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Normally convergent products define holomorphic functions with the expected zeros

Statement

Let $Ω \subseteq C$ be open, and let $(f_{n})_{n \geq 0}$ be holomorphic functions on $Ω$ whose product is normally convergent in the sense of Normal convergence of holomorphic products. Assume moreover that no factor $f_{n}$ is identically zero on $Ω$ . Then the partial products

$P_{m} (z) : = \prod_{n = 0}^{m} f_{n} (z)$

converge locally uniformly on $Ω$ to a holomorphic function $F$ .

Moreover, on every compact set $K \subseteq Ω$ , all but finitely many factors $f_{n}$ are zero-free and the tail limit is zero-free; therefore the zeros of $F$ on $K$ , counted with multiplicity, are exactly those contributed by the finitely many exceptional factors.

Facts & Assumptions

Given: An open set $Ω$ and a normally convergent holomorphic product $\prod f_{n}$ on $Ω$ , with no factor $f_{n}$ identically zero on $Ω$ .

[F1]

Normal convergence means that on each compact $K \subseteq Ω$ there is an index $N$ such that $f_{n}$ has no zero on $K$ for $n \geq N$ and $\sum_{n \geq N} \sup_{K} ∣ f_{n} - 1 ∣ < \infty$ (Normal convergence of holomorphic products).

[F2]

If $\sum ∣ a_{n} ∣$ converges, then $\prod (1 + a_{n})$ converges and has nonzero value (Absolute convergence criterion for complex infinite products).

[F3]

Locally uniform limits of holomorphic functions are holomorphic (Locally uniform limits of holomorphic functions are holomorphic and their derivatives converge locally uniformly).

[F4]

Multiplication by a holomorphic factor that is nonzero at a point does not change the order of a zero there (The order of a zero is the exponent in its local holomorphic factorization).

Proof

technique · direct

1.1

Fix a compact set $K \subseteq Ω$ . By [F1], choose $N$ so that $f_{n}$ has no zero on $K$ for $n \geq N$ and $M_{n} : = \sup_{K} ∣ f_{n} - 1 ∣$ satisfies $\sum_{n \geq N} M_{n} < \infty$ ; enlarging $N$ if needed, assume also $M_{n} < 1 / 2$ for $n \geq N$ .

F1givenchoose

2.1

For $m \geq ℓ \geq N$ and $z \in K$ , one has $∣ \prod_{n = ℓ}^{m} f_{n} (z) - 1 ∣ \leq \prod_{n = ℓ}^{m} (1 + M_{n}) - 1$ , while $∣ f_{n} (z) ∣ \geq 1 - M_{n}$ for $n \geq N$ . By [F2], the real products $\prod_{n \geq N} (1 + M_{n})$ and $\prod_{n \geq N} (1 - M_{n})$ converge, so the tail partial products of $\prod f_{n}$ are uniformly bounded above and uniformly bounded away from $0$ on $K$ .

F2step 1.1algebra

3.1

The estimate of step 2.1 implies that the tail partial products are uniformly Cauchy on $K$ , hence converge uniformly there to a continuous zero-free limit $Q_{K}$ ; multiplying by the finite holomorphic prefix $\prod_{n < N} f_{n}$ gives uniform convergence of the full partial products on $K$ . Because $K$ was arbitrary, the convergence is locally uniform on $Ω$ , and [F3] makes the limit function $F$ holomorphic.

F2F3step 2.1algebra

4.1

On the fixed compact set $K$ , write $F = (\prod_{n < N} f_{n}) Q_{K}$ with $Q_{K}$ holomorphic and zero-free by step 3.1. Because no factor $f_{n}$ is identically zero on $Ω$ , the finitely many prefix factors have only isolated zeros, and [F4] shows that every zero of $F$ on $K$ , with its multiplicity, comes from that finite prefix and no tail factor contributes a new zero.

F4step 1.1step 3.1givenalgebra∎

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Sources

Elias M. Stein and Rami Shakarchi, Complex Analysis, Ch. 5 Infinite products (standard reference, not scraped)
Matthias Weber, Complex Analysis, Ch. 3 §3.2 (standard reference, not scraped)