$f$ differentiable at $0\iff\lim_{x\to 0}\frac{f(2x)-f(x)}{x}$ exists 
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.
 A: 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.
A: Hint Assume $f(0)=0$ and start with $\frac{f(x)-f(\frac{x}{2})}{x}$ than try to find the limit of $$\frac{f(x)-f(\frac{x}{2^n})}{x}$$
At last I am behind my desktop.
Let's assume that the ratio $\frac{f(2x)-f(x)}{x}$ has got a limit $a$ when $x\to 0$. By changing the function to $f(x)-ax$ we could assume $a=0$. Set
$$\epsilon(x)=\begin{cases}\frac{f(2x)-f(x)}{x},&\text{if }x\neq 0\\0,&\text{if }x=0\end{cases}$$
$\epsilon$ is continuous.
We have $\forall n\geq 1$
$$ f(x)-f(\frac{x}{2^n})
=\sum_{k=1}^n\left( f(\frac{x}{2^{k-1}})-f(\frac{x}{2^k})\right)
=\sum_{k=1}^n \frac{x}{2^k} \epsilon(\frac{x}{2^k})$$
Let $n\to\infty$ we have formally
$$f(x)-f(0)=x\underbrace{\sum_{k=1}^{+\infty}\frac{\epsilon(x/2^k)}{2^k}}_{\tau(x)}$$
$\epsilon$ being continuous at $0$ allows to find an interval $[-\alpha,\alpha]$ where it is bounded and the series $\tau(x)$ converges normally and is therefore continuous; the ratio $\frac{f(x)-f(0}{x}\to\tau(0)=0$
A: Suppose that $f$ is differentiable at $0$. Then $g(x)=f(2x)$ is also differentiable at $0$. It follows that the following limits exist
$$
\lim_{x\rightarrow 0}\frac{f(2x)-f(0)}{x-0}=\lim_{x\rightarrow 0}\frac{g(x)-g(0)}{x-0}; \quad \lim_{x\rightarrow 0}\frac{f(x)-f(0)}{x-0}
$$
Hence
$$
\lim_{x\rightarrow 0}\frac{f(2x)-f(x)}{x}=\lim_{x\rightarrow 0}\left(\frac{f(2x)-f(0)}{x-0}-\frac{f(x)-f(0)}{x-0}\right)
$$
exists. 
Now, we give a counterexample to show that only the existence of the limit
$$
\lim_{x\rightarrow 0}\frac{f(2x)-f(x)}{x}
$$ 
(without the assumption that $f$ is continuous at $0$) does not imply that $f$ is differentiable at $0$. 
Consider the function
$$
f(x)=
\begin{cases}
1 &\text{if} \quad x\in\mathbb{Q},\\
0 &\text{if} \quad x\notin\mathbb{Q}.
\end{cases}
$$
Then $f(2x)=f(x)$ for all $x\in\mathbb{R}$ and so
$$
\lim_{x\rightarrow 0}\frac{f(2x)-f(x)}{x}=0.
$$ 
But $f$ is not continuous and differentiable at $0$
A: Based on the idea of @marwalix, write 
\begin{eqnarray}
f(x) - f(x/2) &=& k \,x/2 + \epsilon(x/2) \cdot x/2 \\
f(x/2) - f(x/4) &=& k\, x/4 + \epsilon(x/4) \cdot x/4 \\
\ldots
\end{eqnarray}
add up on both sides, use $f$ continuous at $0$ and get 
$$ f(x) - f(0) = k\, x + \eta(x) \cdot x$$
