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.

  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ @Filburt Why your second application of FTC is not correct? $\endgroup$ – GaC Feb 21 '17 at 15:46

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.