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The logarithmic derivative of a normally convergent product

Statement

Let $\prod_{n \geq 0} f_{n}$ be a normally convergent holomorphic product on an open set $Ω$ , and let $F$ be its holomorphic limit. On every compact set $K \subseteq Ω$ disjoint from the zero set of $F$ , the series

$\sum_{n \geq 0} \frac{f_{n}^{'} (z)}{f_{n} (z)}$

converges uniformly on $K$ and

$\frac{F^{'} (z)}{F (z)} = \sum_{n \geq 0} \frac{f_{n}^{'} (z)}{f_{n} (z)} (z \in K) .$

Facts & Assumptions

Given: A normally convergent product $\prod f_{n}$ on $Ω$ with limit $F$ .

[F1]

A normally convergent product has a holomorphic limit, and on each compact set only finitely many factors contribute zeros (Normally convergent products define holomorphic functions with the expected zeros).

[F2]

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

Proof

technique · direct

1.1F1givenchoose

Fix a compact set $K \subseteq Ω$ disjoint from the zero set of $F$ . By [F1], choose $N$ so that $f_{n}$ has no zero on $K$ for $n \geq N$ , and write $P_{m} = \prod_{n = 0}^{m} f_{n}$ . Then $P_{m} \to F$ uniformly on $K$ , and for $m \geq N$ the functions $P_{m}$ are zero-free on $K$ .

2.1F2step 1.1algebra

By [F2], the derivatives $P_{m}^{'}$ converge uniformly on $K$ to $F^{'}$ . Since $F$ has no zero on the compact set $K$ , the uniform convergence of $P_{m}$ to $F$ makes $P_{m}$ uniformly bounded away from $0$ on $K$ for all large $m$ , so $P_{m}^{'} / P_{m} \to F^{'} / F$ uniformly on $K$ .

3.1step 2.1algebra∎

For each finite product, ordinary differentiation gives $P_{m}^{'} / P_{m} = \sum_{n = 0}^{m} f_{n}^{'} / f_{n}$ on $K$ . Passing to the uniform limit from step 2.1 yields the stated series identity and uniform convergence.

Depends on

Used by

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Sources

Matthias Weber, Complex Analysis, Ch. 3 §3.2 (standard reference, not scraped)