Compact convergence (locally uniform convergence) of sequence of functions.

Question

Let $D\in \mathbb{C}$ be open, and let $f,f_1,f_2,...:D\rightarrow \mathbb{C}$ be functions.

We say the sequence $\{f_n\}$ is compactly convergent to $f$ (equivalent to locally uniformly convergent), if for each compact set $K\subset D$, the sequence $\{f_{n\restriction{K}}\}_{n=1}^{\infty}$ converges to $f_{\restriction{K}}$ uniformly on $K$

Let $f_n(z)=\prod_{i=1}^n(1+\frac {z}{i^2})$, with $z\in \mathbb{C}$

Show $f_n(z)$ converges compactly on $\mathbb{C}$.

Here is my thought process.

I know that $\prod_i^{\infty}(1+\frac {z}{i^2})$ converges if and only if $\sum_{i=1}^{\infty}log(1+\frac {z}{i^2})$ converges.

So showing $f_n(z)$ is compactly convergent on $\mathbb{C}$, amounts to showing for all $\epsilon>0$ there exists a sequence $a_n>0$, such that $|log(1+\frac {z}{n^2})|\leq a_n$ and $\sum_{n=1}^{\infty}a_n<\infty$ in $B_{\epsilon}[z]$.

If I can show that, then that implies $\sum_{i=1}^{\infty}(1+\frac {z}{i^2})$ converges on compact $B_{\epsilon}[z]$ and thus $\prod_{i=1}^{\infty}(1+\frac {z^2}{i^2})$ converges on compact $B_{\epsilon}[z]$

Therefore $f_n(z)$ converges compactly on $\mathbb{C}$ correct?

Any help would be much appreciated. Thanks!

Answer 1

A rigorous argument is as follows: let log denote the principle branch of logarithm. Let $g(z)=\frac {log(1+z)} z$ for $0<|z|<1$ and $1$ for $z=0$. Then $g$ is a continuous function on $\{z:|z| \leq \frac 1 2\}$. Hence it is bounded. Let $|g(z)| \leq M$ for $|z| \leq \frac 1 2$. Now, for $n$ sufficiently large $|\frac z {n^{2}}|\leq \frac 1 2$ for all $z$ in our compact set. This gives uniform convergence of $\sum log(1+\frac z {n^{2}})$ because $|log(1+\frac z {n^{2}})| \leq M|\frac z {n^{2}}|$ and $|z|$ is bounded on the compact set. Uniform convergence of this series implies uniform convergence of the infinite product.