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I learned in my real analysis class that if $f_n:[a,b] \to \mathbb{R}$ is a sequence of differentiable functions such that $f_n \to f$ uniformly and $f_{n}^\prime \to g$ uniformly then $f$ is differentiable and f' = g.

Can the assumptions be weakened? In particular I'd like to dispense of the condition that $f_{n}^\prime$ converges uniformly to some function $g$.

This question is motivated by the following: Let K = \{f:[a,b] \to \mathbb{R}: f \text{ is differentiable and } \|f\|_{\infty} + \|f'\|_{\infty} \le 1\}. By the Arzelà-Ascoli theorem, the closure of this set is compact in $C([a,b])$. But I want it to actually be compact possibly after imposing some additional conditions on the possible functions $f$. This happens when $K$ is closed, hence the question.

Thanks in advance.

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    @sdcvvc that article states the theorem I just quoted. I was wondering if the uniform convergence of $f_n'$ can be replaced by something else. For example, if $f_n' \to g$ pointwise and $f_n'(x)$ is monotonic for all $x$ the result follows from Dini's Theorem (although of course this particular example is trivial since in this case $f_n' \to g$ uniformly).2011-12-04

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It is enough for $f_n$ to be $C^1$, $g$ continuous, f'_n \to g in $L^1$ (i.e. \int_a^b |f'_n(x) - g(x)|\ dx \to 0 as $n \to \infty$), and $f_n \to f$ pointwise. For taking the limit of f_n(x) = f_n(a) + \int_a^x f'_n(x)\ dx as $n \to \infty$ you get $f(x) = f(a) + \int_a^x g(x)\ dx$, which implies (by the Fundamental Theorem of Calculus) that $f$ is differentiable with f' = g.