# Talk:Derivative of Infinite Product of Analytic Functions

I think this proof should be improved. The proof seems to suggest (when it appeals to Uniformly Convergent Sequence on Dense Subset) that if $D$ is an open subset of a Banach space, and $E\subseteq D$ is dense, and $f_n\to f$ locally uniformly on $E$, then $f_n\to f$ locally uniformly on $D$. But this is false. For instance take $f_n(x)=x^n$ for $x\in(0,1)$, $f_n(x)=1$ for $x\geq 1$, and $f(x)=0$ for $x\in(0,1)$, $f(x)=1$ for $x\geq 1$. Then $f_n\to f$ locally uniformly on $\mathbb{R}\setminus\{1\}$, but it's not the case that $f_n\to f$ locally uniformly on $\mathbb{R}$. I believe special properties of analytic functions are needed to justify the inference here.
But does $x^n$ converge locally uniformly? --prime mover (talk) 14:02, 14 October 2021 (UTC)
Yes, $x_n\to 0$ locally uniformly on $[0,1)$. Indeed if $0\leq r<1$ then given $\epsilon>0$ we can find $n$ with $r^n<\epsilon$, and then for all $s\in [0,r]$ we have $|s^n|<\epsilon$. Thus $x^n\to 0$ uniformly on $[0,r]$.
Yes the problem is that Uniformly Convergent Sequence on Dense Subset only works for uniform convergence, not local uniform convergence. Thus the inference from "By Uniformly Convergent Sequence Multiplied with Function, the series converges locally uniformly in $E$" to "By Uniformly Convergent Sequence on Dense Subset, the series converges locally uniformly in $D$" is not valid. I think I can see how the theorem can be established, but I'm new to the site and not familiar with how you're supposed to write out proofs etc
Examples of how to write out proofs are everywhere. Just copy the form and style. Custom $\LaTeX$ constructs can be found here: Symbols:LaTeX Commands/ProofWiki Specific and their use is encouraged. --prime mover (talk) 16:40, 15 October 2021 (UTC)