Uniform convergence, differentiability theorem's converse

Question

Suppose $\{f_n \}$ is a sequence of differentiable functions on $[a,b]$ such that $f_n \to f$ uniformly, and $f'(x) = \lim_{n \to \infty} f_n'(x)$ (pointwise). Then is it true that $f_n'$ converges uniformly?

My initial reaction is that the answer is yes, because all the differentiable $f_n$'s will be able to fit inside an "$\epsilon \text{-thick}$ tube" around $f$, but intuition for sequences of functions rarely seem to be correct. I'd appreciate some help with this; thank you!

Answer 1

This is not technically a counterexample, since it does not satisfy the second condition, but I believe it gives the right idea of why it will not work (I am not sure).

Consider the sequence $$ f_{n}(x) = \frac{1}{n} \sin(n^{2}x). $$ This converges uniformly to $f(x) = 0$, but clearly the derivative is not going to converge uniformly to zero. Of course, it will not converge at all; the point, however, is that even if the change over an interval is bounded, the rate of change can be unbounded if the size of the interval shrinks fast enough.

So probably a counterexample could be constructed by replacing the $n^{2}x$ with something more complicated. I think perhaps that $nx^{-1}$ might work (basically the topologist's sine function but with the scaling factor).

Answer 2

Let $f_n(x)=\frac {x^{n}} n$ on $[0,1]$ Then $f_n \to 0$ uniformly and $f_n'(x) \to 0$ for $x <1$, $f_n'(x) \to 1$ for $x=1$. Since the limit is not continuous it follows that $f_n'$ does not converge uniformly.