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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?

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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.

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  • $\begingroup$ As an idea, try looking at the sequence $f+\frac{1}{n}\phi(nx)$, where $\phi$ is a bounded function with derivative $1$ at $0$. $\endgroup$ Apr 29 '14 at 17:54
  • $\begingroup$ What if we flipped $f$ and $f'$ ? $$$$ Let $\{f'_n\}$ be a Cauchy sequence. By Cauchy criterion for uniform convergence, $f'_n$ converges uniformly to some $f'$. Does that imply that $f_n \rightarrow f$ uniformly? $\endgroup$ Apr 29 '14 at 18:12
  • $\begingroup$ Up to a constant, yes. This is just fundamental theorem of calculus + interchanging integration and uniform limits. Generally, integration behaves well numerically, differentiation does not. $\endgroup$ Apr 29 '14 at 18:21
  • $\begingroup$ @Katestrophical see math.stackexchange.com/q/214218/99325 $\endgroup$
    – derivative
    Apr 29 '14 at 18:23
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This is a counterexample. Take e.g. $f_n(x) = \frac{1}{n} (\sqrt{1+(n x)^2} - 1)$. We have that $f_n(0) = 0$, and $f'_n(x) = \dfrac{nx}{\sqrt{1+(nx)^2}}$ is continuous. We also have that $\lim_{n\to\infty} f_n(x) = |x|$, and $$ |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.

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