Hadamard factorization theorem explained
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
have
Hadamard's canonical representation:
where
are those
roots of
that are not zero (
),
is the order of the zero of
at
(the case
being taken to mean
),
a polynomial (whose degree we shall call
), and
is the smallest non-negative integer such that the series
converges. The non-negative integer
is called the genus of the entire function
. In this notation,
In other words: If the order
is not an integer, then
is the integer part of
. If the order is a positive integer, then there are two possibilities:
or
.
Furthermore, Jensen's inequality implies that its roots are distributed sparsely, with critical exponent
.
For example,
,
and
are entire functions of genus
.
Critical exponent
Define the critical exponent of the roots of
as the following:
where
is the number of roots with modulus
. In other words, we have an asymptotic bound on the growth behavior of the number of roots of the function:
It's clear that
.
Theorem:[1] If
is an entire function with infinitely many roots, then
Note: These two equalities are purely about the limit behaviors of a real number sequence
that diverges to infinity. It does not involve complex analysis.
Proposition:
,
[2] by
Jensen's formula.
Proof
Since
is also an entire function with the same order
and genus, we can
wlog assume
.
If
has only finitely many roots, then
with the function
of order
. Thus by
an application of the Borel–Carathéodory theorem,
is a polynomial of degree
, and so we have
.
Otherwise,
has infinitely many roots. This is the tricky part and requires splitting into two cases. First show that
, then show that
.
Define the function
where
. We will study the behavior of
.
Bounds on the behaviour of
In the proof, we need four bounds on
:
- For any
,
when
.
- For any
, there exists
such that
when
.
- For any
, there exists
such that
when
.
for all
, and
as
.These are essentially proved in the similar way. As an example, we prove the fourth one.
where
is an entire function. Since it is entire, for any
, it is bounded in
. So
inside
.
Outside
, we have
is well-defined
Source:
For any
, we show that the sum
converges uniformly over
.
Since only finitely many
, we can split the sum to a finite bulk and an infinite tail:
The bulk term is a finite sum, so it converges uniformly. It remains to bound the tail term.
By bound (1) on
,
|1-Ep(z/ak)|\leq
\leqB/Kp+1
. So if
is large enough, for some
,
Since
p+1=floor(\alpha)+1>\alpha
, the last sum is finite.
As usual in analysis, we fix some small
.
Then the goal is to show that
is of order
. This does not exactly work, however, due to bad behavior of
near
. Consequently, we need to pepper the complex plane with "forbidden disks", one around each
, each with radius
. Then since
by the previous result on
, we can pick an increasing sequence of radii
that diverge to infinity, such that each circle
avoids all these forbidden disks.
Thus, if we can prove a bound of form
ln\left|
\right|=O(|z|\rho)=o(|z|\rho)
for all
large
that avoids these forbidden disks, then by the same application of Borel–Carathéodory theorem,
deg(Q)\leqfloor(\rho+2\epsilon)
for any
, and so as we take
, we obtain
.
Since
ln\left|f(z)\right|=o(|z|\rho)
by the definition of
, it remains to show that
-ln\left|g(z)\right|=O(|z|\rho)
, that is, there exists some constant
such that
for all large
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
with modulus
and infinitely many
with modulus
. So we have to bound:
The upper-bounding can be accomplished by the bounds (2), (3) on
, and the assumption that
is outside every forbidden disk. Details are found in.
This is a corollary of the following:
If
has genus
, then
.
Split the sum to three parts:
The first two terms are
. The third term is bounded by bound (4) of
:
By assumption,
, so
. Hence the above sum is
Applications
With Hadamard factorization we can prove some special cases of Picard's little theorem.
Theorem:[3] If
is entire, nonconstant, and has finite order, then it assumes either the whole complex plane or the plane minus a single point.
Proof: If
does not assume value
, then by Hadamard factorization,
for a nonconstant polynomial
. By the
fundamental theorem of algebra,
assumes all values, so
assumes all nonzero values.
Theorem: If
is entire, nonconstant, and has finite, non-integer order
, then it assumes the whole complex plane infinitely many times.
Proof: For any
, it suffices to prove
has infinitely many roots. Expand
to its Hadamard representation
f(z)-w=eQ(z)zm\prodkEp(z/ak)
. If the product is finite, then
is an integer.
Notes and References
- Web site: Dupuy . Taylor . Hadamard’s Theorem and Entire Functions of Finite Order — For Math 331 .
- Web site: Kupers . Alexander . April 30, 2020 . Lectures on complex analysis . Lecture notes for Math 113.., Theorem 12.3.4.ii.
- Book: Conway, John B. . Functions of One Complex Variable I . 1978 . Springer New York . 978-0-387-94234-6 . Graduate Texts in Mathematics . 11 . New York, NY . 10.1007/978-1-4612-6313-5. Chapter 11, Theorems 3.6, 3.7.