Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I know that there is a theorem which permits us to interchange the derivative with the integral in some cases. I was wondering if there is a known theorem which permits us to interchange the derivative with a limit. For example, under which regularity conditions for the function $f$ we can have something like

$$ \frac{\partial}{\partial a}\lim_{n \to \infty} f(n,a) =\lim_{n \to \infty} \frac{\partial}{\partial a} f(n,a)?$$

share|cite|improve this question
up vote 1 down vote accepted

A more "real-variable" style condition is this. Let me write $f_n(\alpha)$ instead of $f(n,\alpha)$, and $f(\alpha) = \lim_{n \to \infty} f_n(\alpha)$. Suppose in some closed interval $I$, $f$ and all $f_n$ are $C^2$, $f_n \to f$ pointwise, and all $f_n''$ are uniformly bounded. Then $f_n' \to f'$ uniformly on $I$.

It suffices to prove this in the case $f = 0$ (in general, take $g_n = f_n - f$ which satisfies similar conditions to $f_n$ but converges to $0$, so if $g_n' \to 0$ we have $f_n' \to f$).

Let $\alpha, \beta$ be distinct members of $I$, and let $B$ be a uniform bound for $f_n''$ on $I$. By Taylor's theorem, $f_n(\alpha) - f_n(\beta) = f_n(\alpha) + f_n'(\alpha) (\beta - \alpha) + f_n''(\xi_n) (\beta - \alpha)^2/2$ for some $\xi_n \in I$. Write this as $$f_n'(\alpha) = \frac{f_n(\beta) - f_n(\alpha)}{\beta - \alpha} - f_n''(\xi_n) \frac{\beta - \alpha}{2} $$ Given $\epsilon > 0$ and $\alpha$, take $\beta$ so that $0 < |\beta - \alpha| < 2\epsilon/(3 B)$. Take $n$ large enough that $|f_n(\alpha)| < |\beta - \alpha| \epsilon/3$ and $|f_n(\beta)| < |\beta - \alpha| \epsilon/3$. Then we have $|f_n'(\alpha)| < \epsilon/3 + \epsilon/3 + \epsilon/3 = \epsilon$. Since this works for all $\epsilon$, conclude that $f_n'(\alpha) \to 0$ as $n \to \infty$.

share|cite|improve this answer
Great answer. Thank you :) – Beni Bogosel Mar 11 '12 at 12:10

This is true, for example, if for each $n$ $f(n,z)$ is analytic in a (complex) neighbourhood $D$ of $a$ and $f(n,z)$ converges to its limit as $n \to \infty$ uniformly on compact subsets of $D$.

share|cite|improve this answer
Yes, that's right. There is a theorem for holomorphic functions which says that. Thank you. :) – Beni Bogosel Mar 10 '12 at 9:31

Again, write $f_n(x) = f(n,x)$. If $f_n \in \mathcal C^1$, $f_n$ converges and $f_n'$ converges uniformly, $(\lim_{n \to \infty} f_n)' = \lim_{n \to \infty} f_n'$.

It can be proved by first showing that $$\lim_{n \to \infty} f_n(x) - f_n(0) = \lim_{n \to \infty} \int_0^x f'(t) dt = \int_0^x (\lim_{n \to \infty} f_n'(t)) dt$$ (this uses that the $f_n$ are $\mathcal C^1$ in the first step and uniform convergence in the second step) and then differentiating on both sides.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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