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Weierstrass factorization for entire functions

Statement

Let $f$ be an entire function, not identically zero, and let $m$ be the order of its zero at $0$ . If $f$ has infinitely many nonzero zeros, let $(a_{n})_{n \geq 1}$ list them with multiplicity and without finite accumulation point. Then there is an entire function $g$ such that

$f (z) = z^{m} e^{g (z)} \prod_{n \geq 1} E_{p_{n}} (z / a_{n})$

for suitable integers $p_{n} \geq 0$ .

If $f$ has finitely many nonzero zeros $a_{1}, \dots, a_{N}$ , the corresponding conclusion is

$f (z) = z^{m} e^{g (z)} \prod_{j = 1}^{N} E_{0} (z / a_{j}),$

where the product is $1$ when $N = 0$ .

Facts & Assumptions

Given: A nonzero entire function $f$ .

[F1]

The zero at $0$ has finite order $m$ , and locally one can factor off $(z - a)^{m}$ from a holomorphic function according to its zero multiplicity (The order of a zero is the exponent in its local holomorphic factorization).

[F2]

The Weierstrass product theorem constructs an entire product with any prescribed discrete zero divisor on $C$ (Weierstrass product theorem on the complex plane).

[F3]

The plane $C$ is star-shaped and therefore homologically simply connected (Star-shaped plane domains are homologically simply connected).

[F4]

A nowhere-zero holomorphic function on a homologically simply connected domain has a holomorphic logarithm (A nonvanishing holomorphic function on a homologically simply connected domain has a holomorphic logarithm).

[F5]

The elementary factor $E_{0} (w) = 1 - w$ has its unique zero at $w = 1$ . (Weierstrass elementary factors)

Proof

technique · direct

1.1

Let $m$ be the order of the zero of $f$ at $0$ , with $m = 0$ if $f (0) \neq 0$ . If the nonzero zero multiset is infinite, [F2] gives an entire function $P (z) = z^{m} \prod_{n \geq 1} E_{p_{n}} (z / a_{n})$ with exactly the same zeros as $f$ , counted with multiplicity. If it is finite, list it as $a_{1}, \dots, a_{N}$ and put $P (z) = z^{m} \prod_{j = 1}^{N} E_{0} (z / a_{j})$ , with the empty product equal to $1$ ; [F5] gives the same zero-divisor conclusion.

F1F2F5givenconstructcases

2.1

The quotient $h = f / P$ is holomorphic on $C$ : away from the common zeros this is immediate, and at each zero the matching multiplicities from step 1.1 and [F1] remove the singularity. Moreover $h$ has no zero anywhere, because every zero of $f$ was already cancelled by $P$ .

F1F2step 1.1algebra

3.1

By [F3], the whole plane is homologically simply connected, so [F4] gives an entire function $g$ with $e^{g} = h$ . Substituting the definition of $h$ from step 2.1 yields the required infinite or finite factorization of $f$ .

F3F4step 2.1algebra∎

Depends on

Used by

FALSE: the order of an entire function always equals its canonical genus False statement
FALSE: Weierstrass factorization is unique False statement
Hadamard factorization for finite-order entire functions Theorem

Dependency tree · two levels

44 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Elias M. Stein and Rami Shakarchi, Complex Analysis, Ch. 5 The Weierstrass product theorem (standard reference, not scraped)