In mathematics , and particularly in the field of complex analysis , the Hadamard factorization theorem asserts that every entire function with finite order can be represented as a product involving its zeroes and an exponential of a polynomial. It is named for Jacques Hadamard .
The theorem may be viewed as an extension of the fundamental theorem of algebra , which asserts that every polynomial may be factored into linear factors, one for each root. It is closely related to Weierstrass factorization theorem , which does not restrict to entire functions with finite orders.
Formal statement [ edit ]
Define the Hadamard canonical factors
E
n
(
z
)
:=
(
1
−
z
)
∏
k
=
1
n
e
z
k
/
k
{\displaystyle E_{n}(z):=(1-z)\prod _{k=1}^{n}e^{z^{k}/k}}
Entire functions of finite
order
ρ
{\displaystyle \rho }
have
Hadamard 's canonical representation:
[1]
f
(
z
)
=
z
m
e
Q
(
z
)
∏
n
=
1
∞
E
p
(
z
/
a
n
)
{\displaystyle f(z)=z^{m}e^{Q(z)}\prod _{n=1}^{\infty }E_{p}(z/a_{n})}
where
a
k
{\displaystyle a_{k}}
are those
roots of
f
{\displaystyle f}
that are not zero (
a
k
≠
0
{\displaystyle a_{k}\neq 0}
),
m
{\displaystyle m}
is the order of the zero of
f
{\displaystyle f}
at
z
=
0
{\displaystyle z=0}
(the case
m
=
0
{\displaystyle m=0}
being taken to mean
f
(
0
)
≠
0
{\displaystyle f(0)\neq 0}
),
Q
{\displaystyle Q}
a polynomial (whose degree we shall call
q
{\displaystyle q}
), and
p
{\displaystyle p}
is the smallest non-negative integer such that the series
∑
n
=
1
∞
1
|
a
n
|
p
+
1
{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{|a_{n}|^{p+1}}}}
converges. The non-negative integer
g
=
max
{
p
,
q
}
{\displaystyle g=\max\{p,q\}}
is called the genus of the entire function
f
{\displaystyle f}
. In this notation,
g
≤
ρ
≤
g
+
1
{\displaystyle g\leq \rho \leq g+1}
In other words: If the order
ρ
{\displaystyle \rho }
is not an integer, then
g
=
[
ρ
]
{\displaystyle g=[\rho ]}
is the integer part of
ρ
{\displaystyle \rho }
. If the order is a positive integer, then there are two possibilities:
g
=
ρ
−
1
{\displaystyle g=\rho -1}
or
g
=
ρ
{\displaystyle g=\rho }
.
Furthermore, Jensen's inequality implies that its roots are distributed sparsely, with critical exponent
α
≤
ρ
≤
g
+
1
{\displaystyle \alpha \leq \rho \leq g+1}
.
For example,
sin
{\displaystyle \sin }
,
cos
{\displaystyle \cos }
and
exp
{\displaystyle \exp }
are entire functions of genus
g
=
ρ
=
1
{\displaystyle g=\rho =1}
.
Critical exponent [ edit ]
Define the critical exponent of the roots of
f
{\displaystyle f}
as the following:
α
:=
lim sup
r
log
r
N
(
f
,
r
)
{\displaystyle \alpha :=\limsup \limits _{r}\log _{r}N(f,r)}
where
N
(
f
,
r
)
{\displaystyle N(f,r)}
is the number of roots with modulus
<
r
{\displaystyle <r}
. In other words, we have an asymptotic bound on the growth behavior of the number of roots of the function:
∀
ϵ
>
0
N
(
f
,
r
)
≪
r
α
+
ϵ
,
and exists a sequence
r
k
such that
N
(
f
,
r
k
)
>
k
r
k
α
−
ϵ
{\displaystyle \forall \epsilon >0\quad N(f,r)\ll r^{\alpha +\epsilon },{\text{ and exists a sequence }}r_{k}{\text{ such that }}N(f,r_{k})>kr_{k}^{\alpha -\epsilon }}
It's clear that
α
≥
0
{\displaystyle \alpha \geq 0}
.
Theorem: [2] If
f
{\displaystyle f}
is an entire function with infinitely many roots, then
α
=
inf
{
β
:
∑
k
|
a
k
|
−
β
<
∞
}
=
1
lim inf
k
log
k
|
a
k
|
{\displaystyle \alpha =\inf \left\{\beta :\sum _{k}|a_{k}|^{-\beta }<\infty \right\}={\frac {1}{\liminf \limits _{k}\log _{k}|a_{k}|}}}
Note: These two equalities are purely about the limit behaviors of a real number sequence
|
a
1
|
≤
|
a
2
|
≤
⋯
{\displaystyle |a_{1}|\leq |a_{2}|\leq \cdots }
that diverges to infinity. It does not involve complex analysis.
Proposition:
α
(
f
)
≤
ρ
{\displaystyle \alpha (f)\leq \rho }
,[3] by Jensen's formula .
Since
f
(
z
)
/
z
m
{\displaystyle f(z)/z^{m}}
is also an entire function with the same order
ρ
{\displaystyle \rho }
and genus, we can wlog assume
m
=
1
{\displaystyle m=1}
.
If
f
{\displaystyle f}
has only finitely many roots, then
f
(
z
)
=
e
Q
(
z
)
∏
k
=
1
n
E
0
(
z
/
a
k
)
{\displaystyle f(z)=e^{Q(z)}\prod _{k=1}^{n}E_{0}(z/a_{k})}
with the function
e
Q
(
z
)
{\displaystyle e^{Q(z)}}
of order
ρ
{\displaystyle \rho }
. Thus by an application of the Borel–Carathéodory theorem ,
P
{\displaystyle P}
is a polynomial of degree
d
e
g
(
Q
)
≤
floor
(
ρ
)
{\displaystyle deg(Q)\leq {\text{floor}}(\rho )}
, and so we have
ρ
=
d
e
g
(
Q
)
=
g
{\displaystyle \rho =deg(Q)=g}
.
Otherwise,
f
{\displaystyle f}
has infinitely many roots. This is the tricky part and requires splitting into two cases. First show that
g
≤
floor
(
ρ
)
{\displaystyle g\leq {\text{floor}}(\rho )}
, then show that
ρ
≤
g
+
1
{\displaystyle \rho \leq g+1}
.
Define the function
g
(
z
)
:=
∏
k
=
1
∞
E
p
(
z
/
a
k
)
{\displaystyle g(z):=\prod _{k=1}^{\infty }E_{p}(z/a_{k})}
where
p
=
floor
(
α
)
{\displaystyle p={\text{floor}}(\alpha )}
. We will study the behavior of
f
(
z
)
/
g
(
z
)
{\displaystyle f(z)/g(z)}
.
Bounds on the behaviour of |E p | [ edit ]
In the proof, we need four bounds on
|
E
p
|
{\displaystyle |E_{p}|}
:
For any
r
∈
(
0
,
1
)
{\displaystyle r\in (0,1)}
,
|
1
−
E
p
(
z
)
|
≤
O
(
|
z
|
p
+
1
)
{\displaystyle |1-E_{p}(z)|\leq O(|z|^{p+1})}
when
|
z
|
≤
r
{\displaystyle |z|\leq r}
.
For any
r
∈
(
0
,
1
)
{\displaystyle r\in (0,1)}
, there exists
B
>
0
{\displaystyle B>0}
such that
ln
|
E
p
(
z
)
|
≥
−
B
|
z
|
p
+
1
{\displaystyle \ln |E_{p}(z)|\geq -B|z|^{p+1}}
when
|
z
|
≤
r
{\displaystyle |z|\leq r}
.
For any
r
∈
(
0
,
1
)
{\displaystyle r\in (0,1)}
, there exists
B
>
0
{\displaystyle B>0}
such that
|
E
p
(
z
)
|
≥
|
1
−
z
|
e
−
B
|
z
|
p
{\displaystyle |E_{p}(z)|\geq |1-z|e^{-B|z|^{p}}}
when
|
z
|
≥
r
{\displaystyle |z|\geq r}
.
ln
|
E
p
(
z
)
|
≤
O
(
|
z
|
p
+
1
)
{\displaystyle \ln |E_{p}(z)|\leq O(|z|^{p+1})}
for all
z
{\displaystyle z}
, and
ln
|
E
p
(
z
)
|
≤
O
(
|
z
|
p
)
{\displaystyle \ln |E_{p}(z)|\leq O(|z|^{p})}
as
|
z
|
→
∞
{\displaystyle |z|\to \infty }
.
These are essentially proved in the similar way. As an example, we prove the fourth one.
ln
|
E
n
(
z
)
|
=
R
e
(
−
z
n
+
1
∑
k
=
0
∞
z
k
k
+
n
+
1
)
≤
|
z
|
n
+
1
|
g
(
z
)
|
{\displaystyle \ln |E_{n}(z)|=Re\left(-z^{n+1}\sum _{k=0}^{\infty }{\frac {z^{k}}{k+n+1}}\right)\leq |z|^{n+1}|g(z)|}
where
g
(
z
)
:=
∑
k
=
0
∞
z
k
k
+
n
+
1
{\displaystyle g(z):=\sum _{k=0}^{\infty }{\frac {z^{k}}{k+n+1}}}
is an entire function. Since it is entire, for any
R
>
0
{\displaystyle R>0}
, it is bounded in
B
(
0
,
R
)
{\displaystyle B(0,R)}
. So
ln
|
E
n
(
z
)
|
=
O
(
|
z
|
n
+
1
)
{\displaystyle \ln |E_{n}(z)|=O(|z|^{n+1})}
inside
B
(
0
,
R
)
{\displaystyle B(0,R)}
.
Outside
B
(
0
,
R
)
{\displaystyle B(0,R)}
, we have
ln
|
E
n
(
z
)
|
=
ln
|
1
−
z
|
+
R
e
(
Q
(
z
)
)
=
o
(
|
z
|
)
+
O
(
|
z
|
n
)
<
O
(
|
z
|
n
+
1
)
{\displaystyle \ln |E_{n}(z)|=\ln |1-z|+Re(Q(z))=o(|z|)+O(|z|^{n})<O(|z|^{n+1})}
g is well-defined[ edit ]
Source:[2]
For any
R
>
0
{\displaystyle R>0}
, we show that the sum
∑
k
=
1
∞
ln
|
E
p
(
z
/
a
k
)
|
{\displaystyle \sum _{k=1}^{\infty }\ln |E_{p}(z/a_{k})|}
converges uniformly over
B
(
0
,
R
)
{\displaystyle B(0,R)}
.
Since only finitely many
|
a
k
|
<
K
R
{\displaystyle |a_{k}|<KR}
, we can split the sum to a finite bulk and an infinite tail:
∑
k
=
1
∞
ln
|
E
p
(
z
/
a
k
)
|
=
∑
|
a
k
|
<
K
R
ln
|
E
p
(
z
/
a
k
)
|
+
∑
|
a
k
|
≥
K
R
ln
|
E
p
(
z
/
a
k
)
|
{\displaystyle \sum _{k=1}^{\infty }\ln |E_{p}(z/a_{k})|=\sum _{|a_{k}|<KR}\ln |E_{p}(z/a_{k})|+\sum _{|a_{k}|\geq KR}\ln |E_{p}(z/a_{k})|}
The bulk term is a finite sum, so it converges uniformly. It remains to bound the tail term.
By bound (1) on
|
E
p
|
{\displaystyle |E_{p}|}
,
|
1
−
E
p
(
z
/
a
k
)
|
≤
B
|
z
/
a
k
|
p
+
1
≤
B
/
K
p
+
1
{\displaystyle |1-E_{p}(z/a_{k})|\leq B|z/a_{k}|^{p+1}\leq B/K^{p+1}}
. So if
K
{\displaystyle K}
is large enough, for some
B
′
>
0
{\displaystyle B'>0}
,[nb 1]
∑
|
a
k
|
≥
K
R
ln
|
1
−
(
1
−
E
p
(
z
/
a
k
)
)
|
≤
∑
|
a
k
|
≥
K
R
B
′
B
|
z
/
a
k
|
p
+
1
<
|
z
|
p
+
1
B
′
B
∑
k
1
|
a
k
|
p
+
1
{\displaystyle \sum _{|a_{k}|\geq KR}\ln |1-(1-E_{p}(z/a_{k}))|\leq \sum _{|a_{k}|\geq KR}B'B|z/a_{k}|^{p+1}<|z|^{p+1}B'B\sum _{k}{\frac {1}{|a_{k}|^{p+1}}}}
Since
p
+
1
=
floor
(
α
)
+
1
>
α
{\displaystyle p+1={\text{floor}}(\alpha )+1>\alpha }
, the last sum is finite.
g ≤ floor(ρ ) [ edit ]
As usual in analysis, we fix some small
ϵ
>
0
{\displaystyle \epsilon >0}
.
Then the goal is to show that
f
(
z
)
/
g
(
z
)
{\displaystyle f(z)/g(z)}
is of order
≤
ρ
+
ϵ
{\displaystyle \leq \rho +\epsilon }
. This does not exactly work, however, due to bad behavior of
f
(
z
)
/
g
(
z
)
{\displaystyle f(z)/g(z)}
near
a
k
{\displaystyle a_{k}}
. Consequently, we need to pepper the complex plane with "forbidden disks", one around each
a
k
{\displaystyle a_{k}}
, each with radius
1
|
a
k
|
ρ
+
ϵ
{\displaystyle {\frac {1}{|a_{k}|^{\rho +\epsilon }}}}
. Then since
∑
k
1
|
a
k
|
ρ
+
ϵ
<
∞
{\displaystyle \sum _{k}{\frac {1}{|a_{k}|^{\rho +\epsilon }}}<\infty }
by the previous result on
α
{\displaystyle \alpha }
, we can pick an increasing sequence of radii
R
1
<
R
2
<
⋯
{\displaystyle R_{1}<R_{2}<\cdots }
that diverge to infinity, such that each circle
∂
B
(
0
,
R
n
)
{\displaystyle \partial B(0,R_{n})}
avoids all these forbidden disks.
Thus, if we can prove a bound of form
ln
|
f
(
z
)
g
(
z
)
|
=
O
(
|
z
|
ρ
+
ϵ
)
=
o
(
|
z
|
ρ
+
2
ϵ
)
{\displaystyle \ln \left|{\frac {f(z)}{g(z)}}\right|=O(|z|^{\rho +\epsilon })=o(|z|^{\rho +2\epsilon })}
for all large
z
{\displaystyle z}
[nb 2] that avoids these forbidden disks, then by the same application of Borel–Carathéodory theorem,
d
e
g
(
Q
)
≤
floor
(
ρ
+
2
ϵ
)
{\displaystyle deg(Q)\leq {\text{floor}}(\rho +2\epsilon )}
for any
ϵ
>
0
{\displaystyle \epsilon >0}
, and so as we take
ϵ
→
0
{\displaystyle \epsilon \to 0}
, we obtain
g
≤
floor
(
ρ
)
{\displaystyle g\leq {\text{floor}}(\rho )}
.
Since
ln
|
f
(
z
)
|
=
o
(
|
z
|
ρ
+
ϵ
)
{\displaystyle \ln \left|f(z)\right|=o(|z|^{\rho +\epsilon })}
by the definition of
ρ
{\displaystyle \rho }
, it remains to show that
−
ln
|
g
(
z
)
|
=
O
(
|
z
|
ρ
+
ϵ
)
{\displaystyle -\ln \left|g(z)\right|=O(|z|^{\rho +\epsilon })}
, that is, there exists some constant
B
>
0
{\displaystyle B>0}
such that
∑
k
=
1
∞
ln
|
E
p
(
z
/
a
k
)
|
≥
−
B
|
z
|
ρ
+
ϵ
{\displaystyle \sum _{k=1}^{\infty }\ln |E_{p}(z/a_{k})|\geq -B|z|^{\rho +\epsilon }}
for all large
z
{\displaystyle z}
that avoids these forbidden disks.
As usual in analysis, this infinite sum can be split into two parts: a finite bulk and an infinite tail term, each of which is to be separately handled. There are finitely many
a
k
{\displaystyle a_{k}}
with modulus
<
|
z
k
|
/
2
{\displaystyle <|z_{k}|/2}
and infinitely many
a
k
{\displaystyle a_{k}}
with modulus
≥
|
z
k
|
/
2
{\displaystyle \geq |z_{k}|/2}
. So we have to bound:
∑
|
a
k
|
<
|
z
k
|
/
2
ln
|
E
p
(
z
/
a
k
)
|
+
∑
|
a
k
|
≥
|
z
k
|
/
2
ln
|
E
p
(
z
/
a
k
)
|
{\displaystyle \sum _{|a_{k}|<|z_{k}|/2}\ln |E_{p}(z/a_{k})|+\sum _{|a_{k}|\geq |z_{k}|/2}\ln |E_{p}(z/a_{k})|}
The upper-bounding can be accomplished by the bounds (2), (3) on
|
E
p
|
{\displaystyle |E_{p}|}
, and the assumption that
z
{\displaystyle z}
is outside every forbidden disk. Details are found in.
[2]
ρ ≤ g + 1 [ edit ]
This is a corollary of the following:
If
f
(
z
)
=
e
Q
(
z
)
z
m
∏
k
E
p
(
z
/
a
k
)
{\displaystyle f(z)=e^{Q(z)}z^{m}\prod _{k}E_{p}(z/a_{k})}
has genus
g
{\displaystyle g}
, then
ln
|
f
(
z
)
|
=
O
(
|
z
|
g
+
1
)
{\displaystyle \ln |f(z)|=O(|z|^{g+1})}
.
Split the sum to three parts:
ln
|
f
(
z
)
|
=
R
e
(
Q
(
z
)
)
+
m
R
e
(
ln
z
)
+
∑
k
ln
|
E
p
(
z
/
a
k
)
|
.
{\displaystyle \ln |f(z)|=Re(Q(z))+m\,Re(\ln z)+\sum _{k}\ln |E_{p}(z/a_{k})|.}
The first two terms are
O
(
|
z
|
p
)
=
o
(
|
z
|
g
+
1
)
{\displaystyle O(|z|^{p})=o(|z|^{g+1})}
. The third term is bounded by bound (4) of
|
E
p
|
{\displaystyle |E_{p}|}
:
∑
k
ln
|
E
p
(
z
/
a
k
)
|
≤
B
|
z
|
q
+
1
∑
k
1
|
a
k
|
q
+
1
.
{\displaystyle \sum _{k}\ln |E_{p}(z/a_{k})|\leq B|z|^{q+1}\sum _{k}{\frac {1}{|a_{k}|^{q+1}}}.}
By assumption,
p
=
floor
(
α
)
{\displaystyle p={\text{floor}}(\alpha )}
, so
∑
k
1
|
a
k
|
q
+
1
<
∞
{\displaystyle \sum _{k}{\frac {1}{|a_{k}|^{q+1}}}<\infty }
. Hence the above sum is
O
(
|
z
|
q
+
1
)
=
O
(
|
z
|
g
+
1
)
.
{\displaystyle O(|z|^{q+1})=O(|z|^{g+1}).}
Applications [ edit ]
With Hadamard factorization we can prove some special cases of Picard's little theorem .
Theorem:[4] If
f
{\displaystyle f}
is entire, nonconstant, and has finite order, then it assumes either the whole complex plane or the plane minus a single point.
Proof: If
f
{\displaystyle f}
does not assume value
z
0
{\displaystyle z_{0}}
, then by Hadamard factorization,
f
(
z
)
−
z
0
=
e
Q
(
z
)
{\displaystyle f(z)-z_{0}=e^{Q(z)}}
for a nonconstant polynomial
Q
{\displaystyle Q}
. By the fundamental theorem of algebra ,
Q
{\displaystyle Q}
assumes all values, so
f
(
z
)
−
z
0
{\displaystyle f(z)-z_{0}}
assumes all nonzero values.
Theorem:[4] If
f
{\displaystyle f}
is entire, nonconstant, and has finite, non-integer order
ρ
{\displaystyle \rho }
, then it assumes the whole complex plane infinitely many times.
Proof: For any
w
∈
C
{\displaystyle w\in \mathbb {C} }
, it suffices to prove
f
(
z
)
−
w
{\displaystyle f(z)-w}
has infinitely many roots. Expand
f
(
z
)
−
w
{\displaystyle f(z)-w}
to its Hadamard representation
f
(
z
)
−
w
=
e
Q
(
z
)
z
m
∏
k
E
p
(
z
/
a
k
)
{\displaystyle f(z)-w=e^{Q(z)}z^{m}\prod _{k}E_{p}(z/a_{k})}
. If the product is finite, then
ρ
=
g
{\displaystyle \rho =g}
is an integer.
References [ edit ]
^ Conway, J. B. (1995), Functions of One Complex Variable I, 2nd ed. , springer.com: Springer, ISBN 0-387-90328-3
^ a b c Dupuy, Taylor. "Hadamard's Theorem and Entire Functions of Finite Order — For Math 331" (PDF) .
^ Kupers, Alexander (April 30, 2020). "Lectures on complex analysis" (PDF) . Lecture notes for Math 113 . , Theorem 12.3.4.ii.
^ a b Conway, John B. (1978). Functions of One Complex Variable I . Graduate Texts in Mathematics. Vol. 11. New York, NY: Springer New York. doi :10.1007/978-1-4612-6313-5 . ISBN 978-0-387-94234-6 . Chapter 11, Theorems 3.6, 3.7.
^ so that
B
/
K
p
+
1
<
1
/
2
{\displaystyle B/K^{p+1}<1/2}
, then we can use the bound
ln
|
1
−
w
|
≤
B
′
|
w
|
∀
|
w
|
<
1
/
2
{\displaystyle \ln |1-w|\leq B'|w|\forall |w|<1/2}
to get
^ That is, we fix some yet-to-be-determined constant
R
{\displaystyle R}
, and use "
z
{\displaystyle z}
is large" to mean
|
z
|
>
R
{\displaystyle |z|>R}
.