Let $f: \mathbb R \to \mathbb R$ be continuously differentiable function.

Define $g_n$ as $g_n(x)= 2^n \{f(x+2^{-n})-f(x)\}$.

Prove that the sequence $\{g_n\}$ converges uniformly to $f'$ on the interval $[0, 1]$.

I already know that $\{g_n\}$ converges pointwise to $f'$, but I have trouble figuring out the uniform convergence. I'd like to use the compactness of $[0, 1]$ and continuity of $f'$, but I don't know how to apply them. Does anyone have any ideas?

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    $\begingroup$ Uniform continuity of $f'$ on $[0,2]$ and mean value theorem. $\endgroup$ – Daniel Fischer Jul 7 '17 at 16:04

Let $\epsilon>0$ be given. Since $f'$ is continuous on a compact set, it is uniformly continuous, so there exists $\delta>0$ such that $|f'(x)-f'(y)|<\epsilon$ whenever $|x-y|<\delta$. Then, by the mean value theorem, $f(x+2^{-n})-f(x) = 2^{-n}f'(\theta)$ for some $\theta\in(x,x+2^{-n})$. Now if $n$ is such that $2^{-n}<\delta/2$, we have $$ \big|f'(x)-2^n[f(x+2^{-n})-f(x)]\big| = |f'(x)-f'(\theta)|<\epsilon, $$ so that the convergence is uniform, as desired.

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