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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?

amWhy
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Ant
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2 Answers2

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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.

Spenser
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    @Ant https://notendur.hi.is/vae11/%C3%9Eekking/principles_of_mathematical_analysis_walter_rudin.pdf – Spenser Mar 17 '15 at 17:26
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    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 – Ant Mar 17 '15 at 17:32
  • @Ant The other implication is false. This Theorem is pretty much the best you can say about limit of functions and derivatives. – Spenser Mar 17 '15 at 17:34
  • I am currently looking for this statement to prove the fourier inversion theorem. But assuming the Fourier inversion theorem, you might actually get the other direction if the fourier transform is continous (not sure about that). Because convergence of cumulative distribution functions $F_n$ implies convergence of characteristic functions you would get that the inverse fourier transform of the characteristic functions converges (i.e. the densities) if you started out with densities (and the fourier transform is continous). – Felix B. Apr 27 '22 at 16:54
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This is a partial answer, there is a gap at the end.

Assume that $F_n$, $F$ are cumulative distribution functions (or can be scaled to be such), then convergence implies convergence in distribution, which in turn implies convergence of characteristic functions. $$ \varphi_{f_n} \to \varphi_{f} $$ Now the characteristic function of a distribution with a density is simply its inverse Fourier transform (up to constants). So slapping the Fourier transform over the characteristic functions $\varphi_{f_n}$ results in

$$ f_n = \hat{\varphi}_{f_n} = \int e^{-itx} \varphi_{f_n}(t) dt \overset{?}{\to} \int e^{-itx} \varphi_f(t) dt = f $$

Unfortunately we only have pointwise convergence of the characteristic functions, so the convergence of the integrals is not a given. But if you can move the limits into the integral you would be done. Maybe there is a nice criterion for that which translates back to a regularity condition on $F$

Felix B.
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