TITLE: Solving functional equation $2f(x) = f(2x)$
QUESTION [7 upvotes]: $f(x)$ is a $\mathbb{R} \rightarrow \mathbb{R}$ differentiable function satisfying the following equation: 
$$2f(x) = f(2x).$$
Can it be proved that $f(x) = kx$ for some $k$?
Note that if $f(x)$ is in $\mathcal{C}^1$, it can be proved in the following way:
$$g(x) := \bigg\{
\begin{array}{ll}
f(x)/x & x \in \mathbb{R}\backslash\{0\} \\
\lim_{x\rightarrow0}f(x)/x = f'(0) &x=0
\end{array}
\bigg.
$$
is a continuous function in $\mathbb{R}$, satisfying
$$ g(\ln x) = g(\ln x^2) \text{ for } x\in(0,+\infty).$$
Therefore, $\forall x \in(0,+\infty)$
$$ g(\ln x) = g(\ln x^{1/2}) = \lim_{n\rightarrow\infty}g(\ln x^{1/2^n}) = g(0),$$
which means $g(x)$ is a constant and $f(x) = kx$.

REPLY [2 votes]: Yes, and your proof already works: you only use the continuity of $g$ at 0, which  holds by the definition of $g$, without needing to assume $f \in \mathcal C^1$.