if $f_n(x)$ converges uniformly to a function $f(x)$ does $f_n'(x)$ converge uniformly to $f'(x)$?

Question

Let [a,b] denote a finite interval and consider a sequence $\{f_n(x)\}_{n=0}^\infty$ in $C^1([a,b])$. if $f_n(x)$ converges uniformly to a function $f(x)$ on $[a,b]$, does $\{f_n'(x)\}$ converge uniformly to $f'(x)$?

My intuition for this problem is that the converse is true, but I'm not sure how to justify this.

Any help is appreciated!

Answer 1

Consider:

$$f_n(x) = \frac1{n} \sin(nx)$$

$(f_n)$ converges uniformly to $0$ (on $\Bbb R$), but $f_n'$ doesn't even converge pointwise

Answer 2

The answer is no.

One may consider $$ f_n(x):=\sqrt{x^2+\frac1n},\quad x \in [0,1], $$ we have, for $n\geq1$, $$ \left|f_n(x)-|x|\right|=\frac{\frac1n}{\sqrt{x^2+\frac1n}+|x|}\leq \frac1{\sqrt{n}} $$ thus the convergence of $\left\{ f_n \right\}$ is uniform over $[0,1]$.

On the other hand, we have $$ f'_n(x)=\frac{x}{\sqrt{x^2+\frac1n}} \longrightarrow f'(x)= \begin{cases} 0 & \text{if $\,x=0$,} \\[2ex] 1 & \text{if $\,0<x\leq1$,} \end{cases} $$ the latter convergence can't be uniform since each $f'_n$ is continuous over $[0,1]$ but $f'$ is not.