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Analysis Qualifying Exam Solutions
Thomas Goller
September 4, 2011
Contents
1 Spring 2011 21.1 Real Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Complex Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . 4
2 Fall 2010 62.1 Real Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.2 Complex Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . 8
3 Spring 2010 103.1 Real Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103.2 Complex Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . 13
4 Fall 2009 154.1 Real Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154.2 Complex Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . 17
5 Spring 2009 205.1 Real Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205.2 Complex Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . 21
6 Fall 2008 246.1 Real Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
7 Spring 2008 277.1 Real Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
1
Chapter 1
Spring 2011
1.1 Real Analysis
A1. (a) `1(Z) is separable. A countable set whose finite linear combinationsare dense is enn∈Z, where en has a 1 in the nth position and is 0 everywhere
else. If x ∈ `1(Z), then the sums∑Nk=−N xkek approximate x arbitrarily well
in the norm as N →∞ since∑k |xk| <∞.
(b) `∞(Z) is not separable. This is proved by finding an uncountable num-ber of disjoint open sets, whence every dense set must have an element in eachinterval and thus must be uncountable. Recalling that `∞(Z) consists of allbounded sequences, consider all sequences containing just 0’s and 1’s. Thesemap surjectively onto the real numbers by considering binary expansions, henceare uncountable. Any two distinct sequences of 0’s and 1’s have distance 1 fromeach other in the norm, so for any r < 1
2 , taking a ball of radius r around eachof these sequences gives an uncountable number of disjoint open sets.
A2. (a) If f is measurable, then so is |f |. We need to show that S(a) :=|f |−1
((−∞, a)
)is measurable for each a ∈ R. The case a ≤ 0 is trivial since
then S(a) = ∅ is measurable. Note that when a > 0, we have
|f |−1((−∞, a)
)= f−1
((−a, a)
).
Now, f measurable implies f−1((−∞, a)
)is measurable for each a ∈ R, so each
f−1([a,∞)
)is measurable by taking complements. Therefore each f−1
((a,∞)
)is measurable since
f−1((a,∞)
)=
∞⋃n=1
f−1([a+
1
n,∞)
),
so f−1((−a, a)
)= f−1
((−∞, a)
)∩ f−1
((−a,∞)
)is measurable, as desired.
2
(b) |f | measurable does not imply that f is measurable, as long as Xcontains non-measurable sets. For if N is a non-measurable set, define
f(x) =
−1 x ∈ N ;
1 otherwise.
Then |f | ≡ 1 is measurable, but f−1((−∞, 0)
)= N is not measurable.
A3. Haven’t reviewed Fourier series.
A4. f ∈ L1 implies∫|f | < ∞, and f uniformly continuous means that
for any ε > 0, there is a δ > 0 such that |f(x) − f(y)| < ε whenever |x − y| <δ. Suppose for contradiction that x is a sequence with |xk| → ∞ for whichf(xk) 6→ 0. Then there is an ε > 0 such that for each N > 0, there exists n ≥ Nsuch that |f(xn)| ≥ ε. Thus x has a subsequence xnk such that |f(xnk)| ≥ εfor each k ≥ 1.
We may assume the xnk are distinct, deleting repeats if necessary. Thenby uniform continuity, choose δ > 0 such that |f(x) − f(y)| < ε
2 whenever|x− y| < δ. Also, make δ small enough so that |xnk − xnj | ≥ δ whenever k 6= j.
Then taking a ball of radius δ2 around each xk and letting E be the union of
these balls, we see that ∫|f | ≥
∫E
|f | ≥∞∑k=1
δ · ε2
=∞,
contradicting the fact that f ∈ L1.
A5. Without (1), consider the sequence
fn =
χ[0,1],
1
2χ[0,2],
1
4χ[0,4], . . .
,
which converges pointwise to the function f ≡ 0 for all x. Then (1) fails and
limn→∞
∫|fn − f | = lim
n→∞
∫|fn| = 1 6= 0.
With (1), we can prove the result. Set gn = inffn, fn+1, . . . , which ismeasurable since the fn are measurable. Then 0 ≤ gn ≤ fn and gn f , so∫|gn−f | → 0 by the dominated convergence theorem, whence also
∫gn →
∫f .
Now we use (1) and the fact that |fn − gn| = fn − gn to deduce that∫|fn − f | ≤
∫|fn − gn|+
∫|gn − f | =
∫fn −
∫gn +
∫|gn − f | → 0.
3
1.2 Complex Analysis
B6. g is holomorphic on U . Let z0 ∈ U . Since f is holomorphic at z0 ∈ U , wecan write f(z) =
∑∞n=0 an(z − z0)n for z near z0. Thus
g(z) := f(z) =
∞∑n=0
an(z − z0)n
is a power series expansion for g near z0.
B7. Since |f(z)| ≤ C|z|k for z outside DR, all the poles of f are inside DR,and there can be only finitely many such poles since DR is compact. Let f(z)be the sum of the principal parts of f at each of these poles. Then g := f − fis an entire function that still satisfies a bound of the form |g(z)| ≤ C ′|z|k for|z| ≥ R since the principal parts are bounded away from their poles.
We claim that g is a polynomial of degree ≤ k, which will imply that f is arational function. Since g is entire, it suffices to prove that g(j)(0) = 0 for allj > k (look at the power series expansion). For each R′ ≥ R > 0, the Cauchyinequalities imply
|g(k+1)(0)| ≤ (k + 1)!
R′k+1C ′R′
k → 0,
as R′ →∞, giving the desired result.
B8. (i) =⇒ (ii): Suppose a is a zero of f of order l ≥ k. Let f(z) =(z−a)l
∑n≥0 an(z−a)n be the power series expansion of f near a, so that a0 6= 0.
Let C = |a0| and choose 0 < ε < 1 sufficiently small such that Bε(a) ⊆ U and|∑n≥1 anε
n| ≤ |a0|. Then for all z ∈ Bε(a),
|f(z)| ≤ |z − a|l(|a0|+ |a0|) ≤ 2|a0||z − a|k.
(ii) =⇒ (i): Suppose (ii) holds and let f(z) =∑n≥0 an(z − a)n be the
power series expansion for f near a. Then for all z ∈ Bε(a) with z 6= a,∣∣∣∣ f(z)
(z − a)k
∣∣∣∣ ≤ Cimplies a0 = a1 = · · · = ak−1 = 0.
B9. The entire function sin z := 12i (e
iz − e−iz) has zeroes at πk, k ∈ Z,an essential singularity at∞, and no other zeroes or singularities. The functionsin 1
z therefore has zeroes at 1πk and an essential singularity at 0, with no other
zeroes or singularities.
4
Since sin z = z − z3
3! + z5
5! − · · · we also have sin 1z = z−1 − z−3
3! + z−5
5! − · · · .Thus the residue at 0 of the product of these two series, namely the coefficientof z−1 of the Laurent series, is 0.
Is this the right answer?
B10. Set
f(z) =zeiz
z2 + 2z + 10.
The denominator factors into (z + 3i + 1)(z − 3i + 1), so there are poles at3i − 1 and −3i − 1. We will integrate f over the upper-half-plane semicircleconsisting of γ, the real-axis interval [−R,R], and the semicircle γR of radiusR. The imaginary part of the integral over γ gives the real integral we want toevaluate. Since we are taking R large, we pick up the residue at 3i− 1, which is
limz→3i−1
(z − 3i+ 1)zeiz
(z + 3i+ 1)(z − 3i+ 1)=
(3i− 1)e−3−i
6i
=1
6e3(3 cos 1 + sin 1 + i(cos 1− 3 sin 1)).
What a mess! The integral over γR vanishes thanks to the 1/z decay of fnear the real axis, together with the better decay of f due to eiz away from thereal axis.
5
Chapter 2
Fall 2010
2.1 Real Analysis
A1. Yes, the fact that f−1((r,∞)
)is measurable for each r ∈ Q implies that
f is measurable. Taking complements, we see that f−1((−∞, r]
)is measurable
for each r ∈ Q. Now for any s ∈ R, let rk be an increasing sequence of rationalnumbers < s such that rk s. Then
f−1((−∞, s)
)=
∞⋃k=1
f−1 ((−∞, rk]) ,
so f is measurable.
A2. Suppose 1 ≤ p ≤ q ≤ ∞ such that 1p + 1
q = 1, where we allowp = 1, q = ∞. Since f, fn ∈ Lp and g, gn ∈ Lq, we also have f − fn ∈ Lp andg − gn ∈ Lq. Holder’s inequality implies fngn and fg are in L1. Note that∣∣|fg| − |fngn|∣∣ =
∣∣|fg| − |fng|+ |fng| − |fngn|∣∣ ≤ |g||f − fn|+ |fn||g − gn|.Thus using Holder’s inequality,∫ ∣∣|fg| − |fngn|∣∣ ≤ ∫ |g||f − fn|+ ∫ |fn||g − gn|
= ‖g(f − fn)‖1 + ‖fn(g − gn)‖1≤ ‖g‖q‖f − fn‖p + ‖fn‖p‖g − gn‖q.
Since ‖g‖q and ‖fn‖p are bounded but ‖f − fn‖p → 0 and ‖g − gn‖q → 0, wesee that ‖fg − fngn‖1 → 0, which gives the desired result since∣∣∣∣∫ |fg| − ∫ |fngn|∣∣∣∣ ≤ ∫ ∣∣|fg| − |fngn|∣∣→ 0.
6
(Maybe I’m missing something, but I don’t think the case p = 1, q = ∞ isany different than p > 1.)
A3. Let HTn // C denote the bounded linear functional defined by
Tn(y) = (y, xn). Then the countable collection Tn is pointwise bounded byassumption, hence the norms are bounded by the uniform boundedness property.So there exists M ≥ 0 such that
|(y, xn)| ≤M‖y‖ for all y ∈ H.
In particular, choosing y = xn gives ‖xn‖ ≤M , establishing the desired result.
A4. Suppose f is measurable such that∫|f |p <∞ and
∫|f |r <∞. Set
E = x ∈ X : |f(x)| ≥ 1,
which is measurable. Then∫X
|f |q =
∫E
|f |q +
∫X−E
|f |q
≤∫E
|f |p +
∫X−E
|f |r
≤∫X
|f |p +
∫X
|f |r
<∞,
so f ∈ Lq as desired.
A5. I will only sketch the proof, which is quite long (see Stein 175-177 fordetails). If x ∈M , then take y = x. Otherwise, set d := infy∈M ‖x− y‖. SinceM is closed and x /∈ M , d > 0. Now choose a sequence yn in M such that‖x − yn‖ → d, which we will prove is Cauchy, and whose limit in M (since Mis closed) will be the desired unique element y ∈M closest to x. For
‖x− y‖ ≤ ‖x− yn‖+ ‖yn − y‖ → d as n→∞,
and uniqueness is shown by first proving that x − y is orthogonal to M (takeperturbations of y in M and look at the norm of the difference with x) and thenapplying the Pythagorean theorem to x−y and y−y′, where y′ ∈M is anotherelement minimizing distance to x.
To show the sequence is Cauchy, we use the parallelogram law
‖A+B‖2 + ‖A−B‖2 = 2(‖A‖2 + ‖B‖2
)for all A,B ∈ H
with A = x− yn and B = x− ym.
7
2.2 Complex Analysis
B6. We write
f(z) = 1− cos z = 1− eiz + e−iz
2=(eiz2 − e− iz2
)2
,
so the zeroes of f are at z = 2π · k for k ∈ Z and are each of order 2 since
limz→0
eiz − 1
z= limz→0
ieiz
1= 1
by L’Hospital’s rule (or apply L’Hospital to limz→2πk(1− cos z)/(z− 2πk)2 andnote that you get a constant).
B7. (i) We have
f(z) =1
2i
(eizz+1 − e−
izz+1
).
Clearly the only singularity of f is at z = −1, and we claim that the singularityis essential. Consider the sequence xn := −1 + i
n. Then xn → −1 and
|f(xn)| =∣∣∣∣ 1
2i
(e−n+i − en−i
)∣∣∣∣ ≥ en
4→∞.
On the other hand, let yn := −1 + 1n, which also satisfies yn → −1, but
|f(yn)| =∣∣∣∣ 1
2i
(ei(−n+1) − e−i(−n+1)
)∣∣∣∣ ≤ 1.
So z = −1 is an essential singularity of f .
(ii) Recall that
sin z = z − z3
3!+z5
5!− · · ·
and
cos z = 1− z2
2!+z4
4!− · · · ,
as well as the identity
sin(a+ b) = sin a cos b+ cos a sin b,
which holds for all a, b ∈ C (since the proof relies on the properties of exponen-tials). So
sinz
z + 1= sin
(1− 1
z + 1
)= sin 1 cos
1
z + 1− cos 1 sin
1
z + 1
= sin 1
(1− 1
2!(z + 1)2+ · · ·
)− cos 1
(1
z + 1− 1
3!(z + 1)3+ · · ·
)
8
is the Laurent series for f(z).
(iii) The residue of f at z = −1 is the coefficient of 1z+1 in the Laurent
series, namely − cos 1.
B8. We integrate
f(z) =zeiz
(z + 3i− 1)(z − 3i− 1)
on a semicircle of radius R in the upper half plane, picking up the residue atz = 3i+ 1. The real part of the integral along the real axis goes to the desiredintegral as R→∞, while the integral along the curved portion goes to 0 by thedecay of eiz away from the real axis. The residue is
res3i+1f = limz→3i+1
(z − 3i− 1)f(z) =1
6e3(3 cos 1 + sin 1 + i(− cos 1 + 3 sin 1)).
The real part of 2πi · res3i+1f , which is the value of the desired integral, istherefore π
3e3 (cos 1− 3 sin 1).
B9. (i) If f is entire and |f(z)| ≤ C|z|n for all z ∈ C, then f is a polynomialof degree ≤ n. To see this, it suffices to prove that f (k)(0) = 0 for all k > n(look at the power series). This in turn can be proved by the Cauchy inequality:for any R > 0 we have
|f (k)(0)| ≤ (k)!
RkCRn → 0
as R→∞ when k > n.
(ii) The dimension is n+ 1 since a basis is given by 1, z, z2, . . . , zn.
B10. The polynomials f(z) := 2z5 − z3 + 3z2 − z and g(z) = 8 are bothentire and have no zeroes on the circle |z| = 1. Indeed, |f(z)| ≤ 2+1+3+1 = 7on this circle, while |g(z)| = 8 on the circle. Thus by Rouche’s theorem, thenumber of zeroes of f + g inside the circle is the same as the number of zeroesof g inside the circle, namely zero. Thus f + g has all 5 of its zeroes (countedwith multiplicities) outside the circle.
9
Chapter 3
Spring 2010
3.1 Real Analysis
1. (a) Since the sequence is Cauchy, we can pick a subsequence fnk suchthat
‖fnk − fnk+1‖p <
1
2k
for each k ≥ 1. We claim that this subsequence converges pointwise almosteverywhere. To prove this, we consider the series
f := fn1+
∞∑k=1
(fnk+1− fnk)
and
g := |fn1|+
∞∑k=1
|fnk+1− fnk |
with partial sums SK(f) and SK(g), respectively. Then the triangle inequalityimplies
‖SK(g)‖p ≤ ‖fn1‖p +
K∑k=1
‖fnk+1− fnk‖p < ‖fn1
‖p + 1,
and we have |SK(g)|p |g|p, so the monotone convergence theorem implies∫|g|p = lim
K→∞
∫|SK(g)|p <∞.
So g ∈ Lp([0, 1]), and since |f | ≤ g, we have f ∈ Lp([0, 1]) as well. In particular,the series defining f converges (is less than infinity) almost everywhere. Byconstruction, SK−1(f) = fnk+1
, so fnk(x)→ f(x) for almost every x. (See alsoStein 159-60.)
10
(b) Consider the sequence χ[0,1], χ[0, 12 ], χ[ 12 ,1], χ[0, 14 ], . . . . The sequence isCauchy and converges to the 0 function in the norm. But the pointwise limitfails to exist anywhere since every x ∈ [0, 1] occurs in infinitely many intervalsof the form [ m2n ,
m+12n ].
2. (a) The key thing is to show that if fn → 0 in Lp, then the same istrue in Lq (continuity). Here we must use the fact that Lp ⊂ Lq; it is not truein general (e.g. look at R). I don’t know how to do this.
(b) Suppose for contradiction that infE′∈M µ(E′) = 0, so that we can chooseE1, E2, . . . inM′ with µ(En)→ 0. Then set fn = µ(En)−1/pχEn , which is sim-
ple and hence measurable, so that ‖fn‖p = 1, but ‖fn‖q =[µ(En)1−q/p]1/q =
µ(En)−ε for some ε > 0. But this means that ‖fn‖q → ∞, contradicting theboundedness from (a).
3. Note that πV is the map that sends x ∈ H to the unique closest elementin V in the norm of H. One shows that x− πV (x) ⊥ V .
(a) For any x ∈ H, x− πV (x) ⊥ V , so by the Pythagorean theorem
‖x‖2 = ‖(x− πV (x)) + πV (x)‖2 = ‖x− πV (x)‖2 + ‖πV (x)‖2.
Thus ‖πV (x)‖ ≤ ‖x‖, whence ‖πV ‖ ≤ 1. It follows that ‖πV ‖ = 1 since‖πV (x)‖ = ‖x‖ for nonzero x ∈ V .
πV is the identity on V since the unique closest element in V of an elementin V is itself. Thus πV is idempotent.
We need to show that (πV x, y) = (x, πV y) for all x, y ∈ H. This followseasily from
0 = (x− πV x, πV y) = (x, πV y)− (πV x, πV y)
and likewise0 = (πV x, y − πV y) = (πV x, y)− (πV x, πV y).
(b) Set V := im(P ), which is a subspace since it is the image of a linearoperator, and which we claim is closed. For if xn is a Cauchy sequence in V ,with limit y ∈ H, then P idempotent and continuous implies xn = Pxn →Py, whence Py = y proves that y ∈ V . So V is closed.
Using the fact that P and πV are self-adjoint,
(Px, y) = (x, Py) = (x, πV Py) = (πV x, Py) = (PπV x, y) = (πV x, y)
for all x, y ∈ H. Thus ((P − πV )x, y) = 0 for all x, y, so (P − πV )x is perpen-dicular to all of H, hence 0, and this holds for each x, whence P = πV .
11
(Did we not need the assumption that the operator norm is 1?)
4. (a) The map `1Λ // (`∞)∗ defined by
Λ(ξ)(η1, η2, . . . ) =∑i
ξiηi
is well defined since
|Λ(ξ)(η)| ≤∑i
|ξi||ηi| ≤ ‖η‖∞∑i
|ξi| = ‖η‖∞‖ξ‖1 <∞.
This also shows that the operator norm ‖Λ(ξ)‖ ≤ ‖ξ‖1, which is in fact equalitysince if η = (ξ1/|ξ1|, ξ2/|ξ2|, . . . ), which has ‖η‖∞ = 1, then
|Λ(ξ)(η)| =
∣∣∣∣∣∑i
ξiξi/|ξi|
∣∣∣∣∣ =
∣∣∣∣∣∑i
|ξi|
∣∣∣∣∣ = ‖ξ‖1.
So Λ is norm-preserving. It is clearly injective since if ξ 6= 0, then some ξi 6= 0,hence the corresponding η with a 1 in the ith position and 0’s everywhere elsesatisfies Λ(ξ)(η) = ξi 6= 0, hence Λ(ξ) 6= 0.
For the failure of surjectivity, I need to show that (`∞)∗ contains functionalsthat vanish on all η such that ηn → 0 as n→∞. But I don’t know how to findsuch a functional. Perhaps as some sort of limit? Heeeeeelp!!!
(b) We need 1p + 1
q = 1, but I’m not sure about other conditions.
5. Suppose for contradiction that there is a set E such that the subsetΣ := x ∈ E : fn(x) /∈ E for all n ≥ 1 has positive. We will prove that
Σ, f(Σ), . . . , fn(Σ)
are almost disjoint (their pairwise intersections have measure 0) for each n ≥ 0,which will contradict the fact that µ(X) < ∞ since then the fact that f ismeasure preserving will imply that
µ(Σ ∪ f(Σ) ∪ · · · ∪ fn(Σ)) = (n+ 1)µ(Σ)→∞.
(Note that if µ(X) =∞, for instance X = R with the Lebesgue measure, thenf can be a shift by a constant and the claim fails.)
We use induction. When n = 0, the claim is trivial. Now suppose forinduction that for some k ≥ 0,
Σ, f(Σ), . . . , fk(Σ)
are almost disjoint. Since f is measure preserving,
f(Σ), f2(Σ), . . . , fk+1(Σ)
12
must then also be almost disjoint. But Σ has empty intersection with eachf j(Σ) by the definition of Σ, whence
Σ, f(Σ), f2(Σ), . . . , fk+1(Σ)
are almost disjoint.
3.2 Complex Analysis
6. (a) The singularities of f inside the circle of radius 2 are simple poles atz = ±1. The corresponding residues are
res1(f) = limz→1
(z − 1) · f = 1
andres−1(f) = lim
z→−1(z + 1) · f = 1.
Thus∫γf = 2πi(1 + 1) = 4πi.
(b) In the annulus 1 < |z| < 3, we have
1
z + 1=
1/z
1 + 1/z=
1
z
(1− 1
z+
1
z2− · · ·
);
1
z − 1=
1/z
1− 1/z=
1
z
(1 +
1
z+
1
z2+ · · ·
);
1
z + 3=
1/3
1 + z/3=
1
3
(1− z
3+z2
9− · · ·
).
Expanding f by partial fractions, we get
f(z) =6z + 2
(z2 − 1)(z + 3)=
1
z + 1+
1
z − 1− 2
z + 3,
so the coefficient of 1z of the Laurent series of f is 1 + 1 = 2, which agrees with
the sum of the residues in (a). To check the integral, note that every term ofthe Laurent series has a primitive (and hence its integral vanishes) except theone corresponding to 1
z , and integrating 1z over any circle centered at the origin
gives 2πi. Multiplying by the coefficient, namely the residue, yields 4πi.
7. Skip.
13
8. (a) If f is analytic and injective on U , then f ′ is nonzero on U . Assumefor contradiction that f ′(z0) = 0. Let the power series for f about z0 be
f(z) = a0 + a1(z − z0) + a2(z − z0)2 + · · · .
Then a1 = 0 since the derivative vanishes, so
f(z)− a0 = ak(z − z0)k +G(z),
where ak 6= 0, k ≥ 2, and G vanishes to order at least k+1 at z0. Now for smallw write
f(z)− a0 − w = F (z) +G(z), where F (z) := ak(z − z0)k − w.
The idea is that we can choose a small enough circle C around z0 and w smallenough so that |F (z)| > |G(z)| on C and F has k roots inside C. Then byRouche’s theorem, F (z)+G(z) has k roots inside C, contradicting the injectivityof f .
One way to pick C and w is the following. By the vanishing of G, choose
ε > 0 so small that |G(z)| < |ak|2 εk on the circle Cε(z0). Then choose 0 < δ < 1
2εk
and set w = δak, so that
|F (z)| = |ak||(z − z0)k − δ| ≥ |ak|1
2εk > |G(z)|
on Cε(z0). Then F has its zeroes inside Cε(z0) since δ1/k < ε, so we are done.
(b) f ′ never 0 on U does not ensure that f is injective on U . Let U = C andf(z) = ez. Then f ′(z) = ez has no roots, but e0 = e2πi, so f is not injective.
Another example is f(z) = z2, on the punctured disc centered at the origin.
9. There is a mistake: the upper half disc should be defined as all z with|z| < 2 and Im z ≥ 0. First scale this map by 1
2 , then map to the upper half
plane by z 7→ −(z + 1z ), and finally map to the unit disc with z 7→ i−z
i+z . Allautomorphisms of the unit disc are rotations composed with Blaschke factors.Chasing i, we see that i 7→ i
2 7→32 i 7→
−15 . So to map −1
5 to 0, we use theBlaschke factor ψα(z) = α−z
1−αz , with α = −15 . Thereafter we can compose with
arbitrary rotations since they fix the origin.
10. Skip.
14
Chapter 4
Fall 2009
4.1 Real Analysis
1. All the fn, f are integrable since they are bounded by f1, which is integrable.By assumption ∫
|fn − f | =∫
(fn − f) =
∫fn −
∫f → 0.
SetEδ = x ∈ X : fn(x)− f(x) ≥ δ for all n,
which is measurable since it is the intersection⋂n(fn − f)−1([δ,∞]). Let E be
the set of all x ∈ X for which fn(x) 6→ f(x), which is measurable since it is theunion
⋃k E1/k. Then∫
|fn − f | ≥∫E1/k
|fn − f | ≥m(E1/k)
k
for all n, so m(E1/k) = 0. Thus m(E) = 0 as well.If we don’t assume f1 is integrable, then all bets are off. For instance, if
X is a set with infinite measure (e.g. R), fn ≡ 2 for all n, and f ≡ 1, then∫fn →
∫f since both sides are infinite, but fn 6→ f for all x.
2. For boundedness, simply note that
|f(x)| ≤∫ ∞−∞|f(y)| dy = ‖f‖1 <∞.
For continuity, note that f(x)e−2πixy → f(x)e−2πix0y for all y ∈ R as x → x0
by the continuity of the exponential, so f(x) → f(x0) by the dominated con-vergence theorem.
15
3. The function Hf // C sending y 7→ limn→∞ (xn, y) = limn→∞(y, xn)
is well-defined and linear. We will show that f is bounded/continuous. Thenby the Riesz representation theorem, there exists x ∈ H such that f(y) = (y, x)for all y ∈ H. Thus
limn→∞
(xn, y) = f(y) = (y, x) = (x, y)
gives the desired result.To prove boundedness, define the collection of bounded linear functionals
HTn // C by
Tny = (y, xn).
By assumption, supn |Tny| < ∞ for each y, so by the uniform boundednessprinciple, supn ‖Tn‖ < ∞. Thus there is some M ≥ 0 such that |Tny| ≤ M‖y‖for all y ∈ H, whence
|f(y)| =∣∣∣ limn→∞
(y, xn)∣∣∣ ≤ lim
n→∞|(y, xn)| ≤M‖y‖
proves that f is bounded.
4. Since f ∈ L∞(X), |f | ≤ ‖f‖∞ a.e. Thus∣∣∣∣∫ fg
∣∣∣∣ ≤ ∫ |fg| ≤ ‖f‖∞ ∫ |g|by monotonicity and the fact that we can ignore a set of measure 0 when inte-grating. This proves that T is bounded and that ‖T‖ ≤ ‖f‖∞. Next we willprove that ‖T‖ ≥ ‖f‖∞. We need to show that given ε > 0, there is a g suchthat ∣∣∣∣∫ fg
∣∣∣∣ > (‖f‖∞ − ε)∫|g|.
By definition of ‖f‖∞, the set
Eδ := x ∈ X : |f(x)| > ‖f‖∞ − δ, δ > 0
has positive measure, and it is measurable since f is measurable. Set g(x) :=χE(x)f(x)/|f(x)|, which is in L1(X) since m(Eδ) ≤ m(X) <∞. Then∣∣∣∣∫ fg
∣∣∣∣ =
∫Eδ
|f | > m(Eδ)(‖f‖∞ − δ) = (‖f‖∞ − δ)∫|g|,
so taking δ = ε gives the desired result. Note that the strict inequality dependson the fact that m(Eδ) > 0.
16
5. The claim is false. Consider the sequence of intervals
In :=
[0, 1],
[0,
1
2
],
[1
2, 1
],
[0,
1
3
],
[1
3,
2
3
],
[2
3, 1
],
[0,
1
4
], . . .
and let χIn be the corresponding sequence of characteristic functions. Then‖χIn‖2 =
√µ(In) → 0 but the sequence χIn(x) has infinitely many 0’s and
1’s for each x, and therefore does not converge.
4.2 Complex Analysis
6. We prove Schwarz’s lemma.
Lemma 1 (Schwarz’s lemma). Let ∆f // ∆ be holomorphic such that f(0) =
0, where ∆ is the open unit disc. Then
(i) |f(z)| ≤ |z| for all z ∈ ∆;
(ii) if equality holds in (i) for some z0 6= 0, then f is a rotation;
(iii) |f ′(0)| ≤ 1, and if equality holds then f is a rotation.
Proof. (i): Since f(0) = 0, f(z)/z is holomorphic in ∆. If |z| = r < 1, we have∣∣∣∣f(z)
z
∣∣∣∣ ≤ 1
r,
and by maximum modulus this 1r bound holds for all |z| ≤ r. Letting r → 1
shows that 1 is an upper bound for all z ∈ ∆.
(ii): By assumption |f(z)| = |z|, so by maximum modulus f is constant.Thus f(z) = cz, where |c| = 1, namely f is a rotation.
(iii): Write g(z) := f(z)/z. Then
g(0) = limz→0
f(z)− f(0)
z= f ′(0),
and we already showed in (i) that g is bounded by 1 in absolute value. If|g(0)| = 1, then by (ii) we see that f is a rotation.
7. We will compute the Laurent series of f(z) = (1 + z2)e1/z to find theresidue of f at the essential singularity z = 0. Since
ez = 1 + z +z2
2!+z3
3!+ · · · ,
17
we have
e1/z = 1 +1
z+
1
2!z2+
1
3!z3+ · · · .
Thus
f(z) = · · ·+ (1
2!+
1
4!)
1
z2+ (1 +
1
3!)1
z+ (1 +
1
2!) + z + z2,
so by the residue theorem the integral of f on the ccw. unit circle is
2πi · (1 +1
3!) =
7πi
3.
8. We begin by proving that f is also holomorphic on the real axis. ByMorera’s theorem, it is enough to check that the integral of g over every triangleis 0. If the triangle T has a side on the real axis, then slightly shifting it ensuresthat T lies entirely in one of the open half planes, hence the integral over anarbitrarily close contour is 0. Continuity then ensures the integral over T is 0.Similarly, if the triangle crosses the real axis, then splitting the triangle intosmaller triangles and shifting them slightly gives the result. So f is entire.
Now define a new function
g(z) :=
f(z) Im(z) ≥ 0;
f(z) otherwise.
Then g is holomorphic on the open set Im(z) < 0 since for z near z0 in the lowerhalf plane we can use the power series expansion for f in the upper half planeto write
f(z) =∑
an(z − z0)n,
whenceg(z) = f(z) =
∑an(z − z0)n
is a power series expansion for g. So g is holomorphic except possibly on thereal axis, and continuous on the real axis since f is continuous and real-valuedthere. Replacing f by g in the above argument thus shows that g is entire.Since f and g agree on the entire upper half plane, they must agree everywhereby the uniqueness of analytic continuation.
9. Skip.
10. Let F (z) = z4 and G(z) = −6z − 3. Then for all |z| = 2, we have
|F (z)| = 16 > 15 = 6 · 2 + 3 ≥ |G(z)|,
18
whence by Rouche’s theorem the functions F and F +G have the same numberof solutions within the unit circle, namely all 4.
19
Chapter 5
Spring 2009
5.1 Real Analysis
1. There are typos in the problem, which I assume is meant to be the follow-ing. Let xn be a sequence of measurable functions with 0 ≤ fn ≤ 1 for each
n such that∫ 1
0fn → 0. Then is it necessarily true that fn → 0 a.e.?
No, this is false. Consider the sequence of intervals
In :=
[0, 1],
[0,
1
2
],
[1
2, 1
],
[0,
1
3
],
[1
3,
2
3
],
[2
3, 1
],
[0,
1
4
], . . .
and let χIn be the corresponding characteristic functions. Then the assumptionshold but χIn → 0 nowhere.
2. If f ∈ L∞, then for each p ≥ 1,
‖f‖pp =
∫|f |p ≤
∫‖f‖p∞ ≤ ‖f‖p∞
since µ(X) = 1. Thus f ∈ Lp for each p.For any ε > 0, we wish to show that there is an N such that for all p ≥ N ,
‖f‖∞−‖f‖p < ε. Choose δ such that ε > δ > 0 and set Eδ := x ∈ X : |f(x)| >‖f‖∞ − δ, which has positive measure. Then
‖f‖pp ≥∫Eδ
|f |p ≥ (‖f‖∞ − δ)pµ(Eδ),
which implies‖f‖p ≥ (‖f‖∞ − δ)µ(Eδ)
1/p.
Since µ(Eδ) has positive measure ≤ 1, µ(Eδ)1/p → 1 as p→∞, whence we get
the desired result since δ < ε.
20
3. First, we may assume T is injective. (Assuming the axiom of choice,we can extend a basis of ker(T ) to a basis for X, obtaining a decompositionX = ker(T ) ⊕ X ′. T |X′ is a compact bijective linear map from X ′ to Y .)Suppose for contradiction that Y is infinite dimensional. Let y1, y2, . . . be anormal basis of Y . Since T is bijective, there exist unique x1, x2, . . . such thatTxn = yn. Then xn cannot be bounded since if it were, compactness ofT would ensure the existence of a convergent subsequence Txnk, which isimpossible. So we may assume |xn| → ∞
This basis method does not seem so good. I can’t even prove that the yksdon’t converge!
Help!
4. Define bounded linear functionals HTn // C by Tny = 〈y, vn〉 for
all y ∈ H. By assumption, this collection of linear functionals is bounded foreach y ∈ H, hence bounded in the norm by the uniform boundedness principle.Thus there exists M ≥ 0 such that |Tny| ≤ M‖y‖ for all n, y. In particular,‖vn‖2 = |Tnvn| ≤M‖vn‖ implies ‖vn‖ ≤M for all n.
5. Skip.
5.2 Complex Analysis
6. Since f(z) sin z → 1 as z → 0 and sin z = z− z3/3! + · · · has a zero of order1 at 0, the limit implies that f has a pole of order 1 at 0, with residue 1. Thusthe residue theorem shows that the integral of f on the ccw. unit circle is 2πi.
7. Since f is bounded by B outside DR, all its poles must be in DR,which is compact, so there can be only finitely many poles, at a1, . . . , an. Theng(z) = f(z)(z − a1) · · · (z − an) is entire and satisfies
|g(z)| ≤ C|z|n
for some constant C ≥ 0. We claim that this implies that g is a polynomial ofdegree ≤ n, for which it suffices to show that g(k)(0) = 0 for all k > n (look atthe power series). This is proved by using the Cauchy inequalities:
|g(k)(0)| ≤ k!
RkCRn → 0
21
as R→∞ when k > n. Thus
f(z) =g(z)
(z − a1) · · · (z − an)
expresses f as a rational function.
8. Fix a point z0 ∈ Ω. Define F (z) =∫γf(z) dz, where γ is any curve con-
sisting of line segments parallel to the axes, starting at z0 and ending at z. Thisis well-defined since our assumption ensures that the definition is independentof the curve chosen. We now prove that F is holomorphic, with derivative f .
For h so small that z + h ∈ Ω, note that
F (z + h)− F (z) =
∫ξ
f(w) dw,
where ξ is any curve from z to z + h composed of line segments parallel to theaxes. Actually, we insist that ξ is of minimal length (as direct as possible fromz to z + h, to ensure so that we can use the continuity of f to bound part ofthe integral, as follows. Since f is continuous, write f(w) = f(z) +ψ(w), whereψ(w)→ 0 as w → z. Thus∫
ξ
f(w) dw = f(z)
∫ξ
dw +
∫ξ
ψ(w) dw.
Since∫ξdw = h and ∣∣∣∣∫
ξ
ψ(w)
∣∣∣∣ ≤ |h| supw∈ξ|ψ(w)|,
with the supremum tending to 0 as h→ 0, we see that
limh→0
F (z + h)− F (z)
h= f(z),
as desired.
9. As stated, the problem is false since, for instance, f(z) = z + z2 hasf(0) = 0, f ′(0) = 1, but f is not the identity map. Clearly we are meant toassume that f maps Q to Q. Let z0 ∈ Q denote the point at which f(z0) = z0
and f ′(z0) = 1. I want to use an argument similar to the Schwarz lemma but Idon’t know how. Riemann mapping theorem to map Q to the unit disc? Butthen the derivative may change...
10. The idea is to rewrite the integral as a complex integral that reducesto the desired integral when we use the parametrization z = eiθ. Let γ be the
22
ccw. unit circle and consider the integral
1
i
∫γ
1
z(a+
z− 1z
2i
) dz =1
i
∫γ
2i
2aiz + z2 − 1dz
= 2
∫γ
1
(z + i(a−√a2 − 1))(z + i(a+
√a2 − 1))
dz.
From the left side, we see that taking z = eiθ and dz = ieiθ dθ gives the desiredintegral. On the other hand, the right side shows that the integrand has polesat −i(a ±
√a2 − 1). The one corresponding to (+) clearly does not lie in the
unit circle, but we claim that −i(a −√a2 − 1) always does for a > 1, namely
that f(a) := a −√a2 − 1 < 1. For this it suffices to argue that f(1) = 1, and
that f ′(a) < 0 for a > 1, hence f is strictly decreasing for a > 1. This is easy:
f ′(a) = 1− 2a
2√a2 − 1
= 1− a√a2 − 1
< 1− 1 = 0.
Now we proceed to use the residue theorem. The residue at the pole −i(a−√a2 − 1) is
limz→−i(a−
√a2−1)
2
z − i(a+√a2 − 1)
=2
−2ai=i
a,
so the residue theorem shows that the integral is
2πi(i
a) = −2π
a.
23
Chapter 6
Fall 2008
6.1 Real Analysis
1. (a) Ta is clearly linear since multiplication in R distributes over addition.Tax ∈ `∞ since
‖Tax‖∞ = sup |(Tax)n| = sup
∣∣∣∣∣n∑i=1
aixi
∣∣∣∣∣ ≤∞∑i=1
|aixi| ≤ ‖x‖∞‖a‖1 <∞,
which also shows that ‖Ta‖ ≤ ‖a‖1.
(b) To prove equality, note first that if a = 0, then Ta = 0 and the resultis trivial. So suppose a 6= 0 and let x be the sequence defined by
xn =
an/|an| when an 6= 0;
0 when an = 0.
(xn is the sign of an or 0). Then ‖xn‖∞ = 1 since a 6= 0, and
sup
∣∣∣∣∣n∑i=1
aixi
∣∣∣∣∣ = sup
∣∣∣∣∣n∑i=1
|ai|
∣∣∣∣∣ = ‖a‖1 = ‖a‖1‖xn‖∞.
2. (a) Suppose f ∈ L2. Let E be those points x ∈ X for which |f(x)| ≤ 1.Then∫
|f | =∫E
|f |+∫X−E
|f | ≤ µ(E) +
∫X−E
|f |2 ≤ µ(E) + ‖f‖22 <∞,
so f ∈ L1.
24
Since |f | is measurable and non-negative, there is a sequence of simple func-tions ϕn such that ϕn |f |. Thus the monotone convergence theorem guar-antees that
∫ϕn →
∫|f |, and likewise
∫ϕ2n →
∫|f |2. If we can prove that
‖ϕ‖1 ≤√µ(X)‖ϕ‖2
holds for all non-negative simple functions ϕ, then we get the inequality for allf ∈ L2 since
‖ϕn‖1 ≤√µ(X)‖ϕn‖2 ≤
√µ(X)‖f‖2
by monotonicity, and the inequality is preserved upon taking the limit of theleft side.
So let ϕ be a simple function, with canonical form ϕn =∑Nk=1 akχEk , where
the Ek are disjoint finite measurable sets and the ak are distinct positive realnumbers. Then(∫
ϕ
)2
=
(N∑k=1
akµ(Ek)
)2
=
N∑k=1
a2kµ(Ek)2 +
∑1≤j<k≤N
ajakµ(Ej)µ(Ek). (∗)
On the other hand,(N∑k=1
µ(Ek)
)∫ϕ2 =
(N∑k=1
µ(Ek)
)(N∑k=1
a2kµ(Ek)
)
=
N∑k=1
a2kµ(Ek)2 +
∑1≤j<k≤N
(a2j + a2
k)µ(Ej)µ(Ek). (∗∗)
Since ajak ≤ a2j+a2
k for non-negative real numbers aj , ak, we see that (∗) ≤ (∗∗).Thus
‖ϕn‖1 =
∫ϕn ≤
√√√√ N∑k=1
µ(Ek) · ‖ϕn‖2 ≤√µ(X)‖ϕn‖2,
as desired.To see that ‖i‖ ≥
√µ(X), simply take f ≡ 1.
(b) Suppose for contradiction that there is a sequence En of measur-able sets such that µ(En) → 0. Taking a subsequence if necessary, we mayassume µ(En) ≤ 1
4n . Then define a sequence of simple functions ϕn byϕn = 1√
µ(En)χEn . Then∥∥∥∥∥
N∑n=1
ϕn
∥∥∥∥∥1
≤N∑n=1
∫ϕn =
N∑n=1
√µ(En) ≤
∞∑n=1
√µ(En) ≤
∞∑n=1
1
2n≤ 1.
Taking the limit of the left side, we see that f :=∑∞n=1 ϕn is well-defined and
in L1. But
‖f‖22 ≥N∑n=1
∫ϕ2n =
N∑n=1
1 = N
25
for each N , whence f /∈ L2, contradicting the assumption.
3. For only if, suppose f = zg for some z ∈ C∗. If h ∈ H such that〈g, h〉 = 0, then
〈f, h〉 = 〈zg, h〉 = z〈g, h〉 = 0,
so there is no h with the stated properties.For if, suppose that for every h ∈ H orthogonal to g, 〈f, h〉 6= 1. Then in
fact 〈f, h〉 = 0 is necessary since otherwise we could scale h appropriately. LetS be the subspace spanned by g, which is closed, and consider H = S ⊕ S⊥.Choose λ such that f = λg+h, where h ∈ S⊥. Then h ⊥ g and therefore h ⊥ fimply the equalities
λ〈g, g〉 = 〈f, g〉 =1
λ〈f, f〉,
whence ‖f‖ = |λ|‖g‖. But the Pythagorean theorem gives
‖f‖2 = |λ|2‖g‖2 + ‖h‖2,
whence h = 0, so f = λg as desired.
4. Define g(x, y) = f(x)χ[y,1]×[0,1](x, y), which is non-negative and mea-surable since f(x) is measurable on [0, 1] × [0, 1] (see Stein 85). Note that g isreally just
g(x, y) =
f(x) if x ≥ y;
0 otherwise.
By Fubini’s theorem, we obtain∫[0,1]
(∫[0,1]
f(x)χ[y,1]×[0,1](x, y) dy
)dx =
∫[0,1]
(∫[0,1]
f(x)χ[y,1]×[0,1](x, y) dx
)dy,
in the extended sense, where the left side yields∫[0,1]
xf(x) dx
and the right side yields ∫[0,1]
(∫[0,y]
f(x) dx
)dy.
This proves the claim.
5. Skip.
26
Chapter 7
Spring 2008
7.1 Real Analysis
1. Set fn := fχEn , where
En := x ∈ X : f(x) ≤ n.
Then fn f , so∫fn →
∫f . Thus given ε > 0, we can choose N > 0 such that∫
f −∫fn < ε/2 whenever n ≥ N . Now if we choose δ > 0 such that δ < ε/2N ,
then for any measurable E ⊆ X with µ(E) < δ, we compute∫E
f =
∫E
(f − fn) +
∫E
fn ≤∫
(f − fn) + n · µ(E) <ε
2+
nε
2N≤ ε,
which proves the result.
2. Since K is continuous on a compact subset of R2, K is bounded, namelyK ∈ L∞([0, 1]× [0, 1]), with |K| having supremum ‖K‖∞. Then
|(Tf)(x)| =
∣∣∣∣∣∫
[0,1]
K(x, y)f(y) dy
∣∣∣∣∣ ≤∫
[0,1]
|K(x, y)||f(y)| dy ≤ ‖K‖∞‖f‖1,
so Tf is in fact uniformly bounded, hence square integrable on [0, 1], a set withfinite measure. Since in fact [0, 1] has measure 1, the above inequality impliesthat
‖Tf‖1 :=
∫[0,1]
|Tf | ≤∫
[0,1]
‖K‖∞‖f‖1 ≤ ‖K‖∞‖f‖1,
whence ‖T‖ ≤ ‖K‖∞.
27
3. We can write T = kerT ⊕ (kerT )⊥, where T |(kerT )⊥ gives a bijectionwith im(T ). Then T |(kerT )⊥ is bounded since linear transformations betweenfinite-dimensional Hilbert spaces are bounded, whence T has the same bound.
Bounded linear transformations map closed sets to closed sets since if xk →x, then Txk → Tx. This holds because
‖Tx− Txk‖ = ‖T (x− xk)‖ ≤ ‖T‖‖x− xk.‖
Now, every bounded closed set of a finite-dimensional normed space is com-pact. For this it suffices to show that every bounded sequence in a finite-dimensional vector space has a convergent subsequence. Choosing a normal ba-sis e1, . . . , en, we have ‖a1e1 + · · ·+ anen‖ ≤ |a1|+ · · ·+ |an|, so convergenceis analogous to convergence in Cn, which is guaranteed by Bolzano-Weierstrass.
Then if T is bounded and xk is bounded, Txk is bounded, hence has aconvergent subsequence, so T is compact.
28
