Does uniform convergence of $f$ imply convergence of derivatives?

Question

Let $X$ denote the collection of all differentiable functions $f : [0, 1] \rightarrow \Bbb R$, such that $f(0)=0$ and $f'$ is continuous.

Let $\{f_n\}$ be a Cauchy sequence. By Cauchy criterion for uniform convergence, $f_n$ converges uniformly to some $f$.

Does that imply that $f'_n \rightarrow f'$ uniformly?

Answer 1

No! Imagine the $f_n$ getting close to $f$ uniformly, but getting bumpier and bumpier. You should be able to use this idea to come up with a counterexample.

Answer 2

This is a counterexample. Take e.g. $f_n(x) = \frac{1}{n} (\sqrt{1+(n x)^2} - 1)$. We have that $f_n(0) = 0$, and $f'_n(x) = \dfrac{nx}{\sqrt{1+(nx)^2}}$ is continuous. We also have that $\lim_{n\to\infty} f_n(x) = |x|$, and $$ |x| - f_n(x) = \frac{1}{n} \left(1 - \frac{1}{n|x| + \sqrt{1 + (nx)^2}}\right) $$ hence $0 \leq |x| - f_n(x) < \frac{1}{n}$ for every $x$, which implies that $f_n(x) \to |x|$ uniformly. On the other hand $$ \lim_{n\to\infty} f'_n(x) = \begin{cases} 1 & \text{ if $x>0$} \\ 0 & \text{ if $x=0$} \\ -1 & \text{ if $x<0$} \end{cases} $$ So we have that $f'_n$ is a sequence of continuous functions that converge to a discontinuous function, so its convergence cannot be uniform.