Suppose that $\{f_n\}_{n \in \mathbb{N}}$ is a family of sufficiently smooth functions, defined on a domain $D \subseteq \mathbb{R}^m$, where they converge to a function $f$ in the topology of compact convergence. Then, suppose that their partial derivatives $\left\lbrace\frac{\partial f_n}{\partial i}\right\rbrace_{n \in \mathbb{N}}$ also converge to a function $g$ uniformly on compacts on the domain $D$. Can we conclude that $\frac{\partial f}{\partial i} = g$?
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Hi. Take a look at this – Sgg8 May 08 '23 at 19:03
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Does this answer your question? Can I exchange limit and differentiation for a sequence of smooth functions? – Sgg8 May 08 '23 at 19:03
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Rudin proves it for $\Bbb{R}$, but the reasoning for your metric space is essentially the same – Sgg8 May 08 '23 at 19:07
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For $\Bbb{R}^n$ you can take a look at a proof here in case you struglle to generalize Rudin's arguments.
Sgg8
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Hi, thanks for the reply. I was aware of the result you linked in the comment. It also makes sense that the limit function $f$ in my question would be partially differentiable on every compact $C \subseteq D$. It's difficult to properly formulate it, but my question is: if $f$ has a partial derivative on every compact $C \subseteq D$, does that mean that $f$ has a partial derivative on the entirety of $D$ and is that partial derivative continuous on $D$? – the_dude May 08 '23 at 19:41
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1Nevermind, the partial derivative $g$ would be continuous since the functions $\frac{\partial f_n}{\partial i}$ are continuous, they converge on compacts to $g$ and $D$ is compactly generated since it's a metric space. – the_dude May 08 '23 at 19:51