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

Define the Hadamard canonical factors

Entire functions of finite order have Hadamard's canonical representation:[1]

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:[2] 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: ,[3] 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 |Ep|

In the proof, we need four bounds on :

  1. For any , when .
  2. For any , there exists such that when .
  3. For any , there exists such that when .
  4. 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

g is well-defined

Source:[2]

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 , . So if is large enough, for some ,[nb 1]

Since , the last sum is finite.

g floor(ρ)

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 for all large [nb 2] that avoids these forbidden disks, then by the same application of Borel–Carathéodory theorem, for any , and so as we take , we obtain .

Since by the definition of , it remains to show that , 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.[2]

ρ g + 1

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:[4] 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:[4] 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 . If the product is finite, then is an integer.

References

  1. Conway, J. B. (1995), Functions of One Complex Variable I, 2nd ed., springer.com: Springer, ISBN 0-387-90328-3
  2. 1 2 3 Dupuy, Taylor. "Hadamard's Theorem and Entire Functions of Finite Order — For Math 331" (PDF).
  3. Kupers, Alexander (April 30, 2020). "Lectures on complex analysis" (PDF). Lecture notes for Math 113., Theorem 12.3.4.ii.
  4. 1 2 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.

Notes

  1. so that , then we can use the bound to get
  2. That is, we fix some yet-to-be-determined constant , and use " is large" to mean .
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