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The question actually came to me from series of functions. Suppose $f_n(x)$ is a sequence of continuous differentiable functions, that $$\sum_{n=1}^\infty f_n'(x)$$ converges uniformly on any closed subinterval $[a,b]\subseteq \mathbb{R}$, and that $\sum_{n=1}^\infty f_n(x)$ converges pointwise for any $x$.

Then necessarily on any subinterval the sum of $f_n(x)$ converges to a function $f(x)$ such that the sum of $f_n'(x)$ converges to $f'(x)$, and it seems somewhat intuitive to go from this to saying that the sum of $f_n(x)$ converges to a differentiable limit function on all of $\mathbb{R}$ as differentiation is a pointwise property.

But I'm still not sure about it.. I wasn't really able to prove it myself, and I'm not sure that the function I get for one interval will even match the function I get on another.

So is the claim true? Or is there an obvious counter-example?

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  • $\begingroup$ Are you missing some bound on $\sum_n f_n$? $\endgroup$ – user99914 Nov 28 '15 at 9:55
  • $\begingroup$ @JohnMa don't really think I am.. The theorem in Wikipedia doesn't mention anything on a bound for any closed interval. Would it have any affect on all of $\mathbb{R}$ to have such a bound? en.wikipedia.org/wiki/Uniform_convergence#To_differentiability $\endgroup$ – Nescio Nov 28 '15 at 10:06
  • $\begingroup$ Well it seems I misread your question. So it seems that you want to show $f(x) = \sum f_n(x)$ is differentiable. Then like what you said, $f(x)$ is diff on each closed and bounded intervals. Then $f$ has to be bound (note that $f$ has nothing to do with $[a,b]$: it is given that $\sum_n f_n$ converge pointwisely to $f$). $\endgroup$ – user99914 Nov 28 '15 at 10:13
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What you have to show is that for any given $x \in \mathbb{R}$ the sequence converges to a differentiable function the derivative of which is given by the series involving the derivatives. But since any $x$ is contained in a suitably chosen compact interval (e.g. $[-|x|-1, |x| + 1$) where your assumption applies, this is, of course, true.

You cannot, of course, expect uniform convergence (of the series of derivatives) on all of of $\mathbb{R}$

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  • $\begingroup$ Well, that should have been pretty obvious to me looking back at it.. Thanks for clearing it all up anyway $\endgroup$ – Nescio Nov 28 '15 at 21:20

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