TITLE: $f$ differentiable at $0\iff\lim_{x\to 0}\frac{f(2x)-f(x)}{x}$ exists
QUESTION [6 upvotes]: Let $f$ be a real function that is continuous at $0$.
Prove that $f$ differentiable at $0\iff\lim_{x\to 0}\frac{f(2x)-f(x)}{x}$ exists

The $\Rightarrow$ part is trivial, and $\lim_{x\to 0}\frac{f(2x)-f(x)}{x}=f'(0)$
What about $\Leftarrow$ ? I find it surprisingly hard, as I have made no progress toward a proof.
Any suggestion is welcome.

REPLY [5 votes]: Concerning the difficult part: After substracting a suitable linear function we may assume
$$\lim_{x\to0}f(x)=0,\qquad \lim_{x\to0}{f(2x)-f(x)\over x}=0\ ,$$
and we have to prove that $\lim_{x\to0}{f(x)\over x}=0$.
Given an $\epsilon>0$ there is a $\delta>0$ such that
$$\left|f(x)-f\left({x\over2}\right)\right|<{\epsilon\over2}|x|\qquad\bigl(|x|<\delta\bigr)\ .$$
Now let an $x$ with $|x|<\delta$ be given. Then for arbitrary $N\geq1$ we have
$$f(x)=\sum_{k=0}^{N-1}\left(f\left({x\over 2^k}\right)-f\left({x\over 2^{k+1}}\right)\right)+f\left({x\over 2^N}\right)$$
and therefore
$$\bigl|f(x)\bigr|\leq{\epsilon\over2}|x|\sum_{k=0}^{N-1}{1\over 2^k} +f\left({x\over 2^N}\right)\ .$$
Letting $N\to\infty$, keeping $x$ fixed, we obtain
$$\bigl|f(x)\bigr|\leq\epsilon|x|\ .$$
Since $\epsilon$ was arbitrary and this is true for all $x$ with $|x|<\delta$ the claim follows.