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Weierstrass product theorem on the complex plane

Statement

Let $m \geq 0$ be an integer, and let $(a_{n})_{n \geq 1}$ be a sequence of nonzero complex numbers with no finite accumulation point, where each value may repeat according to its intended multiplicity. Then there are integers $p_{n} \geq 0$ such that

$P (z) : = z^{m} \prod_{n \geq 1} E_{p_{n}} (z / a_{n})$

converges normally on $C$ and is entire, with zero set exactly ${0}$ of order $m$ together with the nonzero zeros $a_{n}$ counted with multiplicity.

Facts & Assumptions

Given: The integer $m \geq 0$ and the sequence $(a_{n})$ .

[F1]

A Weierstrass product is a product of elementary factors with varying orders (Weierstrass products, canonical products, and genus).

[F2]

For every integer $p \geq 0$ and $∣ w ∣ \leq 1$ , the elementary factor satisfies $∣ 1 - E_{p} (w) ∣ \leq ∣ w ∣^{p + 1}$ (The unit-disc estimate for Weierstrass elementary factors).

[F3]

A normally convergent holomorphic product defines an entire function whose zeros on each compact set are exactly those contributed by the finitely many exceptional factors (Normally convergent products define holomorphic functions with the expected zeros).

Proof

technique · direct

1.1

Set $p_{n} : = n$ for every $n \geq 1$ . Fix $R > 0$ . Because $(a_{n})$ has no finite accumulation point, only finitely many terms satisfy $∣ a_{n} ∣ \leq 2 R$ ; choose $N$ so that $∣ a_{n} ∣ > 2 R$ for all $n \geq N$ .

givenchooseconstruct

2.1

For $∣ z ∣ \leq R$ and $n \geq N$ , one has $∣ z / a_{n} ∣ \leq R / ∣ a_{n} ∣ < 1 / 2 \leq 1$ , and $E_{n} (z / a_{n}) \neq 0$ because its unique zero occurs at $z = a_{n}$ , outside the closed disc $∣ z ∣ \leq R$ . Using [F2] with $p = n$ gives $∣ 1 - E_{n} (z / a_{n}) ∣ \leq ∣ z / a_{n} ∣^{n + 1} \leq (R / ∣ a_{n} ∣)^{n + 1} \leq 2^{- n - 1} .$ Therefore $\sum_{n \geq N} \sup_{∣ z ∣ \leq R} ∣ 1 - E_{n} (z / a_{n}) ∣$ converges.

F2step 1.1algebra

3.1

Step 2.1 is exactly the normal-convergence condition on the closed disc $∣ z ∣ \leq R$ , and $R$ was arbitrary. Hence [F3] makes $\prod_{n \geq 1} E_{n} (z / a_{n})$ an entire function whose zeros are exactly the points $a_{n}$ , counted with multiplicity. Multiplying by $z^{m}$ contributes precisely the order- $m$ zero at $0$ and no other zeros, so $P (z) = z^{m} \prod_{n \geq 1} E_{n} (z / a_{n})$ has exactly the stated zero divisor. Since $p_{n} = n \geq 0$ , this is a Weierstrass product in the sense of [F1].

F1F3step 2.1algebra∎

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Sources

Elias M. Stein and Rami Shakarchi, Complex Analysis, Ch. 5 The Weierstrass product theorem (standard reference, not scraped)
Matthias Weber, Complex Analysis, Ch. 3 §3.2 (standard reference, not scraped)