Let $f \colon (0,+\infty) \to \mathbb R$ be a continuous function such that $ f(x)=f(2x),\qquad \forall x \in \mathbb R. $ What can we say about $f$?
An easy induction shows that $ f(x)=(2^{r}x), \qquad \forall r \in \mathbb Z $
From this I think we can show that:
if $\displaystyle \lim_{x \to 0} f(x)=l \in \mathbb R$ then $f$ is constant. Indeed, take $x \in \mathbb R$: then $ f(x)=f(2^{-n}x) $ Letting $n \to +\infty$, by continuity, we get $f(x)=l$.
If $\displaystyle \lim_{x \to +\infty} f(x)=l \in \mathbb R$ then $f$ is constant (same proof, just replace $-n$ with $n$).
If $f$ is uniformly continuous then it is constant: take a $\varepsilon>0$ and aribitrary $x,y \in \mathbb R$. Then $ \vert f(x) -f(y) \vert = \vert f(2^{-n}x)-f(2^{-n}y) \vert < \varepsilon $ if we take $n$ big enough s.t. $\displaystyle \frac{\vert x-y \vert}{2^n}<\delta$.
Do you agree with this? Do you think it is correct?
What in general case (without further assumptions on $f$)? Is $f$ bounded? If yes, how can we prove it?
What if we replace the number $2$ (which has nothing to special for me in this question) with an arbitrary positive real number $a \ge 1$?
Note: I took ispiration from this question and especially from Davide Giraudo's answer.