(Princeton lectures in analysis 2) Elias M Stein, Rami Shakarchi - Complex analysis Vol 2 -Princeton University Press (2003)

follows because n(2|z|) ≤ c|z|s
by Theorem 2.1. Since we restricted s to satisfy s > ρ0, we can take
an initial s sufficiently close to ρ0, so that the assertion of the lemma is
established (with s being replaced by s′).
Corollary 5.4 There exists a sequence of radii, r1, r2, . . ., with
rm →∞, such that∣∣∣∣∣
∞∏
n=1
Ek(z/an)
∣∣∣∣∣ ≥ e−c|z|s for |z| = rm.
Proof. Since
∑ |an|−k−1 <∞, there exists an integer N so that
∞∑
n=N
|an|−k−1 < 1/10.
Therefore, given any two consecutive large integers L and L + 1, we can
find a positive number r with L ≤ r ≤ L+ 1, such that the circle of
radius r centered at the origin does not intersect the forbidden discs of
Lemma 5.3. For otherwise, the union of the intervals
In =
[
|an| − 1|an|k+1 , |an|+
1
|an|k+1
]
5. Hadamard’s factorization theorem 151
(which are of length 2|an|−k−1) would cover all the interval [L,L+ 1].
(See Figure 1.) This would imply 2
∑∞
n=N |an|−k−1 ≥ 1, which is a con-
tradiction. We can then apply the previous lemma with |z| = r to con-
clude the proof of the corollary.
I2
I3
a1
I1
a2
In
a3
an
Figure 1. The intervals In
Proof of Hadamard’s theorem
Let
E(z) = zm
∞∏
n=1
Ek(z/an).
To prove that E is entire, we repeat the argument in the proof of Theo-
rem 4.1; we take into account that by Lemma 4.2
|1−Ek(z/an)| ≤ c
∣∣∣∣ zan
∣∣∣∣k+1 for all large n,
and that the series
∑ |an|−k−1 converges. (Recall ρ0 < s < k + 1.) More-
over, E has the zeros of f , therefore f/E is holomorphic and nowhere
vanishing. Hence
f(z)
E(z)
= eg(z)
152 Chapter 5. ENTIRE FUNCTIONS
for some entire function g. By the fact that f has growth order ρ0, and
because of the estimate from below for E obtained in Corollary 5.4, we
have
eRe(g(z)) =
∣∣∣∣ f(z)E(z)
∣∣∣∣ ≤ c′ec|z|s
for |z| = rm. This proves that
Re(g(z)) ≤ C|z|s, for |z| = rm.
The proof of Hadamard’s theorem is therefore complete if we can estab-
lish the following final lemma.
Lemma 5.5 Suppose g is entire and u = Re(g) satisfies
u(z) ≤ Crs whenever |z| = r
for a sequence of positive real numbers r that tends to infinity. Then g
is a polynomial of degree ≤ s.
Proof. We can expand g in a power series centered at the origin
g(z) =
∞∑
n=0
anz
n.
We have already proved in the last section of Chapter 3 (as a simple
application of Cauchy’s integral formulas) that
(6)
1
2π
∫ 2π
0
g(reiθ)e−inθ dθ =
{
anr
n if n ≥ 0
0 if n < 0.
By taking complex conjugates we find that
(7)
1
2π
∫ 2π
0
g(reiθ)e−inθ dθ = 0
whenever n > 0, and since 2u = g + g we add equations (6) and (7) to
obtain
anr
n =
1
π
∫ 2π
0
u(reiθ)e−inθ dθ, whenever n > 0.
For n = 0 we can simply take real parts of both sides of (6) to find that
2Re(a0) =
1
π
∫ 2π
0
u(reiθ) dθ.
6. Exercises 153
Now we recall the simple fact that whenever n �= 0, the integral of e−inθ
over any circle centered at the origin vanishes. Therefore
an =
1
πrn
∫ 2π
0
[u(reiθ)− Crs]e−inθ dθ when n > 0,
hence
|an| ≤ 1
πrn
∫ 2π
0
[Crs − u(reiθ)] dθ ≤ 2Crs−n − 2Re(a0)r−n.
Letting r tend to infinity along the sequence given in the hypothesis of
the lemma proves that an = 0 for n > s. This completes the proof of the
lemma and of Hadamard’s theorem.
6 Exercises
1. Give another proof of Jensen’s formula in the unit disc using the functions
(called Blaschke factors)
ψα(z) =
α− z
1− αz .
[Hint: The function f/(ψz1 · · ·ψzN ) is nowhere vanishing.]
2. Find the order of growth of the following entire functions:
(a) p(z) where p is a polynomial.
(b) ebz
n
for b 
= 0.
(c) ee
z
.
3. Show that if τ is fixed with Im(τ ) > 0, then the Jacobi theta function
Θ(z|τ ) =
∞∑
n=−∞
eπin
2τe2πinz
is of order 2 as a function of z. Further properties of Θ will be studied in Chap-
ter 10.
[Hint: −n2t + 2n|z| ≤ −n2t/2 when t > 0 and n ≥ 4|z|/t.]
4. Let t > 0 be given and fixed, and define F (z) by
F (z) =
∞∏
n=1
(1− e−2πnte2πiz).
Note that the product defines an entire function of z.
154 Chapter 5. ENTIRE FUNCTIONS
(a) Show that |F (z)| ≤ Aea|z|2 , hence F is of order 2.
(b) F vanishes exactly when z = −int + m for n ≥ 1 and n,m integers. Thus,
if zn is an enumeration of these zeros we have∑ 1
|zn|2 =∞ but
∑ 1
|zn|2+� <∞.
[Hint: To prove (a), write F (z) = F1(z)F2(z) where
F1(z) =
N∏
n=1
(1− e−2πnte2πiz) and F2(z) =
∞∏
n=N+1
(1− e−2πnte2πiz).
Choose N ≈ c|z| with c appropriately large. Then, since( ∞∑
N+1
e−2πnt
)
e2π|z| ≤ 1 ,
one has |F2(z)| ≤ A. However,
|1− e−2πnte2πiz| ≤ 1 + e2π|z| ≤ 2e2π|z|.
Thus |F1(z)| ≤ 2Ne2πN|z| ≤ ec′|z|2 . Note that a simple variant of the function F
arises as a factor in the triple product formula for the Jacobi theta function Θ,
taken up in Chapter 10.]
5. Show that if α > 1, then
Fα(z) =
∫ ∞
−∞
e−|t|
α
e2πizt dt
is an entire function of growth order α/(α− 1).
[Hint: Show that
−|t|
α
2
+ 2π|z||t| ≤ c|z|α/(α−1)
by considering the two cases |t|α−1 ≤ A|z| and |t|α−1 ≥ A|z|, for an appropriate
constant A.]
6. Prove Wallis’s product formula
π
2
=
2 · 2
1 · 3 ·
4 · 4
3 · 5 · · ·
2m · 2m
(2m− 1) · (2m + 1) · · · .
[Hint: Use the product formula for sin z at z = π/2.]
7. Establish the following properties of infinite products.
6. Exercises 155
(a) Show that if
∑ |an|2 converges, then the product ∏(1 + an) converges to a
non-zero limit if and only if
∑
an converges.
(b) Find an example of a sequence of complex numbers {an} such that ∑ an
converges but
∏
(1 + an) diverges.
(c) Also find an example such that
∏
(1 + an) converges and
∑
an diverges.
8. Prove that for every z the product below converges, and
cos(z/2) cos(z/4) cos(z/8) · · · =
∞∏
k=1
cos(z/2k) =
sin z
z
.
[Hint: Use the fact that sin 2z = 2 sin z cos z.]
9. Prove that if |z| < 1, then
(1 + z)(1 + z2)(1 + z4)(1 + z8) · · · =
∞∏
k=0
(1 + z2
k
) =
1
1− z .
10. Find the Hadamard products for:
(a) ez − 1;
(b) cos πz.
[Hint: The answers are ez/2z
∏∞
n=1(1 + z
2/4n2π2) and
∏∞
n=0(1− 4z2/(2n + 1)2),
respectively.]
11. Show that if f is an entire function of finite order that omits two values, then
f is constant. This result remains true for any entire function and is known as
Picard’s little theorem.
[Hint: If f misses a, then f(z)− a is of the form ep(z) where p is a polynomial.]
12. Suppose f is entire and never vanishes, and that none of the higher derivatives
of f ever vanish. Prove that if f is also of finite order, then f(z) = eaz+b for some
constants a and b.
13. Show that the equation ez − z = 0 has infinitely many solutions in C.
[Hint: Apply Hadamard’s theorem.]
14. Deduce from Hadamard’s theorem that if F is entire and of growth order ρ
that is non-integral, then F has infinitely many zeros.
15. Prove that every meromorphic function in C is the quotient of two entire
functions. Also, if {an} and {bn} are two disjoint sequences having no finite limit
156 Chapter 5. ENTIRE FUNCTIONS
points, then there exists a meromorphic function in the whole complex plane that
vanishes exactly at {an} and has poles exactly at {bn}.
16. Suppose that
Qn(z) =
Nn∑
k=1
cnk z
k
are given polynomials for n = 1, 2, . . .. Suppose also that we are given a sequence of
complex numbers {an} without limit points. Prove that there exists a meromorphic
function f(z) whose only poles are at {an}, and so that for each n, the difference
f(z)−Qn
(
1
z − an
)
is holomorphic near an. In other words, f has a prescribed poles and principal
parts at each of these poles. This result is due to Mittag-Leffler.
17. Given two countably infinite sequences of complex numbers {ak}∞k=0 and
{bk}∞k=0, with limk→∞ |ak| =∞, it is always possible to find an entire function