Suppose we have a sequence of differentiable functions $f_n$ from a closed interval $I\subseteq\mathbb{R}$ to $\mathbb{R}$ with the following properties:

  • $f_n$ converges uniformly to some function $f$;
  • $f_n'$ converges in $L^1$, so that a pointwise limit is defined a.e. on $I$;
  • $f_n$ has bounded variation.

Is it necessarily true that $f$ is differentiable and $f' = \lim f_n'$ on all of $I$?

I have spend considerable time thinking of how to prove this and trying to come up with a counter-example to no avail. In particular, I have checked several well-known counter-examples to intuitive but false statements on the convergence of functions and none seems to work here, which gives me some hope that the answer to my question is affirmative.

There is one candidate for a counter-example that I have not been able to fully work out. Let $f_n$ be the sequence generating devil's staircase, rounded slightly to be differentiable. Then $f_n$ converges uniformly to the devil's staircase, $f_n'$ converges to 0 a.e., but $f_n'$ goes to infinity on the Cantor set. But I am not convinced that $f_n$ can be constructed so that $f_n'$ converges in $L^1$ because while $f_n'$ are zero on larger and larger sets, where they are nonzero, they grow larger and larger.


1 Answer 1


Maybe I don't get it correctly but I guess the following provides a counterexample: Consider the standard mollification of the function $f(x)=|x|$ on $[-1,1]$. This yields a sequence of functions that converges uniformly to $f$ and the sequence of derivatives converges to $$g(x)=\begin{cases}1\text{ if }x\geqslant 0\\ -1\text{ else}\end{cases}$$ in $L^1$, but $f$ is not differentiable.

EDIT: Typo.


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