ANALYSIS QUALIFYING EXAM PROBLEMSSOLUTIONS BY DAVE GAEBLER

Contents

Fall 2001 1Winter 2002 8Spring 2002 11Fall 2002 15Winter 2003 18Fall 2003 22Winter 2004 24Fall 2004 26Winter 2005 28Fall 2005 29Spring 2006 31Fall 2006 35

Fall 2001

Problem R1: Consider real numbers an,m for n = 1, 2, . . . and m = 1, 2, . . .and assume that the inner and outer sums in the expressions

A =∞∑n=1

[ ∞∑m=1

an,m

]

B =∞∑m=1

[ ∞∑n=1

an,m

]are absolutely convergent.(a) Give an example that shows that we may have A 6= B.(b) Under what reasonable additional assumptions on an,m can we con-

clude A = B?

Solution.(a) Let

an,m =sign(n−m)

2|n−m|.

Then the inner and outer sums are absolutely convergent, but one wayall the inner sums are positive, and the other way they’re negative.

(b) If all the terms are nonnegative, or if∑∑

|an,m| converges, Fubini’stheorem guarantees that the two iterated sums are equal.

1

2 ANALYSIS QUALS

Problem R2: Let n ≥ 1. Let O(n) denote the set of all real n× n matricesG which satisfy

GTG = I

where I is the identity matrix.(a) Prove that O(n) is compact.(b) Prove that O(n) is not connected.

Solution.(a) The columns of an orthogonal matrix are orthonormal; in particular,

they each have norm 1. This implies that the Euclidean norm of anorthogonal matrix is

√n. Thus, O(n) is a bounded subset of Rn2

, andit is closed because it’s the inverse image under the continuous functionG 7→ GTG of the closed set I. By the Heine-Borel theorem, O(n) iscompact.

(b) If O(n) were connected, its image under the continuous function det :O(n) → R would be connected as well. But det(G) = ±1 for all G ∈O(n), and there are matrices with determinant 1 (e.g. the identity) aswell as those with determinant −1 (e.g. negate the last column of theidentity). Hence O(n) is not connected.

Problem R3: Prove the mean value theorem: Let f : R → R be a differen-tiable function. For every a < b there exists a < ξ < b such that

f(b)− f(a)b− a

= f ′(ξ).

Solution. Let

g(x) = f(x)− f(a)− f(b)− f(a)b− a

(x− a).

Then g is differentiable and g(a) = g(b) = 0. Since g is continuous on thecompact set [a, b], it attains a maximum and a minimum. If these are bothat the endpoints, g is constant on [a, b], so g′(x) = 0 for all x ∈ (a, b).On the other hand, if g attains either a max or a min on the interval,it must occur at a point where g′(x) = 0. Either way, ∃ξ ∈ (a, b) withg′(ξ) = 0⇒ f ′(ξ) = f(b)−f(a)

b−a .

Problem R4: The Fourier transform f of a function f in L1(R) is defined as

f(ξ) =∫

Re−2πixξf(x)dx.

(a) State how the Fourier transform of a function in L2(R) is defined. (Youdo not need to prove claims which you use to state this. Do not usethe fact (b) below.)

(b) Let f be in L2(R). Define

Mf(ξ) = supr>0

∣∣∣∣∫ r

−rf(x)e−2πixξdx

∣∣∣∣ .A deep fact proved by Carleson and Hunt states that

‖Mf‖2 ≤ C‖f‖2

ANALYSIS QUALS 3

for some universal constant C. Use this theorem to prove

f(ξ) = limr→∞

∫ r

−re−2πixξf(x)dx

for almost every ξ ∈ R.

Solution.(a) For f ∈ L1∩L2 the Fourier transform is defined already. Since CC(R) is

dense in L2 and is contained in L1, this defines the Fourier transformon a dense subset of L2. It turns out that the Fourier transformrestricted to L1 ∩L2 is an isometry, i.e. ‖f‖2 = ‖f‖2. An isometry ona dense subspace can always be extended to an isometry on the wholespace, which defines the Fourier transform L2 → L2.

(b) Let

Eα =ξ ∈ R : lim sup

r→∞

∣∣∣∣∫ r

−re−2πixξf(x)dx− f(ξ)

∣∣∣∣ > 2α.

For any ε > 0, ∃g ∈ L1 ∩ L2 with ‖f − g‖2 < ε. Then∫ r

−re−2πixξf(x)dx−f(ξ) =

(∫ r

−re−2πixξg(x)dx− g(ξ)

)+∫ r

−re−2πixξ(f(x)−g(x))dx+(g − f(ξ)).

Since

lim supr→∞

∣∣∣∣∫ r

−re−2πixξg(x)dx− g(ξ)

∣∣∣∣ = 0

for all ξ, this implies Eα ⊂ Fα +Gα, where

Fα = ξ ∈ R : M(f − g)(ξ) > αand

Gα = ξ ∈ R : |f − g(ξ)| > α.Now by Chebyshev’s inequality,

m(ξ : M(f − g)(ξ) > α) < ‖M(f − g)‖22α2

<C2ε2

α2

and

m(ξ : |f − g(ξ)| > α) < ‖f − g‖22

α2=‖f − g‖22

α2<ε2

α2,

so

m(Eα) <(C2 + 1)ε2

α2.

This is true for any ε, so m(Eα) = 0. But this is true for any α, so

m

(ξ :∫ r

−re−2πixξf(x)dx 6→ f(ξ)

)= m

( ∞⋃n=1

E1/n

)≤∞∑n=1

m(E1/n) = 0.

Problem R5: For n ≥ 1 let `1(n), `2(n), `∞(n) be the Banach space Rnequipped with the norm

n∑i=1

|xi|,

(n∑i=1

|xi|2)1/2

, supi=1,...,n

|xi|

4 ANALYSIS QUALS

respectively. Answer the following questions and prove your assertions:(a) Which of the Banach spaces `1(2), `2(2), `∞(2) are isometrically iso-

morphic?(b) Which of the Banach spaces `1(3), `2(3), `∞(3) are isometrically iso-

morphic?

Solution. First, `2 is not isometrically isomorphic to either `1 or `∞ ineither 2 or 3 dimensions (or, in fact, any dimension higher than 1). This isbecause the parallelogram law

‖a+ b‖2 + ‖a− b‖2 = 2‖a‖2 + 2‖b‖2

holds in `2, but not in `1 or `∞. (A simple counterexample for both `1 and`∞ is a = (1, 0) or (1, 0, 0) and b = (0, 1) or (0, 0, 1).)(a) In 2 dimensions, `1 ' `∞. One isomorphism is given by

(a, b) 7→(a+ b

2,a− b

2

).

One can easily check by breaking into cases that

∣∣∣∣a+ b

2

∣∣∣∣+∣∣∣∣a− b2

∣∣∣∣is |a| if |a| ≥ |b|, and |b| if |b| ≥ |a|. So this is an isometry.

(b) In 3 dimensions, `1 6' `∞. To prove this we will count the numberof extreme points on the unit ball. An extreme point of a convex setS is one that is not in the interior of any line segment lying in S. Itis easy to verify that the property of being an extreme point of theunit ball is preserved by an isometric isomorphism, so if `1 ' `∞ thenthe unit balls of `1 and `∞ have the same number of extreme points.However, the unit ball of `1 is an octahedron, which has six extremepoints, whereas the unit ball of `∞ is a cube, which has eight.

Problem R6: Let V be the complex Banach space `∞(N), i.e. the spaceof all sequences x = (xn)n=1,2,... with the norm ‖x‖ = supn |xn|. Everysequence f in `1(N) gives rise to a linear functional φf : V → C by theformula φf (x) =

∑∞n=1 fnxn.

a) Prove that φf is continuous for each f ∈ `1(N).b) Prove that there are elements in the dual space of V which are not of

the form φf for any f ∈ `1(N).

Solution.

ANALYSIS QUALS 5

a) Since φf is a linear functional, its continuity follows from its bound-edness. Since

|φf (x)| =

∣∣∣∣∣∞∑n=1

fnxn

∣∣∣∣∣≤∞∑n=1

|fn||xn|

≤∞∑n=1

|fn|‖x‖∞

= ‖x‖∞∞∑n=1

|fn|

= ‖f‖1‖x‖∞,φf is bounded.

b) One can easily check that the set of convergent sequences forms asubspace of `∞ and that lim is a bounded linear functional on thissubspace. By the Hahn-Banach theorem there exists an extension toa linear functional f : `∞ → C such that f(x) = lim(x) whenever x isconvergent. But for f ∈ `1, φf cannot agree with lim on all convergentsequences, because changing a finite number of components of x willnot affect its limit but can affect φf (x) (e.g. change xi for some iwhere fi 6= 0).

Problem C1: Find an explicit conformal mapping from the region

|z| < 1 \ [0, 1)

onto the upper half plane Imz > 0.

Solution. The mapping ζ = i w+w−1 takes this region to the UHP minus the

slit from i to i∞. The mapping ξ = − 1ζ takes this to the UHP minus the

finite slit from 0 to i. Finally, the mapping z =√ξ2 + 1 (found by invert-

ing the formula ξ =√z2 − 1 from the Schwarz-Christoffel transformation)

sends this to the UHP. Explicitly, ξ = 2√w

w+1 . (Handy’s solution is nicer.)

Problem C2: Evaluate the integral∫ ∞−∞

cosx dx9 + 10x2 + x4

.

Solution. Let f(z) = eiz

9+10z2+z4 . Since 9 + 10z2 + z4 = (z2 + 9)(z2 + 1) =(z + i)(z − i)(z + 3i)(z − 3i), the only singularities are simple poles at ±iand ±3i. The residues in the upper half-plane are

Res(f, i) =ei·i

ddz (9 + 10z2 + z4)|z=i

=e−1

16i

and

Res(f, 3i) =ei·3i

ddz (9 + 10z2 + z4)|z=3i

=e−3

−48i.

6 ANALYSIS QUALS

By the Residue Theorem applied to a semicircular contour Γ from −R toR in the upper half plane,

2πi(e−1

16i+

e−3

−48i

)=∫

Γ

f(z)dz =∫ R

−Rf(z)dz +

∫C

f(z)dz

where C is the curved part of the contour. Since |eiz| ≤ 1 on the UHP,|f(z)| ≤ 1

R4−10R2−9 for large R; since the length of C is πR,∫Cf(z)dz → 0

as R→∞ by the ML estimate. So∫ ∞−∞

f(z)dz = limR→∞

∫ R

−Rf(z)dz =

π(3e−1 − e−3)24

.

Now cos x10+9x2+1 = Ref(z), so∫ ∞−∞

cosx10 + 9x2 + 1

dx = Re

∫ ∞−∞

f(z)dz =π(3e−1 − e−3)

24.

Problem C3: Define

∂

∂z=

12

(∂

∂x− i ∂

∂y

)and

∂

∂z=

12

(∂

∂x+ i

∂

∂y

).

(a) For m and n positive integers, calculate

∂

∂zznzm.

(b) Let P (x, y) be a polynomial in the two real variables x and y. ThenP has the form

P (x, y) =N∑n=0

N∑m=0

an,mznzm

but you do not need to prove it. Prove, howover, that if

∂

∂zP (x, y) = 0,

then an,m = 0 for all m > 0.

Solution.

ANALYSIS QUALS 7

(a)

∂

∂xznzm =

∂

∂x(x+ iy)n(x− iy)m

= n(x+ iy)n−1(x− iy)m +m(x− iy)m−1(x+ iy)n

= (x+ iy)n−1(x− iy)m−1(n(x− iy) +m(x+ iy)

)∂

∂yznzm = i(x+ iy)n−1(x− iy)m−1

(n(x− iy)−m(x+ iy)

)∂

∂zznzm =

12

(∂

∂x+ i

∂

∂y

)znzm

=12

2m(x+ iy)n(x− iy)m−1 = mznzm−1.

(b) By linearity,

∂

∂zP (x, y) =

∑n

∑m

an,mmznzm−1 =

∑m

mzm−1

(∑n

an,mzn

).

The m = 0 term is automatically zero, but for m > 0 this can onlybe identically zero if

∑n an,mz

n = 0. (We are assuming here that apolynomial in x and y has a unique representation as a polynomialin z and z, i.e. that terms with different powers of z and z cannotsomehow cancel each other; I take this to be part of what we’re allowedto assume.) But a polynomial in z is zero only if each coefficient iszero, so an,m = 0 for m > 0 and for all n.

Problem C4: Let u(z) be a harmonic function on the entire plane C suchthat ∫ ∫

C|u(z)|2dxdy <∞.

Prove that u(z) = 0 for all z.

Solution. For any z, by the mean value property,

u(z) =1

πR2

∫ ∫DR

u(z)dz,

where DR is the disk of radius R centered at z. Then

|u(z)| ≤ 1πR2

∫ ∫DR

|u(z)|dxdy.

By the Cauchy-Schwarz inequality applied to L2(DR), this is at most

1πR2

√(∫ ∫DR

|u(z)|2dxdy)

(πR2) ≤ 1R√π

√∫ ∫C|u(z)|2dxdy.

Letting R→∞, we see that |u(z)| = 0.

8 ANALYSIS QUALS

Winter 2002

Problem 1: Suppose fn is a sequence of continuous functions on [0, 1] whichconverges to a continuous function f on [0, 1]. Does it follow that fn con-verges uniformly? Give a proof or provide a counterexample.

Solution. No, it does not follow. Pick your favorite of the standard coun-terexamples. For example, the triangular spikes

fn(x) =

2nx 0 ≤ x ≤ 1

2n

1− 2n(x− 1

2n

)1

2n ≤ x ≤1n

0 1n ≤ x ≤ 1

converge to 0 on [0, 1], but not uniformly because they all have uniformnorm 1.

Problem 2: Let an be a sequence of positive numbers converging to 0. Showthat given any x > 0 there exist non-negative integers k1, k2, . . . such that∑knan = x.

Solution. The obvious “greedy algorithm” works, i.e. just pick each kn tobe as large as possible without making the sum exceed x. Formally, define

Sn =n∑j=1

kjaj

and inductively define

kn+1 =⌊x− Snan+1

⌋.

Then

0 ≤ x− Snan+1

− kn+1 < 1⇒ 0 ≤ x− Sn+1 < an+1.

Since an → 0, we can take the limit of this inequality to conclude thatSn → x.

Problem 3: Let f and g be Lebesgue integrable functions on the interval[0, 1]. Set

F (x) =∫ x

0

f(y)dy, G(y) =∫ y

0

g(x)dx

where dy and dx denote Lebesgue measure. Assume h(x, y) = f(y)g(x) isLebesgue measurable on [0, 1]× [0, 1]. Prove∫ 1

0

F (x)g(x)dx = F (1)G(1)−∫ 1

0

f(y)G(y)dy.

Solution. By the definition of F ,∫ 1

0

F (x)g(x)dx =∫ 1

0

g(x)∫ x

0

f(y)dydx =∫

[0,1]2f(y)g(x)χy≤xdA =

∫[0,1]2

f(y)g(x)(1−χy>x)dA.

ANALYSIS QUALS 9

Here we have used Fubini’s theorem to turn the iterated integral into atwo-dimensional integral; this is legitimate because∫

[0,1]2|f(y)g(x)χy≤xdA ≤

∫[0,1]2

|f(y)g(x)| =(∫ 1

0

|f(y)|dy)(∫ 1

0

|g(x)|dx)<∞

by Tonelli’s theorem. Returning to our series of integrals above, the last isequal to∫

[0,1]2f(y)g(x)dA−

∫[0,1]2

f(y)g(x)χy>xdA =(∫ 1

0

f(y)dy)(∫ 1

0

g(x)dx)−∫ 1

0

∫ y

0

f(y)g(x)dx

= F (1)G(1)−∫ 1

0

f(y)G(y)dy.

Problem 4: Let F (x) be a bounded real function on R such that

limx→−∞

F (x) = 0

and such that for all ε > 0 there exists δ > 0 such that whenever (aj , bj),1 ≤ j ≤ n is a finite family if pairwise disjoint open intervals,∑

(bj − aj) < δ ⇒∑|F (bj)− F (aj)| < ε.

Prove that there is f ∈ L1(R) such that for all x ∈ R,

F (x) =∫ x

−∞f(t)dt.

Solution. This is the standard theorem on absolutely continuous functions;see for example page 106 of Folland, or pp. 146-148 of Rudin. Dependingon how much you’re allowed to assume, writing this up could be a realpain (e.g. if you can assume that BV functions are differentiable a.e. itmakes life easier, but if you want to prove that from scratch it could takea while).

Problem 5: Let f : [0, 1] → C be Lebesgue measurable. Assume fg ∈L2([0, 1]) for every g ∈ L2([0, 1]). Prove that f ∈ L∞([0, 1]).

Solution. (From Handy.) Let

Ek = x ∈ [0, 1] : 2k ≤ |f(x)| < 2k+1,

and let mk = m(Ek). Suppose f /∈ L∞; then mk > 0 for infinitely many k.Now let

g =∞∑k=0

2−k1√mk

χEk

where the summand is understood to be zero if mk = 0. Then∫ 1

0

g2(x)dx =∞∑k=0

2−2k <∞

10 ANALYSIS QUALS

so g ∈ L2. However,

f ≥∑

2kχEk

⇒ fg ≥∑mk 6=0

1√mk

χEk

⇒ f2g2 ≥∑mk 6=0

1mk

χEk

⇒∫ 1

0

f2g2 ≥∑mk 6=0

1

which is infinite because there are infinitely many k with mk 6= 0. Hencefg /∈ L2.

Problem 9: Let D be the domain in the complex plane C that is the in-tersection of the two open disks centered at ±1 whose boundary circlespass through ±i. Find a conformal map f of D onto the open unit disk∆ = |w| < 1 such that f(i) = 1 and f(−i) = −1. (You may express fas a composition of other specific maps.) What are the images of arcs ofcircles passing through ±i under your map f? (Justify your answer.)

Solution. The map ζ = z−iz+i sends the region D to z : θ < arg(z) < θ+ 2π,

where θ = 2 arctan(√

2 + 1). To see this, consider that the fractional linearmapping z 7→ z−i

z+i maps i to 0 and −i to ∞, so circles through i and −imust go to circles through 0 and∞, i.e. rays. We can find the specific raysthat are the images of our circles by plugging in z = ±(

√2 − 1), resulting

in angles of ±2 arctan( 1√2−1

) = ±2 arctan(√

2− 1). (News flash: Turns outθ = π

4 .) To find which of the two sectors bounded by these rays is ourimage, we note that 0 maps to −1. We now compose this map with themaps ξ = e−iθz, which rotates this to the sector 0 < arg(z) < 2(π−θ), andη = ξπ/2(π−θ) which maps this to the upper half plane. We thus obtain amap from D to the upper half plane. We now compose this with the mapw = λ z−iz+i , where λ is a unimodular constant to be chosen. This maps tothe unit disk; so far i 7→ 0 7→ −1 and −i 7→ ∞ 7→ 1, so we choose λ = −1.

Circles through ±i are mapped to rays, which are mapped to other rays,which are mapped to other rays; the final transformation which maps into∆ sends these to circular arcs through ±1.

Problem 11: Show by contour integration that∫ 2π

0

dθ

x+ cos θ=

2π√x2 − 1

, x > 1.

Determine for which complex values of z the integral∫ 2π

0

dθ

z + cos θ

exists and evaluate the integral. Justify your reasoning.

ANALYSIS QUALS 11

Solution. Let z = eiθ, so cos θ = z+1/z2 and dz = ieiθdθ ⇒ dθ = dz

iz . Then∫ 2π

0

dθ

x+ cos θ= −2i

∮dz

z2 + 2xz + 1,

where the integral is taken around the unit circle. Of the poles −x ±√x2 − 1, only −x+

√x2 − 1 lies inside the unit circle; it has residue

1ddz (z2 + 2xz + 1)|z=−x+

√x2−1

=1

2√x2 − 1

,

so by the Residue Theorem,∫ 2π

0

dθ

x+ cos θ= −2i

(2πi

12√x2 − 1

)=

2π√x2 − 1

.

On the region C \ [−1, 1], both∫ 2π

0dθ

z+cos θ and 2π√z2−1

are analytic func-tions of z, and since we have shown them to be equal for z ∈ (1,∞), theuniqueness principle guarantees that they are equal on C \ [−1, 1]. (Weuse the principal branch of the square root, with the branch cut along thenegative real axis, and the square root of a positive real number taken tobe positive.) For z ∈ [−1, 1] the integral diverges since the denominatorequals zero for at least one value of θ.

Problem 12: Let S be a sequence of points in the complex plane that con-verges to 0. Let f(z) be defined and analytic on some disk centered at 0except possibly at the point of S and at 0. Show that either f(z) extends tobe meromorphic in some disk containing 0, or else for any complex numberw there is a sequence ζj such that ζj → 0 and f(ζj)→ w.

Solution. Assume WLOG that f has no removable singularities, since oth-erwise we may redefine it appropriately at those points. Suppose that f ismeromorphic at 0 and is eventually meromorphic at the points zj of S, sayfor j ≥ N . Then if r = minj<N |zj |, f is meromorphic on the disk |z| < r.Conversely, suppose that either 0 is an essential singularity, or infinitelymany of the zj are essential singularities. If 0 is an essential singularity, theCasorati-Weierstrass theorem immediately implies the result. Suppose nowthat there is a subsequence zjk of essential singularities. Assume WLOGthat |zjk | < 1

2k(since otherwise we may take a sub-subsequence). Then

for any w, the Casorati-Weierstrass theorem guarantees a point ζk with|ζk − zjk | < 1

2kand |f(ζk) − w| < 1

2k. Then |ζk| < 1

2k−1 , so ζk → 0, andf(ζk)→ w.

Spring 2002

Problem 1: Let V be a finite dimensional real vector space, and let ‖‖V bea norm on V . Let P be the set of one-dimensional linear subspaces of V(P is called a real projective space). For W1,W2 ∈ P define

d(W1,W2) = inf‖v1 − v2‖ : vj ∈Wj and ‖vj‖V = 1.

Prove that d is a metric on P and that P is compact with respect to thismetric.

12 ANALYSIS QUALS

Solution. What the crap. This is so a geometry problem, not an analysisproblem. Not to mention that using inf instead of min is totally unnecessarysince each subspace only has two unit vectors. Whatever.

Problem 4: Let f(x) be a sequence of Borel measurable real-valued functionson the interval [0, 1] such that fn(x)→ 0 for all x ∈ [0, 1]. Prove there is aBorel set A ⊂ [0, 1] such that(i) m([0, 1] ∩A′) < ε;(ii) fn(x)→ 0 uniformly on A.

Solution. This is just Egorov’s theorem, of course, which you can look upin any real analysis book; in case you’re lazy, here’s a proof: Let

EN,m =∞⋂n=N

x : |fn(x)| < 1m.

This is the set for which fn(x) stays less than 1m . Since ∪NEN,m is the

set of points where fn → 0, which is all of [0, 1] by hypothesis, and sinceEN ⊂ EN+1, limN→∞m(EN,m) = 1 for all m. Choose K with ε > 1

2Kand

N1, N2, . . . such that

m (ENm,m) > 1− 12K+m

, m = 1, 2, . . . .

Let A = ∩mENm,m. Then

m(A′) = m (∩ENm,m)′ = m(∪E′Nm,m

)≤∑

m(ENm,m)′ ≤∑ 1

2K+m=

12K

< ε.

Moreover, for any δ > 0, choose m such that 1m < δ; then because A ⊂

ENm,m, |fn(x)| < 1m < δ for n > Nm, for all x ∈ A. Hence fn → 0

uniformly on A.

Problem 6: On the two-point space 0, 1 let µ be the measure such thatµ(0) = µ(1) = 1

2 . Let X be the infinite product space

X =∞∏j=1

Xj

where each Xj = 0, 1, and let σ be the product measure on X (defined byσ(∪nj=1x : xj = aj) = 2−n). Find an explicit mapping f : X → [0, 1] suchthat there exists X ′ ⊂ X, with µ(X ′) = 0, and Y ′ ⊂ [0, 1] with Lebesguemeasure zero such that

f : X \X ′ → Y \ Y ′

is one-to-one and measure preserving, i.e. m(F (E)) = µ(E) where m de-notes Lebesgue measure.

Solution. This is one of those problems where it’s clear what’s going on (apoint in X corresponds to a binary expansion of a point in [0, 1]), but it’sa headache to write up. The map is

f(xn) =∞∑n=1

xn2n

ANALYSIS QUALS 13

and the sets areX = eventually constant sequences and Y = dyadic rational points q/2k.The latter has measure zero because it’s countable; the former has measurezero because it can be explicitly written as

∞⋃N=1

∞⋂n=N

x : xn = 0

plus a corresponding set where xn is eventually 1. Now ∩N+kn=Nxn = 0 is

a union of 2N−1 sets where the first N + k terms are specified, which hasmeasure 2−N−k by definition; this measure approaches zero as k →∞.

Finally, we need to show that f is one-to-one and measure preservingonce the appropriate sets are excised. Letxn and yn ∈ X, and let mbe the first point at which they differ; suppose WLOG xm = 1 and ym = 0Then

f(xn)− f(yn) =1

2m+

∞∑j=m+1

xj − yj2j

> 0

because xj − yj ∈ −1, 0, 1 and cannot always be −1 if x and y are noteventually constant. Thus f is one-to-one. It’s measure-preserving becauseeach of the generating sets ∩nj=1x : xj = aj is mapped to a dyadic intervalwith the same measure.

Problem 9: Let f be any conformal mapping from the strip S = z ∈ C :−1 < Iz < 1 onto the unit disc D = z : |z| < 1 such that uniformly iny ∈ (−1, 1),

limx→∞

f(x+ iy) = 1, and limx→−∞

f(x+ iy) = −1.

Find the images in D of the set of horizontal lines in S and the set of verticalline segments in S.

Solution. The conformal map ζ = eπ/2(z+i) maps S to the UHP; composingthis with the map w = z−i

z+i maps to D. Any other conformal map mustcome from composing with a conformal self-map of the unit disk, which hasthe form w′ = λ w−a

1−aw for some |λ| = 1 and |a| < 1. One can easily checkthat our initial map has the desired uniform property; the self-maps of theunit disk are all uniformly continuous because they are continuous on theclosed disk, so we need to find which ones fix ±1. The equations

λ1− a1− a

⇒ λ− λa = 1− a

and

λ−1− a1 + a

= −1⇒ −λ− λa = −1− a

together imply λ = 1 and λ = aa , so w′ = z−a

1−az for some real a < 1.The exponential map takes horizontal lines to rays through the origin;

these in turn are mapped to circular arcs through ±1, since they are “cir-cles” through 0 and ∞. The subsequent conformal self-map of the disksends them to other circles through ±1 since these points are fixed. Thevertical line segments are sent to orthogonal trajectories of such arcs, whichare circular arcs through ±i.

14 ANALYSIS QUALS

Problem 11: Let u(z) be a harmonic function on the complex plane C suchthat ∫ ∫

C|u(z)|2dxdy <∞.

Prove u(z) = 0 for all z ∈ C.

Solution. This is identical to Fall 2001 Problem 4, above.

Problem 12: Evaluate ∫ ∞0

x2dx

x4 + 6x2 + 13.

Solution. Let f(z) = z2

z4+6z2+13 = z2

(z2+3)2+4 . Then f has simple poles at±√−3± 2i. Of these,

√−3 + 2i and

√−3− 2i are in the upper half plane.

To compute the residues, we use the fact that fg has residue f(z0)

g′(z0) at asimple pole z0. So

Res(f ;√−3 + 2i) =

−3 + 2i4(2i)

√−3 + 2i

=18i

√−3 + 2i

and

Res(f ;√−3− 2i) =

−3− 2i4(−2i)

√−3− 2i

= − 18i

√−3− 2i.

Hence∫Γ

f(z)dz = 2πi∑

Res(f ; z) = 2πi18i

(√−3 + 2i−

√−3− 2i

)=π

4

π√

2(√

13− 3)

4where Γ is a semicircular contour in the upper half-plane with radius R.(Here the square roots are both taken to be in the upper half plane.) Sincef(z) ∼ 1

R2 on the curved part of the contour, the ML estimate guaranteesthat the integral over the curved part goes to zero. Hence∫ ∞

0

f(x)dx =12

∫ ∞−∞

f(x)dx =π

4

√13− 3

√13

2.

Problem 14: Let S = [0, 1]× [0, 1] be the unit square in C and let f : S → Cbe a continuous map such that f(z) 6= 0 for all z ∈ S. Prove that thereexists continuous g : S → C such that

f(z) = eg(z)

for all z ∈ S.

Solution. Write f(x, y) = r(x, y)eiθ(x,y) where r and θ are continuous realfunctions of x and y. Then since r 6= 0 one can define g(x, y) = log r(x, y)+iθ(x, y). The problem thus reduces to showing that r and θ can be definedin a continuous fashion. For r this is clear since r = |f |. For θ, we notethat φf : S → T, where φ(z) = z

|z| and T is the unit circle, is a continuousmap. Since S is simply connected, and is path connected and locally pathconnected, the Lifting Criterion (Prop 1.33 in Hatcher) guarantees the ex-istence of a lift f : S → R. (Alternatively, simple connectedness implies

ANALYSIS QUALS 15

that the closed differential d(arg(f(z))) (which can be defined locally) isexact.)

Fall 2002

Problem 1: Let f, g be two absolutely continuous functions on the interval[0, 1] which are everywhere positive. Show that the pointwise quotient f/gis also absolutely continuous on [0, 1].

Solution. Since f, g are continuous on a compact domain, they are bounded,so ∃M such that |f(x)| ≤ M and |g(x)| ≤ M for all x ∈ [0, 1]. Moreover,g attains a minimum m which is positive by hypothesis. Now given ε > 0,choose δ > 0 such that

k∑n=1

|bn − an| < δ ⇒k∑

n=1

|f(bn)− f(an)| < εm2

Mand

k∑n=1

|g(bn)− g(an)| < εm2

2M.

Now∣∣∣∣f(bn)g(bn)

− f(an)g(an)

∣∣∣∣ =|f(bn)g(an)− f(an)g(bn)|

|g(bn)g(an)|

≤ |f(bn)g(an)− f(an)g(bn)|m2

=|f(bn)(g(an)− g(bn)) + g(bn)(f(bn)− f(an))|

m2

≤ |f(bn)||g(an)− g(bn)|+ |g(bn)||f(bn)− f(an)|m2

≤ M

m2(|g(an)− g(bn)|+ |f(bn)− f(an)|) .

Summing over n, this impliesk∑

n=1

∣∣∣∣f(bn)g(bn)

− f(an)g(an)

∣∣∣∣ ≤ M

m2

k∑n=1

|f(bn)−f(an)|+ M

m2

k∑n=1

|g(bn)−g(an)| < ε

2+ε

2= ε.

Hence f/g is absolutely continuous.

Problem 10: Let (X,M, µ) be a measure space, and let f1, f2, . . . be a se-quence of complex-valued functions in L1(X,M, µ) such that

‖fn‖L1(X,M,µ) ≤ 2−n

and

‖fn‖L∞(X,M,µ) ≤12

for all n = 1, 2, . . . . Show that the infinite product∞∏n=1

(1 + fn(x))

is convergent for µ-a.e. x, and that the product is a measurable function.

Solution. Since∫ ∑|fn(x)|dµ =

∑∫|fn(x)|dµ ≤

∑2−n = 1,

16 ANALYSIS QUALS

where we have used the monotone convergence theorem to interchange thesum and the integral, we see that

∑|fn(x)| ∈ L1, so it must be finite a.e.

Since the product∏

(1 + fn) converges absolutely exactly where∑|fn|

does, this product is absolutely convergente (hence convergent) a.e. Nowthe product of finitely many measurable functions is measurable (this iseasily proved by induction), and

∞∏n=1

(1 + fn(x)) = limN→∞

N∏n=1

(1 + fn(x))

is a pointwise limit of measurable functions, and therefore measurable.(The condition |fj | ≤ 1

2 comes in handy if you want to prove fromscratch that the convergence of the sum is equivalent to the convergence ofthe product. Since

| log(1 + z)| =∣∣∣∣z − z2

2+ . . .

∣∣∣∣≤ |z|+

∣∣∣∣z2

2− z3

3+ . . .

∣∣∣∣≤ |z|+

∣∣∣∣z2

2

∣∣∣∣+∣∣∣∣z3

3

∣∣∣∣+ . . .

≤ |z|+ |z|2

2+|z|3

2+|z|4

2+ . . .

= |z|+ |z|2

1− |z|≤ 2|z|

for |z| ≤ 12 . Hence

∑| log(1 + z)| converges if

∑|z| converges.

Problem 11: Find the linear fractional transformation f : C→ C such thatf(0) = 1, f(1) = 0, and f(∞) = i. What is the image of the line z :Re(z) = 1 under f?

Solution. Writing the transformation in the form z+abz+c and applying the

three conditions, we succesively discover that c = a, a = −1, and b = −i.So

z 7→ z − 1−iz − 1

is the desired transformation. To find the image of the given line, we notethat it must be a line or circle through the images of 1 and ∞, which are0 and i. Checking a third point, we find that 1 + i 7→ −1, so the image isthe circle through these points (centered at − 1

2 + 12 i, radius 1√

2.)

Problem 12: Suppose the function f(z) is continuous on the closed unit disk|z| ≤ 1 and analytic on the open disk |z| < 1. Assume f(z) = 0 for allz in the semi-circle z : |z| = 1, Im(z) > 0. Prove that f(z) = 0 on theclosed disk.

Problem 13: Let Un(z) be a sequence of positive harmonic functions on aconnected open set Ω containing the origin. Show that if

limn→∞

Un(0) = 0,

ANALYSIS QUALS 17

thenlimn→∞

‖Un‖L∞(K) = 0

for all compact subset K ⊂ Ω.

Solution. Harnack’s inequality says that for a positive harmonic function uon a domain D,

R− rR+ r

u(z0) ≤ u(z) ≤ R+ r

R− ru(z0)

for z in the compact subdisk |z − z0| ≤ r of the disk |z − z0| < R ⊂ D.(This follows from the positivity of u and a straightforward estimate of thePoisson kernel.) Now let G = z ∈ D : Un(z) → 0. Then Harnack’sinequality implies that G is open, because around any z0 ∈ G there existsa disk contained in D, on which Un(z) → 0 uniformly. Similarly, D \ G isopen, because if z0 ∈ D \G, there exists ε > 0 and a subsequence Unj withUnj (z0) > ε; then there is a disk on which u(z) ≥ R−r

R+r ε > 0 so un(z) 6→ 0on this disk. Since both G and D \G are open and D is connected, one ofthem is empty; but 0 ∈ G, so D = G. Thus, un(z) → 0 for all z ∈ D. Toprove that this convergence is uniform on compact subsets of D, cover acompact subset K ⊂ D by finitely many open disks D1, . . . , Dm ⊂ D withcenters z1, . . . , zm; then un(z)→ 0 uniformly on each disk, and since thereare finitely many, un(z)→ 0 uniformly on all of K.

Problem 14: Evaluate the integral

limR→∞

∫ R

0

cos(x2)dx

and justify your reasoning.

Solution. We use the eighth-of-a-circle contour bounding the circular sectorof radius R from θ = 0 to π

4 . Then∫Γ

eiz2dz = 0

since eiz2

is entire. Now the integral over the part with θ = π4 is∫ 0

R

e−r2eiπ/4dr = −eiπ/4

∫ R

0

e−r2dr

and the integral along the axis is∫ R

0eix

2dx. On the curved part,

z = Reiφ, dz = Rieiφdφ,

∫eiz

2dz =

∫ π/4

0

RieiφeiR2e2iφdφ.

Now|RieiφeiR

2e2iφ | = R|eiR2e2iφ | = Re−R

2 sin 2φ.

Since sinα ≥ 2απ for 0 ≤ α ≤ 2

π , the integral over the curved part is at most∫ π/4

0

Re−4R2φ/πdφ =π

4R(1− e−R

2).

18 ANALYSIS QUALS

As R→∞, this goes to zero. Hence∫ ∞0

eix2dx = eiπ/4

∫ ∞0

e−r2dr =

√π

2eiπ/4.

Taking real parts, ∫ ∞0

cos(x2)dx =√π

8.

Problem 15: Let 1 ≤ p <∞ and U(z) be a harmonic function on the entirecomplex plane such that∫ ∫

C|u(z)|p <∞.

Prove that u(z) = 0 for all z.

Solution. This is a slightly more general form of the familiar problem (Fall2001 Problem C4 and Spring 2002 Problem 11) that uses the L2 version ofthis statement. The only difference is that now we use Holder’s inequalityinstead of Cauchy-Schwarz. For 1 < p <∞, let q be the conjugate exponentpp−1 ; then

|u(z0)| = 1πR2

∣∣∣∣∣∫ ∫

DR(z0)

u(z)dxdx

∣∣∣∣∣≤ 1πR2

∫ ∫DR(z0)

|u(z)|dxdy

=1

πR2‖uχDR(z0)‖1

≤ 1πR2‖u‖p‖χDR(z0)‖q

= (πR2)1/q−1‖u‖p= (πR2)−1/p‖u‖p

which tends to 0 as R→∞. (Here DR(z0) is the disk of radius R centeredat z0.) In the case p = 1, we simply have

|u(z0)| ≤ 1πR2

∫ ∫DR

|u(z)|dxdy ≤ 1πR2

∫ ∫C|u(z)|dxdy

which tends to zero as R→∞.

Winter 2003

Problem 1: Let µ be a finite, positive, regular Borel measure on R2, and letG be the family of finite unions of squares of the form

S = j2n ≤ x ≤ (j + 1)2n; k2n ≤ y ≤ (k + 1)2,

where j, k, and n are integers. Prove that the set of linear combinations ofcharacteristic functions of elements of G are dense in L1(µ).

ANALYSIS QUALS 19

Solution. First, we note that this problem is easy if we are allowed to replaceeither of the ≤’s with a strict inequality. (On the qual I might just do that.)In that case every open set is a countable disjoint union of squares in S; onecan see this by taking the collection of squares in S which are contained inthe open set G, but which are not contained in any larger square containedin G. This collection can be straightforwardly shown to be disjoint andcover G. Since µ is regular, any measurable set E can be enclosed in anopen set G such that µ(G \ E) < ε. Then χG =

∑χSn approximates

χE within ε. Once we can approximate characteristic functions, we canof course approximate simple functions, then nonnegative functions, thencomplex functions.

If neither inequality is allowed to be strict, the same technique will workif µ is absolutely continuous with respect to Lebesgue measure: any openset can be written as an almost disjoint union of closed rectangles, and thecorresponding sum of characteristic functions will be equal a.e. to χG. If µhas a singular part, I think the theorem is still true, but I don’t know howto prove it.

Problem 4: Prove or disprove: If F is a strictly increasing continuous mapfrom the real line R onto itself and if A ⊂ R is Lebesgue measurable, thenf−1(A) is Lebesgue measurable.

Problem 5: Is the Banach space `∞ of bounded complex sequences a =an∞n=1 with the supremum norm ‖a‖∞ = sup |an| separable? Prove youranswer is correct.

Solution. No. There are uncountably many sequences all of whose termsare 0 or 1, but they are all a distance 1 apart.

Problem 6: Let x ∈ R have decimal expansion

x = a0.a1a2a3 . . .

and assume this expansion is unique (it is unique for almost all x). Define

An(x) =

1, if an is even−1, if an is odd.

Let f ∈ L1(R). Prove

limn→∞

∫Rf(x)An(x)dx = 0.

Solution. Note that An(x) = −An(x+ 110n ). So∫

f(x)An(x)dx = −∫f(x− 1

10n)An(x)dx.

Suppose f ∈ CC(R). Let supp(f) ⊂ [−M,M ]. Then f is uniformly contin-uous, so given ε > 0 there exists δ > 0 such that |f(x) − f(y)| < ε

2M+2 if|x− y| < δ. Let

λ =∫ (

f(x− 110n

)− f(x))An(x)dx.

20 ANALYSIS QUALS

Now for sufficiently large n, 110n < δ so the integrand is bounded by ε

M+2

and vanishes outside [−M − 1,M + 1], so |λ| < ε. Now

−∫f(x)An(x)dx =

∫f(x− 1

10n)dx

=(∫ (

f(x− 110n

)− f(x))An(x)dx

)+∫f(x)An(x)dx

= λ+∫f(x)An(x)dx

⇒∫f(x)An(x)dx =

λ

2.

But |λ| < ε, so∫f(x)An(x)dx → 0. Now let f ∈ L1(R). Given ε > 0,

choose g ∈ CC(R) with ‖f − g‖1 < ε. Then∣∣∣∣∫ f(x)An(x)dx∣∣∣∣ =

∣∣∣∣∫ (f(x)− g(x))An(x)dx+∫g(x)An(x)dx

∣∣∣∣ ≤ ∣∣∣∣∫ g(x)An(x)dx∣∣∣∣+∫ |f(x)−g(x)|dx

since |An(x)| = 1. This is eventually less than 2ε. Since this is true for anyε,∫f(x)An(x)dx→ 0.

Problem 9: Find the Laurent series expansion for1

z(z + 1)

valid in 1 < |z − 1| < 2.

Solution. Note that 1z(z+1) = 1

z −1z+1 . Since

1z

=1

z − 11

1 + 1z−1

=1

z − 1

(1− 1

z − 1+

1(z − 1)2

− . . .)

and1

z + 1=

12

11 + z−1

2

=12

(1− 1

2(z − 1) +

14

(z − 1)2 − . . .)

are both convergent in this annulus, the Laurent series expansion is

1z(z + 1)

=∞∑n=0

(−1)n

2n+1(z − 1)n +

−1∑n=−∞

(−1)n−1(z − 1)n.

Problem 10: Prove, by using the residue calculus, that∫ ∞0

√x

1 + x2dx = π

√2.

Solution. We use the keyhole contour. By the ML estimate, the integralover the small circle is at most 2πε

√ε

1+ε2 → 0 and the integral over the large

circle is at most 2πR√R

1+R2 → 0. THe integral on the bottom part of the

contour is −I and on the top is I, where I =∫∞

0

√x

1+x2 dx. The residues are

Res(f ; i) =√i

2i=

1− i2√

2

ANALYSIS QUALS 21

and

Res(f ;−i) =√−i−2i

=−1− i2√

2so

I = πi∑

Res(f ; z) =π√2.

Problem 11: Let f and g be continuous functions on the real line related bythe Fourier transform,

f(x) =1√2π

∫ ∞−∞

g(x)e−ikxdx, g(x) =1√2π

∫ ∞−∞

f(x)eikxdx.

Prove that both f and g cannot be compactly supported (i.e. vanish outsidesome finite interval of the real line).

Solution. Actually they can be, if they’re both identically 0. :-) Otherwise,suppose f is compactly supported. Then

h(k) =1√2π

∫ ∞−∞

f(x)eikxdx

defines an analytic function of k on the entire complex plane. (It is easy toshow the integral converges everywhere, and one may differentiate underthe integral sign to obtain a derivative.) This function is equal to g on thereal line. But if g is compactly supported, then it vanishes for real x oflarge magnitude. Thus the set of zeros of h has an accumulation point,so h must be everywhere zero, which then implies that both g and f areeverywhere zero.

Problem 12: Let f(z) be an entire function which is not a constant. Provethat ef(z) has an essential singularity at z =∞.

Solution. Liouville’s Theorem for harmonic functions states that a har-monic function on the whole plane which is bounded above or below mustbe constant. (The proof of this follows easily from Harnack’s inequality;see Spring 2006 Problem 7.) The Cauchy-Riemann equations imply thatif Re(f) is constant, then f is constant. Hence Re(f) must be unboundedin both directions; in particular, there are values of z with arbitrarily largemodulus for which Re(f(z)) < 0. This implies |ef(z)| < 1 at these points,so |ef(z)| 6→ ∞ as z → ∞. Hence ef(z) does not have a pole at ∞. Onthe other hand, the unboundedness of Re(f) guarantees that there existsa sequence zn → ∞ with Re(zn) → ∞ and therefore with |ef (zn)| → ∞,so ef(z) cannot have a removable singularity at ∞ (or be analytic there).Hence ef(z) has an essential singularity at ∞.

Problem 13: Explain why there is a unique analytic function g in the domainD roughly sketched below (shaded region) with the property that eg(z) = zfor all z ∈ D and g(1) = 2πi. Compute g(i+ 1).

22 ANALYSIS QUALS

Solution. Let γ be any closed curve lying in the given domain. By theargument principle,

12πi

∮γ

1zdz = N0 −N∞,

where N0 is the number of poles of f(z) = z enclosed by γ, and N∞ isthe number of poles. But f has no poles anywhere and no zeros in D, so∮dzz = 0 around any closed curve in D. Since D is simply connected, every

closed form is exact, so dzz = dg for some analytic function g on D. In

fact there is a family of such g, differing by an additive constant. Now leth(z) = eg(z). Then h′(z) = h(z)g′(z) = h(z)

z which implies ddz

h(z)z = 0, so

h(z) = cz for some nonzero constant c. By adding an appropriate constantto our choice of g (namely any value of − log(c)), we may find a g such thateg(z) = z. This proves existence. Now suppose there are two such analyticfunctions g and g′. Then eg−g

′= eg

eg′= z

z = 1 so g − g′ = 2πin for someinteger n (since g−g′ is continuous and D is connected, the value of n is thesame throughout D). Specifying the value of g at one point thus uniquelydetermines it throughout D. To compute g(1 + i), we use

g(1 + i)− g(1) =∫ 1+i

1

dg =∫ 1+i

1

dz

z.

This integral is taken along a path lying in D, but since dzz is closed on the

plane minus the negative real axis, the integral is path independent and theanswer is just Log(1 + i) = 1

2 log 2 + iπ4 . Thus

g(1 + i) =12

log 2 +9πi4.

Fall 2003

Problem 9: Let A = z : Im(z) > 0 \ z : Re(z) = 0 and 0 ≤ Im(z) ≤ 1.Find a conformal map that maps A one-to-one onto the upper half planez : Im(z) > 0, or show that no such map exists.

Solution. The map is w =√z2 + 1. One can get this from the Schwarz-

Christoffel transformation with angles of π2 , 2π, and π

2 ; it is also easy toverify directly that z 7→ z2 + 1 maps A onto the plane minus the real axis,so taking the (principal branch of the) square root maps to the UHP.

Problem 10: Use a contour integral to evaluate∫ ∞0

11 + x2n

dx, n ≥ 1.

Solution. We use the circular sector of radius R with 0 ≤ θ ≤ πn . Let I be

the desired integral. Then the integral along the axis is I and the integralalong the other radius is

z = reiπ/n, dz = eiπ/ndr,

∫f(z)dz =

∫ 0

R

11 + r2n

eiπ/ndr = −eiπ/nI.

ANALYSIS QUALS 23

The integral along the curved part is bounded by 2πRn(1+R2n) → 0 by the ML

estimate. The only singularity within the countour is at eiπ/2n with residue1

ddz (1 + z2n)|z0

=1

−2ne−iπ/2n.

Hence

(1− eiπ/n)I = 2πi1

−2ne−iπ/2n⇒ I =

π

2n sin(π2n

) .

Problem 11: Let p(z) = zn + an−1zn−1 + · · ·+ a0 be a complex polynomial.

Show that there must be at least one point with |z| = 1 and |p(z)| ≥ 1.

Solution. Suppose to the contrary that |p(z)| < 1 = |zn| on the unit circle.By Rouche’s Theorem, zn and zn − p(z) have the same number of zeros(counting multiplicity) inside the unit circle. But zn has n zeros inside theunit circle (an n-fold zero at the origin), whereas zn−p(z) is a degree n−1polynomial and hence can have at most n− 1.

Problem 12: Let D be the open unit disk in the complex plane. Endow Dwith the Lebesgue measure λ. Let A ⊂ L2(D,λ) be the subspace consistingof those L2 functions which are analytic on the disk.(a) Show that A is infinite-dimensional.(b) Show that A is a closed subspace of L2(D,λ).

Solution.(a) The analytic functions 1, z, z2, . . . are all linearly independent because

they are orthogonal:

intDznzmdxdy =

∫D

rn+mei(n−m)θrdrdθ = 0

since∫ 2π

0ei(n−m)θdθ = 0 for n 6= m. Since A contains infinitely many

linearly independent functions, it is infinite-dimensional.(b) Let f ∈ A. Let z ∈ D; then for any r < d(z,Dc),

f(z)2 =1πr2

∫B(z,r)

f(z)2dxdy

⇒ |f(z)|2 =1πr2

∣∣∣∣∣∫B(z,r)

f(z)2dxdy

∣∣∣∣∣≤ 1πr2

∫B(z,r)

|f(z)|2dxdy

≤ 1πr2‖f‖2.

Hence |f(z)| ≤ 1√πd(z,Dc)

‖f‖2. Then for any compact subset K ⊂ D,d(K,Dc) > 0; this implies that if fn is Cauchy, then fn convergesuniformly on K, since

|fn(z)− fm(z)| ≤ 1√πd(K,Dc)

‖fn − fm‖2

24 ANALYSIS QUALS

for all z ∈ K. Now any z ∈ D is contained in a compact subset of D,and since the uniform limit of analytic functions is analytic, lim fn isanalytic at z.

Winter 2004

Problem CA1: Let f(z) = u(x, y) + iv(x, y) denote a non-constant analyticfunction on some open domain D ⊂ C. Show that at each point of D, the“level curves” u(x, y) =constant and v(x, y) =constant intersect at rightangles.

Solution. This is actually false as stated; the level curves intersect at rightangles at any point z for which f ′(z) 6= 0. A counterexample that violatesthis condition is f(z) = z2; at the origin the real axis is a level curve ofboth u and v, as is the line y = x.

With this additional assumption, we note that ∇u = (ux, uy) and ∇v =(vx, vy) are orthogonal to the curves (it is a general fact that the gradient isorthogonal to the level sets); by the Cauchy-Riemann equations, ∇u ·∇v =uxvx + uyvy = ux(−uy) + uy(ux) = 0. Since the (nonzero) normal vectorsof the curves are orthogonal, the curves are orthogonal.

Problem CA2: Let K(z) denote a real-valued function of the complex vari-able z define in some open domain D ⊂ C. Then K(z) is said to be strictlysubharmonic if the inequality

K(z0) <1

2π

∫ 2π

0

K(z0 + ρeiφdφ

holds for all ρ which are smaller than the distance from z0 to the boundaryof D. Let f(z) denote a non-constant analytic function on D. Show thatf(z) is strictly subharmonic.

Proof. This is a simple application of the triangle inequality for integrals:

|f(z)| = 12π

∣∣∣∣∫ 2π

0

f(z + ρeiφ)dφ∣∣∣∣ ≤ 1

2π

∫ 2π

0

|f(z + ρeiφ)dφ.

The equality condition is that the integrand have constant argument. Butfor an analytic function this implies the function itself is constant: WLOGwe assume f is real on the circle (else multiply it by an appropriate eiθ);then by the maximum-modulus principle for harmonic functions, Im(f)is zero on the disc, which by the Cauchy-Riemann equations implies f isconstant on the disc, which implies f is constant on the domain. Hence theinequality is strict.

Problem CA3: Using contour (or any other) methods, compute the integral∫ ∞0

xdx

1 + x3.

Solution. A partial fraction decomposition yields

x

1 + x3=−1/3x+ 1

+1/3(x+ 1)x2 − x+ 1

ANALYSIS QUALS 25

and∫x

1 + x3dx =

16

ln(x2 − x+ 1(x+ 1)2

)+

1√3

tan−1

(2x− 1√

3

)+ C.

Evaluating between 0 and ∞ yields the answer 2π3√

3.

Alternatively, we use a third-of-a-circle contour. The integral along theaxis is

∫ R0

x1+x3 dx. The integral along the curved part goes to 0 by the ML

estimate. The integral along the other radius is

z = re2πi/3, dz = e2πi/3dr,

∫f(z)dz = −e4πi/3

∫ R

0

r

1 + r3dr = −e4πi/3

∫ R

0

r

1 + r3dr.

The sum of the integrals is thus

(1− e4πi/3)∫ ∞

0

xdx

1 + x3= −√

3ie2πi/3I

where I is the desired integral. The residue at eπi/3 (the only singularitywithin the contour, a simple pole) is

zddz (1 + z3)

=13z

=13e−πi/3

so

2πie−πi/3

3= −√

3ie2πi/3I ⇒ I =2π

3√

3.

Problem CA6: Let f(z) denote a function which is analytic at all points ona simple closed curve γ and everywhere inside γ except for the (isolated)points b1, . . . , br where it has poles of order β1, . . . , βr. Furthermore, fdoes not vanish on γ but does have zeros at the points a1, . . . , as whicheare inside γ ande these zeros are of multiplicity α1, . . . , αs. Then, accordingto a well known formula,

12πi

∮γ

f ′(z)f(z)

dz =s∑

k=1

αk −r∑

k=1

βk.

Now suppose that g(z) is analytic inside and on γ. What is the general-ization of the above formula when the integrand on the left hand side isreplaced by g(z) f

′(z)f(z) ? Provide justification for your answer.

Solution. In a neighborhood of ak, f(z) = (z − ak)αkh(z), where h(z) isanalytic and nonzero at ak. Then f ′(z) = h′(z)(z − ak)αk + αkh(z)(z −ak)αk−1 so

g(z)f ′(z)f(z)

= g(z)(h′(z)h(z)

+ αk1

z − ak

)in a neighborhood of ak. Thus, the residue of g(z) f

′(z)f(z) at ak is αkg(ak).

Similarly, the residue at bk is −βkg(bk), so by the Residue Theorem,

12πi

∮γ

g(z)f ′(z)f(z)

dz =s∑

k=1

αkg(ak)−r∑

k=1

βkg(bk).

26 ANALYSIS QUALS

Fall 2004

Problem 8: Evaluate the integral

I =∫ ∞−∞

cosx(1 + x2)2

dx.

You will need to briefly justify any limits you take.

Solution. We use a semicircular contour of radius R to evaluate∫ ∞−∞

eiz

(1 + z2)2dz.

Since |eiz| ≤ 1 on the upper half plane, the integral over the curved parttends to 0 as R→∞ by the ML estimate. The only singularity inside thecontour is at i; since it is a pole of order 2, the residue is

limz→i

d

dz

((z − i)2 eiz

(1 + z2)2

)=−i2e.

By the Residue Theorem,∫ ∞−∞

eiz

(1 + z2)2dz = 2πi

−ie

=π

e.

Taking real parts, ∫ ∞−∞

cosx(1 + x2)2

dx =π

e.

Problem 9: Find the number of zeros of the polynomial z6+12z4+z3+2z+6in the first quadrant inside the unit circle (the domain z = x + iy : x >0, y > 0, |z| < 1), and also in the first quadrant outside the unit circle (thedomain z = x+ iy : x > 0, y > 0, |z| > 1).

Solution. First, p(z) has no real zeros. It is obvious that p(x) > 0 for x > 0;for x < 0 this follows from the fact that 6+2x+x3 > 0 for −1 ≤ x ≤ 0 and12x4 + 2x + x3 > 0 for x ≤ −1. Now consider a quarter-disk DR of largeradius R lying in the first quadrant. As z goes from 0 to R along the realaxis, arg p(z) doesn’t change because p(z) is a positive real number. As zgoes from R to Ri along the curved part, p(z) ≈ z6 so arg p(z) changes byabout 6π2 = 3π. Finally, consider what happens as z goes from Ri to 0down the imaginary axis. Here z = iy as y goes from R to 0 in the reals;then p(z) = (−y6 + 12y4 + 6) + i(2y − y3). At y = R this has argumentslightly greater than −π. It stays in the lower half plane until it intersectsthe real axis at y =

√2, at which point p(z) = 46 and the argument has

increased by π. Finally, p(iy) stays in the upper half plane from y =√

2 toy = 0, where p(z) = 6, so the argument does not change in this segment.Thus, arg p(z) changes by a total of about 4π (and hence exactly 4π) aroundthe disk, so there are two zeros inside it, i.e. p has two zeros in the firstquadrant. Because the coefficients of p are real, the roots occur in conjugatepairs. Hence there are two zeros in the fourth quadrant, and therefore oneeach in the second and third.

ANALYSIS QUALS 27

(At this point we might note that Rouche’s theorem guarantees that fourof these zeros lie inside the unit circle, but we don’t know which four.)

We can repeat the process for a quarter-circle of radius one. Once againthere is no change in argument along the positive real axis. Along theimaginary axis, p(z) stays in the upper half plane as noted above (since0 ≤ y ≤

√2), and goes from 17+i to 6, a small negative change in argument.

Finally, the change in argument along the curved sector is 2kπ − ε, whereε = arctan(1/17). To find k, we note that |12z4| > |z6 + z3 + 2z+ 6| on theunit circle. In general, if |g| < |f |, then arg(f)−π

2 < arg(f+g) < arg(f)+π2 ,

as may be seen by drawing a circle of radius |g| centered at f . This impliesthat the change in arg(f + g) is within π

2 of the change in arg(f). Now thechange in arg(z4) along the sector is 2π, so the change in arg p(z) must bewithin π

2 of 2π. Hence k = 1 and the total change in argument along thequarter of the unit circle is 2π, so there is one zero inside the first quadrantinside the unit circle.

Problem 10: Let D = z = x+ iy : y > 0 \ [i, i∞] be the domain obtainedfrom the open upper half-plane by excising the vertical ray from i to i∞.Find a conformal map w = f(z) of D onto the open unit disk |w| < 1.You map represent the map f(z) as a composition of other maps. Includea sketch or diagram with your solution.

Solution. First, note that z 7→ −1z maps this to the UHP minus the closed

slit from 0 to i. Then z 7→√z2 + 1 maps this to the UHP (see Fall 2003

Problem 9).

Problem 12: Let D be a domain, i.e. a connected open set. Let f : D → Rbe a continuous function with the property that whenever B(z0, r) is aclosed disk contained in D, and h : B(z0, r) → R is a harmonic functionsuch that f(z) ≤ h(z) for all z ∈ ∂B(z0, r), f(z0) ≤ h(z0). (Such functionsare called subharmonic.)

Show that f cannot attain its maximum at any point in D unless it is aconstant function.

Solution. First we show that f has the property that

f(z0) ≤ 12π

∫ 2π

0

f(z0 + reiφ)dφ

for any r such that B(z0, r) ⊂ D; this is often used instead as the definitionof subharmonic. Given such a ball, let u be a harmonic function whichequals f on ∂B(z0, r); such a u may be explicitly found from the Poissonkernel, viz.

u(z0 + ρeiθ) =1

2π

∫ 2π

0

1− ρ2

1− 2ρ cos(θ − φ) + ρ2f(z0 + reiφ)dφ

for 0 ≤ ρ < r. Then

f(z0) ≤ u(z0) =1

2π

∫ 2π

0

u(z0 + reiφ)dφ =1

2π

∫ 2π

0

f(z0 + reiφ)dφ.

Now suppose f attains a maximum value M on D at some point z0. LetG = z ∈ D : f(z) = M. We will show that G = D. Clearly G is closed

28 ANALYSIS QUALS

in D since it is the inverse image under a continuous function of the closedset M ⊂ R. Now let z ∈ G and choose any r > 0 such that B(z, r) ⊂ D.Then for any ρ < r,

M = f(z) ≤ 12π

∫ 2π

0

f(z + ρeiφ)dφ.

But the integrand is at most M everywhere, so this average is at mostM with equality iff the integrand is M everywhere (since the integrand iscontinuous). Hence f(ζ) = M for |ζ − z| < r. Thus, G is open as well asclosed; since D is connected, G = D or ∅, but z0 ∈ G, so G = D.

Winter 2005

Problem 2: Let Ω be a connected open subset of the plane and let z0, z∈Ω.Prove there is a constant M = M(z0, z1) > 0 such that whenever U(z) is apositive harmonic function on Ω,

1M≤ U(z1)U(z0)

≤M.

Solution. By Harnack’s inequality, such a constant exists provided z0 andz1 are contained in a disk contained in Ω. Since Ω is connected, it is pathconnected. Let Γ be a path from z0 to z1. Since Γ is a compact subset ofΩ, it is covered by finitely many open disks. This allows us to find an Mfor z0 and z1.

Problem 4: Let −1 < a < 0 < b < 1 and let Ω be the domain

z : |z| < 1 \ ((−1, a] ∪ [b, 1)) .

Find a conformal map from Ω onto the open unit disc. You do not need towrite the full expression; just tell us the sequence of maps you are taking.

Solution. First, the map z 7→ i 1+z1−z maps to the UHP minus the slit from

0 to 1+a1−a i and the infinite slit from 1+b

1−b i to i∞. Scaling via z 7→ 1−a1+az

maps this to the UHP minus the slits from 0 to i and from αi to i∞, whereα = (1+b)(a−1)

(1−b)(1+a) > 1. Then z 7→√z2 + 1 maps this to the UHP minus the

slit from√α2 − 1i to i∞. Then z 7→ − 1

z/√α2−1

takes this to the UHP

minus the slit from 0 to i; the map z 7→√z2 + 1 takes this to the UHP,

and the map z 7→ z−iz+i takes this to the open unit disc.

Problem 5: Prove that the limit

limN→∞

∫ N

−N

x2 sin(πx)x3 − 1

dx

exists and compute its value.

Solution. We use an indented semicircular contour to evaluate p.v.∫∞−∞

z2eπiz

z3−1 dz.The integral of the curved part goes to zero by Jordan’s lemma. Theresidues at simple poles are Res(f ; z0) = z20e

πiz0

3z20= 1

3eπiz0 , so

Res(f ; 1) =−13, Res(f ; e2πi/3) = eπi(−1/2+

√3/2i)/3 =

−ieπ√

3/2

3.

ANALYSIS QUALS 29

By the fractional residue theorem, the integral around the indentation ap-proaches

−πi(−13

)=πi

3so by the Residue Theorem,

p.v.

∫ ∞−∞

z2eπiz

z3 − 1= −πi

3+ 2πi

−ie−π√

3/2

3=

2π3e−π√

3/2 − πi

3.

Comparing imaginary parts,

p.v.

∫ ∞−∞

x2 sin(πx)x3 − 1

= −π3.

Problem 6: Let P (z) and Q(z) be polynomials and consider the function

f(z) = P

(1z

)+Q

(1

z − 1

)on the domain Ω = C \ 0, 1. Give conditions on P and Q necessary andsufficient for the differential equation

F ′(z) = f(z)

to have a single valued solution on Ω.

Solution. The condition is that the linear terms of both P and Q are zero.Suppose this is the case. Then by the Residue Theorem,

∮γf(z)dz = 0 for

any closed curve γ ∈ Omega, so the integral∫ zz0f(z)dz is path independent

and provides an antiderivative for f on Ω. Conversely, if a single-valuedantiderivative F exists, then

∮γf(z)dz =

∮γdF = 0 around any closed

curve, which implies that the residues at 0 and 1 are both zero.

Fall 2005

Problem C1: Suppose that f(z) is analytic and non-constant on a connectedopen set G in the complex plane. Prove that f(G) is an open subset of thecomplex plane.

Solution. Let w0 ∈ f(G), so w0 = f(z0) for some z0 ∈ G. Suppose f(z)−w0

has a zero of order m > 0 at z0. Because the zeros of the nonconstantanalytic function f(z) − w0 are isolated, there exists ρ > 0 such thatB(z0; ρ) ⊂ G and f(z0 + reiθ) 6= w0 for 0 < r ≤ ρ. Now because thecircle z : |z− z0| = ρ is compact, its image under f is compact and henceis a positive distance δ > 0 from w0. Now consider the integral

N(w) =1

2πi

∮|z−z0|=ρ

f ′(z)f(z)− w

dz

on the region |w−w0| < δ. Then the denominator is nonzero on the contourof integration, so N(w) is an analytic function of w (in particular, we canfind a derivative for it by differentiating under the integral sign). But thislogarithmic integral is integer-valued, so it must be constant on the region|w − w0| < δ. Since f(z) − w0 has a zero of order m at z0 and no other

30 ANALYSIS QUALS

zeros or poles on or inside the contour (i.e. for |z − z0| ≤ ρ), N(w0) = m,so N(w) = m throughout the region |w−w0| < δ. Since f(z)−w never hasany poles, this means that f(z)−w must have m zeros inside the contour,the interior of which lies in G. In particular, w ∈ f(G). This is true forany w0 ∈ f(G), so f(G) is open.

Problem C2: Find an explicit conformal mapping from the upper half-planeslit along the vertical segment

z ∈ C : Im(z) > 0 \ (0, i]

to the unit disk z ∈ C : |z| < 1.

Solution. The map z 7→√z2 + 1 takes the given region to the UHP (cf. Fall

2003 Problem 9); the map z 7→ z−iz+i then takes this to the unit disk.

Problem C3: Consider the meromorphic function

f(z) =1− z2

2i(z2 − (a+ 1a )z + 1)

, |a| < 1.

Find the Laurent series expansion for f(z) valid in a neighborhood of theunit circle |z| = 1.

Solution. Since

f(z) =1− z2

2i(z − a)(z − 1a )

=1

2ia1

z − a− 1

2ia1

z − 1a

,

and1

z − a=

1z

11− a

z

=1z

(1 +

a

z+a2

z2+ . . .

)and

1z − 1

a

=11a

11− az

=1a

(1 + az + a2z2 + . . .

)are convergent for |a| < |z| < 1

|a| , the Laurent expansion on this annulus is

f(z) =∞∑n=0

ian

2zn +

∞∑n=1

−ian−2

21zn.

Problem C4: Using the residue calculus, evaluate the integral∫ ∞0

log x(x2 + 1)2

dx.

Solution. We use the principal branch of log z, with the branch cut alongthe positive real axis. Integrating along a semicircular contour indented atthe origin, the integral over the large curved part is bounded by 2 logR

(R2+1)2 (πR)→0 since | log(Reiθ)| ≤ 2 logR for R large. Similarly, the integral over thesmall curved part is bounded by 2| log ε|

(ε2+1)2 (πε)→ 0 since | log(εeiθ)| ≤ 2| log ε|for ε small. The integral from −R to −ε is∫ −ε

−R

log z(z2 + 1)2

dz =∫ R

ε

iπ + log x(x2 + 1)2

dx

ANALYSIS QUALS 31

so

2∫ ∞

0

log x(x2 + 1)2

dx+ iπ

∫ ∞0

1(x2 + 1)2

dx = 2πiRes(f ; i).

The residue can be computed as

limz→i

d

dz(z − i)2f(z) = lim

z→i

d

dz

log z(z + i)2

= limz→i

z + i− 2z log zz(z + i)3

=π + 2i

8.

Comparing real and imaginary parts,∫ ∞0

1(x2 + 1)2

dx =π

4

and, what we are interested in,∫ ∞0

log x(x2 + 1)2

dx = −π4.

Spring 2006

Problem 1: Let µ and µn, n ∈ N, be finite positive Borel measures on Rsuch that ∫

f(x)dµn(x) →n→∞

∫f(x)dµ(x)

for all continuous functions f with compact support.(1) Show that then for each compact set K ⊂ R,

lim supn→∞

µn(K) ≤ µ(K).

(2) Give an example that shows µ(R) < lim supn→∞ µn(R) is possible.

Solution.(1) Let ε > 0. Since R has the property that every open set is σ-compact,

and since µ is finite on compact sets, µ is regular (Theorem 2.18 fromRudin). Hence ∃G ⊃ K open such that µ(G \ K) < ε. G can alsobe chosen to be bounded and hence have compact closure (if K ⊂[−M,M ], replace G with G∩ (−M−1,M+1)). By Urysohn’s lemma,∃fε continuous such that 0 ≤ f ≤ 1, f = 1 on K and f = 0 outside G.Then

µn(K) ≤∫fεdµn →

∫fεdµ ≤ µ(K) + ε,

so µn(K) is eventually less than, say, µ(K) + 2ε. This is true for anyε, so lim supµn(K) ≤ µ(K).

(2) Let An = [0, 1]∪ [n, n+1]. Define µn(E) = m(E∩An) for all Lebesguemeasurable E, where m indicates Lebesgue measure. Let µ(E) =m(E ∩ [0, 1]). Then if f is compactly supported, f = 0 on [n, n + 1]for sufficiently large n, so

∫fdµn =

∫ 1

0f(x)dx =

∫fdµ. However,

µn(R) = 2 for all n whereas µ(R) = 1.

32 ANALYSIS QUALS

Problem 2: Let fn be a sequence of L(R) functions such that

limn→∞

∫Rfn(x)g(x) = g(0),

for each g ∈ C0(R), that is, continuous functions vanishing at infinity. Showthat ‖fn‖L1 is uniformly bounded but that fn is not Cauchy in L1.

Solution. In functional lingo, we’re given that the absolutely continuousmeasures µn corresponding to fn are weakly convergent to the Dirac deltameasure in the space M(R) = C0(R)∗. The fact that the sequence is uni-formly bounded is then an immediate consequence of, guess what, the uni-form boundedness principle. However, the sequence cannot be Cauchy be-cause L1 is complete, so if µn were Cauchy it would have a limit in thenorm topology and this limit would be absolutely continuous. But normconvergence implies weak convergence, so this norm limit would have to bethe same as the weak limit, which is not absolutely continuous.

Problem 3: Consider the space L∞([0, 1], λ) where λ denotes the Lebesguemeasure. Let

d(f, g) = infε>0

(ε+ λ(x : |f(x)− g(x)| > ε)

).

Prove that d is a metric and that fn → f if and only if fn → f in measure.

Solution. Nonnegativity and symmetry are obvious. To prove positivity,suppose f 6= g. Then there must be a set E of positive measure and a δ > 0such that |f − g| > δ on E. Then for any ε ≤ δ, x : |f(x) − g(x)| > ε ≥λ(E), which implies d(f, g) ≥ min(δ, λ(E)) > 0. For the Triangle Inequality,let δ > 0. Choose εfh such that εfh + λ(x : |f(x) − h(x)| > εfh) <d(f, h) + δ and similarly for εhg. Now at least one of |f − h| > εfh and|h−g| > εhg must be true if |f−g| > εfh+εhg, so that λ(x : |f(x)−g(x)| >εfh + εhg) ≤ λ(x : |f(x) − h(x)| > εfh) + λ(x : |h(x) − g(x)| > εhg).Thus, if we let ε = εfh + εhg, we have

ε+ λ(x : |f(x)− g(x)| > ε) ≤ d(f, h) + d(h, g) + 2δ.

Taking the infimum over ε can only decrease the expression on the left, sod(f, g) ≤ d(f, h) + d(h, g) + 2δ. This is true for all δ > 0, so d(f, g) ≤d(f, h) + d(h, g). Hence d is a metric.

Suppose fn → f in measure. Let η > 0. Then λ(x : |fn(x) − f(x)| >η2) <

η2 for sufficiently large n. This implies d(fn, f) ≤ η for sufficiently

large n. This is true for any η > 0, so d(fn, f)→ 0.Conversely, suppose fn 6→ f in measure. Then there exists δ > 0 and

a number α > 0 such that λ(x : |fn(x) − f(x)| > δ) > α for infinitelymany n. The same computation used above to show positivity shows thatd(fn, f) ≥ min(δ, α) > 0 for these values of n, so that d(fn, f) 6→ 0.

Problem 4: Prove (one direction of) the Arzela-Ascoli Theorem: Supposefn : [0, 1]→ R is a sequence of functions such that(1) ∃M : ∀x ∈ [0, 1],∀n ≥ 1, |fn(x)| ≤M(2) ∀ε > 0, ∃δ > 0 : ∀n ≥ 1, |x− y| < δ ⇒ |fn(x)− fn(y)| < ε.

Then fn has a subsequence which converges uniformly.

ANALYSIS QUALS 33

Solution. First, note that a subsequence of the fn must converge at everyrational point in [0, 1]. To see this, let r1, r2, . . . be an enumeration of therationals. Then for each n, the real number fn(r1) lies in the compact set[−M,M ], so a subsequence of these numbers converges. Take the corre-sponding subsequence of the functions fn and consider their values at r2.These values also lie in the compact set [−M,M ], so a subsequence mustconverge. Continuing in this manner, we then take the “diagonal” subse-quence containing the first term of the first subsequence, the second termof the second, and so on. This sequence will be convergent at every rationalpoint.

Now let fjk denote a subsequence of the fn which converges at everyrational point. We will prove that this subsequence is uniformly Cauchy.Let ε > 0 and choose δ such that |x− y| < δ ⇒ |fj(x)− fj(y)| < ε/3 for allj. Let L be a positive integer with 1

L < δ. Since fjk(x) is Cauchy at eachof the rational points 0, 1

L ,2L , . . . , 1, ∃N such that |fjm( pL )− fjn( pL )| < ε/3

for p = 0, . . . , L. Then for any x ∈ [0, 1], choose p such that |x− pL | < ε/3.

For this p and for m,n > N ,

|fjm(x)−fjn(x)| ≤ |fjm(x)−fjm(p/L)|+|fjm(p/L)−fjn(p/L)|+|fjn(p/L)−fjn(x)| < ε/3+ε/3+ε/3 = ε.

Problem 5: Suppose f ∈ Lp(Rn, dx) and g ∈ Lq(Rn, dx) where1p

+1q

= 1

and 1 < p <∞. Show that

f ∗ g(x) =∫

Rnf(x− y)g(y)dy

is a continuous function with

lim|x|→∞

f ∗ g(x) = 0.

Solution. (Folland page 241). It follows from Holder’s inequality that ‖h1 ∗h2‖∞ ≤ ‖h1‖p‖h2‖q. Let τyf(x) = f(x − y) denote the translates of f .Using the Lp-continuity of translation, we have

‖τy(f ∗ g)− f ∗ g‖u = ‖(τyf − f) ∗ g‖∞ ≤ ‖τyf − f‖p‖g‖q → 0

as y → 0. Thus f ∗ g is continuous (in fact, uniformly continuous). Nowlet fn and gn be continuous functions with compact support such that‖fn − f‖p → 0 and ‖gn − g‖q → 0. The convolution of two CC functions isCC , so f ∗ g ∈ CC(R). Now

‖fn ∗ gn‖u ≤ ‖fn − f‖p‖gn‖q + ‖f‖p‖gn − g‖q → 0

so that fn ∗ gn → f ∗ g uniformly. But the uniform limit of CC functions isa C0 function, so f ∗ g vanishes at infinity.

Problem 7: Prove that any nonnegative harmonic function on R2 is constant.

Solution. Let z = reiθ ∈ C. By Harnack’s inequality,R− rR+ r

u(0) ≤ u(reiθ) ≤ R+ r

R− ru(0)

34 ANALYSIS QUALS

for any R. Letting R→∞, u(reiθ) = u(0).(Since Harnack’s inequality almost makes this problem trivial, it might

be well to at least sketch a proof of it; simply express the harmonic functionin question in terms of its Poisson kernel and bound the Poisson kernel byR−rR+r from below and R+r

R−r from above.)

Problem 8: Compute the limit

limR→∞

∫ R

0

eix2dx.

You may use the fact that

limR→∞

∫ R

0

e−x2dx =

12√π

without proof.

Solution. See Fall 2002 Problem 14. The answer is√π

8+ i

√π

8.

Problem 11: Let ε = 1100 and let U be an ε-neighborhood of the spiral θeiθ :

0 ≤ θ ≤ 4π and let O be an ε-neighborhood of the spiral 2θeiθ : 0 ≤θ ≤ 2π. Let f : U → C and g : O → C be the corresponding analyticcontinuations of

z 7→ log(

cos(z)1− z2

)from the ε-neighborhood of the origin on the real axis such that f(0) = 0 =g(0). Find the imaginary part of f(4π)− g(4π).

Problem 12: Prove that the infinite product

∞∏n=0

(1 + z2n)

converges and equals (1− z)−1 for all z in the open unit disc.

Solution. Since ∑|z|2

n

≤∑|z|n

converges in the unit disc (the inequality comes from the fact that it’s asubsum), the infinite product is absolutely convergent for |z| < 1. Then

∞∏n=0

(1 + z2n) = limN→∞

N∏n=0

(1 + z2n) = limN→∞

2N+1−1∑j=0

zj =∞∑n=0

zn =1

1− z.

ANALYSIS QUALS 35

Fall 2006

Problem 1: Let θ ∈ [0, 1] be an irrational number. Let X by the unit circlee2πit : t ∈ [0, 1] endowed with the arclength measure, and let α : X → Xbe the rotation by 2πθ, given by α(e2πıt) = e2πi(t+θ).(1) Show that if f ∈ L2(X) and f α = f , then f is a.e. constant.(2) Use part 1 to show that if Y ⊂ X is a Lebesgue-measurable subset so

that α(Y ) = Y , then either Y has measure zero, or X \Y has measurezero.

Solution.(1) By definition of the Fourier transform,

f(n) =∫ 1

0

f(e2πix)e−2πinxdx

and

f α(n) =∫ 1

0

f(e2πi(x+θ))e−2πinxdx

= e2πinθ

∫ 1

0

f(e2πi(x+θ))e−2πin(x+θ)dx

= e2πinθf(n)

by the periodicity of f . Since f α = f , this implies e2πinθf(n) = f(n).But since θ is irrational, e2πinθ 6= 1 for n 6= 0, so all the Fourier coef-ficients of f are zero except the constant term. Since f has the sameFourier transform as a constant, it must be constant a.e.(I’m not sure what’s the most immediate way to see that Fourier in-version must apply; one way would seem to be that L2 ⊂ L1 for spacesof finite measure, and inversion always works on L1 ∩ L2.)

(2) Applying part (1) to χY (which is obviously in L2), we see that χY isa.e. constant. If it’s 0 a.e, m(Y ) = 0; if χY = 1 a.e., m(X \ Y ) = 0.

Problem 2: Let E,F be two Lebesgue-measurable subsets of R, and letχE , χF be their respective characteristic functions.(1) Show that the convolution χE ∗ χF defined by

χE ∗ χF (x) =∫

RχE(y)χF (x− y)dy

is a continuous function.(2) Show that

n(χE ∗ χ[0,1/n])→ χE

as n→∞ pointwise a.e.

Solution.(1) For this part I will assume that at least one of E and F has finite

measure, since otherwise the convolution might not be finite-valued.

36 ANALYSIS QUALS

WLOG, m(F ) <∞, so χF ∈ L1. We have

|χE ∗ χF (x+ δ)− χE ∗ χF (x)| =∣∣∣∣∫

RχE(y)χF (x+ δ − y)dx−

∫RχE(y)χF (x− y)dx

∣∣∣∣≤∫

R|χE(y) (χF (x+ δ − y)− χF (x− y))| dy

≤∫

R|χF (x+ δ − y)− χF (x− y)| dy

= ‖χF (x+ δ)− χF (x)‖L1

by translation and reflection invariance. Since translation acts contin-uously on L1 (as is easily proved by approximating L1 functions withcontinuous functions of compact support), this tends to 0 as δ → 0.

(2) Using the fact that f ∗g = g∗f (which follows from Fubini’s theorem),

n(χE ∗ χ[0,1/n])(x) = n

∫Rχ[0,1/n](y)χE(x− y)dy

=1

1/n

∫ 1/n

0

χE(x− y)dy

By the definition of Lebesgue points, this will tend to χE(x) for every xwhich is a Lebesgue point of χE . But almost every point is a Lebesguepoint, so the result follows.(Note: The fact that almost every point is a Lebesgue point is generallytrue only for L1 functions, and χE may not be in L1 because wedon’t know that m(E) <∞. However, we can write E as a countableunion of sets with finite measure (say, its intersection with the intervals[n, n+ 1)) and apply the result to them, then take a countable union.)

Problem 3: Let (X,M, µ) be a measure space, µ(X) = 1. Let f ∈ L∞(X,M, µ).Prove that

limp→∞

(∫|f |pdµ

)1/p

= ‖f‖∞.

Solution. Let M = ‖f‖∞. For each p < ∞, |f |p ≤ Mp so f ∈ Lp and‖f‖p ≤ M . Now for any n, let En = x : |f(x)| ≥ M − 1

n. Defineδn = µ(En) > 0 since otherwise ‖f‖∞ would be less than M . Now f(x) ≥(M − 1

n )χEn so

‖f‖p ≥(∫

En

(M − 1n

)dµ)1/p

=(M − 1

n

)δ1/pn .

Since 0 < δn ≤ 1, δ1/pn → 1 as p→∞. Thus,

‖f‖p ≥M −2n

for sufficiently large p. Since this is true for any n, and since we alreadyknow ‖f‖p ≤M for all p, we have ‖f‖p → ‖f‖∞.

ANALYSIS QUALS 37

Problem 4: Assume that f is a continuously differentiable 2π periodic func-tion on R. Show that the Fourier series

∞∑n=−∞

f(n)eint, t ∈ R

is absolutely convergent, where

f(n) =1

2π

∫ 2π

0

f(t)e−intdt.

Solution. Since f ′ is continuous on the compact set R/2πZ, it is in Lp forevery p ∈ [1,∞]; in particular, it is in L2, so

∑|f ′(n)|2 <∞ by Parseval’s

theorem. Now f ′(n) = −inf(n) (integrate by parts), so∑|f(n)| =

∑|f ′(n)| 1

n≤√∑

|f ′(n)|2√∑ 1

n2<∞

by the Cauchy-Schwarz inequality.

Problem 5: Let `2 be the space of all square-summable sequences of complexnumbers, and let T : `2 → `2 be a linear operator. Let en be the sequence

en = (00 . . . 010 . . . ),

where 1 is in the nth position. Let anm = 〈Tem, en〉 be the “matrix coeffi-cients” of T .(1) Assume that

∑∞n,m=1 |anm|2 <∞. Show that T is a bounded operator.

(2) Assume instead that sup|anm| : 1 ≤ n,m < ∞ is finite. Must T bebounded?

Solution.(1) Let v =

∑n cnen, so ‖v‖2 =

∑n |cn|2. Then

‖Tv‖2 =

∥∥∥∥∥∑m,n

cnamnen

∥∥∥∥∥2

=∑m,n

|cnamn|2 =∑n

|cn|2(∑

m

|amn|2)< M

∑n

|cn|2 = M‖v‖2

where M =∑m,n |amn|2 <∞.

(2) In this case T need not be bounded. Let

anm =

1 m ≤ n0 else

.

I.e. T (en) = e1 + e2 + · · · + en. In this case sup |amn| = 1 < ∞, but‖T (en)‖ =

√n is not bounded.

Problem 6: Let D = z : |z| < 1 be the unit disk in the complex plane,endowed with the usual Lebesgue measure. Let H = L2(D) be the space ofsquare-integrable complex-valued functions on D.(1) Show that the functions zn : n ≥ 0 are orthogonal in H.(2) Do the functions ‖zn‖−1

L2(D) : n ≥ 0 form an orthonormal basis forH?

Solution.

38 ANALYSIS QUALS

(1) In polar form, zn = rneinθ. Then

〈zn, zm〉 =∫∫

D(rneinθ)(rme−imθ)rdrdθ =

(∫ 1

0

rm+ndr

)(∫ 2π

0

ei(n−m)θdθ

).

For n 6= m the latter integral is zero. Hence zn and zm are orthogonalfor m 6= n.

(2) No. Consider the function f(reiθ) = r. Then 〈f, zn〉 = 0 for n > 0,but f is not equal a.e. to a constant function.

Problem 7: Let f : C → C be entire and injective (i.e. univalent). Provethat f is linear: f(z) = az + b.

Solution. If f has infinitely many nonzero terms in its power series expan-sion, then it has an essential singularity at ∞; by Picard’s theorem, f thentakes on all values (except possibly one) in every neighborhood of ∞. Butthen it cannot be injective (take any neighborhood of ∞, choose any twopoints outside the neighborhood, and at least one of their images must behit by something in the neighborhood as well). Hence f has only finitelymany nonzero terms in its Taylor series, so f is a polynomial. If f hasdegree n > 2, f has n roots; unless they’re all the same, this contradictsinjectivity. If they are all the same, f(z) = c(z − z0)n for some c and z0.But then f(z0 + 1) = f(z0 + e2πi/n) so f is not injective. Hence the degreeof f must be at most 1. It cannot be zero because constant functions arenot injective, so f(z) = az + b with a 6= 0.

Problem 8: Suppose that f is holomorphic on the open unit disk D = z :|z| < 1. Suppose also that for z ∈ D, Re(f(z)) > 0 and that f(0) = 1.Prove that for all z ∈ D,

|f(z)| ≤ 1 + |z|1− |z|

.

Solution. Define

g(z) =f(z)− 1f(z) + 1

.

Then since Re(f) > 0, g is analytic on D and |g(z)| ≤ 1 on D. Also notethat g(0) = 0. By the Schwarz lemma, |g(z)| ≤ |z| on D. Then

|f(z)| − 1 ≤ |f(z)− 1| ≤ |z||1 + f(z)| ≤ |z|+ |z||f(z)| ⇒ |f(z)| ≤ 1 + |z|1− |z|

by the Triangle Inequality.

Problem 9: Evaluate ∫ π

0

dθ

2 + cos θusing the residue theorem.

Solution. By the periodicity of the cosine, this is half of∫ 2π

0

dθ

2 + cos θ.

ANALYSIS QUALS 39

To evaluate this integral, let z = eiθ. Then dθ = dziz and the integral

becomes a contour integral around the unit circle, namely∮dz

iz

1

2 + z+1/z2

= −2i∮

dz

(z + 2)2 − 3.

The singularity inside the unit circle is at z = −2 +√

3, with residue1

(z +√

3 + 2)=

12√

3,

so by the Residue Theorem,∫ π

0

dθ

2 + cos θ=

12

(−2i)(

2πi1

2√

3

)=

π√3.

Problem 10: Find an explicit conformal mapping from the slit disk

S = z : |z| < 1 and z /∈ [12, 1)

onto the disk D = z : |z| < 1. (You may express your answer as acomposition of explicit maps.)

Solution. The map z 7→ i 1+z1−z takes this region to the UHP minus the slit

from 3i to i∞; then z 7→ − 1z/3 takes this to the UHP minus the slit from 0

to i. Then z 7→√z2 + 1 takes this to the UHP, and z 7→ z−i

z+i takes this tothe unit disc.

Problem 11: Show that if α ∈ C satisfies 0 < |α| < 1 and n ∈ Z+ =1, 2, 3, . . . then the equation

ez(z − 1)n = α

has exactly n simple roots in the right half-plane z : Re(z) > 0.

Solution. First, we note that ddz e

z(z − 1)n = ez(z − 1)n−1(z + n − 1) 6= 0in the right half-plane, so all roots must be simple. Next, for z in theRHP but on or outside the circle |z − 1| = 1, |ez(z − 1)n| = |ez||z − 1|n =eRe(z)|z − 1|n > 1 > |α| so there can be no roots in this region. Finally, onthe circle |z − 1| = 1, since |ez| ≥ 1, |ez(z − 1)n| ≥ 1, so Rouche’s theoremguarantees that ez(z − 1)n and ez(z − 1)n − α have the same number ofzeros inside the circle; the former clearly has a zero of order n at 1 and noother zeros, so ez(z−1)n−α has n simple zeros inside the circle |z−1| = 1and no others in the RHP.

Problem 12: For an = 1− 1n2 let

f(z) =∞∏n=1

an − z1− anz

.

(1) Show that f defines a holomorphic function on D = z : |z| < 1.(2) Prove that f does not have an analytic continuation to any larger diskz : |z| < r where r > 1.

Solution.

40 ANALYSIS QUALS

(1) A little algebra reveals thatan − z1− anz

= 1− 1 + z

(1− z)n2 + z.

Now∞∑n=1

|1 + z||(1− z)n2 + z|

converges (one way to see this is that the denominator is eventuallymore than |1 − z|n2 − 1, and then applying the integral test), so theproduct converges absolutely.

(2) Note that f(an) = 0 and an → 1. If f extends to an analytic functionon a region including 1, the set of zeros has an accumulation point,so the function would have to be constant. But f(0) = 1 6= 0, acontradiction.