Application of Infinite Sums and Products
Entire function of finite order
Munsik Kim
Mathematical Science Department
KAIST
Nov 2021
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Table of Contents
1
Overview
2
Main
3
Discussion
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Overview
In Ch 8.2, we learn that an entire function can be written as
m g (z)
f (z) = z e
∞
Y
En−1
n=1
z an
(1)
where g is entire and En (z)’s are Weierstrass elementary factors so
that f vanishes to orderm at 0 and aj ’s are zeros of f listed with
multiplicities.
Note that the behavior of zeros of a holomorphic function is arbirary.
In Ch 9, we will learn that some hypotheses are imposed about the
general behavior of the function can control the behavior of zeros.
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Overview
In Ch 9.1, we learned that a bounded holomorphic function f on
D(0, 1) which has zero at aj counted according to their multiplicities
and vanishes to order m ≥ 0 at z = 0 can be written as
" ∞
#
Y āj
f (z) = z m ·
−
Ba (z) · F (z)
(2)
|aj | j
j=1
where F is a bounded holomorphic function on D(0, 1), zero free, and
sup |f (z)| =
z∈D(0,1)
sup |F (z)|.
z∈D(0,1)
Here, Ba (z) is the Blaschke factor.
In Ch9.3, we want to extend the domain D(0, 1) to C.
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Finite Order
First, we look at the definition of the ”order” of the entire function :
Definition
Suppose f is an entire function. Then we say that f is of finite order if
there exist a, r > 0 such that
|f (z)| ≤ exp(|z|a )
for |z| > r .
The finite order is one of the hypotheses.
Let λ = λ(f ) be the infimum of all possible a satisfying above.
We call such λ the order of f .
i.e., sin z is of order 1, while exp(exp z) is not of finite order.
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Relation between the zeros and order of f
Suppose {aj } are the zeros of f . Then, to see the relation between
the zeros and order of f , consider
X
|an |−(k+1) .
(3)
n
The infimum of all K ’s for which (3) is finite is an indicator of how
many zeros of f are in D(0, r ) when r is large.
For example, aj = j for j = 1, 2, √
· · ·, then for any K > 0, the sum
converges, that is finite. If aj = j, then K would have to be greater
than 1.
Let n(r ) be the number of zeros in D(0, r ) of our entire function f .
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Main Theorem 1
Theorem 1
Let f be an entire function of finite order λ and satisfying f (0) = 1. If aj
are the zeros of f listed with multiplicities, then
∞
X
|an |−λ−1 < ∞
(4)
n=1
To prove the Theorem 1, we need some technical Lemmas.
Note that we assume that f (0) = 1 6= 0. This can be extended later.
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Lemmas
Lemma 1
If f is an entire function with f (0) = 1, let M(r ) = max|f (z)| where
|z|=r
0 < r < ∞. Then
(log 2) · n(r ) ≤ log M(2r )
Proof of Lemma 1
By Jensen’s formula in Ch 9.1,
n(2r )
0 = log |f (0)| = −
X
k=1
log
Z 2π
2r 1
+
log |f (2re iθ )|dθ,
|ak |
2π 0
(5)
where ak ’s are the zeros of f in D(0, 2r ) counted with multiplicity. Let
|ak−1 | ≤ |ak | for all k = 1, 2, · · · so that a1 , · · · an(r ) ∈ D(0, r ).
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Lemmas
Proof of Lemma 1 - continued
Then
n(r )
X
k=1
log 2 ≤
n(r )
X
k=1
Z 2π
n(2r )
X
2r
2r
1
log
≤
log
=
log |f (2re iθ )|dθ.
ak
ak
2π 0
k=1
Using the equation (5), we can conclude that
1
n(r ) · log 2 ≤
2π
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Z
2π
log |f (2re iθ )|dθ ≤ log M(2r )
(6)
0
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Main Theorem 1 - Proof
Proof
Since f has finite order λ,
M(r ) ≤ exp(r λ+/2 )
for any small > 0 and sufficiently large r > 0.. Using the Lemma 1
n(r ) ≤
Note that
1
1
log M(2r ) ≤
(2r )λ+ 2
log 2
log 2
(7)
n(r )
1 (2r )λ+ 2
≤
log 2 r λ+
r λ+
(8)
converges to 0 as r → ∞ since λ is finite.
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Main Theorem 1 - Proof
Proof
Therefore we can write for sufficiently large r ,
n(r ) ≤ r λ+
(9)
By assumptions, {a1 , a2 , · · · aj } ⊆ D̄(0, |aj |) and we have
j ≤ n(|aj | + δ) ≤ (|aj | + δ)λ+
(10)
for large j and arbitrarily small δ > 0. Letting δ → 0, we get
|aj |−(λ+1) ≤ j −(λ+1)/(λ+)
(11)
P
Then by setting < 1, from the fact that n n1p converges when p > 1,
X
|an |−(λ+1) converges.
(12)
n
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Main Theorem 1
Using the Weierstrass factorization, we want to factorize the entire
function f as follows:
Y
f (z) = exp(g (z)) · z m
E[λ] (z/an )
(13)
n
with Theorem 8.2.2 and Theorem 1. That is, for an entire function of
finite order λ , E[λ] (z/an ) is enough.
Let pn = [λ](:= greatest integer such that ≤ λ) for all n. We can
easily check that it satisfies the assumption
8.2.2 ; (i) |aj |
P in Theorem
−p−1 < ∞.
is not bounded (ii) ∃m ∈ N such that ∞
|a
|
n
n=m+1
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Theorem 8.2.2
Theorem 8.2.2
Let {aj }∞
j=1 be a sequence of nonzero complex numbers with no
accumulation point in C. If {pj } are positive integers that satisfy
∞ X
r pn +1
<∞
|an |
n=1
for every r > 0, then
∞
Y
n=1
Ep n
z an
converges uniformly on compact subsets of C to an entire function F .
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Main Theorem 2
We have examined what λ says about the zeros of f . Now we consider the
exponent g (z). Here, we will prove that an entire function f of finite order
can be written as products as mentioned before.
Theorem 2
If f is an entire function of finite order λ and f (0) = 1, then the
Weierstrass canonical product
f (z) = e g (z) P(z)
(14)
has the property that g is a polynomial of degree less than or equal to λ.
Before prove the Theorem 2, we need some Lemmas.
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Lemmas
Lemma 2
Let f be an entire function of finite order λ with f (0) = 1. Let p > λ − 1
and let {aj } be the zeros of f listed with multiplicities. Suppose that
|a1 | ≤ |a2 | ≤ · · ·. Then, for any z ∈ C,
lim
n(r )
X
r →+∞
ākp+1 · (r 2 − āk z)−p−1 = 0
(15)
k=1
Proof of Lemma 2
Let z be fixed and r > 2|z|. Consider a1 , · · · an(r ) so that they are in
D(0, r ). Then for those ak ’s
|r 2 − āk z| ≤ r 2 −
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r2
r2
=
2
2
Entire function of finite order
(16)
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Lemmas
Proof of Lemma 2 - continued
Hence
2 p+1 2 p+1
=
(17)
r2
r
From the definition of the order and Lemma 1, for any > 0 small and r
large enough,
|āk |p+1 |r 2 − āk z|−p−1 ≤ r p+1
log 2 · n(r )r −p−1 ≤ (log M(2r ))r −p−1
(18)
≤ (2r )λ+ r −p−1
Then
n(r )
X
k=1
2 p+1
1 λ+p++1 λ+−p−1
≤
2
r
. (19)
ākp+1 (r 2 −āk z)−p−1 ≤ n(r )
r
log 2
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Lemmas
Proof of Lemma 2 - continued
Take = p+1−λ
, then the exponent of r is negative so that it goes to zero
2
as r → +∞
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Lemmas
Lemma 3
If f is an entire function of finite order λ and f (0) = 1, then, for
p > λ − 1, p an integer and fixed z ∈ C,
Z 2π
1
2re iθ (re iθ − z)−p−2 log |f (re iθ )|dθ → 0
2π 0
(20)
as r → +∞
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Lemmas
Proof of Lemma 3
Consider the function
φ(w ) =
1
(w − z)p+2
which has residue 0 at z ∈ D(0, r ) and satisfies
I
φ(w )dw = 0.
∂D(0,r )
In parametrized form,
Z
2π
φ(re iθ )ire iθ dθ = 0
0
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Lemmas
Proof of Lemma 3 - continued
Then, we can get
1
2π
Z
2π
2re iθ (re iθ − z)−p−2 log |f (re iθ )|dθ
0
Z 2π
1
=
2re iθ (re iθ − z)−p−2 (log |f (re iθ )| − log M(r ))dθ
2π 0
Z 2π
r −p−2
1
≤
2r ·
(log M(r ) − log |f (re iθ )|)dθ
2π 0
2
(21)
≤ 2p+3 r −p−1 log M(r ) ≤ 2p+3 r λ+−p−1
using the fact that log M(r ) is constant with respect to θ, Lemma 1, and
Jensen’s inequality. Take = (p + 1 − λ)/2.
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Lemmas
Lemma 4
Let f be a non constant entire function of finite order λ and f (0) = 1. Let
{a1 , · · · } be zeros of f listed with multiplicities. Suppose that
|a1 | ≤ |a2 | ≤ · · ·. If p > λ − 1 is an integer, the
X
1
d p h f 0 (z) i
=
−p!
dz p f (z)
(aj − z)p+1
(22)
j
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Lemmas
Proof of Lemma 4
Let r > 2|z|. From Poisson-Jensen formula, we get
log |f (z)| = −
n(r )
X
j=1
r 2 − āj z
1
log
+
r (z − aj )
2π
Z
2π
Re
0
re iθ + z re iθ − z
log |f (re iθ )|dθ.
(23)
By logarithmic differentiation in z
n(r )
n(r )
X
f 0 (z) X
−1
=
(z − aj ) +
āj (r 2 − āj z)−1
f (z)
j=1
j=1
Z 2π
1
2re iθ (re iθ − z)−2 log |f (re iθ )|dθ.
+
2π 0
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(24)
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Lemmas
Proof of Lemma 4
Differentiate both sides of the equation p times to get
n(r )
n(r )
X
X
d p h f 0 (z) i
−p−1
= − p!
(aj − z)
+ p!
ājp+1 (r 2 − āj z)−p−1
( )
dz
f (z)
j=1
j=1
(25)
Z 2π
1
2re iθ (re iθ − z)−2 log |f (re iθ )|dθ.
+ (p + 1)!
2π 0
By Lemma 2, the second term goes to zero as r → ∞. By Lemma 3, the
integral term goes to zero as r → ∞.
d
Note that dz
log |f (z)| =
q
¯
|f (z)| = f (z) · f (z).
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f 0 (z)
f (z) .
We can check by setting f = u + iv and
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Lemmas
Lemma 5
If f is an entire function of finite order λ, f (0) = 1, and if
P(z) =
∞
Y
E[λ]
n=1
z an
is the associated product, then for any integer p > λ − 1 we have
X
d p h P 0 (z) i
1
=
−p!
dz p P(z)
(aj − z)p+1
(26)
j
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Lemmas
Proof of Lemma 5
If PN is the Nth partial product, then it is clear that
N
X
1
d p h PN0 (z) i
=
−p!
+ (error)
p
dz PN (z)
(an − z)p+1
(27)
n=1
By Lemma 4, the first term on the RHS converges to equation (26) and
the error becomes 0 when p > λ − 1, since it is the (p + 1)st derivative of
a polynomial of degree p. Since PN → P normally, then PN0 → P 0
normally so that the LHS converges to the one in the equation (26)
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Main Theorem 2 - Proof
Proof
Differentiate
f (z) = e g (z) P(z)
so that
P 0 (z)
f 0 (z)
= g 0 (z) +
.
f (z)
P(z)
Let p > λ − 1. Differentiating both sides p times and applying Lemmas 4
and Lemma 5 gives
−p!
X
n
X
1
1
d p+1
=
g (z) − p!
p+1
p+1
(an − z)
dz
(an − z)p+1
n
(28)
so that
d p+1
g (z) = 0
(29)
dz p+1
and hence g is a holomorphic polynomial of degree that does not exceed p.
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Main Theorem 2
Note that the hypothesis f (0) = 1 in Theorems and Lemmas is in
effect superfluous since if f vanishes to order m at 0, then
f (z)
f˜(z) = m
z
still have finite order λ and it can be suitably normalized so that
f˜(0) = 1.
Theorem 1 and 2 are called Hadamard factorication theorem
If f is entire of finite order and {aj } the zeros of f , then we define the
rank p if f to be the least non negative integer such that
X
|an |−p−1 < ∞
an 6=0
Using the Theorem 1, such p exists.
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Main Theorem 3
From Theorem 2, we can write
f (z) = c · z m · e g (z) · P(z)
(30)
where g is a polynomial.
Let q be the degree of g . Then we define the genus µ of f to be the
maximum of p and q.
Main Theorem 3
An entire function of finite order λ is also of finite genus µ and µ ≤ λ.
The proof of Theorem 3 is trivial by definition of the genus.
Now we derive a few interesting consequences.
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Main Theorem 4
Main Theorem 4
Let f be an entire function of finite order λ ∈
/ Z. Then there are infinitely
many distinct zj ∈ C such that f (zj ) = c for any c ∈ C.
Proof
Without loss of generalizy, assume c = 0. Now suppose that the
statement is false. Then f −1 ({0}) = {a1 , · · · , aN } for some N ∈ Z and
f (z) = e g (z) · (z − a1 ) · · · (z − aN ).
From the previous Theorem, g is a polynomial of degree not exceeding λ.
However, if > 0 and |z| |aj | for all j, as |z| → ∞,
e |z|
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λ−
|f (z)|
λ+
≤ |e g (z) | = QN
≤ e |z|
| j=1 (z − aj )|
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Main Theorem 4- Proof
Proof - continued
RHS is from the definition of finite order and LHS is trivial. Therefore the
order of e g (z) is λ. However, since the order of e g is the degree q of g and
q = λ, contradicting with λ ∈
/ Z.
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Main Theorem 5
Theorem 5
Let f be a non constant entire function of finite order. Then the image of
f contains all complex numbers except possibly one. If, in addition, f has
non integer order, then f assumes each of these values infinitely many
times.
Proof
If the image of f omits distinct α1 and α2 , then consider f − α1 . Since
this function never vanishes, we can write f (z) − α1 = e g (z) for some
entire function g (z) and Hadamard’s theorem guarantees that g is
polynomial. It is the same for α2 . Since f (z) − α1 omits the value
α2 − α1 , g omit the value log(α2 − α1 ). So if β ∈ C with e β = α1 − α2 ,
then g (z) − β never vanishes, contradicts the Fundamental theorem of
algebra unless g is constant.
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Discussion
In Ch 9.3, we check that, if f is an entire function of finite order, we
can write f as
f (z) = c · e g (z) z m P(z)
(31)
where
P(z)
∞
Y
n=1
E[λ]
z an
where ai ’s are the zeros of f and f vanishes at 0 in order m and g (z)
is a polynomial of degree less than or equal to λ.
For a finite order entire functions, the image of the function contains
all complex number except possibly one.
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