# Uniformly convergent subsequence of continuously differentiable functions with non-differentiable limit?

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 $$(0,1)$$.

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 $$f$$.
• 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? Commented Sep 12, 2022 at 23:34
• Try $0$ on $[0,1/2]$, and $(x-1/2)/2$ on $[1/2, 1]$. Commented Sep 13, 2022 at 1:08
• Does $(x-1)^2 / 2$ on $[1/2, 1]$ do the trick? Commented Sep 13, 2022 at 1:30
• Sure, if you use a suitable function on $[0,1/2]$. Commented Sep 13, 2022 at 4:37