describe more recent results related

Supported by NSF grant DMS 0140068. 2000 Mathematics Subject Classification. Primary 37; Secondary 28.


145




146 ISAAC KORNFELD

to multitowers, Bratteli diagrams, etc, which played the cenral role in the recent remarkable progress in topological orbit theory.

Throughout these notes S : X -> X denotes a homeomorphism of a compact metric space X (usually a Cantor set), and J1 stands for one of its Borel probabil-ity invariant measures; T : Y -> Y denotes an automorphism or endomorphism of a nonatomic Lebesgue probability space (Y, !3, v). An automorphism (endo-morphism) T is called aperiodic if v( { x E Y : Tnx = x for some n}) = 0.

The statement of RL can be found in most of the ergodic theory books, and it is as follows.

Rokhlin Lemma. Let T be an aperiodic measure preserving automorphism of

the space (Y, !3, v). Then for every E > 0 and every natural n there exists a set

B E !3 such that

(i) Ti B, 0 ~ i < n are disjoint;

(ii) v(U~,:- 01 Ti B) > 1 -E.

T n-lB • • •

T 2 B •

TB i B n

Figure 1

Geometrically, this means that the dynamics of the transformation T, "up to c", can be described as in Figure 1. We call the set T = UZ:01 Ti B, as well as the collection {Ti B : 0 ~ i < n} a tower, or an (n, c)-tower forT, and call the set B

a base of this tower. Every point x E T moves under the action of T vertically one step up until it reaches the top level of the tower; then it either goes to the



SOME OLD AND NEW ROKHLIN TOWERS 147

"remainder" n =X\ T, or returns to the base B, but not necessarily vertically down; a point from n may go either to n, or to B.

RL became well known to many people from the classical little book [H] by P. Halmos, where it is given with the proof due to D. Ornstein. An immediate consequence of it is the possibility to approximate aribtrarily well an aperiodic automorphism T by periodic ones in the sense that there exist automorphisms Tn

of period n with v( { x : Tx =/:. Tnx}) ---+ 0 as n ---+ oo. This approximation was used by V.A. Rokhlin [R2] in his proof that generically (in the Baire category sense) a measure preserving transformation is not mixing. The paper [R2] appeared in 1948, soon after V.A. Rokhlin was released from a GULAG camp (see [R1], pp. 428-429).

Remark on endomorphisms. Usually, RL is formulated for the invertible trans-formations (automorphisms). In fact, a version of it is also true for endomor-phisms. One only needs to use the preimages y-i B instead of the images of B; so, in a sense, we are dealing here with "pits" instead of towers. Several (but not all) known proofs do not require invertibility; one of them (which works for ergodic endomorphisms) can be extractedfrom [EP]; a short argument for reducing the general aperiodic case to the ergodic one (which works for the endomorphisms too) can be found in [W1].

Remark on zd-actions. J.-P. Conze [C1] andY. Katznelson- B. Weiss [KW] independently generalized RL from the case of a single automorphism (or action of Z) to the case of free zd actions, d ~ 1. We say that a measure preserving zd

action on the spaceY, generated by the (commuting) automorphisms T1, ... , Td,

is free if v({x E Y: Tf 1 T;' 2 ••• TJdx = x}) = 0 unless i 1 = i 2 = · · · = id = 0.

zd-Rokhlin Lemma. Let T1 , ... , Td be commuting measure preserving auto-

morphisms of the space Y, and the zd action generated by them is free. Then

for every E: > 0 and every natural n there exists a set B E B such that

(i} Tf 1 • • • TJd B, 0:::; i1, ... , id < n are disjoint;

(ii} v(U~~~.,id=o Tf 1 • • • TJd B) > 1 -c.

Geometrically, ford= 2, a Z2 tower T can be visualized as ann X n square whose "cells" represent the sets Tf 1 T~ 2 B, and the action of T1 moves these cells in, say, horizontal direction, while the action of T2 moves them in the vertical direction.

A variant of RL is true even for the free actions of general countable amenable groups [OW].

There is a huge number of results and topics related to RL which are not discussed in these notes. Let us just mention a few of them. The possibility



148 ISAAC KORNFELD

of approximating general automorphisms by periodic ones, which is an immedi-ate corollary of RL, allows one to compare it with the Weierstrass Theorem of analysis and makes it a starting point of the approximation theory of measure preserving transformations (see [KS], [KSS], [K]) which produced many deep results and examples in ergodic theory. The well known cutting and stacking method [Fr] is, in a sense, a convenient form of the Rokhlin Tower construction, which proved to be very useful. These results and methods, as well as RL itself, have been extended from the measure preserving case to the case of nonsingular transformations (C. Linderholm, see [IT], [W2]).

Rokhlin Lemma (more precisely, its strengthened version, in which the levels of a tower are required to be independent of a given partition) is one of the main tools in the famous Ornstein solution of the isomorphism problem for the Bernoulli transformations and in the entire isomorphism theory (see [01], [02]), [Ka]).

A very nice account of many results and ideas in this area was given by B. Weiss in [Wl].

Remark on old and new applications. Since the statement of Rokhlin Lemma involves parameters c: and n, it is natural to expect that in the applications we will put c:-+ 0, or/and n-+ oo. This means that we will be dealing with a sequence Tm of towers which become taller and taller and/ or their remainders become smaller and smaller in measure. This does not necessarily mean that the towers Tm themselves increase monotonically with n as sets, nor does this mean that the levels of the consecutive towers are "nicely related". For some applications we just do not need any geometric relationship between the consecutive towers in a sequence. In a rather arbitrary manner, we will call such applications "old", as opposed to the applications in which the consecutive towers need to be geometrically related ("nested"), which will be called "new". We do not want to prescribe any formal meaning to this "classification" (in particular, some chronologically old applications may fall into the "new" category and vice verso). There is a feeling, however, that, nonetheless, it may help to organize somehow the material around Rokhlin towers.

1. Towers in problems on ergodic averages and coboundaries.

Moving ergodic averages. One of the first examples showing the contrast between the asymptotic behavior of the ergodic averages in the almost every-where sense and in the £ 2-sense was given in [AJ]. This example is a prototype of more general results in this direction ([JR], [BJR]), and gives a simple illustration of Rokhlin Lemma in action.




A well known feature of the von Neumann mean ergodic theorem is that, for

a measure preserving automorphism T: Y---+ Y and for f E L2 (Y) not only the

standard time averages ~I:?,:~ f o Ti converge, as n ---+ oo, to the projection, j, off onto the subspace ofT-invariant functions, but also

1 k+n-1 . _

lim II- " foTJ -!II =0 n~oo n L....,; j=k

uniformly in k. In other words, time averages taken over a time window of a

large size n are close to the space average uniformly over the positions of these

windows. Following [AJ], let us show that the situation in the a.e. sense is very

different. Namely, for every ergodic automorphism T: Y---+ Y there is an f E L2

(moreover, one can take f = XE, an indicator function of some measurable set

E) for which the "moving averages"

(1) 1 j=n+[vfnl

L f(Tix) j=n

[vnJ + 1

do not converge a.e.

To do this, construct a sequence Tm of "very tall" towers forT, whose heights

Nm grow exponentially and errors em (measures of the remainders) are small

enough. It is convenient to take Nm = 4m +2m, and take any em with em < /0 .

Denote by Em the union of the upper 2m levels of the tower Tm, and by Dm the

union of the lower 4m- 2m levels of Tm; there are 2m levels ("gap") between Dm

and Em (see Figure 2). Then, if E = Um Em, we have v(E) < ~· To show that

we can put f = XE (the indicator function of E), take any x E D := lim sup Dm

(a point belonging to infinitely many of Dm 's). Then we can place x in the

lower part Dm of an arbitrarily tall tower Tm in Figure 2. Consider the "time

window" for x starting at the moment n = n(x) such that Tnx just reaches

the first level of the upper part of Tm, and let the size of this window be [ vnJ. Note that n ::::=: 2m, but n :S 4 m. The last inequality implies that all points Ti x,

n :S j :S n + [vnJ, will be in Em and, hence, in E. This means that, for f = XE,

the time average ( 1) is 1, while the space average J f < ~. Since n can be made

arbitrarily large and, clearly, v(D) ::::=: 190 · ~'the construction is complete.



150 ISAAC KORNFELD

E2

El • • •

Dl D2 Dm

Figure 2

Averaging along subsequences. Without giving the details, we will de-scribe the connection between Rokhlin towers and problems on the behavior of ergodic averages taken along subsequences. U. Krengel [Kre] showed that there are sequences K = { kj} of positive integers such that for every aperi-odic automorphism T : Y --+ Y there is an f E L1 (Y) for which the averages an,K,T f(x) = ~ ""£j<n f(Tk1 x) do not converge a.e. as n--+ oo. Such sequences have been called "universally bad". J.-P. Conze [C2] used general results of E. Stein [St] and S. Sawyer [Sa] on the necessity of maximal inequalities for almost everywhere convergence to observe that the behavior of the averages taken along a subsequence K is closely related to the so called maximal operator M = Mx,r,

associated with the subsequence K and an automorphism T, which is defined, for every f E L1 (Y), by Mf(x) = supn>Oan,K,rf(x). Namely, if K is not uni-versally bad, then for every aperiodic T the operator Mx,T must be a bounded operator from L1 to the so called weak-L1 , or (more directly) there must exist a constant C = C ( K, T) such that

(2) sup [v({x: Mx,rf(x) > >.})]:::; C(~,T)_ /EL"IIJII,=l

This fact allows one to use Rokhlin towers in order to make the construction of universally bad sequences simpler and more transparent. Namely, the towers




can be used to show that the left-hand side of (2) actually does not depend on T; therefore, the constant C(K, T) depends only on K: C = C(K). Rouhgly speaking, the reason for this is that, according to RL, up to arbitrary c any two aperiodic transformations look the same. A little more precisely (see [Jo] for the details), given two aperiodic automorphisms T1 and T2 , acting on Y1 and Y2 respectively, one can construct sufficiently tall towers Ti(N,c) and ~(N,c) for them, with the same height N and the same small error c, and use them to copy any function h on Y1 onto a function h on Y2 by identifying, in a measure preserving way, the corresponding levels of Ti and ~, and then defining h as the copy of h under this identification. Then llhll1 = l!hl!1, and the distributions of the averages an,K,TJ1 and an,K,T2 h are the same for each n < N, which explains why the constant C does not depend on T.

For a concrete T, say, for an irrational rotation, one can construct sequences K with various properties (say, having the growth arbitrarily close to linear) for which the maximal operator MK,T is unbounded (see, e.g., (C2]). The above argument shows that these sequences are actually universally bad.

The cohomological equation. For an automorphism T : Y --+ Y, a measur-able function f : Y--+ R is called a (real-valued) measurable coboundary forT if the equation

(3) f(x) = g(x)- g(Tx) a.e.

(which is called the cohomological equation forT) admits a measurable solution g (which, in this case, is called a transfer-function for f). It is clear that, for an ergodic T, the solution of (3), if exists, is unique up to a constant. Two measurable functions, h and h, are called cohomologous if their difference is a measurable coboundary.

Remark on terminology. It may be noted that a natural setting for introducing (group-valued) coboundaries, cocycles, and cohomology of dynamical systems is not that of an individual transformation, but of general group actions (S2]. The terminology (cocycles, coboundaries) is influenced by this setting, to which, in turn, it came from homological algebra. We will not discuss this general setting here.

The equation (3), as well as its multiplicative analog ((f(Tx) = w(x)f(x))

appears in many different problems of dynamics and has a long history. The cohomological equation for the irrational rotations of the circle, i.e., for the transformations T : 8 1 --+ 8 1, defined by Tx = x +a (mod 1), x E (0, 1)), a is irrational, has been mentioned by D.Hilbert (Hi] as an example of an analytical problem which may have non-analytical solutions. To see this, one can use



152 ISAAC KORNFELD

(formal) Fourier expansions f = l::fnexp(27rinx) and g = L9nexp(27rinx) for the known function f and the unknown function g. Then the Fourier coefficients f n and 9n for f and g are connected by the formula

9n = 1 (2 . ) ' - exp 1rzna

which may involve "small denominators" for infinitely many n's. This shows that the properties of the solution g, if it exists, may depend on the diophantine properties of a. In particular, the equation may not admit an analytic, or smooth, or even L2 solution g even iff is analytic. D.V. Anosov [An] gave an example of an irrational a and an analytic function f for which (3) does have a measurable solution g, which is not integrable.

It is much easier to see that this phenomenon ("wild" transfer-function g for a "good" function f) can occur if f is assumed to be just continuous, instead of analytical. Moreover, a simple Rokhlin tower argument shows that for every homeomorphism S : X --+ X of a compact metric space which is aperiodic with respect to one of its invariant measures 11 there exists a continuous coboundary with measurable, but non-integrable transfer-function.

Indeed, take a sequence of towers Sm for S, whose levels are closed sets and whose heights Nm are big enough and errors em are small enough. One can take Nm = m 5 and arbitrary em --+ 0. For each m take an auxiliary continuous function 'Pm on X such that 'Pm(x) = ~ 2 on the set Em, which is the union of the top m 3 levels of Sm; let 'Pm = 0 outside a small <~"m-neghborhood of Em

(enough <~"m = m-8 ), and 0 ~ 'f!m(x) ~ ~ 2 elsewhere. Then put fm(x) = 'f!m(x)- 'f!m(Tm3 x), and 9m(x) = l::;'~ 0- 1 'f!m(Tix).

It is easy to check that (i) fm(x) = 9m(x)- 9m(Tx) for all x EX; (ii) lfm(x)l ~ ,;_2 for all x EX; (iii) !L({x EX: 9m(x) # 0}) ~~for some c1 > 0; (iv) fx 9m ?: ~ for some c2 > 0.

This implies that f = L~ fm is a continuous coboundary for S whose transfer function g = L~ 9m is finite a.e., due to (iii) and the Borel-Cantelli Lemma. However, is not integrable, due to (iv).

By modifying slightly the parameters of this construction, one can get, for any fixed p ?: 1, a continuous function f whose transfer-function belongs to LP, but not to any LP', p' > p.

The set of measurable coboundaries and Z2-towers. For a measure preserving automorphism T : Y --+ Y let Cob(T) denote the set of all measurable coboundaries forT. Iff E Cob(T), and f is integrable, then a short argument




using the ergodic theorem shows that J f = 0, even if the transfer-function for f is not integrable (see [An]). A necessary and sufficient condition for f E Cob(T)

was given by K. Schmidt [S1]. It says that f E Cob(T) if and only if the sequence of functions Sn,r(x) = ~~=~ f(Tkx), n = 1, 2, ... satisfies the property, which is sometimes called tightness and means that for every E > 0 there is N > 0 such that

v({x E Y: lsn,r(x)l > N}) <E.

It is not hard to show, using the towers and the Schmidt criterion, that if v

is nonatomic, for any T there are functions with zero integrals which are not in Cob(T).

The set Cob(T) contains a lot of information about T, and one can expect that two transformations, T1 and T2 , whose sets of coboundaries are related in some way, must be related themselves. A trivial observation is that if T1 = T.j' 1 ,

then Cob(T1 ) = Cob(T2). It turns out that for two ergodic commuting trans-formations the converse is true. Moreover, if T1 and T2 are ergodic commuting transformations on a nonatomic spaceY, and Cob(TI) ~ Cob(T2), then T1 = T:f

for some n E z. The proof is based on the Z2 Rokhlin lemma, and its idea is as follows. First,

assuming that T1 and T2 have the same coboundaries, we rule out the possibility that the Z2 action generated by them is free. Namely, we show that if this action were free, we would be able to construct a function f which is in Cob(TI) but not in Cob(T2). We will be looking for this function in the form f = ~:_'= 1 fm,

where, for each m, the behavior of the ergodic T1-sums sn,TJm = ~~=~ fm oTf

is tight, while the behavior of the ergodic T2 sums for fm is not tight. Then the Schmidt criterion, applied to f, will say that f has the desired properties.

To do this, we construct inductively the sequence Tm of large Nm x Nm towers for the Z2 action, and denote their bases by Em· Then, for each m, define fm

on the tower Tm by making it (an appropriately chosen) constant on each "cell" T1Td Bm, 0 :::; i, j < N m, of the tower. Since T1 moves the cells horizontally, and T2 moves them vertically, we will achieve our goal by alternating the positive and negative values of fm horizontally and, at the same time, accumulating the values of the same sign vertically. Some precautions are needed to guarantee the convergence of the series ~ fn; these are technical details which can be found in [Ko]. This takes care the case of the free action.

If the action generated by the commuting ergodic transformations T1 , T2 is not free, then one can immedeately see that this action cannot be faithful, i.e., for some integers p, q =/=- 0 the automorphism TfT:j is the identity. In this case a similar, but slightly more complicated Z2-tower argument shows that, unless p, q = ±1, one can still construct a function which satisfies the Schmidt criterion



154 ISAAC KORNFELD

(and therefore is a coboundary) for one of Ti, i = 1, 2, but not for another. It remains to observe that in the case p, q = ±1 we have either T1 = T2, or T1 = T2- 1.

If one drops the assumption that T1 and T2 commute, the above argument breaks down, primarily because the tower construction becomes non-available. It is an open problem whether or not Rokhlin Lemma can be generalized to, say, actions of the free group of two generators. This makes it natural to ask what happens with the above mentioned result itself if one drops or weakens the assumption of commutativity.

It turns out (see [KL]) that if the group G generated by the automorphisms T1 and T2 is nilpotent and acts ergodically, then it is still true that Cob(T1 ) = Cob(T2) implies T1 = Tf1, and Cob(TI) 5;;; Cob(T2) implies that T1 = T:f for some n. On the other hand, if the group G is solvable, these statements, in general, become false. A counterexample is given in [KL].

An important difference between the commutative case and the nilpotent case is that, for T1 and T2 generating a nilpotent group G which acts ergodically, it may well happen (in contrast with the commutative case) that the action of G

is faithful, but not free. Therefore, the proof for the nilpotent case requires an appropriate "non-free version" of the nilpotent Rokhlin Lemma (see [KL]).

Monotone sequences of towers and co boundaries. According to the in-formal "classification" mentioned in the introduction (old and new applications), all facts that we have discussed so far, fall into the category of" old applications" in the sense that their proofs involve the sequences of towers with no requrements on the geometrical rel11-tions between the levels of the consecutive towers in these sequences. In some situations, such requirements become highly desirable, and it seems that the mildest requirement of this sort is the "monotonicity" of tow-ers. Namely, for a given automorphism T, we may want to construct a sequence {7m}, m = 1, 2, ... of towers whose errors em tend to zero, and 7m 5;;; 7m+1 for all m. A short argument shows that this is possible to do for every aperiodic T

(see [AOW] or [L]). Indeed, fix a sequence em ~ 0 of the errors and fix a height N. Choose a

sequence {8m}, 8m > 0 and 8m «:::em· Start with a not necessarily monotone sequence {7~ 0 )} of (N, 8m)-towers. It will be modified recursively to get a se-quence of sequences {7~ 1 )}, {7~ 2 )}, ... {7~ 8 )}, ••• which converge, ass~ oo, in a natural sense to a certain sequence {7m}· Also, on the s-th step of this process we will achieve the monotonicity of the first s + 1 towers, which will guarantee that the limit sequence is monotone.

On the first step we "shave off'' a part of the tower 71(o) by removing all x E 71(o) \ 7;(o), as well as all points in 71(o) above and below such x's. Denote




the remaining tower ~( 1 ). The total measure removed is :::::; N 81 , hence by making 81 sufficiently small, we will still have v(~( 1 )) > 1- '¥-· For all m 2: 2 we put T/nl) = T/n°). On the second step we remove, in a similar way, points from two towers, ~(l) and 7;,( 1), to obtain the monotonicity of the first three towers, and so on. In the limit we will get a monotone sequence {Tm} of (N, Em)-towers. P. Arnoux, D. Ornstein, and B. Weiss used this procedure in [AOW] to show that every ergodic automorphism is isomorphic to an interval exchange transformation of countably many intervals.

Remark onE-free Rokhlin Lemma. For some transformations Tone can get monotone sequences of towers without using the above infinite procedure. This is certainly possible to do for any T : Y --+ Y having the property that for every set A C Y, whose complement is of positive measure, and for any n, there is an n-level tower forT, which covers A. E. Lehrer and B. Weiss [LW] showed that in fact this property holds for every automorphism T which is completely ergodic,

i.e. rn is ergodic for all n # 0. They called this statement the E-free Rokhlin Lemma. More precisely, they proved that, given an automorphism T, for any n

for which rn is ergodic, and for any A C Y with v(A) < 1, there is an n-level tower T forT such that T :J A. V. Ryzhikov showed [Ry] that, in general, E-free Rokhlin lemma does not hold for the Z2 actions. In this regard, it may worth to notice that the inductive "monotonization" procedure is still valid for the zd actions, d 2: 2.

An interesting application of monotone sequences of Z2 towers was found by D. Lind [L], who gave a short and, in a sense, elementary proof of a somewhat mysterious fact about coboundaries which has been originally proved by J. Feld-man and C. Moore [FM] by using the spectral sequences method and other tools of homological algebra. A special case of this fact is the following.

Let T1 and T2 be two commuting measure preserving automorphisms of the same space Y, which generate a free Z2 action. Then any measurable function f can be represented as a sum of two coboundaries, one for T1 and one for T2 .

In other words, the equation

(4)

has a measurable solution (g1 , g2 ) for every measurable f. Even in the case when Ti is a rotation of the circle by o:i, i = 1, 2, and o:1 , 0:2 are rationally independent, this fact is nontrivial (probably, as nontrivial as the general case). Moreover, not only is it unclear in this case how to represent some "wild" functions as sums of two coboundaries; it also seems unclear how to represent in this form, say,



156 ISAAC KORNFELD

the function f = 1 (a concrete question in real analysis!). Recall that every integrable coboundary for a single transformation must have a zero integral.

We will explain only the idea of the proof; the details can be found in [L]. First, let us observe a significant difference between the "one dimensional" equation ( 3) and "two-dimensional" equation ( 4). Let T be an n-level Rokhlin tower for an automorphism T with base B. Suppose that, for a given left-hand side f of (3), its solution g is known on the set B. Then one can extend g uniquely to all levels ofT, one level at a time, using (3). Therefore, if there are points x E rn- 1 B whose image Tx is in B (they always exist if the error of the tower is small enough), the presence of such points is a clear obstacle to defining g

arbitrarily on B. This is, of course, consistent with the fact that the solution of (3), if exists, is unique up to an additive constant.

In contrast with this, the solution (g1 , g2 ) of ( 4) is highly non-unique, and, in fact, can be defined arbitrarily on some cells of the two-dimensional towers for the Z2 action. Roughly speaking, this is because, working with a two-dimensional tower, every time when we encounter an obstacle as above in, say, the verti-cal direction, the presence of the second, horizontal direction gives us enough flexibility to overcome it.

Take Nm ----> oo, Em > 0, and construct a sequence {Tm} of two-dimensional towers for the Z2 action with bases Bm and errors Em. These towers are visual-ized as Nm x Nm squares, on which T1 acts horizontally, and T2 vertically. We can assume that

(i) Tm ~ Tm+l for all m; (ii) if s < m and x E Bm, then all points TfTdx, 0:::; i,j < Ns, are in Tm, or,

informally, ym intersects the T 8 -image of any x E Bm in complete N 8 -squares. Moreover, these N 8 -squares are required to be non-adjacent, i.e., there is at least one row (resp., column) between the horizontal (resp., vertical) sides of any two such N 8 -squares.

Note that property (ii) can be achieved by a procedure similar to the mono-tonization procedure described earlier.

The idea of the construction of a solution (g1, g2) is as follows. First, define g1 arbitrarily (but measurably) on Ti, and define g2 arbitrarily on the bottom row of Ti, except the rightmost cell. Then use the equation ( 4) to lift g2 to the entire Ti, except the rightmost column. After this, the equation ( 4) will hold on all of Ti, except the top row and the rightmost column.

Starting from the next tower, 72, one should work more carefully, since the problem arises of fitting together the previously defined values of (g1 , g2 ) with the new values. We extend g1 and g2 to the columns of 72, one column at a time, starting from the left. The "dangerous" cells for g2 are those lying in the




lowest levels of the N 1 x N 1 squares, where 92 has already been defined on the first step. It turns out that property (ii) can be used to arrange the procedure in such a way that every time when we need to define 92 on a "dangerous" cell, the value of 91 on its neighbor from the right is still not defined, and, therefore, can be chosen to make ( 4) work. Proceeding inductively, we can define the solution (91, 92) on all towers Tm, hence, almost everywhere.

2. Multitowers, nested sequences of multitowers, Bratteli diagrams and Vershik maps.

In this section we will be dealing with both measurable and topological sys-tems. In the topological situation, we will be mostly concerned with minimal homeomorphisms S : X ......_. X of a Cantor set X, which are often called Cantor minimal systems.

There are various problems, especially in the measure-theoretical and topolog-ical orbit equivalence theory, in which it is desirable to work with the sequences {Tm} of towers, which are not only monotone, but satisfy several additional reg-ularity conditions. For instance, if ~m denotes the partition of the space into the individual levels and the remainder ofTm, we may want the sequence {~m} to be increasing and to converge to the partition E into separate points; this means that every partition ~m+ 1 refines the previous partition ~m, and the entire a-algebra of measurable sets is generated by the elements of all ~m 's.

The so called odometer transformation, which is described below, illustrates, in a sense, the best possible scenario of the behavior of sequences of towers.

Example: the 2-adic odometer transformation T. Formally, this transforma-tion can be defined as a piecewise isometric mapping of Y = [0, 1) with count ably many intervals of continuity: Tx = x+3·2-n (mod 1) for x E [1-2-n+1 , 1-2-n), n E N, (see, e.g., [P], p. 211). A more transparent way to define the same transformation T is the following. First, introduce a partition 6 of Y into two subintervals, ~6 1 ) = [0, ~) and ~i 1 ) = [~, 1), and letT map ~6 1 ) linearly onto ~i 1 ). This means that ~ = { ~6 1 ), ~i 1 )} is a tower of height n1 = 2 for T with error c = 0. Next, cut the tower ~ vertically into two columns of width i, and letT map the top level of the left column, i.e., the interval [~, ~),linearly onto the bottom level of the right column, i.e., onto the interval [ t, ~). This gives us the tower 'I2 for T, consisting of four intervals, each of length i, and so on. For each m, the tower Tm consists of 2m sets, which are the elements of the partition of [0, 1) into 2m equal dyadic subintervals. The sequence of towers {Tm : m = 1, 2, ... } actually defines Ton the entire space.

The transformation T preserves the Lebesgue measure, but, of course, is not a homeomorphism. A very natural way to "make it a homeomorphism", i.e., to



158 ISAAC KORNFELD

get a topological model S for the transformation T is to do basically the same construction, but on the standard middle-third Cantor set X, instead of the interval [0, 1). Namely, we define the necessary homeomorphismS: X-+ X by a sequence Sm of towers. The first tower S 1 consists of two clopen (=closed and open) sets, ~~ 1 =X n [0, ~] and ~i 1 =X n [~, 1]. We cut this tower vertically into two clopen columns, the left one, based on X n [0, ~], and the right part, based on X n [~, n To get the tower S2 , we put, as before, the right part of each column of on top of the left part, and so on.

One more interesting model for the same odometer transformation (which justifies its name) can be obtained if we take X to be the group of all 2-adic integers, and define S : X -+ X as the group translation by the element 1 = 1 · 2° + 0 · 21 + 0 · 21 + ... , i.e., Sx = x + 1, where+ is the 2-adic addition.

Unfortunately, one cannot expect that such a nice sequence of towers can be constructed for every transformation. In particular, it may be impossible to get a sequence of towers whose level sets and remainder sets generate the entire a-algebra. In fact, this property characterizes the class of the rank 1 transfor-

mations (see [B], [F]), each of which has zero entropy and simple spectrum. In light of this (and for a number of other reasons), it is often more convenient to work with the so called multi towers rather than with usual Rokhlin· towers.

A (usual) n-level tower T = {B, TB, ... rn- 1 B} with remainder R can be visualized as consisting not of one column, but of two columns, one of height n,

and the other (remainder) of height one. The idea of a multitower is to allow several columns, each of its own height. There may be certain restrictions for the heights of the columns of a multitower for a concrete transformation T. Say, if all heights n 1 , n 2 , ... , nk have a common divisor d > 1, then Td cannot be ergodic. S. Alpern [A] proved a theorem which basically says that there are no other restrictions on the "architecture" of multi towers for general transformations.

The Alpern multitower theorem. For any k 2: 2, let n 1 , ... , nk be relatively

prime positive integers, and let q1, ... , qk be positive numbers such that n1 q1 + · · · + nkqk = 1. Then for any aperiodic measure preserving transformation T of a

Lebesgue probability space (Y, v) there exist sets B 1 , ... , Bk with v(B;) = q; and

such that {TiE;}: 1::::; i::::; k, 0::::; j::::; n; -1} is a partition ofY.

Remark on the multitower terminology. We call the collection T = {Ti Bi} : 1 ::::; i ::::; k, 0 ::::; j ::::; n; - 1} from the Alpern theorem a multitower forT, defined by the collections of parameters { n;} and { qi}. The set Uf= 1 B; is called the base of the multi tower T, and the set Uf= 1 Tni - 1 ( B;) is called the roof of T. For each i, 1 ::::; i ::::; k, the set Uj~ 0 1 B; is called the ith column of the multitower. If Tr and 72 are multitowers for T1 and T2 respectively, and their collections




of parameters are the same, we say that 1i and 72 are isomorphic. If only the collections { ni} for two multi towers are the same, we say that these multi towers have the same architecture (this can be applied in the topological situation too, where no measure is mentioned). If the collection { ni} is relatively prime, we say that the corresponding tower is relatively prime. Sometimes Alpern's theorem is called a copying lemma, because, given two aperiodic transformations, it allows us to "copy" the multi towers for one of them to the multi towers of the same shape for another. This copying technique plays a very important role in the proof of N. Ormes's orbit realization theorem (see §3). The multitower theorem can also be visualized as a discrete analog of Rudolph's two-step coding theorem [Ru], which states that a continuous-time system (flow) can be represented as a special flow, or flow under a function, with the roof function having a prescribed two-valued distribution.

Multitowers appear very naturally in the topological situation, for the minimal homeomorphisms of a Cantor set. If S : X ---+ X is a Cantor minimal system, and B C X is a clopen set, we can construct the Kakutani skyscraper for S

over B (see, e.g., [P]). Namely, for any x E B define the first return moment

r(x) = min{k 2: 1 : Skx E B}, and let En= r- 1(n), n = 1, 2, ... This allows us to represent X as the disjoint union of the sets Si En, n = 1, 2, ... , 0 ~ i < n.

In general, n may take infinitely many values ( = there may be arbitrarily tall columns in a skyscraper). For a Cantor minimal system and clopen B, a standard compactness argument shows that the number of columns is finite (i.e., n(x) is bounded by, say, N), hence the skyscraper is actually a multi tower. In this case the partition ~ = { Si En : n = 1, 2, ... , N, 0 ~ i < n} is called the Kakutani-Rokhlin partition corresponding to the set B. Moreover, by taking an appropriately chosen decreasing sequence Bm of clopen sets, converging to a singleton, one can get an increasing ( = more and more refined) sequence ~m of clopen partitions, such that all their elements generate the topology on X. The corresponding sequence of multitowers is called a nested sequence. A (short) proof of the existence of nested sequences of clopen multitowers for arbitrary Cantor minimal system can be found in a paper [HPS] by R. Herman, I. Putnam, and C. Skau.

Bratteli diagrams associated with multitowers. The clopen multitowers constructed for a Cantor minimal system, or, equivalently, the Kakutani-Rokhlin partitions, allow one to relate to this system a combinatorial graph-like object called ordered Bratteli diagram. The Bratteli diagrams appeared first in the the-ory of operator algebras [Br], and later the ordered Bratteli diagrams have been very successfully used in the study of orbit structure of topological dynamical



160 ISAAC KORNFELD

systems. There are several excellent expositons of this subject (see, e.g., [Sk]), so we will only give main definitions and mention the central facts.

Definition 1. Let (V, E) be a pair of countable sets, consisting of the "vertex set" V and the "edge set" E. Let both V and E be decomposed into disjoint unions of finite nonempty sets: E = {Em}m;:::o, and V = {Vm}m;:::l, with Vo

consisting of exactly one element, v0 . A Bratteli diagram is a pair (V, E), together with two maps , r : E --+ V, and s : E --+ V ( r stands for range, s stands for source), satisfying the following conditions: 1) r(Em) ~ Vm, s(Em) ~ Vm-1, m = 1, 2, ... ; 2) s-1(v) is non-empty for all v E V and r- 1(v) is non-empty for all v E V \ V0 .

Definition 2. An ordered Bratteli diagram is a Bratteli diagram associated with a pair (V, E) and maps r, s, together with a partial order ::S: on the set E such that the edges e' and e" are comparable if and only if r(e') = r(e").

Traditionally an ordered Bratteli diagram is denoted by (V, E, :::;), without mentioning the range and source maps explicitly. Note that, despite the fact that a Bratteli diagram is a more general object than a graph (in particular, multiple edges are allowed) we can speak about the paths in a Bratteli daigram B. Namely, a finite path in B is a finite sequence { e1, e2 , ••• , ek} of edges such that r(ei) = s(ri+l) for all i, 1 ::S: i ::S: k -1. If s(e1) E Vm, and r(ek) E Vm+k, we say that this path is from Vm to Vm+k· The infinite paths are defined similarly.

In the early 1980's A.M. Vershik employed Bratteli diagrams (defined in a slightly different way) in the context of measurable dynamics to prove that every ergodic automorphism T of a nonatomic Lebesgue space Y is measure-theoretically isomorphic to a so called adic transformation [V2]. Roughly speak-ing, the class of adic transformations is a natural generalization of the 2-adic odometer transformation, where instead of 2-adic integers ( = k-adic integers with the constant base k = 2) the naturally defined "km-adic integers" for vari-ous sequences {km} are used in a similar way. So, every adic transformation can be visualised as a Cantor minimal system.

More recently, Bratteli diagrams and the so called dimension groups, closely related to them, played a central role in a series of the remarkable results by H. Herman, T. Giordano, E. Glasner, I. Putnam, C. Skau, B. Weiss and others on topological orbit equivalence (see [HPS], [GPS], [GW] and references therein).

Let us explain how Bratteli diagrams appear in topological dynamics. We start with a nested sequence { Sm}m>O of clopen multitowers, or, equivalently, an increasing sequence of clopen Kakutani-Rokhlin partitions for a Cantor minimal systemS. Suppose that the multitower Sm consists of k = km columns cim), i =




1, ... , k, and the tower So is trivial, i.e., consists of one column with just one level in it, this level being all of X.

For each m = 0, 1, ... define Vm as (or identify Vm with) the set of all columns of Sm; this means that Vm consists of km elements. Also, form the set Em, m?: 1 by taking an edge e E Em each time when a column, say, c;;n) of Sm intersects a column, say, c;;-1) of Sm_1 . For a more precise definition of the sets Em, as well as for the formal definitions of the range and source maps and the order on each Em, we refer to [HPS], pp. 841-842.

Example: Bratteli diagram for the 2-adic odometer. Consider a nested se-quence { Sm}m?:O of clopen multitowers for the 2-adic odometer transformation S of the Cantor ternary set X. This sequence {Sm}m>O is a slight modification of the sequence { Sm} which was described above. We take So to be the trivial tower whose only set is all of X, and for m ?: 1 the tower Sm consists of two columns of height 2m- 1 , the first of which is the bottom half of the tower Sm,

and the second is the top half of Sm. The corresponding Bratteli diagram looks as in Figure 3.

--------vo El

----VI

E2

----V2

E3

----V3 Figure 3

The Bratteli diagram associated with a Cantor minimal system S depends not only on S, but also on the choice of a sequence of multitowers for S. However, different Bratteli diagrams obtained by the above construction from the same transformation S are equivalent in a natural sense. The corresponding equiv-alence relation is the one generated by the following notion of contraction of



162 ISAAC KORNFELD

Bratteli diagram. Given a (non-ordered) Bratteli diagram (V, E) and given a subsequence { mk}k;::::o of integers, the contraction of (V, E) to this subsequence is, by definition, the Bratteli diagram (V', E'), where, for any k 2:: 0, the vertex set V~ is defined by Vk = Vmk, and, for any k 2:: 1, the edge set Ek is the set of all paths in E connecting some vertex in Vmk-l with some vertex in Vmk (more formally, see [HPS], definition 2.2).

It turns out that there is a procedure to go back, from a Bratteli diagram to a Cantor minimal system. This is just the procedure that was used by A.M. Vershik in [V2] to construct the adic topological models for ergodic measure preserving transformations (we have already mentioned this fact before). The topological mapping obtained by this procedure from a Bratteli diagram is called the Vershik map. R. Herman, I. Putnam, and C. Skau [HPS] applied a very similar procedure to prove that every Cantor minimal system is topologically conjugate to a Vershik map associated with an ordered Bratteli diagram.

The description of this procedure is as follows. First, given an ordered Bratteli diagram B = (V, E, ::;), we denote by X = Xs the set of all infinite paths x = {x1 X2 X3 ... } with Xm E Vm and r(xm) = s(xm+l)· We generate the product-topology on X by the collection of the cylinder sets of the form Cm,y1 ,y2 , ... ,ym = {x E X : xi = Yi, 1 ::; i ::; m}. The partial order ::; on B induces the natural lexicographic order on X, and the Vershik map S : X --+ X maps an element x E X to its immediate successor, which is well defined unless x = Xmax is maximal in the lexicographic order; we also put S(xmax) = Xmin, where Xmin is minimal in the lexicographic order. For this construction to work, we need, of course, to assume that the maximal and minimal paths exist and unique. Next, let us assume that the diagram B is simple , which means that for every m there is M, M > m, such that for each v E V m and for each v' E V M there are paths connecting them. It can be checked that for a simple Bratteli diagram B the corresponing topological space X 8 is either finite, or is a Cantor set. If X is a Cantor set, then the simplicity of B implies the minimality of the Vershik map S. Going back to our example with the 2-adic odometer S, we can now see the reason why we have replaced the sequence of towers {Sm} for it by {Sm}; it was done just to avoid the case of the finite space X (actually, the space X for the original sequence of towers is a singleton), and to be able to recover by the above construction the Cantor set X, where S was originally defined.

3. Towers and Bratteli diagrams in orbit equivalence theory. Dye's theorem, Jewett-Krieger theorem, N. Ormes's theorem.

The technique of Bratteli diagrams. Vershik maps, etc, plays a very important role in the theory of topological orbit equivalence.




Definition 3. Let 8 : X ~ X and S : X ~ X be two topological dynamical systems. We say that 8 an S are topologically orbit equivalent if there is a homeomorphism h : X ~ X which takes orbits of 8 onto orbits of S, i.e., there are functions m, n : X ~ Z such that h o 8m(x)x = So h(x) and h o 8(x) = §n(x) o h(x) for every x EX.

The following strengthening of the notion of topological orbit equivalence may look somewhat artificial at first glance, but, in fact, is natural from several points of view. One of them is related to Bratteli diagrams, and some comments about this will be made below.

Definition 4. The topological dynamical systems 8 an S are called strongly topologically orbit equivalent if they are topologically orbit equivalent and the corresponding functions m and n (which are sometimes called orbit cocycles) have at most one point of discontinuiuty each.

It turns out that the strong orbit equivalence of two Cantor minimal systems, (X1, 81) and (X2, 82), admits a nice interpretation in terms of Bratteli diagrams associated with them (more precisely, associated with some sequences of Rokhlin multitowers, which, in turn, are associated with them).

The connection between the strong orbit equivalence and the Bratteli diagrams is established in the following theorem, the statement of which involves the notion of the contraction of a Bratteli diagram to a subsequence, defined in §2.

Theorem. ([GP8j and {GW, theorem 1.1). The Cantor minimal systems

(X1, 81) and (X2, 82) are strongly topologically orbit equivalent if and only if

there exist simple ordered Bratteli diagrams B1 and B2 constructed for some

sequences of nested clopen multitowers for 81 and 82 respectively, as well as a

non-ordered Bratteli diagram B, such that both B1 and B2 are contractions of B

(each for its own subsequence).

There is an analog of this theorem which gives an interpretation of usual (not strong) orbit equivalence in terms of Bratteli diagrams, but it is stated in a slightly more complicated way ([GPS]).

Historically, the study of the topological orbit equivalence was preceded by the measure theoretical orbit equivalence theory which took its origins in the pioneering work of H. Dye [Dl], [D2].

Definition 5. Let (Yi, Ai, vi), i = 1, 2, be two nonatomic Lebesgue probability measure spaces, and Ti : Yi ~ Yi be two measure preserving automorphisms. They are called orbit equivalent if there are null sets Ni C Yi, (vi(Ni) = 0), and a measure preserving one-to-one correspondence() : Y1 \ N1 ~ Y2 \ N2 , taking T1 orbits onto T2 orbits, i.e., such that for every x1 E Y1 \ N1 the image under () of the orbit {Tfx1 : n E Z} is the orbit under T2 of the point ()(x 1 ).



164 ISAAC KORNFELD

Dye's Theorem on Orbit Equivalence. Any two ergodic measure preserving

transformations of nonatomic Lebesgue probability spaces are orbit equivalent.

A number of different proofs of Dye's theorem are known (see, e.g., [HIK], [W2]). Several years ago, N. Ormes [Or] obtained remarkable results, called Strong Orbit Realization Theorem (SORT) and Orbit Realization Theorem (ORT), which contain a new proof of Dye's theorem, and also, much more im-portantly, give a simultaneous generalization of this theorem and another fun-damental fact of ergodic theory - Jewett-Krieger theorem on uniquely ergodic realizations of measure preserving transformations [J], [Kr].

Jewett-Krieger Theorem on uniquely ergodic realizations. Let T be

an ergodic measure preserving transformation of a Lebesgue probability space

(Y, !3, v). Then there is a compact metric space X and its minimal homeomor-

phism S : S __, S having exactly one Borel probability measure J-L, such that the

measure preserving systems (X, S, J-L) and (Y, T, v) are (measure theoretically)

isomorphic.

What N. Ormes's Orbit Realization Theorem (ORT) says is that not only do the uniquely ergodic models for a given ergodic automorphism T exist, but they can be found in the orbit equivalence class of a given Cantor uniquely ergodic system S. (In Strong Orbit Realization Theorem the same is stated for strong orbit equivalence classes, under a certain spectral condition for a system.

N. Ormes's Orbit Realization Theorem (ORT). LetS be a minimal home-

omorphism of a Cantor set X, and let J-l be an ergodic S -invariant Borel proba-

bility measure on X. Let T be an ergodic automorphism of a nonatomic Lebesgue

probability space (Y, v). Then there exists a minimal homeomorphism S' of X,

where S and S' have the same orbits, and (Y, T, v) is measurably isomorphic to

(X,S',J-L).

To see that the Jewett-Krieger theorem is contained in ORT, we can apply ORT to a homeomorphism S, which is uniquely ergodic, and observe that S'

must also be uniquely ergodic in this case, since the set of invariant measures is the same for all homeomorphisms in an orbit equivalence class.

It is well known that, in general, minimal homeomorphisms may have many invariant probability measures. Moreover, it is known [D] that, given any metriz-able Choquet simplex n, there is a minimal homeomorphism of a Cantor set such that the simplex of all its Borel probablity invariant measures is affinely home-omorphic to n.

From this fact and Theorem 1 one gets




Corollary. ([Or}, corollary 7.4) Let 0 be a metrizable Choquet simplex, and let

p be any extreme point of 0. LetT be an ergodic automorphism of a nonatomic

Lebesgue probability space (Y, v). Then there exists a minimal homeomorphism

S of a Cantor set X, whose simplex Ms of Borel probablity invariant measures

is affinely homeomorphic to 0 via a mapping f : 0 -+ Ms, and (Y, T, v) is

measurably isomorphic to (X, S, f(p)).

In other words, in the class of minimal homeomorphisms with a given simplex 0 of Borel probablity invariant measures, one can always find a homeomorphism, which models (with respect to one of its invariant measures) an arbitrary pre-scribed ergodic automorphism.

Now, the following natural question arises. What if, instead of one ergodic automorphism, we have a family :F of such automorphisms. Can we model all automorphisms of this family by one single Cantor minimal system S with a prescribed simplex 0 of invariant measures, i.e., is it possible to find a Can-tor minimal system S whose simplex of Borel probability invariant measure is affinely homeomorphic to 0 and such that there is a one-to-one correspondence between the family :F and the set of all extreme (=ergodic) points of n) under which every T E :F is measurably isomorphic to the corresponding ergodic sys-tem (S, J.L)? If the family :F is infinite, then this question must be formulated more accurately (some regularity conditions for the above one-to-one correspon-dence are certainly needed). However, for finite families :F the question is still interesting and nontrivial. Under some conditions on the family (existence of certain common factors) a positive answer has been obtained in the papers ofT. Downarowicz andY. Lacroix [DL] and T. Downarowicz and F. Durand [DD]. We should also mention B. Weiss's work [W3], where the so called universal minimal system is constructed. However, even the following "elementary" special case of our question seems open. Let a 1 and a 2 be two irrational numbers (which may or may not be rationally independent; for instance, the case a 1 = a 2 is not excluded). Does there exist a minimal homeomorphism S of a compact metric space X having exactly two Borel ergodic invariant measures, /-Ll and J.L2 , such that (S, J.Li) is measurably isomorphic to the rotation of the circle by ai, for i = 1, 2? We are not going to study this question in the present notes, but plan to do this in a subsequent paper (in preparation).

Acknowledgements. I thank D. Akimov and V. Malakhov for help in producing the diagrams for this paper.



166 ISAAC KORNFELD

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DEPARTMENT OF MATHEMATICS, NORTH DAKOTA STATE UNIVER-SITY, FARGO ND 58105, USA E-mail address: [email protected]



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This volume grew out of two ergodic theory workshops held at the University of North Carolina at Chapel Hill. Included are research and survey articles devoted to various topics in ergodic theory. The book is suitable for graduate students and researchers interested in these and related areas.

ISBN 0- 8218- 3313- 8

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