4
$\begingroup$

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?

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Thanks! Can you provide an online source though? :) $\endgroup$ – Ant Mar 17 '15 at 17:23
  • 1
    $\begingroup$ @Ant notendur.hi.is/vae11/%C3%9Eekking/… $\endgroup$ – Spenser Mar 17 '15 at 17:26
  • 2
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.