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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

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

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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 $n \to \infty$.

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Great answer. Thank you :) –  Beni Bogosel Mar 11 '12 at 12:10
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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 $D$.

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

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