Let $F_n$ be a sequence of differentiable real valued functions.

Suppose that $$\lim_{n \to \infty} F_n(x) = F(x)$$ and that $F(x)$ is differentiable.

Under which conditions does that imply

$$\lim_{n \to \infty} F'_n(x) = F'(x)$$?

Do I need some regularity, or maybe that the $F_n$ converges uniformly?


You need to add the assumption that $F_n'$ converges uniformly on a closed interval $[a,b]$. In fact:

Theorem: Suppose $\{f_n\}$ is a sequence of functions, differentiable on $[a,b]$ and such that $\{f_n(x_0)\}$ converges for some point $x_0$ on $[a,b]$. If $\{f_n'\}$ converges uniformly on $[a,b]$, then $\{f_n\}$ converges uniformly on $[a,b]$, to a function $f$, and $$f'(x)=\lim_{n\to\infty}f_n'(x),\quad(a\leq x\leq b).$$

Source: Rudin, Principles of Mathematical Analysis, Theorem 7.17.

  • 1
    $\begingroup$ Thanks! Can you provide an online source though? :) $\endgroup$ – Ant Mar 17 '15 at 17:23
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    $\begingroup$ @Ant notendur.hi.is/vae11/%C3%9Eekking/… $\endgroup$ – Spenser Mar 17 '15 at 17:26
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    $\begingroup$ thank you! But doesn't that give the converse implication? I mean you assume that the $F_n'$ converge uniformly and conclude that also $F_n$ converge uniformly (if they converge in a single point, which I think it's just a way to make sure that the indefinite constant are the same). I was interested in the other implication $\endgroup$ – Ant Mar 17 '15 at 17:32
  • $\begingroup$ @Ant The other implication is false. This Theorem is pretty much the best you can say about limit of functions and derivatives. $\endgroup$ – Spenser Mar 17 '15 at 17:34

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