s.v. astashkin and f.a. sukochev- banach–saks property in marcinkiewicz spaces

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J. Math. Anal. Appl. 336 (2007) 12311258

www.elsevier.com/locate/jmaa

BanachSaks property in Marcinkiewicz spaces

S.V. Astashkin a,, F.A. Sukochev b

a Department of Mathematics, Samara State University, Samara, Russiab School of Informatics and Engineering, Flinders University, Bedford Park, SA 5042, Australia

Received 6 June 2006Available online 23 March 2007

Submitted by Steven G. Krantz

Abstract

We catalogue all Marcinkiewicz function and sequence spaces with the BanachSaks property and

present necessary and sufficient conditions for a wide subclass of spaces to possess the p-BanachSaks

property, 1 < p < . We apply our results to several open problems.

2007 Elsevier Inc. All rights reserved.

Keywords: Rearrangement invariant spaces; Marcinkiewicz spaces; BanachSaks property; Boyd indices; Spaces with

mixed norm; Rademacher type of Banach spaces

1. Introduction

We recall that a Banach space X has the BanachSaks property (respectively p-Banach

Saks property, 1 < p < ) if every weakly null sequence (xn)n=1 in X has a BanachSaks(respectively p-BanachSaks) subsequence (xnk )k=1, i.e.

limm m

1

mk=1

xnk

= 0;respectively

Research of both authors was partially supported by the ARC.* Corresponding author.

E-mail addresses: [email protected] (S.V. Astashkin), [email protected] (F.A. Sukochev).

0022-247X/$ see front matter 2007 Elsevier Inc. All rights reserved.

doi:10.1016/j.jmaa.2007.03.040


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1232 S.V. Astashkin, F.A. Sukochev / J. Math. Anal. Appl. 336 (2007) 12311258

limsupm

m1/p

mk=1

xnk

< .

For example, if 1 < p 1 implies p-BanachSaks prop-

erty [26] (see also [18,25]). However, the study of BanachSaks properties in Banach spaces

with no (nontrivial) Rademacher type is far from being easy. Among rearrangement invariant

function and sequence spaces (which are generalizations of Lp(0, 1) and lp -spaces), the most

important class of spaces, which do not have type p for any p > 1 is given by (separable)

Marcinkiewicz spaces M0() and m0() associated with an arbitrary concave increasing func-

tion : [0, ) [0, ), (0) = 0. The main objective of the present paper is to completelycharacterize Marcinkiewicz spaces M0( ) and m0() with BanachSaks properties in terms

of the function . This characterization is given in terms of dilation indices of the function

(see e.g. [20] or next section). It may be also restated in terms of (so-called) Boyd indices ofMarcinkiewicz spaces and from this viewpoint our investigation complements and extends re-

cent results of [4] and [28]. An interesting feature of our results is a substantial difference in

BanachSaks properties between function and sequence Marcinkiewicz spaces. Exploiting this

difference, we are able to contribute to the description of isomorphic types of such spaces and (in

many cases) to show that a (function) Marcinkiewicz space M0() does not isomorphically em-

bed into (Marcinkiewicz) ideal Cm0() of compact operators. These results complement earlier

results of [2] that Lp(0, 1) does not embed into the Schatten von Neumann ideal Cp, 1 p < .Another application of our study is related to the theory of vector-valued rearrangement invariant

spaces and spaces with mixed norm. A.V. Bukhvalov [8] asked for which rearrangement invari-

ant spaces E = E[0, 1] there exists an isomorphism E E(E) or E E[E]. It is of interestto note that a similar question, in the setting of symmetric sequence spaces, was also studied by

C. Read [27]. In [8], A.V. Bukhvalov announced the result that such an isomorphism (obviously)

holds for E = Lp , but fails for E = Lp,q when 1 < p = q < , p = 2. Our contribution to thisproblem consists in showing that such an isomorphism does not exist in the case when q = ,i.e. for separable parts of the weak-Lp -spaces, Lp,, 1 < p < . In particular, we also showthat the separable parts of function and sequence weak-Lp-spaces are not isomorphic. This

latter result is in sharp contrast with the result from [21] asserting that the nonseparable weak-

Lp-spaces Lp,(0, 1) and lp, are isomorphic.Another interesting feature of our technique in this paper is the active usage of so-called sym-

metric functionals on Marcinkiewicz spaces [1113]. The latter are a commutative counterpartof Dixmier traces appeared in noncommutative geometry (see e.g. [9] and references therein)

and their application to the study of BanachSaks property in (separable) Marcinkiewicz spaces

is novel and may be of independent interest.

Finally, we are compelled to compare our results to earlier publications in this area. The

theme of BanachSaks properties in Marcinkiewicz spaces was discussed in [30] (no proofs

were given), where it were announced that every function Marcinkiewicz space M0()(0, 1)

and every sequence Marcinkiewicz space m0() have the BanachSaks property. These claims

were supported by references to the earlier paper [29]. However, they are demonstrably false

as is shown by our present results: (1) the space M0()(0, 1) has the BanachSaks property

if and only if either Boyd indices of M0()(0, 1) are nontrivial or M0()(0, 1) = L1(0, 1);(2) the space m0() has the BanachSaks property if and only if either the lower Boyd index of

m0() is greater than 1 or m0() = l1.


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S.V. Astashkin, F.A. Sukochev / J. Math. Anal. Appl. 336 (2007) 12311258 1233

The part (1) above substantially extends and strengthens a recent result from [15] that any

Marcinkiewicz space M0( ) which is simultaneously (the separable part of) an Orlicz space

fails the BanachSaks property.

We treat the cases of Marcinkiewicz spaces on (0, 1), (0,

) andN in Sections 3 and 4 below.

The following section is reserved for definitions and auxiliary results. Applications are presentedin the final section.

2. Preliminaries: Marcinkiewicz spaces

Consider a Banach space (X, X) of real-valued Lebesgue measurable functions (withidentification a.e.) on the interval J = (0, ), or else on J = (0, 1), or on J =N. Let x denotethe nonincreasing, right-continuous rearrangement of|x| given by

x(t) = infs 0: |x| > s t (t > 0)where denotes Lebesgue measure. Then X will be called rearrangement invariant (or r.i., forbrevity) space if

(i) X is an ideal lattice, that is if y X, and x is any measurable function on J with0 |x| |y|, then x X and xX yX;

(ii) ify X and if x is any measurable function on J with x = y, then x X and xX =yX.

In the case J = N, it is convenient to identify x with the rearrangement of the sequence |x| =(

|xn

|)n

=1 in the decreasing order. For basic properties of r.i. spaces we refer to the monographs

[20,23,24]. We note that for any r.i. space X = X(J) the following continuous embeddings hold(L1 L)(J) X (L1 + L)(J).

We will denote by L0(0, ) the closure of(L1 L)(0, ) in (L1 + L)(0, ). It is easy toshow that x L0(0, ) if and only if limt x(t) = 0. An r.i. space X has the Fatou property ifX = X, where X is the second Kthe dual ofE [20,24], i.e. if it follows from (xn)n1 X,xn x a.e. and supn xnX < that x X and x liminfxnX.

The r.i. space X is said to be fully symmetric Banach space if it has the additional property

that ify X and (L1 + L)(J) x y, then x X and xX yX. Here, x y denotessubmajorization in the sense of HardyLittlewoodPlya:

t0

x(s)ds t

0

y(s)ds, t > 0.

A classical example of nonseparable fully symmetric function and sequence spaces X(J) is

given by Marcinkiewicz spaces.

2.1. Marcinkiewicz function and sequence spaces

Let denote the set of all concave increasing functions

: [0,

)

[0,

) with (0)

=0

and

limt0+

t

(t)= 0. (2.1)


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Let denote the set of all such that limt0+ (t) = 0 and limt (t) = . Importantfunctions belonging to include the functions t t , (log(1 + t)) for 0 < < 1. Let .Define the weighted mean function

a(x,t) = 1(t)

t0

x(s) d s, t > 0,

and denote by M()(0, ) (respectively M()(0, 1)) the Marcinkiewicz space of measurablefunctions x on (0, ) (respectively (0, 1)) such that

xM() := supt>0

a(x,t) <

respectively xM() := sup0 0 and (0) = 0,then M() = M0( ) = L1.

For brevity, when the discussion does not depend on whether J = (0, ) or J = (0, 1), weshall write simply M0( ). We shall also frequently write to denote the norm M().

Finally, we shall also use Marcinkiewicz spaces M() (respectively M0()) of all measurable

functions f on (0, 1) (0, 1) such that f M()(0, 1) (respectively f M0()(0, 1)) withthe norm fM() := fM(). Here, the decreasing rearrangement f computed with respectto the product measure on (0, 1) (0, 1). Clearly, the spaces M( ) (respectively M0()) andM()(0, 1) (respectively M0()(0, 1)) are isometric.

It is sometimes convenient to identify the space M0()(0, 1) (respectively m0()) witha subspace of M0()(0, ) spanned by the functions supported on (0, 1) (respectively by thefunctions taking constant values on intervals [n 1, n], n 1).

2.2. c0-Copies in Marcinkiewicz spaces M0()

Recall that a sequence of elements (xn)n=1 of a Banach space (X, X) is said to be semi-

normalized if there exist constants 0 < a, b < such that a xn b, n 1. Concerning thefollowing assertion see also [29, Lemma 2 and Proposition 1].

Proposition 2.1. Every semi-normalized sequence (xn)n=1 M0( ), , converging to 0in measure contains a basic subsequence (xnk )

k=1 equivalent to the standard unit vector basis

(ek )k=1 in c0.


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Proof. First, we show that for any x M0() and > 0 the inequalityx + xn (1 + ) max

x, xn (2.3)holds for all sufficiently large n. Indeed, since x

M0() then by [20, Lemma 2.5.4] there exist

0 < h1 < h2 < such that

1

(h)

h0

x(s)ds x, h [0, h1] [h2, ).

Therefore, for every h [0, h1] [h2, ) and n 1, we have

1

(h)

h0

x + xn

(s)ds x + xn. (2.4)

Since xn 0 in measure, , and h1, h2 are fixed, we may select 0 < h0 < h1 satisfying(h0)

(h1)

2(2.5)

and such that for all sufficiently large n 1, we have

1

(h1)

h2h0

xn (s)ds

2xn. (2.6)

Combining (2.5) and (2.6), we see that for any h (h1, h2) we have1

(h)

h0

x + xn

(s)ds x + 1

(h0)

h00

xn (s)ds (h0)

(h1)+ 1

(h1)

h2h0

xn (s)ds

x + xn. (2.7)Inequality (2.3) follows from (2.4) and (2.7), if we observe that

x + xn x + xn= sup0


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2.3. Dilation indices of and Boyd indices ofM()

For every , we set

M (t) := sup0


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If there is a constant C > 0 such that t(t) C (t > 1) then limt (t) = 0 and again formula(2.10) implies

limsupt

ln( (ts(t))(s(t))

)

ln t lim

t

ln(M (t1(t)))

ln(t1(t)) +lim

t

ln(M0 (C))

ln t

.

Similarly, in the case when t1(t) C (t > 1) we have limt (t) = 1 and

limsupt

ln( (ts(t))(s(t))

)

ln t lim

tln(M (C))

ln t+ lim

tln(M0 (t

(tn)n ))

ln(t (tn)n )

0 .

Thus, (2.9) is proved and the proof is completed. 2

Recall (see [24, p. 130]) that for every s > 0 the dilation operator Ds on M()(0, ) (re-spectively M()(0, 1)) is given by

(Ds f)(t) = f ts

, f M()(0, );

respectively

(Ds f)(t) =

f

ts

, tmin{1, s},

0, s < t 1 (s < 1),f M()(0, 1).

In the case of sequence spaces m(), the operators Ds are defined only if s is an integer or the

reciprocal of an integer. Iff = (aj )j=1 m(), then

Dnf:=

(

n times

a1, a1, . . . , a1,n times

a2, a2, . . . , a3, . . . )D 1

nf := n1

n

j=1aj ,

2nj=n+1

aj , . . .

.

The Boyd indices pX and qX of an r.i. space X are defined by (see [7,24])

1 pX = lims

log s

log DsXX qX = lim

s0+log s

log DsXX.

It is well known that the Boyd indices of the spaces M() and M0() coincide and that (see

e.g. [20, Section II.4]) for X = M()(0, ) (respectively M()(0, 1); m()) we have

pX =1

1 =1

X, qX =

1

1 =1

X;

respectively

pX =1

1 0= 1

0X

, qX =1

1 0= 1

0X

;

pX =1

1 = 1

X, qX =

1

1 = 1

X.

Here, X(t) := t(t) , t > 0, is the fundamental function for X, i.e. for every measurable setA

(0,

) (respectively A

(0, 1); A

N), we have

X

mes(A)= AX,

where throughout by A we will denote the characteristic function of a set A.


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Proposition 2.3. (See [20, Section II.1].) If X = M0()(0, ) (respectively M0()(0, 1);m0()) and , then the conditions (i)(iv) and(i)(iv) below are equivalent

(i) pX=

1;

(ii) X = 1 (respectively 0X = 1; X = 1);(iii) = 0 (respectively 0 = 0; = 0);(iv) t (0, 1)

(sn)n1 (0, ): limn

(tsn)

(sn)= 1

respectively (sn)n1 (0, 1): limn

(tsn)

(sn)= 1;

(sn)n1

N: lim

n

(tsn)

(sn) =1

and

(i) qX = ;(ii) X = 0 (respectively 0X = 0; X = 0);

(iii) = 1 (respectively 0 = 1; = 1);(iv) 1

(sn)n1

(0,

): lim

n

(sn )

(sn) =

respectively (sn)n1 (0, 1): limn

(sn )

(sn)= ;

(sn)n1 N: limn

(sn )

(sn)=

.

2.4. A variant of subsequence splitting principle

IfE has the Fatou property, the following result follows, in fact, from the first part of the proof

of [15, Proposition 3.3]. In the general case the proof follows along the same lines (see also [4,Lemma 3.6]) and is therefore omitted.

Proposition 2.4. Let E = E(0, ) be a separable r.i. space and let (xn)n1 be a sequence inE such that xn 1, n 1. Then there exist a subsequence (yn)n1 (xn)n1, sequences(un)n1, (vn)n1 in E and an elementu E, with uE 2, such that

yn = un + vn, n 1,where un u

(n 1) and (vn)n1 is a bounded sequence which tends to 0 in measure on(0, ).

If, in addition, the sequence (yn)n1 is weakly null and E L0(0, ), then the sequences(un)n1, (vn)n1 may be chosen to be weakly null as well.

IfE has the Fatou property, the sequence (un)n1 may be chosen equimeasurable.


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3. Main results

Definition 3.1. Let X be a Banach space and 1 < p < . A weakly null sequence (xj )j=1 Xis a p-BanachSaks sequence (respectively a BanachSaks sequence) if

limsupn

n1/p

nk=1

xk

<

respectively limn n

1

nk=1

xk

= 0

.

We note that X has the p-BanachSaks property (respectively BanachSaks property) if and

only if every weakly null sequence in X contains a p-BanachSaks subsequence (respec-

tively BanachSaks subsequence). We write X (pBS) (respectively X (BS)) if X has thep-BanachSaks property (respectively the BanachSaks property).

3.1. The Marcinkiewicz spaces M0(),

The following theorem completely characterizes Marcinkiewicz function spaces M0(),

, which have the BanachSaks property.

Theorem 3.2. LetX be M0()(0, ) orM0()(0, 1), where . The space X (BS) if andonly if1 < pX qX < .

For convenience, we split the proof into a number of ingredients some of which are of in-

dependent interest. The following proposition significantly strengthens [15, Theorem 5.9]. For

convenience, we use similar notations.

Proposition 3.3. If is a function from satisfying (+0) = 0 and 0 = 0 , then the spaceX = M0()(0, 1) does not have the BanachSaks property. Furthermore, there exist a constant > 0 and a weakly null sequence of independent functions (xn)n1 X, with

10 xn(t)dt= 0,

n 1, satisfyingkB

xk

nfor every subsetB N, card(B) = n, n 1.

Proof. Since 0 = 0, we have by Proposition 2.3 that for every t (0, 1) there exists a sequence(sn)n1 (0, 1) satisfying limn (tsn)(sn) = 1. Passing to a subsequence, if necessary, we mayassume that sn s0 = s0(t) [0, 1] as n . By the assumption, the function is absolutelycontinuous (see [20, Lemma II.1.1]). Therefore, if it were that s0 > 0, then (ts0) = (s0)which is a contradiction with the assumption that is increasing. Thus, s0 = 0 and without lossof generality, we may (and shall) assume that for every t (0, 1), we have

limsups0+

(ts)

(s)= 1. (3.1)

For an arbitrary

(0, 1), we set

qn :=

n(1 ), n 1,where [z] stands for the integral part ofz.


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Clearly,

1 qnn

, n 1, (3.2)

and it follows from (3.1) that there exists a sequence (tn)n1 (0, 1), tn 0, such thattn

tn+1

1

1 , n 1, (3.3)

and

tqn

n

(tqn ), n 1. (3.4)

Now, we set

yn(s)

:=n ( tn

n)

tn

[0, tn

n ](s)

+ (s)( tn

n,1

](s), n 1, s

(0, 1). (3.5)

Since (s) (s)s

, the function yn, n 1, is a nonnegative decreasing bounded function such

that

yn = 1,t

0

yn(s)ds = (t), t

tn

n, 1

. (3.6)

Let (yn)n1 be an arbitrary sequence of positive independent functions on [0, 1] such that(yn)

= yn, n 1. We set

zn(u,v) := yn(u)rn(v), (u,v) (0, 1) (0, 1),where (rn)n1 denotes the sequence of Rademacher functions, rn(t ) := signsin(2 t ), t (0, 1).Clearly, the system (zn)n1 M0() is a system of independent functions satisfying the equality1

0

10 zn(u,v)dudv = 0, n 1. By [19, Theorem 1, inequality (3)], there exists a constant > 0

such that for every B N,iB

zi

M0()

iB

zi

Z2

M0()

. (3.7)

Here, the sequence (zn)n1 consists of pairwise disjointly supported functions on (0, ) such

that (zn) = zn, n

1, and Z

2

M0() is the space of all measurable functions z on (0, ) for whichthe (quasi)-normzZ2

M0()= z(0,1)M0()(0,1) + z[1,)L2(1,)

is finite. We claim that for every n 1n

k=1zk

Z2

M0()

n. (3.8)

Indeed, from the definition of the quasi-norm in Z2M0()

above, we haven

k=1zk

Z2

M0()

n

k=qnzk

Z2

M0()

n

k=qnzk

(0,1)

M0()(0,1)

.


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Further, from the definition of the norm in M()(0, 1) and since (zk ) = zk = (yk ) = yk , k 1,

we have

n

k=qnzk

(0,1)M0()(0,1)

(nqn+1) tqnn

0

(nk=qn zk)(s)ds((n qn + 1) tqnn )

nk=qn

tqnn

0 (zk)(s)ds

((n qn + 1) tqnn )=n

k=qntqn

n

0 yk(s)ds

((n qn + 1) tqnn ).

Since tn 0, it follows from (3.3) that for every k such that qn + 1 k n, we havetk

k

tqn+1qn + 1

tqn+1

tqn (1

) tqn

n

tqn

n,

whencetqn

n0

yk (s)ds =

tqn

n

(see (3.6)). Therefore, using (3.2) and (3.4), we conclude

nk=qn

tqnn

0 yk(s)ds

((n

qn

+1)

tqn

n

)

(n qn) ( tqnn ) (tqn )

= n

1 qnn

(tqnn

)

(tqn )

(3.2),(3.4) n.

This yields (3.8).

Now, we observe that the space Z2M0()

admits an r.i. norm Z2

M0()

equivalent to the quasi-

norm Z2M0()

(see e.g. [24, Theorem 2.f.1] or [5, Section 6]) such that (Z2M0()

, Z2M0()

) is

a separable r.i. function space on (0, ). Combining this fact with [20, Theorem II.4.3], we inferthat there exists a constant > 0 such that for every f g, g Z2M0() , f L1 + L(0, ),it follows that f Z2M0() and fZ2M0() gZ2M0() .

Note that it follows from the definition of the sequence (yn)n1 above that yp yq , p < q.Therefore, for any t > 0 and every B

N, card B

=n, we have

t0

n

k=1zk

(s)ds =

t0

n

k=1yk(s k + 1)(k1,k](s)

ds

t0

kB

yk (s k + 1)(k1,k](s)

ds =t

0

kB

zk

(s)ds,

in other words, nk=1 zk kB zk . This guarantees that

n

k=1zk

Z2

M0()

kB

zk

Z2

M0()

.


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Combining this inequality with (3.7) and (3.8) and setting := , we obtainkB

zk

M0()

n.

Fixing an arbitrary measure preserving transformation : (0, 1) (0, 1) (0, 1) and settingxk (u) := zk

(u)

, k 1, u (0, 1),

we rewrite the inequality above askB

xk

n, B N, card(B) = n.Since (xk)k1 M0() is a sequence of independent random variables on (0, 1) such that10 xk(s)ds = 0, it remains only to show that (xn)n1 is weakly null, or, equivalently, that(zn)n1 is weakly null in M0(). Since the sequence (zn)n1 is uniformly bounded in M0(),

it is sufficient to verify that

zn, y 0, n ,for all y from a dense subset of the space (M0())

, which may be identified with the Lorentzspace ((0, 1) (0, 1)) (see [20, Lemma II.5.4 and Theorem II.5.4]). The latter space is a sepa-rable r.i. space and it is easy to see that the linear span of the set W = {w(s,t) = u(s)v(t): u, v L(0, 1)} is dense in ((0, 1) (0, 1)). For every w W, we have

10

10

zk(s,t)w(s,t) ds dt=1

0

u(s)yk (s)ds

10

v(t)rk(t)dt 0, k ,

since 1

0

u(s)yk (s)ds

u yk1 u yk = u

and since limk 10 v(t)rk(t)dt= 0 for every v L1(0, 1). This completes the verification thatthe sequence (xn)n1 M0()(0, 1) is weakly null and the proof of Proposition 3.3. 2To treat the situation when = 0, we recall the concept of symmetric functionals on

Marcinkiewicz spaces M()(0, ) introduced in [13].

Definition 3.4. (See [13, Definition 2.1].) A positive linear functional f on M()(0, ) (re-spectively on m()) is said to be symmetric if f(x) f(y) for all 0 x, y M()(0, )such that x y. Such a functional is said to be supported at infinity if f (x[0,s]) = 0 for allx

M()(0,

) and every s > 0.

The following theorem completely characterizes Marcinkiewicz spaces admitting nontrivial

symmetric functionals supported at infinity.


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Theorem 3.5. (See [1113].) If () = , then a nonzero symmetric functional onM()(0, ) (respectively m()) supported at infinity exists if and only if

liminft

(2t )

(t) =1. (3.9)

For more information on symmetric functionals (and their prototype, so-called Dixmier

traces used in noncommutative geometry) we refer to [9] and references therein.

Proposition 3.6. If() = and = 0, then m0() / (BS).

Proof. If = 0, then (see Proposition 2.3) condition (3.9) holds, and therefore there existsa nonzero symmetric functional f m(). Fix a positive element x m(), x = 1 such thatf(x) = , 0 < 1 (without loss of generality, we assume that fm() 1). Let (xj )j=1 m() be a sequence of pairwise disjoint positive elements such that xj = x, j 1. For anysubsequence (xjk )

k=1 (xj )j=1, we have

Nk=1

xjk

f

Nk=1

xjk

= N , N 1. (3.10)

We are going to construct a weakly null semi-normalized sequence (yj )j=1 m0() such that

0 yj xj , j 1, which is not a BS-sequence. First of all, observe that if z is an arbitrary

element from Lorentz sequence space () = m0() (see e.g. [20,23]), then |z|, xn |z, yn|and |z|, xn 0 as n . Thus, the fact that any sequence (yj )j=1 m0( ) satisfying thecondition yj xj , j 1, is weakly null is immediate. We shall select yj , j 1, to be finitelysupported (the latter guarantees yj m0(), j 1) and such that

Nj=1

yj

N 4 , N 1. (3.11)To see that such a selection is possible, recall first that any Marcinkiewicz space M(),

(and, hence, every m()), possesses a Fatou norm. Fix n 1 and consider the ele-ment

2n1j=2n1 xj . We select yj , 2

n1 j 2n 1, to be equimeasurable, finitely supported andsatisfying the inequality

2n1j=2n1

yj

2n1j=2n1

xj

4 , n 1. (3.12)We also require that for every m n 1

yp yq , for all 2

n1 p 2n 1, 2m1 q 2m 1. (3.13)

Now fix an arbitrary N N and let n N satisfy 2n1 N 2n 1, n 1. Combining (3.12)and (3.10), we arrive at (3.11). Note that if (yj )

j=1 is a subsequence of (yj )

j=1 then condi-

tion (3.13) guarantees thatN

j=1yj

Nj=1

yj

, N 1.


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This shows that (yj )j=1 m0() is a weakly null semi-normalized sequence which is not

a BS-sequence and completes the proof of Proposition 3.6. 2

The next important ingredient of the proof of Theorem 3.2 treating the case of Marcinkiewicz

spaces M0()(0, 1) with 0 = 1 is contained in Proposition 3.9 below. The following lemmaprovides technical information concerning the behavior of dilation operators in such spaces.

Lemma 3.7. Let (k)k1 0. If 0 = 1 (respectively = 1), then there exists a pairwise disjoint sequence (xk)k1 M0()(0, 1) (respectively M0()(0, )) which is equivalent to thestandard unit vector basis of the space c0 and which satisfies the condition

1

2 Dn xn xn 1, n 1. (3.14)

Proof. Assume that 0

= 1. Arguing as in the proof of Proposition 3.3 and referring to Proposi-tion 2.3, we infer from the assumption on that

limsups0+

(s)

(s)= , 1. (3.15)

Fix (0, 1). We claim that there exists (0, 1) such that for x := [0,] () , we have1

2 Dx x = 1.

Indeed, noting that Dx = [0,] () , we see that

Dx =()

()= ()

().

It follows from (3.15) that there exists s (0, ) such that(s 1

)

(s)

1

2.

Setting := s 1

, we see that (0, 1) and

Dx

=

(s 1

)

(s)

1

2.

Thus, the sequence (xn)n1 satisfying (3.14) may be chosen as a sequence of (normalized) scalar

multiples of indicator functions of pairwise disjoint intervals en (0, 1) with m(en) = n, n 1,where ns are determined by ns as above. The condition

0 = 1 implies (+0) = 0. There-

fore, in view of (3.15) we may assume that n 0 and (n+1) 12 (n) (n 1). If now n Nand for some 1 i n we have

nj=i+1 j t 0, and j : (0, ) (0, ) is a measure preserv-ing transformation, and fj (s) := f (j (s)), s (0, ), j 1. To estimate TNXX const max{TNLpLp , TNLq Lq }, we recall that Lp , 1 < p < , has the (Rademacher)type min{p, 2} and therefore for any f Lp, we have

TN(f )LpLp =

Nj=1

uj Qj fj rj ()

Lp((0,1),Lp )

const N

j=1 Qj fj

min

{p,2

}Lp 1/ min{p,2}

const Nmax{1/p,1/2} fLp .


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Since X is an interpolation space for the couple (Lp, Lq ), we infer from the above estimate

that TNXX const Nmax{1/p,1/q,1/2} where the constant does not depend on N. Note, thatthe assumption on X also guarantees that the space M()(0, ) is also an interpolation spacefor the couple (Lp, Lq ) and therefore the same argument yields that

TN

M (0,

)

M (0,

)

const Nmax{1/p,1/q,1/2}. These estimates are sufficient to show that (3.21) holds for N 1with nk = k, k 1. Indeed, recall that since in (3.20) f M()(0, ) and aj f it fol-lows from [20, Theorem 2.2.1] that for every aj there exist a measure preserving transformation

j : (0, ) (0, ) and functions uj L(0, ) and vj L1 L such that uj L 1,vj L1L 2j and aj = uj fj + vj (j 1). Therefore,

Nj=1

aj

=TN(f )M (0,) fM (0,) const Nmax{1/p,1/q,1/2} + sup

j1

Qj XX

and this gives (3.21). 2

Proof of Theorem 3.2. If 0 < < 1, then by [24, Theorem 2.b.11] M0()(0, ) is aninterpolation space for the couple (Lp, Lq ) if 0 0) and since

y1 L1 and limt (t) = , then for every > 0 there is M > 0 such that

suptM

1

(t)

t0

fN(s)ds .

Moreover, by (3.28)

sup0 1,

where [0 , 0 ]. It is easy to see that quasi-concave function satisfies 0 < = 0 =0 < 1 and so M0() = M0()(0, ) (BS) by Proposition 3.10. It is also straightforward that

(up to a norm equivalence)M0() = M0()(0, ) = M0() L1.

If(xn)n1 M0() is weakly null, then for every y M0() = () = () + L

y, xn =

0

y(t)xn(t)dt 0, n .

Since (xn)n1 is simultaneously weakly null in M0() and L1 and since M0() (BS),L1

(BS), then there is a BanachSaks sequence (xnk )

(xn) in both spaces. Therefore,

1

n

n

k=1xnk

M0()

1n

n

k=1xnk

M0()

+ 1n

n

k=1xnk

1

0. 2


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Case 5. If () < and either 0 = 0 and (+0) = 0 or 0 = 1, then M0()(0, ) /(BS).

Proof. Since M0()(0, 1) may be identified with subspace of all functions from M0()(0,

)

supported at (0, 1), then the assertion follows from Propositions 3.3 and 3.9. 2

We have proved the following result, which complements and extends Theorem 3.2.

Theorem 3.11. If , then M0()(0, ) (BS) if and only if one of the following conditionsis satisfied:

(i) 0 < < 1;(ii) limt0 ( t ) > 0 and0 <

< 1;

(iii) limt

( t ) 0 andlimt (t ) < (i.e. M()(0, ) = M0()(0, ) = L1).

4. BS and pBS properties in Marcinkiewicz sequence spaces m0()

For technical reasons we shall also consider the following property:

Definition 4.1. We say that an r.i. space X = X(J) has a disjoint p-BanachSaks prop-erty (respectively disjoint BanachSaks property) and write X disjoint (pBS) (respectivelyX disjoint (BS)) if every weakly null disjointly supported sequence in X has a p-BanachSakssubsequence (respectively a BanachSaks subsequence).

We begin our discussion here by a simple remark that m0() (BS) (respectively m0() (pBS)) if and only ifm0() disjoint (BS) (respectively m0() disjoint (pBS)). Indeed, froma weakly null sequence (xj )

j=1 in m0() one can extract a subsequence (yj )

j=1 such that

yj zj 2j , j 1, for some disjointly supported sequence (zj )j=1 m0().As before, (ej )

j=1 is the standard unit vector basis in m0(). We fix b = (bj )j=1 =

i=1((i) (i 1))ei and denote by (bk )k=1 a disjointly supported sequence of elementsequimeasurable with the element b.

Lemma 4.2. If(xk)k=1

,

xk

1, k 1, is a disjointly supported sequence of elements in m(),

, thenn

k=1xk

2n M

1

n

, n 1. (4.1)

Proof. Using [10, Lemma 2.3], it is easy to see that for every n 1, we have

nk=1

xk n

k=1bk

and therefore by [20, 2.5.2] we haven

k=1xk

nk=1

bk

. (4.2)


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Fixing n and setting c :=nk=1 bk =i=1 ci ei , we havem

i=1ci = n(j 1) + (m (j 1)n)

(j) (j 1)

= (j n m)(j 1) + m (j 1)n(j) (4.3)for every (j 1)n + 1 m j n, j 1. If j = 1, i.e. 1 m n, then (4.3) implies thatm

i=1 ci = m(1), whence

1

(m)

mi=1

ci =m(1)

(m)

n(1)

(n).

In the case when j 2 we have, by (4.3),m

i=1 ci n(j) and therefore

1(m)

mi=1

ci n (j)((j 1)n) n (j)(jn/2) 2n (j)(jn) .

So, n

k=1bk

2n supj1(j)

(jn) 2n M

1

n

, n 1. 2

Lemma 4.3. For all n 1, we have

n

k=1

bk n2M 1

n. (4.4)Proof. Using the notation from the proof of Lemma 4.2 and (4.3), we have

nk=1

bk

supjN1

(nj)

nji=1

ci n supjN

(j)

(jn). (4.5)

For every s n there exists j N such that nj s < n(j + 1). Since is concave, we have(s/n)

(s)

(j + 1)(nj)

(2j )

(nj) 2

(j)

(nj),

which, in combination with (4.5), yieldsn

k=1bk

n2 supsn(s/n)

(s)= n

2M

1

n

. 2

The main result of this section is

Theorem 4.4. If , then for all 1 p < ,

(i) m0

( )

(BS)

> 0;

(ii) m0( ) (pBS) supnN n11/pM ( 1n ) < ;(iii) if1 < p < 1

1, then m0() (pBS). Ifp > 11 , then m0() / (pBS).


25/29


26/29



Note, that the assumption supnN n1/2M (1n

) < implies (but is not equivalent to) theassertion that

12

. Note also that the assumption M0()(0, 1) L2(0, 1) relates only to thebehavior of at zero and in many cases may be verified directly. For example, let 1, 2 begiven (for sufficiently small ts) by

1(t) = t1/2, 2(t) = t1/2 ln

e

t

.

It is obvious that M0(2)(0, 1) M0(1)(0, 1) and it is well known (see e.g. [20, Theo-rem II.5.7]) that L2(0, 1) M0(1)(0, 1). Thus, M0(1) and M0(2) do not embed isomor-phically into m0(t

1/2).

We mention also a noncommutative extension of Proposition 5.2.Let E be a separable r.i. sequence space and CE be (associated with E) separable ideal of

compact operators on l2 (see e.g. [17]). Recall, that a compact operator x on l2 belongs to CE if

and only if the sequence of s-numbers s(x)

=(sn(x))n1 (counted with multiplicities) belongs

to E and that xCE = s(x)E .

Proposition 5.3. If and satisfy the assumptions of Proposition 5.2 , thenM0()(0, 1) does not embed isomorphically into ideal Cm0() .

Proof. It follows from Theorem 4.4(ii) combined with [3, Theorem 2.4] that Cm0() (2BS). 2

In particular, M0(t1/2) and M0(t

1/2 ln( 1t

)) do not embed isomorphically into Cm0(t1/2).

These results complement those in [25] and [14] (see also [2, Theorem 6], where it is estab-

lished that Lp(0, 1), 1 p = 2 < , does not embed isomorphically into Clp ).

5.2. Marcinkiewicz spaces with mixed norm

Now, we shall present an application of our results to the problem (suggested by A.V. Bukhva-

lov [8]) whether an (separable) r.i. space E, a vector-valued space

E(E) := E[0, 1], E= f: [0, 1] E, f is strongly measurableand fE(E) :=

t

f (t)

E

E<

,

and the space E[E

]with mixed norm are pairwise isomorphic. To recall the definition of spaces

with mixed norm, fix (for simplicity) separable Banach function spaces E and F on -finite

measure spaces (T1, 1, 1) and (T2, 2, 2), respectively, and let (T,,) be the product

measure space. The set E[F] consisting of all functions K = K(s,t) : T S(T,,) (= theset of all -measurable -a.e. finite functions on T) such that the function t K(, t)F ,s Eand

KE[F] :=K(s,t)

F,s

E,t

< .Our main contribution to the problem of comparison of the spaces E, E(E) and E[E] is thatif E = M0(), 0 < 1, then E does not isomorphically embed into E(E) and E[E].In particular, the spaces L0p,

:=M0(t

11/p) and L0p,

(L0p,

) are not isomorphic. This com-

plements Bukhvalovs result (announced in [8]) that the spaces Lp,q and Lp,q (Lp,q ) are notisomorphic for 1 < p = q < , p = 2. We begin with the following remark complementingcorresponding results in [1], [15, Theorem 5.1].


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Proposition 5.4. IfX is a separable r.i. space on [0, 1], then X[c0] / (BS).

Proof. We denote the standard unit vector basis in c0 by (ei )i1 and set

xk := xk (t) = 2k

i=1r2k+i (t)ei , k 1.

Fix y = (yk(t))k=1 (X[c0]) = X[l1]. We have

xk , y

= 2ki=1

10

r2k+i (t)yi (t)dt. (5.1)

Due to the assumption y X[l1], for every > 0 there exists N 1 such that

i=N+1

10

yi (t )dt < 2

.

Combining this inequality with (5.1), we have for all k log2 N

xk, y Ni=1

1

0

r2k+i (t)yi (t)dt

+ 2 .Since for the Rademacher system (rk)

k=1 we obviously have

10 rk(t)y(t)dt 0 for every

y L1(0, 1), we infer that xk

0 weakly.Using the same arguments as in the proof of Proposition 3.9, we see that for every m 1m

n=1xn(t) =

mn=1

2ni=1

r2n+i (t)ei =2m

i=1

nQi

rn(t)

ei ,

where Qi {1, 2, . . . , 2m+1}, Qi Qj = (i = j ), and card(Qi ) m for all i = 1, 2, . . . , 2m.Furthermore, if 1 i 2m/2, then card(Qi )

m2 . As in the proof of Proposition 3.9, using

Lemma 3.8, we have (see inequality (3.19)) that

t: max1i2m/2nQi rn(t)m

4 1 e1/8. (5.2)

This immediately implies that for every m 1m

n=1xn

X[c0]

m

n=1xn

L1[c0]

m

4

t: max1i2m/2

nQi

rn(t )

m4

m

4 1 e1/8. (5.3)

Since (5.2) also holds for an arbitrary subsequence of (xk )k1, we now see that (5.3) implies

that the weakly null sequence (xk )k1 X[c0] is not a (BS)-sequence. 2


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Remark 5.5. The result of Proposition 5.4 is similar to [15, Theorem 5.1] (see also [22, Exam-

ple 5.6.8]) asserting that Lp(c0) = Lp([0, 1], c0) / (BS). Repeating the proof of that theorem,one can easily show that M0()(c0) = M0()([0, 1], c0) / (BS).

Theorem 5.6. For any concave increasing function on [0, 1] such that (+0) = 0 we haveM0()[M0()] / (BS).

Proof. Take a subspace in M0( ) spanned by a sequence of normalized pairwise disjoint ele-

ments which is isomorphic to c0 (see Proposition 2.1). It is easy to see that we may now identify

the mixed norm space M0()[c0] with a subspace in M0( )[M0( )]. Theorem 5.6 now followsfrom Proposition 5.4. 2

Corollary 5.7. If is a concave increasing function on [0, 1] such that0 < < 1, then

(i) the spaces M0( ) andM0()[M0()] are not isomorphic;(ii) the spaces M0( ) andM0()(M0()) are not isomorphic.

Proof. Part (i) follows from Theorem 5.6 and Proposition 3.10. To show (ii) it is suffi-

cient to prove that M0()(M0()) / (BS). To this end it suffices to note that the vector-valued space M0()([0, 1], c0) may be identified with a subspace of M0()(M0()) and thatM0()([0, 1], c0) / (BS). The latter assertion follows from Remark 5.5. 2

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