ANALYSIS QUALIFYING EXAM PROBLEMS BRIAN LEARY ...
ANALYSIS QUALIFYING EXAM PROBLEMS BRIAN LEARY ...
ANALYSIS QUALIFYING EXAM PROBLEMS BRIAN LEARY ...
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<strong>ANALYSIS</strong> <strong>QUALIFYING</strong> <strong>EXAM</strong> <strong>PROBLEMS</strong><br />
<strong>BRIAN</strong> <strong>LEARY</strong><br />
Contents<br />
Spring 2011 1<br />
Fall 2010 7<br />
Spring 2010 13<br />
Fall 2009 19<br />
Spring 2009 23<br />
Fall 2008 28<br />
Spring 2008 33<br />
Fall 2007 38<br />
Winter 2007 43<br />
I make no claims about the infallibility of these solutions. In addition to my own<br />
solutions, I also received considerable aid from Ben Hayes and Quinn Maurmann,<br />
as well as the previous work by William Meyerson, Dave Gaebler, and Jon Handy.<br />
Spring 2011<br />
Problem 1: (a) Define what it means to say that fn → f weakly in L 2 ([0, 1]).<br />
(b) Suppose fn ∈ L 2 ([0, 1]) converge weakly to f ∈ L 2 ([0, 1]) and define<br />
‘primitive’ functions:<br />
Fn(x) :=<br />
x<br />
0<br />
fn(t)dt and F (x) =<br />
x<br />
0<br />
f(t)dt.<br />
Show that Fn, F ∈ C([0, 1]) and that Fn → F uniformly on [0, 1].<br />
Proof. (a) We say that fn → f weakly in L 2 ([0, 1]) if for all g ∈ L 2 ([0, 1]),<br />
we have<br />
1<br />
0<br />
fn(x)g(x)dx →<br />
1<br />
0<br />
1<br />
f(x)g(x)dx as n → ∞.
2 <strong>ANALYSIS</strong> QUALS<br />
(b) WLOG, assume 0 ≤ y ≤ x ≤ 1. Then<br />
<br />
x y <br />
<br />
|Fn(x) − Fn(y)| = <br />
fn(t)dt − fn(t)dt<br />
<br />
0<br />
0<br />
<br />
x <br />
<br />
= <br />
fn(t)dt<br />
<br />
y<br />
<br />
1<br />
<br />
<br />
= <br />
fn(t)χ [y,x](t)dt<br />
<br />
≤<br />
≤<br />
0<br />
1<br />
0<br />
1<br />
Hence, Fn ∈ C([0, 1]).<br />
Similarly,<br />
0<br />
|fn(t)| χ [y,x](t)dt<br />
|fn(t)| 2 dt<br />
1/2 1<br />
χ [y,x](t)<br />
0<br />
2 1/2<br />
dt<br />
= fn L 2 (x − y) 1/2 → 0 as x − y → 0.<br />
|F (x) − F (y)| ≤ f L 2 (x − y) 1/2 ,<br />
so F ∈ C([0, 1]) as well. It is left to show that Fn → F uniformly on<br />
[0, 1].<br />
Note that x ↦→ |Fn(x) − F (x)| is a continuous function on [0, 1], so<br />
as [0, 1] is compact, it attains a maximum. Thus, there exists y ∈ [0, 1]<br />
such that<br />
|Fn(x) − F (x)| ≤ |Fn(y) − F (y)|<br />
for every x ∈ [0, 1]. Let ε > 0. Note that χ [0,y] ∈ L 2 ([0, 1]). As fn → f<br />
weakly, we may choose N such that<br />
<br />
1<br />
1<br />
<br />
<br />
<br />
fn(t)χ [0,y](t)dt − f(t)χ [0,y](t)dt<br />
< ε when n ≥ N.<br />
0<br />
0<br />
Let x ∈ [0, 1] and suppose n ≥ N. Then we have that<br />
<br />
1<br />
1<br />
<br />
<br />
|Fn(x) − F (x)| ≤ |Fn(y) − F (y)| = <br />
fn(t)χ [0,y](t)dt − f(t)χ [0,y](t)dt<br />
< ε.<br />
Thus, Fn → F uniformly.<br />
0<br />
Problem 5: (a) Show that l ∞ (Z) contains continuum many functions xα :<br />
Z → R obeying<br />
xα l ∞ = 1 and xα − xβ l ∞ ≥ 1 whenever α = β.<br />
(b) Deduce (assuming the axiom of choice) that the Banach space dual of<br />
l ∞ (Z) cannot contain a countable dense subset.<br />
(c) Deduce that l 1 (Z) is not reflexive.<br />
0
<strong>ANALYSIS</strong> QUALS 3<br />
Proof. (a) First, we note that P(Z), the power set of Z, contains continuum<br />
many elements. For α ∈ P(Z), α = ∅, let xα = χα, the<br />
characteristic function of α. Then |xα L ∞ = 1. But for α = β, there<br />
exists n ∈ α∆β. Then |xα(n) − xβ(n)| = 1, so xα − xβ L ∞ ≥ 1.<br />
(b) Assume for sake of contradiction that l ∞ (Z) ∗ has a countable dense<br />
subset {fn}. That is, l ∞ (Z) ∗ is separable.<br />
For each n, choose yn ∈ l ∞ (Z) such that |fn(yn)| ≥ 1<br />
2 fn. Let S be<br />
the set of linear combinations of {yn} with rational coefficients. Then<br />
S is countable. We claim S = l ∞ (Z). If not, S is a closed proper<br />
subspace of l ∞ (Z). By the Hahn-Banach theorem, we may choose<br />
f ∈ l ∞ (Z) ∗ such that f(y) = 0 for all y ∈ S, but f = 1.<br />
Choose {fnj } for our countable dense set such that fnj − f → 0 as<br />
j → ∞. But<br />
<br />
fnj − f ≥ fnj (ynj ) − f(ynj ) = fnj (ynj ) ≥ 1<br />
<br />
fnj .<br />
2<br />
Thus, we must have <br />
fnj → 0, so f = 0, which contradictions the<br />
assumption that S was a proper subspace. Hence, l∞ (Z) is separable.<br />
Thus, we let {yn} be a countable dense set in l∞ (Z). We have that<br />
for every ε > 0 and for every non-empty α ∈ P(Z), there exists n<br />
such that yn − xαL∞ < ε. But there are uncountably many xα with<br />
xα − xβ ≥ 1 if α = β, so by the Pigeonhole Principle (and Axiom of<br />
Choice), the xα’s cannot all be approximated by the yn’s, so we have<br />
a contradiction to the assumption that l∞ (Z) ∗ has a countable dense<br />
subset.<br />
(c) By definition, l1 (Z) reflexive means that l1 (Z) ∼ = (l1 (Z)) ∗∗ . But l1 (Z)<br />
is separable, and (l1 (Z)) ∗∗ ∼ = (l∞ (Z)) ∗ is not separable by part (b).<br />
Hence, l1 (Z) is not reflexive.<br />
Note: I lost a point on the qual with this solution for not elaborating<br />
on the fact that l1 (Z) is separable. I don’t know whether they wanted<br />
me to prove this, or just mention a countable dense set. For the record,<br />
the set of linear combinations with rational coefficients of the sequences<br />
ek consisting of all zeroes with 1 in the kth spot is a countable dense<br />
set.<br />
<br />
Problem 6: Suppose µ and ν are finite positive (regular) Borel measures on<br />
R n . Prove the existence and uniqueness of the Lebesgue decomposition:<br />
There are a unique pair of positive Borel measures µa and µs so that<br />
µ = µa + µs, µa
4 <strong>ANALYSIS</strong> QUALS<br />
<br />
X fndν → a. Let gn = max{f1, . . . , fn}. Then gn ∈ F for all n. Furthermore,<br />
gn increases pointwise to f = sup{fn}. Then <br />
X gndν → a, and by<br />
the monotone convergence theorem, a = <br />
fdν and f ∈ F.<br />
X<br />
Define dµs = dµ − fdν, and dµa <br />
= fdν. Then if ν(E) = 0, we have that<br />
E fdν = 0, so µa 0 and E ∈ M such that ν(E) = 0<br />
and E is a positive set for µs−εν. That is, µs ≥ εν on E. But then εχEdν ≤<br />
χEdµs, so εχEdν ≤ χE(dµ−fdν), which implies that (εχE +f)dν ≤ µ(E).<br />
But this contradicts our selection of a as the supremum of all such integrals.<br />
Therefore, we conclude that µs ⊥ ν. This completes the proof. <br />
Problem 7: Prove Gorsat’s theorem: If f : C → C is complex differentiable<br />
(and so continuous), then for every triangle ∆ ⊂ C<br />
<br />
f(z)dz = 0<br />
∂∆<br />
where line integral is over the three sides of the triangle.<br />
Solution. Let ∆ be a closed triangle in C and let a, b, c be the vertices of ∆.<br />
Let a ′ , b ′ , c ′ be the midpoints of [b, c], [a, c], [a, b], respectively. Consider the<br />
four triangles ∆ j , j = 1, 2, 3, 4, with vertices {a, c ′ , b ′ }, {b, a ′ , c ′ }, {c, b ′ , a ′ },<br />
and {a ′ , b ′ , c ′ }. Let<br />
Then<br />
<br />
J =<br />
J =<br />
Hence, <br />
∂∆<br />
4<br />
<br />
j=1<br />
∂∆ j<br />
f(z)dz.<br />
f(z)dz.<br />
<br />
<br />
f(z)dz<br />
≥ |J/4|<br />
∂∆k for some k. Define ∆1 to be this ∆k . Replace ∆ with ∆1 and repeat. This<br />
yields a sequence of triangles {∆n} such that<br />
∆ ⊃ ∆1 ⊃ ∆2 ⊃ . . . ⊃ ∆n ⊃ ∆n+1 ⊃ . . .<br />
and the length of ∂∆n is 2−nL, where L is the length of ∂∆. Thus, we have<br />
that<br />
|J| ≤ 4 n<br />
<br />
<br />
<br />
<br />
<br />
<br />
f(z)dz<br />
for all n.<br />
∂∆n<br />
Choose z0 ∈ ∞<br />
n=1 ∆n, and we note that in fact this z0 is unique. Let<br />
ε > 0. As f is complex differentiable, we may choose r > 0 such that if<br />
|z − z0| < r, then<br />
for z ∈ ∆.<br />
|f(z) − f(z0) − f ′ (z0)(z − z0)| ≤ ε |z − z0|<br />
Choose n such that |z − z0| < r for all z ∈ ∆n. Note also that |z − z0| ≤<br />
2 −n L for all z ∈ ∆n.<br />
We use the fact that <br />
γ zn dz = 0 for any closed path γ ⊂ C and n =
∂∆n<br />
<strong>ANALYSIS</strong> QUALS 5<br />
0, 1, 2, . . ., which can be seen considering antiderivatives, parametrization,<br />
and FTC. Then<br />
<br />
<br />
f(z)dz = f(z) − f(z0) − f ′ (z0)(z − z0)dz,<br />
∂∆n<br />
so<br />
<br />
<br />
f(z)dz<br />
≤<br />
<br />
∂∆n<br />
∂∆n<br />
|f(z) − f(z0) − f ′ <br />
(z0)(z − z0)| dz ≤<br />
∂∆n<br />
ε |z − z0| ≤ ε(2 −n L) 2 .<br />
Then |J| ≤ εL 2 . Hence, J = 0. <br />
Problem 8: (a) Define upper-semicontinuous for functions f : C → [−∞, ∞).<br />
(b) Define what it means for such an upper-semicontinuous function to be<br />
subharmonic.<br />
(c) Prove for refute (with a counterexample) each of the following:<br />
The pointwise supremum of a bounded family of subharmonic functions<br />
is subharmonic.<br />
The pointwise infimum of a family of subharmonic functions is subharmonic.<br />
(d) Let A(z) be a 2×2 matrix-valued holomorphic function (i.e., the entries<br />
are holomorphic). Show that<br />
z ↦→ log (A(z)) is subharmonic<br />
where A(z) is the norm as an operator on the Hilbert space C 2 .<br />
Proof. (a) f is defined to be upper-semicontinuous if for every α ∈ R, the<br />
set {z ∈ C : f(z) < α} is an open set.<br />
(b) f is defined to be subharmonic if for any closed ball B and every<br />
continuous h : B → R that is harmonic on the interior of B such that<br />
f(z) ≤ h(z) on the boundary of B, we have that f(z) ≤ h(z) for all<br />
z ∈ B.<br />
(c) First, let {fi}i∈I be a bounded family of subharmonic functions and<br />
define f(z) = supI fi(z). Let B be a closed ball. Suppose that h is<br />
continuous on B, harmonic on the interior, and f(z) ≤ h(z) on the<br />
boundary. Then for all i ∈ I, fi(z) ≤ h(z) on the boundary as well, so<br />
since fi is subharmonic, we conclude that fi(z) ≤ h(z) on all of B for<br />
all i ∈ I. Thus, we have that f(z) ≤ h(z) on B as well.<br />
The second claim is false. Consider fn(z) = Re(zn ), with f(z) =<br />
infn fn(z). Let B be the closed unit ball. For almost every z ∈ ∂B,<br />
f(z) = −1, but f(0) = 0, so the sub-mean-value-property is violated:<br />
(d) We write<br />
f(0) ≤ 1<br />
2π<br />
<br />
∂B<br />
f(z)dz.<br />
A(z) = sup{Re 〈A(z)v, w〉 : v, w ∈ C 2 , v = w = 1}.<br />
Thus, we have the family of functions fv,w defined by fv,w(z) =<br />
log Re 〈A(z)v, w〉 for v, w unit vectors of C 2 . For each fixed v, w, this<br />
is the real part of a holomorphic function, so it is harmonic, and thus<br />
subharmonic. By part c, this means that z ↦→ log (A(z)) is subharmonic,<br />
since it is the supremum of subharmonic functions.
6 <strong>ANALYSIS</strong> QUALS<br />
Problem 10: Let D = {z : |z| < 1} and let Ω = {z ∈ D : Imz > 0}.<br />
Evaluate<br />
sup{Ref ′ ( i<br />
)|f : Ω → D is holomorphic}.<br />
2<br />
Proof. Note: When I did this on the qual, I misread the problem and<br />
thought that Ω was the upper half-plane instead of merely the upper halfdisc,<br />
which caused the already awful calculations to get worse. Here is my<br />
attempt at the correct calculations, but it seems too ugly to be correct.<br />
Follow at your own peril.<br />
First, note that by rotation, it suffices to find the sup of f ′ ( i<br />
2 ) instead<br />
of Re f ′ ( i<br />
2 ).<br />
The idea is to use Schwarz Lemma. That is, we want g : D → Ω in order<br />
to have f ◦ g map D to D, and then tweak by automorphisms of the disc to<br />
get the map to send 0 to 0.<br />
We have that z ↦→ i 1+z<br />
1−z takes D to the upper half-plane, z ↦→ √ z maps<br />
the upper half-plane to the first quadrant, and z ↦→ z−1<br />
z+1 maps the first<br />
quadrant to Ω. Thus, let<br />
<br />
1+z i 1−z − 1<br />
g(z) = .<br />
i + 1<br />
1+z<br />
1−z<br />
Let f1 be the automorphism of the disc taking g(0) to i<br />
2 , and f2 be the<br />
automorphism of the disc taking f( i<br />
2<br />
explicitly; for example we recall that f2(z) =<br />
) to 0. These guys can be written out<br />
i z−f( 2 )<br />
1−f( i<br />
2<br />
Let h = f2 ◦ f ◦ f1 ◦ g. Then h : D → D with h(0) = 0. By Schwarz<br />
Lemma, |h ′ (0) ≤ 1. But we note that<br />
By calculation, we have that<br />
and<br />
and<br />
where we note that<br />
h ′ (0) = g ′ (0)f ′ 1(g(0))f ′ ( i<br />
2 )f ′ 2(f( i<br />
2 )).<br />
f ′ 2(f( i 1<br />
)) =<br />
2 1 − f( i<br />
2 ) ,<br />
g ′ (0) =<br />
f ′ 1(g(0)) =<br />
|g(0)| =<br />
1<br />
√ i( √ i + 1) 2 ,<br />
i 1 + 2<br />
2 ,<br />
1 − |g(0)|<br />
<br />
√ i − 1 <br />
<br />
√ i + 1 .<br />
)z .
<strong>ANALYSIS</strong> QUALS 7<br />
Thus, as |h ′ (0)| ≤ 1, we have that<br />
<br />
<br />
<br />
f ′ ( i<br />
2 )<br />
<br />
<br />
<br />
≤ (1 − i f( 2 ) √ 2 ) i + 1 2<br />
(1 − |g(0)| )<br />
<br />
1 + i<br />
<br />
<br />
,<br />
and so<br />
<br />
<br />
<br />
f ′ ( i<br />
2 )<br />
<br />
<br />
<br />
≤ (1 − i f( 2 ) √ 2 √ 2 )( i + 1 − i − 1 )<br />
<br />
i 1 + <br />
.<br />
We have that equality holds if and only if h is a rotation. To maximize<br />
the right-hand-side of the above, we may take f( i<br />
2 ) = 0 (which of course<br />
eliminates the need for f2). Then we conclude that<br />
<br />
<br />
sup <br />
f ′ ( i<br />
2 )<br />
<br />
<br />
<br />
=<br />
<br />
√ i + 1 2 − √ i − 1 2<br />
<br />
1 + i<br />
<br />
<br />
.<br />
This completes the proof. <br />
Fall 2010<br />
Problem 1: For this problem, consider just Lebesgue measurable functions<br />
f : [0, 1] → R together with the Lebesgue measure.<br />
(a) State Fatou’s lemma (no proof required).<br />
(b) State and prove the Dominated Convergence Theorem.<br />
(c) Give an example where fn(x) → 0 a.e., but fn(x)dx → 1.<br />
Solution.<br />
(a) Fatou: Let {fn} be a sequence of nonegative functions such that f =<br />
lim fn exists a.e. Then f(x)dx ≤ lim inf fn(x)dx.<br />
(b) DCT: Let {fn} be a sequence of functions such that fn → f pointwise<br />
a.e., and suppose there exists Lebesgue integrable g with fn} ≤ g a.e.<br />
for every n. Then fn(x)dx → f(x)dx.<br />
2<br />
2<br />
Proof: g + fn → g + f a.e., and g + fn is nonnegative a.e., so by<br />
Fatou, g + f ≤ lim inf (g + fn). Then<br />
<br />
<br />
g + f ≤ g + lim inf<br />
so we have that<br />
<br />
<br />
f ≤ lim inf<br />
Similarly, g − fn → g − f a.e. and g − fn is nonnegative a.e., so by<br />
Fatou, g − f ≤ lim inf (g − fn). Then<br />
<br />
<br />
g − f ≤ g − lim sup<br />
so we have that<br />
<br />
lim sup<br />
fn ≤<br />
Hence, f = lim fn.<br />
(c) Consider the pointwise-linear bump functions fn defined to be 0 at 0,<br />
n at 1/n, 0 at 2/n, and linear between.<br />
<br />
2<br />
fn<br />
f<br />
fn<br />
fn
8 <strong>ANALYSIS</strong> QUALS<br />
Problem 2: Prove the following form of Jensen’s inequality:<br />
If f : [0, 1] → R is continuous, then<br />
1<br />
e f(x) 1 <br />
dx ≥ exp f(x)dx .<br />
0<br />
Moreover, if equality occurs then f is a constant function.<br />
Solution. Let a = min [0,1] f, b = max [0,1] f, and t = 1<br />
f(x)dx. Let<br />
0<br />
Then for a < s < t, we have that<br />
e<br />
β = sup<br />
a
<strong>ANALYSIS</strong> QUALS 9<br />
But by the geometric series, the LHS is equal to<br />
1<br />
N<br />
and since α /∈ Q, the denominator is never 0. Hence, this approaches<br />
0 as N → ∞, and 1<br />
0 e2πikxdx = 0, so the statement is true for all<br />
trigonometric polynomials.<br />
1 − e 2πiNαk<br />
1 − e 2πiαk<br />
Now let ε > 0, f a continuous function from R/Z to R. As trigonometric<br />
polynomials are uniformly dense in the space of continuous periodic<br />
functions, we may choose P (x) trigonometric such that<br />
sup |f(x) − P (x)| <<br />
x∈R/Z<br />
ε<br />
3 .<br />
Then for N sufficiently large, we have that<br />
<br />
N−1<br />
1 <br />
<br />
1 <br />
ε<br />
P (nα) − P (x)dx<br />
<<br />
N<br />
3 .<br />
n=0<br />
Then <br />
N−1<br />
1 <br />
<br />
1 <br />
<br />
f(nα) − f(x)dx<br />
N<br />
n=0<br />
0 <br />
≤ 1<br />
<br />
N−1 <br />
N−1<br />
1 <br />
<br />
1 <br />
<br />
|f(nα) − P (nα)| + P (nα) − P (x)dx<br />
N<br />
N<br />
+<br />
0<br />
n=0<br />
n=0<br />
0<br />
0<br />
< ε. Hence, the statement holds for general f.<br />
(b) Choose f + ε , f − ε continuous on the torus so that f ( εx) ≤ χ [a,b](x) ≤<br />
f + ε (x) and<br />
1<br />
b − a − 2ε ≤<br />
0<br />
f − 1<br />
ε (x)dx,<br />
0<br />
f + ε (x)dx < b − a + 2ε.<br />
Let SN = 1<br />
N<br />
N−1<br />
n=0 χ [a,b](nα). Then<br />
N−1<br />
1 <br />
f<br />
N<br />
− ε (nα) ≤ SN ≤ 1<br />
N−1 <br />
N<br />
n=0<br />
n=0<br />
f + ε (nα).<br />
1<br />
|P (x) − f(x)| dx<br />
But the left side approaches 1<br />
0 f − ε (x)dx and the right side approaches<br />
1<br />
0 f + ε (x)dx. Thus<br />
b − a − 2ε ≤ lim<br />
N→∞ SN ≤ b − a + 2ε.<br />
As this holds for all ε > 0, we have that<br />
lim<br />
n→∞ SN = b − a =<br />
1<br />
0<br />
χ [a,b](x)dx.<br />
Problem 6: Let D := {z ∈ C : |z| ≤ 1} and consider the (complex) Hilbert<br />
space<br />
<br />
∞<br />
H := f : D → C|f(z) = ˆf(k)z k with f 2 ∞<br />
:= (1 + |k| 2 )| ˆ f(k)| 2 <br />
< ∞<br />
k=0<br />
k=0
10 <strong>ANALYSIS</strong> QUALS<br />
(a) Prove that the linear functional L : f ↦→ f(1) is bounded.<br />
(b) Find the element g ∈ H representing L.<br />
(c) Show that f ↦→ ReL(f) achieves its maximal value on the set<br />
B := {f ∈ H : f ≤ 1 and f(0) = 0} ,<br />
that this maximum occurs at a unique point, and determine this maximal<br />
value.<br />
Solution. (a) Using Cauchy-Schwarz,<br />
<br />
∞<br />
<br />
<br />
<br />
|L(f)| = ˆf(k)<br />
<br />
<br />
<br />
k=0<br />
≤<br />
∞ <br />
<br />
<br />
k=0<br />
ˆ <br />
<br />
f(k) <br />
∞<br />
<br />
1 + |k| 2 <br />
<br />
= <br />
1 + |k| 2<br />
k=0<br />
ˆ <br />
∞<br />
<br />
1<br />
f(k) ≤<br />
1 + |k|<br />
k=0<br />
2<br />
1/2 <br />
∞<br />
(1 + |k|<br />
k=0<br />
2 <br />
<br />
) ˆ <br />
<br />
f(k)<br />
<br />
∞<br />
1<br />
=<br />
1 + |k| 2<br />
1/2 f .<br />
k=0<br />
2<br />
1/2 (b) Let ˆg(k) = 1<br />
1+|k| 2 , g(z) = ∞ k=0 ˆg(k)zk . Then g(z) converges for all<br />
z ∈ D, and<br />
∞<br />
k=0<br />
so g ∈ H. Then<br />
(1 + |k| 2 )(1 + |k| 2 ) −2 < ∞,<br />
〈f, g〉 =<br />
=<br />
∞<br />
(1 + |k| 2 ) ˆ f(k)ˆg(k)<br />
k=0<br />
∞<br />
k=0<br />
1 + |k| 2<br />
1 + |k| 2 ˆ f(k)<br />
= f(1) = L(f)<br />
(c) B is a compact, convex set in a Hilbert space, so a bounded linear<br />
functional on it must achieve a maximum at a unique point. We claim<br />
the maximizer is f(z) = z √ 2 . This is in B, and the handwavey argument<br />
of why this is the maximizer is that we want all possible weight on the<br />
smallest possible k, since the computation of the norm gives increasing<br />
penalty to larger k. Since f(0) = 0, ˆ f(0) = 0, so we cannot put any<br />
weight on k = 0, and thus we put all possible weight on k = 1.<br />
<br />
Problem 7: Suppose that f : C → C is continuous on C and holomorphic on<br />
C − R. Prove that f is entire.<br />
Solution. Let Γ be a triangle in C. If Γ does not cross the real axis, then as f<br />
is holomorphic off R, Cauchy integral theorem guarantees that <br />
f(z)dz =<br />
Γ<br />
0.<br />
Now suppose Γ crosses the real axis. For k > 0, we cut a 1/k neighborhood<br />
out about the real axis and let Γ (1)<br />
k and Γ(2)<br />
k be the resulting polygons above<br />
and below the real axis, respectively, as shown in the figure.
Then <br />
Γ (j)<br />
k<br />
<strong>ANALYSIS</strong> QUALS 11<br />
f(z)dx = 0 for all k, j = 1, 2. But Γ (j)<br />
k converges uniformly<br />
to Γ (j) with <br />
Γ (1) f(z)dz + <br />
Γ (2) f(z)dz = <br />
<br />
f(z)dz. Hence, f(z)dz = 0.<br />
Γ Γ<br />
Thus, by Morera’s theorem, f is entire. <br />
Problem 9: Let<br />
f(z) =<br />
∞<br />
anz n<br />
n=0<br />
be a holomorphic function in D. Show that if<br />
∞<br />
n|an| ≤ |a1|<br />
n=2<br />
with a1 = 0 then f is injective.<br />
Solution. Note that f ′ (z) = ∞ n=1 nanzn−1 . Then<br />
|f ′ <br />
∞ <br />
(z) − a1| = nanz<br />
<br />
n−1<br />
<br />
<br />
<br />
<br />
≤<br />
∞<br />
n|an||z| n−1 ∞<br />
< n|an| ≤ |a1|.<br />
n=2<br />
n=2<br />
Hence, |f ′ (z) − a1| < |a1|. Thus, we may choose θ so that Re(e iθ f ′ (z)) > 0<br />
for all z ∈ D. Then if z1 = z2,<br />
<br />
z1 <br />
|f(z1) − f(z2)| = <br />
f ′ <br />
<br />
(z)dz<br />
> |z2 − z1|<br />
z2<br />
1<br />
0<br />
n=2<br />
e iθ Re(f ′ (z1 + t(z2 − z1)))dt > 0.<br />
Thus, f(z2) = f(z1), so f is injective. <br />
Problem 10: Prove that the punctured disc {z ∈ C : 0 < |z| < 1} and the<br />
annulus given by {z ∈ C : 1 < |z| < 2} are not conformally equivalent.
12 <strong>ANALYSIS</strong> QUALS<br />
Problem 11: Let Ω ⊂ C be a non-empty open connected set. If f : Ω → C is<br />
harmonic and f 2 is also harmonic, show that either f or f is holomorphic<br />
on Ω.<br />
Proof. Write f(x + iy) = u(x, y) + iv(x, y). By the Cauchy-Riemann equations,<br />
it suffices to show that either<br />
∂u ∂v<br />
=<br />
∂x ∂y<br />
and ∂v<br />
∂x<br />
∂u<br />
= −∂u , or = −∂v<br />
∂y ∂x ∂y<br />
∂v ∂u<br />
and =<br />
∂x ∂y .<br />
Then as ∆f = 0, we have that ∆u = ∆v = 0. Note that f 2 = u 2 (x, y) −<br />
v2 (x, y) + 2iu(x, y)v(x, y). Thus,<br />
∆(f 2 ∂u 2 ) = 2 +<br />
∂x<br />
∂2 2 u<br />
∂v<br />
u(x, y) − −<br />
∂x2 ∂x<br />
∂2 <br />
v<br />
v(x, y)<br />
∂x2 ∂u 2 +2 +<br />
∂y<br />
∂2 2 u<br />
∂v<br />
u(x, y) − −<br />
∂y2 ∂y<br />
∂2 <br />
v<br />
v(x, y)<br />
∂y2 <br />
+2i 2 ∂u ∂v<br />
∂x ∂x + ∂2u ∂x2 v(x, y) + ∂2v ∂v<br />
u(x, y) + 2∂u<br />
∂x2 ∂y ∂y + ∂2u ∂y2 v(x, y) + ∂2 <br />
v<br />
u(x, y)<br />
∂y2 But using ∆u = ∆v = 0, we get that<br />
∆(f 2 ∂u 2 2 2 <br />
2 <br />
∂u ∂v ∂v ∂u ∂v ∂u ∂v<br />
) = 2 + − − + 4i +<br />
∂x ∂y ∂x ∂y ∂x ∂x ∂y ∂y<br />
We set this equal to 0 and separate real and imaginary parts to find that<br />
∂u ∂v ∂v<br />
= −∂u<br />
∂x ∂x ∂y ∂y and<br />
2 2 2 2 ∂u ∂u ∂v ∂v<br />
+ = + .<br />
∂x ∂y ∂x ∂y<br />
Thus, ∂u<br />
∂x<br />
<br />
∂v 2<br />
− i ∂x =<br />
∂v<br />
∂y<br />
+ i ∂u<br />
∂y<br />
2 , so ∂u ∂v<br />
∂x − i ∂x<br />
<br />
∂v = ± ∂y<br />
+ i ∂u<br />
∂y<br />
<br />
. This<br />
yields the result. <br />
Problem 12: Let F be the family of functions f holomorphic on D with<br />
<br />
|f(x + iy)| 2 dxdy < 1.<br />
x 2 +y 2
<strong>ANALYSIS</strong> QUALS 13<br />
= 1<br />
<br />
√ |f(w)|<br />
πr D<br />
2<br />
1/2 < 1<br />
√ ,<br />
πr<br />
since f ∈ F. Thus, let A = 1 √ , which depends only on K and does not<br />
πr<br />
depend on x or f. <br />
Spring 2010<br />
Problem 1: (a) Let 1 ≤ p < ∞. Show that if a sequence of real-valued<br />
functions {fn}n≥1 converges in L p (R), then it contains a subsequence<br />
that converges almost everywhere.<br />
(b) Give an example of a sequence of functions converging to zero in L 2 (R)<br />
that does not converge almost everywhere.<br />
Solution. (a) By Chebyshev’s inequality,<br />
µ ({x : |f(x)| > α}) ≤ fp p<br />
.<br />
αp Then we have that<br />
µ ({x : |fn(x) − f(x)| > ε}) ≤ fn − f p<br />
p<br />
εp → 0 as n → ∞,<br />
so fn → f in measure.<br />
Then we may choose a subsequence {fnj } such that<br />
µ x : fnj (x) − fnj+1 (x) > 2 −j ≤ 2 −j .<br />
Let Ej = {x : |fnj (x) − fnj+1 (x)| > 2−j }. Let Fk = ∞<br />
j=k Ej. Then<br />
µ(Fk) ≤<br />
∞<br />
µ(Ej) ≤ 2 1−k .<br />
j=k<br />
For x /∈ Fk and k ≤ j ≤ i, we have that<br />
<br />
fnj (x) − fni (x) ≤<br />
i−1<br />
<br />
fnl (x) − fnl+1 (x) i−1<br />
≤<br />
l=j<br />
l=j<br />
2 −l ≤ 2 1−j ,<br />
so {fnj } is pointwise Cauchy on F C k . Let F = ∞<br />
k=1 Fk. Then µ(F ) =<br />
0 and fnj converges on F C , so fnj<br />
converges almost everywhere.<br />
(b) We use the example of the moving dyadic intervals. Let f1 = χ [0,1],<br />
f2 = χ [0,1/2], f3 = χ [1/2,1], and in general, for n = 2 k + j < 2 k+1 ,<br />
let fn = χ [j/2 k ,(j+1)/2 k ]. Then fn → 0 in L 2 , but for every x ∈ [0, 1],<br />
fn(x) → 0 as n → ∞.<br />
<br />
Problem 3: For an f : R → R belonging to L1 (R), we define the Hardy-<br />
Littlewood maximal function as follows:<br />
1<br />
(Mf)(x) := sup<br />
h>0 2h<br />
x+h<br />
x−h<br />
|f(y)|dy.<br />
Prove that it has the following property: There is a constant A such that<br />
for any λ > 0,<br />
|{x ∈ R (Mf)(x) > λ}| ≤ A<br />
λ f L 1
14 <strong>ANALYSIS</strong> QUALS<br />
where |E| denotes the Lebesgue measure of E. If you use a covering lemma,<br />
you should prove it.<br />
Solution. We do employ the following covering lemma, due to Weiner in<br />
the style of Vitali.<br />
Lemma: Let C be a collection of open balls in Rn , and let U = <br />
B∈C B.<br />
k<br />
If c < m(U), there exist disjoint B1, B2, . . . , Bk ∈ C with m(Bj) > 3 −n c.<br />
Pf: Choose compact K ⊂ U with m(K) > c. Then finitely many balls<br />
A,A2, . . . , Am ∈ C cover K. Let B1 be the largest of the Ai, let B2 be the<br />
largest of the Ai disjoint from B1, and inductively, let Bk be the largest of<br />
the Ai disjoint from the previously chosen Bjs.<br />
Then for any i, if Ai is not one of the Bj, it has nonempty intersection<br />
with a Bj. Choosing the smallest such j, we have diam(Bj) ≥diam(Ai).<br />
Let B ∗ j be the ball concentric with Bj with thrice the radius. Then Ai ⊂ B ∗ j ,<br />
so K ⊂ <br />
j B∗ j . Hence<br />
j=1<br />
c < m(K) ≤ m( B ∗ j ) = m(B ∗ j ) = 3 n m(Bj).<br />
This proves the lemma.<br />
From this lemma, the problem is solved easily: Let Eλ = {x : (Mf)(x) ><br />
λ}. For each x ∈ Eλ, choose rx > 0 such that<br />
<br />
1<br />
|f(y)|dy > λ.<br />
m(B(x, rx))<br />
j<br />
B(x,rx)<br />
The balls B(x, rx) cover Eλ, so by the lemma, if C < m(Eλ), some finite<br />
collection of disjoint balls {Bj = B(xj, rxj )} satisfies m(Bj) > 1<br />
3c. But<br />
then<br />
c < 3 <br />
m(Bj) ≤ 3 <br />
<br />
|f(y)|dy ≤<br />
λ<br />
3<br />
λ fL1 .<br />
j<br />
Bj<br />
Letting c → m(Eλ) yields the result. <br />
Problem 4: Let f be continuous on D and analytic on D with f(0) = 0.<br />
(a) Prove that if 0 < r < 1 and inf |z|=r |f(z)| > 0, then<br />
1<br />
2π<br />
2π<br />
0<br />
log |f(re iθ )|dθ ≥ log |f(0)|.<br />
(b) Use (a) to prove that |{θ ∈ [0, 2π] : f(e iθ ) = 0}| = 0.<br />
Solution. (a) This is a version of Jensen’s Theorem, a result of which it is<br />
said that the proof is easier to remember than the theorem.<br />
Let zj enumerate the zeroes of f in |z| ≤ r with multiplicity. Consider<br />
the Blashke factor<br />
B(z) = r(z − zj)<br />
r2 − zjz .<br />
j
<strong>ANALYSIS</strong> QUALS 15<br />
Then B has the same zeroes as f, and |B(z)| = 1 when |z| = r. Then<br />
<br />
log f(z)<br />
<br />
<br />
B(z) is harmonic, so by the mean value principle, we have that<br />
<br />
<br />
log <br />
f(0) <br />
2π<br />
<br />
1<br />
B(0)<br />
= log<br />
2π<br />
<br />
iθ<br />
f(re ) dθ.<br />
But<br />
0<br />
<br />
<br />
log <br />
f(0) <br />
<br />
<br />
<br />
B(0)<br />
= log |f(0)| − log zj<br />
<br />
<br />
,<br />
r<br />
and thus<br />
log |f(0)| = 1<br />
2π<br />
log<br />
2π 0<br />
f(re iθ ) dθ + <br />
<br />
log<br />
j<br />
But zj <br />
<br />
r < 1, so <br />
j log zj <br />
<br />
r ≤ 0, so we have that<br />
1<br />
2π<br />
2π<br />
0<br />
log |f(re iθ )|dθ ≥ log |f(0)|.<br />
j<br />
<br />
<br />
.<br />
r<br />
(b) If |{θ ∈ [0, 2π] : f(e iθ ) = 0}| > 0, then 2π<br />
0 log f(re iθ ) dθ → −∞ as<br />
r → 1, which contradicts the inequality in (a).<br />
<br />
Problem 5: (a) For f ∈ L 2 (R) and {xn} converging to zero, define fn(x) =<br />
f(x + xn). Show that {fn} converges to f in L 2 .<br />
(b) Let W ⊂ R be Lebesgue measurable of positive measure. Show that<br />
W − W = {x − y : x, y ∈ W } contains an open neighborhood of the<br />
origin.<br />
Solution. (a) We want to show that fn − f 2 → 0. Since Cc(R) is dense<br />
in L 2 , choose g ∈ Cc(R) such that f − g 2 < ε. Then<br />
fn − f 2 ≤ fn − gn 2 + gn − g 2 + g − f 2 < 2ε + gn − g 2<br />
Chose δ so that |g(x) − g(y)| < ε when |x − y| < δ, and choose N so<br />
that if n > N, |xn| < δ. Then for n > N, we have that<br />
<br />
gn − g 2<br />
2 =<br />
zj<br />
|g(x + xn) − g(x)|<br />
R<br />
2 dx ≤ ε 2 2supp(g).<br />
Thus, fn − f2 → 0 as n → ∞.<br />
(b) Here are two solutions, neither of which uses part (a).<br />
(1) We use the fact that<br />
<br />
f(x) = χW (y)χW (y − x)dy<br />
R<br />
is continuous, which is Q3 in Fall 2004. Then since W has positive<br />
measure, f(0) > 0. As f is continuous, there exists δ > 0 such<br />
that f(t) > 0 for t ∈ (−δ, δ). Then for each t in this interval, we<br />
have that <br />
R χW (y)χW (y − t)dy > 0, so there exists yt such that<br />
χW (yt) = χW (yt − t) = 1. Thus yt ∈ W and yt − t ∈ W . Hence<br />
t = yt − (yt − t) ∈ W − W . Hence (−δ, δ) ⊂ W − W .
16 <strong>ANALYSIS</strong> QUALS<br />
(2) As W has positive measure, then by the definition of outer measure,<br />
there exists a collection of closed intervals In such that W ⊂ In<br />
and<br />
∞<br />
|In| ≤ (21/20)m(W ).<br />
n=1<br />
There exists n such that m(W ∩ I) ≥ (9/10)m(In), for if not, then we<br />
have<br />
m(W ) ≤ m(W ∩ In) < 9<br />
10 m(In) ≤ 9<br />
2120m(W ) < m(W ),<br />
10<br />
which is a contradiction.<br />
Thus, let In = I = [a, b], so m(W ∩I) ≥ (9/10)m(I). Let W0 = W ∩I.<br />
Then it suffices to show that W0 −W0 contains an open neighborhood.<br />
Suppose not. Then for all n, choose an ∈ (−1/n, 1/n) \ (W0 − W0).<br />
Consider W0 ∩ (W0 + an). If x ∈ W0 ∩ (W0 + an), then x = y + an for<br />
y ∈ W0, so an ∈ W0 − W0. Hence, W0 ∩ (W0 + an) = ∅ for all n. But<br />
W0 ∪ (W0 + an) ⊂ I ∪ (I + an) ⊂ [a − |an|, b + |an|],<br />
and thus<br />
m(W0 ∪ (W0 + an)) ≤ b − a + 2|an| → m(I)<br />
as n → ∞. However, since an ∈ W0 − W0, we have that<br />
m(W0 ∪ (W0 + an)) = 2m(W0) ≥ 18<br />
10 m(I)<br />
so we must have m(I) = 0, which is a contradiction. This completes<br />
the proof.<br />
<br />
Problem 7: Let H be a Hilbert space and let E be a closed convex subset of<br />
H. Prove that there exists a unique element x ∈ E such that<br />
x = inf y .<br />
y∈E<br />
Solution. Let d = infy∈E y. Choose {yn} ∈ E such that yn → d. Then<br />
yn − ym 2 = 2 yn 2 + 2 ym 2 2<br />
yn<br />
− 4 <br />
+ ym <br />
<br />
2 .<br />
As E is convex, we have that yn+ym<br />
2 ∈ E, so yn+ym<br />
<br />
<br />
2<br />
2 ≥ d2 . Thus,<br />
yn − ym 2 ≤ 2 yn 2 + 2 ym 2 − 4d 2 → 0 as n, m → ∞.<br />
Thus, {yn} is Cauchy, and since H is complete, the sequence converges to<br />
a point x. Then x ∈ E since E is closed, and x = d. This proves existence.<br />
If y also satisfies y = d, then<br />
2<br />
<br />
<br />
x + y <br />
<br />
2 = 4d 2 <br />
<br />
− 4 <br />
x + y <br />
<br />
2 <br />
0 ≤ x − y 2 = 2 x 2 + 2 y 2 − 4<br />
2<br />
≤ 4d 2 − 4d 2 = 0,<br />
so x = y. This proves uniqueness.
∂Q<br />
<strong>ANALYSIS</strong> QUALS 17<br />
Problem 8: Let F (z) be a non-constant meromorphic function on the complex<br />
plane C such that for all z ∈ C,<br />
F (z + 1) = F (z) and F (z + i) = F (z).<br />
Let Q be a square with vertices z, z + 1, z + i and z + (1 + i) such that F<br />
has no zeros and no poles on ∂Q. Prove that inside Q the functions F has<br />
the same number of zeros as poles (counting multiplicities).<br />
Solution. As F has no zeroes or poles on ∂Q, by the argument principle,<br />
we have that<br />
<br />
F ′ (w)<br />
dw = 2πi(#zeroes − #poles).<br />
F (w)<br />
∂Q<br />
By periodicity, F (w) = F (w + 1) = F (w + i) and F ′ (w) = F ′ (w + 1) =<br />
F ′ (w + i). Thus, we have that<br />
z+1<br />
and z+i<br />
Thus, we have that<br />
F ′ (w)<br />
dw =<br />
F (w)<br />
z+1<br />
so we conclude that<br />
z<br />
z<br />
z<br />
F ′ (w)<br />
F (w) dw+<br />
F ′ (w)<br />
dw =<br />
F (w)<br />
z+i+1<br />
z+i<br />
F ′ (w)<br />
F (w) dw,<br />
F ′ z+1+i<br />
(w)<br />
F<br />
dw =<br />
F (w) z+1<br />
′ (w)<br />
F (w) dw.<br />
<br />
z+1+i<br />
∂Q<br />
z+1<br />
F ′ (w)<br />
F (w) dw−<br />
z+i+1<br />
F<br />
z+i<br />
′ (w)<br />
F (w) dw−<br />
z+i<br />
F<br />
z<br />
′ (w)<br />
F (w) dw<br />
F ′ (w)<br />
dw = 0,<br />
F (w)<br />
and hence the number of zeroes is equal to the number of poles. <br />
Problem 10: Let Ω ⊂ C be a connected open set, let z0 ∈ Ω, and let U be<br />
the set of positive harmonic functions u on Ω such that u(z0) = 1. Prove<br />
for every compact set K ⊂ Ω there is a finite constant M (depending on<br />
Ω, z0, and K) such that<br />
sup sup u(z) ≤ M.<br />
u∈U z∈K<br />
You may use Harnack’s inequality for the disk without proving it, provided<br />
you state it correctly.<br />
Solution. We will use Harnack’s inequality for the disk without proving, so<br />
let’s hope we state it correctly. Here I am using the version from Conway.<br />
Harnack’s inequality states that if u is a non-negative harmonic function<br />
on B(a, R), then for every 0 ≤ r < R and every θ, we have<br />
R − r<br />
R + r u(a) ≤ u(a + reiθ R + r<br />
) ≤<br />
R − r u(a).<br />
First, we note that it suffices to prove that for every z ∈ Ω, there exists an<br />
open neighborhood Uz of z and a constant Cz such that<br />
sup<br />
sup u(w) ≤ Cz.<br />
u∈U w∈Uz
18 <strong>ANALYSIS</strong> QUALS<br />
If this holds, then since any compact K can be covered by finitely many<br />
Uz, we can let M be the maximum of the resulting finitely many Cz, and<br />
we will be done.<br />
Let E = {z ∈ Ω : there exists Uz, Cz as above}. We claim that E is<br />
nonempty, open, and closed. Since Ω is connected, this will show that<br />
E = Ω.<br />
First, E is obviously open: for z ∈ E, every w ∈ Uz is also in E with, for<br />
example, Cw = Ck and appropriately chosen Uw ⊂ Uz.<br />
E is nonempty: Choose ε such that B(z0, ε) ⊂ Ω. We apply Harnack<br />
with a = z0, R = ε, r = |w − z0|, and see that for all u ∈ U, w ∈ B(z0, ε/2),<br />
we have that<br />
ε + |w − z0|<br />
u(w) ≤<br />
ε − |w − z0| u(z0) ≤ Cz0,<br />
for some constant Cz0 . Thus, z0 ∈ E.<br />
E is closed: let z ∈ Ω, zn<br />
B(z, ε) ⊂ Ω.<br />
∈ E such that zn → z. Choose ε so that<br />
Choose n large enough so that |zn − z| < ε<br />
4 . Let Uz = B(z, ε<br />
4 ). Let u ∈ U<br />
and w ∈ Uz. Then |w − zn| < e<br />
2 . We apply Harnack’s inequality with<br />
a = zn, R = 3ε<br />
4 , and r = |w − zn|, and we see that<br />
u(w) ≤<br />
3ε<br />
4 + |w − zn|<br />
3ε<br />
4 − |w − zn| u(zn) ≤ KCzn,<br />
where K is some constant. We let Cz = KCzn . Thus, z ∈ E, so E is closed.<br />
This completes the proof. <br />
Problem 11: Let φ : R → R be a continuous function with compact support.<br />
(a) Prove there is a constant A such that<br />
f ∗ φ L q ≤ A f L p for all 1 ≤ p ≤ q ≤ ∞ and all f ∈ L p .<br />
If you use Young’s (convolution) inequality, you should prove it.<br />
(b) Show by example that such a general inequality cannot hold for p > q.<br />
Problem 13: Let X and Y be two Banach spaces. We say that a bounded<br />
linear transformation A : X → Y is compact if for every bounded sequence<br />
{xn}n≥1 ⊂ X, the sequence {Axn}n≥1 has a convergent subsequence in Y .<br />
Suppose X is reflexive (that is, (X ∗ ) ∗ = X) and X ∗ is separable. Show<br />
that a linear transformation A : X → Y is compact if and only if for every<br />
bounded sequence {xn}n≥1 ⊂ X, there exists a subsequence {xnj } and a<br />
vector φ ∈ X such that xnj = φ + rnj and Arnj → 0 in Y .<br />
Solution. The reverse direction is easy: Let {xn} be bounded. Choose<br />
{xnj } such that xnj = φ + rnj and Arnj → 0 in Y .<br />
Aφ + Arnj → Aφ as j → ∞, which proves this direction.<br />
Then Axnj =<br />
For the forward direction, suppose A is compact and let {xn} be bounded.<br />
WLOG, pass to a subsequence and assume Axn → y ∈ Y . Suffices to show<br />
y ∈ A(X).
<strong>ANALYSIS</strong> QUALS 19<br />
Consider ˆxn ∈ X∗∗ . This is a bounded sequence, so by Banach-Alaoglu,<br />
there is a weak* convergent subsequence ˆxnj → ˆx. That is, for every<br />
f ∈ X∗ , ˆxnj (f) → ˆx(f). But then f(xnj ) → f(φ), where φ is the canonical<br />
preimage of ˆx in X, using reflexivity. Hence, xnj → φ weakly in X.<br />
We want A(xnj ) → A(φ); that is, A(φ) = y. Assume for sake of contradiction<br />
that this does not hold. Then by the Hahn-Banach theorem,<br />
we may choose g ∈ Y ∗ separating A(φ) from y. Then g ◦ A ∈ X∗ , and<br />
g ◦ A(xnj ) → g(y) = g(A(φ)).<br />
But xnj → φ weakly, so (g ◦ A)(xnj ) → (g ◦ A)(φ), a contradiction.<br />
Thus, A(xnj ) → A(φ), and we may let rnj = xnj − φ, and we have that<br />
A(rnj ) → 0. This completes the proof. <br />
Fall 2009<br />
Problem 1: Find a non-empty closed set in the Hilbert space L 2 ([0, 1]) that<br />
does not contain an element of smallest norm. Prove your assertion.<br />
Solution. Let en(x) = (1 + 2 −|n| )e 2πinx . Then en → 1 as |n| → ∞, but<br />
en = 1 for all n.<br />
Also, en − em 2<br />
2 = (1 + 2−|n| ) 2 + (1 + 2 −|m| ) 2 ≥ 2, so the set is discrete,<br />
and therefore any L 2 convergent sequence must be constant. Hence,<br />
the set is closed. <br />
Problem 4: Prove the following variant of the Lebesgue Differentiation theorem:<br />
Let µ be a finite Borel measure on R, singular with respect to Lebesgue<br />
measure. Then for Lebesgue-almost every x ∈ R,<br />
µ([x − ε, x + ε])<br />
lim<br />
= 0.<br />
ε→0 2ε<br />
Proof. Let m be Lebesgue measure. Since µ ⊥ m, we may choose A Borel<br />
such that µ(A) = m(Ac ) = 0. Note that<br />
<br />
<br />
<br />
<br />
µ([x − ε, x + ε]) <br />
<br />
|µ| ([x − ε, x + ε])<br />
2ε ≤ ,<br />
2ε<br />
so it suffices to assume µ is positive. Let<br />
Fk = {x ∈ A : lim sup<br />
ε→0<br />
µ([x − ε, x + ε])<br />
2ε<br />
≥ 1<br />
k }.<br />
Then F = Fk is the set of x in which the conclusion fails. Thus, it suffices<br />
to show that m(Fk) = 0 for all k.<br />
For δ > 0, choose Uδ ⊃ A open such that λ(Uδ) < δ. For every x ∈ Fk,<br />
there exists a ball Bx = B(x, rx) ⊂ Uδ such that kµ(Bx) > m(Bx). Let<br />
Vδ = <br />
x∈Fk Bx. By the Weiner covering lemma, for c < m(Vδ), there exist<br />
disjoint Bx1, . . . , Bxl such that m(Bxj ) > 3 −n c. Thus, we have that<br />
c < 3 n<br />
l<br />
m(Bxj ) ≤ 3nk j=1<br />
l<br />
µ(Bxj ) ≤ 3nkµ(Uδ) ≤ 3 n kδ.<br />
j=1<br />
Thus, m(Vδ) ≤ 3 n kδ, so we must have that m(Fk) = 0.
20 <strong>ANALYSIS</strong> QUALS<br />
Problem 5: Construct a Borel subset E of the real line R such that for all<br />
intervals [a, b] we have<br />
0 < m(E ∩ [a, b]) < |b − a|<br />
where m denotes Lebesgue measure.<br />
Proof. We will use the Baire Category Theorem on the complete metric<br />
space<br />
X = {χE ∈ L 1 : E Borel .<br />
First, to see that X is complete, it suffices to show that X is closed in L1 .<br />
Suppose fn → f in L1 , where fn ∈ X. Then there is a subsequence fnk<br />
that converges to f pointwise a.e. (cf. another common qual problem).<br />
But as characteristic functions converge pointwise a.e. to f, we must have<br />
that f is almost everywhere a characteristic function of some set F . As<br />
every set differs by a measure zero set from a Borel set, we can assume that<br />
F is Borel. Thus, f ∈ X, so X is complete.<br />
Now for a non-trivial interval I = [a, b], we define the set UI = {χE ∈ X :<br />
0 < m(E ∩ I) < m(I)}. We claim that UI is open and dense in X. First,<br />
UI is dense in X: if χE ∈ UI, then either m(E ∩I) = 0 or m(E ∩I) = m(I).<br />
In the former case, we can add a small amount of measure to E, and in the<br />
latter case, we can remove a small amount of measure from E to obtain E ′<br />
with E − E ′ L1 < ε, and E ′ ∈ X. This gives density. To see that UI is<br />
open in X, we note that χE ↦→ m(E ∩ I) is a continuous function on L1 restricted to X. Thus, if 0 < m(E ∩ I) < m(I), then 0 < m(E ′ ∩ I) < m(I)<br />
for all E ′ such that χE ′ is L1-close to χE, so UI is open.<br />
Thus, by the Baire Category Theorem,<br />
<br />
UI is dense in X,<br />
I=[a,b],a,b∈Q<br />
so in particular there exists a Borel F such that χF is in the intersection.<br />
Since every non-trivial interval contains an interval with rational endpoints,<br />
this F does the job, which completes the proof.<br />
<br />
Problem 7: a) Define unitary operator on a complex Hilbert space.<br />
b) Let S be a unitary operator on a complex Hilbert space. Using your<br />
definition, prove that for every complex number |λ| < 1 the operator<br />
S − λI is invertible. Here I denotes the identity operator.<br />
c) For a fixed vector v in the Hilbert space and all {λ ∈ C : |λ < 1}, we<br />
define<br />
h(λ) = (S + λI)(S − λI) −1 v, v .<br />
Show that Re h is a positive harmonic function. [You may not invoke<br />
the spectral theorem - this is part of a proof of that theorem.]<br />
Solution. a) An operator U : H → H is a unitary operator if U is<br />
bounded, linear, and U ∗ U = UU ∗ = I.
<strong>ANALYSIS</strong> QUALS 21<br />
b) Suppose S − λI is not invertible. Choose v ∈ H such that |v| = 1, and<br />
(S − λI)v = 0. Then Sv = λv. Hence<br />
1 = 〈v, v〉 = 〈(S ∗ S)v, v〉 = 〈Sv, Sv〉 = 〈λv, λv〉 = |λ| 2 〈v, v〉 = |λ| 2 .<br />
Hence, |λ| = 1. Thus, conversely, |λ| < 1 implies that S − λI is<br />
invertible.<br />
c) We will prove that h is holomorphic, which means that Re h is harmonic,<br />
and then we will prove that Re h > 0. First, if v = 0, the<br />
statement is false. Let v = 0. Then<br />
(S + λI)(S − λI) −1 = (S + λI)(I − λS −1 ) −1 S −1<br />
= (S + λI)(I + λS −1 + λ 2 S −2 + . . .)S −1<br />
= (S + λI)(S −1 + λS −2 + λ 2 S −3 + . . .)<br />
= I + 2λS −1 + 2λ 2 S −2 + . . .<br />
Hence,<br />
(S + λI)(S − λI) −1 v, v = 〈v, v〉 + 2λ S −1 v, v + 2λ 2 S −2 v, v + . . .<br />
This is a power series expansion of h about 0, so h is analytic in the<br />
disc. Hence, h is holomorphic on |λ| < 1, and thus Re h is harmonic.<br />
Now let w = (S − λI) −1 v, and as v = 0, we have that w = 0 by<br />
part b. Then<br />
h(λ) = 〈(S + λI)w, (S − λI)w〉<br />
= 〈Sw, Sw − λw〉 + 〈λw, Sw − λw〉<br />
= 〈Sw, Sw〉 − 〈Sw, λw〉 + 〈λw, Sw〉 − 〈λw, λw〉<br />
= Sw 2 − |λ| 2 w 2 + 〈Sw, λw〉 − 〈Sw, λw〉<br />
= Sw 2 − |λ| 2 w 2 + 2iIm 〈Sw, λw〉<br />
= w 2 − |λ| 2 w 2 + 2iIm 〈Sw, λw〉<br />
Hence, Re h(λ) = w 2 (1 − |λ| 2 ) > 0 as |λ| 2 < 1.<br />
Problem 8: Let Ω be an open convex region in the complex plane. Assume f<br />
is a holomorphic function on Ω and the real part of its derivative is positive:<br />
Re(f ′ (z)) > 0 for all z ∈ Ω.<br />
a) Prove that f is one-to-one.<br />
b) Show by example that the word “convex”’ cannot be replaced by “connected<br />
and simply connected”’.<br />
Solution. a) Suppose z1 = z2 ∈ Ω. Then as Ω is convex, the line between<br />
z1 and z2 is contained in Ω. Thus,<br />
<br />
<br />
z1<br />
z1<br />
<br />
|f(z1) − f(z2)| = <br />
<br />
z2<br />
f ′ <br />
(z)dz<br />
=<br />
<br />
<br />
<br />
<br />
<br />
≥ <br />
<br />
z1<br />
z2<br />
z2<br />
Re(f ′ (z))dz + i<br />
Re(f ′ <br />
<br />
(z))dz<br />
> 0.<br />
z1<br />
z2<br />
Im(f ′ <br />
<br />
(z))dz
22 <strong>ANALYSIS</strong> QUALS<br />
Thus, f(z1) = f(z2), so f is injective.<br />
b) Let Ω = {z : − 3π<br />
4<br />
3π<br />
< arg(z) < 4 }. This is clearly connected and<br />
simply connected. Let f(z) = z 5/3 , with an appropriate branch cut so<br />
that f is holomorphic on Ω.<br />
Then f ′ (z) = 5<br />
3 z2/3 . Let re iθ ∈ Ω. Then f ′ (re iθ ) = 5<br />
3 r2/3 e 2iθ/3 .<br />
But π<br />
2<br />
< 2<br />
3<br />
θ < π<br />
2 , and so Re(f ′ (z)) > 0 for all z ∈ Ω.<br />
However, f(e 7πi/10 ) = e 7πi/6 = e −5πi/6 = f(e −π/2 ). Hence f is not<br />
injective on Ω.<br />
<br />
Problem 9: Let f be a non-constant meromorphic function on the complex<br />
plane C that obeys<br />
f(z) = f(z + √ 2) = f(z + i √ 2).<br />
(In particular, the poles of these three functions coincide.) Assume f has<br />
at most one pole in the closed unit disc D.<br />
a) Prove that f has exactly one pole in D.<br />
b) Prove that this is not a simple pole.<br />
Solution. a) Let z = e 5πi/4 . Then by the periodicity, f takes all its<br />
possible values on the square with corners at z, z + √ 2, z + i √ 2, and<br />
z + √ 2 + √ 2. This square is then contained in D, so it suffices to show<br />
that there is a pole in this square, P . If not, then f is entire. But since<br />
is compact, f is bounded, so by Liousville, f would be constant, a<br />
contradiction. Thus, f has a pole in P .<br />
b) Using the periodicity, we have that<br />
and<br />
z+ √ 2<br />
z<br />
z+i √ 2<br />
z<br />
Thus, we have that <br />
f(w)dw =<br />
f(w)dw =<br />
∂P<br />
z+ √ 2+i √ 2<br />
z+i √ 2<br />
z+ √ 2+i √ 2<br />
z+ √ 2<br />
f(w)dw = 0.<br />
f(w)dw<br />
f(w)dw.<br />
But by the residue theorem, we have that<br />
<br />
∂P f(w)dw = 2πi <br />
Res(f, z0).<br />
poles z0<br />
Thus, Res(f) = 0. Thus, counting multiplicities, f has at least two<br />
poles. But as f has only one pole, it cannot be simple.
<strong>ANALYSIS</strong> QUALS 23<br />
Problem 12: Let f be a non-constant meromorphic function in the complex<br />
plane. Assume that if f has a pole at the point z ∈ C, then z is of the<br />
form nπ with an integer n ∈ Z. Assume that for all non-real z we have that<br />
estimate<br />
|f(z)| ≤ (1 + |Im(z)| −1 )e −|Im(z)|<br />
Prove that for every integer n ∈ Z, f has a pole at the point πn.<br />
Proof. By the estimate, we have that f(πn + ir) = O(r −1 ) for r real, so all<br />
poles are simple poles with residue of norm at most one. Note that<br />
|sin(z)| ≤ e |Im(z)|<br />
Then letting g(z) = f(z) sin(z), it suffices to show that g is constant, which<br />
would imply that f is a constant multiple of cosecant, which has poles at<br />
nπ for all n.<br />
First, since sin(z) has simple zeroes for all nπ, we have that g is entire.<br />
Secondly, using our estimates for f and sin, we have that<br />
|g(z)| ≤ 1 + |Im(z)| −1<br />
for z ∈ C\R. Thus, |g| ≤ 2 off the strip |Im(z)| ≤ 1.<br />
We consider g on this strip, and in particular, on each rectangle with vertices<br />
±nπ ± i. On the horizontal segments of the rectangles, we have the same<br />
estimate as before, |g| ≤ 2. For the vertical segments, we use the fact that<br />
sin(x + iy) = sin(x) cosh(y) + i cos(x) sinh(y). Then we have that<br />
|sin(nπ + iy)| = |sinh(y)| ≤ C|y| = C|Im(z)|<br />
where the constant C does not depend on n. Thus,<br />
g ≤ C |Im(z)| (1 + |Im(z)| −1 )e −|Im(z)| < K<br />
on the vertical segments. Therefore, g is bounded on the edges of the rectangle,<br />
and the bound depends not on n. Thus, by the maximum modulus<br />
principle, g is bounded on the interior of this rectangle, and as this bound<br />
does not depend on n, we may conclude that g is bounded on the strip.<br />
Thus, g is bounded on C, so by Liousville, g is constant. This completes<br />
the proof. <br />
Spring 2009<br />
Problem 1: Let f and g be real-valued integrable functions on a measure<br />
space (X, B, µ), and define<br />
Prove<br />
Ft = {x ∈ X : f(x) > t}, Gt = {x ∈ X : g(x) > t}.<br />
<br />
∞<br />
|f − g|dµ = µ((Ft\Gt) ∪ (Gt\Ft))dt.<br />
−∞
24 <strong>ANALYSIS</strong> QUALS<br />
<br />
Solution. Let E1 = {(x, t) : x ∈ X, t ∈ R, f(x) > t and g(x) < t}.<br />
Let E2 = {(x, t) : x ∈ X, t ∈ R, f(x) < t and g(x) > t}. Then<br />
<br />
<br />
|f − g|dµ = |f − g|dµ + |f − g|dµ<br />
<br />
=<br />
{f>g}<br />
{f>g}<br />
<br />
f − gdµ +<br />
{ff}<br />
g − fdµ<br />
∞<br />
<br />
=<br />
{f>g}<br />
∞<br />
χE1(x, t)dt<br />
−∞<br />
= (χE1 + χE2 )(x, t)dtdµ(x),<br />
−∞<br />
∞<br />
<br />
dµ(x) +<br />
{g>f}<br />
χE2(x, t)dt<br />
−∞<br />
dµ(x)<br />
since E1 and E2 are disjoint, and χE1 vanishes off {f > g} in X, and χE2<br />
vanishes off {g > f} in X. Then by Fubini, we have that<br />
∞<br />
∞ <br />
(χE1 + χE2 )(x, t)dt, dµ(x) =<br />
−∞<br />
(χE1 + χE2 )(x, t)dµ(x)dt.<br />
But <br />
χE1dµ = µ(Ft\Gt) and<br />
−∞<br />
<br />
χE2dµ = µ(Gt\Ft),<br />
and as these sets are disjoint, µ((Ft\Gt)∪(Gt\Ft)) = µ(Ft\Gt)+µ(Gt\Ft).<br />
Thus,<br />
∞ <br />
∞<br />
(χE1 + χE2)(x, t)dµ(x)dt = µ((Ft\Gt) ∪ (Gt\Ft))dt.<br />
−∞<br />
Problem 2: Let H be an infinite dimensional real Hilbert space.<br />
a) Prove the unit sphere S of H is weakly dense in the unit ball B of H.<br />
b) Prove there is a sequence Tn of bounded linear operators from H to<br />
H such that Tn = 1 for all n but lim Tn(x) = 0 for all x ∈ H.<br />
Proof. a) Let x0 ∈ B, ε > 0. Let Uε(x0) be an ε-neighborhood of x0 in<br />
the weak topology. That is,<br />
<br />
<br />
Uε(x0) =<br />
−∞<br />
x ∈ H : sup |〈x − x0, yi〉| < ε for k ∈ N, yi ∈ H<br />
i=1,...,k<br />
Let fi(x) = 〈x, yi〉. Then fi is a linear functional. As H is infinite<br />
dimensional, the intersection of the kernels of a finite number of linear<br />
functionals is a nontrivial linear subspace. Thus,<br />
x0 +<br />
k<br />
ker(fi)<br />
i=1<br />
is contained in Uε(x0), so there is a line L through x0 contained in<br />
Uε(x0). This line intersects S, so choose xε ∈ L ∩ S. Then<br />
〈x0, y〉 = lim<br />
ε→0 〈xε, y〉 for all y ∈ H.<br />
Thus, S is weakly dense in B.<br />
.
<strong>ANALYSIS</strong> QUALS 25<br />
b) I present the proof in the (simple) case in which H is a separable<br />
Hilbert space (which is how Stein-Shakarchi defines their Hilbert spaces).<br />
In this case, let {en} be an orthonormal basis for H.<br />
Define Tn(y) = 1<br />
y 〈en, y〉 y. Then<br />
Tn(y) ≤ y , and Tn(en) = 1, so Tn = 1.<br />
But by Bessel’s inequality,<br />
<br />
n<br />
|〈en, y〉| 2 ≤ y 2 , so 〈en, y〉 → 0 for all y.<br />
Thus, lim Tn(y) = 0 for all y ∈ H.<br />
Problem 3: Let X be a Banach space and let X ∗ be its dual Banach space.<br />
Prove that if X ∗ is separable the X is separable.<br />
Solution. Let {fn} be a countable dense set of X ∗ . For each n, choose<br />
xn ∈ X such that xn = 1 and |fn(xn)| ≥ 1<br />
2 fn.<br />
Let S be the set of all linear combinations of the xn with rational coefficients<br />
(rational real and imaginary parts if X is a Banach space over C).<br />
Thus, S is countable. We aim to show that S is dense in X.<br />
Let Y = S and assume for sake of contradiction that Y = X. Then by<br />
Hahn-Banach theorem, we may choose f ∈ X∗ such that f(y) = 0 for all<br />
y ∈ Y and f = 0. Then by density of {fn}, choose a sequence fin → f.<br />
Then<br />
1<br />
f − fin ≥ |(f − fin )(xin )| = |fin (xin )| ≥ fin 2<br />
for all n ∈ N. But then as fin → f, we must have fin → 0, which means<br />
that f = 0, a contradiction. Thus Y = X. <br />
Problem 4: Let f(x) be a non-decreasing function on [0, 1]. You may assume<br />
the theorem that f is differentiable almost everywhere.<br />
a) Prove that 1<br />
0 f ′ (x)dx ≤ f(1) − f(0).<br />
b) Let {fn} be a sequence of non-decreasing functions on the unit interval<br />
[0, 1], such that the series F (x) = ∞<br />
n=1 fn(x) converges for all x ∈<br />
[a, b]. Prove that F ′ (x) = ∞<br />
n=1 f ′ n(x) almost everywhere on [0, 1].<br />
Proof. a) Define the difference quotient<br />
gn(x) =<br />
f(x + 1/n) − f(x)<br />
,<br />
1/n<br />
and then by the theorem we are permitted to assume, gn(x) → f ′ (x)<br />
as n → ∞ for a.e. x. As f is non-decreasing, we have that gn is<br />
non-negative, so by Fatou, we have that<br />
1<br />
0<br />
f ′ (x)dx ≤ lim inf<br />
1<br />
0<br />
gn(x)dx.
26 <strong>ANALYSIS</strong> QUALS<br />
1<br />
0<br />
But then<br />
gn(x)dx = 1<br />
1/n<br />
= 1<br />
1/n<br />
1<br />
f(x + 1/n)dx − 1<br />
1/n<br />
0<br />
1+1/n<br />
1<br />
f(x)dx − 1<br />
1/n<br />
1<br />
0<br />
1/n<br />
0<br />
f(x)dx<br />
f(x)dx → f(1) − f(0) as n → ∞.<br />
b) We let rn(x) = <br />
k≥n fk(x). Hence, for a.e. x, we can write<br />
and so<br />
n−1 <br />
F (x) = fj(x) + rn(x),<br />
j=1<br />
F ′ n−1 <br />
(x) =<br />
j=1<br />
f ′ j(x) + r ′ n(x).<br />
Thus, it suffices to show that r ′ n(x) → 0 as n → ∞. Note that as<br />
fn is non-decreasing for all n, we have that rn is also non-decreasing<br />
for all n, and thus r ′ n is non-negative. Note further that since F is<br />
a convergent series, we have that rn(x) → 0 for all x ∈ [0, 1]. Thus,<br />
rn(1) − rn(0) → 0. But by the first part, we have that<br />
1<br />
0<br />
r ′ n(x)dx ≤ rn(1) − rn(0).<br />
Thus, 1<br />
0 r′ n(x)dx → 0 as n → ∞, so r ′ n(x) → 0 for a.e. x ∈ [0, 1].<br />
Problem 5: Let I = I0,0 = [0, 1] be the unit interval, and for n = 0, 1, 2, . . . ,<br />
and 0 ≤ j ≤ 2 n − 1 let<br />
In,j = [j2 −n , (j + 1)2 −n ].<br />
For f ∈ L 1 (I, dx) define Enf(x) = 2 n −1<br />
j=0<br />
<br />
n 2 In,j fdt<br />
<br />
χIn,j .<br />
Prove that if f ∈ L 1 (I, dx) then limn→∞ Enf(x) = f(x) almost everywhere<br />
on I.<br />
Solution. Let S = {f ∈ L 1 (I, dx) : Enf → f a.e. on I}. Then S contains<br />
the characteristic functions of dyadic intervals, as EmχIn,j = χIn,j for<br />
m > n.<br />
But we can approximate a positive L 1 function f by a sequence of positive<br />
linear combinations of characteristic functions of dyadic intervals. Thus,<br />
such an f is in S by monotone convergence. S is a vector space, so writing<br />
f as the difference of its positive part and negative part yields the<br />
result. <br />
Problem 8: Let f be an entire non-constant function that satisfies the functional<br />
equation<br />
f(1 − z) = 1 − f(z)<br />
for all z ∈ C. Show that f(C) = C.
<strong>ANALYSIS</strong> QUALS 27<br />
Solution. Picard’s Big Theorem states that if f is entire and non-constant,<br />
its range omits at most one value.<br />
Suppose w ∈ f(C). Then 1 − w ∈ f(C). Note that t = 1 − t implies<br />
t = 1/2, and f(1/2) = 1 − f(1/2), so f(1/2) = 1/2 ∈ f(C). Thus, t = 1 − t,<br />
and neither is in the range, so the range of f omits at least two values,<br />
which gives a contradiction of Picard. Hence, f(C) = C. <br />
Problem 10: Let D be the open unit disc and µ be Lebesgue measure on<br />
D. Let H be the subspace of L 2 (D, µ) consisting of holomorphic functions.<br />
Show that H is complete.<br />
Solution. First, since L2 (D, µ) is complete, it suffices to show that H is a<br />
closed subspace of L2 (D, µ). Let f ∈ H. Let K be a compact subset of<br />
D. Let r = dist(K, U c ). Let z ∈ K, B = B(z, r/2) ⊂ U. By mean value<br />
principle and Cauchy-Schwarz, we have that<br />
<br />
<br />
|f(z)| = <br />
1<br />
<br />
<br />
f(w)dw<br />
Area(B)<br />
<br />
B<br />
≤ 4<br />
πr2 <br />
|f(w)| dw<br />
<br />
B<br />
= 4<br />
πr2 |f(w)| χBdw<br />
U<br />
≤ 4<br />
πr2 <br />
|f(w)|<br />
U<br />
2 1/2 <br />
dw<br />
U<br />
<br />
4<br />
=<br />
πr2 1/2 1/2 |f|<br />
U<br />
χ 2 1/2 Bdw<br />
Thus, we let CK = 4<br />
πr2 1/2, and note this depends only on K.<br />
Now suppose {fn} ∈ H and fn → f in L 2 (D, µ). Then for each compact<br />
K ⊂ D and each z ∈ K, we have that<br />
|fn(z) − f(z)| ≤ CK fn − f L 2 (D,µ) → 0.<br />
Hence, fn → f uniformly on compact subsets of D, and thus, f is holomorphic<br />
on D. Therefore, H is complete.<br />
<br />
Problem 11: Suppose that f : D → C is holomorphic and injective in some<br />
annulus {z : r < |z| < 1}. Show that f is injective in D.<br />
Solution. Let Ar be the annulus, and let r < R < 1 be arbitrary. Let γ be<br />
the closed curve<br />
γ = {Re iθ : −π ≤ θ < π}<br />
so that γ ⊂ Ar.<br />
Using the Jordan curve theorem and the injectivity of f, we have that<br />
f(γ) is a simple, closed curve in C, and in particular it is non-intersecting,<br />
and f(int(γ)) = int(f(γ)).
28 <strong>ANALYSIS</strong> QUALS<br />
Since the winding number is constant on connected components, the winding<br />
number of f(z) must be 1 for all z ∈ int(γ), again using injectivity of f<br />
in Ar.<br />
Let b ∈ f(int(γ)). Then by the Argument Principle, the number of points<br />
in int(γ) where f takes the value b is the winding number, 1.<br />
Hence, f is injective on int(γ). As this holds for every R < 1, f is injective<br />
on D. <br />
Fall 2008<br />
Problem 1: Fix 1 ≤ p < ∞ and let {fn} be a sequence of Lebesgue measurable<br />
functions fn : [0, 1] → C. Suppose there exists f in L p ([0, 1]) so that<br />
fn → f in the L p sense, that is,<br />
1<br />
0<br />
|fn(x) − f(x)| p dx → 0.<br />
(a) Show that fn → f in measure, that is,<br />
lim<br />
n→∞ µ ({x : |fn(x) − f(x)| ≥ ε}) = 0<br />
for all ε > 0.<br />
(b) Show that there is a subsequence fnk such that fnk (x) → f(x) almost<br />
everywhere.<br />
Solution. (a) By Chebyshev’s inequality,<br />
µ ({x : |f(x)| > α}) ≤ fp p<br />
.<br />
αp Then we have that<br />
µ ({x : |fn(x) − f(x)| > ε}) ≤ fn − f p<br />
p<br />
εp → 0 as n → ∞,<br />
so fn → f in measure.<br />
(b) As fn → f in measure, we may choose a subsequence {fnj } such that<br />
µ x : fnj (x) − fnj+1 (x) > 2 −j ≤ 2 −j .<br />
Let Ej = {x : |fnj (x) − fnj+1 (x)| > 2−j }. Let Fk = ∞<br />
j=k Ej. Then<br />
µ(Fk) ≤<br />
∞<br />
µ(Ej) ≤ 2 1−k .<br />
j=k<br />
For x /∈ Fk and k ≤ j ≤ i, we have that<br />
<br />
fnj (x) − fni(x) ≤<br />
i−1<br />
<br />
fnl (x) − fnl+1 (x) i−1<br />
≤<br />
l=j<br />
l=j<br />
2 −l ≤ 2 1−j ,<br />
so {fnj } is pointwise Cauchy on F C k . Let F = ∞ k=1 Fk. Then µ(F ) =<br />
0 and fnj converges on F C , so fnj converges almost everywhere.
<strong>ANALYSIS</strong> QUALS 29<br />
Problem 2: Is every vector space isomorphic as a vector space to some Banach<br />
space? Prove your answer. (Banach space = complete normed vector<br />
space, as usual).<br />
Solution. No. Let V be the vector space consisting of all infinite sequences<br />
(over R, for example) with only finitely many non-zero terms. Assume for<br />
sake of contradiction that V can be given a Banach space structure.<br />
Let Vn be the subspace of V consisting of the sequences whose components<br />
are all zero after the nth. Then<br />
∞<br />
V = Vn.<br />
n=1<br />
Note that Vn is finite dimensional, so it is isomorphic as a Banach space<br />
to R n , and thus is a closed subspace. However, Vn is nowhere dense, as<br />
Vn contains no open balls in V . Thus, by the Baire Category Theorem,<br />
Vn = V , a contradiction.<br />
<br />
Problem 3: Prove: If f : [0, 1] → R is an arbitrary function, not necessarily<br />
measurable, then the set of points at which f is continuous is a Lebesguemeasurable<br />
set.<br />
Proof. Let<br />
Sx(δ) = sup{|f(x1) − f(x2)| : x1, x2 ∈ (x − δ, x + delta)},<br />
φ(x) = inf{Sx(δ) : δ > 0}.<br />
Note that<br />
<br />
1<br />
1<br />
Sx ≥ Sx<br />
k k + 1<br />
for every k ∈ N. We claim that f is continuous at x if and only if φ(x) = 0.<br />
First suppose that φ(x) = 0. Let ε > 0 and choose δ > 0 such that<br />
Sx(δ) < ε. Thus, for any y ∈ (x − δ, x + δ), we have that |f(y) − f(x)| < ε,<br />
so f is continuous at x.<br />
Now assuming f is continuous at x, we have that for all ε > 0 we may choose<br />
δ such that |x − y| < δ implies |f(x) − f(y)| < ε, and hence Sx(δ) < 2ε.<br />
Thus, φ(x) = 0.<br />
Next, we claim that {x : φ(x) < α} is an open set for every α (that is, φ is<br />
upper-semicontinuous). Suppose φ(x) < α and choose N such that<br />
<br />
1<br />
Sx < α whenever k > N.<br />
k<br />
Suppose k > N. Let y ∈ x − 1<br />
<br />
1<br />
1 1<br />
2k <br />
, x<br />
<br />
+ 2k <br />
. Note<br />
<br />
that then<br />
<br />
y − 2k , y + 2k<br />
⊂<br />
1<br />
1<br />
1 1<br />
(x−1/k, x+1/k), so φ(y) ≤ Sy 2k ≤ Sx k < α. Thus, x − 2k , x + 2k ⊂<br />
{x : φ(x) < α}, so the set is open.<br />
But then we have that<br />
∞<br />
{x : φ(x) = 0} = {x : φ(x) < 1<br />
k },<br />
k=1<br />
and thus the set of points of continuity of f is a Gδ set, and hence is<br />
measurable.
30 <strong>ANALYSIS</strong> QUALS<br />
Problem 6: Define for each n = 1, 2, 3, . . . , the Cantor-like set Cn as [0, 1]<br />
with its central open interval of length 1<br />
2n · 1<br />
3 removed, then with the two<br />
central open intervals of length 1<br />
2n · 1<br />
32 removed from the remaining two<br />
closed intervals and so on (at the jth stage, 2j−1 intervals of length 1<br />
2n · 1<br />
3j are removed), continuing with j = 1, 2, 3, . . ..<br />
(a) With µ=Lebesgue measure, show that µ([0, 1]\ ∞<br />
n=1 Cn) = 0.<br />
(b) Show that if E is a subset of [0, 1] which is not Lebesgue measurable<br />
(you may assume such an E exists without proof), then for some n ≥ 1,<br />
E ∪ Cn fails to be Lebesgue measurable.<br />
(c) Use part (b) to show that there is a continuous, strictly increasing<br />
function f : R → R with f(R) = R and a Lebesgue measurable set<br />
A ⊂ R such that f(A) is not Lebesgue measurable.<br />
Solution. (a) Cn is constructed by removing intervals of total length 2−n 2<br />
3<br />
at step j. But<br />
1<br />
3<br />
∞<br />
j=0<br />
j 2<br />
=<br />
3<br />
1<br />
1 − 2<br />
3<br />
= 3,<br />
so Cn has a total of 2 −n removed. Thus, Cn has total measure 1−2 −n ,<br />
so <br />
n Cn has measure 1, and hence, µ([0, 1]\ ∞ n=1 Cn) = 0.<br />
(b) E not measurable implies that there exists ε > 0 and a set A such that<br />
µ(E) < µ(E∩A)+µ(E∩A c )−ε. Choose n such that µ(Cn) > 1−ε/10.<br />
Then<br />
µ(E ∩ Cn) < µ(E) < µ(E ∩ Cn ∩ A) + µ(E ∩ Cn ∩ A c ) − 4ε/5,<br />
so E ∩ Cn is not measurable.<br />
(c) Let f be a strictly increasing homeomorphism from C0 to Cn, where<br />
n as in part (b). Such a function exists: both sets are homeomorphic<br />
to {0, 1} N by considering at each point in a Cantor set, whether it is<br />
in the left or right side at each stage of removing central intervals.<br />
Thus, just as the strictly monotonic Cantor-Lebesgue function can be<br />
constructed for C0 onto [0, 1], so too can such a function be created<br />
for Cn, and taking an appropriate composition yields f.<br />
Extend f linearly to all of R. Then f −1 (E∩Cn) is a subset of C0, which<br />
has measure zero. Therefore f −1 (E ∩ Cn) is measurable. However, by<br />
part (b), E ∩ Cn is not measurable.<br />
<br />
Problem 8: Suppose f : U → C is a holomorphic function with intU |f| 2 < ∞<br />
where U = {z ∈ C : 0 < |z| < 1} and the integral is the usual R 2 -area<br />
integral. Prove that f has a removable singularity at z = 0.<br />
Solution. Let w be a point in the annulus around z = 0 of radii ε and 1/2,<br />
for 0 < ε < 1/2. Then by Cauchy integral formula, we have that<br />
<br />
<br />
f(ξ)<br />
f(ξ)<br />
f(w) =<br />
dξ −<br />
ξ − w ξ − w dξ.<br />
|ξ|=1/2<br />
|ξ|=ε<br />
j−1·
δ<br />
0<br />
Then<br />
<br />
f(ξ)<br />
δf(w) = δ<br />
dξ −<br />
|ξ|=1/2 ξ − w<br />
and thus<br />
<br />
<br />
<br />
f(w)<br />
−<br />
<br />
|ξ|=1/2<br />
<strong>ANALYSIS</strong> QUALS 31<br />
f(ξ)<br />
ξ − w dξ<br />
<br />
<br />
<br />
<br />
<br />
≤ 1<br />
δ<br />
<br />
<br />
<br />
<br />
<br />
δ<br />
0<br />
δ<br />
0<br />
<br />
<br />
|ξ|=ε<br />
|ξ|=ε<br />
f(ξ)<br />
ξ − w dξdε<br />
f(ξ)<br />
ξ − w dξdε<br />
<br />
<br />
<br />
<br />
.<br />
We claim that this tends to 0 as δ → 0, which would prove that f(w) =<br />
f(ξ)<br />
1<br />
|ξ|=1/2 ξ−w dξ for |w| < 2 , and this is holomorphic, so any singularity at 0<br />
would be removable.<br />
But using Fubini, Cauchy-Schwarz, and basic estimations, we have that<br />
<br />
f(ξ)<br />
ξ − w dξ<br />
<br />
<br />
<br />
<br />
=<br />
<br />
<br />
δ 2π<br />
f(εe<br />
<br />
<br />
iθ )<br />
εeiθ − w iεeiθ <br />
<br />
<br />
dθdε<br />
<br />
|ξ|=ε<br />
≤<br />
0 0<br />
δ 2π<br />
0<br />
≤ 2<br />
|w|<br />
≤ 2<br />
|w|<br />
≤ 2<br />
|w|<br />
0<br />
2π δ<br />
0<br />
2π<br />
<br />
0<br />
|f(εeiθ )|<br />
|εeiθ − w| εdθdε<br />
0<br />
δ<br />
0<br />
|f(εe iθ )|εdεdθ<br />
|f|<br />
{0
32 <strong>ANALYSIS</strong> QUALS<br />
form e iθ z for some θ.<br />
Thus, we have that<br />
iθ z − i<br />
g(z) = h ◦ f(z) = e<br />
z + i<br />
for some θ.<br />
(b) Since f is holomorphic, its imaginary part is a harmonic function on<br />
the disk. Then by Harnack’s inequality, we have that<br />
1 − |z|<br />
1 + |z|<br />
Im(f(0)) ≤ Im(f(z)) ≤<br />
1 + |z| 1 − |z| Im(f(0)).<br />
Hence, as Im(f(0)) = 1, we have that Im(f(x)) ≥ 1−x<br />
1+x .<br />
Problem 10: Suppose U is a bounded connected open set in C and z0 ∈ U.<br />
Let F = {f : U → D, f holomorphic, f(z0) = 0}.<br />
(a) Show that if K is a compact subset of U, then there is a constant<br />
MK > 0 such that |f ′ (z)| ≤ MK for all z ∈ K, f ∈ F .<br />
(b) Use part (a) to show that if {fn} is a sequence in F , then there is a<br />
subsequence {fnj } which converges uniformly on every compact subset<br />
of U to a function f0 ∈ F .<br />
(Note: Part of this is to show f0(U) ⊂ D.)<br />
Solution. (a) Let r = dist(K, U c ). Let γ be a closed curve in U around K<br />
with dist(K, γ) ≥ r<br />
2 . Then by Cauchy Integral formula, we have that<br />
f ′ (w) = 1<br />
<br />
f(z)<br />
dz.<br />
2πi (z − w) 2<br />
Hence for w ∈ K, we have that<br />
|f ′ (w)| ≤ 1<br />
<br />
1 1<br />
dz ≤<br />
2π |z − w| 2 2π<br />
γ<br />
γ<br />
4<br />
4M<br />
length(γ) ≤<br />
r2 r2 where M is the bound on |z| in U. Thus, let MK = 4M<br />
r2 .<br />
(b) As f : U → D, |f(z)| < 1 for all z ∈ U and f ∈ F . Hence by Montel’s<br />
theorem, there exists a subsequence {fnj } converging uniformly on<br />
compact subsets of U to a function f0.<br />
But then fnj (z0) → f0(z0), so f0(z0) = 0. Since |fnj (z)| < 1 for all<br />
z ∈ U, we must have that |f0(z)| ≤ 1 for all z ∈ U. But as f0 is<br />
holomorphic, it is an open mapping by the Open Mapping theorem,<br />
so as U is open, f0(U) is open. Thus, f0(U) ⊂ D, so f0 ∈ F .<br />
<br />
Problem 11: Let D denote the open unit disk in the complex plane let D<br />
denote its closure. Suppose f : D → C is continuous on D and analytic in<br />
its interior. Show that if f takes only real values on ∂D, then f must be<br />
constant.<br />
Solution. Let g(z) be defined to be f(z) for z ∈ D, and f <br />
1 1<br />
z for z ∈ D.<br />
Note that this is well-defined: if both conditions are satisfied, then z must<br />
be of the form eiθ , in which case f <br />
1<br />
z = f(eiθ iθ ) = f(e ) since f takes real<br />
values on the circle.<br />
Thus, by Schwarz Reflection Principle, g is entire.
<strong>ANALYSIS</strong> QUALS 33<br />
But as D is compact, g is bounded on D, and hence is bounded on C<br />
by definition of g. Thus g is entire and bounded, and hence constant by<br />
Liousville. As g is constant, we conclude that f is constant. <br />
Problem 12: Evaluate<br />
for all real numbers a > 0.<br />
π<br />
0<br />
dθ<br />
a 2 + sin 2 θ<br />
Solution. First, note that sin 2 θ = − 1<br />
4 (e2iθ − 2 + e −2iθ ). Let z = e 2iθ , so<br />
as θ ranges from 0 to 2π, z will range around the unit circle T. Then<br />
dz = 2izdθ, and the integral becomes<br />
<br />
T<br />
dz<br />
2iz(a2 − (z − 2 + z−1 1<br />
=<br />
)/4) 2i<br />
= π <br />
residues of poles of<br />
poles<br />
The poles are at<br />
4a 2 + 2 ± √ 16a 4 + 16a 2<br />
2<br />
<br />
T<br />
4dz<br />
(4a 2 + 2)z − z 2 − 1<br />
−4<br />
z 2 − (4a 2 + 2)z + 1<br />
inside T.<br />
= 2a 2 + 1 ± 2a a 2 + 1.<br />
The + root is outside the unit circle, so it is of no interest to us. We<br />
calculate the residue of the other root, z0 = 2a 2 + 1 − 2a √ a 2 + 1.<br />
so<br />
Res(f, z0) =<br />
z 2 − (4a 2 + 2)z + 1<br />
z − z0<br />
−4<br />
= 2z − 2 − 4a 2 ,<br />
2(2a2 + 1 − 2a √ a2 =<br />
+ 1) − 2 − 4a2 −4<br />
−4a √ a 2 + 1 =<br />
1<br />
a √ a 2 + 1 .<br />
Thus,<br />
π<br />
dθ<br />
0 a2 + sin 2 θ =<br />
π<br />
a √ a2 + 1 .<br />
Warning: I have little faith in my calculations on this one. <br />
Spring 2008<br />
Problem 2: Let R/Z be the unit circle with the usual Lebesgue measure.<br />
For each n = 1, 2, 3, . . . , let Kn : R/Z → R + be a non-negative integrable<br />
function such that <br />
R/Z Kn(t)dt<br />
<br />
= 1 and limn→∞ ε
34 <strong>ANALYSIS</strong> QUALS<br />
Proof. First, since f is continuous on a compact set, it is uniformly continuous,<br />
and the image is compact. Let M be the maximum value of f. Let<br />
ε > 0. Choose δ > 0 such that<br />
|f(x − t) − f(x)| < ε whenever |t| < δ for every x ∈ R/Z.<br />
Then for any x ∈ R/Z, we have that<br />
<br />
<br />
<br />
<br />
<br />
<br />
|f ∗ Kn(x) − f(x)| = [f(x − t) − f(x)]Kn(t)dt<br />
R/Z<br />
<br />
<br />
<br />
<br />
<br />
= [f(x − t) − f(x)]Kn(t)dt +<br />
|t|
so y ∈ B(x, ε), so y ∈ B(x (n1)<br />
i , 1<br />
n1<br />
B(x (2N)<br />
k<br />
, 1<br />
2N<br />
) ⊂ B(x(n1)<br />
i<br />
<strong>ANALYSIS</strong> QUALS 35<br />
, 1<br />
n1<br />
) ∪ B(x(n2)<br />
j<br />
) ∪ B(x(n2) j , 1 ). Thus, we have that<br />
n2<br />
, 1<br />
n2<br />
). Hence, B is a base for the<br />
topology, so X is second countable.<br />
(c) Let {x1, x2, . . .} be the dense subset of X found in (a). By Urysohn’s<br />
Lemma, for n, m ≥ 1, there exists ψn,m ∈ C(X) such that ψn,m = 1<br />
on B(xn, 1<br />
m ) and ψn,m is supported on B(xn, 2<br />
Then {ψn,m} separates points, as the balls are a base for the topology<br />
by (b). By the Stone-Weierstrass theorem, the algebra of rational<br />
polynomial combinations of ψn,m is dense, so C(X) is separable.<br />
<br />
Problem 4: Let f, g ∈ L2 (R) be two square-integrable functions on R (with<br />
the usual Lebesgue measure). Show that the convolution<br />
<br />
f ∗ g(x) := f(y)g(x − y)dy<br />
of f and g is a bounded continuous function on R.<br />
Solution. First, we show it is bounded:<br />
<br />
<br />
<br />
<br />
|f ∗ g(x)| = <br />
f(y)g(x − y)dy<br />
<br />
R<br />
1/2 <br />
≤<br />
R<br />
|f(y)|<br />
R<br />
2 dy<br />
= f2 g2 .<br />
m ).<br />
|g(x − y)|<br />
R<br />
2 1/2 dy<br />
Hence, f ∗ g is bounded. Now we show that f ∗ g is continuous.<br />
<br />
<br />
<br />
<br />
<br />
|f ∗ g(x1) − f ∗ g(x2)| = f(y)g(x1 <br />
− y)dy − f(y)g(x2 − y)dy<br />
<br />
R<br />
R<br />
<br />
<br />
<br />
<br />
= <br />
f(y) [g(x1 − y) − g(x2 − y)] dy<br />
<br />
R<br />
<br />
<br />
<br />
<br />
= f(x1 + y) [g(y) − g(x2 − x1 − y)] dy<br />
<br />
R<br />
<br />
<br />
≤ f <br />
2 g − g(x2−x1) ,<br />
2<br />
where g (x2−x1) is the translation. But <br />
g − g(x2−x1) → 0 as |x2 − x1| →<br />
2<br />
0. Hence f ∗ g is continuous. <br />
Problem 5: Let H be a Hilbert space, and let T : H → H be a bounded<br />
linear operator on H.<br />
(a) Show that if the operator norm T of T is strictly less than 1, then<br />
the operator 1 − T is invertible.<br />
(b) Let σ(T ) denote the set of all complex numbers z such that T − zI is<br />
not invertible. (This set is known as the spectrum of T .) Show that<br />
σ(T ) is a compact subset of C.
36 <strong>ANALYSIS</strong> QUALS<br />
Solution. (a) Let (1 − T ) −1 =<br />
n=0<br />
∞<br />
T n . This converges, as<br />
n=0<br />
<br />
∞ <br />
T<br />
<br />
n<br />
<br />
<br />
<br />
<br />
≤<br />
∞<br />
T n < ∞.<br />
n=0<br />
But then (1 − T )(1 − T ) −1 = I, so 1 − T is invertible.<br />
(b) We claim σ(T ) is closed and bounded. First, we prove bounded. Let<br />
|α| > T . Then αI − T = α(I − T<br />
α ), and <br />
T <br />
α < 1, so by part (a), we<br />
have that (I − T<br />
α ) is invertible, so α ∈ σ(T ). Thus,<br />
σ(T ) ⊂ {α ∈ C : |α| ≤ T },<br />
so σ(T ) is bounded. Next, we aim to show that σ(T ) is closed. Let G<br />
be the set of invertible operators on H. We claim that G is open. Let<br />
S ∈ G and suppose that T − S < S −1 1 .<br />
Then S −1 T − I = S −1 (T − S) < 1, so by part (a), S −1 T is invertible,<br />
and therefore T −1 = (S −1 T ) −1 S −1 , so T is invertible. Therefore,<br />
G is open.<br />
Define f : C → B(H) by α ↦→ (αI − T ). Then f is continuous, so<br />
f −1 (G) is open. But f −1 (G) = {z ∈ C : T − zI invertible} = C\σ(T ),<br />
so σ(T ) is closed. Thus, σ(T ) is compact.<br />
<br />
Problem 8: Let H be a real Hilbert space, let K be a closed non-empty<br />
subset of H, and let v be a point in H. Show that there exists a unique<br />
w ∈ K which minimizes the distance to v in the sense that v − w <<br />
v − w ′ for all w ′ ∈ K\{w}. (Hint: you may find the parallelogram law<br />
to be useful.)<br />
Solution. Note that the problem is incorrect as stated, as can be seen by<br />
Problem 1 on Fall 2009. We employ the additional assumption that K is<br />
convex.<br />
Let d = infw∈K v − w. Choose {wn} ∈ K such that v − wn → d. Then<br />
wn − wm 2 = 2 wn 2 + 2 wm 2 2<br />
wn<br />
− 4 <br />
+ wm <br />
<br />
2 .<br />
As K is convex, we have that wn+wm<br />
2<br />
∈ K, so wn+wm<br />
2<br />
<br />
2 ≥ d2 . Thus,<br />
wn − wm 2 ≤ 2 wn 2 + 2 wm 2 − 4d 2 → 0 as n, m → ∞.<br />
Thus, {wn} is Cauchy, and since H is complete, the sequence converges to<br />
a point w. Then w ∈ K since K is closed, and v − w = d. This proves<br />
existence.<br />
If y also satisfies v − y = d, then<br />
2<br />
<br />
<br />
w + y <br />
<br />
2 = 4d 2 <br />
<br />
−4 <br />
w + y <br />
<br />
2 <br />
0 ≤ w − y 2 = 2 w − v 2 +2 y − v 2 −4<br />
so w = y. This proves uniqueness.<br />
2<br />
≤ 4d 2 −4d 2 = 0,
Problem 10: Let the power series f(z) =<br />
<strong>ANALYSIS</strong> QUALS 37<br />
<br />
∞<br />
anz n have radius of conver-<br />
gence r > 0. For each ρ with 0 < ρ < r let Mf (ρ) := sup{|f(z)|; |z| = ρ}.<br />
Show that the following holds for each such ρ:<br />
∞<br />
n=0<br />
n=0<br />
|an| 2 ρ 2n ≤ Mf (ρ) 2 .<br />
Solution. Let φ be a function on the unit circle T defined by φ(z) = f(ρz).<br />
Then the Fourier expansion of φ is anρnz n , so φ 2<br />
2 = 2π |an| 2ρ2n , by<br />
Parseval. But we have that<br />
φ 2<br />
2 ≤ 2π φ2 ∞<br />
and since φ ∞ = Mf (ρ), we have that<br />
∞<br />
n=0<br />
|an| 2 ρ 2n ≤ Mf (ρ) 2 .<br />
Problem 11: Let f : D → D be a holomorphic map having two unequal fixed<br />
points a, b ∈ D. Show that f(z) = z for all z ∈ D. (Hint: use Schwarz’s<br />
lemma.)<br />
Solution. Let φ(z) = z−a<br />
1−az . Then φ is an automorphism of the disc. Let<br />
g = φ ◦ f ◦ φ−1 . Then g : D → D, g(0) = φ ◦ f(a) = φ(a) = 0, and<br />
g(φ(b)) = φ ◦ f ◦ φ−1 (φ(b)) = φ(b). Furthermore, g is holomorphic. Thus,<br />
we may apply Schwarz Lemma, and as g(φ(b)) = φ(b), we may conclude<br />
that g(z) = z for all z ∈ D.<br />
Then f = φ−1 ◦ g ◦ φ is also the identity, f(z) = z. <br />
Problem 12: Consider the annulus A := {z ∈ C : r < |z| < R}, where<br />
0 < r < R. Show that the function f(z) = 1/z cannot be uniformly<br />
approximated in A by complex polynomials.<br />
Solution. Let P (z) =<br />
arbitrary. Then<br />
2π<br />
0<br />
n<br />
ckz k be a complex polynomial. Let r < η < R be<br />
k=0<br />
f(ηe it )P (ηe it )dt = 1<br />
η<br />
= c0<br />
η<br />
= 0<br />
2π<br />
e<br />
0<br />
it (c0 + ηc1e it + c2η 2 e 2it + . . .)dt<br />
2π 2π<br />
0<br />
e it dt + c1<br />
0<br />
e 2it dt + . . .
38 <strong>ANALYSIS</strong> QUALS<br />
But as 2π<br />
0 f(ηeit )f(ηeit )dt = 2π<br />
<br />
2π<br />
2π<br />
=<br />
η2 ≤<br />
<br />
<br />
<br />
0<br />
2π<br />
Thus, f − P ∞ ≥ 1<br />
η<br />
0<br />
η2 , we have that<br />
f(ηeit ) f(ηe it ) − P (ηe it ) <br />
<br />
dt<br />
<br />
<br />
f(ηe it ) f − P ∞ dt<br />
= 2π<br />
η f − P ∞<br />
for every such choice of η, so f − P ≥ 1<br />
R<br />
for any<br />
polynomial P , and hence there can be no sequence of polynomials that<br />
converges uniformly to f.<br />
<br />
Problem 13: Let Ω ⊂ C be an open set containing the closed unit disk D,<br />
and let fn : Ω → C be a sequence of holomorphic functions on Ω which<br />
converge uniformly on compact subsets of Ω to a limit f : Ω → C. Suppose<br />
that |f(z)| = 0 whenever |z| = 1. Show that there is a positive integer N<br />
such that for n ≥ N, the functions fn and f have the same number of zeros<br />
in the unit disk D.<br />
Solution. As |f(z)| = 0 and {|z| = 1} is compact, we can let m = min |z|=1 |f(z)|.<br />
As fn → f uniformly, we may choose N so that n ≥ N implies that<br />
|fn(z) − f(z)| < m<br />
2 for all z ∈ ∂D.<br />
Then |fn − f| < |f| on ∂D, so by Rouche’s Theorem, fn and f have the<br />
same number of zeros inside D.<br />
<br />
Fall 2007<br />
Problem 1: Let f : R → R be a function of class C 1 , periodic of period 2π,<br />
and satisfying<br />
Prove that π<br />
π<br />
−π<br />
|f(t)|<br />
−π<br />
2 dt ≤<br />
f(t)dt = 0.<br />
π<br />
|f<br />
−π<br />
′ (t)| 2 dt.<br />
Find all such functions for which equality holds.<br />
Proof. Write ˆ f(n) for the nth Fourier coefficient of f. Then we have that<br />
ˆf ′ (n) = in ˆ f(n).<br />
By Parseval, we know that<br />
π<br />
|f(t)| 2 ∞ <br />
<br />
dt = 2π ˆ <br />
<br />
f(n) 2<br />
,<br />
and<br />
2π<br />
∞<br />
n=−∞<br />
−π<br />
<br />
<br />
ˆ <br />
<br />
f(n)<br />
2<br />
≤ 2π<br />
∞<br />
n=−∞<br />
n=−∞<br />
<br />
<br />
n ˆ <br />
<br />
f(n)<br />
2<br />
=<br />
π<br />
|f<br />
−π<br />
′ (t)| 2 dt,
<strong>ANALYSIS</strong> QUALS 39<br />
where the inequality follows since ˆ f(0) = 0 by assumption, and the equality<br />
also comes from Parseval.<br />
If equality holds, then ˆ f(n) = 0 for all n, so f is identically zero, as seen<br />
below in problem 8 of this exam.<br />
<br />
Problem 2: Is there a closed uncountable subset of R which contains no<br />
rational numbers? Prove your answer.<br />
Solution. Let {qn} be an enumeration of the rational numbers. Then we<br />
have that<br />
∞<br />
Q = B(qn, 2 −n )<br />
n=1<br />
is a countable union of open sets, and hence is open. Thus, R\Q is closed<br />
and contains no rational numbers. Furthermore,<br />
∞<br />
µ(Q) ≤ 2 1−n < ∞,<br />
n=1<br />
so R\Q has strictly positive Lebesgue measure, and hence is uncountable.<br />
<br />
Problem 3: (a) Prove that a complete normed vector space (a Banach space)<br />
is either finite-dimensional or has uncountable dimension in the vector<br />
space sense, i.e., it is not generated as finite linear combinations of<br />
elements of some countable subset.<br />
(b) Use part (a) to give an example of a vector space that cannot be given<br />
the structure of a Banach space, that is, it is not complete in any norm.<br />
Solution. (a) Assume X is a an infinite-dimensional Banach space, and<br />
assume for sake of contradiction that (xi)is a countable Hamel basis<br />
for X. Let Vn = span{x1, . . . , xn}. Then X = ∞<br />
i=1 Vn. Note that<br />
Vn is finite dimensional, so it is isomorphic as a Banach space to R n ,<br />
and thus is a closed subspace. However, Vn is nowhere dense, as Vn<br />
contains no open balls in V . Thus, by the Baire Category Theorem,<br />
Vn = X, a contradiction.<br />
(b) Let V be the vector space consisting of all infinite sequences with<br />
finitely many non-zero terms. Then V has a countable Hamel basis:<br />
let xi be the sequence of all zeros with a 1 in the ith coordinate. Every<br />
element of V can be written uniquely as a finite linear combination of<br />
the xi. Thus, by part (a), V cannot be a Banach space.<br />
<br />
Problem 4: Prove that non every subset of [0, 1] is Lebesgue measurable.<br />
Solution. Consider the relation x ∼ y is x − y ∈ Q. This is an equivalence<br />
relation. For α ∈ [0, 1], let Eα be the equivalence class of α. Then [0, 1] ⊂<br />
<br />
α Eα. By the axiom of choice, we may choose exactly one element xα ∈<br />
[0, 1] from each equivalence class, and let N be the set of these elements.<br />
We claim that N is not Lebesgue measurable.<br />
Assume for sake of contradiction that N is measurable. Let {rk} be an
40 <strong>ANALYSIS</strong> QUALS<br />
enumeration of Q ∩ [0, 1]. Let Nk = N + rk for each k. Then by definition<br />
of the equivalence relation, Nk ∩ Nj = if k = j. Then<br />
∞<br />
[0, 1] ⊂ Nk ≤ [−1, 2].<br />
k=1<br />
If N is measurable, then Nk is measurable, and µ(N) = µ(Nk). Thus, we<br />
have<br />
∞<br />
1 ≤ µ(Nk) ≤ 3,<br />
and so<br />
1 ≤<br />
k=1<br />
∞<br />
µ(N) ≤ 3,<br />
k=1<br />
which yields a contradiction, as µ(N) can be neither 0 nor strictly positive.<br />
<br />
Problem 7: Suppose H is a separable Hilbert space.<br />
(a) Prove: If T : H → H is a linear mapping such that I − T < 1,<br />
where I is the identity map of H to itself, then T is invertible.<br />
(b) Suppose {en} is a complete, orthonormal set in H (a Hilbert space<br />
basis). Suppose also that {fn} is an orthonormal set in H such that<br />
en − fn 2 < 1. Prove that fn is a complete orthonormal set in H.<br />
Solution. (a) Since I − T < 1, we have that<br />
∞<br />
(I − T ) n is convergent<br />
in the norm topology to a map which is the inverse of I − (I − T ) = T .<br />
Thus, T is invertible.<br />
(b) An invertible bounded linear map T sends orthonormal bases to orthonormal<br />
bases, so it suffices to find such a map T with T (ej) = fj.<br />
Or rather, we define T by T (ej) = fj, and we show that T is invertible.<br />
T can be considered as an infinite matrix where the i, j entry is 〈fj, ei〉.<br />
Then let D = I − T , considered as an infinite matrix. The jth column<br />
represents the vector ej − fj, so the sum of the squares of the norms<br />
of the entries is strictly less than 1, by assumption. Let αi be the sum<br />
of squares of norms of elements in the ith row.<br />
Write x = anen. (I − T )x = bnen. Then bn = <br />
j Dn,jajej<br />
and<br />
bn 2 <br />
≤ |Dn,j| 2 <br />
|aj| 2<br />
= αn x 2 .<br />
n−0<br />
Hence, (I − T )x 2 ≤ x 2 <br />
i αi, so we have that<br />
<br />
<br />
I − T ≤ αi < 1.<br />
Thus by part (a), T is invertible.<br />
i
<strong>ANALYSIS</strong> QUALS 41<br />
Problem 8: Let f : R → R be a continuous function that is periodic with<br />
period 2π. Show that if all the Fourier coefficients of f are 0, then f is<br />
identically 0.<br />
Proof. By Parseval, we know that<br />
π<br />
−π<br />
|f(t)| 2 dt = 2π<br />
∞<br />
n=−∞<br />
<br />
<br />
ˆ <br />
<br />
f(n)<br />
By assumption, all Fourier coefficients are 0, so the right-hand side is 0.<br />
Thus,<br />
π<br />
−π<br />
|f(t)| 2 dt = 0.<br />
But this implies that f = 0 a.e., so since f is continuous, we have that f is<br />
identically 0. <br />
Problem 10: Let U be a bounded connected open set in C. Prove that if K is<br />
a compact subset of U, then there is a constant CK such that for every point<br />
z ∈ K and every holomorphic L2 function f on U, |f(z)| ≤ CK<br />
.<br />
2<br />
.<br />
U |f|2 1/2<br />
Solution. Let r = dist(K, U c ). Let z ∈ K, B = B(z, r/2) ⊂ U. By mean<br />
value principle and Cauchy-Schwarz, we have that<br />
<br />
<br />
|f(z)| = <br />
1<br />
<br />
<br />
f(w)dw<br />
Area(B)<br />
<br />
B<br />
≤ 4<br />
πr2 <br />
|f(w)| dw<br />
B<br />
= 4<br />
πr2 <br />
|f(w)| χBdw<br />
U<br />
≤ 4<br />
πr2 <br />
|f(w)|<br />
U<br />
2 1/2 <br />
dw χ<br />
U<br />
2 1/2 Bdw<br />
<br />
4<br />
=<br />
πr2 1/2 1/2 |f|<br />
U<br />
Thus, we let CK = 4<br />
πr2 1/2, and note this depends only on K. This<br />
completes the proof. <br />
Problem 11: Let U be a bounded connected open set in C. Suppose that 0<br />
belongs to U and that F : U → U is a holomorphic function with F (0) =<br />
0 and with F ′ (0) = 1. Prove that F is the identity map. [Suggestion:<br />
Consider the power series of F composed with itself many times].<br />
Solution. In a neighborhood of 0, we can write F (z) = z +a2z 2 +a3z 3 +. . ..<br />
Let F (n) (z) = F ◦ F ◦ · · · ◦ F (z), where there are n compositions. Then<br />
F (n) : U → U. We have that<br />
F (2) (z) = F (z) + <br />
akF (z) k = z + 2a2z 2 + 2a3z 3 + . . . .<br />
Similarly, we have that<br />
k≥2<br />
F (n) (z) = z + na2z 2 + na3z 3 + . . . .
42 <strong>ANALYSIS</strong> QUALS<br />
As U is bounded, there exists M with |z| ≤ M for all z ∈ U. Then by the<br />
Cauchy estimates, for B(0, r) ⊂ U, we have that<br />
<br />
<br />
<br />
d<br />
<br />
k<br />
dzk (F (n) <br />
<br />
(z)) <br />
M<br />
<br />
<br />
≤ sup<br />
rk F (n) <br />
<br />
(z) = Cr,nM<br />
rk .<br />
B(0,r)<br />
Thus, |nak| ≤ Cr,nM<br />
rk . But since Cr,n ≤ M, if we have ak = 0, we get a<br />
contradiction by fixing r and taking n → ∞. Thus, ak = 0 for all k ≥ 2,<br />
which means that F (z) = z. <br />
Problem 12: Find a conformal map to the unit disc of the half disc {z :<br />
|z| < 1, Rez > 0}. You may write your answer as a composition of simpler<br />
conformal maps.<br />
Solution. Let f1(z) = z−1<br />
z+1 . This maps the half disc to the upper left quadrant.<br />
Let f2(z) = z2 . This maps the upper left quadrant to the lower half<br />
plane. Let f3(z) = z+i<br />
z−i . This maps the lower half plane to the unit disc.<br />
Thus, f = f3 ◦ f2 ◦ f1 is the desired conformal map. <br />
Problem 13: Suppose R is a positive number and U = {z : |z| < R}.<br />
(a) Show that for each holomorphic function f on U, there is a power<br />
∞<br />
series anz n which converges at each point z ∈ U to f(z).<br />
n=0<br />
(b) Show that there is only one such power series.<br />
Proof. (a) We have that for some fixed ε,<br />
<br />
<br />
<br />
z <br />
<br />
ξ < 1 − ε<br />
for all ξ ∈ ∂D. Thus, the geometric series yields<br />
∞<br />
n z 1<br />
= =<br />
ξ 1 −<br />
ξ<br />
ξ − z .<br />
n=0<br />
z<br />
ξ<br />
But then by the Cauchy integral formula, we have that<br />
f(z) = 1<br />
<br />
2πi<br />
<br />
f(ξ) 1<br />
dξ =<br />
ξ − z 2πi<br />
f(ξ) 1<br />
∞<br />
n z<br />
,<br />
ξ ξ<br />
∂D<br />
n=0<br />
∂D<br />
and since we have uniform convergence in the geometric series, we may<br />
switch the sum and integral and get<br />
∞<br />
<br />
<br />
1 f(ξ)<br />
f(z) =<br />
dξ z<br />
2πi (ξ) n+1 n<br />
=<br />
∞<br />
n=0<br />
∂D<br />
f (n) (0)<br />
z<br />
n!<br />
n .<br />
(b) If we had anz n = bnz n for all z ∈ D, then by matching coefficients,<br />
we have an = bn for all n.<br />
<br />
n=0
<strong>ANALYSIS</strong> QUALS 43<br />
Winter 2007<br />
Problem 1: Let fn : R → R be non-negative integrable functions with<br />
fn1 = 1. Suppose fn → f pointwise a.e. with f = 1. Show that<br />
<br />
<br />
fn(x)dx → f(x)dx<br />
uniformly in the choice of Borel set A ⊂ R.<br />
A<br />
Solution. By Fatou’s lemma, we have that<br />
<br />
<br />
lim inf (−|fn − f| + |fn| + |f|) ≤ lim inf −|fn − f| + |fn| + |f|.<br />
Hence,<br />
A<br />
<br />
2 f1 = 2 ≤ 2 − lim sup<br />
|fn − f|<br />
and thus fn − f1 → 0 as n → ∞. Then<br />
<br />
<br />
<br />
<br />
<br />
fn(x)dx − f(x)dx<br />
<br />
A<br />
A<br />
≤<br />
<br />
|fn(x) − f(x)| dx ≤ fn − f1 .<br />
A<br />
Thus, <br />
A fn(x)dx → <br />
f(x)dx uniformly. <br />
A<br />
Problem 2: Consider the function f : (0, ∞) × (0, ∞) → R defined by<br />
∞<br />
f(x, y) =<br />
n=0<br />
x<br />
x2 .<br />
+ yn2 Show that the limit g(y) := limn→∞ f(x, y) exists for all y > 0 and compute<br />
g(y).<br />
Solution. We have that<br />
f(x, y) =<br />
∞<br />
n=0<br />
= x<br />
y<br />
x<br />
x2 1/y<br />
·<br />
+ yn2 1/y<br />
∞<br />
n=0<br />
= 1 x<br />
+<br />
x y<br />
But we also have that<br />
∞<br />
∞<br />
1<br />
1<br />
≤<br />
dn =<br />
+ n2 + n2<br />
n=1<br />
x 2<br />
y<br />
0<br />
x 2<br />
y<br />
x 2<br />
y<br />
1<br />
+ n2<br />
∞ 1<br />
.<br />
+ n2<br />
n=1<br />
x 2<br />
y<br />
√ y<br />
x arctan<br />
n √ y<br />
x<br />
n=∞ <br />
n=0<br />
Thus, f(x, y) ≤ 1 π<br />
x + 2 √ π<br />
y . Similarly, f(x, y) ≥ 2 √ y . Thus,<br />
Problem 3: Let f ∗ g(x) = <br />
π<br />
lim f(x, y) =<br />
x→∞ 2 √ y .<br />
√<br />
y<br />
=<br />
x<br />
· π<br />
2 .<br />
f(x − y)g(y)dy denote the convolution of f<br />
R<br />
and g. Fix g ∈ L1 (R). Do the following:<br />
(1) Show that Ag(f) := f ∗ g is a bounded operator L1 (R) → L1 (R).<br />
(2) Suppose in addition g ≥ 0. Find the corresponding norm Ag.
44 <strong>ANALYSIS</strong> QUALS<br />
(3) Show that the only f ∈ L 1 (R) for which f ∗ f = f is f = 0.<br />
Solution. (1) We have that<br />
<br />
<br />
<br />
<br />
<br />
Ag(f) = <br />
f(x − y)g(y)dydx<br />
≤ f1 g1 ,<br />
R<br />
R<br />
so as g ∈ L1 , we have that Ag is a bounded operator.<br />
(2) By part (1), we have that Ag ≤ g. We have that<br />
<br />
<br />
<br />
Ag(g) = g(x−y)g(y)dydx = g(y) g(x−y)dxdy = g(y) g dy = g 2 ,<br />
R<br />
R<br />
R<br />
R<br />
and so Ag ≥ g, and hence Ag = g.<br />
(3) Suppose that f ∗ f = f. Then by taking Fourier transforms, we have<br />
that f ˆ∗<br />
f = ˆ f. But using properties of Fourier transforms, we have<br />
f ˆ∗<br />
f = ˆ f · ˆ f. Thus, ˆ f 2 − ˆ f = 0 a.e. Hence ˆ f = 0 or 1 a.e.<br />
But since f ∈ L1 , we have that ˆ f is continuous and vanishes at ∞ by<br />
Riemann-Lebesgue theorem, so we conclude that ˆ f = 0 everywhere.<br />
Then by Fourier inversion, we conclude that f = 0.<br />
<br />
Problem 4: Let α ∈ R\Q and let T : L 2 ([0, 1]) → L 2 ([0, 1]) be defined by<br />
(T f)(x) = f(x + α mod 1).<br />
Denote Snf = f + T f + T 2f + · · · + T n−1f. Do the following:<br />
(1) For any f ∈ L2 ([0, 1]), prove that 1<br />
nSnf converges in L2 . Identify the<br />
limit.<br />
(2) Suppose f : [0, 1] → R is continuous with f(1) = f(0). Show that the<br />
convergence in (1) is uniform.<br />
Solution. We prove both at once. We claim that 1<br />
nSnf(x) → 1<br />
f(t)dt for<br />
0<br />
every x ∈ [0, 1], and that the convergence is uniform for continuous periodic<br />
functions.<br />
First, let f(x) = e2πikx . If k = 0, the statement is clearly true. Suppose<br />
k = 0. Then the statement is that<br />
N−1<br />
1 <br />
e<br />
N<br />
2πik(x+nα) →<br />
n=0<br />
1<br />
0<br />
e 2πikt dt.<br />
But by the geometric series, the LHS is equal to<br />
2πikx 1 1 − e<br />
e<br />
N<br />
2πiNαk<br />
1 − e2πiαk and since α /∈ Q, the denominator is never 0. Hence, this approaches 0 as<br />
N → ∞, and 1<br />
0 e2πikxdx = 0, so the statement is true for all trigonometric<br />
polynomials.<br />
Now let ε > 0, f a continuous function from R/Z to R. As trigonometric<br />
polynomials are uniformly dense in the space of continuous periodic<br />
functions, we may choose P (x) trigonometric such that<br />
sup |f(x) − P (x)| <<br />
x∈R/Z<br />
ε<br />
3 .<br />
R
Then for N sufficiently large, we have that<br />
<br />
N−1<br />
1 <br />
<br />
1 <br />
<br />
P (x + nα) − P (t)dt<br />
N<br />
<br />
Then<br />
n=0<br />
<br />
<br />
1<br />
<br />
N<br />
N−1 <br />
n=0<br />
<strong>ANALYSIS</strong> QUALS 45<br />
0<br />
f(x + nα) −<br />
≤ 1<br />
N−1 <br />
|f(x + nα) − P (x + nα)| +<br />
N<br />
n=0<br />
+<br />
1<br />
0<br />
<br />
<br />
1<br />
<br />
N<br />
1<br />
0<br />
N−1 <br />
n=0<br />
<br />
<br />
<br />
f(t)dt<br />
<br />
|P (t) − f(t)| dt < ε.<br />
< ε<br />
3 .<br />
P (x + nα) −<br />
1<br />
0<br />
<br />
<br />
<br />
P (t)dt<br />
<br />
Thus, the convergence is uniform, which proves part (2). Part (1) follows<br />
by noting that continuous periodic functions are dense in L2 ([0, 1]).<br />
<br />
n<br />
Problem 5: Let An(f) = 1<br />
n f(x)dx. Show that there exists a continuous<br />
0<br />
linear functional A : L∞ (R+) → R such that<br />
A(f) = lim<br />
n→∞ An(f)<br />
whenever the limit exists. Here, R+ = (0, ∞).<br />
Solution. We define A as above, and prove that it is a continuous linear<br />
functional on L ∞ (R+). Clearly, A is a linear functional, as for c, d ∈ R,<br />
A(cf + dg) = lim<br />
n→ An(cf + dg) = lim<br />
n→∞ cAn(f) + dAn(g) = cA(f) + dA(g).<br />
Thus, it suffices to show that A is bounded. But we have that<br />
<br />
<br />
|A(f)| = <br />
lim<br />
n <br />
1<br />
<br />
f(x)dx<br />
n→∞ n<br />
<br />
0<br />
<br />
n <br />
= lim <br />
1<br />
<br />
n→∞ f(x)dx<br />
n<br />
<br />
0<br />
n<br />
1<br />
≤ lim |f(x)| dx<br />
n→∞ n 0<br />
n<br />
1<br />
≤ lim f<br />
n→∞<br />
L∞ dx<br />
n<br />
0<br />
= lim<br />
n→∞ f L ∞<br />
= f L ∞ .<br />
Thus, A is a bounded linear functional, with A ≤ 1. <br />
Problem 6: Let X be a Banach space and let A : X → X be a linear map.<br />
Define<br />
ρ(A) = {λ ∈ C : (λ − A) maps X onto X}<br />
Show that ρ(A) is an open subset of C.
46 <strong>ANALYSIS</strong> QUALS<br />
<br />
γR<br />
Solution. The problem is false as stated, as some assumption of boundedness<br />
must be made. A counterexample is to let X be l ∞ , the space of<br />
bounded sequences (x1, . . . , xn, . . .), and let<br />
A : (x1, . . . , xn, . . .) ↦→ (x1, x2/2, . . . , xn/n, . . .).<br />
Then A is linear and surjective, so 0 ∈ ρ(A), but 1/n ∈ ρ(A) for any n ∈ N,<br />
so ρ(A) is not open.<br />
That said, we alter the problem and change the definition of ρ(A). Let<br />
ρ(A) = {λ ∈ C : (λ − A) −1 exists and is bounded}<br />
Let G be the set of operators on X with bounded inverse. We claim that<br />
G is open. Let S ∈ G and suppose that T − S < S −1 1 .<br />
Then S −1 T − I = S −1 (T − S) < 1, so we have that<br />
(S −1 T ) −1 = (I − (I − S −1 T )) −1 = (I − S −1 T ) i exists.<br />
Therefore T −1 = (S −1 T ) −1 S −1 , so T is invertible. Therefore, G is open.<br />
Define f : C → B(X) by α ↦→ (αI − T ). Then f is continuous, so f −1 (G)<br />
is open. But f −1 (G) = ρ(A), so ρ(A) is open.<br />
<br />
Problem 8: Determine the number of zeros of the polynomial<br />
p(z) = z 4 + z 3 + 4z 2 + 2z + 3<br />
in the right half-plane {z : Re z > 0}.<br />
Solution. Let ΓR be the contour that is the right half of the circle of radius<br />
R centered at the origin. Let γR be the semi-circle part of the contour, and<br />
let γR ˆ be the segment from Ri to −Ri.<br />
Note that on {Re z = 0}, we have that Re p(it) = t 4 − 4t 2 + 3 = (t 2 −<br />
3)(t 2 − 1), and Im p(it) = −t 3 + 2t = −t(t 2 − 2). Thus, there are no zeros<br />
on {Re z = 0}.<br />
Thus, we can use the argument principle in this contour, which says that<br />
p<br />
ΓR<br />
′ (z)<br />
p(z) dz = 2πi(number of zeros in the interior of ΓR). By taking R → ∞,<br />
we can calculate the total number of zeros in {z : Re z > 0}.<br />
On γR, p(z) = z 4 + O(|z| 3 ), and p ′ (z) = 4z 3 + O(|z| 2 ), so p′ (z)<br />
p(z) = 4z−1 +<br />
O(|z| −2 ). Thus,<br />
p ′ (z)<br />
dz =<br />
p(z)<br />
π<br />
0<br />
Ri · 4 · e−θ<br />
R dθ + O(|R|−1 ) = 4πi + O(|R| −1 ) → 4πi as R → ∞.<br />
On ˆ<br />
γR, note that critical points (points where p crosses either the real or<br />
imaginary axes) are at t = 0, ±1, ± √ 2, ± √ 3 for z = it. Suppose R > √ 3.<br />
Then on ( √ 3i, Ri), p takes values in the 4th quadrant (by noticing that<br />
the real part is positive and the imaginary part is negative). Similarly, we
<strong>ANALYSIS</strong> QUALS 47<br />
see that on ( √ 2i, √ 3i), p is in the 3rd quadrant; on (i, √ 2i), p is in 2nd<br />
quadrant; on (0, i), p is in 1st quadrant; on (−i, 0), p is in 4th quadrant; on<br />
(− √ 2i, −i), p is in 3rd quadrant; on (− √ 3i, − √ 2i), p is in 2nd quadrant;<br />
and on (−Ri, √ 3i), p is in the 1st quadrant.<br />
Thus, moving from Ri to −Ri gives two clockwise revolutions around the<br />
origin, meaning the winding number is 2, which means that p<br />
γR ˆ<br />
′ (z)<br />
p(z) dz →<br />
−4πi as R → ∞.<br />
Hence, <br />
ΓR<br />
p ′ (z)<br />
p(z) dz → 4πi − 4πi = 0, so there are no zeros in {Re z > 0}.<br />
<br />
Problem 9: Let f(z) be analytic for 0 < |z| < 1. Suppose there are C > 0<br />
and m ≥ 1 such that<br />
<br />
<br />
f (m) <br />
<br />
(z) ≤ C<br />
, 0 < |z| < 1.<br />
|z| m<br />
Show that f(z) has a removable singularity at z = 0.<br />
Solution. Suppose that f(z) has Laurent series<br />
∞<br />
cnz n =<br />
−1<br />
cnz n ∞<br />
+ cnz n .<br />
Then<br />
f (m) (z) =<br />
n=−∞<br />
−1<br />
n=−∞<br />
cn<br />
n=−∞<br />
n=0<br />
n!<br />
(n − m)! zn−m ∞<br />
+<br />
n=0<br />
cn<br />
n!<br />
(n − m)! zn−m .<br />
Then as f (m) (z) ≤ C<br />
|z| m for 0 < |z| < 1, we have that cn = 0 for n < 0.<br />
∞<br />
Thus, f(z) = cnz n , so f is analytic on D.<br />
n=0<br />
Problem 10: Let J = {iy : 1 ≤ y < ∞} and H be the open upper half<br />
plane. Consider the domain D = H\J. Find a bounded harmonic function<br />
u : D → R such that u(x + iy) → 0 as y ↓ 0 and u(z) → 1 as z → J. It is<br />
fine to represent the solution in terms of a composition of conformal maps.<br />
Proof. Let’s map this slit half plane to something better, keeping track of<br />
where we’re sending J. Let f1(z) = z−i<br />
z+i . Then f1 takes D to the slit disc,<br />
and takes J to the slit.<br />
Let f2(z) = √ z, under the branch cut that makes sense (positive real axis).<br />
Then f2 takes the slit disc to the upper half disc, taking the image of J to<br />
R ∩ D.<br />
Let f3(z) = z−1<br />
z+1 . Then f3 takes the upper half disc to the second quadrant<br />
of the plane and takes the image of J to the negative real axis.<br />
Let f4(z) = iz2 . This takes the second quadrant to the right half plane,<br />
taking the image of J to the positive imaginary axis.<br />
Let f5(z) = Log(z), with the negative real axis branch cut. Then f5 takes<br />
the right half plane to the horizontal strip between iπ/2 and −iπ/2, which<br />
takes J to the line Im(z) = π/2.
48 <strong>ANALYSIS</strong> QUALS<br />
Let f = f5 ◦ f4 ◦ f3 ◦ f2 ◦ f1. Then f is a conformal map taking D to<br />
a strip, which takes J to the line Im(z) = π/2, and takes R to the line<br />
Im(z) = −π/2.<br />
Let u(z) = 1 1<br />
2 + π Im(f(z)). Then since the imaginary part of a holomorphic<br />
function is harmonic, we have that u is harmonic on D, and u(x + iy) → 0<br />
as y ↓ 0 and u(z) → 1 as z → J.<br />
<br />
Problem 11: Prove that a meromorphic function f(z) in the extended complex<br />
plane C ∗ = C ∪ {∞} is the sum of the principal parts at its poles.<br />
Solution. In a neighborhood of each pole a of f, we have<br />
f(z) = c−m<br />
(z − a)<br />
c−1<br />
+ · · · +<br />
(z − a) m−1 z − a + c0 + c1(z − a) + · · · ,<br />
m + c−m+1<br />
c−m<br />
where the principal part is (z−a) m + · · · + c−1<br />
z−a . Either f vanishes at infinity<br />
or not. That is, either f(1/z) has a pole at 0 or a removable singularity at<br />
0. If f(1/z) has a pole at 0, then f has a pole at ∞.<br />
Including possibly ∞, let z1, . . . , zm be the poles of f. Near each non-infinite<br />
zk, as above, we may write f(z) = fk(z) + gk(z), where fk is the principal<br />
part and gk is holomorphic If zk = ∞, we can write f(1/z) = ˜ f∞(z)+g∞(z),<br />
where ˜ f∞(z) is the principal part at 0 of f(1/z), and g∞ is holomorphic in<br />
a neighborhood of 0. Define fk(z) = ˜ f∞(1/z).<br />
Let H = f −<br />
m<br />
fk. Near each pole, we have subtracted the principal<br />
k=1<br />
part of f, so we see that H is entire (the singularities at the points zk are<br />
all removable). But since we also subtracted ˜ f∞(1/z), we have that H is<br />
bounded. Hence by Liousville, H is constant. But by the definition of H,<br />
we must have that H = 0 everywhere. Thus, f is the sum of the principal<br />
parts of its poles.<br />
<br />
Problem 12: Let D be a domain (connected open set) in C and let (un) be a<br />
sequence of harmonic functions un : D → (0, ∞). Show that if un(z0) → 0<br />
for some z0 ∈ D, then un → 0 uniformly on compact subsets of D.<br />
Solution. As un(z0) → 0, we have that {un(z0)} is Cauchy. Thus, for every<br />
ε > 0 there exists N such that 0 ≤ un(z0) − um(z0) < ε for all N ≤ n ≤ m.<br />
Let K ⊂ D be compact. Harnack’s inequality states that there exists a constant<br />
CK > 0 so that maxx∈K u(x) ≤ CK minx∈K u(x) for u : D → (0, ∞)<br />
harmonic.<br />
Applying this to un − um, we have that<br />
max<br />
x∈K (un(x) − um(x)) ≤ CK min<br />
x∈K (un(x) − um(x)).<br />
If z0 ∈ K, then minx∈K(un(x) − um(x)) < ε, and thus un → 0 uniformly<br />
on K.
<strong>ANALYSIS</strong> QUALS 49<br />
If z0 ∈ K, choose K ′ such that K ′ ∩ K = ∅ and z0 ∈ K ′ , and choose<br />
z1 ∈ K ′ ∩K. Since un(z1) → 0, the argument can be repeated to show that<br />
un → 0 uniformly on K.<br />
Thus, un → 0 uniformly on compact subsets of D.