Notes on Complex Analysis

4.4.5 Hadamard Factorization Theorem

Theorem 4.4.18.

Let be the Weierstrass canonical factorization of , where is entire with finite order and . Then is a polynomial of degree .

Proof.

By logarithmic differentiation and by taking subsequent derivatives, we have

By applying Proposition 4.4.3 and Lemma 4.4.8, we have

Hence, is a polynomial of degree . Choosing so that , the assertion follows.

Corollary 4.4.4.

Let be the Weierstrass canonical factorization of , where is entire with finite order . Then is a polynomial of degree .

Proof.

Let , where and are entire, , and has finite order and for . For , such that implies

Thus, by letting . Additionally, for any , , such that

because for sufficiently large ,

Hence, . Letting implies . Let where is a constant, so that . It is also trivial that . Explicitly, we have .

By Theorem 4.4.18 on , is a polynomial of degree , and so is .

Then the results of Corollary 4.4.4 and Theorem 4.4.16 may be consolidated into a single statement:

Theorem 4.4.19 (Hadamard Factorization Theorem).

Let be the genus of and let be the order of , where is entire with finite order. Then .

Theorem 4.4.20.

The factorization

defines an entire function and uniformly converges on any compact disk .

Proof.

The zeros of are simple at each of . Aside from the simple zero at , let

enumerate the zeros of . By Example 4.4.2, and the Hadamard Factorization Theorem (Theorem 4.4.19), the order of is 1, the genus does not exceed 1, and

where is a polynomial (and where the product locally uniformly converges in ). Since the partial products , where

have a single accumulation point, the subsequence converges to the same point. Since

we have

Then from we have

Since , we have

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