This is a two part question. I'm done the first part but a bit stumped with the second. Just posting the whole thing for clarity.

For each $n$, let $f_n: [0,1] \to \mathbb{R}$ be a continuous function which satisfies $f_n(0) = 0$ and $f_n$ is continuously differentiable on (0,1) with $|f'_n(x)| \leq x$ for $x \in (0,1)$.

  1. Prove that there exists a subsequence of $(f_n)$ which converges uniformly to a continuous function $f$.
  2. Must the limit function $f$ in Part 1 be differentiable on $(0,1)$? Prove or provide a counterexample.

The first part was straightforward to prove. I see some questions here on MSE that are related to part 2 and show certain sequences of continuous functions converging uniformly to a continuous function that is not differentiable. But it seems like I have a greater number of restrictions on my functions $f_n$ and I cant figure out whether these restrictions are enough to force $f$ to be differentiable on $(0,1)$.


1 Answer 1


Hint: start by finding an $f$ that is continuous, differentiable except at $1/2$, and satisfies $|f'(x)| \le x$ on $(0,1/2)$ and $(1/2, 1)$. Then find a sequence $f_n$ of differentiable approximations of $f$.

  • $\begingroup$ Hm. To me, the "obvious" function $f$ which is continuous and differentiable except at $1/2$ would be $|x-1/2|$, but it doesn't mean the requirement $|f'(x)| \leq x$. Is there a way I can modify it to make it work? $\endgroup$
    – MathFrak96
    Commented Sep 12, 2022 at 23:34
  • $\begingroup$ Try $0$ on $[0,1/2]$, and $(x-1/2)/2$ on $[1/2, 1]$. $\endgroup$ Commented Sep 13, 2022 at 1:08
  • $\begingroup$ Does $(x-1)^2 / 2$ on $[1/2, 1]$ do the trick? $\endgroup$
    – MathFrak96
    Commented Sep 13, 2022 at 1:30
  • $\begingroup$ Sure, if you use a suitable function on $[0,1/2]$. $\endgroup$ Commented Sep 13, 2022 at 4:37

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