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The problem is the following:

Let $\{f_k\}$ be an uniformly bounded sequence of absolutely continuous differentiable functions on $[0,1]$. Suppose that $f_k\to f$ in $L^1[0,1]$ and that $\{f'_k\}$ is Cauchy in $L^1[0,1]$. Prove that $f$ is absolutely continuous on $[0,1]$.

  • There is a subsequence of $\{f_k\}$ that converges almost everywhere to $f$. But even if the convergence is uniform, I would have nothing;

  • Since $\{f_k'\}$ is Cauchy and $L^1$ is complete, there is $g\in L^1$ such that $f'_k\to g$ in $L^1$. It seems that there is no way to guarantee that $f'\to g$ uniformly. So I don't know what to do with this.

  • What to do with the uniform boundedness of $\{f_k\}$?

What is the starting point to solve this problem? Any hint will be really appreciated.

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    $\begingroup$ What do you know about absolutely continuous functions? Is there a connection between the derivative and the function? If you have $L^1$ convergence, this tells you something about certain integrals. Is there a characterization of absolutely continuous functions using integrals? $\endgroup$ – PhoemueX Feb 21 '17 at 14:00
  • $\begingroup$ @PhoemueX the FTC: (1) $f$ is A.C if and only if there exists $f'\in L^1$ a.e and $f(x)-f(0)=\int_0^xf'd\lambda$ for $x\in[0,1]$; (2) The total variation is A.C. too; $\endgroup$ – Filburt Feb 21 '17 at 14:03
  • $\begingroup$ @PhoemueX (3) it maps sets of measure 0 to sets of measure 0 $\endgroup$ – Filburt Feb 21 '17 at 14:14
  • $\begingroup$ Nice! Now, can you transfer property (1) from the $f_n $ to $f$? $\endgroup$ – PhoemueX Feb 21 '17 at 14:14
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    $\begingroup$ @Filburt Why your second application of FTC is not correct? $\endgroup$ – GaC Feb 21 '17 at 15:46
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Here is a more complete answer, elaborating on the hints I gave above.

You already know (upon switching to a subsequence, which I will assume in the following) that $f_n (x) \to f(x)$ almost everywhere. Furthermore, $f_n ' \to g$ in $L^1$. In particular, $f_n(x_0) \to f(x_0)$ for some $x_0$. But we have $$ f_n(x_0) - f_n (0) = \int_0^{x_0} f_n '(t) dt \to \int_0^{x_0} g(t) dt. $$ Hence, $(f_n (x_0) - f_n (0))_n$ and $(f_n(x_0))_n$ are convergent, and hence so is $(f_n (0))_n$, say $f_n (0) \to y$. But this again yields $$ f_n (x) = f_n (0) + \int_0^x f_n ' (t) dt \to y + \int_0^x g(t) dt. $$ Since $f_n (x) \to f(x)$ almost everywhere, this means $$ f(x) = y + \int_0^x g(t) dt $$ almost everywhere. Now redefine $f$ so that $f(x) = y + \int_0^x g(t) dt$ everywhere and conclude that this redefined $f$ is absolutely continuous.

This redefinition can not be avoided, since the $L^1$ convergence $f_n \to f$ only determines $f$ almost everywhere.

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