combined muh ly exams

7/30/2019 Combined Muh Ly Exams

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SOLUTIONS TO COMBINED MUHLY EXAMS

DAVE GAEBLER

Fall 2007 Real Midterm 1

(1) Definitions (25 points):(a) The lim sup of a sequence of numbers.(b) What it means for an infinite sequence of real-valued functions all

defined on the same set to converge uniformly to a given function.(c) What it means for a set to be open in a metric space (X, d).

(d) What it means for a set to be dense in a metric space.(e) A totally bounded metric space.

Solution.

(a) The lim sup is the least number which is eventually greater than orequal to the terms of the sequence. Symbolically,

limsup xn = limn supmn

xm.

(b) Uniform convergence means that the supremum of the difference be-

tween the functions tends to zero. That is, fnu f on a set A if

limn supxA

|fn(x) f(x)| = 0.

More explicitly, fnu f on A if

( > 0)(N N)(n > N)(x A)|fn(x) f(x)| < .(c) An open subset of a metric space is a set each of whose points has a

nontrivial neighborhood which lies in the set. That is, a set A X isopen if

(x A)(r > 0)(y X)(d(x, y) < r y A).(d) A subset A X of a metric space is dense if its closure is the whole

space, i.e. A = X. Alternatively, it means that for any x X and any > 0, there is a point a A with d(a, x) < . Still a third definition isthat for any point x

X there is a sequence

{an

} A with an

x.

(e) A metric space X is totally bounded if, for any > 0, there is a finitecovering of X by balls with radius . That is, there is a finite set{x1, . . . , xn} X such that Bx1() Bxn() = X, where Br(x)denotes the open ball of radius r centered at x.

(2) True or false: If f(x) =

x, then f is uniformly continuous on {x | x 1}.(15 points)

1


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2 GAEBLER MUHLY EXAMS

Solution. True. For any x, y 1,

|

y

x|

=|y x|

y + x |y x|

2.

Thus, we can make |y x| < if we require |y x| < 2. (3) Let f be a continuous real-valued function defined on a closed interval [ a, b].

Show that the graph of f is a compact subset ofR2. (15 points)

Solution. Because [a, b] is compact, and the continuous image of a compactset is compact, it suffices to prove that the function F : R R2 definedby F(x) = (x, f(x)) is continuous. Fix x. Let > 0. By the continuity off, there exists > 0 such that |y x| < |f(y) f(x)| < /2. Choose = min(/2, ). Then

|yx| < d((x, f(x)), (y, f(y))) =

(x y)2 + (f(x) f(y))2 0. Because f is uniformly continuous, there exists > 0such that |x y| < |f(x) f(y)| < for all x, y R. Choose N > 1/.Then for any n > N and any x R, |fn(x)f(x)| = |f(x+1/n)f(x)| < because 1/n < . Thus fn

u f. (5) True or false: If fn(x) = xn, x [0, 1], then there is a subsequence of

{fn}n=1 that converges uniformly on [0, 1]. (15 points)

Solution. False. If a subsequence fnk converged uniformly to some f, thenf would have to be continuous, because the uniform limit of continuousfunctions is continuous. Because uniform convergence implies pointwiseconvergence, the subsequence fnk would therefore converge pointwise toa continuous function. But the sequence fn converges pointwise to thefunction

g(x) =

0 x < 1

1 x = 1

and therefore any subsequence converges pointwise to this limit, and notto a continuous function.

(6) IfA is a subset of a metric space (X, d), and if x is a point of X, then thedistance from x to A is defined to be inf

{d(x, y)

|y

A

}. Show that if A

is a closed subset in R, if x R and if r = d(x, A), then there is a pointx0 A so that r = d(x, x0). (15 points)Solution. Since d(x, A) = r, there exists a sequence {xn} A such that|xn x| r. Since every real number is either greater or less than x, thisimplies either a decreasing sequence xn x + r or an increasing sequencexn xr. In these two cases, x + r and xr respectively are limit pointsof A, and therefore are in A since A is closed.


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GAEBLER MUHLY EXAMS 3

(7) Bonus Problem: Define a sequence {xn}n=1 as follows: x1 = 1 and for alln 1, xn+1 = xn + 1x2n . Is {xn}

n=1 a bounded sequence?

Proof. Solution No. The sequence is monotonically increasing; if it werebounded, it would therefore have a limit L. But then

L = lim xn+1 = lim xn +1

x2n= lim xn + lim

1

x2n= L +

1

L2

which is impossible. Hence the sequence increases without bound.An alternate statement of the solution: Because the sequence is increas-

ing, it has a limit L in the extended real numbers. Then L = L + 1L2 1L2 = 0 L = .


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(1) Definitions (25 points):

(a) What it means for a family of functions between two metric spaces tobe equicontinuous at a point.

(b) What it means for a real-valued function on R to be lower semicon-tinuous.

(c) What it means for a real-valued function on a measurable space (X,M)to be measurable.

(d) What it means for two sets from a measure space (X,M, ) to be equala.e.

(e) Let f be a nonnegative function defined on a measure space (X,M, ).Define the symbol

X

f(x)d(x).

Solution.

(a) A family

Fof functions from X to Y is equicontinuous at a point

x X if( > 0)( > 0)(y X)(f F)(d(y, x) < d(f(y), f(x)) < ).

(b) A function f : R R is lower semicontinuous at a point x if( > 0)( > 0)(|y x| < f(y) > f(x) )

and is lower semicontinuous if it is lower semicontinuous at x for eachx R.Alternatively, f is LSC if the set f1((a, )) is open for all a R.

(c) A function f : X R is M-measurable iff1((a, )) M for all a R. This is equivalent to the more general definition that f1(U) Mfor every open set U.

(d) Two sets A, B from a measure space (X,M, ) are equal a.e. if(AB) =0, where AB = (A \ B) (B \ A). (???)(e) Let S be the set of nonnegative measurable simple functions on X.

For s S define X s(x)d(x) by writings(x) =

nj=1

cjEj (x)

with the cj distinct; this can be done uniquely since Ej = s1({cj})for each point cj in the range of s. We then define

X

s(x)d(x) =n

j=1cj(Ej).

Having defined the integrals of nonnegative measurable simple func-tions, we now define the integral of any nonnegative measurable func-tion to be

X

f(x)d(x) = supsSsf

X

s(x)d(x).


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(2) True or false: Let F be an equicontinuous family of functions defined onthe unit interval [0, 1], let f be a bounded continuous function defined onR, and let f

Fdenote the set of all f

,

F. Then f

Fhas compact

closure in C([0, 1]). (15 points)

Solution. False. Consider the family F of triangle bump functions de-fined by

a(x) =

0 x ax a a < x a + 1a + 2 x a + 1 < x a + 20 x > a + 2

for a R. (The graph of such a function is an isosceles triangle withbase extending from a to a + 2 and with height 1.) The family

Fis

equicontinuousin fact, uniformly equicontinuousbecause for any > 0and x,y,a R, |x y| < |a(x) a(y)| < . Let f be the truncatedidentity function

f(x) =

0 x 0x 0 < x 11 x > 1

which is bounded and continuous. Then f a = a for any a R, sofF= F. However, Fdoes not have compact closure in C([0, 1]), becausethe sequence 0, 1, 2, . . . are all a distance 1 apart and hence has noCauchy subsequence.

(3) Suppose that I1 and I2 are disjoint open intervals and that Aj is a subsetofIj, j = 1, 2. Show directly from the definition of Lebesgue outer measurem that m(A1 A2) = m(A1) + m(A2). (15 points)

Solution. As always, m(A1 A2) m(A1) + m(A2), so we need onlyshow the reverse inequality holds. For this, it suffices to show that for anycover C ofA1 A2 there is a pair (C1, C2), where Ci is a cover of Ai, suchthat |C1| + |C2| |C|. To accomplish this, let x be any point which liesbetween I1 and I2. Given any cover C of A1 A2, form the cover C bysubdiving any interval which straddles x; that is, if the interval [a, b] is in Cand x (a, b), then C contains instead the intervals [a, x] and [x, b]. Notethat |C| = |C| and that any interval in C intersects at most one ofA1 andA2. Hence there is a partition of C into a cover C1 of A1 and a cover C2ofA2 (the decision of where to put intervals which intersect neither A1 norA2 is arbitrary). Since |C1| + |C2| = |C|, we are done.

(4) Let A be the set of all points x in the unit interval [0, 1] such that the decimalexpansion of x does not contain the digit 1. Show that A is measurableand calculate its measure. (15 points)


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Solution. Define the sets Bn inductively by B0 = [0, 1], B1 = [0, 0.1] [0.2, 1], and

Bn+1 = Bn \9

j1=1j1=1

. . .9

jn=1j1=1

[0.j1j2 . . . jn1, 0.j1j2 . . . jn2].

Then each Bn contains all numbers without a 1 in their first n digits, andis therefore a cover of A. Moreover, |Bn+1| = 910 |Bn|, so by induction|Bn| =

9

10

n. Since |Bn| 0 and each Bn is a cover of A, m(A) = 0.

Hence A is measurable and m(A) = 0.

(5) For what values of > 0 is the function f(x) = x Lebesgue integrableon the interval [0, 1]? (15 points)

Solution. This function is integrable on [0, 1] iff < 1. We can evaluate theintegral using Riemann integration, because of the Monotone Convergence

Theorem: Let fn = f [1/n,1], so that fn f. Thus10

f(x)dx = limn

10

fn(x)dx = limn

11/n

f(x)dx.

In the case < 1 we have11/n

f(x)dx =1

1 x1

11/n

=1

1

1

1

n

1 1

1 as n ; in the case = 1 we have1

1/n

f(x)dx = ln(x)1

1/n= ln(n) ,

and in the case > 1 we have1

1/n

f(x)dx = 1 1 x

11

1/n= n

1 1 1 .

(6) Suppose {fn}n=0 is a sequence of Lebesgue measurable functions on R suchthat

0 f1 f2 f3 . . .and let f(x) = limn fn(x). Show that if lim

R

fn(x)dm(x) is finite,then f must be finite except on a set of measure 0. (15 points)

Solution. By the Monotone Convergence Theorem,r

ef(x)dm(x) = limR

fn(x)dm(x)is finite. But iff were infinite on a set of positive measure, its integral wouldbe infinite. Hence f is finite a.e.

(7) Bonus Problem: Let f be the non-decreasing function on R defined by theformula

f(x) =

0 x < 0

x2 0 x 11 x > 1

and let = f be the Lebesgue-Stieltjes measure determined by f. Showthat every subset of [0, 1] that is a Lebesgue null set is also a -null set.


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Is the converse true? That is, if A is a -null set of [0, 1], must A be aLebesgue null set?

Solution. One can easily check that the -measure of an interval is at mostits Lebesgue measure:

([a, b]) = f(b) f(a) |b a|.(This can easily be done by dividing into cases based on the location of aand b.) Any Lebesgue null set has covers with arbitrarily small Lebesguemeasure, and their -measure must be even smaller by the above observa-tion, making it also a -null set.

The converse is false. The interval [1, 2] is -null because ([1, 2]) =f(2) f(1) = 0, but of course it is not Lebesgue null.


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Fall 2007 Real Final


(a) A Vitali covering of a set of real numbers.(b) Let f be defined on an open interval (a, b) and let x (a, b) be given.

Define D+f(x).(c) What it means for a function to be of bounded variation.(d) What it means for a function to be absolutely continuous.(e) A singular monotone function.

Solution.

(a) A Vitali covering of a set E R is a collection C of intervals such thatfor every x E and for every > 0 there is an interval I C suchthat x I and |I| < .

(b)

D+f(x) = liminfh0

f(x + h) f(x)h

.

(c) Given an interval [a, b], and a partition P = {x0, . . . , xn} of the interval(so xi < xi+1 and x0 = a, xn = b), define the variation of f withrespect to P to be

VP(f) =n

k=0

|f(xk+1) f(xk)|.

Then the variation of f on the interval [a, b] is defined to be

V(f) = supP

VP(f).

If V(f) < , then f is of bounded variation on [a, b]. Similarly, using(finite) partitions ofR we may define a function of bounded variation

on R.(d) A function is absolutely continuous if for every > 0 there exists > 0

such that for any (finite) collection of intervals [a1, b1], . . . , [an, bn],

ni=1

(bi ai) < n

i=1

|f(bi) f(ai)| < .

(e) A singular monotone function is one whose derivative is zero a.e. butwhich is not constant. OR: A strictly monotone function is calledsingular if its derivative is zero a.e.

(2) True or false: Let f be a nonnegative function on R that is integrable withrespect to Lebesgue measure. Let be the measure on R defined by the

formula (A) =A f(x)dx. If {An}nN is a decreasing sequence of sets

with intersection A, then limn (An) = (A).

Solution. True. Let n be the characteristic function of An. The sequencef n is dominated by the integrable function f, so by the Dominated Con-vergence Theorem,

(A) =R

limn

f n = limn

R

f n = limn

(An).


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(3) True or false: Let f be a non-decreasing function defined on R and let

be the associated Lebesgue-Stieltjes measure. Suppose every subset of aninterval [a, b] is a null set for . Then the derivative of f exists at everypoint of (a, b) and is identically zero there.

Solution. True. Suppose to the contrary that there is a point x (a, b)at which is is not true that f(x) = 0. That means there is a value ofh such that f

(x+h)f(x)h = 0. (In fact, there are arbitrarily small such h;

in particular, we can choose one small enough that x + h (a, b).) Thusthere is a point y (a, b) such that f(y) = f(x). But then ([x, y]) = 0, acontradiction. So f(x) = 0 at every x (a, b).

(4) Let f be a function defined on the interval (a, b) and assume that there is apositive constant M such that |D+f(x)| M for all x in (a, b). Show thatfor each > 0, the set of all intervals {[x, y] | x (a, b) and |f(y) f(x)| (M + )(y x)} is a Vitali covering of (a, b).Solution. Let S denote the given set of intervals.

Let x (a, b). Let > 0. Because D+f(x) < M, there exists > 0such that f

(x+h)f(x)h

< M + for 0 < h < . Thus [x, x + h] S for eachh (0, ). Let 0 < < min(, ). Then [x, x + ] S because < , and|[x, x + ]| < because < . Hence S contains arbitrarily small intervalsaround every point of (a, b).

(5) True or false: Let E be a measurable subset of the interval [0, 1] and foreach x [a, b] let f(x) = m(E [0, x]), where m is Lebesgue measure. Thenf is absolutely continuous.

Solution. Possible erratum: Should [a, b] be [0, 1] in the problem statement?(The answer is actually still true if [a, b] is an arbitrary interval, however.)

True. For any 0 x < y 1, note that[0, y] = [0, x](x, y] E[0, y] = (E[0, x])(E(x, y]) m(E[0, y]) = m(E[0, x])+m(E(x, y]).

Hence

f(y) f(x) = m(E (x, y]) m((x, y]) = y x.Thus, given any > 0, if [a1, b1], . . . , [an, bn] is any collection of interval withtotal length less than , then

f(bi) f(ai) < . Hence f is absolutely

continuous.Note that we can also deduce the absolute continuity off by expressing

it as an integral:

f(x) =

x

0

E(t)dt.

We know E is integrable because E [0, 1] and hence has finite measure.So f is absolutely continuous.

(6) Suppose that for each positive integer n, Fn(x) is a nondecreasing, abso-lutely continuous function on the interval [a, b] satisfying Fn(a) = 0. Sup-pose further that the derivatives of the Fn(x) increase in n; i.e. assume


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that Fn(x) Fn+1(x) for each n and for almost all x. Finally, assume thatthe sequence {Fn(b)}nN is bounded. Show(a) There is a real-valued function F : [a, b]

R such that lim

nFn

(x) =F(x) for each x [a, b].

(b) F(x) is absolutely continuous on [a, b].(c) The sequence {Fn}nN converges uniformly to F on [a, b].

Solution.

(a) At any fixed x, the values of Fn(x) are increasing and bounded above:They are increasing because

Fn+1(x) =

xa

Fn+1(t)dt xa

Fn(t)dt

and are bounded above because Fn(x) Fn(b) M where M is abound on {Fn(b)}n1. Hence Fn(x) has a limit, which we call F(x).

(b) At each x, the sequence Fn(x) is increasing, so the extended-real-

valued function G(x) = limn Fn(x) is well defined. By the MonotoneConvergence Theorem, at each point x we have

F(x) = limn

xa

Fn(t)dt =xa

limn

Fn(t)dt =xa

G(t)dt.

Moreover, G is integrable, becauseba

G(t)dt =

ba

lim Fn(t)dt = limba

Fn(t)dt = lim Fn(b) = F(b) < .

Thus, F is equal to the integral of an integrable function, so it isabsolutely continuous.

(c) For a fixed n, at any x [a, b] we have

F(x) Fn(x) = x

aG(t) Fn(t)dt

b

aG(t) Fn(t)dt = F(b) Fn(b).

Since Fn(b) F(b), this implies that Fn u F.

(7) Bonus Problem: Let f be a function of bounded variation defined on aninterval [a, b]. For each x in [a, b] let Txa be the total variation of f on theinterval [a, x]. Show that f is absolutely continuous if and only if Txa isabsolutely continuous as a function of x.

Solution. Suppose Ta is absolutely continuous. For any a x < y b, weclearly have

Tya Txa + |f(y) f(x)|since if P is a partition of [a, x] with variation within of T

xa , then the

partition P {y} of [a, y] has variation at least Txa + |f(y) f(x)| .Using this, for a given > 0, choose by the absolute continuity ofTa suchthat for any intervals [ai, bi] of total length less than ,

|Tbia Taia | < .Then for those same intervals,

|f(bi) f(ai)|

|Tbia Taia | < so that f is absolutely continuous as well.


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Conversely, suppose f is AC. Given > 0, choose according to theabsolute continuity of f. We will show that the same works for Ta.Suppose not. Let [a

1, b

1], . . . , [a

n, b

n] be a collection of intervals with total

length less than but with

Tbia Taia = > . Making use of the lemmathat Tya Txa = V([x, y]), this means

V([bi, ai]) = . Then for each

i = 1, . . . , n let Pi = {ai = x(i)0 , x(i)1 , . . . , x(i)n = bi} be a partition of [ai, bi]with variation at least V([bi, ai]) 2n . But then if we take all the intervals[x

(i)k , x

(i)k+1], their total length is still less than but

|f(x(i)k+1)f(x(i)k )| > n2n = +2 > , a contradiction.


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(a) The least upper bound property of an ordered field.(b) The limsup (of a sequence of numbers)(c) an open set (in a metric space)(d) a separable metric space(e) a complete metric space

Solution.

(a) An ordered field F has the least upper bound property if every nonemptysubset ofF which is bounded above has a least upper bound. In sym-bols,

(A F)

(A = ) (x F)(y A)(y x)

(x F)

(y A)(y x) (z F)(y A)(y z) x z.(b) The limsup of a sequence of numbers is the least number which is

eventually greater than or equal to the terms of the sequence. Insymbols,

limsupn

an = limn supmn

am.

(c) An open set in a metric space is one in which every point is containedin a ball contained in the set. That is, U X is open if

(x U)(r > 0)Br(x) Uwhere

Br(x) = {y X | d(x, y) < r}.(d) A separable metric space is one with a countable dense subset. That

is, X is separable if there exists a countable set {x1, x2, . . . } X suchthat for any x X and any > 0 there exists n N with d(x, xn) < .

(e) A metric space X is complete if every Cauchy sequence has a limit.(A Cauchy sequence xn is one for which ( > 0)(N N)(m, n >N)d(xm, xn) < . A limit of a sequence xn is a point x X such thatd(xn, x) 0.)

(2) Let {an}n=1 be a nondecreasing sequence of real numbers that is boundedabove. Show that the liminf of the sequence is the same as the sup of thesequence, i.e. show that

limnan = sup an.

Solution. Let S = sup an and L = liminfan. Clearly S L, becauseS an for any an, so S infmn am for any n, so S limn infmn am.For the converse, let > 0. Then S is not an upper bound for thesequence {an}, because S is the least upper bound. Hence there existsM with aM > S . Because the sequence is nondecreasing, this impliesan > S for all n M. Therefore infmn am > S for any n M,


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whence limn infmn am > S . Since L S for any > 0, itfollows that L S.

(3) True or false? If (X, ) is a metric space with the property that n=0Fn = whenever {Fn}n0 is a decreasing sequence of nonempty closed subsets ofX, then X is compact.

Solution. True. Let {xn} be any sequence in X. Let Fn be the closure ofthe set {xm}mn. Since n=0Fn does not contain any point xn, there mustbe another point in F0, which is therefore the limit of some subsequence.Thus X is sequentially compact, and therefore compact.

(4) Let (X, d) be a metric space and let {xn}n=0 be a sequence in X with theproperty that the sum

n=0 d(xn, xn+1) is finite. Show that {xn}n=0 is a

Cauchy sequence.

Solution. Let > 0. Choose N such that the tail

n=M

d(xn, xn+1) < .

Let n > m > N. Then, by the triangle inequality,

d(xm, xn) n1k=m

d(xk, xk+1)

k=N

d(xk, xk+1) < .

Hence {xn} is Cauchy. (5) Let A and B be two nonempty compact subsets of a metric space (X, d)

with the property that for each > 0 there is an a A and a b B with theproperty that d(a, b) < . Show that the intersection A B is nonempty.

Solution. For each n 1, let an A and bn B be such that d(an, bn) 0, there exists K such thatd(ank , a) < /2 and

12nk

< /2 are simultaneously true for k > K. Thenfor k > K,

d(bnk , a) d(bnk , ank) + d(ank , a) fn(x) 0.


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Solution. For any x X, choose Nx such that fNx(x) < 2 . By continuity,there exists a neighborhood Ux x such that fNx < on Ux. The Ux forman open cover ofX; by compactness, there is a finite subcover U

x1, . . . , U

xn.

Let N = max(Nx1 , . . . , N xn). Then fk(x) < on each Uxn , i.e., on all of

X, whenever k > N. Thus fnu 0.


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(a) An equicontinuous family of functions from one metric space to an-other.

(b) A measurable function (defined on R).(c) A G set.(d) A Lebesgue null set.(e) Let be an outer measure defined on all the subsets ofR. Define

what it means for a set E R to be measurable with respect to .

Solution.

(a) A family F of functions from X to Y is equicontinuous at a pointx X if

( > 0)( > 0)(f F)(y X)d(x, y) < d(f(x), f(y)) < .The family F is said to be equicontinuous if it is equicontinuous ateach point of X.

(b) A function f : R R is measurable if f1((a, ]) is measurable forevery a R. Equivalently, if f1(U) is measurable for every open setU.

(c) A subset of a topological space is G is it is a finite intersection ofopen sets.

(d) A subset ofRn is a Lebesgue null set if it has coverings by cubes witharbitrarily small total volume.

(e) A subset E R is -measurable if for any subset A R, (A) =(A E) + (A Ec).

(2) True or false? If A is a dense open subset of the interval [0, 1], then theLebesgue measure of A is equal to 1.

Solution. False. Let {qn}n1 be an enumeration of the rationals in [0, 1].Let In be an interval of width

13n centered at qn. Let A =

In. Then A is

open and dense, but

m(A) = m

n=1

In

n=1

m(In) =n=1

1

3n=

1

2.

(3) True or false? A subset ofR is measurable if and only if its characteristicfunction is a measurable function.

Solution. True. Suppose E R is measurable. Then

1E ((a, )) =

a 1E 0 a < 1R a < 0


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which is measurable in all three cases. Conversely, suppose E R has ameasurable characteristic function E. Then

E = 1E

12

, is measurable.

(4) True or false? Every bounded measurable function on the interval [0, 1] inR is the uniform limit of step functions.

Solution. False. Consider the characteristic function of the rational num-bers Q. Any step function must be constant on some interval, which willcontain both rational and irrational numbers. Hence no step function canbe uniformly closer to Q than

12 .

(5) True or false? Let T be a not-necessarily-countable set and for each t Tlet ft be a bounded continuous function defined on R. Define f to be

suptT ft. Then f is a measurable function.Solution. True. For any a R,

f1((a, ]) =tT

f1t ((a, ])

is a union of open sets, hence open, hence measurable.

(6) True or false? Let f be a bounded measurable function on the interval[0, 1]. For each integer n, let Fn(x) :=

x0

f(t) cos(nt)dt. Then {Fn}n=0 isan equicontinuous family.

Solution. True. Let M be a bound for |f|. Let > 0. Then for anyx, y [0, 1],

|x y| < M

|Fn(y) Fn(x)| =y

x

f(t) cos(nt)dt

yx

|f(t)|| cos(nt)|dt

yx

M 1dt= M|y x|< .

Hence the family {Fn} is equicontinuous (in fact, uniformly equicontinu-ous).

(7) Bonus Problem: Calculate the limit

limn

10

exp tan

2

x

e sin2(nx)

xndx.

Solution. For any x [0, 1] and any n, tan 2 x > 0 and e sin2(nx) > 0, soexp

tan 2 x e sin2(nx) [0, 1]. This implies that all the functions

fn(x) = exp tan

2

x

e sin2(nx)

xn


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are dominated by the constant function 1, and that the limit functionlimn fn(x) is 0 everywhere. Hence, by the Dominated ConvergenceTheorem,

limn

10

fn(x)dx =

10

limn

fn(x)dx = 0.


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Fall 2008 Real Final


(a) A Vitali covering of a set of real numbers.(b) Let f be defined on an open interval (a, b) and let x (a, b) be given.

Define D+f(x).(c) An indefinite integral.(d) What it means for a function to be absolutely continuous.(e) A singular monotone function.

Solution.

(a) A Vitali covering of a subset E R is a collection C of intervals suchthat for any x E, there is an interval I C with x I and |I| < .

(b)

D+f(x) = liminfh0

f(x + h) f(x)h

.

(c) An indefinite integral is a function of the form

F(x) =

xa

f(t)dt

where f is an integrable function.(d) A function F is absolutely continuous if for every > 0 there exists a

> 0 such that for any finite collection of intervals [a1, b1], . . . , [an, bn],

nk=1

(bk ak) < n

k=1

|F(bk) F(ak)| < .

(e) A singular monotone function is one whose derivative is zero a.e.

(2) True or false? Iff is a continuous function on a closed interval [ a, b] andif f is continuously differentiable on the open interval (a, b), then f is ofbounded variation on [a, b]. (15 points)

Solution. False. Consider the example

f(x) =

0 x = 0

x sinx

0 < x 1.

It is easy to verify that f is continuous on [0, 1] and continuously differen-tiable on (0, 1). However, let x2k =

12k and x2k+1 =

24k+1 for k = 1, 2, . . . .

Then f(x2k) = 0 whereas f(x2k+1) = x2k+1 =2

4k+1 . Then

|f(x

2k+1)

f(x2k

)|

=2

4k + 1=

|f(x

2k+2)

f(x2k+1

)|

for each k. Thus,

2Kj=2

|f(xj+1)f(xj)| =Kk=1

(|f(x2k+1) f(x2k)| + |f(x2k+2) f(x2k+1)|) =Kk=1

4

4k + 1

which can be made arbitrarily large. Hence f has unbounded variation on[0, 1].


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(3) True or false? A homeomorphism of an interval [a, b] is differentiable almosteverywhere. (15 points)

Solution. True. By the Intermediate Value Theorem, a continuous 1-1 func-tion is monotonic, and monotonic functions are differentiable a.e.

(4) True or false? Let f be a non-decreasing function defined on R and let be the associated Lebesgue-Stieltjes measure. Suppose every subset of aninterval [a, b] is a null set for . Then the derivative of f exists at everypoint of (a, b) and is identically zero there. (15 points)

Solution. Identical to Fall 2007 Real Final, Problem 3.

(5) Let f be a function defined on the interval (a, b) and assume that there is apositive constant M such that |D+f(x)| M for all x in (a, b). Show thatfor each > 0, the set of intervals {[x, y] | x (a, b) and |f(y) f(x)| (M + )(y x)} is a Vitali covering of (a, b). (15 points)Solution. Identical to Fall 2007 Real Final, Problem 4.

(6) True or false? Let E be a measurable subset of the interval [0, 1] and foreach x [a, b] let f(x) = m(E [0, x]), where m is Lebesgue measure. Thenf is absolutely continuous. (15 points)


(7) Bonus Problem: Suppose that for each positive integer n, Fn(x) is a non-decreasing, absolutely continuous function on the interval [a, b] satisfyingFn(a) = 0. Suppose further that the derivatives of the Fn(x) increase in n;i.e. assume that Fn(x) Fn+1(x) for each n and for almost all x. Finally,assume that the sequence {Fn(b)}nN is bounded. Show(a) There is a real-valued function F : [a, b] R such that limn Fn(x) =

F(x) for each x [a, b].(b) F(x) is absolutely continuous on [a, b].(c) The sequence {Fn}nN converges uniformly to F on [a, b].



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Spring 2007 Complex Final

(1) Calculate the integral 11+z2 dz, where is the circle of radius 2 centeredat the origin with the usual parametrization.

Solution. Since z2 + 1 = (z i)(z + i), the poles are at i. The residuesare

Ri(f) =1

i + i=

1

2iand Ri(f) =

1

i i =1

2iso the integral is 2i

Reszi(f) = 0.

Alternately, one may use partial fractions to write

1

z2 + 1=

1

2i

1

z i 1

z + i

.

By deformation of contours,

1

zi is the same as the integral around acircle of radius 1 centered at i, which is well known to be 2i. Similarly forthe other integral. So the integral becomes 12i (2i 2i) = 0.

(2) We showed in class that ifF is a normal family of analytic functions definedon a region G, then the collection F consisting of the derivatives of thefunctions in F is a normal family. Is the converse statement true?

Solution. No. Consider the silly example where F consists of all constantfunctions on some region. The family F consists only of the zero functionand therefore is obviously normal. However, F is not normal, since thesequence fn(x) = n does not have any normally convergent subsequences.

(3) Let f be analytic in the region B(a; r)

\{a

}and suppose there is a function

g analytic in the same region such that g(z) = f(z) for all z B(a; r)\{a}.Show that the residue of f at z = a is zero.

Solution. Since g is analytic on the punctured disc, it converges to its Lau-rent series

g(z) =

n=an(z a)n =

1n=

an(z a)n + a0 +n=1

an(z a)n

which can be differentiated termwise to obtain

f(z) = g(z) =1

n=nan(z a)n1 +

n=1

nan(z a)n

which by inspection has no 1za term. Hence the residue of f at a iszero.

(4) Let f be an analytic function on the open unit disc D. Suppose f mapsD into itself and has two fixed points. Show that f must be the identityfunction.

(5) Find a Mobius transformation that maps the real axis onto itself and mapsthe line y = x onto the circle |w + i| = 2.


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Solution. A Mobius transformation maps R to R iff it can be expressed withreal coefficients, so we will seek such a representation. Note that the twointersections of the line y = x and R (namely 0 and

) must be mapped

to the two intersections of the circle with R (namely 1). We choose to setf(0) = 1 and f() = 1. Now we can arbitrarily fix one of the coefficients,so we set d = 1. The condition f(0) = 1 then implies b = 1. The equationf() = 1 implies c = a. Thus,

f(z) =1 + az

1 az .It turns out that any value of a will work (proof omitted). For specificity,take a = 1. We can verify that this works by noting that f(1 + i) = i2+i =1+2i

5 which is on the circle because |1+2i5 + i| = | 1+7i5 | =

2.

(6) Bonus Problem: Let f be analytic in a neighborhood of the closed unitdisc, |z| 1, and suppose |f(z)| = 1 for all z with |z| = 1. Show that f isa rational function. (Hint.)


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Spring 2008 Complex Midterm 1

(1) Let f(z) = n=0 anzn be a power series with positive radius of convergenceR. Suppose z0 is a point on the circle of radius R centered at the originand suppose that the series

n=0 anz

n converges absolutely when z = z0.Show that the series converges absolutely for every z with |z| = R.Solution. Let z be any number with |z| = R. Then

n=0

|an||z0|n =n=0

|z|n

so if the first sum converges, so does the second.

(2) Find all solutions to the equation sin(z) =

2.

Solution. Let u = eiz. Then sin(z) = (u + 1/u)/(2i). Setting this equalto

2 yields the quadratic equation u2

2

2iu

1 = 0 with solutionsu = (2 1)i. This then yields

log u = log(

2 + 1) +

2+ 2k

i

in the first case and

log u = log(

2 1) +

3

2+ 2k

i

in the second. We then have

z =log u

i=

2 + 2k log(

2 + 1)i or

32 + 2k log(

2 1)i.

(3) Let f(z) = |z|, z C. For what values of z does the derivative f(z) exist?Solution. The derivative exists nowhere. The real and imaginary parts off can be expressed as u(x, y) =

x2 + y2 and v(x, y) = 0. By the Cauchy-

Riemann equations, f is differentiable when ux = uy = 0, which neverhappens.

Alternatively, let z = rei. If we let h = tei then

limh0

f(z + h) f(z)h

= ei,

but if we let h = tei(+/2) then

limh

0

f(z + h) f(z)h

= 0.

(4) Let R > 0 and let (t) = Reit, 0 t 2 be the standard parametrizationof the circle centered at the origin with radius R. Calculate the integral

ez

zndz,

for each n, n = 0, 1, . . . .


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Solution. The only pole inside the contour (or anywhere else, for that mat-ter) is at 0. To calculate the residue, we note that

ezzn

= k=0

zknk!

(which is entire except at 0, by the Ratio Test) so that the coefficient ofz1 is 1(n1)! . Hence, by the Residue Theorem, the integral is

2i(n1)! .

(5) Let f be a function analytic in a region G and assume the closed disc B(a; r)is contained in G. Show that

f(a) =1

2

20

f(a + rei)d.

Solution. Parametrize the circle |z a| = r by z = rei, 0 2. Thenz a = rei and dz = irei. By Cauchys Theorem,

f(a) = 12i

|z|=r

f(z)z a dz = 12i

20

f(a + rei

)rei

ireid = 12

20

f(a + rei)d.

(6) (Bonus) Let f(z) be analytic in the region G and assume that |f(z)1| < 1for all z in G. Show that

f(z)f(z)

dz = 0

for every closed, continuous, rectifiable path in G.

Solution. (Identical to 2005 CM1 Problem 4.) Because the range off omits

the negative real axis, log f(z) is analytic on g, so f

f has a primitive and

the result follows from Cauchys Theorem.

Alternatively,

f(z)f(z)

=

f(z)1 (1 f(z)) =

k=0

(1 f(z))kf(z) =k=0

(1 f(z))kf(z)

and each of the integrals is zero because each integrand has a primitive.Here we are justified in interchanging the sum and the integral because1 f is bounded away from zero on the compact set and hence the seriesconverges uniformly.


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(1) True or false: There is an entire function f such that f(x) = 1x for all x in

the interval [1, 2].

Solution. False. If true, then f and 1z are both analytic on the puncturedplane and are equal on a set with an accumulation point, hence are equalon the whole punctured plane. But then

f(z)dz = 0 on a circular contour

around the origin, a contradiction.

(2) Let f(z) = z3 + z2. Calculate the maximum of the modulus of f on theclosed unit disc |z| 1 and determine where the maximum is attained.Solution. By the Triangle Inequality,

|f(z)| = |z3 + z2| |z3| + |z2|with equality iff z3 and z2 have the same argument. Since

|z3| + |z2| = |z|3 + |z|2 1 + 1 = 2,we see that the maximum is at most 2. Since |f(1)| = 2 the maximum isexactly 2. Moreover, the maximum is attained nowhere else; clearly it canonly be attained on the unit circle, and the requirement that e3i and e2i

have the same argument becomes 3i = 2i + 2ik = 2k z = ei =1.

(3) Let f be an entire function and suppose its range is not dense. Show thatf is constant.

Solution. Suppose the closure of the range of f omits a point a. Let > 0such that B(a) is disjoint from the range of f. Then

|f(z) a| 1f(z) a 1so that 1f(z)a is a bounded entire function, and therefore constant byLiouvilles Theorem. Hence f is constant.

(4) Let R(z) = 2z3(z1)(z2) . Find the Laurent series for R in the annulus 1 0 and r < 1, choose N such that form,n > N, we have both |fn(0) fm(0)| < 2 and |fn fm| < 2 uniformlyon Br(0). Then for any z Br(0),

|fn(z) fm(z)| =fn(0) fm(0) +

z0

(fn(w) fm(w))dw

|fn(0) fm(0)| +z

0

|fn(w) fm(w)|dw

2

+ r

2<

where the integral is taken along a radial path. Hence {fn}n=0 is uniformlyCauchy and therefore uniformly convergent on Br(0).

(3) Let f be analytic in the region B(a; r) \ {a}. Show that f has a primitivein B(a; r) \ {a} if and only if the residue of f at z = a is zero.Solution. One direction is identical to Spring 2007 Complex Final Problem3. For the converse, suppose Resa(f) = 0. Since f is analytic on the


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punctured disk, it converges there to a Laurent series

f(z) =

n= an(z a)

n

.

We are given that a1 = 0. Hence we can define a new series by termwiseintegration:

F(z) =

n=

an1n

(z a)n.

This series converges absolutely on the same punctured disk, and uniformlyon compact subdisks (by e.g. the Ratio Test), so it defines an analyticfunction. Therefore its Laurent series can be differentiated termwise toobtain F(z) = f(z).

(4) For n = 0, 1, 2, . . . let Sn(z) = nk=0

zk

k! . Show that for each R > 0 there is

an N such that for all n > N, Sn has no zeros in the disc B(0; R).Solution. Because ez has no zeros, it attains a positive minimum m onthe closed bisc BR(0). Because e

z is entire, its power series converges to

it uniformly on any compact disc, so Sn(z)u ez on BR(0). Hence, for

sufficiently large n, we have by the extended triangle inequality|Sn(z)| |ez| |Sn(z) ez| < m2

|Sn(z)| > |ez| m2

>m

2

on BR(0). Hence Sn(z) has no zeros on BR(0).Alternatively, quoting either Rouches theorem or Hurwitzs theorem

kills the problem immediately.

(5) Consider the function

S(z) = exp

1 + z1 z

,

which is analytic in the plane except for an essential singularity at z = 1.(a) Show that |S(z)| = 1, if |z| = 1 and z = 1, and that |S(z)| > 1 for all

z, |z| < 1.(b) Explain why part (a) does not violate the maximum modulus theorem

and the Casorati-Weierstrass theorem.

Solution.

(a) Let g(z) = 1+z1z . Then

g(z) =1 + z

1

z

1 z1

z

=1 |z|2

|1

z

|2

+z z

|1

z

|2

.

The first term is real and the second pure imaginary, so g(z) = 1|z|2|1z|2 .This is zero when |z| = 1 and z = 1, whence eg(z) has modulus 1 forthese values, and is positive for |z| < 1, whence |eg(z)| > 1 for thesevalues.

(b) This does not violate the maximum modulus theorem because S isnot continuous on the closed disc, and does not violate Casorati-Weierstrass because we havent calculated the values in punctured


28/40


neighborhoods of 1, only on the portions of those neigborhoods whichlie inside the disc.

(6) Bonus Problem: Let f be a function that is meromorphic on the extendedplane C. Show that f must be a rational function.

Solution. Suppose f is not identically zero (if it is, the problem is trivial). First, f has only finitely many zeros and poles on D. If it had infinitely

many zeros, they would have an accumulation point, whence f =0. Similarly, if f had infinitely many poles in the disc, 1f (which is

also meromorphic) would have infinitely many zeros in the disc andtherefore be identically zero, an impossibility.

Letg(z) =

f

1z

z = 0

f(

) z = 0.

Then g is also meromorphic. Moreover, if f has infinitely many zerosor poles outside D, then g has infinitely many zeros or poles inside D,forcing it to be constant 0 or which is again a contradiction.

Thus, f has finitely many zeros and poles on all of C. Let a1, . . . , anbe the zeros (listed with multiplicity) and b1, . . . , bm the poles. Then

h(z) =(z b1) . . . (z bm)(z a1) . . . (z an) f(z)

is meromorphic with no zeros or poles. (Here a factory of z onthe top or bottom is understood to be a factor of z on the bottom ortop, respectively.) Since it does not have a pole at , it is a boundedentire function and therefore constant. Hence f is rational.


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(1) Find the power series representation of the function f(z) = 11z2 at the

origin and calculate its radius of convergence.

Solution. Since f has singularities at 1, its radius of convergence is atmost 1. For |z| < 1, we can expand the geometric series to obtain

f(z) =k=0

z2k

which is absolutely convergent. Hence the radius of convergence is 1.

(2) True or false: There is an entire function f(z) such that for all real numbersx > 1, f(x) = 1x .

Solution. False. See 2008 CM2 Problem 1.

(3) Let f(z) = f(x, y) = u(x, y) + iv(x, y) be an entire function. Suppose thatu is a function of x alone and that v is a function of y alone. Show that fis a polynomial of degree at most 1.

Solution. From the Cauchy-Riemann equations and the expression f =ux + ivx = vy iuy, we see that f is real everywhere. This implies that itis constant (since eif

is a bounded entire function), so f is a polynomialof degree 1 or less.

(4) Let be a closed, continuous, rectifiable path in a region G such thatn(; a) = 0 for all a C \ G and suppose f is analytic in G. Show that forall a G \ ,

f(z)z

a

dz =

f(z)

(z

a)2

dz.

Solution. Ifa is outside then both integrals are 0 because the integrandsare analytic on and inside . Suppose now that a is inside . Then byCauchys formula,

f(a) =1

2i

f(z)

z a dz.

Differentiating both sides with respect to z and applying Leibnizs Rule,

0 =1

2i

d

dz

f(z)

z a dz

=1

2i

d

dz

f(z)

z a dz

= 12i

f(z)z a f(z)(z a)2

dz

=1

2i

f(z)z a dz

f(z)

(z a)2

whence we deduce the equality of the two integrals as desired.

(5) True or false: Iff is an entire function with the property that |f(z)| > 1for all z, then f is constant.


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Solution. True. Since f has no zeros, 1f is an entire function, and since

1f < 1 everywhere, 1f is constant by Liouvilles Theorem. Hence f isconstant.

(6) Bonus problem: Let f(z) be analytic in the region G and assume that|f(z) 1| < 1 for all z G. Show that it is possible to define an analyticbranch of log(f(z)) in G.

Solution. Since the range ofg omits the branch cut of the log function, thecomposition log(f(z)), using the main branch of log, is analytic on G. (To

do this directly, one could define the branch of log(f(z)) by integrating f

f ,

which can be done by expanding the denominator 1 (1f) as a geometricseries.)


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(1) Evaluate the integral 1

x4+1dx.

Solution. Let R be a semicircular contour of radius R in the upper halfplane whose diameter extends from R to R. Let CR denote the curvedpart of this contour. The singularities of the integrand f(z) = 11+z4 inside

the contour occur at ei/4 and e3i/4, with residues

Res(f; ei/4) =1

4z3

z=ei/4

=e3i/4

4

and

Res(f; e3i/4) =1

4z3

z=e3i/4

=ei/4

4

which add up to i

24 . By the Residue Theorem,

RR

dx

x4 + 1 +CR

f(z)dz =R f(z)dz =

2

2 .

On CR , the integrand is less than1

R41 0 in aneighborhood ofa. Since 1(za)m takes on arbitrarily large positive values,so does

f. If

g(a) = 0 then

g(z) is bounded away from zero (WLOG

> > 0) in a neighborhood of a, in which 1(za)m can be made to takearbitrarily large negative imaginary values, making f large. Hence, iffis bounded near a, it must be a removable singularity.

(3) Let f(z) = e1

(z1)2 (z 1)3 for all z = 1. Determine whether z = 1 is aremovable singularity of f, a pole of f, or an essential singularity of f. Ifz = 1 is either a pole or an essential singularity of f, calculate the residueof f at z = 1.


32/40


Solution. The singularity type (and residue, if applicable) are the same as

those of g(z) = z3e1/z2

at 0, so we consider the latter. On the punctured

plane, the seriesz3

k=0

z2k

k!=

k=0

z32k

k!

converges to g (by e.g. the Ratio Test). Since the Laurent series for gnear 0 contains infinitely many negative-order terms, it has an essentialsingularity.

(4) True or false? Iff(z) = 1z then there is a sequence of polynomials {pn} thatconverge to f uniformly on the circle |z| = 1, i.e. limn sup|z|=1 |pn(z) f(z)| = 0.Solution. False. Suppose such a sequence exists. Because polynomials areentire,

pn(z)dz = 0 for all n. But by the uniformity of the convergence,

2i =

1z

dz =

limnpn(z)dz = limn

pn(z)dz = 0,

a contradiction.

(5) There are two regions in which the function f(z) = 1z(1+z2) has a Lau-

rent series in powers of z. Determine those regions and the correspondingLaurent series.

Solution. The singularities of f are at z = 0, i, i. Hence the two regionsare 0 < |z| < 1 and |z| > 1. By expanding geometric series, we have on theformer

f(z) =1

z(1 (z2)) =1

z

k=0(z2)k =

k=0(1)kz2k1

and on the latter

f(z) =1

z3(1 1/z2) =1

z3

k=0

1

z2

k=

k=0

(1)kz2k3.

(6) Bonus Problem: True or false? Iff is analytic in the domain C \ {0}, andif |f(z)| |z|+1|z| for all z = 0, then f must be constant.

Solution. True. If f had a pole at 0 then its growth near the origin wouldbe at least 1|z| . If it had an essential singularity at 0, then zf(z) would aswell, but the latter is bounded near 0. Hence the singularity at the originis removable, so f can be extended to a bounded entire (hence constant)

function.


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Spring 2009 Complex Final

(1) Calculate the integral 12i |z|=r cot zdz as a function ofr, for r in the rangeN < r < (N + 1), where N N.Solution. This integral is in the form 12i

f(z)f(z) dz, where f(z) = sin(z).

By the Argument Principle, this is just N0 N where N0 is the numberof zeros and N the number of poles inside the contour. Since sin(z) hasno poles and has zeros at z = k for k Z, there are 2N + 1 zeros and nopoles inside the given contour. So the answer is 2N + 1.

(2) Iff is analytic in a region G, then a point z G is called a fixed point off in case f(z) = z. If f is analytic in a neighborhood of the closed unitdisc D, and if |f(z)| < 1 when |z| = 1, then how many fixed points must fhave in the open unit disc D?

(3) Suppose f is analytic in a neighborhood of the closed unit disc D, and does

not vanish on D. Show that

log |f(0)| = 12

log |f(eit)|dt.

Solution. This is a direct consequence of the mean-value property of har-monic functions plus the fact that log |f(z)| is harmonic. (It is the real partof log f(z); even though the latter must be defined on two different sets tocover the punctured plane, it is sufficient because log |f(z)| is still harmonicat each point.) Alternatively, because the disc is simply connected and f(z)is nonzero on it, there is a function g(z) such that f(z) = eg(z), from whichthe result follows.

(4) Suppose{

fn}n=0

is a sequence of analytic functions defined in a region G.Suppose, also, that f is analytic in G and that the sequence of real parts{fn}n=0 converges uniformly on compact subsets of G to the real part off. Finally, suppose that at a point z0 G the sequence {fn}n=0 convergesto f(z0). Show that {fn}n=0 converges uniformly to f on compact subsetsof G.

Solution. WLOG suppose f = 0. Let Br(z) be some compact subdisk ofG. By the Schwarz integral formula,

fn(z) =1

2i

Br(z)

+ z

z un()d

+ ivn(z0)

for any z Br(z0) and for any n. Since +z(z) is uniformly bounded, andsince un 0 uniformly, we have fn(z) ivn(z0) 0 uniformly throughoutthe disc. Note in particular that if vn 0 at any point of the disc, thenvn 0 uniformly on the disc. Next, show that vn 0 at all points ofG: If not, draw a path from a point that does to a point that doesnt,cover it with finitely many open discs, and none can be the first where thisfails. Finally, cover compact subsets of G by finitely many discs, on eachof which vn 0 uniformly. Hence vn 0 (and thence fn 0) uniformlyon compact subsets.


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Solution not using Schwarz, by Katya and Jonas. Let gn = efn . Let A

G be compact. Then the gn are uniformly bounded on A, hence normal.That is, any subsequence has a further subsequence that converges in H(G).These sub-sub-limits have modulus everywhere, so theyre constant. Thusefn 1 in H(G). Then

( > 0)(N N)(z A)|efn 1| < .For each k let

logk z = ln |z| + i(arg z + 2k).Then

exp1(B(1)) =

k=logk B(1)

so

fn(A)

k=logk(B(1)).

The real parts are betweer ln(1 ) and ln(1 + ), and the imaginary partsbetween 2k sin1() and 2k + sin1(). Thus, we get a union of stripsin which fn(A) is contained. For small the strips are disjoint, but fn(A)is connected and contains fn(z0), so fn ( sin1 , sin1 ) for large n.(For this argument, we may need to use As which are compact, connected,and contain z0; any other A can be enclosed in some such.)

(5) Suppose f and g are two analytic functions mapping the open unit disc Dinto a region G. Suppose f(0) = g(0), f is one-to-one, and f maps D ontoG. Show that for every r, 0 < r < 1,

g(B(0, r)) f(B(0, r)),and that if equality holds for any r, 0 < r < 1, then equality holds for allr, 0 < r < 1.

Solution. Define : D D by (z) = f1(g(z)). Then is a conformalself-map of the disk; in particular, |(z)| < 1 for |z| < 1. Note also that(0) = 0. By the Schwarz lemma, |(z)| |z| for |z| < 1.

Now let w g(B(0, r)) so that there is z B(0, r) with g(z) = w. Let t =f1(w) = (z). By the above, |t| < |z| so that t B(0, r). Thus z = f(t) f(B(0, r)). Hence g(B(0, r)) f(B(0, r)). Suppose equality holds for somer (0, 1). Applying f1 to both sides, we have (B(0, r)) = B(0, r), whichby continuity extends to the closure so that (B(0, r)) = B(0, r). Thismeans there exists z B(0, r) such that |(z)| = |z|. By the Schwarzlemma again, this implies that (z) = z for some unimodular , whence(B(0, r)) = B(0, r) for all r and thus g(B(0, r)) = f(B(0, r)).

(6) Bonus Problem: This is a continuation of problem 3. Drop the assumptionthat f is nonzero throughout a neighborhood of D and suppose only thatf(0) = 0, that for 0 < r < 1 the zeros of f in a neighborhood of B(0; r)


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counted with multiplicity are a1, . . . , an, and that f does not vanish on thecircle |z| = r. Show then that

log |f(0)| = n

k=1

log

r|ak|

+ 1

2

log |f(reit)|dt,

for all r such that f does not vanish on the circle |z| = r. (Hint.)


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Summer 2007 Qual

Part I.

(1) Let A and B be two subsets of [0, 1] whose union is all of [0, 1]. Show thatm(A) 1 m(B).

(2) Define the following function on C([0, 1]) C([0, 1]): d(f, g) = 10 |f(x) g(x)|dx. Show that d is a metric on C([0, 1]) and determine whetherC([0, 1]) is complete in this metric.

(3) Let A be a subset ofR with the property that for each > 0 there are(Lebesgue) measurable sets B and C such that

B A Cand m(C Bc) < . Show that A is measurable.

(4) True or false: Let f be a nonnegative continuous function on R and suppose

Rf(x)dx < . Then lim|x| f(x) = 0.

(5) True or false: If

{fn

}n=1 is a sequence of Lebesgue measurable functions

such that0 f1 f2 . . . ,

if supR

fn(x)dx < , and if f(x) = lim fn(x) for all x, then {x | f(x) =} has measure zero.

Part II.

(1) Calculate the radius of convergence of the power seriesn=0

zn2

.

(2) Suppose f is analytic in the region 0 < |z| < 1 and suppose there is aconstant K such that

|f(z)| K|z|1/2

there. What kind of isolated singularity does f have at zero?(3) For what values of z does the series

n=0

zn

1 znconverge? Is the sum an analytic function of z?

(4) True or false: There is a nonconstant entire function f such that f(z +1) =f(z) and f(z + i) = f(z) for all z in C.

(5) True or false: Suppose {fn}n=0 is a sequence of functions defined andanalytic in the open unit disc D. Suppose also that the values of each fnare contained in the upper half-plane. Then {fn}n=0 is a normal family.


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Summer 2008 Qual

Part I.

(1) If{En}n=1 is a sequence of sets, the limsup of the sequence is defined tobe lim sup En =

n1

kn Ek. True or false: If{En}n=1 is a sequence

of Lebesgue measurable subsets ofR such that (En) < for all n, then(lim sup En) = limsup (En).

(2) Let E be the subset of [0, 1] consisting of numbers whose decimal expansioncontains at least one 4. Calculate the Lebesgue measure of E.

(3) Let f be a nonnegative measurable real-valued function defined on all ofR. Show that f is Lebesgue integrable on R if and only if the series

k=0 k(Ek) is convergent, where Ek is the set of points x such thatk f(x) < k + 1.

(4) True or false: Iff is a continuous function on [0, 1] and iff is continuouslydifferentiable on the open interval (0, 1) with bounded derivative, then f is

absolutely continuous.(5) True or false: Suppose that fn is the continuous function on [0, 2] definedby fn(x) = sin(nx). Then {fn}n=1 is an equicontinuous family.

Part II.

(1) True or false: If fn(z) = sin(nz), then the sequence {fn}n=1 is a normalfamily on the disc B(0;2).

(2) Let G be the domain obtained by deleting from the open unit disc B(0; 1)the origin and points 2n, n = 1, 2, . . . and suppose f is analytic on G.Suppose further that each of the isolated singularities of f, 2n, is a pole.Show that while 0 is not an isolated singularity of f, the conclusion of theCasorati-Weierstrass theorem still holds: The range of f on any set of theform G

B(0; ) is dense in C.

(3) Suppose that f is analytic in B(0; 1) and that lim sup |f(zn)| 1 for everysequence {zn}n=1 in B(0; 1) that converges to a point on the boundary ofB(0; 1). Show that |f(z)| < 1 for all z B(0; 1).

(4) Let f(z) = z(z2+4)sin(z) . Find all the singularities of f in the complex

plane. (They are all isolated.) Classify them.

(5) Find the Laurent expansion of f(z) = z2

(z1)(z2) in the region 1 < |z| < 2.

Winter 2009 Qual

Part I.

(1) True or false? Let [a, b] and [c, d] be two nonempty closed bounded intervalsin R, and suppose that f is a strictly increasing function mapping [a, b] onto

[c, d]. Then f is a homeomorphism.

Solution. True. We first note that f must be continuous; because it ismonotonic, it can only have jump discontinuities, but if it had any thenit would not be surjective. Hence f is continuous. The same reasoningapplies to f1 (which exists because f is increasing); the only possiblediscontinuities are jumps, which would cause it not to be surjective. Hencef is homeomorphic.


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(2) True or false? A set E R is Lebesgue measurable if and only if there isa G set G containing E such that m

(G Ec) = 0, where m is Lebesgueouter measure.

Solution. True. Suppose E is measurable, and let Gn be open sets (i.e.unions of open intervals) containing E with m(Gn) m(E) + 1n . Sinceopen sets are measurable, m(GN Ec) = m(Gn) m(E) 1n . LetG =

Gn. Then m

(G Ec) = m(G Ec) = lim m(Gn Ec) = 0.Conversely, suppose E is a set such that there is a G set G with m

(GEc) = 0. Let A be any set. Then

m(A G) = m((A E) (A (G \ E))) m(A E) + m(A (G \ E)) m(A E) + m(G \ E)= m(A

E)

and since the reverse inequality obviously holds, we have m(A G) =m(A E). Similarly,

m(A \ E) = m(A Ec)= m(A (Gc G \ E))= m((A Gc) (A (G \ E))) m(A Gc) + m(A (G \ E)) m(A Gc) + m(G \ E)= m(A \ G)

and the reverse is clearly true, so that m(A\

G) = m(A\

E).Now because G is measurable, we have

m(A) = m(A \ G) + m(A G).Substituting the two equalities derived above, this becomes

m(A) = m(A \ E) + m(A E).Thus E is measurable.

(3) True or false? Suppose f is a nonnegative measurable function defined onR and that {En}n=0 is a sequence of measurable sets with the propertythat n0 En is a Lebesgue null set. Then lim infn En f(x)dx = 0.Solution. False. Let En = [0, 1] for all odd n and [1, 2] for all even n, andlet f = [0,2].

(4) Let f be a measurable function defined on R. A point x R is called aLebesgue point of f in case limh0 1h

x+hx

|f(t) f(x)|dt = 0. Show thatiff is Lebesgue integrable on the interval [0, 1] and ifF(x) :=

x0

f(t)dt foreach x [0, 1], then F exists and equals f(x) at each Lebesgue point x.


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Solution. We have

F(x + h) F(x)h

f(x) = 1h x+h

x

f(t)dt

f(x)=

1hx+hx

f(t) f(x))dt

1

h

x+hx

|f(t) f(x)|dt

and since the latter integral tends to 0, the first expression does as well.Thus

F(x + h) F(x)h

f(x) F(x) = f(x).

(5) True or false? Let f be a Lebesgue integrable function defined on the

interval [0, 1], and define Fn(x) =1

0f(t)cos

xtn

dt. Then {Fn}n=1 is an

equicontinuous family on [0, 1].

Solution. True. Let > 0. By the uniform continuity of cos on [0, 1], thereexists > 0 such that | cos(u) cos(v)| < f1 when |u v| < . For this and for any n N and x, y [0, 1], if |y x| < then |ytn xtn| =tn |y x| < and therefore

|Fn(y) Fn(x)| =1

0

f(t)

cos

yt

n

cos

xt

n

1

0

|f(t)|cos

yt

n

cos

xt

n

1

0

|f(t)| f1 dt= .

Hence {Fn} is equicontinuous. Part II.

(1) Identify the largest disc on which the series

n=0 n2(2z 1)n converges.

Solution. By the Ratio Test, this sequence converges for |2z 1| < 1 anddiverges for |2z 1| > 1. Hence the largest disc on which it converges is

{z | |2z 1| < 1} = B

1

2,

1

2.

(It converges only on the open disc, not closed; indeed, it diverges at everypoint on the boundary of this disc, because the terms tend to in absolutevalue.)

(2) Let f be an entire function with the property that |f(z)| |z| for all z.Show that f is a polynomial of degree at most two.

(3) True or false: Iffn(z) = einz, then the sequence {fn}n=1 is a normal family

on the upper half-plane.


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Solution. True. All the functions are analytic, and the family is uniformlybounded on compact subsets. (In fact, its uniformly bounded on all of theUHP, since

|fn

(z)|

= en

(z) < 1 there.)

(4) Find the maximum and minimum of the modulus of the function f(z) =z2 z on the closed unit disc.Solution. The minimum modulus is 0, which is attained both at 0 and 1.By the Maximum Modulus Principle, the maximum is attained on the unitcircle, on which we have

|f(z)|2 = |z2 z|2= (z2 z)(z2 z)= z2z2 + zz zz2 z2z= 1 + 1 zz(z + z)= 2(1

z)

2with equality iff z is pure imaginary. Thus, |f(z)| 2, which is attainedwhen z = i.

(5) Evaluate the integralI

ez

(z+1)4 dz, where I is the imaginary axis parametrized

from i to i.Solution. The only pole of f(z) = e

z

(z+1)4 is at 1, a pole of order 4 with aresidue

limz1

1

3!

d3

dz3ez =

1

6e.

Let R denote the semicircular contour in the LHP with diameter from Rito Ri. Let C

rdenote the curved part of

R. By the Residue Theorem,

i

3e=

R

f(z)dz =

RiRi

f(z)dz +

CR

f(z)dz.

On CR, |ez| 1 so |f(z)| 1(R1)4 < 24

R4 for large R. HenceCR

f(z)dz 0, so that in the limit of large R we obtaini

if(z)dz =

i

3e.

combined muh ly exams

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