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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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    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? – 2017-02-21
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    @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; – 2017-02-21
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    @PhoemueX (3) it maps sets of measure 0 to sets of measure 0 – 2017-02-21
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    Nice! Now, can you transfer property (1) from the $f_n $ to $f$? – 2017-02-21
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    @PhoemueX yes! Then f is a indefinite integral and, then, A.C. Is this correct? – 2017-02-21
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    @PhoemueX Let me explain what I did. Since $f_k\to f$ in $L^1$, there is a subsequence of $\{f_k\}$ that converges a.e to $f$. For simplicity, say $f_k\to f$ a.e. Since $\{f'_k\}$ is Cauchy in $L^1$, it converges to $g$ in $L^1$. For each $k$, by FTC, $f_k(x)-f_k(0)=\int_0^xf'_kd\lambda$. So $$\lim_{k\to\infty} (f_k(x)-f_k(0))=\lim_{k\to\infty}\int_0^xf'_kd\lambda=\int_0^xgd\lambda.$$ Then $f$ is a indefinite integral. By the FTC again, $f$ is A.C. **do we need $\{f_k\}$ uniformly bounded? is there some error in my argument?** – 2017-02-21
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    @PhoemueX Of course there is something wrong. The above equality wolds only almost everywhere. Further, the second application of FTC is not correct (although seems valid). – 2017-02-21
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    @Filburt Why your second application of FTC is not correct? – 2017-02-21
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    @SpettroDiA The FTC don't state this, thought. Maybe it is a corollary, (actually I don't know, yet). We can conclude from the above that $f$ is a.e an indefinite integral. I will try to show that $f$ is A.C from this. – 2017-02-21
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    An idea, can we show that $f$ is continuous ? – 2017-02-21
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    @SpettroDiA Every indefinite integral is A.C. (I have a proof). The problem is just this, $f$ is a indefinite integral only a.e., and since I can't see a way to prove that $f$ is continuous, I can't conclude that $f$ is an indefinite integral for every $x$. – 2017-02-21
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    @Filburt: The indefinite integral is absolutely continuous and hence continuous. Furthermore, if $(f_n (x_0))$ converges for **some** $x_0$, then it follows that $(f_n (0))_n$ converges (why?) and then that $(f_n (x))_n$ converges *everywhere*. Then you get $f(x) = \int_0^x g d \lambda$ everywhere and you are done. – 2017-02-21
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    @PhoemueX I can't see why the convergence of $\{f_k(x_0)\}$ implies that of $\{f_k(0)\}$ much less $\{f_k(x)\}$ for all $x$. Is there a relation to the fact that $f_k$ is an indefinite integral? (although the integral indefinite is continuous, we still don't know if $f$ is continuous too) – 2017-02-21

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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.