# Justifying differentiation of infinite product

Let $(z_n)_{n\in\mathbb{Z}}$ be a sequence of complex numbers s.t. the product $P(z):=\prod_{n=1}^{\infty}{\left(1-\frac{z}{z_{-n}}\right)\left(1-\frac{z}{z_{n}}\right)}$ is absolutely convergent for every $z\in\mathbb{C}$, and hence defines an entire function with zeros at every $z_k$. Is there some nice way to justify $\frac{\mathrm{d}}{\mathrm{d} z}P(z_k)=\frac{-1}{z_k}\prod_{|n|\geq1, n\neq k}{\left(1-\frac{z_k}{z_{n}}\right)}$? I guess one is not allowed to simply use the product rule for differentiation here?

• Although justification requires a little effort, of course the outcome of application of the product rule must be correct if the thing makes sense at all... :) The issue is about the limit of derivatives being the derivative of the limit. For $z$ in a compact, if a sequence $f_n$ has a sup-norm limit, and if the sequence of derivatives $f_n'$ has a sup-norm limit, then the derivative of the limit is the limit of the derivatives. If you mean to restrict to holomorphic functions, then Cauchy's integral formulas give a simpler mechanism. Is this what you are asking about? – paul garrett Sep 24 '15 at 22:05
• Take the derivative of its logarithm it will transform it to a sum that is easily differentiatable – Oussama Boussif Sep 24 '15 at 22:15
• If you know that $P$ is entire, then presumably you know that the product converges uniformly on compact sets, and then differentiation commutes with the limit. – Eric Wofsey Sep 24 '15 at 22:20
• Thank's for that answer. So I guess it would be completely justified given your statement about the sup-norm limits since the product converges absolutely? – Nikaka Sep 24 '15 at 22:29
• Something isn't quite right with your last expression. The left-hand side appears to be the derivative of $P$ evaluated at $z_k$. The right-hand side has terms that embed $z$, not $z_k$. – Mark Viola Sep 24 '15 at 22:33

## 1 Answer

Yes, this can be done. You can find this on ProofWiki: Derivative of Infinite Product of Analytic Functions

You don't even need the absolute convergence of the product $f=\prod f_n$, locally uniform convergence (=uniform convergence on compact sets, modulo some details, depending on the author, but many authors do not define it) suffices:

Theorem. Let $D\subset\mathbb C$ be open, $(f_n)$ analytic on $D$ and $f=\prod f_n$ locally uniformly convergent (thus analytic). Then $f'=\sum f_n'\prod_{k\neq n}f_k$ locally uniformly.

A practical sufficient condition is the locally uniform convergence of $\sum|f_n-1|$, which is satisfied by any reasonable sequence I can think of.

Outline of proof

First establish $$\frac{f'}f=\sum\frac{f_n'}{f_n}$$ locally uniformly, multiply everything by $f$, and check that both sides coincide at the zeroes of $f$.

To obtain this, we'd like $\log f = \sum \log f_n$, locally uniformly in order to differentiate term-wise. This is true, or at least locally:

Theorem. If $f=\prod f_n$ locally uniformly and $z_0\in D$, then $$\log \left(\prod_{n=n_0}^\infty f_n\right) = \sum_{n=n_0}^\infty \log f_n + 2k\pi i$$ uniformly on some neighborhood $U$ of $z_0$, for some $n_0\in\mathbb N$, $k\in\mathbb Z$ constant on $U$.

The proof consists of very carefully taking logarithms. The surprising part (to me) is that $k$ can be taken constant, i.e. eventually the product does not switch branches anymore.