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 $[0,r]$.

Par for the course. We had a contributor once (he's no longer around) who was much cleverer than everybody else. So clever that he decided he did not need to cite his sources because he knew everything. This appears to be one of his pages.

This is all way over my head, and I would need to study it and I lack the patience.

Hence someone else is going to need to read this proof and determine in what way it is fallacious. Demonstrating a counterexample is one thing, but what we really need is for someone to point out exactly where in the argument an invalid step has been made. Since you seem to know about this, can you say precisely where the mistake is? --prime mover (talk) 11:26, 15 October 2021 (UTC)

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

Gotcha, yes, that makes sense.

Can we craft a proof that fills the gap -- basically that a locally uniform sequence is also locally uniform on a dense subset?

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)