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Let $f_n$: $(0,1)\to (0,1)$ be given, where $f_n$ is a.e. differentiable, absolutely continuous, and monotone increasing. Suppose $f_n\to f$ uniformly. Then would it implies that $f_n'\to f'$ a.e.?

I know generally uniformly convergence has no way to imply convergence of derivative, but I somehow remembered that monotone increasing would help, but I can't recall where I saw this statement.


PS: I also have $f_n'$ are uniformly bounded (but may not continuous). So I think my statement should hold, right?

Thank you!

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2 Answers 2

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Sketchy Proof: By absolute continuity, we have \begin{align} f_n(x) = f_n(a) + \int^x_a f'_n(t)\ dt \end{align} then it follows \begin{align} &\lim_{n\rightarrow \infty} f_n(x) = \lim_{n\rightarrow \infty} f_n(a) +\lim_{n\rightarrow \infty}\int^x_a f_n'(t)\ dt \\ \Rightarrow &\ \ f(x) = f(a) +\lim_{n\rightarrow \infty} \int^x_a f'_n(t)\ dt. \end{align} Hence if $\lim_{n\rightarrow \infty} f_n'(t)$ converges pointwise to some function $g$, then you can show that \begin{align} f(x) = f(a) +\int^x_a g(t)\ dt \ \ a.e. \end{align} which means $f$ is absolutely continuous and \begin{align} f'(x) = g(x) \ \ a. e. \end{align} However, if we don't know apriori that $\lim_{n\rightarrow \infty} f_n'(t)$ exist, then we might not be able to conclude anything. Think construction of Cantor function.

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Consider for any interval $I\subset (0,1)$ the function $f_I$ which is zero before $I$, grows linearly with derivative one in $I$, and then is constant. Consider the sequence of intervals obtained considering for every $n\in \mathbb N$ the intervals $(k/2^n,(k+1)/2^n)$ for $k=0,...,2^n-1$. The sequence of the related functions $f_I$ satisfies the properties and converges uniformly to zero, but the derivative has nowhere limit.

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  • $\begingroup$ Thank you sir, but my function has uniformly bounded derivative $\endgroup$
    – spatially
    Sep 27, 2016 at 12:09
  • $\begingroup$ @JumpJump they have all derivative at most 1 $\endgroup$
    – Del
    Sep 27, 2016 at 14:12
  • $\begingroup$ I see. My bad, understand. $\endgroup$
    – spatially
    Sep 27, 2016 at 16:18
  • $\begingroup$ So I need to know apriori that $f_n'$ converge a.e. as @Jacky Chong comment above. $\endgroup$
    – spatially
    Sep 27, 2016 at 16:19
  • $\begingroup$ @Jumpjump Yes, exactly. I put this example to show the necessity of such assumption, because the proposed cantor approximating functions do not have bounded derivatives $\endgroup$
    – Del
    Sep 27, 2016 at 16:28

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