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Kansas State University
Department of Mathematics
Real and Complex Analysis
Qualifying Exams
(New System)
Solution Manual
Compiled by Peter Nguyen
Edited by R. B. Burckel
How to Use this Manual
This solution manual has been constructed so that each chapter represents a
specific semester’s qualifying exam. Within each chapter, the problems are listed in
the order in which they originally appeared, and moreover, they have been transcribed
almost verbatim. The primary exception to this is the omission of hints.
Following the statement of each problem, the reader will find a list of “key terms.”
These are the important definitions and theorems relevant to the given solution. In
most cases, these are concepts or results that the test-taker is presumed to be knowl-
edgable of. As such, these concepts have been used freely throughout the solutions.
After the list of key terms, the reader will find a solution to the given problem.
These solutions should not be preferred over others, although some are canonical.
iii
Notation
Z := Integers
N := n ∈ Z : n ≥ 1N0 := N ∪ 0R := Real numbers
C := Complex numbers
S+ := S ∩ (0,∞) (S ⊆ C)
D(z, r) := w ∈ C : |z − w| < r (z ∈ C, r > 0)
D(z, r) := w ∈ C : |z − w| ≤ r (z ∈ C, r > 0)
D′(z, r) := w ∈ C : 0 < |z − w| < r (z ∈ C, r > 0)
D := D(0, 1)
D := D(0, 1)
D′ := D′(0, 1)
C(z, r) := w ∈ C : |z − w| = r (z ∈ C, r > 0)
T := C(0, 1)
| | := Lebesgue measure on Rk (k ∈ N)
Ec := x ∈ X : x 6∈ E (E ⊆ X)
1E(x) :=
1 x ∈ E
0 x ∈ Ec
(E ⊆ X)
∐:= disjoint union
B(x, r) := y ∈ X : d(x, y) < r (d a metric on X, r > 0)
iv
Contents
How to Use this Manual iii
Notation iv
Chapter 1. Spring 2004 1
Chapter 2. Fall 2004 9
Chapter 3. Spring 2005 18
Chapter 4. Fall 2005 27
Chapter 5. Spring 2006 38
Index 47
Bibliography 49
v
CHAPTER 1
Spring 2004
Problem 1.1. Let f : [0, 1] → C be continuous and define F : Ω := C\[0, 1] → C
by
F (z) :=
∫ 1
0
f(t)
t− zdt.
Prove that F is holomorphic in Ω.
Key terms: Morera’s Theorem, Fubini’s Theorem, Lebesgue’s Dominated Conver-
gence Theorem.
Solution. Let a1, a2, a3 ∈ C be such that the closed convex hull ∆ of the ori-
ented triangle T := [a1, a2, a3, a1] is contained in Ω. (Here we are using the notation
[a1, ..., an] :=⋃n−1
m=1[am, am+1], where [ai, aj] is the oriented complex interval.) Then
by Fubini’s Theorem, we have
∫
T
F (z)dz =
∫
T
[∫ 1
0
f(t)
t− zdt
]dz =
∫ 1
0
[∫
T
1
t− zdz
]f(t)dt
=
∫ 1
0
[2πiIndT (t)] f(t)dt =
∫ 1
0
0 · f(t)dt = 0.
As this holds for every triangle whose closed convex hull is contained in Ω, Morera’s
Theorem implies that F ∈ H(Ω).
Alternative Solution. This proof does not involve the use of Morera’s Theorem.
First note that since f is continuous and [0, 1] is compact, we may find M ∈ N such
that |f(t)| ≤ M , for every t ∈ [0, 1]. Now, fix z0 ∈ Ω. Since Ω is open, we may find
r > 0 such that D := D(z0, r) ⊂ Ω, and therefore, D∩ [0, 1] = ∅. In consequence, we
see that
(1.1.1) |t− z| ≥ r (z ∈ D, t ∈ [0, 1]).
1
For z ∈ D, define gz : [0, 1] → C by gz(t) := f(t)/[(t − z)(t − z0)]. By (1.1.1), we
conclude that the collection each gz is well-defined, and moreover, |gz(t)| ≤ M/r2,
for every z ∈ D and t ∈ [0, 1]. From this, it follows that gz ∈ L1([0, 1]), for every
z ∈ D. Finally, as it is obvious that limz→z0
gz(t) = gz0(t), for every t ∈ [0, 1], Lebesgue’s
Dominated Convergence Theorem ensures that
(1.1.2) limz→z0
∫ 1
0
f(t)
(t− z)(t− z0)dt =
∫ 1
0
f(t)
(t− z0)2dt.
For each z ∈ D′(z0, r), observe that
F (z)− F (z0)
z − z0
=1
z − z0
[∫ 1
0
f(t)
t− zdt−
∫ 1
0
f(t)
t− z0
dt
]
=1
z − z0
∫ 1
0
f(t)(t− z0)− f(t)(t− z)
(t− z)(t− z0)dt(1.1.3)
=
∫ 1
0
f(t)
(t− z)(t− z0)dt.
Letting z → z0 in (1.1.3) and appealing to (1.1.2) shows that F ′(z0) exists, and in
fact,
F ′(z0) =
∫ 1
0
f(t)
(t− z0)2dt.
As z0 is an arbitrary point of Ω, we are finished.
¤
Problem 1.2. Let Ω := z ∈ C : 1 < |z| < 2. Show that there does not exist a
sequence Pn of polynomials in z such that Pn(z) → 1/z uniformly, for every z ∈ Ω.
Key terms: uniform convergence, index of closed curve.
Solution. Assume, with a view to reach a contradiction, that Pn is a sequence
of polynomials converging to 1/z uniformly. Let γ : [0, 2π] → C be the closed curve
given by γ(t) = 32eit. Since each polynomial is a continuous derivative in Ω, we see
that∫
γPn(z)dz = 0, for every n. On the other hand,
∫γ(1/z)dz = 2πiIndγ(0) =
2πi. The uniform convergence of the Pn on γ to 1/z ensures that 2πi =∫
γdzz
=
limn→∞
∫
γ
Pn(z)dz = 0. This is the desired contradiction.
¤
2
Problem 1.3. Evaluate
∫ ∞
0
dx
1 + x7.
Key terms: contour integration, pole, Residue Theorem, residue.
Solution. Let R > 1 and CR be the path given as the sum of the path1 t 7→ Reit,
for 0 ≤ t ≤ 2π7
and the oriented intervals [Re2πi7 , 0], and [0, R].
Put f(z) := 11+z7 . Note that f has exactly one simple pole lying inside the region
bounded by CR at z0 := eπi7 . Since z7
0 = −1, we have
∫
CR
f(z)dz = 2πiRes(f, z0) = 2πi limz→z0
(z − z0)f(z)(1.3.1)
= 2πi limz→z0
z − z0
z7 − z70
= 2πi1
ddz
z7∣∣z=z0
=2πi
7z60
.
Now, since R > 1, we have∣∣∣ R1+(Reit)7
∣∣∣ ≤ RR7−1
, for all t, whence
∣∣∣∣∣∫ 2π
7
0
iReitdt
1 + (Reit)7
∣∣∣∣∣ ≤∫ 2π
7
0
Rdt
R7 − 1=
2πR
7(R7 − 1).
Thus,∫ 2π
7
0iReitdt
1+(Reit)7→ 0, as R →∞. Furthermore,
∫
CR
f(z)dz =
∫ 2π7
0
iReitdt
1 + (Reit)7−
∫ R
0
z20dr
1 + (z20r)
7+
∫ R
0
dx
1 + x7(1.3.2)
=
∫ 2π7
0
iReitdt
1 + (Reit)7+
(1− z2
0
) ∫ R
0
dx
1 + x7.
Combining (1.3.1) with (1.3.2) and taking the limit as R →∞ gives
∫ ∞
0
dx
1 + x7=
2πi
7z60 (1− z2
0)=
π
7 sin(
π7
) .
The final equality being obtained from the computation:
z60 − z8
0 = e6πi7 − e
8πi7 = −e−
πi7 + e
πi7 = 2i sin(π/7).
¤
1See [RUD, 217 - 218] for a brief discussion on what is meant by a sum of paths.
3
Problem 1.4. Let f : [0,∞) → C be a Lebesgue measurable function satisfying
(†) |f(x)| ≤ aekx,
for some a, k ∈ (0,∞). Put Ω := z ∈ C : Im z > k.(a) Show that for every z ∈ Ω, the function gz : [0,∞) → C defined by gz(t) :=
eitzf(t) is in L1([0,∞)).
(b) Prove that the map F : Ω → C defined by F (z) :=∫∞
0gz(t)dt is holomorphic
in Ω.
Key terms: Lebesgue measure, Lebesgue’s Dominated Convergence Theorem, com-
plex differentiability, (Lebesgue) measurable function.
Solution. (a) Clearly, each gz is measurable, being the product of measurable func-
tions. Now, fix z := x + iy ∈ Ω and observe∫ ∞
0
|gz(t)| dt =
∫ ∞
0
∣∣eit(x+iy)f(t)∣∣ dt ≤
∫ ∞
0
e−ty · aektdt
= a
∫ ∞
0
e−t(y−k)dt =a
y − k< ∞,
since y = Im z > k. This completes the proof of (a).
(b) Fix z0 = x0 + iy0 ∈ Ω. Choose r > 0 such that D := D(z0, r) ⊂ Ω. Let
t ∈ [0,∞) and z ∈ D(z0, r/2) with z 6= z0. Since it∫
[z0,z]eitξdξ = eitz − eitz0 , we have
∣∣∣∣eitz − eitz0
z − z0
− iteitz0
∣∣∣∣ =
∣∣∣∣it
z − z0
∫
[z0,z]
eitξdξ − it
z − z0
∫
[z0,z]
eitz0dξ
∣∣∣∣
=
∣∣∣∣it
z − z0
∫
[z0,z]
(eitξ − eitz0
)dξ
∣∣∣∣
≤ t
|z − z0| · length[z0, z] · supξ∈[z0,z]
∣∣eitξ − eitz0∣∣
≤ t supξ∈[z0,z]
(∣∣eitξ∣∣ +
∣∣eitz0∣∣)(1.4.1)
≤ 2t supξ∈[z0,z]
∣∣eitξ∣∣
= 2t supξ∈[z0,z]
∣∣e−tIm ξ∣∣
= 2te−tk0 ,
4
for some k0 > k, since [z0, z] ⊂ D(z0, r/2) is a compact subset of Ω.2 By the hypothesis
(†), we find that
(1.4.2)∣∣te−tk0f(t)
∣∣ ≤ ate−(k0−k)t ∈ L1([0,∞)).
Combining (1.4.1) and (1.4.2) with Lebesgue’s Dominated Convergence Theorem, we
find that
limz→z0
[F (z)− F (z0)
z − z0
−∫ ∞
0
iteitz0f(t)dt
]
= limz→z0
∫ ∞
0
(eitz − eitz0
z − z0
− iteitz0
)f(t)dt
=
∫ ∞
0
(limz→z0
eitz − eitz0
z − z0
− iteitz0
)f(t)dt
= 0.
Thus, F ′(z0) exists and, in fact, F ′(z0) =∫∞0
iteitz0f(t)dt.
¤
Problem 1.5. A Lebesgue measurable subset of R is said to be negligible if it has
Lebesgue measure zero. Prove that a subset A of R is negligible if and only if there
exists a sequence Un of open subsets of R such that limn→∞
|Un| = 0 and A ⊆ ⋂n Un.
Key terms: Lebesgue measure, Lebesgue measurable set, negligible set, complete
measure space.
Solution. (⇒) As A has Lebesgue measure zero, it follows from the very definition
of Lebesgue (outer) measure that, for every n ∈ N, we may find a sequence In,m∞m=1
of open intervals such that A ⊂ ⋃m In,m and
∑m |Im,n| < 1
n. Put Un :=
⋃m In,m.
Then Un is open, being the union of open sets. Since A ⊆ Un, for each n, it must be
that A ⊆ ⋂n Un. Now, by countable subadditivity, |Un| ≤
∑m |In,m| < 1
n, for each
n ∈ N, whence |Un| → 0, as n →∞.
(⇐) Let Un be the sequence of open sets given by the hypothesis of this impli-
cation. Observe that 0 ≤ |⋂m Um| ≤ |Un|, for every n. Since |Un| → 0, as n →∞ (by
2This is a standard way to circumvent the Mean-Value Theorem, which is unavailable for
complex-valued functions.
5
assumption), it must be that |⋂m Um| = 0. Now,⋂
m Um is Lebesgue measurable, be-
ing the countable intersection of open (and therefore Borel) sets. Thus, A is Lebesgue
measurable, since it is the subset of a set of measure zero and since Lebesgue measure
is complete. Finally, A has measure zero, by monotonicity.
¤
Problem 1.6. Let 1 ≤ q < p ≤ ∞.
(a) Prove that Lp([0, 1]) ⊂ Lq([0, 1]).
(b) Show (by example) that the inclusion in (a) is strict.
(c) Give an example of a measure space (X, A , µ) for which the inclusion Lp(X) ⊇Lq(X) holds.
Key terms: Lebesgue measure, Lp-space, counting measure.
Solution. (a) See the solution for Problem 5.1(a).
(b) Let us first handle the case when p = ∞. Consider the function f(x) :=
x−12q ∈ Lq([0, 1]). For M > 0, let bM := min M−2q, 1. It is easy to check that
x ∈ (0, 1) : |f(x)| > M = (0, bM),
from which it follows that ess sup f = ∞, since |(0, bM)| = bM > 0. Therefore,
f 6∈ L∞([0, 1]). So, the inclusion of (a) is strict in this case.
Now, fix 1 ≤ q < p < ∞ and let r ∈(
1p, 1
q
). Notice that 1 − qr > 0, while
1− pr < 0. So,∫ 1
0
∣∣∣∣1
xr
∣∣∣∣q
dx = limt→0+
∫ 1
t
x−qrdx = limt→0+
1
1− qrx1−qr
∣∣1t
= limt→0+
1
1− qr(1−t1−qr) =
1
1− qr,
while∫ 1
0
∣∣∣∣1
xr
∣∣∣∣p
dx = limt→0+
∫ 1
t
x−prdx = limt→0+
1
1− prx1−pr
∣∣1t
= limt→0+
1
pr − 1(t1−pr − 1) = ∞.
Thus, 1/xr ∈ Lq([0, 1])\Lp([0, 1]), showing the inclusion is strict, in this case as well.
(c) One can construct a multitude of trivial examples easily: merely take X :=
any set, A := any σ-algebra on X (one may pick the power set of X for concreteness),
and finally, µ := the zero measure on X (i.e., µ(A) = 0, for all A ∈ A ).
6
Of course, a nontrivial example is to be sought out. Perhaps, the simplest such
example is to take X := N, A := the power set of N, and µ := counting measure.
Here if f ∈ Lq(X), then by definition∑∞
n=1 |f(n)|q < ∞. Consequently, we may
find N ∈ N such that |f(n)|q ≤ 1, for each n ≥ N . So, for n ≥ N , we have
|f(n)|p ≤ |f(n)|q, whence
∞∑n=1
|f(n)|p ≤N−1∑n=1
|f(n)|p +∞∑
n=N
|f(n)|q < ∞.
This shows that f ∈ Lp(X), and therefore, Lp(X) ⊆ Lq(X).
¤
Problem 1.7. Let p ∈ [1,∞) and suppose fn∞n=1 ⊂ Lp(R) is a sequence that
converges to 0 in the p norm. Prove that we may find a subsequence fnk such that
fnk→ 0 a.e.
Key terms: pointwise convergence, p-norm convergence, almost everywhere, com-
plete metric space, Beppo-Levi Theorem.
Solution. Choose n1 ∈ N such that ‖fn‖p < 12, for all n ≥ n1. Having found this n1,
we may find n2 ∈ N such that n2 > n1 and ‖fn‖p < 122 , for all n ≥ n2. Continuing in
this manner, we may inductively find nk ∈ N such that nk > nk−1 and ‖fn‖p < 12k ,
whenever n ≥ nk. Set fn0 = 0. Note that ‖fnk‖p < 1
2k , for all k ≥ 1.
By the Beppo-Levi Theorem (or Lebesgue’s Monotone Convergence Theorem), we
have ∫
R
[ ∞∑
k=1
|fnk(x)|p
]dx =
∞∑
k=1
‖fnk‖p
p ≤∞∑
k=1
(2−k
)p=
1
2p − 1< ∞.
Consequently,∑∞
k=1 |fnk(x)|p is finite for a.e x ∈ R, and therefore, convergent for
a.e. x ∈ R. For these x, it follows from the convergence of the above series that
|fnk(x)|p → 0, as k →∞, or equivalently, that fnk
(x) → 0, as k →∞.
¤
Problem 1.8. Let f be an entire function having the property that
f(z + m + ni) = f(z) (z ∈ C, m, n ∈ Z).
7
Prove that f is constant.
Key terms: entire function, Liouville’s Theoerem.
Solution. Put S := [0, 1]× [0, 1]. The periodicity condition on f gives that f(C) =
f(S). Since S is compact and f is continuous (it is holomorphic), it follows that f is
bounded on S, and therefore, f is bounded on C. By Liouville’s Theorem, we deduce
that f is constant.
¤
8
CHAPTER 2
Fall 2004
Problem 2.1. Compute
(a)
∫ ∞
0
e−[x]dx, where [x] := max n ∈ Z : n ≤ x, the integer part of x.
(b)
∫ π/2
0
f(x)dx, where f(x) :=
cos x x ∈ R\Qsin x x ∈ Q
.
Key terms: Lebesgue’s Monotone Convergence Theorem, Beppo-Levi Theorem.
Solution. (a) Applying the Beppo-Levi Theorem (or Lebesgue’s Monotone Conver-
gence Theorem) and the fact that [x] = n, on [n, n + 1), for n ∈ N0, we get
∫ ∞
0
e−[x]dx =
∫ ∞
0
e−[x]
[ ∞∑n=0
1[n,n+1)(x)
]dx =
∞∑n=0
∫ ∞
0
e−[x]1[n,n+1)(x)dx
=∞∑
n=0
∫ ∞
0
e−n1[n,n+1)(x)dx =∞∑
n=0
e−n = e/(e− 1).
(b) Put I := [0, π/2]. Since integrals over sets of measure zero are 0, we find that
∫ π/2
0
f(x)dx =
∫
I\Q+
∫
I∩Qf(x)dx =
∫
I\Qcos xdx +
∫
I∩Qsin xdx
=
∫
I\Qcos xdx =
∫
I\Qcos xdx +
∫
I∩Qcos xdx
=
∫ π/2
0
cos xdx = 1.
Note that the above functions are measurable, since in a complete measure space we
may arbitrarily redefine functions over sets of measure zero and retain measurability.
¤
9
Problem 2.2. (a) Does the function
f(x) :=
x sin 1x
0 < x ≤ 1
0 x = 0
have bounded variation?
(b) Compute Var[0,50](ex), the total variation of ex on the interval [0, 50].
Key terms: bounded variation, total variation.
Solution. (a) No. For each integer N ≥ 3, let PN denote the partition 0 := x0 < x1 <
... < xN := 1, where xn := 1π2+π(N−n)
, for n = 1, ..., N − 1. Since sin 1xn
= (−1)N−n,
for each n = 1, ..., N − 1, the total (or absolute) variation TN corresponding to this
partition satisfies
TN ≥N−1∑n=2
∣∣∣ 1π2+π(N−n)
(−1)N−n − 1π2+π[N−(n−1)]
(−1)N−n+1∣∣∣
=N−1∑n=2
1π2+π(N−n)
+ 1π2+π[N−(n−1)]
≥N−1∑n=2
1π
(1
N−n+1+ 1
N−n+2
)(2.2.1)
≥N−1∑n=2
2π(N−n+2)
=N∑
n=3
2πn
.
Now, the total variation T satisfies T ≥ TN , for all N ∈ N. Hence, letting N →∞ in
(2.2.1) verifies that f is not of bounded variation.
(b) Let PN := 0 = x0 < x1 < ... < xN = 50 be a partition of [0, 50]. By the
Mean-Value Theorem, we may find x∗n ∈ [xn−1, xn] such that ex∗n = exn−exn−1
xn−xn−1. Hence,
N∑n=1
|exn − exn−1| =N∑
n=1
ex∗n(xn − xn−1).
Letting our partitions become finer and finer, one version of the Riemannian theory
of integration ensures the sum on the right converges to∫ 50
0exdx = e50 − 1. On the
other hand, the left sum converges to the total variation.
More generally, the above argument shows that one has Var[a,b]f =∫ b
a|f ′(x)| dx,
for every continuously differentiable function f on a compact interval [a, b].
¤
10
Problem 2.3. Suppose f ∈ H(Ω\ 0) and that 0 is either a pole of order m for
f or a removable singularity whose removal results in 0 being a zero of order m for
f . Show that 0 is a first order pole of f ′/f having residue either −m or m.
Key terms: pole, removable singularity, residue
Solution. In either case, we may write f(z) = zng(z), where n := ±m, g ∈ H(Ω)
and g(0) 6= 0. Since g(0) 6= 0 and g is continuous there (it is differentiable), we may
find r > 0 such that g is zero-free on D := D(0, r). Thus,
(2.3.1)f ′(z)
f(z)=
zng′(z) + nzn−1g(z)
zng(z)=
g′(z)
g(z)+ nz−1 (z ∈ D\ 0).
As g ∈ H(D) and g is zero-free on D, it follows that g′/g ∈ H(D), whence (2.3.1)
ensures that f ′/f has a simple pole at 0 with residue n = ±m.
¤
Problem 2.4. Show that a nonconstant entire function maps the plane onto a
dense subset of the plane.
Key terms: Casorati-Weierstrass Theorem, entire function
Solution. The proof to follow mimics the canonical proof of the Casorati-Weierstrass
Theorem. Let f be a nonconstant entire function. And suppose, in order to reach a
contradiction, that f(C) is not dense in C. Then we may find z0 ∈ C and r > 0 such
that f(C) ∩D(z0, r) = ∅. Consequently,
(2.4.1) |f(z)− z0| ≥ r,
for every z ∈ C. This implies that the function g : C → C defined by g := 1f−z0
is entire. Now, (2.4.1) also implies that |g| ≤ 1r
< ∞, and so, Liouville’s Theorem
ensures that g is constant. This, in turn, forces f to be constant. (Specifically,
f = 1c
+ z0, where g = c.) This contradiction implies f(C) is dense in the plane.
¤
Problem 2.5. Let Ω be a bounded region. Suppose that f ∈ C(Ω) ∩ H(Ω) is
zero-free and |f | = C on ∂Ω.
11
(a) Prove that f must be constant.
(b) Is the boundedness condition imposed on Ω essential?
Key terms: Maximum Modulus Principle.
Solution. (a) First note that since f is zero-free, it must be that C > 0. By the
Maximum Modulus Principle,
(2.5.1) |f(z)| ≤ ‖f‖∂Ω := max |f(z)| : z ∈ ∂Ω = C (z ∈ Ω).
Now, since f is zero-free, we evidently have 1/f ∈ C(Ω) ∩H(Ω). Consequently, the
Maximum Modulus Principle applies to 1/f , and therefore,
(2.5.2) |1/f(z)| ≤ ‖1/f‖∂Ω = 1/C (z ∈ Ω).
Inequalities (2.5.1) and (2.5.2) combine to show that |f(z)| = C on Ω. Thus, |f |attains its maximum at every point of Ω, and so, the Maximum Modulus Principle
forces f to be constant on Ω.
(b) Yes. Consider the region Ω := z ∈ C : 0 < Re z. The exponential map ez is
zero-free and holomorphic on Ω, as well as, continuous and identically equal to 1 in
modulus on the boundary of Ω. Yet, this map fails to be constant on Ω. In fact, the
continuity of f shows that this constant must be C.
¤
Problem 2.6. Does there exists a Lebesgue measurable subset E of R such that
|E∩I||I| = 1
2, for every interval I ⊂ R.
Key terms: Lebesgue measure, Lebesgue point, almost everywhere
Solution. No. To verify this, we will need the following general results applied to
the one-dimensional scenario:
Lemma 2.6.1. Let E be Lebesgue measurable subset of Rk. Then for a.e. x ∈ Rk,
the limit
limr→0
|E ∩B(x, r)||B(x, r)|
exists and is equal to 1 for a.e. point of E and 0 for a.e. point of Ec.
12
Proof. Assume for a moment that the lemma holds for measurable sets having
finite measure. For each n ∈ N, put En := E ∩ B(0, n). Each En has finite measure,
being a subset of a set of finite measure, namely B(0, n). Thus, the lemma applies to
each En. For each n, let An be the set of points x ∈ En such that limr→0
|En ∩B(x, r)||B(x, r)|
exists and is equal to 1. Likewise, let Bn be the set of points x ∈ Rk\En for which
the limit exists equal to 0. Since the lemma is assumed to hold for finite measured
sets, we have that∣∣Rk\An
∣∣ = 0 and∣∣Rk\Bn
∣∣ = 0, for each n. Put A :=⋂∞
n=1 An and
B :=⋂∞
n=1 Bn and notice that∣∣Rk\A∣∣ =
∣∣Rk\B∣∣ = 0, being the countable union of
null sets.
Fix x ∈ A. Since En ∩ B(x, r) ⊆ En+1 ∩ B(x, r), for all n, and since⋃∞
n=1 En ∩B(x, r) = E ∩B(x, r), we conclude that
(2.6.1)|E ∩B(x, r)||B(x, r)| = lim
n→∞|En ∩B(x, r)||B(x, r)| .
Let ε > 0 be given. By (2.6.1), we may find N ≥ 1 so large so that
(2.6.2)
∣∣∣∣|E ∩B(x, r)||B(x, r)| − |EN ∩B(x, r)|
|B(x, r)|
∣∣∣∣ < ε/2.
Notice that the quantity on the left is well-defined, since 0 < |B(x, r)| < ∞. As
x ∈ A ⊆ AN , we may find r0 > 0 such that 0 < r < r0 implies∣∣∣ |EN∩B(x,r)|
|B(x,r)| − 1∣∣∣ < ε/2.
This fact along with (2.6.2) and the triangle inequality shows that∣∣∣ |E∩B(x,r)||B(x,r)| − 1
∣∣∣ < ε,
whenever 0 < r < r0. Hence, limr→0
|E ∩B(x, r)||B(x, r)| exists and is equal to 1 for every x ∈ A.
In a similar (in fact, easier) fashion, we see that limr→0
|E ∩B(x, r)||B(x, r)| exists and is equal
to 0 for every x ∈ B. Let F and G be for E and Rk\E as An and Bn are for En
and Rk\En, respectively. Then we have just proved that A ⊆ F and B ⊆ G so that
Rk\F ⊆ Ac and Rk\G ⊆ Bc, and therefore,∣∣Rk\F
∣∣ =∣∣Rk\G
∣∣ = 0.
The above argument guarantees that it suffices to prove the lemma with the extra
condition that |E| < ∞. In this case, put
vr(x) :=1
|B(x, r)|∫
B(x,r)
|1E(y)− 1E(x)| dy
13
for 0 < r and x ∈ Rk. Since E has finite measure, 1E ∈ L1(Rk), whence almost every
point of Rk is a Lebesgue point of 1E; i.e., limr→0
vr(x) = 0, for a.e. x ∈ Rk.
Case 1: x ∈ E. Then |1E(y)− 1E(x)| = 1Ec(y), and therefore,
vr(x) =|Ec ∩B(x, r)||B(x, r)| =
|B(x, r)| − |E ∩B(x, r)||B(x, r)| = 1− |E ∩B(x, r)|
|B(x, r)| (∀r > 0).
Case 2: x ∈ Ec. Then we directly have
vr(x) =|E ∩B(x, r)||B(x, r)| (∀r > 0).
Since limr→0
vr(x) exits and is equal 0 for a.e. x ∈ Rk, the proof is complete.
¤
Proposition 2.6.2. Let E ⊂ Rk be Lebesgue measurable and let ε > 0. Then
there exists x ∈ Rk and r > 0 such that either ε > |E∩B(x,r)||B(x,r)| or 1− ε < |E∩B(x,r)|
|B(x,r)| .
Proof. Since Rk is the disjoint union of E and Ec, either |E| > 0 or |Ec| > 0.
In the former case, the preceding lemma implies the existence of an x ∈ E such that
limr→0
|E ∩B(x, r)||B(x, r)| = 1. Thus, we may find some r > 0 such that 1− ε < |E∩B(x,r)|
|B(x,r)| . So,
we are finished with this situation.
If, however, |Ec| > 0, then applying the preceding lemma again shows that there
exists an x ∈ Ec such that limr→0
|E ∩B(x, r)||B(x, r)| = 0. So, we may find some r > 0 such
that ε > |E∩B(x,r)||B(x,r)| . This finishes the proof.
¤Returning to Problem 2.6, the answer is no. To see this, we simply take k = 1
and ε = 12
in the preceding proposition.
¤
Problem 2.7. Let f(x, y) :=
x2 − y2
(x2 + y2)2(x, y) 6= (0, 0)
0 (x, y) = (0, 0)
.
(a) Show that∫ 1
0
dx
∫ 1
0
f(x, y)dy 6=∫ 1
0
dy
∫ 1
0
f(x, y)dx.
14
(b) To save Fubini’s Theorem, use polar coordinates to verify that
∫ 1
0
∫ 1
0
|f(x, y)| dxdy = ∞.
Key terms: polar coordinates, Fubini’s Theorem
Solution. (a) First, some clarification is due regarding the meaning of the iterated
integrals. The quantity
∫
A
dx
∫
B
f(x, y)dy,
is the Lebesgue integral of the function given by F (x) :=∫
Bf(x, y)dy, provided this
function is well-defined and Lebesgue integrable. Similarly, for the other iterated
integrals in (a). In our present situation,
F (x) :=
∫ 1
0
∂
∂y
(y
x2 + y2
)dy =
y
x2 + y2
∣∣y=1
y=0=
1
x2 + 1,
so that the measurability of F is evident. (F is, in fact, continuous on (0, 1].) Thus,
∫ 1
0
F (x)dx = arctan x∣∣10
= π/4.
On the other hand, since f(x, y) = −f(y, x), we see that the remaining iterated
integral exists and evaluates to −π/4.
15
(b) Note that the circular sector S =reiθ : 0 ≤ r ≤ 1, 0 ≤ θ ≤ π/2
is contained
in the rectangle R = (x, y) : 0 ≤ x, y ≤ 1. Thus,
∫ 1
0
∫ 1
0
|f(x, y)| dxdy =
∫ ∫
R
|f(x, y)| dxdy
≥∫ ∫
S
|f(x, y)| dxdy
=
∫ 1
0
[∫ π/2
0
|f(r cos θ, r sin θ)| dθ
]rdr
=
∫ 1
0
(∫ π/2
0
∣∣∣∣cos 2θ
r2
∣∣∣∣ dθ
)rdr
=
∫ 1
0
(∫ π/4
0
−∫ π/2
π/4
cos 2θdθ
)1
rdr
=
∫ 1
0
1
rdr
= ∞.
Problem 2.8. Prove that f ≡ 0, whenever f ∈ H(C) ∩ L1(R2).
Key terms: Cauchy’s Formula, polar coordinates, Lp-space.
Solution. For r > 0 and z ∈ C, let γr,z be the parameterization of the boundary of
the circle centered at z with radius r given by γr,z(t) := z + reit, 0 ≤ t ≤ 2π. Since f
is entire and C is convex, we may apply Cauchy’s formula to find that
f(z) =1
2πi
∫
γr,z
f(ξ)
ξ − zdξ =
1
2πi
∫ 2π
0
f(z + reit)
(z + reit)− z(rieitdt)
=1
2π
∫ 2π
0
f(z + reit)dt,
and so,
(2.8.1) |f(z)| ≤ 1
2π
∫ 2π
0
∣∣f(z + reit)∣∣ dt.
16
Integrating both ends of (2.8.1) with respect to rdr over [0, R] and appealing to
Fubini’s Theorem gives
R2
2|f(z)| ≤ 1
2π
∫ R
0
[∫ 2π
0
∣∣f(z + reit)∣∣ dt
]rdr
=1
2π
∫
E
|f(z + w)| dλ2(w)(2.8.2)
≤ 1
2π
∫
R2
|f | = 1
2π‖f‖1 ,
where E := (x, y) ∈ D(0, r) : 0 < x, y and λ2 is Lebesgue measure in R2. Since
(2.8.2) is clearly equivalent to |f(z)| ≤ ‖f‖1πR2 , which holds for all z ∈ C and R > 0, we
may deduce that |f(z)| = 0, for all z. Equivalently, f ≡ 0.
¤
17
CHAPTER 3
Spring 2005
Problem 3.1. Compute
(a)
∫ 2π
0
dt
cos t− 2.
(b) Res(zne10/z,∞), n ∈ N.
(c) Res (exp(z2)/z2n+1, 0), n ∈ N.
Key terms: residue, contour integration, Residue Theorem, pole.
Solution. (a) Let γ be the curve defined by tγ7→ eit, for 0 ≤ t ≤ 2π. If we write
z = γ(t), then cos t = (z + z−1)/2, dt = −idz/z, and so,
(3.1.1)
∫ 2π
0
dt
cos t− 2=
∫
γ
(−idz/z)
(z + z−1)/2− 2= −2i
∫
γ
dz
z2 − 4z + 1.
Let f(z) be the function in the integrand of (3.1.1). The quadratic formula shows
that f has simple poles at z0 = 2 − 2√
3 and z1 = 2 + 2√
3. Among these, z0 lies
in the region bounded by the image of γ, while z1 lies outside this region. Thus, the
integral on the far right of (3.1.1) is equal to
2πiRes(f, z0) = 2πi limz→z0
(z − z0)f(z) = 2πi limz→z0
1
z − z1
=2πi
z0 − z1
= −√
3πi
6.
Substitution of this into (3.1.1) yields
∫ 2π
0
dt
cos t− 2= −
√3π
3.
(b) Recall that Res(f,∞) := Res(f(1/z), 0). Now,
(1/z)ne10/(1/z) = e10z/zn = (1/zn)∞∑
k=0
(10z)k/k! =∞∑
k=0
(10k/k!)zk−n.
Hence, the coefficient of the z−1 term is Res(zne10/z,∞) = 10n−1/(n− 1)!.
18
(c) Observe that
exp(z2)/z2n+1 = (1/z2n+1)∞∑
k=0
(z2)k/k! =∞∑
k=0
z2k−2n−1/k!.
So, the coefficient of the z−1 term is Res(exp(z2)/z2n+1, 0) = 1/n!.
¤
Problem 3.2. Let Ejmj=1 be a collection of measurable subsets of [0, 1]. Assume
that this collection has the property that every x ∈ [0, 1] belongs to at least n of the
Ej, where n ≤ m. Prove that |Ej| ≥ nm
, for some j.
Key terms: Lebesgue measure.
Solution. Since every x ∈ [0, 1] belongs to at least n of the Ej, it must be that∑m
j=1 1Ej(x) ≥ n, for every x ∈ [0, 1]. Now, suppose, in order to reach a contradiction,
that |Ej| < nm
, for every j. (As all of our integrals will be over [0, 1] we will suppress
the integrating set.)
n = n · |[0, 1]| =∫
n ≤∫ m∑
j=1
1Ej=
m∑j=1
∫1Ej
=m∑
j=1
|Ej| <m∑
j=1
n
m= n.
The appearance of the strict inequality is the sought contradiction, and therefore, the
desired conclusion follows.
¤
Problem 3.3. Given 1 ≤ p < q < r < ∞, prove that
Lp(X,µ) ∩ Lr(X, µ) ⊆ Lq(X).
Key terms: Lp-space.
Solution. Let f ∈ Lp(X) ∩ Lr(X). Set E := x : |f(x)| > 1 and notice that
|f |p < |f |q < |f |r on E, while |f |r ≤ |f |q ≤ |f |p on Ec. Therefore,
∫
X
|f |q =
∫
E
|f |q +
∫
Ec
|f |q ≤∫
E
|f |r +
∫
Ec
|f |p ≤∫
X
|f |r +
∫
X
|f |p < ∞.
¤
19
Problem 3.4. Let f ∈ L1(X, µ). Prove that for every ε > 0, there exists δ > 0
such that ∣∣∣∣∫
A
fdµ
∣∣∣∣ < ε,
whenever A is a measurable subset of X with µ(A) < δ.
Key terms: Lp-space, Lebesgue’s Monotone Convergence Theorem.
Solution. Let ε > 0 be given. We will actually prove the existence of δ > 0 such
that ∫
A
|f | dµ < ε,
for every measurable A ⊂ X with µ(A) < δ. Once this is shown, we will be finished
because the inequality∣∣∫
Afdµ
∣∣ ≤ ∫A|f | dµ holds. Since we are only concerned with
|f |, we may assume that f = |f |; that is, we assume f is nonnegative.
For n ∈ N, define fn := min f, n. Notice that 0 ≤ fn(x) ≤ fn+1(x) ≤ f(x) and
fn(x) ↑ f(x) holds for every x ∈ X and n ∈ N. So, by Lebesgue’s Monotone Conver-
gence Theorem, we have limn→∞
∫X
fndµ =∫
Xfdµ, or equivalently,
∫X
(f − fn)dµ → 0,
as n → ∞. Hence, we may choose N so large so that∫
X(f − fN)dµ < ε
2. Having
chosen this N , we now choose δ > 0 such that Nδ ≤ ε2. So, for measurable A ⊂ X
with µ(A) < δ, we get∫
A
fdµ =
∫
A
(f − fN)dµ +
∫
A
fNdµ ≤∫
X
(f − fN)dµ +
∫
A
Ndµ
<ε
2+ Nµ(A) <
ε
2+ Nδ < ε.
¤
Problem 3.5. Is there an f ∈ H(D) such that mn(f) → ∞, as n → ∞, where
mn(f) = inf|f(z)| : 1− 1
n< z < 1
?
Key terms: zero-set, Bolzano-Weierstrass, Maximum Modulus Principle.
Solution. No. To see this, assume the contrary. That is, suppose there exists
f ∈ H(D) such that mn(f) → ∞, as n → ∞. Choose n so large so that mn(f) > 0.
We conclude from the definition of mn(f) that f is zero-free on the annulus A :=
20
z : 1− 1
n< z < 1
. Thus, all the zeros of the nonconstant function f must lie in
the compact set D(0, 1− 1
n
). Since the zero-set of f contains no limit point of D, we
conclude (by Bolzano-Weierstrass) that f has only finitely many zeros, say z1, ..., zk.By iterated use of Theorem 10.18 in [RUD], we may write
f(z) = (z − z1)m1 · · · (z − zk)
mkg(z),
where g ∈ H(D) and g has no zero in D. Observe that
|f(z)| = |(z − z1)m1 · · · (z − zk)
mkg(z)| = |z − z1|m1 · · · |z − zk|mk |g(z)|
≤ 2m1+···+mk |g(z)| ,
for every z ∈ D. From this it follows that mn(g) →∞, as n →∞. Since∣∣∣∣g
[(1− 1
2n
)eiθ
]∣∣∣∣ : 0 ≤ θ ≤ 2π
⊂
|g(z)| : 1− 1
n< z < 1
,
for each n, it must be that
min0≤θ≤2π
∣∣∣∣g[(
1− 1
2n
)eiθ
]∣∣∣∣ ≥ mn(g),
for each n. By the corollary to Theorem 10.24 in [RUD] (i.e. the Minimum Modulus
Principle, which is just the Maximum Modulus Principle applied to 1/g, g being
zero-free), we may deduce that |g(0)| ≥ mn(g), for each n, which is impossible if
mn(g) →∞. This is the desired contradiction. ¤
Problem 3.6. Let f : Ω → C, where Ω is open.
(a) True or false: (i) If f is holomorphic and ef is constant, then f is constant.
(ii) Does the answer change if f is only assumed continuous?
(b) Is a bounded function that is harmonic on all of C constant?
Key terms: harmonic function, entire function, Liouville’s Theorem
Solution. Recall that we are assuming Ω to be a region, and so, it is connected.
(a) It suffices to show the following: If f is continuous and ef is constant, then f is
constant, since holomorphic functions are continuous. The non-zero constant function
ef is ew, for some w ∈ C. Hence, ef(z)−w = 1, or equivalently, f(z) − w ∈ 2πiZ, for
21
all z ∈ Ω. As f(z) − w is a continuous mapping of the connected region Ω into the
discrete space 2πiZ, it must be that f − w = 2πin, for some fixed integer n, whence
f(z) = w + 2πn. That is, f is constant.
(b) As the real and imaginary parts of f satisfy the same conditions as f does, we
may without loss of generality assume that f is real-valued. As such, on each closed
disk D(0, N), N ∈ N, f is the real part of a holomorphic function that is defined
uniquely up to a pure imaginary additive constant (see [RUD, 235 - 236]). Adjusting
these constants at each radius, we may conclude that f is the real part of an entire
function, say F = f + iv. Now, since the exponential function is nonnegative and
strictly increasing on the real axis, we find that
∣∣eF∣∣ =
∣∣ef+iv∣∣ =
∣∣ef∣∣ ∣∣eiv
∣∣ =∣∣ef
∣∣ = ef ≤ e|f | ≤ eM ,
where M is a (finite) bound for f . Consequently, we may apply Liouville’s Theorem
to the entire function eF to conclude that eF is constant. From part (a), F must be
constant, and therefore, f is constant.
¤
Problem 3.7. Let I be a nonempty (compact) interval in R and let f : I → C
be real-analytic, by which it is meant that in a neighborhood of every point of I, f is
represented by a convergent power series. Show that we may extend f to a holomorphic
function on some open subset of C that contains I. That is, prove that there exists
open Ω ⊂ C and F ∈ H(Ω) such that I ⊂ Ω and F |I ≡ f .
Key terms: power series, Uniqueness Theorem for Holomorphic Functions, uniform
convergence.
Solution. It turns out that the assumption that I be compact is not essential. So, we
will dispense with it. Now, for each p ∈ I, let∑∞
n=0 cn(p)(x− p)n be the convergent
power series representation of f about p, given by the hypothesis, and let Rp denote
the radius of convergence of this power series. Define fp(z) :=∑∞
n=0 cn(p)(z − p)n,
for z ∈ D(p,Rp). By assumption, Rp > 0. Choose rp such that 0 < rp < Rp. By
22
elementary power series theory, we have that the series for fp converges uniformly and
absolutely on Dp := D(p, rp). Note that fp(x) = f(x), for every x ∈ I∩ [p−rp, p+rp].
Put Ω :=⋃
p∈I Dp and define F : Ω → C by F (z) := fp(z), whenever p ∈ I is such
that z ∈ Dp. Provided that F is well-defined, it is obvious that F ∈ H(Ω), since F is
representable by a power series about every point. Furthermore, that F |I ≡ f is also
self-evident, by construction. Hence, we seek to show that F is well-defined.
Suppose p, q ∈ I are such that z ∈ Dp ∩Dq. Without loss of generality, we may
assume that p < q. Note that the triangle inequality ensures that q− p < rq + rp. So,
if there were an x with p < x < q and x 6∈ Dp ∪Dq, then we would have
q − p = (q − x) + (x− p) ≥ rq + rp > q − p.
This contradiction forces (p, q) ⊂ Dp ∪ Dq. Hence, if (p, q) ∩ Dp ∩ Dq = ∅, then
Dp, Dq would constitute a separation of the connected interval (p, q), see [RUD,
196]. From this impossibility, we may conclude that there exists x ∈ (p, q)∩Dp ∩Dq.
It follows that we may find ε > 0 so small so that (x− ε, x + ε) ⊂ Dp ∩Dq. Since the
interval (x− ε, x+ ε) lies inside I, we see that fp and fq agree on (x− ε, x+ ε). As this
interval obviously has a limit point in the region Dp ∩Dq, the Uniqueness Theorem
for Holomorphic functions guarantees that fp and fq agree on all of Dp ∩Dq, whence
F is well-defined.
¤
Problem 3.8. Let rk∞k=1 be one-to-one enumeration of the rational numbers in
(0, 1]. For each k ∈ N, let pk, qk ∈ N be relatively prime positive integers such that
rk = pk/qk. Define fk : [0, 1] → R by fk(x) := exp[−(pk − xqk)2]. Prove that fk → 0
in measure, as k →∞, yet, for each x ∈ [0, 1], the pointwise limit of the the sequence
fk(x)∞k=1 does not exist.
Key terms: convergence in measure, pointwise convergence
Solution. Let us prove some helpful lemmas:
Lemma 3.8.1. Suppose a, b, c, d ∈ N are such that gcd(a, b) = gcd(c, d) = 1 and
a/b = c/d. Then a = c and b = d.
23
Proof. Since gcd(a, b) = 1, we may find m,n ∈ Z such that am + bn = 1. Now,
combined with the (assumed) fact that a/b = c/d, we get
c = c(am + bn) = acm + bcn = acm + adn = a(cm + dn).
Thus, a divides c. Symmetrically, we find that c divides a, whence a = c. Analogously,
we get that b = d, completing the proof of the lemma.
¤
Corollary 3.8.2. Define ρ : Q+ → N × N by the condition that ρ(r) :=
(ρ1(r), ρ2(r)) if and only if r = ρ1(r)/ρ2(r) and gcd(ρ1(r), ρ2(r)) = 1. Then ρ is
a well-defined injective function.
Proof. For r, s ∈ Q+, the following are equivalent:
(1) r = s. (2) ρ1(r)/ρ2(r) = ρ1(s)/ρ2(s).
(3) ρ1(r) = ρ1(s) and ρ2(r) = ρ2(s). (4) (ρ1(r), ρ2(r)) = (ρ1(r), ρ2(r)).
(5) ρ(r) = ρ(s).
Hence, ρ is well-defined and injective.
¤
Lemma 3.8.3. For each N ∈ N, put SN := r ∈ Q ∩ (0, 1] : ρ2(r) = N. Then∐
N SN = Q ∩ (0, 1] and |SN | ≤ N , for each N .
Proof. That∐
N SN is all of Q ∩ (0, 1] is obvious, since every positive rational
r has a natural number as its denominator ρ2(r). That the SN are disjoint follows
just as easily, since ρ2 is a function and so must assign only one value for any given
input. To see that the cardinality of each SN is at most N , suppose r ∈ SN . In order
that r ≤ 1, we must have ρ1(r) ≤ ρ2(r) = N . There are precisely N natural numbers
having this property, so that there are at most N possible numerators for r. Hence,
there are at most N possible values for r. That is, there at most N elements in SN .
¤
Corollary 3.8.4. limk→∞
qk = ∞.
24
Proof. It is a matter of definition (and the uniqueness element of Lemma 3.8.1)
that qk = ρ2(rk). For M ∈ N, put TM := r ∈ Q ∩ (0, 1] : ρ2(r) ≤ M. Then TM =∐M
N=1 SN , where the SN are as above. Since |SN | ≤ N and TM is the disjoint union,
we get |TM | ≤∑M
N=1 N = M(M + 1)/2. In particular, TM has finite cardinality for
each M . So, for each M ∈ N, the fact that k 7→ rk is injective guarantees that we
may set K := max k ∈ N : rk ∈ TM. Thus, for each k > K, rk 6∈ TM , and therefore,
qk > M . This proves the corollary.
¤
Lemma 3.8.5. Given x ∈ (0, 1] and prime n, there exists ln such that |x− rln | <1/qln.
Proof. Put Ij :=(
j−1n
, jn
], for j = 1, ..., n. Then (0, 1] =
∐nj=1 Ij. So, x ∈ Ij, for
some j, and therefore |x− j/n| < 1/n. By hypothesis, the rational number j/n is rln
for a unique ln ∈ N. Since n is prime gcd(j, n) = 1, and so, if rln = j/n, then pln = j
and qln = n. Hence, |x− rln | = |x− j/n| < 1/n = 1/qln .
¤We are finally prepared to tackle the proof of Problem 3.8. Notice that we must
formally verify that each fk is well-defined, but this is handled by Corollary 3.8.2.
Now, fix x ∈ (0, 1]. Choose ε so that 0 < ε < x. Since there are infinitely many
rational numbers in (0, ε), we may find a subsequence rkm ⊂ (0, ε). Notice that
0 < |x− ε| < |x− rkm|, for each m. Multiplying through by qkm , squaring, and then
taking exponentials gives fkm(x) < exp[−qkm(x− ε)2]. Letting m →∞ and appealing
to Corollary 3.8.4 shows that fkm(x) → 0.
On the other hand, Lemma 3.8.5 yields a (different) subsequence rln such that
|x− rln| < 1/qln , for each n. In consequence, −1 < −(pln−xqln)2 so that fln(x) > 1/e.
It follows that the pointwise limit fk(x) cannot exist. And this is true of every
x ∈ (0, 1].
It remains to be seen that the fk converge in measure to the zero function. To this
end, let 0 < ε < 1 be given. Observe that if x ∈ (0, 1] is such that exp[−(pk−xqk)2] =
|fk(x)| > ε, then |x− rk| < (1/qk)√
ln(1/ε). Thus, x : fk(x) > ε ⊆ (rk−δk, rk +δk),
25
where δk := (1/qk)√
ln(1/ε). Hence, |x : |fk(x)| > ε| ≤ |(rk − δk, rk + δk)| = 2δk.
By Corollary 3.8.4, δk → 0 as k → ∞, whence we may choose K so large so that
2δk < ε, for every k ≥ K. And so, fk → 0 in measure.
¤
Problem 3.9. Let (X, A , µ) be a finite measure space and f a nonnegative mea-
surable function on X. Then f ∈ L1(X) if and only if∞∑
n=0
2nµ(Sn) < ∞, where
Sn := x ∈ X : f(x) ≥ 2n.
Key terms: finite measure space, Lp-space.
Solution. (⇐) For n ∈ N0, let Tn := x ∈ X : 2n ≤ f(x) < 2n+1. Also, put A :=
x : 0 ≤ f(x) < 1. Note that X = Aq (∐
n Tn) so that 1X = 1A +∑∞
n=0 1Tn . Also,
notice that 1Tn ≤ 1Sn , since Tn ⊆ Sn. Thus,
f = f ·(
1A +∞∑
n=0
1Tn
)= f · 1A +
∞∑n=0
f · 1Tn ≤ 1A +∞∑
n=0
2n+11Tn
≤ 1X + 2∞∑
n=0
2n1Sn ∈ L1(X),
since µ(X) and∑∞
n=0 2nµ(Sn) are both assumed finite.
(⇒) Let Tm be as above and notice that Sn =∐∞
m=n Tm, for each n ∈ N0. Define
a : N0 × N0 → 0, 1 by an,m :=
0 m < n
1 m ≥ n. Then
∞∑n=0
2n1Sn =∞∑
n=0
2n
[ ∞∑m=n
1Tm
]=
∞∑n=0
∞∑m=0
2n1Tman,m =∞∑
m=0
∞∑n=0
2n1Tman,m
=∞∑
m=0
(m−1∑n=0
2n
)1Tm =
∞∑m=0
(2m − 1)1Tm ≤∞∑
m=0
2m1Tm ≤∞∑
m=0
f1Tm
= f ·( ∞∑
m=0
1Tm
)≤ f · 1X = f ∈ L1(X).
¤
26
CHAPTER 4
Fall 2005
Problem 4.1. Complete the following:
(a) State and prove the Schwarz lemma for holomorphic self-maps of D.
(b) From (a), conclude that any conformal automorphism of D that fixes zero
must be a rotation.
Key terms: Schwarz Lemma, conformal mapping, rotation, removable singularity,
Maximum Modulus Principle, Chain Rule.
Solution. (a) Here essentially is the version found in [RUD, 254] and its proof:
Theorem 4.1.1 (Schwarz Lemma). Suppose f ∈ H(D) is such that ‖f‖∞ ≤ 1
and zero is fixed by f . Then
(4.1.1) |f(z)| ≤ |z| (z ∈ D)
and
(4.1.2) |f ′(0)| ≤ 1.
If equality holds in (4.1.1) for some z ∈ D′ or if equality holds in (4.1.2), then there
exists u ∈ T such that f(z) = uz, for every z ∈ D.
Proof. Since f has a zero at 0, it follows that the map f(z)/z has a removable
singularity there. (Write f as power series about the origin and factor.) Thus, there
exists g ∈ H(D) such that f(z) = zg(z). Now, fix z ∈ D\ 0. Then for any r such
that |z| < r < 1 (such an r exists, as D is open), we have, by the Maximum Modulus
Principle, that
|f(z)||z| = |g(z)| ≤ max
θ
∣∣g(reiθ)∣∣ = max
θ
∣∣f(reiθ)∣∣
r≤ 1
r,
27
since |f(z)| ≤ 1. Letting r → 1 and multiplying through by |z|, gives |f(z)| ≤ |z|,provided z ∈ D\ 0. Since this inequality is evidently also true at z = 0, we get
(4.1.1). Now, (4.1.2) follows from the fact that f ′(0) = g(0), by the product rule.
Finally, if equality holds in (4.1.1) for some z ∈ D′ or if equality holds in (4.1.2),
then application of Maximum Modulus Principle to g again shows that f(z)/z is a
constant of unit modulus.
(b) Since f is conformal automorphism of D, f−1 is as well (see Theorem 10.33 in
[RUD]). Thus, f and f−1 both satisfy the hypothesis of the Schwarz Lemma, and
therefore,
(4.1.3) |f ′(0)| ,∣∣(f−1)′(0)
∣∣ ≤ 1.
Since z = (f f−1)(z), for all z ∈ D, we get
(4.1.4) 1 = f ′[f−1(0)](f−1)′(0) = f ′(0)(f−1)′(0),
by the Chain Rule. Combining (4.1.3) and (4.1.4), we must have |f ′(0)| = 1, allowing
us to deduce from the second part of the Schwarz Lemma that f is a rotation.
¤
Problem 4.2. Let a ∈ D and define fa(z) :=a− z
1− az, for z ∈ D.
(a) Show that fa is an automorphism of D and is its own inverse.
(b) Show that for every conformal automorphism f of D there exists u ∈ T and
a ∈ D such that f = u · fa.
(c) Verify that (b) implies that every conformal automorphism of D extends to
a homeomorphism of D.
(d) Show that the values a and u are uniquely determined by f .
Key terms: conformal mapping, Maximum Modulus Principle, Chain Rule, home-
omorphism.
28
Solution. (a) If a = 0, then fa(z) = −z, and so, (a) is immediate, in this case. So,
we may assume that a ∈ D\ 0. Note that this forces a−1 ∈ C\D. The computation
(fa fa)(z) =a− a−z
1−az
1− a · a−z1−az
=
a(1−az)−(a−z)1−az
1−az−a(a−z)1−az
=a− |a|2 z − a + z
1− az − |a|2 + az
=z − |a|2 z
1− |a|2 = z
shows simultaneously that, on C\ a−1, fa acts as both left and right inverse to itself.
Thus, fa is a bijection on C\ a−1. It remains to show that fa maps D onto D. We
will, in fact, prove a stronger result: fa maps T onto T and D onto D. To see this,
let u ∈ T. Then
|fa(u)| =∣∣∣∣
a− u
1− au
∣∣∣∣ =|a− u|
|−u(a− u−1)| =|a− u|
|−u| · |a− u| = 1.
It follows that fa(T) ⊂ T, and so, T = (fafa)(T) ⊂ fa(T). These inclusions combine
to give fa(T) = T. Since is fa is nonconstant, we may deduce from the Maximum
Modulus Principle that |fa(z)| < sup |fa(w)| : w ∈ ∂D = T = 1, for every z ∈ D.
Thus, fa(D) ⊂ D. Applying fa to both sides reverses the inclusion, so that fa(D) = D.
(b) Put a := f−1(0) ∈ D. Since f and fa are both conformal automorphisms of
D, it follows from the Chain Rule that f fa is also a conformal automorphism of
D. Moreover, (f fa)(0) = f [fa(0)] = f(a) = f [f−1(0)] = 0; i.e., f fixes 0. Part (b)
of Problem 4.1 above yields u ∈ T such that (f fa)(z) = uz, for every z ∈ D. In
consequence, for every z ∈ D, we have
f(z) = f [(fa fa)(z)] = (f fa)[fa(z)] = ufa(z).
(c) fa and multiplication by u are both homemorphisms of D, and therefore, their
composition is a homeomorphism of D.
(d) This amounts to proving that if u, v ∈ T and a, b ∈ D and
(4.2.1) ufa(z) = vfb(z) (∀z ∈ D),
then u = v and a = b. To verify that this is true, first observe that since (4.2.1) holds
for z = b, u a−b1−ab
= 0. Since u 6= 0, we conclude that a− b = 0, and therefore, a = b.
29
Thus, for every z ∈ D, we have ufa(z) = vfa(z), or equivalently, (u − v)fa(z) = 0,
for every z ∈ D. Since fa is not the zero function, it must be that u− v = 0, and so,
u = v.
¤
Problem 4.3. Suppose the series∞∑
n=0
cnzn converges in D and the function f(z)
that it defines vanishes at 1k, for each k ∈ N. Prove that f ≡ 0.
Key terms: power series, Well-Ordering Principle in N0.
Solution. Assume, with a view to reach a contradiction, that there exists n ∈ N0
such that cn 6= 0. By the Well-Ordering Principle, the set n ∈ N0 : cn 6= 0 has a
least element, say N . Put g(z) :=∑∞
n=0 cN+nzn. By the very definition of N , it must
be that cn = 0, for n = 0, 1, ..., N − 1. In consequence,
f(z) =∞∑
n=N
cnzn = zNg(z).
From this and the fact that f(
1n
)= 0, for every n ∈ N, we conclude that g
(1n
)= 0,
for every n ∈ N. As g is represented by a convergent power series in D, it must be
that g ∈ H(D), whence g is continuous. Thus,
cN = g(0) = g(
limn→∞
1n
)= lim
n→∞g
(1n
)= lim
n→∞0 = 0,
contradicting our choice of N . So, our initial assumption is false, from which we infer
that cn = 0, for every n ∈ N0, and so, f ≡ 0.
¤
Problem 4.4. Complete the following:
(a) State the Open-Mapping Theorem for holomorphic functions.
(b) State the Maximum Modulus Principle for holomorphic functions.
(c) Provide a short deduction of (b) from (a).
(d) Show that the image of any closed subset (of the plane) under a nonconstant
polynomial is closed.
(e) From (a) and (d) deduce the Fundamental Theorem of Algebra.
30
Key terms: Open-Mapping Theorem, Maximum Modulus Principle, Fundamental
Theorem of Algebra, Bolzano-Weierstrass Theorem.
Solution. (a) Here is the terse version found in [RUD, 214]:
Theorem 4.4.1 (The Open-Mapping Theorem). If Ω is a region and f ∈ H(Ω),
then f(Ω) is either a region or a point.
(b) Again, we provide Rudin’s version found on [RUD, 253]:
Theorem 4.4.2 (The Maximum Modulus Principle). If f ∈ H(Ω)∩C(Ω), where
Ω is a bounded region, then the inequality
|f(z)| ≤ ‖f‖∂Ω := supw∈∂Ω
|f(w)|
holds for every z ∈ Ω. Moreover, equality occurs at some point in Ω if and only if f
is constant on Ω.
(c) We may deduce (b) from (a) since no nonempty open subset of C contains an
element of maximum modulus. To see this, suppose Ω is an open subset of C and
z0 ∈ Ω. Since Ω is open, we may find r > 0 such that D(z0, r) ⊂ Ω. Put
z1 :=
z0 + z0r2|z0| z0 6= 0
r2
z0 = 0.
Direct computation shows that |z1 − z0| = r2, and therefore, z1 ∈ D(z0, r) ⊂ Ω.
Moreover, we also have that |z1| > |z0|. In summary, for any point z0 ∈ Ω, we may
find another point z1 in Ω that has larger modulus than z0; i.e., Ω does not contain
a point of maximum modulus.
(d) Let p(z) :=∑m
k=0 akzk be a nonconstant polynomial, with am 6= 0 (m > 0),
and let K be a closed subset of the plane. Let w ∈ p(K). Choose a sequence
wn ⊂ p(K) such that wn → w. As, each wn ∈ p(K), we may find zn ∈ K such that
p(zn) = wn. Now, the triangle inequality gives
|am| |z|m =
∣∣∣∣∣p(z)−m−1∑
k=0
akzk
∣∣∣∣∣ ≤ |p(z)|+m−1∑
k=0
|ak| |z|k ,
31
and therefore,
(4.4.1) |z|m(|am| −
m−1∑
k=0
|ak||z|m−k
)≤ |p(z)| (z 6= 0).
It follows easily from (4.4.1) that |p(z)| → ∞, as z → ∞, since am 6= 0. In conse-
quence, the sequence zn, must be bounded, for otherwise there would be infinitely
many n ∈ N such that |wn| = |p(zn)| > |w|+ 1, contradicting that |wn| → |w|. Thus,
by the Bolzano-Weierstrass Theorem, we may find a subsequence znl that converges
to some element, say z, of K, since K is closed. Finally, from the continuity of p, we
get
w = liml→∞
wnl= lim
l→∞p(znl
) = p(
liml→∞
znl
)= p(z),
and so, w ∈ p(K), proving p(K) is closed.
(e) Here is the form of the Fundamental Theorem of Algebra that we shall prove:
Theorem 4.4.3 (The Fundamental Theorem of Algebra). If p(z) is a polynomial
of positive degree m over C, then there are precisely m zeros of p in C, provided one
accounts for multiplicity.
Proof. Write p(z) :=∑m
k=0 akzk. Choose r > 0 such that |am| rm−∑m−1
k=0 |ak| rk >
|a0| = |p(0)|. Then, as in the proof of (d), the triangle inequality ensures that∣∣p(reiθ)
∣∣ > |p(0)|, for all θ. If p does not have any zeros, then the function f(z) :=
1/p(z) would be entire and satisfy |f(0)| >∣∣f(reiθ)
∣∣, for all θ. This contradicts the
Maximum Modulus Principle. Thus p has at least one zero, say z0, and so, we may
write p(z) = (z− z0)q(z), where q is some polynomial of degree m− 1. By induction,
we may deduce that p has exactly m zeros, counting multiplicities.
¤
Problem 4.5. Let X be a Lebesgue measurable subset of R with |X| = ∞. Con-
struct a function f such that f ∈ Lp(X), for every p ≥ 1, but f 6∈ L∞(X).
Key terms: Lebesgue measure, Lp-space, uniform continuity, Lebesgue’s Monotone
Convergence Theorem, Beppo-Levi Theorem, ratio test, essential supremum
Solution. We will use the following two general results:
32
Theorem 4.5.1. Let E be a Lebesgue measurable subset of R. Then the function
defined by fE(x) := |E ∩ (−x, x)|, for x ∈ [0,∞), is a uniformly continuous map
taking [0,∞) onto [0, |E|). Furthermore, limx→∞
fE(x) = |E|.
Proof. Notice that we may write fE(x) =∫R 1E · 1(−x,x). Let x, y ∈ [0,∞).
Without loss of generality, assume x ≤ y. Observe
|fE(x)− fE(y)| =
∣∣∣∣∫
R1E · 1(−x,x) −
∫
R1E · 1(−y,y)
∣∣∣∣ ≤∫
R1E
∣∣1(−x,x) − 1(−y,y)
∣∣
=
∫
R1E
[1(−y,−x) + 1(x,y)
]= |E ∩ (−y,−x)|+ |E ∩ (x, y)|
≤ 2(y − x) = 2 |x− y| .
The uniform continuity of f is now immediate.
Applying Lebesgue’s Monotone Convergence Theorem to the sequence1E1(−n,n)
shows that limn→∞
fE(n) =
∫
R1E = |E|. As fE(0) = 0, we may deduce that fE maps
[0,∞) onto [0, |E|) from the Intermediate-Value Theorem.
¤
Proposition 4.5.2. Let xn ⊂ [0,∞) and X a Lebesgue measurable set of infi-
nite measure. Then there exists a sequence Xn of pairwise disjoint Lebesgue mea-
surable subsets of X such that |Xn| = xn, for each n ∈ N.
Proof. Let fX be as in Theorem 4.5.1. Since fX maps [0,∞) onto [0, |X|) =
[0,∞), we may find x ∈ [0,∞) such that X1 = X ∩ (−x, x) ⊂ X has measure
x1. Assume inductively that we have found pairwise disjoint measurable subsets
X1, ..., Xn of X such that |Xk| = xk, for 1 ≤ k ≤ n. By additivity, we have that
|X| = |X ∩ (⋂n
k=1 Xck)|+
∑nk=1 |Xk|. Since |X| = ∞ and
∑nk=1 |Xk| =
∑nk=1 xk < ∞,
it must be that |X ∩ (⋂n
k=1 Xck)| = ∞. Mimicking the construction of X1, we may
find Xn+1 ⊂ X∩(⋂n
k=1 Xck) ⊆ X such that |Xn+1| = xn+1. That Xn+1 is disjoint from
each Xk, 1 ≤ k ≤ n is obvious. This completes the induction and yields the desired
sequence of sets.
¤
33
Returning to Problem 4.5, by the results above, we may find a sequence Xn of
pairwise disjoint measurable subsets of X such that |Xn| = 2−n, for every n ∈ N.
Put f :=∞∑
n=1
n1Xn . Since f is nonnegative and since the Xn are pairwise disjoint,
it follows that |f |p = f p =∞∑
n=1
np1Xn . Thus, the Beppo-Levi Theorem and the ratio
test for series give∫
X
|f |p =
∫
X
∞∑n=1
np1Xn =∞∑
n=1
∫
X
np1Xn =∞∑
n=1
np |Xn| =∞∑
n=1
np2−n < ∞,
whence f ∈ Lp(X).
To verify that f 6∈ L∞(X), set SM := x ∈ X : |f(x)| > M, for M ≥ 0. Since
we may always find n ∈ N so large so that M < n and since |f | = f , we have that
SM ⊇ x ∈ X : f(x) = n = Xn.
By monotonicity, |SM | ≥ |Xn| = 2−n > 0, and consequently, the set M : |SM | = 0is empty. Hence, the essential supremum ‖f‖∞ = ∞, and therefore, f 6∈ L∞(X).
¤
Problem 4.6. Let (X, A , µ) be a finite positive measure space. Suppose f ∈L∞(X) is such that ‖f‖∞ > 0. For n ∈ N, put αn :=
∫X|f |n dµ = ‖f‖n
n. Show that
(a) ‖f‖n → ‖f‖∞, as n →∞ and
(b) αn+1/αn → ‖f‖∞, as n →∞.
Key terms: essential supremum, finite measure space, positive measure.
Solution. (a) f ∈ L∞(X) means that ‖f‖∞ < ∞, where
‖f‖∞ := inf M : µ (x : |f(x)| > M) = 0 .
So, for every ε > 0, the set Sε := x ∈ X : |f(x)| > ‖f‖∞ − ε is nonempty, and
moreover, µ(Sε) > 0. Since |f | > ‖f‖∞ − ε on Sε, we have the following
0 < (‖f‖∞ − ε)n µ(Sε) =
∫
Sε
(‖f‖∞ − ε)n dµ ≤∫
Sε
|f |n dµ
≤∫
X
|f |n dµ = ‖f‖nn ,
34
for every n ∈ N, provided ε < ‖f‖∞. Taking the nth root gives
[µ(Sε)]1n (‖f‖∞ − ε) ≤ ‖f‖n .
Since 0 < µ(Sε) ≤ µ(X) < ∞, we have that limn→∞
[µ(Sε)]1n = 1, whence
‖f‖∞ − ε ≤ lim infn→∞
‖f‖n .
As this is true for arbitrary ε > 0, we get
(4.6.1) ‖f‖∞ ≤ lim infn→∞
‖f‖n .
For the reverse inequality, we recall that |f | ≤ ‖f‖∞ a.e. Hence, |f |n ≤ ‖f‖n∞ a.e.,
for each n ∈ N. So, we obtain
‖f‖nn =
∫
X
|f |n dµ ≤∫
X
‖f‖n∞ dµ = ‖f‖∞ µ(X).
Taking the nth root, gives
(4.6.2) ‖f‖n ≤ ‖f‖∞ [µ(X)]1n .
Now, since 0 < ‖f‖∞, we must necessarily have 0 < µ(X), and so, limn→∞
[µ(X)]1n = 1.
By taking the limit superior in (4.6.2), we get lim supn→∞
‖f‖n ≤ ‖f‖∞. This with (4.6.1)
proves that limn→∞
‖f‖n exists and is equal to ‖f‖∞.
(b) Combining∫
X|f |n dµ ≤ ‖f‖n
∞ µ(X) < ∞ with (4.6.1), we see that αn ∈(0,∞), for every n ∈ N. As
∫
X
|f |n+1 dµ =
∫
X
|f | · |f |n dµ ≤ ‖f‖∞∫
X
|f |n dµ,
we obtain αn+1 ≤ ‖f‖∞ αn, for every n, so that
(4.6.3) lim supn→∞
αn+1
αn
≤ ‖f‖∞ .
Now, Holder’s Inequality yields
αn =
∫
X
|f |n dµ ≤[∫
X
(|f |n)n+1
n dµ
] nn+1
[∫
X
1n+1X dµ
] 1n+1
= (αn+1)n
n+1 [µ(X)]1
n+1
=αn+1 [µ(X)]
1n+1
‖f‖n+1
,
35
or equivalently,
(4.6.4)‖f‖n+1
[µ(X)]1
n+1
≤ αn+1
αn
(n ∈ N).
Part (a) along with the fact that limn→∞
[µ(X)]1
n+1 = 1 (as µ(X) ∈ (0,∞)) yields
(4.6.5) ‖f‖∞ = limn→∞
‖f‖n+1
[µ(X)]1
n+1
≤ lim infn→∞
αn+1
αn
,
by (4.6.4). Consideration of (4.6.5) along side (4.6.2) shows that limn→∞
αn+1/αn exists,
and moreover, ‖f‖∞ = limn→∞
αn+1/αn.
Problem 4.7. For p ∈ R, define hp :=∑∞
n=1 np1In, where In :=(
1n+1
, 1n
]. Prove
(a) hp ∈ L1(R), provided p < 1.
(b) h1 ∈ L1weak(R)\L1(R).
(c) hp 6∈ L1weak(R), for p > 1.
Key terms: Lp-space, Weak Lp, Beppo-Levi Theorem, p-test.
Solution. (a) Observe that∫
hp =
∫ ∑np1In =
∑np
∫1In =
∑np
(1
n− 1
n + 1
)(4.7.1)
=∑ np
n(n + 1)=
∑ np−1
n + 1≤
∑ 1
n2−p,
where the change of integration and summation is justified by the Beppo-Levi Theo-
rem. The series on the right of (4.7.1) converges by the p-test for series, since p < 1.
(b) For λ ∈ R, put Sλ := x ∈ R : |h1(x)| > λ. Fix λ ≥ 0. Define N :=
max n ∈ N : n ≤ λ. Note that N ≤ λ, while N + 1 > λ.
It follows that SN ⊇ Sλ so that |Sλ| ≤ |SN |. Now, from the definition of h1
we readily see that h1(x) = |h1(x)| > N if and only if x ∈ ∐∞n=N+1 In. Thus,
SN =∐∞
n=N+1 In, and therefore,
|SN | =∞∑
n=N+1
(1
n− 1
n + 1
)=
1
N + 1− lim
n→∞1
n + 1=
1
N + 1<
1
λ.
In summary,
λ |Sλ| ≤ λ |SN | < λ(1/λ) = 1.
36
As λ ≥ 0 is arbitrary, we may take the supremum over all such λ to conclude that
h1 ∈ L1weak(R).
To see that h1 6∈ L1(R), we note that the computation involved in (4.7.1) is valid
for h1. That is,∫
h1 =∑
1n+1
. Since the series is not finite, neither is the integral.
(c) First, let us set some notation. For x ∈ R, we define [x] := max n ∈ Z : n ≤ x.Now, analogous to the above, let Sλ := x ∈ R : |hp(x)| > λ. Fix p and λ and set
N := min n ∈ N : np ≥ λ. By definition, Np ≥ λ, while (N − 1)p < λ. Now,
x ∈ SN if and only if hp(x) = |hp(x)| > N if and only if x ∈ ∐∞n=[N1/n]+1 In so that
SN =∐∞
n=[N1/n]+1 In, and therefore, |SN | = 1[N1/p]+1
. Since N ≥ λ, SN ⊆ Sλ, and so,
λ |Sλ| ≥ λ |SN | > (N − 1)p
[N1p ] + 1
≥ (N − 1)p
[N1p ] + 1
≥ (N/2)p
2N1p
=Np− 1
p
2p+1.
Now, since p > 1, N →∞, as λ →∞. Hence, we conclude that hp 6∈ L1weak(R).
¤
Problem 4.8. Choose intervals Jn ⊂ (0, 1) in such a way that U =⋃
n Jn is
dense in (0, 1) and yet the set K := (0, 1)\U has positive measure.
Key terms: Lebesgue measure
Solution. Let 0 < ε < 1 and put Q := Q ∩ (0, 1). Note that Q is dense in (0, 1),
yet has measure zero, since it is countable. By the definition of Lebesgue (outer)
measure, we may find open intervals In such that ∪nIn covers Q and∑
n |In| < ε.
Put Jn := In ∩ (0, 1). Then each Jn is an interval (or empty),Jn ⊂ (0, 1), for each
n, and moreover, U =⋃
n Jn ⊂⋃
n In. Hence, |U | =≤ |⋃n In| ≤∑
n |In| < ε. Since
(0, 1) has finite measure, we get
|K| = |(0, 1)\U | = |(0, 1)| − |U | = 1− |U | > 1− ε > 0.
¤
37
CHAPTER 5
Spring 2006
Problem 5.1. Let (X, A , µ) be a finite measure space. Prove:
(a) For 1 ≤ r ≤ s ≤ ∞, we have Ls(X) ⊆ Lr(X).
(b) The result of (a) can fail if µ(X) = ∞.
Key terms: finite measure space, Lp-space, Holder’s Inequality, Lebesgue Measure,
conjugate exponent.
Solution. (a) Clearly, we may assume that r < s, so that 1 < sr. Put p := s
rand let
q be its conjugate exponent.
For f ∈ Ls(X), Holder’s inequality gives
∫
X
|f |r dµ =
∫
X
|f |r · 1Xdµ ≤[∫
X
(|f |r)p
]1/p [∫
X
1qXdµ
]1/q
=
[∫
X
|f |rp
]1/p [∫
X
1Xdµ
]1/q
= ‖f‖rs · [µ(X)]1/q < ∞.
Consequently, f ∈ Lr(X). As this is true for each f ∈ Ls(X), the conclusion follows.
(b) Let X := [1,∞), A := Lebesgue measurable subsets of [1,∞), and µ :=
Lebesgue measure. Then for f(x) := 1x, we have
∫
X
|f |2 dµ =
∫ ∞
1
1
x2dx = 1,
while ∫
X
|f | dµ =
∫ ∞
1
1
xdx = ∞.
Hence, in this situation, we see that L2(X) 6⊆ L1(X), even though 1 ≤ 2.
¤
Problem 5.2. Let A,B ⊆ R be Lebesgue measurable and define h(x) := |(A− x) ∩B|.Show that (a) h is a Lebesgue measurable function; and, (b)
∫R h(x)dx = |A| |B|.
38
Key terms: measurable function, measurable set, convolution, Lebesgue’s Monotone
Convergence Theorem, Fubini’s Theorem, translation invariance.
Solution. For n ∈ N, define An := A ∩ (−n, n) and Bn := B ∩ (−n, n). Notice that
|An| , |Bn| ≤ 2n < ∞ so that 1An ,1Bn ∈ L1(R). Consequently, by Theorem 8.14 in
[RUD] (the Convolution Theorem), the function given by
hn(x) :=
∫
R1An(x + y)1Bn(y)dy
is Lebesgue integrable. In particular, each hn is Lebesgue measurable.
Fix x ∈ R and observe that1(An−x)∩Bn
∞n=1
is a monotone increasing sequence
of measurable functions that converge everywhere to 1(A−x)∩B, whence Lebesgue’s
Monotone Convergence Theorem yields∫
R1(A−x)∩B(y)dy = lim
n→∞
∫
R1(An−x)∩Bn(y)dy = lim
n→∞
∫
R1An(x + y)1Bn(y)dy
= limn→∞
hn(x).
On the other hand,
h(x) = |(A− x) ∩B| =∫
R1A−x(y)1B(y)dy =
∫
R1(A−x)∩B(y)dy,
so that h(x) = limn→∞
hn(x). Since x is arbitrary, we conclude that h is the pointwise
limit of a sequence of measurable functions, and so, h is measurable.
To show (b), we first notice that
1A−x(y) = 1A(x + y) = 1A−y(x)
holds for each x, y ∈ R. Combining this with Fubini’s Theorem and the translation-
invariance of Lebesgue measure gives∫
Rh(x)dx =
∫
R
[∫
R1A−x(y)1B(y)dy
]dx =
∫
R
[∫
R1A−x(y)1B(y)dx
]dy
=
∫
R
[∫
R1A−y(x)dx
]1B(y)dy =
∫
R|A− y|1B(y)dy
=
∫
R|A|1B(y)dy = |A|
∫
R1B(y)dy = |A| |B| .
¤
39
Problem 5.3. Let A be a σ-algebra on a set X and assume µ : A → [0,∞] has
the following properties:
(i) If A1, A2 ∈ F with A1 ∩ A2 = ∅, then µ(A1 ∪ A2) = µ(A1) + µ(A2).
(ii) If An∞n=1 is a sequence in A such that An+1 ⊂ An, for all n ∈ N, and∞⋂
n=1
An = ∅, then limn→∞
µ(An) = 0.
Prove that µ is a positive measure on X.
Key terms: σ-algebra, positive measure, finitely additive, countably additive.
Solution. Since A is a σ-algebra, we have ∅ ∈ A . For n ∈ N, let An = ∅. The
sequence An satisfies the hypothesis of (ii), whence
µ(∅) = limn→∞
µ(∅) = limn→∞
µ(An) = 0.
So, µ is not identically infinite.
It remains to be shown that µ is countably additive. Now, condition (i) combined
with induction readily shows that µ is finitely additive; i.e.,
µ
(k⋃
n=1
An
)=
k∑n=1
µ(An),
whenever A1, ..., Ak ∈ A are pairwise disjoint.
To see that µ is, in fact, countably additive, let Bm∞m=1 be a sequence of mea-
surable pairwise disjoint sets. For n ∈ N, put An :=∞⋃
m=n
Bm. Clearly, An+1 ⊂ An, for
each n. Now, suppose, in order to reach a contradiction, that there exists x ∈∞⋂
n=1
An.
Then x ∈ A1 =∞⋃
m=1
Bm. Since the sets Bm are pairwise disjoint, there is a unique
n(x) ∈ N such that x ∈ Bn(x). Hence, x 6∈∞⋃
m=n(x)+1
Bm = An(x)+1. This contradicts
that x ∈∞⋂
n=1
An. Thus,∞⋂
n=1
An = ∅, and therefore, An has the requisite properties
to invoke (ii). In consequence, limn→∞
µ(An) = 0. Now, notice that as the Bm are
40
pairwise disjoint, the setsn−1⋃m=1
Bm and An are disjoint, provided n ≥ 2. So, for each
n ≥ 2, we have
µ
( ∞⋃m=1
Bm
)= µ
[(n−1⋃m=1
Bm
)∪ An
]= µ
(n−1⋃m=1
Bm
)+ µ(An) =
n−1∑m=1
µ(Bm) + µ(An).
Taking the limit as n →∞ completes the proof.
¤
Problem 5.4. Let h be a bounded Lebesgue measurable function on R having the
property that limn→∞
∫
E
h(nx)dx = 0, for every Lebesgue measurable subset E of finite
measure. Show that for every f ∈ L1(R), we have limn→∞
∫
Rf(x)h(nx)dx = 0.
Key terms: Lebesgue measure, Lp-space, abstract integration, simple function.
Solution. Let s be a complex, measurable, simple function on R that vanishes
outside a set of finite measure. Write s =∑m
k=1 αk1Ak, where αk ∈ C\ 0 and Ak
are Lebesgue measurable and pairwise disjoint. Note that since s vanishes outside a
set of finite measure, each Ak has finite measure. Thus, for every such simple function,
we have
limn→∞
∫
Rs(x)h(nx)dx = lim
n→∞
∫
R
[m∑
k=1
αk1Ak(x)
]h(nx)dx
=m∑
k=1
αk
[lim
n→∞
∫
R1Ak
(x)h(nx)dx
](5.4.1)
= 0.
Now, let f ∈ L1(R) and let M ≥ 0 be a finite bound for h. Fix ε > 0. By Theorem
3.13 in [RUD], we may find a complex, measurable, simple function s that vanishes
outside a set of finite measure such that ‖f − s‖1 < ε/M . Now, for each n ∈ N, we
41
have
∣∣∣∣∫
Rf(x)h(nx)dx
∣∣∣∣ =
∣∣∣∣∫
R[f(x)− s(x)]h(nx)dx +
∫
Rs(x)h(nx)dx
∣∣∣∣
≤∫
R|f(x)− s(x)| |h(nx)| dx +
∣∣∣∣∫
Rs(x)h(nx)dx
∣∣∣∣(5.4.2)
≤ M ‖f − s‖1 +
∣∣∣∣∫
Rs(x)h(nx)dx
∣∣∣∣
< ε +
∣∣∣∣∫
Rs(x)h(nx)dx
∣∣∣∣ .
Letting n →∞ in (5.4.2) and appealing to (5.4.1) gives
∣∣∣∣ limn→∞
∫
Rf(x)h(nx)dx
∣∣∣∣ < ε,
by the continuity of the absolute value function. The proof is completed by letting
ε → 0.
¤
Problem 5.5. Let L : Ω → C be a continuous function having the property that
eL(z) = z, for all z ∈ Ω. Prove that
(a) L ∈ H(Ω).
(b) There is no continuous map on D′ that acts as a right inverse for the expo-
nential map on D′.
Key terms: exponential map, complex differentiability, index of a closed curve (with
respect to a point), Fundamental Theorem of Calculus, complex logarithm.
Solution. (a) Note that since ez is never zero, we may deduce that 0 6∈ Ω. Now,
fix z ∈ Ω and h ∈ C\ 0. Notice that L must be injective, for if L(a) = L(b), then
a = eL(a) = eL(b) = b. Thus, L(z+h)−L(z)h
6= 0, and therefore,
L(z + h)− L(z)
h=
[(z + h)− z
L(z + h)− L(z)
]−1
=
[eL(z+h) − eL(z)
L(z + h)− L(z)
]−1
=
[eL(z)+H(h) − eL(z)
H(h)
]−1
,
42
where H(h) := L(z + h)− L(z). Since L is assumed continuous, we have H(h) → 0,
as h → 0. Combining this with the continuity of the map w 7→ w−1 on C\ 0 and
the fact that ez is its own derivative yields
limh→0
L(z + h)− L(z)
h= lim
h→0
[eL(z)+H(h) − eL(z)
H(h)
]−1
=
[limh→0
eL(z)+H(h) − eL(z)
H(h)
]−1
=[eL(z)
]−1= z−1.
This shows that L is differentiable on Ω, and moreover, L′(z) = z−1.
(b) We will actually show that under the assumptions on L, D′(0, r) 6⊆ Ω, for all
r > 0. To verify this, suppose contrarily that, for some 0 < r < ∞, L is a continuous
right inverse for the exponential map in D(0, r), that is, eL(z) = z, for all 0 < |z| < r.
Put ρ := r2. Let γ : [0, 1] → C be the closed curve given by γ(t) := ρe2πit. By part
(a) L is differentiable and L′(z) = 1/z. Therefore,∫
γ
L′(z)dz =
∫
γ
z−1dz = Indγ(0) = 2πi.
On the other hand, by the Fundamental Theorem of Calculus,∫
γL′(z)dz = 0, since
γ is closed. This contradiction completes the proof.
¤
Problem 5.6. Let f and g be entire functions such that |f | ≤ |g|. Prove
(a) If z0 is a zero of g having multiplicity m, then z0 is a zero of f having
multiplicity at least m.
(b) f is a constant multiple of g.
Key terms: entire function, Liouville’s Theorem, zero-set, removable singularity.
Solution. It is enough to prove (b). To this end, note that it is clear that f vanishes
whenever g does. Thus, we may assume that g is not identically equal to zero. As
such, the set of zeros of g, Z(g), contains no limit point in C. In consequence, for
each (fixed) zero z ∈ Z(g), we may find rz > 0 such that D′(z, rz)∩Z(g) = ∅. Hence,
the function fg
is holomorphic and bounded (by 1) on the punctured disk D′(z, rz),
and therefore, has a removable singularity at z. (See Theorem 10.20 in [RUD].) Let
43
hz be an extension of fg
to a holomorphic function on D(z, rz). Note that |hz| ≤ 1,
since this is true for fg
on D′(z, rz) and since hz is continuous (it is holomorphic) on
D(z, rz).
Now, define h : C → C by h(z) =
f(z)g(z)
z 6∈ Z(g)
hz(z) z ∈ Z(g). Clearly, h is differentiable
at every z in the open set C\Z(g). Furthermore, for z ∈ Z(g), h and hz agree in the
neighborhood D(z, rz) of z, where hz is holomorphic, and so, h is also differentiable
at z. We conclude that h is an entire function. Moreover, the conditions on f and
g and the fact that |hz| ≤ 1, for all z ∈ Z(g), ensure that |h| ≤ 1. By Liouville’s
Theorem, we deduce that h is some constant c. The very definition of h thus gives
that f(z) = cg(z), for all z 6∈ Z(g). Yet |f | ≤ |g| implies that f(z) = 0 = c ·0 = cg(z),
for all z ∈ Z(g). So, we actually have f(z) = cg(z), for all z ∈ C.
¤
Problem 5.7. Verify that
(a)
∫ ∞
−∞
dx
1 + x4=
√2
2π.
(b)
∫ 2π
0
e−iθ exp(eiθ)dθ = 2π.
Key terms: contour integration, Residue Theorem, pole, Laurent series.
Solution. (a) Put f(z) := 1z4+1
and note that f has simple poles at zk = exp[i (2k+1)π
4
],
k = 0, ..., 3. For R > 1, let γR : [0, π] → C be the path t 7→ Reit. Among the four
simple poles, only z0 and z1 lie inside the region bounded by γR and the x-axis. Hence,
∫ R
−R
f(x)dx +
∫
γR
f(z)dz = 2πi
3∑
k=0
Res(f, zk),
or equivalently,
(5.7.1)
∫ R
−R
1
x4 + 1dx = −
∫
γR
f(z)dz + 2πi [Res(f, z0) + Res(f, z1)] .
Now, observe that∣∣∣∣∫
γR
f(z)dz
∣∣∣∣ =
∣∣∣∣∫ π
0
1
(Reit)4 + 1Rieitdt
∣∣∣∣ ≤∫ π
0
R
R4 − 1dt =
πR
R4 − 1.
44
Clearly, the above inequality implies that∣∣∣∫
γRf(z)dz
∣∣∣ → 0, as R → ∞. Applying
this fact to (5.7.1) we may conclude that
(5.7.2)
∫ ∞
−∞
1
x4 + 1dx = 2πi [Res(f, z0) + Res(f, z1)] .
At this stage, it is appropriate to calculate Res(f, z0) and Res(f, z1). The proce-
dure is the same for both:
Res(f, zk) = limz→zk
z − zk
z4 + 1=
1
limz→zk
z4 − z4k
z − zk
=1
ddz
z4∣∣z=zk
=1
4z3k
.
So,
Res(f, z0) + Res(f, z1) =exp
(−i3π4
)+ exp
(−i9π4
)
4=−2i sin (π/4)
4= − i
2√
2.
Substitution of this into (5.7.2) gives the desired result.
(b) Let γ be the closed curve given by θ 7→ eiθ, for θ ∈ [0, 2π]. Then
(5.7.3)
∫ 2π
0
e−iθ exp(eiθ)dθ = −i
∫
γ
ez
z2dz.
By the Residue Theorem, one has∫
γ
ez
z2dz = 2πiRes
(ez
z2, 0
)= 2πi,
since ez
z2 has Laurent expansion 1z2 + 1
z+ 1
2+ z
3!+ ... about the point z = 0. Substituting
this into equation (5.7.3) completes the solution.
¤
Problem 5.8. Let S := z ∈ C : 0 < Re z < 2 and for each z ∈ S let lz(t) :=
tz + 1 − t, for 0 ≤ t ≤ 1. Without assuming that a primitive of f exists, show that
F (z) :=∫
lzf defines a holomorphic function in S such that F ′ = f . Explain why the
conclusion fails whenever f is not holomorphic.
Key terms: Cauchy’s Theorem in a Convex Set, complex differentiability.
Solution. This is problem is a specific example of the following more general result,
whose proof follows the statement:
45
Theorem 5.8.1 (Cauchy’s Theorem in a Convex Set). Let f be a function defined
on a convex open subset Ω of C. Show that f ∈ H(Ω) if and only if there exists
F ∈ H(Ω) such that F ′ = f .
Proof. (⇒) Fix a ∈ Ω. Since Ω is convex, it contains the line segment [a, z], for
every z ∈ Ω. Define F : C → C by F (z) :=
∫
[a,z]
f(ξ)dξ. Now, fix z0 ∈ Ω. For any
z ∈ Ω, Cauchy’s Theorem for a triangle gives∫
[z,z0]
f(ξ)dξ = −∫
[z0,a]
f(ξ)dξ −∫
[a,z]
f(ξ)dξ =
∫
[a,z0]
f(ξ)dξ − F (z)
= F (z0)− F (z).
Hence, for z ∈ Ω\ z0, we have∣∣∣∣F (z0)− F (z)
z0 − z− f(z0)
∣∣∣∣ =
∣∣∣∣1
z0 − z
∫
[z,z0]
f(ξ)dξ − 1
z0 − z
∫
[z,z0]
f(z0)dξ
∣∣∣∣
≤ 1
|z0 − z|∫
[z,z0]
|f(ξ)− f(z0)| d |ξ| .(5.8.1)
Let ε > 0 be given. Using the continuity of f , choose δ > 0 so small so that
|ξ − z0| < δ implies ξ ∈ Ω and |f(ξ)− f(z0)| < ε. Now, if 0 < |z − z0| < δ, then
|ξ − z0| < δ, for each ξ ∈ [z, z0]. Thus, from (5.8.1) we may deduce that∣∣∣∣F (z0)− F (z)
z0 − z− f(z0)
∣∣∣∣ <1
|z0 − z|∫
[z,z0]
εd |ξ| = ε,
for all z ∈ Ω with 0 < |z − z0| < δ. This proves that F is differentiable at z0, and
moreover, F ′(z0) = f(z0). As z0 ∈ Ω is arbitrary, the proof of this implication is
complete.
(⇐) As holomorphic functions have complex derivatives of all orders, F ∈ H(Ω)
with F ′ = f implies that f ∈ H(Ω).
¤
46
Index
additivity
countable, 40
finite, 40
almost everywhere (a.e.), 7, 12
Beppo-Levi Theorem, 7, 9, 32, 36
Bolzano-Weierstrass Theorem, 20, 31
Casorati-Weierstrass Theorem, 11
Cauchy’s Formula, 16
Cauchy’s Theorem
in a Convex Set, 45, 46
Chain Rule, 27, 28
complex logarithm, 42
conformal mapping, 27, 28
conjugate exponent, 38
convergence
in measure, 23
p-norm, 7
pointwise, 7, 23
uniform, 2, 22
convolution, 39
counting measure, 6
differentiability
complex, 4, 42, 45
enssential supremum, 32, 34
entire function, 8, 11, 21, 43
exponential map, 42
Fubini’s Theorem, 1, 15, 39
Fundamental Theorem of Algebra, 31, 32
Fundamental Theorem of Calculus, 42
Holder’s Inequality, 38
harmonic function, 21
homeomorphism, 28
index of a closed curve, 2, 42
integration
abstract, 41
contour, 3, 18, 44
Laurent series, 44
Lebesgue
Dominated Convergence Theorem, 1, 4
measure, 4–6, 12, 19, 32, 37, 38, 41
Monotone Convergence Theorem, 9, 20, 32,
39
point, 12
Liouville’s Theorem, 8, 21, 43
Lp-space, 6, 16, 19, 20, 26, 32, 36, 38, 41
47
Maximum Modulus Principle, 12, 20, 27, 28,
31
measurable
function, 4, 39
set, 5, 39
measure
positive, 34, 40
measure space
complete, 5
finite, 26, 34, 38
metric space
complete, 7
Morera’s Theorem, 1
negligible set, 5
Open-Mapping Theorem, 31
p-test, 36
polar coordinates, 15, 16
pole, 3, 11, 18, 44
power series, 22, 30
ratio test, 32
residue, 3, 11, 18
Residue Theorem, 3, 18, 44
rotation, 27
Schwarz Lemma, 27
σ-algebra, 40
simple function, 41
singularity
removable, 11, 27, 43
translation invariance, 39
uniform continuity, 32
Uniqueness Theorem for Holomorphic
Functions, 22
variation
bounded, 10
total, 10
weak Lp, 36
Well-Ordering Principle (in N), 30
zero-set, 20, 43
48
Bibliography
[CAB] R. V. Churchill, J. W. Brown,
Complex Variables and Applications. (Seventh Edition)
McGraw-Hill (2003)
[PAM] M. H. Protter, C. B. Morrey,
A First Course in Real Analysis. (Second Edition)
Springer-Verlag New York, Inc. (1991)
[ROY] H. L. Royden,
Real Analysis. (Third Edition)
Prentice Hall, Inc. (1988)
[RUD] W. Rudin,
Real and Complex Analysis. (Third Edition)
McGraw-Hill. (1987)
49
