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

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?

• hi, Katestrophical, were you able to find the answer to your question? Jan 3 at 16:21

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.
• As an idea, try looking at the sequence $f+\frac{1}{n}\phi(nx)$, where $\phi$ is a bounded function with derivative $1$ at $0$. Apr 29 '14 at 17:54
• What if we flipped $f$ and $f'$ ?  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? Apr 29 '14 at 18:12
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.