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PURE MATHEMATICS ANALYSIS COMPREHENSIVE EXAMSCOMPILATION

STANLEY YAO XIAO

Abstract. Here is a compilation of problems of past exams, organized by year.Only problems that I was able to solve or find a solution for are included.

Date: May 13, 2016.1

2 STANLEY YAO XIAO

1. Spring 2016

1.1. Part I.

Question 1.1. Let f : R → R be a continuous function. Suppose that there is aninfinite countable subset S ⊂ R such that∫ b

a

f(x)dx = 0

whenever a, b 6∈ S. Show that f = 0.

Proof. Suppose, for the sake of a contradiction, that f is not identically zero. Thenthere exists a point x ∈ R such that f(x) 6= 0. Since any scalar multiple and anytranslate of f will also satisfy the hypothesis of our problem, we may assume thatf(0) = 1. Since f is continuous, there exists δ > 0 such that for all y ∈ (−δ, δ) wehave |f(y)− f(0)| < 1/2. In particular, we have that f(y) ≥ 1/2 on (−δ, δ). Since Sis countable, it follows that S ∩ (−δ, δ) is a proper subset of (−δ, δ) so we may finda, b ∈ (−δ, δ) \ S. Then we see that∫ b

a

f(x)dx ≥ 1

2(b− a) > 0,

which contradicts our hypothesis. Thus f must be identically zero.

Question 1.2. Let C([−1, 1]) denote the Banach space of continuous real valuedfunctions on [−1, 1] equipped with the supremum norm. Determine whether each ofthe following sets is dense in C([−1, 1]) and justify your answer:

(a) span1, x2, x4, x6, · · · (b) span1, x171, x172, x173, · · · .

Proof. (a) The answer is no. One sees conspicuously that any properly odd func-tion on [−1, 1] cannot be in the limit of the span of 1, x2, x4, · · · since the limitof this set is necessarily even. For example, f(x) = x+ x3 is not in the limit.

(b) The answer is yes. By the Stone-Weierstrass theorem, since [−1, 1] is compact,it suffices to verify that the given set is an algebra that separates points in[−1, 1] and vanishes nowhere. Since 1 is a generator, we see that the spanvanishes nowhere. Now the function f(x) = x171 is injective on [−1, 1], henceseparates points. It remains to verify that the span is an algebra. By thedefinition of span, we see that it is closed under addition. It is also closedunder multiplication after a quick verification.

Question 1.3. Give an example of a sequence (fn)∞n=1 of non-negative measurablefunctions on R and a measurable function f on R such that

(1) fn+1(x) ≤ fn(x) for all n ≥ 1 and x ∈ R, and(2) limn→∞ fn(x) = f(x) for all x ∈ R, but

limn→∞

∫Rfn(x)dx 6=

∫Rf(x)dx.

PURE MATHEMATICS ANALYSIS COMPREHENSIVE EXAMS COMPILATION 3

Proof. Consider the sequence of non-negative functions on R given by

fn(x) =|x|n.

Clearly, each fn is measurable. Moreover, it is obvious that fn+1(x) ≤ fn(x) identi-cally. Observe that for a fixed x ∈ R, we have

limn→∞

fn(x) = 0,

so the point-wise limit of (fn)∞n=1 is f ≡ 0. But we note that for each n∫Rfn(x) =∞,

so this provides a counterexample, as desired.

Question 1.4. Let (X, d) be a complete countable metric space. Show there is x ∈ Xsuch that the singleton x is open.

Proof. By the Baire category theorem, (X, d) is not the countable union of nowheredense sets. Hence, if we write

X =⋃x∈X

x,

we see that there is some x ∈ X such that x is not nowhere dense. Therefore, thereexists an open subset S ⊂ X such that x is dense in S. But then S has to be equalto x, so x is open, as desired.

Question 1.5. Let X and Y be topological spaces such that X is compact and Y isHausdorff. Let f : X → Y be a continuous bijection. Show that f is homeomorphism.

Proof. Let U be a closed subset of X. To verify that f is a homeomorphism, itsuffices to verify that f(U) is closed in Y . First note that a closed subset of a com-pact topological space is compact. To see this, let Uα be an arbitrary open coverof U in X. Then adjoint the set X \ U , which is open by assumption. Hence theset Uα ∪ X \ K is an open cover for X, hence contains a finite sub-cover, sayU1, · · · ,Uk, X \ U . It is clear that U1, · · · ,Uk is a finite sub cover of Uα of U , so Uis compact, as desired.

Next we note that the continuous image of a compact set is compact. To see this,let Yα be an open cover of f(U) in Y . Then f−1(Yα) is open in X, since f iscontinuous, and f−1(Yα) is an open cover for U in X. Therefore, there is a finitesub cover f−1(Y1), · · · , f−1(Yk) say. Mapping forward, we see that Y1, · · · ,Yk is afinite sub-cover of f(U), so f(U) is compact.

It then remains to show that for any compact subset K of Y that K is closed.Suppose K ⊂ Y is compact. If K = Y , then K is closed in Y and we are done.Otherwise, fix a point x ∈ Y \K. For each y ∈ K, choose open neighborhoods Dy ofy and Ey of x such that Ey ∩Dy = ∅. This is possible because Y is Hausdorff. Thenthe Dy’s form an open cover of K, hence there exists a finite sub-cover Dy1 , · · · , Dyk ,

say. Thus the finite intersection⋂kj=1Eyj is an open neighborhood of x which is

4 STANLEY YAO XIAO

disjoint from K, which shows that Y \ K is open in Y , so K is closed. This showsthat f is a closed map, so it is a homeomorphism.

Question 1.6. Evaluate ∫ 2π

0

1

1 + cos θdθ.

Proof. We will show that the integral is unbounded. We first put t = θ − π, so that∫ 2π

0

dθ

1 + cos θ=

∫ π

−π

dt

1− cos t.

Now note that

1− cos t = 1− 1 +t2

2− t4

24+

t6

720− · · · ,

so there is a pole of order 2 at 0. Then one notes that the function f(x) = 1/x2 isnot integrable in any interval containing 0, so neither is 1/(1− cos t).

Question 1.7. Let X and Y be non-empty sets. let

F = (A,B, f) : A ⊂ X,B ⊂ Y, f : A→ B is a bijection.

Partially order F by (A1, B1, f1) ≤ (A2, B2, f2) if and only if A1 ⊂ A2, B1 ⊂ B2 andf2 restricts to f1 on A1. Use this to show that one of the following possibilities musthold:

(1) There exists a one-to-one function from X into Y .(2) There exists an onto function from X onto Y .

Proof. By construction, every chain (Aα, Bα, fα) has a maximum element (A,B, f)where A =

⋃αAα, B =

⋃αBα, and f is the function which restricts to fα whenever

x ∈ Aα. It is clear that f is a surjection. To see that f is indeed a bijection, supposethat there exist x, y ∈ A such that f(x) = f(y). Then there exist indices α, α′ suchthat x ∈ Aα, y ∈ Aα′ , and so f(x) ∈ Bα, f(y) ∈ Bα′ . Without loss of generality,suppose that Bα ⊂ Bα′ . Then f(x), f(y) map different elements in Aα′ to the sameelement in Bα′ and are equal to fα′(x), fα′(y). This violates the assumption that fα′is a bijection. Thus f is a bijection, as desired.

By Zorn’s lemma, F has a maximal element, say (U, V, g). If U = X, V = Y then wesee that g is a one-to-one and onto function from X to Y and we are done. Otherwise,suppose that U 6= X. Then we cannot have V 6= Y , since otherwise we can lift g bychoosing elements x ∈ X \ U and y ∈ Y \ V and a function h : U ∪ x → V ∪ ysuch that h(x) = y and h(s) = g(s) for all s ∈ U . This contradicts (U, V, g) beingmaximal. Thus, our maximal element must look like (U, Y, g) and so g is a surjectionfrom X onto Y . If V 6= Y , then again we cannot have U 6= X and so our maximalelement must be (X, V, g). But then since g is injective, we see that there exists aninjection from X into Y , as desired.

Question 1.8. (a) Let (X, d) be a metric space and let (fn)∞n=1 be a sequenceof continuous real-valued functions on (X, d) that converges uniformly to afunction f : X → R. Show that f is also continuous.


(b) Let∑∞

n=0 anxn be a power series. Suppose that this series converges at some

point x0 ∈ R with x0 6= 0. Show that the power series converges for everyx ∈ (−|x0|, |x0|).

(c) Define f(x) =∑∞

n=0 anxn for x ∈ (−|x0|, |x0|), where the power series and

x0 ∈ R are as in (b). Show that f is continuous on (−|x0|, |x0|).

Proof. (a) Let (fn) be a sequence of continuous functions real-valued functionson (X, d) which converges uniformly to a function f . Let ε > 0. Choose Nsufficiently large so that for any x ∈ X, we have

|fN(x)− f(x)| < ε.

Now choose δ > 0 so that for any y with d(x, y) < δ we have

|fN(x)− fN(y)| < ε.

Then by the triangle inequality

|f(x)− f(y)| ≤ |f(x)− fN(x)|+ |f(x)− fN(y)|+ |fN(x)− fN(y)| < 3ε,

so f is continuous, as desired.(b) It is well-known that if an, bn are real non-negative sequences such that

an ≤ bn for every n ≥ 0 and∑bn converges, then

∑an converges. For a

given x ∈ (−|x0|, |x0|), put un = anxn and vn = anx

n0 . Then If

∑vn converges

absolutely, we have ∑|un| ≤

∑|vn| <∞,

so∑un converges as well. If

∑vn does not converge absolutely, then it

follows that lim supn→∞ |vn|1/n = 1. This shows that if |x| < |x0|, thenlim supn→∞ |un|1/n ≤ |x/x0|n < 1, so by the root test,

∑un converges.

(c) It suffices to show that the sequence of functions fk(x) =∑k

n=0 anxn converges

to f uniformly on [−|x0|+ η, |x0| − η] for any η > 0. Let x1 = |x0| − η and letε > 0. For any x ∈ [−x1, x1] and k ≥ 1, we have

|f(x)− fk(x)| =

∣∣∣∣∣∞∑

n=k+1

anxn

∣∣∣∣∣ ≤∞∑

n=k+1

|an|xn1 .

Moreover, since we have asserted in part (b) that the series∑anx

n1 converges

absolutely in [−x1, x1], it follows that there exists a positive integer N suchthat

∑∞n=N |an|xn1 < ε. It follows that

|fk(x)− f(x)| < ε

whenever k > N , so that fk converges to f uniformly, as desired.

1.2. Part II.

Question 1.9. (a) (1) Prove Liouville’s theorem that a bounded entire functionf : C→ C is constant.

(2) Let f : C→ C be a non-constant entire function. Show that the range off is dense in C.

(3) Show that a bounded harmonic function u : R2 → R is constant.

6 STANLEY YAO XIAO

(b) (1) Show that for every non-constant polynomial p ∈ C[z],

lim|z|→∞

|p(z)| =∞.

(2) Prove the Fundamental Theorem of Algebra: For every non-constantpolynomial p ∈ C[z], there is z0 ∈ C such that p(z0) = 0.

Proof. (a) (1) We prove Liouville’s theorem as a corollary of our proof of part(c).

(2) We will use the fundamental theorem of algebra, which we will prove alittle later. Let z0 be a fixed non-zero element of C, and let Pk(z) be thek-th order MacLaurin polynomial of f . By the fundamental theorem ofalgebra, the equation

Pk(z) = z0

has a solution in C for every k. Put ε > 0. Now choose N sufficientlylarge so that

|f(z)− Pk(z)| < ε

whenever k ≥ N . Thus, choosing a sequence of ε’s that tends to zero wecan find a sequence of zn’s so that f(zn)→ z0, as desired.

(3) Let x, y be two distinct points on the plane R2. Let D1(R), D2(R) bethe two circles of radius R centred at x and y respectively. Then bythe mean value property of harmonic functions, we see that f(x), f(y)are equal to the average value of f on D1, D2 respectively. Since f isbounded, it follows that as R tends to infinity, the average value of f onD1(R), D2(R) converges to the same value, so f(x) = f(y). This showsthat f is constant.

This also resolves Liouville’s theorem on noting that the real and imagi-nary parts of a complex analytic function are real harmonic functions onR2.

(b) (1) Let P be a non-constant polynomial of degree n, say

P (z) = anzn + · · ·+ a0.

By definition, we see that an 6= 0. Choose R sufficiently large so that|an|R > 2(|an−1|+ · · · + |a0|). Then we see that for |z| > R, the triangleinequality yields

|P (z)| > |an||z|n − |an−1||z|n−1 − · · · − |a0| > |an||z|n−1,

hence the desired claim holds.(2) Let f be a polynomial which vanishes nowhere on C. Then f is entire.

Moreover, it follows that 1/f is entire. By part (b), it follows that 1/f isbounded at infinity and since it is entire, it is bounded away from infinityas well. Therefore 1/f is bounded and by Liouville’s theorem it must beconstant. Therefore f must be constant, so whenever f is a non-constantpolynomial, it must have a zero in C, as desired.

Question 1.10. Let m denote the Lebesgue measure on R.


(a) Let E ⊂ R be a measurable set with 0 < m(E) <∞. Show that the function

F (x) = m((x+ E) ∩ E)

is continuous at x = 0, wgere x+ E = x+ y : y ∈ E.(b) Let E ⊂ R be a measurable set with m(E) > 0. Show that the set

E − E = x− y|x, y ∈ Econtains an open interval (−δ, δ) for some δ > 0.

(c) Let f : R→ R be a measurable function such that f(x) + f(y) = f(x+ y) forall x, y ∈ R. Show that f is continuous.

(d) Let f be as in (c). Show that there is γ ∈ R such that f(x) = γx for everyx ∈ R.

Proof. (a) Since Lebesgue measure is outer regular, for any ε > 0 there exists anopen subset U of R such that m(U) < m(E) + ε/2 and E ⊂ U . Moreover,since U is open in R, it follows that U is the countable union of disjoint openintervals, say

U =∞⋃k=1

(ak, bk).

We see that

m(U) =∞∑k=1

|bk − ak|.

Now choose N minimal so that∞∑

k=N+1

|bk − ak| <ε

2.

Put δ = ε/2N and suppose |x| < δ. Then

U + x =∞⋃k=1

(ak + x, bk + x).

Moreover, we see that

N∑k=1

|bk − ak| < m(U + x) <N∑k=1

|bk − ak|+ ε,

so that

m(U ∩ U + x) < ε.

Finally, note that E ∩ E + x ⊂ U ∩ U + x. Therefore we see that

m(E ∩ E + x) < 2ε,

which shows that F is continuous at 0, as desired.(b) Since Lebesgue measure is both inner and outer regular, for any ε > 0 there

exists an open set U containing E such that m(U) < m(E) + ε/2 and acompact set K which is contained in E such that m(K) > m(E) − ε/2. PutV = R \ U and let d = inf|x − y| : x ∈ K, y ∈ V . Observe that K andV are disjoint, and that the distance between disjoint sets K and V where

8 STANLEY YAO XIAO

K is compact and V is closed is positive; in particular, d > 0. Now choose0 < δ < d. Note that

(−δ, δ) +K ⊂ U.

Suppose that there is s ∈ (−δ, δ) and t ∈ K such that s + t 6∈ U . Then wemust have s+ t ∈ V , by definition. Therefore (s+ t)− t = s and |s| < δ < d,which contradicts the definition of d. Now choose r ∈ (−δ, δ). Suppose that

K ∩ (K + r) = ∅.Note that this implies

m(K ∪ (K + r)) = 2m(K).

Further observe that

m(K) < m(U) < m(K) + ε,

and that K,K + r ⊂ U . But then

2m(K) ≤ m(U) < m(K) + ε,

which is a contradiction when ε < m(K). This implies that K ∩K + r 6= ∅and so there exist y ∈ R such that y = s = s′ + r, so that r = s − s′ withs, s′ ∈ K. Since K is contained in E, we are done.

(c) Note that f(x+ (−x)) = f(x) + f(−x) = f(0) and f(0 + 0) = f(0) + f(0) =2f(0), so f(0) = 0 and f(−x) = −f(x) for all x ∈ R. Let x ∈ R, and letε > 0. Let Uε to be the interval (f(x)− ε, f(x) + ε). Since the preimage of ameasurable set under a measurable function is measurable, we see that f−1(Uε)is measurable. Therefore, by part (b), there is an interval (−δ, δ) containedin f−1(Uε) − f−1(Uε). Now, x + (−δ, δ) ⊂ f−1(Uε), which shows that f iscontinuous at x. Since x was arbitrary, this shows that f is continuous on R.

(d) In part (c) we showed that f is continuous. We now use this to show thatf has to be of the desired form. We had already shown that f(0) = 0. Putγ = f(1). We will show that f(x) = γx whenever x is a rational number. Wefirst verify this for the integers. Since we have shown that f(−x) = −f(x)for all x, it suffices to verify this for positive integers. This follows by induc-tion because f(n + 1) = f(n) + f(1) = nf(1) + f(1) = (n + 1)f(1). Nextlet x = 1/n for n ≥ 1. Then f(1) = f(1/n + · · · + 1/n) = nf(1/n), sof(1/n) = f(1)/n. Finally, let x = m/n with gcd(m,n) = 1 and m,n ≥ 1.Then f(m/n) = mf(1/n) = mf(1)/n, as desired.

Now suppose that f is a continuous function such that f(x) = g(x) on adense subset of R for a continuous function g. We claim that f ≡ g. Let ε > 0,and let x ∈ R. Choose δ > 0 such that |f(x)− f(y)| < ε and |g(x)− g(y)| < εwhenever |x−y| < δ. Now choose y ∈ (x−δ, x+δ) which lies in our prescribeddense set. Then

|f(x)− g(x)| ≤ |f(x)− f(y)|+ |g(x)− g(y)| < 2ε.

Since ε is arbitrary and x is fixed, it follows that f(x) = g(x) and so f ≡ g,as desired.


We then note that f(x) = γx on Q, which is dense in R, and that g(x) = γxis continuous. Therefore f ≡ g, as desired.

Question 1.11. Let (X, ‖·‖X) and (Y, ‖·‖Y ) be normed spaces, and let T : X → Ybe a linear map. Say that T is bounded if the quantity

‖T‖ = supx∈X‖x‖X≤1

‖Tx‖Y <∞

is finite.

(a) Prove that the following are equivalent:(1) T is continuous.(2) T is continuous at 0.(3) T is bounded.

(b) Let (Rn, ‖·‖2) denote the usual Euclidean space. A matrix A ∈ Rn×n givesrise to a linear map A : Rn → Rn in the usual way, so that the norm of A canbe defined as above by

‖A‖ = supx∈Rn‖x‖2≤1

‖Ax‖2 <∞.

(1) Let

D =

d1

d2

. . .dn

∈ Rn×n

be a diagonal matrix. Show that ‖D‖ = max|d1|, · · · , |dn|.(2) Let D be as in (1). Show that

‖D‖ = supx∈Rn‖x‖2≤1

|〈Dx,x〉|.

(3) Let U ∈ Rn×n be an orthogonal matrix, i.e., a matrix satisfying UTU = I,where UT denotes the transpose of U . Show that for every x ∈ Rn,‖Ux‖ = ‖x‖.

(4) Let A ∈ Rn×n be a matrix and let α denote the largest eigenvalue of ATA.

Show that ‖A‖ =√|α|.

(5) Compute ‖A‖ for

A =

[1 12 −1

].

Proof. (a) (1) implies (2): This is obvious.

(2) implies (3): Suppose that T is continuous at 0. Choose δ > 0 so thatfor all x ∈ X with ‖x‖X ≤ δ we have ‖Tx‖Y ≤ 1. Now for an arbitrary z ∈ Xwith ‖z‖X ≤ 1, write z = z′/δ. Then we see that ‖Tz′‖Y = δ‖Tz‖Y ≤ 1, sothat ‖Tz‖Y ≤ δ−1. Hence T is bounded, as desired.

10 STANLEY YAO XIAO

(3) implies (1): Suppose that T is bounded, say with bound M . Let x ∈ Xand let ε > 0. Suppose that x′ ∈ X be such that ‖x− x′‖X ≤ εM−1. Then

‖Tx− Tx′‖Y = ‖T (x− x′)‖Y = εM−1‖T (M(x− x′))‖Y ≤M−1M = ε,

so T is continuous.(b) Let (Rn, ‖·‖2) denote the usual Euclidean space. A matrix A ∈ Rn×n gives

rise to a linear map A : Rn → Rn in the usual way, so that the norm of A canbe defined as above by

‖A‖ = supx∈Rn‖x‖2≤1

‖Ax‖2 <∞.

(1) Let x = (x1, · · · , xn) ∈ Rn with ‖x‖2 ≤ 1. ThenDx = (d1x1, d2x2, · · · , dnxn),

so ‖Dx‖2 =√d2

1x21 + · · ·+ d2

nx2n. Let d = max|di|. Then

‖Dx‖2 ≤ dmax|xi|,

and this can be achieved by setting xi = 0 for all the indices not corre-sponding to d and xj = 1 if d = dj.

(2) We have

|〈Dx, x〉| = |d1x21 + · · ·+ dnx

2n|,

which upon setting xj = 1 again we see that ‖D‖ can be realized by theabove.

(3) We have

〈Ux, Ux〉 = 〈x, UTUx〉= 〈x, x〉

for all x.(4) Note that ATA is symmetric, so that it is orthogonally diagonalizable,

say by U . Now suppose that x is a vector such that ‖A‖ = ‖Ax‖2. Then

〈Ax,Ax〉 = 〈AUx′, AUx′〉= 〈UTATAUx′, x′〉,

where x = Ux′. By part (3), we know that x′ has the same norm asx. Moreover, by our choice of U we know that UTATAU = D for somediagonal matrix D. Now by part (1), we know that ‖A‖ = ‖D‖ = |α|, asdesired.

(5) The characteristic polynomial of A is (1 − x)(−1 − x) − 2 = x2 − 3, sothe largest eigenvalue is

√3. Hence ‖A‖ = 31/4, as desired.

2. Spring 2015

2.1. Part I.

Question 2.1. Let f(x) =

0 if x ∈ Q,1 if x 6∈ Q.

. Does∫ 1

0f(x)dx exist as a Riemann inte-

gral? Does it exist as a Lebesgue integral?


Proof. f(x) is not Riemann integrable, because for any partition P1, P2, · · · , PN of[0, 1], the infimum of f on Pj will be zero and the supremum will be 1, since Q isdense in R. From this it is clear that the lower sum will always be zero and the uppersum will always be 1, so it can never converge.

However, f is Lebesgue integrable because it is equivalent to the constant functionequal to 1, so they only differ on a set of measure zero.

Question 2.2. Suppose that f(x) is a continuous complex valued function on [0,∞)and limx→∞ f(x) = 0. Prove that f can be uniformly approximated on [0,∞) by a

sequence of functions of the form qn(x) =n∑k=1

ake−kx, where ak ∈ C.

Proof. We note that [0,∞) is locally compact, so the Stone-Weierstrass theoremasserts that any algebra which separates points and vanishes nowhere is dense inC0([0,∞),C). Our hypothesis on f allows us to conclude that f ∈ C0([0,∞,C), so itsuffices to check that the algebra generated by functions of the form e−kx separatespoints and vanishes nowhere. Both assertions are clear, since gk(x) = e−kx is injectiveon [0,∞) and does not vanish.

Question 2.3. Suppose that f is an L2 function on the unit disk D in C with respect

to planar Lebesgue measure. Suppose further that f(z) =∞∑n=0

anzn for z ∈ D. Prove

that ‖f‖22 =

∞∑n=0

π|an|2

n+ 1.

Proof. The second hypothesis implies that f is analytic in D, so the series represen-tation converges absolutely in D. Therefore the Fubini-Tonelli theorem applies, andhence

‖f‖22 =

∫ 1

0

∫ 2π

0

(∞∑n=0

anrneint

)(∞∑n=0

anrne−int

)dtrdr

=∞∑n=0

∫ 1

0

∫ 2π

0

|an|2r2n+1drdt+∞∑m=0

∞∑k=0

∫ 1

0

∫ 2π

0

amakrk+m+1ei(m−k)tdtdr

=∞∑n=0

∫ 1

0

∫ 2π

0

|an|2r2n+1drdt+ 0

=∞∑n=0

|an|2(2π)

2n+ 2

=∞∑n=0

π|an|2

n+ 1.

Question 2.4. Find an analytic function f(z) defined on z ∈ C : x > 0, wherez = x+ iy and x, y ∈ R, whose real part is u(x, y) = log(x2 + y2).

12 STANLEY YAO XIAO

Proof. Let f(z) = f(x, y) = u(x, y) + iv(x, y), so that f is analytic on the puncturedplane. Therefore, we require u, v to satisfy the Cauchy-Riemann equations, hence

∂u

∂x=∂v

∂y,∂u

∂y= −∂v

∂x.

Following through, we have∂v

∂y=

2x

x2 + y2

and∂v

∂x= − 2y

x2 + y2.

Taking the anti-derivative of the first, we see that

v(x, y) = 2 arctan(y/x) + C(y),

and by taking the anti-derivative of the second, we see that C(y) ≡ 0. Therefore, wehave

f(z) = f(x, y) = log(x2 + y2) + 2i arctan(y/x).

Question 2.5. How many roots (counting multiplicity) does f(z) = z7 + 5z3− z− 2have in the open unit disc?

Proof. By Rouche’s theorem, it suffices to find an analytic function g such that|f(z) + g(z)| < |f(z)| + |g(z)| for all z ∈ ∂D and such that it is clear how manyzeroes g has in D.

We choose g(z) = −5z3, which plainly has three roots in D. Then |f(z) + g(z)| =|z7 − z − 2| ≤ 2 + 1 + 1 = 4 on ∂D, and |f(z)|+ |g(z)| ≥ 5− 1− 1− 2 + 5 = 6. Thusg satisfies our requirements and f has three roots (with multiplicity) in D.

Question 2.6. Show that every infinite set is the disjoint union of countably infinitesubsets.

Proof.

Question 2.7. Let byc be the integer part of y, let An = x ∈ [0, 1]|b2nxc is even,

and let gn = χAn be the characteristic function of An. Prove that limn→∞

∫ 1

0

fgndx =

1

2

∫ 1

0

fdx for all f ∈ L1(0, 1).

Proof. We obtain an explicit description for the set An. Partition [0, 1] into

[0, 2−n), [2−n, 2−n−1), · · · , [j2−n, (j + 1)2−n), · · · .We index each interval by its left end point, so we obtain the intervals I0, · · · , I2n−1.If x ∈ Ij, then b2nxc = j. Therefore, An = I0 ∪ I2 ∪ · · · ∪ I2n−2. Now, let ε > 0.

Choose a simple function h(x) =∑N

j=1 ajχAj(x) where Aj is a sub-interval of [0, 1]for j = 1, · · · , N such that

‖f − h‖1 < ε/2.


This is possible because simple functions are dense in L1(0, 1). Now, we show thatfor all large n, we have

1

2

∫ 1

0

h(x)dx− ε

2<

∫ 1

0

hgndx <1

2

∫ 1

0

h(x)dx+ε

2.

To do so, it suffices to check the above for characteristic functions of some sub-intervalI = [a, b] ⊂ [0, 1]. It is easy to see that the error

1

2

∫ 1

0

χI(x)dx−∫An

χI(x)dx

is equal to the measure of at most two intervals of length 2−n covering the endpointsof [a, b], so is at most 2−n+1. Therefore, we can make the error as small as we wantby choosing n large enough.

Therefore, by linear extension, there exists N ′ ∈ N such that for all n ≥ N ′, wehave

1

2

∫ 1

0

h(x)dx− ε

2<

∫ 1

0

hgndx <1

2

∫ 1

0

h(x)dx+ε

2.

Now the result follows from Holder’s inequality. That is, we have∣∣∣∣∫ 1

0

(f − h)gndx

∣∣∣∣ ≤ ‖gn‖∞‖f − h‖1 <ε

2.

Question 2.8. Let X be a topological space and ∼ be an equivalence relation on X.Let X/ ∼ be the set of equivalence classes and π : X → (X/ ∼) be the projection.Define the quotient topology on X/ ∼ and prove that f : (X/ ∼)→ Y is continuous

if and only if f = f π is continuous.

Proof. The quotient topology is the one where the open sets are the subsets U of(X/ ∼) such that the union of all equivalence classes in U forms an open set in X.That is, the union ⋃

[a]∈U

[a]

is open in X.

Now suppose that f is continuous. Let V be an open set in Y , so that by conti-nuity, we have f−1(V ) is open in (X/ ∼). Observe that (f)−1(V ) consists of thoseelements in X which lie in some equivalence class [a] ∈ f−1(V ), which by the definition

of quotient topology, is open in X, so by the definition of continuity, f is continu-ous. Conversely, suppose that f is continuous. Again, by the definition of quotienttopology, it follows that f−1(V ) is open in (X/ ∼), whence f is continuous.

2.2. Part II.

Question 2.9. (A1) Let a > 0 and define f(t) = eat for −π ≤ t ≤ π.

(a) Find the Fourier series of f .

14 STANLEY YAO XIAO

(b) Use a computation of ‖f‖2 to evaluate the sum1

a2+ 2

∑n≥1

1

a2 + n2.

Proof. The n-th Fourier coefficient of f is given by

f(n) =

∫ π

−πeate−intdt,

and computing the integral, we obtain

f(n) =eπ(a−in)

a− in− eπ(in−a)

a− in.

If n = 2k, then we have

f(2k) =eπa − e−πa

a− inand if n = 2k + 1, then we have

f(2k + 1) =e−πa − eπa

a− in.

By Parseval’s theorem, it follows that

‖f‖22 =

∑|f(n)|2 = (e−πa − eπa)2

(1

a2+ 2

∑n≥1

1

a2 + n2

).

Explicitly, we have

‖f‖22 =

∫ π

−πe2atdt =

1

2a

(e2aπ − e−2aπ

),

whence1

a2+ 2

∑n≥1

1

a2 + n2=

e2aπ − e−2aπ

2a(eaπ − e−aπ)2.

Question 2.10. (A2) Prove that [0, 1] is not the disjoint union of a countably infinitecollection of non-empty closed sets An. Hint: consider X = [0, 1] \

⋃n≥1 int(An).

Question 2.11. (B1) Let Ω be a simply connected domain properly contained in C,and let z0 ∈ Ω. Suppose that f is holomorphic on Ω, f(Ω) ⊂ Ω and f(z0) = z0.

(a) Prove that |f ′(z0)| ≤ 1.(b) What more can be said when |f ′(z0)| = 1?

Proof. (a) By the Riemann mapping theorem, there exists a conformal mappingg : D → Ω which send 0 to z0. Here D refers to the open unit disk. Leth = g−1 f g. Then h is a conformal self-map of D satisfying h(0) = 0,that is, h satisfies Schwarz’s lemma, so that |h′(0)| ≤ 1. By the chain rule, itfollows that

f ′(z0) = (g(h(g−1)))′(z0)

= g′(h(g−1)(z0))h′(g−1(z0))/g′(g−1(z0))

= g′(h(0))h′(0)/g′(h0)

= h′(0),


hence |f ′(z0)| ≤ 1 as desired.(b) By Schwarz’s lemma again and the above, it follows that |h′(0)| = 1, so that

h(z) = αz for some α ∈ C of norm 1. Therefore, f(z) = g(αg−1(z)).

Question 2.12. (B2)

(a) For which real a does

∫ ∞−∞

cosx

a2 − x2dx make sense as an improper Riemann

integral?(b) Evaluate this integral for those values of a.(c) What meaning can be given to the formula you obtained in (b) for other values

of a?

Proof. (a) If the integrand has a pole on the real line, then the improper integralcannot converge. Therefore, a must be an odd multiple of π/2. We prove thatall such values work.

Let a = (2k+1)π/2. We may assume that a is positive. Then (cosx)(a2−x2)−1

has no poles on the real line, since cos x is analytic over C and vanishes at±a. Let B1, B2 be two real numbers such that B1 < −a and B2 > a. Then,on A = R \ [B1, B2], we have∣∣∣∣∫

A

cosx

a2 − x2dx

∣∣∣∣ ≤ ∫A

1

x2 − a2dx

∫ ∞1

1

x2dx <∞.

Now,

∫ B2

B1

cosx

a2 − x2dx exists and is finite since the integrand is continuous and

the range of integration is an interval of finite length.(b) Suppose that a = (2k + 1)π/2 for some k ∈ N ∪ 0. Then the function

g(x) =cosx

a2 − x2

is continuous on the real line. Now define

f(z) =eiz

a2 − z2.

This function has two simple poles on the real line, namely z = ±a. Weconsider the contour ΓR,ε consisting of the following pieces:(i) The set ΓR

R,ε[−R,−a− ε] ∪ [−a+ ε, a− ε] ∪ [a+ ε, R];(ii) The half circles γε(±a) centered at ±a of radius ε; and

(iii) The half circle ΓR(0) centered at the origin.Then f is analytic on the interior of ΓR,ε for R large and ε small, therefore∫

ΓR,ε

f(z)dz = 0.

The contribution to ΓR,ε from ΓR(0) is negligible by Jordan’s lemma. There-fore, it suffices to calculate the contribution from γε(±a).

16 STANLEY YAO XIAO

By the residue theorem and the simplicity of the zeroes at ±a, we have that∫γε(a)

f(z)dz = πiResz=a f(z)

=πiei(2k+1)/2

2a

=(−1)k+1π

(2k + 1)π

=(−1)k+1

2k + 1,

and similarly, ∫γε(−a)

f(z)dz =(−1)k+1

2k + 1.

By taking limits as ε→ 0 and R→∞, we see that∫ ∞−∞

cosx

a2 − x2= (−1)k+1 2

2k + 1.

(c) The computation can still be made to obtain the principal value of the integral.

Question 2.13. (C1) Find limn→∞

∫ n

0

(1 +

x

n

)ne−2xdx. Justify your arguments

carefully.

We begin with the observation that(1 +

x

n

)n= exp (n log(1 + x/n)) .

Since 0 ≤ x ≤ n, it follows that

log(1 + x/n) =x

n− x2

2n2+ · · · .

The sequence x/n, x2/2n2, · · · is positive and decreasing, hence

x

n− x2

2n2+ · · · ≤ x

n.

Therefore, (1 +

x

n

)n= exp (n log(1 + x/n)) ≤ ex.

In fact, ex is the limit function of (1 + x/n)n. Let χn denote the indicatorfunction on the interval [0, n]. Then the function χn(1 + x/n)ne−2x is domi-nated from above, in lieu of the arguments in the previous paragraphs, by thefunction e−x. Hence, by the dominated convergence theorem, it follows that

limn→∞

∫ n

0

(1 +

x

n

)ne−2xdx =

∫ ∞0

e−xdx = 1.

Question 2.14. (C2) Let I = [0, 1] with Lebesgue measure, and let p ∈[1,∞). Consider a sequence fk ⊂ Lp(I) with ‖fk‖p ≤ 1. Suppose thatf(x) = lim

k→∞fk(x) exists for almost every x. Does f belong to Lp(I)? Prove it

or give a counterexample.


Proof. Yes, the assertion follows from the dominated convergence theorem andthe observation that the function f(x) = 1(x), the function which is identicallyone, is integrable on [0, 1] for all p ≥ 1.

Question 2.15. (D1) Put the lexiographic order on X = [0, 1]2 defined by

(x1, y1) < (x2, y2) if x1 < x2 or x1 = x2 and y1 < y2.

Let T be the order topology generated by the sets

(x, y) : (x1, y1) < (x, y) and (x, y) : (x, y) < (x2, y2).

(a) Show that every subset of X has a least upper bound.(b) What is the induced topology on Y =

(x, y) : y = 1

2

?

(c) What is the closure of Y ?

Proof. (a) Let A ⊂ X, and let α = sup(x,y)∈A

x. If α < 1, then (α, 0) is the least

upper bound for A. If α = 1, then set sup(x,y)∈A

y, and we have that (1, β) is the

least upper bound for A.(b) Notice that for u = (u, 1/2),v = (v, 1/2) ∈ Y , u < v if and only if u < v.

Hence, the lexiographic order induces the typical ordering of the real line onY .

(c)

3. Spring 2014

Question 3.1. Let X and Y be topological spaces, and let f : X → Y be afunction.

(a) Define the closure (denoted in what follows as “cl(A)”) of a subset A ⊂ X.(b) Define what it means for f to be continuous.(c) Suppose that f is continuous. Prove that f(cl(A)) ⊂ cl(f(A)), for all A ⊂ X.(d) Suppose that f(cl(A)) ⊂ cl(f(A)) for all A ⊂ X. Does it follow that f is

continuous? Justify your answer (proof or counter-example).

Proof. (a) For a given set A ⊂ X, the closure cl(A) is equal to the intersection ofall closed sets A containing A. In other words,

cl(A) =⋂

A⊂A⊂XA closed

A.

(b) f is continuous if and only if for all closed subsets B of Y , we have f−1(B) isa closed subset of X.

(c) Let B = cl(f(A)). By definition, B is a closed subset of Y . Since f iscontinuous, it follows that f−1(B) is closed in X. Further, since f(A) ⊂cl(f(A)), it follows that A = f−1(f(A)) ⊂ f−1(cl(f(A)) = f−1(B), hencecl(A) ⊂ f−1(B). This implies that f(cl(A)) ⊂ f(f−1(B)) = B, as desired.

18 STANLEY YAO XIAO

(d) Let B ⊂ Y be a closed set. Write A = cl(f−1(B)), which by definition is aclosed subset of X. By our hypothesis, we have f(A) = f(cl(A)) ⊂ cl(f(A)).Observe that cl(f(A)) = cl(f(f−1(B)) = cl(B) = B. Therefore, A ⊂ f−1(B).Trivially, f−1(B) ⊂ A since the latter is defined as the closure of the former,so A = f−1(B). This implies that f−1(B) is closed. By the definition ofcontinuity, it follows that f is continuous.

Question 3.2. (a) Evaluate the improper integral∫ ∞0

sinx

x(x2 + 9)dx.

Simplify your answer so that it is expressed in terms of real quantities.(b) Prove that every root of the polynomial f(z) = z6−5z2+10 lies in the annulus

1 < |z| < 2.(c) Provide a statement for any major theorem used in your solutions in parts (a)

or (b).

Proof. (a) Note that the integrand is an even function, so the integral is equal tohalf of the integral over the entire real line. We consider the contour integral∫

Γδ,R

eiz

z(z2 + 9)dz,

where Γδ,R is the contour consisting of the following parts: the union of theintervals [−R,−δ)∪(δ, R], the half-circle of radius δ centered at 0, and the half-circle of radius R centered at 0. We label these as C1, C2, C3 respectively, withthe understanding that these paths depend on δ, R. By the residue theorem,we have ∫

Γδ,R

eiz

z(z2 + 9)dz = 2πiRes(f ; 3i),

as z = 3i is the only pole of f(z) = eiz/(z(z2 + 9)) inside Γδ,R. We easilyevaluate this to be

Res(f ; 3i) =ei(3i)

3i(3i+ 3i)=−1

18e.

We now analyze the contribution from C2 = C2(δ), consisting of a small half-circle of radius δ centered at 0. Observe that the contour is clockwise on C2;so our contribution would be negative. In this case, we have∫

C2

eiz

z(z2 + 9)dz = −πiRes(f ; 0),

where the negative sign comes from the orientation of the contour and thefactor of 2 is removed because we are only integrating along half of the circle.Evaluating, we see that∫

C2

f(z)dz = −πi(e0/(02 + 9)) = −πi9.


Now we analyze the contribution from C3 = C3(R), consisting of the half-circle of radius R centered at 0. For this part, we apply Jordan’s lemma toobtain that ∣∣∣∣∫

C3(R)

f(z)dz

∣∣∣∣ ≤ πMR,

where MR is the maximum taken by 1/z(z2 + 9) on C3(R). Plainly, we have

1

|z(z2 + 9)|≤ 1

R(R2 − 9),

hence ∣∣∣∣∫C3(R)

f(z)dz

∣∣∣∣ ≤ π

R(R2 − 9)→R→∞ 0,

hence it follows that

− π

9e=

∫Γδ,R

f(z)dz =

(∫C1

+

∫C2

+

∫C3

)f(z)dz =

∫C1

f(z)dz − πi

9+ 0,

as R→∞. By sending δ to 0 and taking imaginary parts, we see that∫ ∞0

sinx

x(x2 + 9)dx =

π

18− π

18e.

(b) Our aim is to apply Rouche’s theorem, which asserts that for analytic functionsf, g on some simply connected open set Ω, that f, g have the same number ofzeroes if the inequality

|f(z) + g(z)| < |f(z)|+ |g(z)|.

We first consider the roots of f inside the disk D = z ∈ C : |z| < 1. Forthis part, set g(z) = −z6 − 10, which plainly has no zeroes inside D. Observethat on ∂D, we have

|f(z) + g(z)| = |5z2| = 5,

and

|f(z)|+ |g(z)| ≥ 10− |5z2| − |z6|+ 10− |z6| ≥ 13 > 5,

so Rouche’s theorem applies; hence f has no zeroes inside the disk D.

Now consider the disk D′ = z ∈ C : |z| < 2. Now we choose g(z) = −z6,which plainly has all of its roots inside D′. Then we have

|f(z) + g(z)| = |5z2 + 10| ≤ 20 + 10 = 30,

while

|f(z)|+ |g(z)| ≥ |z6| − |5z2| − 10 + |z6| ≥ 64− 20− 10 + 64 > 30,

so Rouche’s theorem again applies and all of the roots of f lie inside D′.Combining these estimates and using the fundamental theorem of algebra toobtain that f has exactly 6 complex roots, it follows that all of the root of flie inside the original annulus, as desired.

20 STANLEY YAO XIAO

(c) The theorems used are the following:

Residue Theorem: Let C be a closed contour and f is a complex meromorphicfunction. Then we have∫

C

f(z)dz = 2πi∑

Res(f ; ak),

where the ak’s are poles of f in the interior of C.

Rouche’s Theorem: Let Ω be a simply connected open set, and f, g are com-plex analytic functions. The number of roots of f in Ω is equal to the numberof roots of g in Ω if the inequality

|f(z) + g(z)| < |f(z)|+ |g(z)|holds identically on ∂Ω.

Fundamental Theorem of Algebra: Let f be a polynomial with complex coef-ficients of degree d. Then f has exactly d roots in C, counting multiplicity.

Question 3.3. Consider the space X = C([0, 1],R) of continuous functions from[0, 1] to R. Let L : X → R be a function which has the following properties:

• L(αf + βg) = αL(f) + βL(g), for all f, g ∈ X and all α, β ∈ R.• L(f · g) = L(f) · L(g), for all f, g ∈ X.• L(1) = 1, where 1 : [0, 1]→ R is the function constantly equal to 1.• L(F ) = 2/3, where F : [0, 1]→ R is defined by F (t) = t, 0 ≤ t ≤ 1.

(a) Let f, g ∈ X be such that f(t) ≤ g(t) for every t ∈ [0, 1]. Prove that L(f) ≤L(g).

(b) Prove that |L(f)| ≤ ‖f‖∞, for all f ∈ X.(c) Prove that L(f) = f(2/3) for every f ∈ X.(d) Provide a complete statement of any major theorem used in your solution to

(c).

Proof. (a) The map L is a linear functional on X, and since X is compact, bythe Riesz representation theorem there exists a Borel measure µ on [0, 1] suchthat

L(f) =

∫ 1

0

f(x)dµ(x)

for all f ∈ X. For f = 1, we have∫ 1

0

1dµ(x) = 1,

and for F , we have ∫ 1

0

xdµ(x) = 2/3.

Now let f, g be as in the problem, and let h = g − f . By hypothesis, h isnon-negative on [0, 1]. It therefore suffices to show that for all non-negativeh ∈ X, we have L(h) ≥ 0. Observe that such h is the square of another


element in X; that is, if h is a non-negative continuous real-valued function,then its square root is also continuous. Therefore, by the second condition,we have

L(h) =(L(√h)2

≥ 0,

as desired.(b) We have, by part (a),

|L(f)| =∣∣∣∣∫ 1

0

f(x)dµ(x)

∣∣∣∣≤∫ 1

0

|f(x)|dµ(x)

≤∫ 1

0

‖f‖∞dµ(x)

= ‖f‖∞,as desired.

(c) By a standard exercise in the theory of Banach spaces, part (b) implies thatL is continuous, since it is bounded (indeed, the exercise says that a linearoperator on a Banach space is continuous if and only if it is bounded). By theStone-Weierstrass theorem, the polynomials are dense in X, and by conditions3 and 4, it follows that for all polynomials P , we have L(P ) = P (2/3). (c) thenfollows by the fact that the polynomials are dense in X and L is continuous.

(d) We used the following theorems:

Riesz Representation Theorem: Suppose that L is a non-negative linear func-tional on a locally compact Hausdorff space X. Then there exists a non-negative Borel measure µ on X such that

L(f) =

∫X

f(x)dµ(x).

Stone-Weierstrass theorem: Suppose that A ⊂ C[0, 1] is an algebra which con-tains the constant functions, separates points, and vanishes nowhere. Then Ais dense in C[0, 1].

Theorem ins Banach Space Theory: Let X be a Banach space, and L a linearoperator on X. Then the following are equivalent:(i) L is bounded;(ii) L is continuous at the origin; and

(iii) L is continuous.

Question 3.4. Let D be the open unit disc in C.

(a) Fix a ∈ D. Construct a conformal mapping ϕ : D→ D such that ϕ(a) = 0.(b) Let f be a holomorphic function on D such that |f(z)| < 1 for all z ∈ D.

Suppose that f has two distinct fixed points. Prove that f(z) = z for all z inD.

(c) Provide a statement of any major theorems used in your solution to (b).

22 STANLEY YAO XIAO

Proof. (a) Consider the map ϕa(z) =z − a1− az

, where w denotes the complex con-

jugate of w. Then it is clear that ϕa(a) = 0. Further, since a ∈ D, it followsthat |a| = |a| < 1, so ϕa does not have any poles in the closure of D. Itremains to show that ϕa(D) = D. It suffices to show, by the analyticity ofϕa in D and the maximum modulus principle, that ϕa(∂D) = ∂D. To wit, letz = eit, t ∈ R, be a point on the boundary of D. Further, write a = a1 + ia2.Then

ϕa(eit)ϕa(eit) =

eit − a1− aeit

e−it − a1− ae−it

=1− aeit − ae−it + |a|2

1− ae−it − aeit + |a|2= 1,

so |ϕa(z)| = 1 for all z ∈ ∂D. Hence, ϕa is a conformal self-map of D with thedesired properties.

(b) Since |f(z)| < 1 for all z ∈ D and f is analytic on D, it follows that f is aconformal self-map of the unit disk. Let z1 be a fixed point of f , and considerthe map

g = ϕz1 f ϕ−1z1.

Note that g is a conformal self map of the disc that satisfies g(0) = 0, so theSchwarz lemma applies to g. In particular, if z2 is a fixed point of f , thenϕz1(z2) is a fixed point of g. Now, we note that z2 6= z1 implies that g(z′) = z′

for some z′ ∈ D, so by Schwarz’s lemma, it follows that g(z) = wz for somefixed w ∈ ∂D. This implies that f(z) = z on D.

(c) The theorem used in part (b) is:

Schwarz’s Lemma: Let f : D → D be a conformal self-map of D, such thatf(0) = 0. Then it follows that |f(z)| ≤ |z| for all z ∈ D and |f ′(0)| ≤ 1.Further, if |f(z)| = |z| for some z ∈ D or if |f ′(0)| = 1, then f = αz for somecomplex α of norm one.

Question 3.5. Let X be a topological space. We consider the following propertiesthat X may have:

• If every open cover of X has a countable sub-cover, then X is called a Lindelofspace.• If every countable open cover of X has a finite sub-cover, then X is called

countably compact.

(a) State what it means for X to be second countable.(b) Prove Lindelof ’s lemma: If X is second countable, then X is a Lindelof space.(c) Suppose that X is a Lindelof space. Prove that X is compact if and only if it

is countably compact.

Proof. (a) A toplogical space X is second countable if there exists a countablecollection Unn≥1 of open sets such that every open set in X can be writtenas the union of some sub-family of the Un’s.


(b) Suppose X is a second countable topological space, and let B be an opencover of X. For each x ∈ X, there exists Bx ∈ B such that x ∈ Bx, by thedefinition of open cover. Now, since Un is a base, there exists Ux ⊂ Bx

which contains x. Then the collection of Ux covers X, and since these areelements of a countable family, there are at most countably many of them,say Uj1 ,Uj2 , · · · . Then the sets Bj1 , Bj2 , · · · are elements of B and a countablesub-cover.

(c) Suppose that X is Lindelof, so that any open cover has a countable sub-cover. Now suppose that X is compact. Then it is trivial that X is countablycompact, so it suffices to prove the converse. Therefore, suppose that X iscountably compact. Since X is Lindelof, every open cover B contains a count-able sub-cover; and since it is countably compact, every countable cover hasa finite sub-cover, therefore X is compact.

Question 3.6. Let µ be the Lebesgue measure on the interval [−1, 1]. Suppose thatfor every n ∈ N we have a non-negative Borel function fn : [−1, 1]→ R such that

(i)∫fndµ = 1, and

(ii) fn(x) = 0 for every x ∈ [−1, 1] such that |x| ≥ 1/n.

(a) Prove that the sequence (fn)∞n=1 does not have any convergent subsequence inthe Banach space (L1(µ), ‖·‖1).

(b) Let g : [−1, 1] → R be continuous, and suppose that g(0) = 14. Give a briefexplanation of why fng is integrable for every n ∈ N, and prove that thesequence of integrals

(∫fngdµ

)∞n=1

is convergent. What is the limit of thissequence of integrals?

(c) Provide a complete statement of any major theorem you may have used inyour solutions to parts (a) or (b).

Proof. (a) Let fnk∞k=1 be a subsequence of fn, and suppose that f is the limitof fnk in L1(µ) with respect to L1 norm. In other words, we must have

limn→∞

∣∣∣∣∫ 1

−1

(f − fn)dµ

∣∣∣∣ = 0.

Therefore, it follows that ∫ 1

−1

fdµ = 1.

Since the fn’s are non-negative, it follows that f is non-negative on [−1, 1].Now, the above equation implies that there exists a subset A of [−1, 1] withpositive Lebesgue measure for which f > 0 on A. Further, since [−1, 1] iscompact, f must be essentially bounded on A, with essential supremum F ,say. Then

∫ 1

−1fdµ =

∫Afdµ. However, as n approaches infinity, we have µ(A∩

[−1/n, 1/n]) tending to zero, and so f must have zero integral, a contradiction.Therefore fn does not contain a convergent subsequence.

(b) Since g is continuous, it can be well-approximated by simple functions (that is,finite linear combinations of indicator functions), and thus fng is measurable.Integrality follows from the fact that g is continuous on a compact set, so it

24 STANLEY YAO XIAO

is bounded.

Let ε > 0, and choose δ > 0 so that |g(x) − g(0)| < ε for all |x| < δ.Now choose N ∈ N sufficiently large so that 1/N < δ. Then for n ≥ N , wehave ∫ 1

−1

fngdµ =

∫ 1/n

−1/n

fngdµ,

and by construction, we have

−ε∫ 1/n

−1/n

fndµ <

∫ 1/n

−1/n

fn(x)(g(x)− 14)dµ < ε

∫ 1/n

−1/n

fndµ.

By our hypothesis on fn and g, it follows that

limn→∞

∫ 1

−1

fngdµ = 14.

(c)

Question 3.7. Consider an infinite dimensional normed vector space (X, ‖·‖).(a) Let Y be a linear subspace of X, such that Y 6= X. Prove that int(Y ) = ∅,

where int(Y ) denotes the interior of Y .(b) Consider the following (true) statement: “If Y is a finite dimensional subspace

of X, then cl(Y ) = Y .” Write one sentence giving the idea of proof of thisstatement.

(c) In this part of the question we assume that (X, ‖·‖) is a Banach space. LetS be a subset of X such that span(S) = X, where span(S) denotes the setof all finite linear combinations of vectors from S. Prove that the set S isuncountable.

(d) Provide a complete statement of any major theorem used in your solution to(c).

(e) Show by example that the conclusion of part (c) no longer holds if we dropthe assumption that the normed vector space (X, ‖·‖) is a Banach space.

Proof. (a) Let y ∈ Y be a point, and let y′ be a vector not in Y . Note that y′

necessarily exists because X 6= Y . Let V be an open set in X containing y.Choose ε > 0 so that B(y, ε) ⊂ V . Now, the vector y + (ε/2)y′ ∈ B(y, ε) \ Y ,and therefore B(y, ε) is not contained in Y . Since y, V were arbitrary, itfollows that Y has empty interior.

(b) Let y1, · · · , yn be a basis of Y , and yN be a sequence in Y which convergesin X. Then it can be checked easily that the coordinates of the yN ’s convergecoordinate wise (with respect to y1, · · · , yn, and hence Y is closed.

(c) Let S be an at most countable subset of X. Then the span of S is countable.It therefore suffices to check that an at most countable normed space is notcomplete. This follows easily from Cantor’s diagonalization argument.

(d)(e) Take, for example, the algebraic closure of Q viewed as a vector space over Q.


Question 3.8. ] Let D be a bounded open set in C with closure cl(D). Let f :cl(D) → C be a continuous function such that the restriction f |D : D → C isholomorphic.

(a) Prove the minimum modulus principle: If f is nowhere zero on D, then theminimum of |f(z)| on cl(D) is attained on the boundary ∂D.

(b) Show by example that the conclusion of part (a) no longer holds if we dropthe assumption that f is nowhere zero on D.

(c) Let f be non-constant. Prove that if |f(z)| is constant on ∂D, then f musthave at least one zero in D.

Proof.(a) We can take for granted the maximum modulus principle, which asserts that for fanalytic on D, the maximum value of |f(z)| is attained on ∂D. Since f is non-zero onD, it follows that 1/f(z) is analytic on D, hence maximum modulus principle appliesto 1/f(z), which is equivalent to the minimum modulus principle for f .

(b) Take D = D, the unit disc, and f(z) = z.(c) Suppose that f has no zeroes in D and is constant on ∂D, say f(z) = K for allz ∈ ∂D. The same must apply to 1/f , since 1/f is analytic as well. By the minimumand maximum modulus principles, it follows that |K| ≤ |f(z)| ≤ |K| for all z ∈ D,so f is constant. Therefore, f must have a zero in D.

Question 3.9. Let (fn)∞n=1 be a sequence of functions from [−1, 1] to R.

(a) Define what it means for the sequence (fn)∞n=1 to be equicontinuous.(b) Suppose that every fn is differentiable with |f ′n(t)| ≤ 1 for every t ∈ [−1, 1].

Prove that (fn)∞n=1 is equicontinuous.(c) Let (fn)∞n=1 be the same as in part (b), and assume in addition that |fn(0)| ≤ 1

for every n ∈ N. Prove that there exist a continuous function f : [−1, 1]→ Rand indices 1 ≤ n(1) < n(2) < · · · < n(k) < · · · such that the subsequence(fn(k))

∞k=1 converges uniformly to f .

(d) Provide a complete statement of any major theorem used in your solution to(c).

(e) In the framework of part (c), show by example that the limit function f maynot be differentiable.

Proof. (a) fn is equicontinuous if for all ε > 0, there exists δ > 0 such that forall |x− y| < δ and all n ∈ N, we have |fn(x)− fn(y)| < ε.

(b) Let ε > 0. Then for any n ∈ N and |x− y| < ε, we have∣∣∣∣fn(x)− fn(y)

x− y

∣∣∣∣ = |f ′n(c)| ≤ 1,

whence

|fn(x)− fn(y)| ≤ |x− y| < ε.

This estimate is independent of x, y and n, so fn is equicontinuous.(c) By the Arzela-Ascoli theorem, fn contains a uniformly convergent subse-

quence if fn is uniformly bounded and equicontinuous. We have alreadyverified that fn is equicontinuous, so it suffices to check that it is uniformly

26 STANLEY YAO XIAO

bounded. For any 0 6= x ∈ [−1, 1], we have∣∣∣∣fn(x)− fn(0)

x− 0

∣∣∣∣ = f ′n(c)

for some c ∈ (0, x), by the Mean Value Theorem. Therefore, we have

|fn(x)| ≤ 1 + |f ′n(c)||x| ≤ 1 + 1 = 2.

Since this bound is independent of x and n, it follows that fn is uniformlybounded, so we are done by Arzela-Ascoli.

(d) We used the following theorems:

Arzela-Ascoli Theorem: Let fn be a sequence of continuous functions whichare uniformly bounded and equicontinuous. Then fn contains a subsequencewhich is uniformly convergent.

Mean Value Theorem: Suppose f : [a, b] → R is continuous on [a, b] anddifferentiable on (a, b). There for a < u < v < b, there exists u < c < v suchthat

f(v)− f(u)

v − u= f ′(c).

(e)

4. Spring 2013

Question 4.1. Let l∞ denote the space of bounded R-valued sequences with norm‖(xn)‖∞ = supn |xn|. a) Show that l∞ is a complete metric space, and that it isnon-separable. b) Show that B = (xn) ∈ l∞ : ‖(xn)‖∞ ≤ 1 is not compact.

Proof. a) Let (x(m)n )m≥0 be a Cauchy sequence in l∞. Thus for any ε > 0 there

exists M ∈ N such that for all m, p > M we have

‖(x(m)n )− (x(p)

n )‖∞ < ε.

In particular, we have

supn|x(m)n − x(p)

n | < ε,

and hence for all n ≥ 0, we have

|x(m)n − x(p)

n | < ε.

This implies that for any fixed n, the sequence (x(m)n )m≥1 is Cauchy and hence con-

verges to some xn ∈ R, since R is complete. Now consider the element (xn). I claim

that this is the limit of the sequence (x(m)n m≥0 ⊂ l∞. Now suppose that (x(m)

n )does not converge to (xn). Then there exists ε > 0 such that for all N ∈ N, there

exists m > N such that ‖(x(m)n ) − (xn)‖ ≥ 4ε. Since (x(m)

n is Cauchy, there ex-

ists M ∈ N such that for all m, p > M we have ‖(x(m)n ) − (x

(p)n )‖ < ε. Now choose

m, p > maxN,M, so we have simultaneously

‖(x(m)n )− (xn)‖ ≥ 4ε


and

‖(x(m)n )− (x(p)

n )‖ < ε.

The first inequality implies that there exists n ∈ N such that |x(m)n − xn| ≥ 2ε, and

for this n, the second inequality implies that for all p > m, we have |x(m)n − x(p)

n | < ε.Combining these we conclude that

|x(p)n − xn| ≥ |x(m)

n − xn| − |x(m)n − x(p)

n | > ε

for all p > m, which contradictions the choice of xn as the limit of x(k)n k≥0. This

shows that (xn) is the limit of (x(m)n , as desired.

Now suppose that (x(m)n m≥0 ⊂ l∞ is any countable subset. Then construct the

element (xn) ∈ l∞ by setting xj = x(j)j + 1 for all j ≥ 1. Then

supn|x(m)n − xn| ≥ |x(m)

m − xm| = 1,

so no subsequence of (x(m)n can converge to (xn). This shows that (x(m)

n is notdense in l∞, and thus, l∞ is not separable.

b) Since l∞ is a metric space, it suffices to check that B is not sequentially com-

pact. Consider the sequence (x(m)n defined by x

(m)m = 1 and x

(m)n = 0 for n 6= m,

for all m ∈ N. Then it is clear that (x(m)n ⊂ l∞, and that this sequence does not

contain a convergent subsequence, since it is not Cauchy. Hence B is not sequentiallycompact and thus not compact by the Borel-Lebesgue theorem.

Question 4.2. A set E ⊂ Rn is called midpoint convex if, for any x, y in E,12(x+ y) ∈ E as well.

a) Give an example of a subset E ⊂ Rn which is midpoint convex, but not con-vex.

b) Suppose E ⊂ Rn is closed. Show that if E is midpoint convex, then it is con-vex.

c) Suppose E ⊂ Rn is open. Show that if E is midpoint convex, then it is con-vex.

Proof. a) Let E = Qn, the set of rational points in Rn. Then I claim that E is mid-point convex but not convex. E is midpoint convex because Q is midpoint convex.Indeed, suppose a/b, x/y ∈ Q where a, x ∈ Z and b, y ∈ N. Then (a/b + x/y)/2 =(ay + bx)/2xy ∈ Q. Hence E is midpoint convex. Now choose λ such that 0 < λ < 1and λ irrational. Now choose the vectors u = (1, 0, · · · , 0) and v = (2, 0, · · · , 0).Then λu + (1− λ)v = (2− λ, 0, · · · , 0), which is not in E by the choice of λ. HenceE is not convex.

b) Suppose that E is closed and midpoint convex. We may suppose that E is non-empty. If E contains only one point, then it is convex vacuously, thus suppose there

28 STANLEY YAO XIAO

exist two distinct points x,y ∈ E. Now, I claim that E contains the set

D =

m

2nx +

2n −m2n

y : 0 ≤ m ≤ 2n, n ∈ Z≥0

.

We do this by strong induction. The case n = 0 is trivial, since we know that x,y ∈ E.The case n = 1 follows from the definition of midpoint convexity. Now suppose that

for some k ∈ N, we have that for all 0 ≤ j ≤ k, thatm

2jx +

2j −m2j

y ∈ E, with

0 ≤ m ≤ 2j. Now for k + 1, consider an element of the form

2m+ 1

2k+1x +

2k+1 − 2m− 1

2k+1y,

where the numerator is chosen to be odd otherwise we would be done already. Wecan rewrite the above as

1

2

(m

2kx +

2k −m2k

y +m+ 1

2kx +

2k −m− 1

2ky

).

Since we have 2m + 1 ≤ 2k+1, it follows that 2m ≤ 2k+1 − 1 and so m ≤ 2k − 1.

Hence m+ 1 ≤ 2k, and that bothm

2kx +

2k −m2k

y andm+ 1

2kx +

2k −m− 1

2ky lie in

E. Hence by midpoint convexity, we have that

2m+ 1

2k+1x +

2k+1 − 2m− 1

2k+1y ∈ E

as well. This shows that D ⊂ E.

Now choose λ such that 0 < λ < 1. Since the dyadic rationals are dense in R,for all ε > 0 there exists m/2k such that |m/2k − λ| < ε. Thus

max1≤j≤n

∣∣∣(λ− m

2k

)xj +

((1− λ)−

(1− m

2k

))yj

∣∣∣ < ε max1≤j≤n

|xj − yj|,

and hence λx + (1 − λ)y is a limit point of E. Since E is closed, it follows thatx + (1− λ)y ∈ E and so E is convex.

c) Let 0 < λ < 1 be given, and set u = λx + (1 − λ)y. Since E is open, thereexist δ > 0 such that B(x, δ), B(y, δ) ⊂ E. Now choose a dyadic rational m/2k sothat 1− δ/2 < |λ2k/m| < 1 + δ/2. Then

λ2k

mx ∈ B(x, δ),

(1− λ)2k

2k −my ∈ B(y, δ).

Then by the fact that D ⊂ E, it follows that

m

2kλ

2k

mx +

(1− m

2k

)(1− λ)

2k

my ∈ E,

which shows that u ∈ E, so E is convex as desired.


Question 4.3. Show that any non-empty open set in a separable metric space (X, d)is the union of a countable family of open balls.

Proof. Let K be a countable dense subset, and let E be a non-empty open subset ofX. Since K is dense, it follows that there exists x ∈ E ∩K. Since E is open, for eachelement x ∈ E ∩K, there exists δx > 0 such that B(x, δx) ⊂ E. For each x ∈ K ∩E,choose the maximal such δx. Since K is countable, we must have E ∩K is countableas well. Now consider the union ⋃

x∈E∩K

B(x, δx).

Clearly, this union is a subset of E. It therefore suffices to show that

E ⊂ U =⋃

x∈E∩K

B(x, δx).

Now let y ∈ E. Since E is open, there exists ε > 0 such that B(y, ε) ⊂ E. Now, byreducing ε if necessary, we can assume that the distance from y to the boundary ofE is at least 3ε. In particular, by the choice of δx, we see that δx ≥ 2ε. In particular,we must have y ∈ B(x, δx). Thus E ⊂ U , as desired.

Question 4.4. Let C[0, 1] denote the space of continuous C-valued functions onthe interval [0, 1] with uniform norm. Let A be the subset of all polynomials withp(0) = p(1).

a) Prove that A is dense in f ∈ C[0, 1] : f(0) = f(1).

b) Let f(t) = |t − 1/2|. Show that any sequence (pn) of elements of A, converginguniformly to f , necessarily has limn→∞ deg pn =∞. Here deg p denotes the degree ofthe polynomial p.

Proof. a) Let S = f ∈ C[0, 1] : f(0) = f(1). Then S is a vector subspace of C[0, 1],and in particular, includes all constant functions on [0, 1]. Thus we may concernourselves only with the case f(0) = f(1) = 0.

Let ε > 0. By the theorem of Weierstrass, we may find a polynomial p such that

sup0≤t≤1

|f(t)− p(t)| < ε

3.

Then it follows that |f(0) − p(0)| = |p(0)| < ε/3, and likewise, |p(1)| < ε/3. Nowconsider the polynomial h(t) = p(t)− (1− t)p(0)− tp(1). Then h(0) = h(1) = 0, andwe have

|f(t)− h(t)| ≤ |f(t)− p(t)|+ |(1− t)p(0)|+ |tp(1)| < ε

3+ε

3+ε

3= ε,

hence A is dense in S, as required.

b) Let Pn denote the set of polynomials of degree at most n

Question 4.5. Let m be Lebesgue measure and L1[0, 1] = L1([0, 1],m) .

a) Let ε > 0 and f ∈ L1[0, 1]. Prove that there exists δ > 0 such that∫A|f |dm < ε

30 STANLEY YAO XIAO

whenever m(A) < δ.

b) Suppose fn, f ∈ L1[0, 1], fn ≥ 0, fn → f pointwise, and∫

[0,1]fn →

∫[0,1]

f . Prove

that∫Efn →

∫Ef for each measurable E ⊂ [0, 1].

Proof. a) We further abbreviate L1[0, 1] by L1. For any fixed positive integer n, let

us write An = x ∈ [0, 1] : |f(x)| ≥ n. If limn→∞

∫An

|f |dm = 0, then we may argue as

follows. First choose N sufficiently large so that∫An|f |dm < ε/2 for all n ≥ N , and

choose δ = ε2N

. Then for any measurable A ⊂ [0, 1] with m(A) < δ we have∫A

|f |dm =

∫A∩AN

|f |dm+

∫A∩ACN

|f |dm

< Nε

2N+ε

2= ε.

Thus, it remains to examine the possibility that limn→∞

∫An

|f |dm 6= 0. This is the

case if and only if there exists an infinite increasing sequence nk and a positivenumber η > 0 such that

∫Ank|f |dm ≥ η for all k ≥ 1. Suppose this is the case. Let

Bn1,n2 = x ∈ [0, 1] : n1 < |f(x)| ≤ n2. Now define a subsequence nkj of nk asfollows. Set k0 = 1, and for j ≥ 1, define kj to be the smallest index l such that∫

Bnkj−1,nkj

|f |dm ≥ η

2.

This is well-defined based on our assumption and the definition of η. Further, it isclear that the Bnkj−1

,nkjare disjoint. Hence we have∫

[0,1]

|f |dm ≥∞∑j=1

∫Bnkj−1

,nkj

|f |dm =∞,

which contradicts the fact that f ∈ L1. Hence this possibility does not exist and weare done.

b) Let ε > 0 be given. Let us write M =∫

[0,1]fdm and Mn =

∫[0,1]

fndm, fur-

ther for a measurable E ⊂ [0, 1] write M(E) =∫Efdm and Mn(E) =

∫Efndm. By

Egoroff’s theorem, there exists B ⊂ [0, 1] such that m(B) ≥ 1 − ε2, and fn → f

uniformly on B. If E is a subset of B or differs by a set of measure zero, then weare done by uniform convergence. Hence it suffices to examine the complement of Bin [0, 1]. Let E = BC and choose N sufficiently large so that for all n ≥ N , we have|M −Mn| < ε/2. Then

|M(E)−Mn(E)| = |M −M(B)−Mn +Mn(B)|≤ |M −Mn|+ |M(B)−Mn(B)|

≤ ε

2+ε

2= ε.


Question 4.6. How many roots does the function g(z) = 4z7 + 7z4 + 1 have withinthe circle |z| = 1?

Proof. Set f(z) = 7z4. By Rouche’s theorem, the functions f, g have the same numberof roots within |z| = 1 if the inequality

|f(z)− g(z)| < |f(z)|+ |g(z)|

holds. Rewriting, it suffices to consider the inequality, for |z| = 1

|4z7 + 1| < |7z4|+ |4z7 + 7z4 + 1|.

An upper bound for the left hand side is

|4z7 + 1| ≤ |4z7|+ 1 = 5,

while a lower bound for the left hand side is

|7z4|+ |4z7 + 7z4 + 1| ≥ 7 + (7− 4− 1) = 9,

thus f, g have the same number of roots in |z| = 1. Plainly, f has four roots, and sog does as well.

Question 4.7. Suppose f : C → C is an entire function with |f(z)| ≤√|z| for all

|z| > R, for some fixed R > 0. Prove that f is a constant.

Proof. Let w1, w2 ∈ C. Choose r > R sufficiently large so that |wi| ≤ r1/3/2. ByCauchy’s integral formula, we have

|f(w1)− f(w2)| =∣∣∣∣ 1

2πi

∫|z|=r

f(z)

z − w1

dz −∫|z|=r

f(z)

z − w2

dz

∣∣∣∣=

1

2π

∣∣∣∣∫|z|=r

f(z)(w1 − w2)

(z − w1)(z − w2)

∣∣∣∣≤ 2πr

2π

r1/2 · r1/3

34r2

≤ 4

3r−

16

Letting r →∞ shows that f(w1) = f(w2). Hence f is constant.

Question 4.8. Evaluate the following integrals.

a)

∫|z|=2

ez

z2 − 2dz.

b)

∫ ∞−∞

cosx

x2 − 2x+ 4dx.

Proof. a) Since ez is holomorphic on C and the roots of z2 − 2 lie on the interior ofthe disc |z| ≤ 2, the integral may be evaluated via the Residue theorem. The residue

at z =√

2 ise√

2

2√

2and the residue at z = −

√2 is

e−√

2

−2√

2. The residue theorem then

32 STANLEY YAO XIAO

yields

2πi

(e√

2

2√

2− e−

√2

2√

2

)=

πi√2

(e√

2 − e−√

2).

b) First note that the polynomial f(x) = x2 − 2x + 4 = (x − 1)2 + 3, hence has no

roots on the real line. Its roots are r1,2 = 2±2i√

32

= 1 ± i√

3. The root r = 1 + i√

3lies in the upper half plane, and thus we may consider the contour ΓR defined as theunion of the line segment [−R,R] and the half circle |z| = R : Im(z) ≥ 0. Denotethe integral we wish to compute as I, and note that I is the real part of the integral∫ ∞

−∞

eix

x2 − 2x+ 4dx,

so it suffices to deal with the latter. By Cauchy’s theorem, we have∫ΓR

eiz

z2 − 2z + 4dz = 2πiRes(1 + i

√3) = 2πi

ei−√

3

2i√

3.

Hence it suffices to consider the integral∫|z|=R,Im(z)≥0

eiz

z2 − 2z + 4dz

and show that this integral goes to zero as R →∞. We parameterize the half-circleby z = Reit, whence∫

|z|=R,Im(z)≥0

eiz

z2 − 2z + 4dz =

∫ π

t=0

e−R sin t+iR cos t

R2e2it − 2Reit + 4Rieitdt.

We can bound the numerator by the modulus, which is e−R sin t, and the denominatorcan be seen to be dominated by the term with R2 via the triangle inequality. Hencethe absolute value of this integral is bounded from above by∫ π

t=0

e−R sin t

(3/4)Rdt.

An application of Jordan’s lemma gives us the result, and hence∫ ∞−∞

cosx

x2 − 2x+ 4dx =

πe−√

3 cos(1)√3

.

Question 4.9. a) Suppose f : C → C is an entire function with f(z + 1) = f(z) =f(z + i) for each z. Show that f is necessarily constant.

b) Let

g(z) =∑m∈Z

∑n∈Z

1

(z − n−mi)4.

Show that g defines an analytic function on C \ a + bi : a, b ∈ Z, with g(z + 1) =g(z) = g(z + i) for each z in its domain.


Proof. a) Since f is doubly periodic, it is defined by the values it take on in the squarez ∈ C : 0 ≤ Re(z) ≤ 1, 0 ≤ Im(z) ≤ 1. Since f has no poles, it is bounded on thissquare and thus, bounded on the entire complex plane. Therefore it is constant byLiouville’s theorem.

b) We note that g has singularities which are separated, and hence in any com-pact subset K ⊂ C, K contains at most finitely many pole of g. Thus by Montel’stheorem it suffices to check that the series defining g converges on K except for thefinitely many excepted values.

It plainly suffices to check only discs centered at the origin of radius R, for all R > 0.For any R > 0, let w be a point in the closed disc of radius R centered at the origin.There at most finitely many pairs m,n ∈ Z such that |w−n−mi| ≤ |n+ im|/2, andthe contribution from those finitely many exceptions is bounded as long as w is nota lattice point itself. Let A(w) denote the set of exceptions, and let B(w) denote thecomplement of A(w) in the Gaussian integers. Then it suffices to show that the sum∑

z∈B(w)

1

|w − z|4

converges.

However, z ∈ B(w) implies that |w − z| > |z|/2, and hence we have∑z∈B(w)

1

|w − z|4≤∑

z∈B(w)

16

|z|4,

and so it suffices to examine the convergence of the sum∑m∈Z\0

∑n∈Z\0

1

(m2 + n2)2.

By the arithmetic-geometric mean inequality, we have m2+n2 ≥ 2mn for all m,n ∈ Z.Hence we can bound this sum from above as∑

mn 6=0

1

4m2n2=

(∞∑n=1

1

2n2

)2

,

and the right hand side plainly converges, so we are done.

Question 4.10. Let A and B be non-empty sets. We say that A has cardinalitygreater than B if there is an injection from B into A, but no bijection.

a) Show that if A has cardinality greater than B and B has cardinality greater thanC, then A has cardinality greater than C.

b) Find a sequence of infinite sets An∞n=1 such that for each n, An+1 has cardi-nality greater than An.

c) Find a set A with cardinality greater than An for each of the sets in b), above.

34 STANLEY YAO XIAO

Proof. a) By hypothesis, there exist injections f : B → A and g : C → B. Thus,by examining the composition f g, we see that we can identify C with a subset ofA, hence the cardinality of C cannot exceed that of A. It suffices to check that theycannot be equal, that is, there does not exist an injection from C into A which isalso a bijection. Suppose otherwise, and let h : A→ C be a bijection. Then considerthe composition g h, which maps A into a proper subset of B. Since this map isan injection, the image of A under g h in B is bijective to A. Hence B contains asubset with the same cardinality as A, which contradictions our hypothesis. Henceno such bijection exists, and A has cardinality strictly greater than C.

b) Let A1 = N, and let An+1 = P(An), where P(S) denotes the power set of theset S. Thus, by Cantor’s theorem, we are done.

c) Take A to be the union of the An’s as above.


5. Spring 2012

Question 5.1. a) Show that there exists a bijection from P(R), the power set of R,onto to the set of all real-valued functions on R.

b) Let C(R,R) = f : R → R, f is continuous. Find a cardinal number α so that|C(R,R)| = 2α.

Proof. a) First note that R and R × R have the same cardinality. Denote RR to bethe set of real valued functions. Then it is clear that RR has the same cardinalityas R × R, since we can specify the domain and range of a real-valued function f ,which uniquely defines it. Further, we note that there is a bijection between the setof indicator functions of subsets of R and P(R), which is then plainly a subset of RR.Hence it follows that

|P(R)| ≤ |RR| = |P(R× R)| = |P(R)|.

Since they have the same cardinality, there is a bijection between the two.

b) We claim that α = ℵ0. This is because each continuous function f is determinedby the values they take on over Q, and therefore the cardinality of the continuousfunctions is bounded from above by the cardinality of subsets Q×S, where S denotesthe set of at most countable subsets of R. We note that S has cardinality equal to R.

Question 5.2. Let H be a complex Hilbert space. Given a non-empty subset E ⊂ H,define E⊥ = v ∈ H : 〈v, w〉 = 0 for all w ∈ E.

a) Let M be a closed subspace of H.

i) Given x ∈ H, show that there exists m0 ∈ M so that ‖x − m0‖ = dist(x,M),where dist(x,M) = inf‖x−m‖ : m ∈M.ii) Show that with x and m0 as in i), x−m0 is orthogonal to M.iii) Conclude that H =M⊕M⊥.

b) Prove that a non-empty subset E ⊂ H satisfies E = (E⊥)⊥ if and only if Eis a closed subspace of H.

Proof. a) i) If x ∈ M, then plainly we may choose m0 = x. Hence assume thatx 6∈ M, and consider the set

K = x−m : m ∈M.

This is a translation ofM, and hence is also closed. Now choose a sequence mn ⊂ Ksuch that lim

n→∞‖x−mn‖ → δ = dist(x,M). For convenience, let us denote kn = x−mn.

Then by the parallelogram law of Hilbert spaces, we have

‖kn − kl‖2 + ‖kn + kl‖2 = 2‖kn‖2 + 2‖kl‖2.

36 STANLEY YAO XIAO

However, we note thatkn + kl

2= x +

mn +ml

2, and since mn + ml ∈ M since M is

a subspace, we have thatkn + kl

2∈ K. Hence it follows that

∥∥∥∥kn + kl2

∥∥∥∥ ≥ δ, and so

‖kn − kl‖2 = 2‖kn‖2 + 2‖kl‖2 − ‖kn + kl‖2

≤ 2(‖kn‖2 + ‖kl‖2)− 4δ2,

and the right hand side tends to zero as n, l approaches infinity. Hence the sequencekn is Cauchy. But a Hilbert space is complete, and hence this sequence has a limit,say k. Recall that K is closed, so k ∈ K. Hence there exists m ∈ M such thatk = x−m. This is the desired choice of m0.

ii) Let us write k0 = x − m0, so that dist(k0,M) = δ. Then for any m ∈ M wehave the inequality

‖k0‖2 ≤∥∥∥∥k0 −

〈k0,m〉‖m‖2

m

∥∥∥∥2

.

Then it follows, by expanding the right hand side, that

‖k0‖2 ≤ ‖k0‖2 − |〈k0,m〉|2

‖m‖2

whence 〈k0,m〉 = 0, as desired.

iii)

Question 5.3. Let h ∈ C[0, 1]. Show that every f ∈ C[0, 1] is a uniform limit ofpolynomials in h if and only if h is strictly monotone.

Proof. Let H denote the polynomial subalgebra of C[0, 1] generated by h. Then by theStone-Weierstrass theorem, H is dense in C[0, 1] under the uniform topology if andonly if H separates points. Let x < y be distinct points in [0, 1]. Suppose h is strictlymonotone. Since h ∈ H, it follows that h(x) 6= h(y), and hence H separates points.Conversely, if h is not monotone, then there exist x < y such that h(x) = h(y), andthis property clearly holds also for any polynomial in h, hence H fails to separatepoints. This proves the required.

Question 5.4. Prove that the following limit exists, and calculate its value.

limn→∞

∫ ∞0

(n∑k=0

(−1)kx2k

(2k)!

)e−2xdx

Proof. Let fn(x) = e−2x

n∑k=0

(−1)kx2k

(2k)!. Now write gn(x) =

n∑k=0

(−1)kx2k

(2k)!. Note that

gn(x) is a polynomial, and by elementary calculus, we know that

limx→∞

gn(x)

ex= 0.


Hence it follows that there exists a constant C > 0 such that |fn(x)| ≤ Ce−x = g(x).Now, it is clear that g(x) is integrable on [0,∞), whence the dominated convergencetheorem tells us that

limn→∞

∫ ∞0

fn(x)dx =

∫ ∞0

( limn→∞

fn(x))dx.

Thus it remains to examine the function

f(x) = e−2x

∞∑n=0

(−1)nx2n

(2n)!.

Now, the series defines the cosine function, whence

f(x) = e−2x cosx.

Finally, it remains to evaluate the integral

J =

∫ ∞0

e−2x cosxdx.

Integrating by parts, we have

J =−1

2e−2x cosx|∞0 −

∫ ∞0

1

2e−2x sinxdx

=−1

2e−2x cosx|∞0 +

1

4e−2x sinx|∞0 −

∫ ∞0

1

4e−2x cosxdx,

whence J =4

5· 1

2=

2

5.

Question 5.5. Recall that

l2 = x = (xk)∞k=1 : xk ∈ R∀k ≥ 1 and ‖x‖2 =

(∞∑k=0

|xk|2)1/2

<∞.

Consider the following subset of l2

H = x = (xk)∞k=1 ∈ l2 : |xk| ≤ 1/k for all k ≥ 1.

a) Consider a sequence (xn)∞n=1 in H, where each xn = (xn,1, xn,2, · · · ). Prove that thesequence (xn)∞n=1 converges in l2 if and only if for each k ≥ 1, the sequence (xn,k)

∞n=1

converges.

b) Prove that H is compact and nowhere dense in l2.

Proof. a) Suppose that (xn) ⊂ H converges in l2. Then there exists x ∈ l2 such thatlimn→∞‖xn − x‖2

2 = 0. In particular, this implies that for all ε > 0, there exists Nε ∈ Nsuch that for all n ≥ Nε, we have ‖xn − x‖2

2 < ε. Hence for n ≥ Nε we have∞∑k=1

|xn,k − xk|2 < ε,

and since this is a sum of non-negative numbers, we must have that each summandis bounded from above by ε as well. Thus lim

n→∞|xn,k − xk| = 0 for each k ≥ 1.

38 STANLEY YAO XIAO

Conversely, suppose for each k ≥ 1 we have limn→∞

|xn,k − xk| = 0. Let ε > 0. Since

we know that |xn,k| ≤ 1/k for all n, k ∈ N, and since we know that xn,k → xk, thereexists N1 ∈ N such that for all n ≥ N1 we have |xn,k−xk| < ε, whence |xk| ≤ 1/k+ εfor any ε > 0, hence |xk| ≤ 1/k. Thus the element x = (x1, x2, · · · ) ∈ H. Hencefor any ε > 0 we can choose N2 so that for all k ≥ N2 and all n ∈ N, we have∞∑

k=N2+1

|xn,k − xk|2 ≤∞∑

k=N2+1

4

k2<

C

N2

<ε

2, where C is a constant independent of the

sequence (xn). Now, decompose the sum below as follows:

∞∑k=0

|xn,k − xk|2 =

N2∑k=1

|xn,k − xk|2 +∞∑

k=N2+1

|xn,k − xk|2.

Now for each 1 ≤ k ≤ N2, choose N3,k sufficiently large so that for n ≥ N3,k, we have|xn,k − xk|2 < ε

2N2. Set N3 = max

1≤k≤N2

N3,k. Then for n ≥ N3, we have

N2∑k=1

|xn,k − xk|2 <ε

2.

Hence, we have that∞∑k=1

|xn,k − xk|2 <ε

2+ε

2,

and we are done.

b) Since H is in a metric space, it suffices to show that H is sequentially compact. Let(xn) be a sequence in H. We first consider the sequence of first coordinates, namely(xn,1)∞n=1. Since we know that |xn,1| ≤ 1 for all n ≥ 1, it follows that (xn,1) ⊂ [−1, 1],which is compact in R. Hence (xn,1) contains a convergent subsequence, correspond-ing to some sequence of indices (nl)

∞l=1 and limit x1. Now consider the subsequence

(xnl) of (xn). Then for this subsequence, repeat the same procedure for the sequenceof second coordinates. Since we know that |xnl,2| ≤ 1/2 for all nl, it follows againthat this sequence contains a convergent subsequenc, corresponding to some limit x2.Then the point (x1, x2, · · · ) generated by this procedure is a limit point of (xn), henceH is sequentially compact and we are done.

Since H is compact in a Hausdorff space, it is closed. Hence it suffices to showthat it has empty interior. To do this it suffices to show that for any x ∈ H andany open neighborhood U of x, U contains something in the complement of H inl2. It plainly suffices, by shrinking the neighborhood U if necessary, to consider thecase of open balls. Now let ε > 0 be given, and consider the open ball B(x, ε).Consider the point y which has coordinates equal to x, except for the index k whichis the smallest positive integer such that ε > 1/k. Then for that coordinate, defineyk = xk ± ε, where the sign is chosen so that xk ± ε has magnitude greater than xk.Then |yk| = |xk ± ε| = |xk| + ε > 1/k, whence y 6∈ H, but is in B(x, ε). Hence H isnowhere dense, as required.


Question 5.6. Let f : [0,∞) → R be a continuous function. Suppose that for allx ∈ [0, 1],

limn→∞

f(nx) = 0.

Prove that limx→∞

f(x) = 0.

Proof. For each ε > 0 and n ∈ N, consider the set An,ε = x ∈ [0, 1] : |f(kx)| ≤ε for all k ≥ n. This set is closed, since it can be written as the intersection overk ≥ n of the sets x ∈ [0, 1] : |f(kx)| ≤ ε, which is closed since f is continuous.Now, for any ε > 0, our hypothesis guarantees that

R+ =∞⋃n=1

An,ε.

The Baire category theorem then tells us that this cannot be a union of nowhere densesets, hence for some n, An,ε contains an open interval, say (a, b) ⊂ An,ε. Then for anyk ≥ n, we have that t ∈ (ka, kb) implies that f(t) ≤ ε. Further, for N sufficientlylarge, we have that

(Na,∞) =∞⋃n=N

(na, nb).

Hence for t > Na, we have that t ∈ (na, nb) for some n ≥ N , whence f(t) ≤ ε. Thisshows that f(x)→ 0 as x→∞, as required.

Question 5.7. Let m denote Lebesgue measure on [0, 1]. Suppose (fn)∞n=1 and fare real-valued, Lebesgue measurable functions on [0, 1] and that (fn)∞n=1 convergespointwise a.e. to f .

a) Prove that for each pair ε, δ > 0 there exist a Lebesgue measurable set A ⊂ [0, 1]and an integer k such that m([0, 1] \ A) < ε and

|fn(x)− f(x)| < δ

for all x ∈ A and n ≥ k.

b) Use this to prove that for each ε > 0 there exists a set B ⊂ [0, 1] such thatm([0, 1] \ B) < ε and (fn)∞n=1 converges uniformly to f on B (This is known as Ego-roff’s theorem).

c) Does Egoroff’s theorem hold if we replace [0, 1] with R? Justify your answer.

Proof. a) Consider the set An = x ∈ [0, 1] : |fn(x)−f(x)| ≥ δ. Since fn → f almosteverywhere, it follows that no x except for possibly a set of measure zero can lie inAn for infinitely many n. That is, the set

B =∞⋂m=1

∞⋃n=m

An

is a set of measure zero. By the continuity of measures, there exists some N ∈ N

such that∞⋃n=N

An has measure at most ε, and we denote this set as A. Then by defi-

nition, in the complement of A, we have for n ≥ N that |fn(x)−f(x)| < δ, as desired.

40 STANLEY YAO XIAO

b) The proof above shows that the bounds are independent of x, and so the con-vergence on A is uniform, which is the conclusion of Egoroff’s theorem.

c) No. Consider the functions fn(x) =x

n, n ≥ 1. Then for all x 6= 0, we have

that fn(x) → 0. However, on the complement of any set of finite measure, the con-vergence cannot be uniform, since n needs to be chosen sufficiently large relative tox.

Question 5.8. Let Ω be a connected open set in R2 ∼= C. Recall that if u : Ω → Ris a C2 function, we say that u is harmonic if uxx + uyy = 0.

a) i) Show that the real and imaginary parts of a holomorphic function are har-monic.

ii) If u : Ω → R, then a function v : Ω → R is a conjugate of u if f = u + iv isholomorphic on Ω. Show that if u is harmonic, then −uy is a conjugate of ux.

b) Prove that a harmonic function u admits a conjugate if and only if the holomorphicfunction g = ux − iuy has a primitive f in Ω. Under what topological conditions isthis guaranteed to hold?

Proof. a) i) Suppose f is holomorphic, and write f = u + iv. The Cauchy-Riemannequations imply that

ux = vy, uy = −vx.Hence we have that

uxx = vxy, uyy = −vxy,since u, v are smooth. Hence

uxx + uyy = 0,

so u is harmonic. Likewise, the Cauchy-Riemann equations imply that

vyy = uxy, vxx = −uxy,whence v is also harmonic.

ii) It suffices to check that ux,−uy satisfy the Cauchy-Riemann equations. We have

uxx = −uyy, (ux)y = −(−uy)x,and we are done.

b) The topological condition is simple connectedness.

Question 5.9. Let f : C→ C be an entire function with power series

f(z) =∞∑n=0

ca,n(z − a)n

about the point a. Suppose that for every a ∈ C, at least one coefficient ca,n is zero.Prove that f is a polynomial.


Proof. Recall that ca,n =f (n)(a)

n!. The hypothesis is then for each a ∈ C, there exists

n ∈ N∪0 such that f (n)(a) = 0. Now, consider the set An = z ∈ C : f (n)(z) = 0.Since f is analytic, An is either all of C (in which case f (n)(a) ≡ 0) or An is a discreteset with no limit points in C. We are told that

C =∞⋃n=0

An,

in which case we must have An = C for some n ≥ 0. But this implies that Am = C forall m ≥ n, in other words f (m) ≡ 0 for all m ≥ n. This implies that f is a polynomialof degree at most n, as desired.

Question 5.10. Let g be an entire function. Show that if g is not a polynomial, thenthere exists a sequence (zn)∞n=1 in C with limn→∞ |zn| =∞ and limn→∞ g(zn) = 0.

Proof. If g is not a polynomial, then g has an essential singularity at infinity. Inparticular, g maps every open neighborhood of infinity to the complex plane less atmost one point, by Picard’s theorem. Hence the result follows.

6. Spring 2010

6.1. Basic Real Analysis.

Question 6.1. a) Let fn : R → R be continuous functions, and suppose that forevery x ∈ R, there exists n ≥ 1 such that fn(x) ∈ Q. Prove that for every c < d in R,one can find some numbers a < b in the interval (c, d) and a positive integer n suchthat the function fn is constant on (a, b).

b) Provide a complete statement of any major theorem used in your solution.

Proof. a) For each q ∈ Q and n ∈ N, set An,q = x ∈ (c, d) : fn(x) = q. Byhypothesis, we have that ⋃

n∈N

⋃q∈Q

An,q = (c, d).

Since the right hand side is uncountable and the left hand is a union over a countableindex set, there exists a pair (n, q) ∈ N×Q such that An,q is uncountable. Then An,qcannot be nowhere dense in (c, d), since any such set must be countable. Thereforethere exists (a, b) ⊂ (c, d) such that An,q ∩ (a, b) is dense in (a, b), so that for allt ∈ (a, b), we have fn(t) = q. However, since fn is continuous, it follows that fn isdetermined by the value it takes on in any dense subset, so fn(t) = q for all t ∈ (a, b),as desired.

b) The theorems used are as follows:

Theorem 6.2. N,Q are uncountable, and for any pair of real numbers c < d, theinterval (c, d) is uncountable.

Theorem 6.3. A countable union of countable sets is countable.

42 STANLEY YAO XIAO

Theorem 6.4. Let I ⊂ R be an interval, and let K ⊂ I be a dense subset of I. Thenfor any function f : K → R, there exists a unique continuous function g : I → Rsuch that g = f on K.

Question 6.5. a) Suppose that fn : [0, 1]→ R are C1 functions such that

|fn(x)|+ |f ′n(x)| ≤ 1

for all x ∈ [0, 1]. Prove that the sequence (fn)n≥1 contains a uniformly convergentsubsequence.

b) Provide a complete statement of any major theorem used in your solution.

Proof. a) By the Arzela-Ascoli theorem, it suffices to check that (fn)n≥1 is uniformlybounded and equicontinuous. It is clear from hypothesis that |fn(x)| ≤ 1 for allx ∈ [0, 1], hence (fn) is uniformly bounded. Thus it suffices to check that it isequicontinuous.

We show that a uniform bound on the derivatives of (fn) is sufficient to prove equicon-tinuity. Let ε > 0. For each n ∈ N, by the mean value theorem we have∣∣∣∣fn(x)− fn(y)

x− y

∣∣∣∣ = |f ′n(c)|

for some x < c < y. This implies that |fn(x) − fn(y)| ≤ |x − y|, and the right handside is independent of n. Hence we may choose δ = ε, in other words, |x − y| < εimplies that |fn(x)− fn(y)| < ε for all n. Thus the family (fn) is equicontinuous, andthus contains a uniformly convergent subsequence by Arzela-Ascoli.

b) The main theorems used are as follows:

Theorem 6.6. (Arzela-Ascoli) Let S denote a countable family of continuous func-tions on some compact interval I ⊂ R. Then S contains a uniformly convergentsubsequence if and only if S is uniformly bounded and equicontinuous.

Theorem 6.7. (Mean Value Theorem) Let f be a function continuous on [a, b] anddifferentiable on (a, b). Then there exists a < c < b such that f ′(c) = (f(b)−f(a))/(b−a).

Question 6.8. Let (X, d) be a compact metric space, and let T : X → X be afunction such that

d(Tx, Ty) = d(x, y)

for all x, y ∈ X.

a) Let a be a point in X. Prove that a is a cluster point of the sequence T na : n ≥ 1.

b) Prove that the function T is surjective.

c) Prove that the function T is a homeomorphism.


Proof. a) Suppose otherwise. Then there exists η > 0 such that d(a, T na) ≥ η for alln ≥ 1. But then d(Tma, T na) ≥ η for all m,n ∈ N, which implies that the sequenceT na : n ≥ 1 has no cluster point. However, this sequence is a subset of a compactmetric space and hence must contain a convergent subsequence, so this is a contra-diction. Hence a is a cluster point, and indeed T na is a cluster point for any n ≥ 1.

b) As we have shown in part a), for any x ∈ X the point x is the limit of a se-quence of points of the form T nkx for some increasing sequence of positive integers(nk)k≥1. Now consider the sequence T nk−1x : k ≥ 1. This sequence contains aconvergent subsequence since X is compact, which by abuse of notation we will alsodenote T nk−1x : k ≥ 1. Then we have simultaneously that lim

k→∞T nk−1x = y for

some y ∈ X and limk→∞

T nkx = x. Hence for all ε > 0, there exists N ∈ N such that

for all k ≥ N , we have simultaneously d(x, T nkx) <ε

2and d(y, T nk−1x) <

ε

2. By the

triangle inequality, we have

d(x, Ty) ≤ d(x, T nkx) + d(Ty, T nkx) = d(x, T nkx) + d(y, T nk−1x) <ε

2+ε

2= ε,

hence x = Ty and T is surjective.

c) We have shown that T is surjective. Now, suppose that x, y ∈ X and Tx = Ty.Then by hypothesis we have d(Tx, Ty) = d(x, y), whence d(x, y) = 0. Hence x = y,so T is also injective. Thus T is a bijection. It remains to show that T, T−1 arecontinuous. Let ε > 0. Then d(Tx, Ty) = d(x, y) implies that it suffices to takeδ = ε, so T is continuous. Since T is surjective, for any x, y ∈ X we may writed(x, y) = d(T (T−1x), T (T−1y)) = d(T−1x, T−1y), whence by the same argument, T−1

is continuous. Hence T is a homeomorphism, as desired.

6.2. Complex Analysis.

Question 6.9. Let 0 < p < 1. Evaluate∫ ∞0

xp

1 + x2dx.

Simplify your answer so that it is expressed in terms of real quantities.

Proof. Define the keyhole contour ΓR,ε by the following pieces: a small circular arcnear the origin of radius ep which we denote γε, a line segment on the line z ∈ C :Im(z) = δ which we call L1 going up to the outer circular segment of radius R,which we denote ΓR, and finally a line segment L2 on the line z ∈ C : Im(z) = −δ.We denote the entire contour by Γ. By the residue theorem, we have∫

Γ

zp

1 + z2dz = 2πi(Res(i) + Res(−i)),

which we evaluate by choosing the branch of the logarithm with branch cut along thepositive real axis. Thus we have that

Res(i) =ip

i+ i=epπi2

2i

44 STANLEY YAO XIAO

and

Res(−i) =(−i)p

−i− i=e

3pπi2

−2i.

Now, the contribution from γε may be bounded by the estimate∫γε

∣∣∣∣ zp

1 + z2

∣∣∣∣ dz ≤ εp

1− ε22πε,

which goes to zero as ε→ 0. Similarly, the contribution from ΓR can be bounded by∫ΓR

∣∣∣∣ zp

1 + z2

∣∣∣∣ dz ≤ πRRp

R2 − 1,

and since 0 < p < 1 it follows that the right hand side tends to zero as R → ∞.Hence we have that

limε→0R→∞

(∫L1

f(z)dz +

∫L2

f(z)dz

)=

∫Γ

f(z)dz,

whence it suffices to examine the integral over L1 and L2. We parameterize L1 withz = x+ iδ, and ∫

L1

zp

1 + z2dz =

∫ R

ε

(x+ iδ)p

1 + (x+ iδ)2dx

while on L2, the integrand is shifted by a factor of e2πip. Hence by taking limits, wesee that ∫ ∞

0

zp

1 + z2dz = lim

ε→0δ→0R→∞

∫ R

ε

(x+ iδ)p

1 + (x+ iδ)2

and ∫L2

zp

1 + z2dz = e2πip

∫ R

ε

(x+ iδ)p

1 + (x+ iδ)2

Hence ∫ ∞0

zp

1 + z2dz =

π(epπi2 − e 3pπi

2 )

1− e2pπi.

Factoring epπi from both the numerator and denominator, we see that the right handside is equal to

π(e−pπi2 − e pπi2 )

e−pπi − epπi.

Now, e−ix − eix for real x is equal to −2i sinx, whence this quantity is reduced to

π sin(pπ/2)

sin(pπ)=

π

2 cos(pπ/2).

Question 6.10. a) State the Schwarz Lemma.

b) Let H = z ∈ C : Im(z) > 0. Suppose that f : H → C is an analytic func-tion with |f(z)| < 1 for all z ∈ H and f(i) = 0. Prove that |f(2i)| ≤ 1/3.


c) Prove that there exists a unique function f satisfying the hypothesis from partb) such that f(2i) = 1/3. Find a formula for f .

Proof. a) Schwarz Lemma: Let D = z ∈ C : |z| < 1 be the open unit disc, andlet f : D → D be a holomorphic map with f(0) = 0. Then we have |f(z)| ≤ |z| forall z ∈ D and |f ′(0)| ≤ 1. Further, if there exists a point z ∈ D such that |f(z)| = |z|or |f ′(0)| = 1, then f = αz for some α ∈ C with unit modulus.

b) By hypothesis, f is an analytic function from H to D. Now consider a Mobius mapg : D→ H that maps D to H, and sends 0 to i. Such a map must map the boundary

|z| = 1 of D to the real line, and map 0 to i. Such a map is g(z) = i1− z1 + z

. Hence

the composition f g is a conformal self map from D to D with f(0) = 0, whenceSchwarz lemma applies. Substituting z = 1/3 and applying the lemma shows that

|f g(1/3)| = |f(2i)| ≤ 1/3,

as desired.

c) By the second part of Schwarz lemma, |f(2i)| = 1/3 implies that f g = αz forsome α of unit modulus. Hence α−1f is a functional inverse of g, which we can de-

termine uniquely. Namely, g−1(z) =i− zi+ z

. Hence f(z) = αi− zi+ z

. Now, f(2i) = 1/3

implies that

f(2i) = αi− 2i

i+ 2i= α−i3i

= α1

3,

whence α = 1.

Question 6.11. Let Ω be a non-empty connected open subset of C, and let f(z) beanalytic on Ω.

a) Define what it means to for Ω to be simply connected.

b) If Ω is simply connected, prove that there is an analytic function g(z) on Ω suchthat g′(z) = f(z).

c) Provide an example of an analytic function f on a non-simply connected domainΩ which does not have a primitive (anti-derivative).

Proof. a) Ω is simply connected if and only if for every closed path γ contained in Ω,γ can be continuously shrunk down to a point while staying within Ω.

b) Let f be an analytic function on Ω. Let a ∈ Ω be fixed, and define

g(z) =

∫γa,z

f(u)du.

Where the integral is over a path γa,z contained in Ω. By path connectedness, we canextend γa,z to a closed path contained in Ω. By simple connectedness and Cauchy’stheorem, for any path Γa,z such that γa,z ∪Γa,z is a closed path with winding number

46 STANLEY YAO XIAO

1, we have ∫γa,z∪Γa,z

f(u)du = 0.

In particular, this shows that the choice of γa,z is irrelevant, and the integral onlydepends on a, z. Thus g is a primitive of f .

c) Take the annulus Ω = z ∈ C : r < |z| < R and the function f(x) = 1/x.Then plainly f is analytic on Ω, but it fails to have a primitive on Ω because nobranch cut of the logarithm avoids Ω.

6.3. Topology and Set Theory.

Question 6.12. a) Prove that every real vector space has a basis.

b) Let A be an infinite set. Prove that the collection F(A) of all finite subsetsof A has the same cardinality as A.

c) Let l∞ denote the Banach space of all bounded real sequences. Prove that thevector space basis of l∞ has cardinality 2ℵ0 .

Proof. a) We recall that a basis is a maximally linearly independent subset, thus itsuffices to show such a set exists. Let V denote a real vector space, and let < denotethe partial order defined for linearly independent subsets of V by A < B if and onlyif A ⊂ B. Let I be a chain under this partial order. Then plainly for each chainI there is a maximal element, and thus by Zorn’s lemma, there exists a maximallylinearly independent set and we are done.

b) Recall that for any finite positive integer n, the set An has the same cardinal-ity as A. Further recall that taking countable unions of infinite sets of the samecardinality preserves the cardinality. Thus, let A(n) denote the set of subsets of A of

size n. Then F(A) =∞⋃n=0

A(n), and since A(n) has cardinality at least A and at most

An, it follows that |A(n)| = |A| for all n ≥ 1. Hence F(A) has the same cardinalityas A, since the union is over a countable index set.

c) First we note that l∞ is not a finite vector space over R, since the existence ofthe vectors of the form (δi,n)∞n=1, where δi,j is the Kronecker delta, shows that infin-itely many linearly independent vectors exist in l∞. Hence a basis is at least countablein cardinality. Further, we note that l∞ lies in RN, whence l∞ itself has cardinalityequal to R, so the cardinality of the basis is at most 2ℵ0 . Thus it suffices to show thatno countable set of vectors can be a basis.

6.4. Measure Theory.

Question 6.13. Let f, fn be bounded Lebesgue measurable functions on [0, 1]. Wesay that fn converge to f in measure if

limn→∞

m(x : |fn(x)− f(x)| ≥ ε) = 0


for all ε > 0. Prove that fn converges to f in L1(0, 1) if and only if fn converges tof in measure and lim

n→∞‖fn‖1 = ‖f‖1.

Proof. Suppose that fn → f in L1. Then from

|‖fn‖1 − ‖f‖1| ≤ ‖fn − f‖1,

we see that limn→∞‖fn‖1 = ‖f‖1. Further, let ε > 0 be given, and let An(ε) = x ∈

[0, 1] : |fn(x)− f(x)| ≥ ε. Then

‖fn − f‖1 =

∫[0,1]

|fn − f |dm =

∫An(ε)

|fn − f |dm+

∫[0,1]\An(ε)

|fn − f |dm.

This implies thatεm(An(ε)) ≤ ‖fn − f‖1

for all n. Now choose N sufficiently large so that ‖fn − f‖1 < ε2 for all n ≥ N ,whence

m(An(ε)) < ε

for n ≥ N , so fn approaches f in measure.

Now suppose that fn approaches f in measure and limn→∞‖fn‖1 = ‖f‖1. Let ε, δ > 0.

It is known that (fn) contains a subsequence (fnj) that converges almost everywhereto some measurable function g. We first show that f = g almost everywhere. Firstwe note that (fnj) approaches g in measure, by part a). Now

|f(x)− g(x)| ≤ |f(x)− fnj(x)|+ |g(x)− fnj(x)|,and so B(ε) = x ∈ [0, 1] : |f(x)− g(x)| ≥ ε ⊂ Anj(2ε) for all j sufficiently large, inparticular, B(ε) has measure zero for all ε > 0. Hence f = g almost everywhere. Now,by Lebesgue’s dominated convergence theorem, since the sequence (fnj) is boundedabove by |f |+ 1 say for all but finitely many indices j, we have fnj → f in L1.

Question 6.14.

7. Spring 2010

7.1. Set Theory and Topology.

Question 7.1. Let X be a non-empty set, and denote by P(X) the power set of X.That is, P(X) = A : A ⊂ X. The cardinality of X is denoted by |X|. Finally, ℵ0

denotes the cardinality of the set N of natural numbers, while c denotes the cardinal-ity of the set R of real numbers.

a) Prove that |X| < |P(X)|.

b) Find a cardinal number α so that |C([0, 1],R)| = 2α, where C([0, 1],R) = f :[0, 1]→ R, f continuous.

c) A function ϕ : P(X) → P(X) is said to be increasing if A,B ∈ P(X) andA ⊂ B implies ϕ(A) ⊂ ϕ(B). Prove that if ϕ : P(X) → P(X) is increasing, thenthere exists a set T ∈ P(X) so that ϕ(T ) = T .

48 STANLEY YAO XIAO

Proof. a) We show that there is no surjective map from X to P(X). Let f : X →P(X) be any map, and define B = x ∈ X : x 6∈ f(x). I claim that B cannot beexpressed as f(y) for any y ∈ X, which shows that f is not surjective. Suppose oth-erwise, and write B = f(y) for some y ∈ X. Suppose that y ∈ f(y). Then y ∈ B, soy 6∈ f(y) by the definition of B, which is a contradiction. Now suppose that y 6∈ f(y).Then y ∈ B = f(y) by the definition of B, which is again a contradiction. Hencey cannot be an element of X, which is a contradiction, which shows that f is notsurjective. Since no surjective map from X to P(X) exists, the latter has cardinalitystrictly greater than the former.

b) I claim that α = ℵ0. First note that R and R × R have the same cardinality.Denote RR to be the set of real valued functions. Then it is clear that RR has thesame cardinality as R×R, since we can specify the domain and range of a real-valuedfunction f , which uniquely defines it. Further, we note that there is a bijection be-tween the set of indicator functions of subsets of R and P(R), which is then plainlya subset of RR. Hence it follows that

|P(R)| ≤ |RR| = |P(R× R)| = |P(R)|.Since they have the same cardinality, there is a bijection between the two.

c) Let I be a chain under the partial order induced by containment. Then ϕ(I)is also a chain. Further, each chain of the form ϕ(I) has an upper bound, namely

the set ϕ

(⋃A∈I

A

). Thus by Zorn’s lemma, there exists a set S such that ϕ(S) is

maximal among all sets of the form ϕ(A), A ∈ P(X). That is, for all A ∈ P(X), wehave ϕ(A) ⊂ ϕ(S). Hence, take the union of all sets A ∈ P(X) with ϕ(A) = ϕ(S),which we will denote T , and note that ϕ(T ) = ϕ(S). Note that if T is strictly con-tained in any set U , we will have ϕ(U) = ϕ(S), whence U would be a subset of T , acontradiction. Hence ϕ(T ) = T .

7.2. Measure Theory.

Question 7.2. Let f : [0, 1]→ R be a function.

a) Define what it means to say that

i) f is absolutely continuous, andii) f is of bounded variation.

b) Prove that if f is absolutely continuous, then f is of bounded variation.

c) Prove that if f is absolutely continuous and E ⊂ [0, 1] has Lebesgue measurezero, then f(E) ⊂ R has Lebesgue measure zero.

Proof. a) i) f is absolutely continuous if for all ε > 0, there exists δ > 0 such that

whenever 0 ≤ xk < yk ≤ 1, 1 ≤ k ≤ n are such thatn∑k=1

|yk − xk| < δ we have


n∑k=1

|f(yk)− f(xk)| < ε.

ii) Define

Tf = sup

n∑k=1

|f(xj)− f(xj−1)| : 0 ≤ x0 < · · · < xn ≤ 1

We say f is of bounded variation if Tf <∞.

b) Set ε = 1, and let δ = δ(1) be the corresponding value of δ. Then let N = δ−1 + 1.For any subdivision 0 ≤ x0 < x1 < · · · < xn = 1, we can collect the intervals(xj−1, xj) into at most N groups, each of total length at most δ. Then the sum∑|f(xj)− f(xj−1)| in each group is at most 1, so the total variation is at most N ,

as desired.

c) If E is measurable and has measure zero, then for any δ > 0, there exists a collec-

tion of at most countably many open intervals Ij = (xj−1, xj) such that E ⊂∞⋃j=1

Ij,

m

(∞⋃j=1

Ij

)< δ, by the outer regularity of Lebesgue measure. Let ε > 0. By absolute

continuity, there exists δ = δ(ε) > 0 such that we haven∑k=1

|f(xj) − f(xj−1)| < ε

whenevern∑k=1

|xj − xj−1| < δ. Thus, we have that m(f(E)) < ε whenever m(E) < δ,

which proves the required.

Question 7.3. a) Suppose that fn : [0, 1] → R and f : [0, 1] → R are Lebesgueintegrable and lim

n→∞‖fn − f‖1 = 0. Show that for every ε > 0 there exists δ > 0 such

that for all A ⊂ [0, 1] with m(A) < δ,∫A|fn|dm < ε for all n.

b) Is the statement in a) true when the condition limn→∞‖fn − f‖1 = 0 is replaced

by limn→∞

fn(x) = f(x) for each x ∈ [0, 1]? Justify your answer.

Proof. a) We first prove the statement for f . We further abbreviate L1[0, 1] by L1.For any fixed positive integer n, let us write An = x ∈ [0, 1] : |f(x)| ≥ n. If

limn→∞

∫An

|f |dm = 0, then we may argue as follows. First choose N sufficiently large

so that∫An|f |dm < ε/2 for all n ≥ N , and choose δ = ε

2N. Then for any measurable

50 STANLEY YAO XIAO

A ⊂ [0, 1] with m(A) < δ we have∫A

|f |dm =

∫A∩AN

|f |dm+

∫A∩ACN

|f |dm

< Nε

2N+ε

2= ε.

Thus, it remains to examine the possibility that limn→∞

∫An

|f |dm 6= 0. This is the

case if and only if there exists an infinite increasing sequence nk and a positivenumber η > 0 such that

∫Ank|f |dm ≥ η for all k ≥ 1. Suppose this is the case. Let

Bn1,n2 = x ∈ [0, 1] : n1 < |f(x)| ≤ n2. Now define a subsequence nkj of nk asfollows. Set k0 = 1, and for j ≥ 1, define kj to be the smallest index l such that∫

Bnkj−1,nkj

|f |dm ≥ η

2.

This is well-defined based on our assumption and the definition of η. Further, it isclear that the Bnkj−1

,nkjare disjoint. Hence we have∫

[0,1]

|f |dm ≥∞∑j=1

∫Bnkj−1

,nkj

|f |dm =∞,

which contradicts the fact that f ∈ L1. Hence this possibility does not exist and weare done.

Now, we show that for any measurable set E ⊂ [0, 1], we have∫Efndm →

∫Efdm.

Let ε > 0 be given. Let us write M =∫

[0,1]fdm and Mn =

∫[0,1]

fndm, further for a

measurable E ⊂ [0, 1] write M(E) =∫Efdm and Mn(E) =

∫Efndm. By Egoroff’s

theorem, there exists B ⊂ [0, 1] such that m(B) ≥ 1− ε2, and fn → f uniformly on B.

If E is a subset of B or differs by a set of measure zero, then we are done by uniformconvergence. Hence it suffices to examine the complement of B in [0, 1]. Let E = BC

and choose N sufficiently large so that for all n ≥ N , we have |M −Mn| < ε/2. Then

|M(E)−Mn(E)| = |M −M(B)−Mn +Mn(B)|≤ |M −Mn|+ |M(B)−Mn(B)|

≤ ε

2+ε

2= ε.

Finally, this implies that ∫A

|fn|dm ≤∫A

|f |dm+ε

2

for all n sufficiently large, and we can control the∫A|f |dm by the first part of the

argument, whence we can give a uniform upper bound for n large. For the possiblyfinitely many exceptions, we merely repeat the argument in the first part for each ofthem.


b) No. Let fn(x) = nχ(0,1/n], where χA is the indicator function of a subset A ⊂ [0, 1].Then fn(x)→ 0 for each x ∈ [0, 1], but for the set An = (0, 1/n] which has arbitrarilysmall measure, we have

∫A|fn|dm = 1, which does not go down to zero.

Question 7.4. A function f : R → R is called additive if it satisfies the functionalequation

f(x+ y) = f(x) + f(y)

for all x, y ∈ R. A function f : R → R is called locally Lebesgue integrable if it isLebesgue integrable over every finite interval.

a) Show that a locally Lebesgue integrable additive function must be linear, i.e.

f(x) = cx

for some real constant c.

b) Assuming the fact stated in a), prove that there are additive functions whichare not locally Lebesgue integrable.

Proof. a) Let I = [a, b] be a given finite interval, so that f is integrable over I.

8. Spring 2008

Question 8.1. a) Prove that if A is a compact subset of a metric space (X, d) andf : X → X is continuous then f is uniformly continuous on A.

b) Let f : R2 → R and suppose f(x, y0) is continuous as a function of x for eachy0 and f(x0, y) is continuous as a function of y for each x0. Is f continuous? Proveor give a counterexample.

Proof. a) Let ε > 0. For each a ∈ A, choose δa so that d(a, x) < δa implies thatd(f(a), f(x)) < ε. Then the collection of balls B(a, δa/3) for a ∈ A is an open cover forA, and since A is compact, there exists a1, · · · , an such that B(a1,

δ13

), · · · , B(an,δn3

)is a finite subcover. Now choose δ = min(δ1/3, · · · , δn/3). Then for x, y ∈ A suchthat d(x, y) < δ, first find ai, aj such that x ∈ B(ai, δi) and y ∈ B(aj, δj). Withoutloss of generality, assume δj ≥ δi. Then d(x, y) < δ implies that d(ai, aj) ≤ d(ai, x) +d(aj, y) + d(x, y) < δj. Hence it follows that d(f(ai), f(aj)) < ε. Finally, we have

d(f(x), f(y)) ≤ d(f(x), f(ai)) + d(f(y), f(aj)) + d(f(ai), f(aj)) < 3ε,

whence f is uniformly continuous on A, since δ is independent of x, y.

b) Consider the function

f(x, y) =x2 − y2

x2 + y2.

By symmetry, it suffices to consider the case when y = y0 is fixed. If y0 6= 0, thenf(x, y0) has no singularities, hence continuous. If y0 = 0, then f(x, 0) = x2/x2 = 1,

52 STANLEY YAO XIAO

which is continuous. Now consider the limit as f approaches (0, 0) along two differentcurves. First suppose that x = y = r, and r → 0. In this case we have

limr→0

r2 − r2

r2 + r2= 0.

Now consider x = 2r, y = r. Then

limr→0

4r2 − r2

4r2 + r2=

3

56= 0,

and so there is no limit as (x, y) → (0, 0) for f(x, y). Hence f cannot be continuousat (0, 0).

Question 8.2. Let (X, d) be a complete metric space. Suppose that Fn is a non-empty closed subset of X for n = 1, 2, · · · and that Fn ⊃ Fn+1 for n = 1, 2, · · · . Putγn = supd(x, y) : x, y ∈ Fn and suppose that lim

n→∞γn = 0.

a) Prove that∞⋂n=1

Fn consists of a single point.

b) Show that if we remove the condition that Fn be closed then∞⋂n=1

Fn need not

be a single point.

Proof. a) For each n, choose xn ∈ Fn. Since limn→∞ γn = 0, it follows that for allε > 0 there exists N ∈ N such that γn < ε for all n ≥ N , whence for m,n > N wehave d(xn, xm) ≤ γN < ε. This implies that the sequence (xn)∞n=1 is Cauchy, and sinceX is a complete metric space, it follows that (xn)∞n=1 has a limit, say x. I claim that

x ∈ F =∞⋂n=1

Fn. This is because (xn)∞n=1 ⊂ F1, which is closed and thus contains all

of its limit points. Further, it is clear that x ∈ Fn for every n. Now suppose thatthere exists y ∈ F . We have d(x, y) ≤ γn for every n ≥ 1, and hence d(x, y) < ε forany ε > 0, so d(x, y) = 0 and x = y. Hence F consists of a single point, as desired.

b) Consider the sets Fn = (0, 1/n). Then γn → 0 and Fn+1 ⊂ Fn, but the inter-section F is empty.

Question 8.3. a) Give the definition of a partially ordered set S.

b) Give the definition of a maximal element of a partially ordered set S.

c) State Zorn’s Lemma.

d) A set T of real numbers is said to be rationally linearly independent if wheneverr1x1 + · · · + rnxn = 0 with distinct x1, · · · , xn ∈ T and rational numbers r1, · · · , rn,we have r1 = · · · = rn = 0. Prove that there is a rationally linearly independent setS of real numbers with the property that each real number x can be expressed in theform x = r1x1 + · · ·+ rmxm with x1, · · · , xm ∈ S and r1, · · · , rm ∈ Q.


Proof. a) A partially ordered set S is a set endowed with a partial order, which is anoperation ≤ which satisfies

1) a ≤ a for all a ∈ S;2) a ≤ b and b ≤ a implies that a = b; and3) a ≤ b and b ≤ c implies that a ≤ c.

b) An element m ∈ S is maximal if there does not exist s ∈ S such that m < s.

c) (Zorn’s Lemma) Let S be a partially ordered set with respect to the partial order≤, and suppose that for every chain C, C has an upper bound in S. Then S containsa maximal element.

d) It suffices to show that R has a maximal linearly independent subset with re-spect to Q. We recall that a basis is a maximally linearly independent subset, thus itsuffices to show such a set exists. Let < denote the partial order defined for Q-linearlyindependent subsets of R by A < B if and only if A ⊂ B. Let I be a chain under thispartial order. Then plainly for each chain I there is a maximal element, and thus byZorn’s lemma, there exists a maximally linearly independent set and we are done.

Question 8.4. Let f be a continuous function on [−1, 1] which satisfies∫ 1

−1

f(x)x2kdx = 0

for k = 0, 1, 1, · · · . Show that f is an odd function, in other words, show thatf(−x) = −f(x) for x ∈ [−1, 1].

Proof. We can write f(x) in terms of its Fourier series. Since f is continuous over acompact interval, it is also uniformly continuous, whence its Fourier series convergesto it uniformly. Write f(x) =

∑∞n=0 (an cos(πnx) + bn sin(πnx)). Then since the

convergence is uniform, we have∫ 1

−1

(∞∑n=0

(an cos(πnx) + bn sin(πnx))

)x2kdx =

∞∑n=0

∫ 1

−1

(an cos(πnx) + bn sin(πnx))x2kdx.

Hence it suffices to check that the hypothesis implies that an = 0 for all n ≥ 1.

We have∫ 1

−1

an cos(πnx)x2kdx =1

πnan sin(πnx)x2k|1−1 −

1

πn

∫ 1

−1

(2k)an sin(πnx)x2k−1dx

Question 8.5. Construct, with justification, a meromorphic function whose onlypoles are poles of order 1 at the points m+ ni|m,n ∈ Z.

54 STANLEY YAO XIAO

Proof. We can obtain the desired function via the Weierstrass factorization theorem,which asserts that for any sequence αn with |αn| → ∞ and αn having no accu-mulation points, we may find an entire function f such that f(αn) = 0 and f is notidentically zero. In our case, by enumerating the countable set m + ni : m,n ∈ Zand taking 1/f , we obtain the desired function. However, we may also give an explicitconstruction following the proof of the Weierstrass factorization theorem.

Consider the function

f(z) =∏m∈Z

∏n∈Z

Em2+n2

(z

m+ ni

)where

En(z) = (1− z) exp

(z +

z2

2+ · · ·+ zn

n

).

By a well-known lemma in complex analysis, which asserts that the product∞∏n=1

(1+an)

of complex numbers converges if and only if the sum∞∑n=1

an converges. Hence we may

examine the sum

Question 8.6. Let f ∈ L1[0, 1] and suppose that∫ 1

−1

f(x)g(x)dx = 0

for all measurable functions g ∈ L1[0, 1]. Show that f = 0 almost everywhere withrespect to Lebesgue measure.

Proof. Without loss of generality we may suppose that f > 0 on a set of positivemeasure. In particular, we choose a δ > 0 such that x ∈ [0, 1] : f(x) ≥ δ haspositive measure. Such a set must necessarily exist since otherwise we can writethe set on which f has positive measure as a countable union of sets of the formx ∈ [0, 1] : f(x) ≥ 1/n, and if they all have zero measure then so would their union,which would contradict our hypothesis. Since f is measurable, the set Aδ = x ∈[0, 1] : f(x) ≥ δ is measurable, whence the indicator function on Aδ is measurable.Then we have ∫ 1

−1

f(x)IAδ(x)dx ≥ δm(Aδ) > 0,

so f would not satisfy the hypothesis of the problem. Thus we must have f = 0 foralmost all x ∈ [0, 1].

Question 8.7. a) Let V be the vector space of continuous complex functions on [0, 1]with inner product

(f, g) =

∫ 1

0

f(t)g(t)dt.

Prove that V is not a Hilbert space with respect to the inner product (·, ·).


b) Let X1, X2, · · · be topological spaces. Define the Cartesian product topology on∞∏n=1

Xn.

c) State Tychonoff’s product theorem.

Proof. a) We show that V is not complete with respect to the norm induced by (·, ·).Define

fn(x) =

1, if 0 ≤ x < 1/2

−nx+ n−22, if 1/2 ≤ x < 1/2 + 1/n

0, if 1/2 + 1/n ≤ x ≤ 1.

Write f = χ[0,1/2]. Then we have

‖fn − f‖22 =

∫ 1/2+1/n

1/2

| − nx+ (n− 2)/2|2dx

=

∫ 1/2+1/n

1/2

(n2x2 − n(n− 2)x+ (n− 2)2/4)dx

=

[n2

3x3 − n(n− 2)

2x2 +

(n− 2)2

4x

]1/2+1/n

1/2

=n2

3

(1

n3+

3

2n2+

3

4n

)− n(n− 2)

2

(1

n+

1

n2

)+

(n− 2)2

4n

=1

3n+

1

2+n

4− n

2− 1

2+ 1 +

1

n+

1

n− 1 +

n

4

=7

3n,

which plainly goes to zero as n→∞. However, f is not continuous on [0, 1], and soV is not complete, hence not a Hilbert space.

b) For each n, define πn : X → Xn to be the projection map from X to Xn. Inother words, if x = (x1, x2, · · · , ) ∈

∏∞n=1 Xn, then πn(x) = xn. The product topology

then is the coarsest topology such that πn is continuous for each n ≥ 1.

c) Let Xα, α ∈ A be a collection of compact topological spaces. Then the prod-uct X =

∏α∈AXα is compact with respect to the product topology.

Question 8.8. a) For any complex number z let Im(z) denote the imaginary part ofz. Find a holomorphic function f : A→ D where A = z ∈ C : 0 < Im(z) < 1 andD = z ∈ C : |z| < 1.

b) Prove that D and C are homeomorphic but that there is no holomorphic func-tion f from C onto D.

Proof. a) We can choose f(z) = eiz. Write z = x + iy. Since z ∈ A, it follows that0 < y < 1, whence f(z) = eiz = ei(x+iy) = e−y+ix, so |f(z)| = e−y < 1.

b) One can take a homeomorphism between the unit disc and the square (−1, 1) ×

56 STANLEY YAO XIAO

(−1, 1), and then consider the map x 7→ tan(πx/2) which is a homeomorphism from(−1, 1) to R, and then consider the fact that C ∼= R2. However, by the Riemannmapping theorem C and D are not conformally equivalent, so there is no analytichomeomorphism between the two sets.

Question 8.9. Let f : R3 → R by f(x, y, z) = x+ y + z − sinxyz.

a) Show that there exists an open ball B in R2 around (0, 0) and a continuous functiong : B → R such that g(0, 0) = 0 and f(x, y, g(x, y)) = 0.

b) Determine the Jacobian of g at (0, 0).

Proof. a) We apply the implicit function theorem. In this case, we consider the partial

derivative∂f

∂z(x, y, z) = 1 − xy cosxyz. When (x, y) = (0, 0), the partial derivative

is equal to 1, hence invertible. Thus the implicit function theorem applies and thereexists a desirable B containing (0, 0) and a corresponding g.

b)

Question 8.10. Let f ∈ L1(R), and continuous. Recall the Poisson summationformula which states that, under certain conditions,

∞∑n=−∞

f(n) =∞∑

n=−∞

f(n),

where f(n) =∫∞−∞ f(t) exp(−2πitn)dt. Sufficient conditions are that the sum on the

RHS converge absolutely and that the sum∞∑

n=−∞

f(n+ t)

converges uniformly for 0 ≤ t ≤ 1. Use the Poisson summation formula to prove that∞∑n=1

1

n2 + 1=π

2

(1 +

2

e2π − 1

)− 1

2.

Proof. We set f(t) = 1/(t2 + 1), and consider the Fourier transform

f(n) =

∫ ∞−∞

f(t) exp(e−2πitn)dt.

This amounts to computing the integral∫ ∞−∞

eait

1 + t2dt, a > 0

which we will do via residue theory.

Consider the contour ΓR consisting of the line segment [−R,R] and the half cir-cle on the upper half plane of radius R with center at the origin, which we denote γR.We denote the integrand by g(t). Then g has only a pole at i on the interior of the


contour, and has no poles or zeroes anywhere on the contour. Hence by the residuetheorem and Cauchy’s integral theorem, we have∫

ΓR

g(z)dz = 2πiRes(g, i).

The residue at i can be readily computed to be

2πiRes(g, i) =2πie−a

2i= πe−a.

The contribution from γR can be controlled via Jordan’s lemma, and hence∫ ∞−∞

g(t)dt = πe−a.

The case when a < 0 can be similarly handled by consider the same contour, butreflected across the real axis. Hence we have

f(n) = πe−2π|n|.

The series∑n∈Z

e−2π|n| = 1 + 2∞∑n=1

e−2πn plainly converges absolutely, since it is a

geometric series. Further, the series∑n∈Z

1

1 + (n+ t)2

plainly converges uniformly for 0 ≤ t ≤ 1 by considering the inequality n2 ≤ (n+t)2 ≤(n+ 1)2 and the comparison test. Hence Poisson summation applies, and we have∑

n∈Z

1

n2 + 1=∑n∈Z

πe−2π|n|.

The left hand side is equal to

1 + 2∞∑n=1

1

n2 + 1,

so we have∞∑n=1

1

n2 + 1=−1

2+π

2

∑n∈Z

e−2π|n|.

Now the sum∑n∈Z

e−2π|n| = 1 + 2∞∑n=1

e−2πn = 1 + 2e−2π

1− e−2π= 1 +

2

e2π − 1,

as desired.

58 STANLEY YAO XIAO

9. Spring 2006

9.1. Set Theory and Toplogy.

Question 9.1. a) Prove that the cardinality of R and R2 are the same.b) Using part a), prove that the cardinality of R2 and R3 are the same.c) Prove that the unit sphere in R3

(x, y, z) ∈ R3 : x2 + y2 + z2 = 1has the same cardinality as R.

Proof. a) We use the Schroeder-Bernstein theorem. First note that R =∞⋃n=1

(−n, n),

and each interval has the same cardinality by a simple argument. Now R has thesame cardinality as any open interval because countable unions of infinite sets havethe same cardinality as the set with largest cardinality in the union. Thus it suf-fices to show that (0, 1) × (0, 1) ⊂ R2 has the same cardinality as (0, 1). Considerthe function f : (0, 1) × (0, 1) → (0, 1) which acts on a point (x, y) ∈ (0, 1) × (0, 1)as follows. Write x = 0.x1x2 · · · and y = 0.y1y2 · · · in base 2 representation, thenf(x, y) = 0, x1y1x2y2 · · · . Plainly f(x, y) ∈ (0, 1), and f(x, y) is surjective. Thisshows that |(0, 1)× (0, 1)| ≥ |(0, 1)|. For the reverse direction, for x ∈ (0, 1) considerthe function g : (0, 1)→ (0, 1)× (0, 1) defined by g(x) = (0.x1x3 · · · , 0.x2x4 · · · ). Wenote that g is also surjective, whence |(0, 1)| ≥ |(0, 1)× (0, 1)|, and so they are equal.

b) We can show that R and R3 have the same cardinality via the same argument,except we consider the map f : (0, 1)3 → (0, 1) by f(x, y, z) = 0.x1y1z1x2y2z2 · · · andthe map g : (0, 1) → (0, 1) by g(x) = (0.x1x4 · · · , 0.x2x5 · · · , 0.x3x6 · · · ). By part a),R2 and R3 have the same cardinality.

c) Recall that the unit sphere may be parameterized by x = sin θ sinϕ, y = sin θ cosϕ, z =cos θ. For a fixed value of ϕ, the map f : [0, 2π)→ S2 given by f(θ) = (x, y, z) wherex = sin θ sinϕ, y = sin θ cosϕ, and z = cos θ. This map is injective, and hence|[0, 2π)| ≤ |S2|.

Question 9.2. a) Prove that there is no surjective continuous map from the closedinterval [0, 1] onto the open interval (0, 1).b) Give an example of a surjective continuous map from the open interval (0, 1) ontothe closed interval [0, 1]c) Prove that no example in part b) can be injective.

Proof. a) If f : [0, 1] → R is a continuous function, then f([0, 1]) must necessar-ily be compact. Indeed, for any compact subset K of a topological space X, withf a continuous function on K, we must have f(K) be compact. To see this, letAα, α ∈ I for some index set I be an open cover for f(K). By continuity, we seethat f−1(Aα) is open for each α, and the set f−1(Aα) : α ∈ I is an open cover forK. By the compactness of K, there exist a finite list of indices 1, · · · , n such thatf−1(A1), · · · , f−1(An) form a finite subcover of K. Then f(f−1(Aj)) = Aj form afinite subcover of f(K), whence f(K) is compact.


Returning to the problem, since (0, 1) is not a compact subset of R with respectto the usual topology, it cannot be the continuous image of a compact set, so no suchmap exists.

b) Let f(x) =1

2+

1

2sin(4πx), then f maps any interval of length 1/2 onto the

interval [0, 1], so in particular it must map (0, 1) onto [0, 1].

c) If the example in b) can be made to be injective, then f−1 would be a contin-uous mapping from [0, 1] onto (0, 1), which is impossible (since both (0, 1) and [0, 1]are connected).

Question 9.3. Let (X1, d1) and (X2, d2) be metric spaces. Let f : X1 → X2 be acontinuous map such that

d1(p, q) ≤ d2(f(p), f(q))

for every pair of points p, q ∈ X1. Assume that f is surjective.

a) Prove that f must be injective.b) If X1 is complete, then must X2 be complete? Give a proof or counterexample.c) If X2 is complete, then must X1 be complete? Give a proof or counterexample.

Proof. a) Suppose that f(p) = f(q). Then d1(p, q) ≤ d2(f(p), f(q)) = 0, so p = q,whence f is injective.

b) Yes; the hypothesis and part a) imply that X1 and X2 must be homeomorphicas metric spaces.

c) Yes; same reasoning as part b).

Question 9.4. Let X be a metrizable topological space. Prove that the followingare equivalent

i) X is bounded under every metric that induces the topology of X;ii) Every continuous function on f : X → R is bounded; andiii) X is compact.

Proof. Suppose that i) holds, and let f : X → R be any continuous function.

9.2. Real Analysis.

Question 9.5. Recall that a sequence fn in a Hilbert space H converges weaklyto f ∈ H if

〈fn, g〉 → 〈f, g〉 for all g ∈ H.Also recall that fn converges strongly to f ∈ H if

‖fn − f‖ → 0.

Give an example of a sequence in L2(R) that converges weakly, but not strongly.Make sure to justify your assertions.

60 STANLEY YAO XIAO

Proof. Let fn(x) =

sin(nx) if x ∈ [0, 2π),

0 if x 6∈ [0, 2π)., for n ≥ 1. Since for any g ∈ L2(R), the

fact that the Fourier series of g converges to g it implies that∫∞

0fn(x)g(x)dx → 0,

whence fn → 0 weakly.

Now, let us examine

‖fn − 0‖22 =

∫ 2π

0

sin2(nx)dx

=

∫ 2π

0

(1− cos(2nx)

2

)dx

= π − (sin(2nx)/(4n))|2π0= π,

which does not tend to zero as n→∞, so fn(x) does not tend to zero strongly.

Question 9.6. Let fn : [0, 1]→ R be a sequence of continuous functions satisfying∫ 1

0

(fn(x))2dx ≤ 1

for all n. Define gn : [0, 1]→ R by

gn(x) =

∫ 1

0

fn(y)√x+ ydy.

a) Find a constant K ≥ 0 such that |gn(x)| ≤ K for all n and x ∈ [0, 1]b) Prove that there exists a subsequence of gn that converges uniformly.

Proof. a) By the Cauchy-Schwartz inequality, we have

|gn(x)|2 ≤(∫ 1

0

(fn(y))2dy

)(∫ 1

0

(x+ y)dy

),

whence |gn(x)|2 ≤ (1)(x+ 1/2) ≤ 3/2, since x ∈ [0, 1]. This bound is uniform in bothn and x.

b) We have shown that gn(x) is uniformly bounded. Now we show that gn(x) isequicontinuous. Let ε > 0, and suppose |x1 − x2| < δ. Further, we can assume thatx1 6= 0, since if both x1 = x2 = 0 then we would be done. Now consider

|gn(x1)− gn(x2)| =∣∣∣∣∫ 1

0

fn(y)(√x1 + y −

√x2 + y)dy

∣∣∣∣ .We can write this as ∣∣∣∣∫ 1

0

fn(y)x1 − x2√

x1 + y +√x2 + y

dy

∣∣∣∣which we can bound above via the Cauchy-Schwartz inequality∣∣∣∣∫ 1

0

fn(y)x1 − x2√

x1 + y +√x2 + y

dy

∣∣∣∣ ≤ |x1−x2|(∫ 1

0

(fn(y))2dy

)1/2(∫ 1

0

dy

x1 + x2 + 2y

)1/2

.


This is further bounded above by

|x1 − x2|√

1

2(log(x1 + x2 + 2)− log(x1 + x2)).

This is further bounded above by C√|x1 − x2|, for some contant C > 0 which is inde-

pendent of n, whence gn(x) is equicontinuous. Thus by the Arzela-Ascoli theorem,there exists a uniformly convergent subsequence.

Question 9.7. Let ϕn : [0, 1] → R be a sequence of non-negative continuousfunctions such that the limit

limn→∞

∫ 1

0

xkϕn(x)dx

exists for every k ≥ 0. Prove that the sequence of real numbers

In(f) =

∫ 1

0

f(x)ϕn(x)dx

converges for every continuous function f : [0, 1]→ R.

Proof. Let f : [0, 1] → bR be a continuous function, and let ε > 0 be given. By theStone-Weierstrass theorem, there exists a polynomial p(x) such that |f(x)− p(x)| <ε/2 for every x ∈ [0, 1]. Set d = deg(p) and M denote the maximum of the absolutevalues of the coefficients of p. Choose N ∈ N sufficiently large so that for all m,n ≥ N ,we have

sup0≤k≤d

∣∣∣∣∫ 1

0

xk(ϕn(x)− ϕm(x))dx

∣∣∣∣ < ε

2M.

Now, consider

|In(f)− Im(f)| ≤∣∣∣∣∫ 1

0

(f(x)− p(x))(ϕn(x)− ϕm(x))dx

∣∣∣∣+

∣∣∣∣∫ 1

0

p(x)(ϕn(x)− ϕm(x))dx

∣∣∣∣<ε

2+ε

2

hence In(f) is Cauchy. Since R is complete, In(f) is convergent, as desired.

9.3. Complex Analysis.

Question 9.8. Use contour integration to evaluate the improper integral∫ ∞0

1

1 + x4dx.

Make sure to justify your steps.

Proof. Let I =∫∞

0dx

1+x4. Then we have

2I =

∫ ∞−∞

dx

1 + x4,

by symmetry. Now consider the contour ΓR defined by the union of the line segment[−R,R] on the real line and the contour γR which is the half circle in the upper half

62 STANLEY YAO XIAO

plane centered at the origin with radius R. Further, the two roots of x4 + 1 in theupper half plane are 1√

2(1 + i) and 1√

2(−1 + i). By the residue theorem, we have∫

ΓR

dz

1 + z4= 2πiRes(f ; (1/

√2)(1 + i)) + 2πiRes(f ; (1/

√2)(−1 + i)).

The two residues are respectively1

4(−1/√

2 + i/√

2)3=

1

2√

2(1 + i),

1

2√

2(−1 + i).

Now, we estimate the integral∫γR

dz

1 + z4=

∫ π

0

Rieitdt

1 +R4e4it.

To estimate this integral, note that∣∣∣∣∫γR

dz

1 + z4

∣∣∣∣ ≤ ∫ π

0

∣∣∣∣ R

R4 − 1

∣∣∣∣ dt < 2π

R3.

In particular, this vanishes as R→∞. Hence

limR→∞

∫ΓR

dz

1 + z4=

∫ ∞−∞

dx

1 + x4= π√

2.

Question 9.9. Suppose f is a bijection from the unit disc to itself such that f isanalytic and fixes the origin. Prove that |f ′(0)| = 1.

Proof. Since f is a bijection, for each z ∈ D there exists u ∈ D such that f(u) = z.This map is invertible. Further, note that f−1 satisfies the same hypothesis as f .Now, if for all u ∈ D we have |f(u)| = |u|, then we must have that f(z) = αz forsome α ∈ C such that |α| = 1. Assume this is not the case. Then, by replacing f withf−1 if necessary, we can assume that there exists u, v ∈ D with |u| < |v| such thatf(u) = v. In that case, we have |f(u)/u| > 1. However, we note that g(z) = f(z)/zfor z 6= 0 and g(0) = f ′(0) is an analytic function on D, and by the proof of theSchwarz lemma we know that |g(z)| ≤ 1 for all z ∈ D, this is a contradiction. Henceno such u, v can exist and we must have f(z) = αz for some |α| = 1, whence theresult holds.

Question 9.10. Let f and g be entire such that |f(z)| < |g(z)| for all z satisfying|z| ≥M for some real constant M ≥ 0. Prove that f/g is rational.

Proof. Let us consider the function h(z) = f(z)/g(z). Since f, g are entire, h canhave at most poles of finite order in C. Thus the only remaining consideration iswhether f/g has an essential singularity at ∞. If this was the case, then by Picard’stheorem there exists a sequence zn with |zn| → ∞ and such that h(zn) → 2.This implies that |h(zn)| > 3/2 for n sufficiently large, or equivalently, |f(zn)| >3|g(zn)|/2 > |g(zn)| for n sufficiently large. This violates the hypothesis, whence f/gis rational.

Question 9.11. Let U and V be open connected subsets in the complex plane. Letf : U → V be analytic. Assume that f−1(K) is compact whenever K is a compactsubset of V .


a) Prove that f is not a constant function.b) Prove that f(U) is a closed subset of V .c) Prove that f(U) = V .

Proof. a) Suppose f is constant, say that f(z) ≡ c for some c ∈ V . Then f−1(c) =U , which is not compact, contradicting our hypothesis. Hence f is not constant.

b) Let zn ⊂ U be a sequence such that f(zn) is a convergent sequence in V .Note that f(zn) ∪ L, where L = limn→∞ f(zn), is a compact subset of V . Thusf−1(f(zn) ∪ L) is compact, whence there exists a subsequence of it, say xm,which is convergent. Write x = limxm. Then we must have lim f(xm) = f(x) = L,so f(U) is closed in V .

c) By the open mapping theorem and part b), f(U) is a clopen subset of V . Thisimplies that f(U) = V since f(U) is non-empty.

University of Waterloo, Dept. of Pure Mathematics, Waterloo, ON, N2L 3G1,Canada

E-mail address: [email protected]

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