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)