# If $f\colon (0+\infty) \to \mathbb R$ and $f(x)=f(2x)$ what can we say about $f$?

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:

1. 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 $f(x)=l$.

2. If $\displaystyle \lim_{x \to +\infty} f(x)=l \in \mathbb R$ then $f$ is constant (same proof, just replace $-n$ with $n$).

3. 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.

-

$f$ is determined by its values on the interval $[1,2)$. First of all, $f(2)=f(1)$. For any $x>0$ there is a unique $k\in\mathbb{Z}$ such that $2^kx\in[1,2)$; then $f(x)=f(2^kx)$. It follows that $f$ is bounded and $$\max_{x>0}f(x)=\max_{1\le x<2}f(x),\quad \min_{x>0}f(x)=\min_{1\le x<2}f(x).$$ Any $y\in[\min_{x>0}f(x),\max_{x>0}f(x)]$ is a limit value of $f$ as $x\to0^+$ and as $x\to+\infty$. The same argument holds for any $a>1$ instead of $2$.
Your reasoning seems correct, but you cannot severely curtail the general case, because of the examples $$f: x\mapsto g(\log_2(x)\bmod1)$$ where $g$ is any continuous function on $[0,1]$ with $g(0)=g(1)$, and "$y\bmod 1$" denotes the fractional part of $y\in\mathbb R$. This is in fact the general form of a solution.
You may conclude that in general $f$ is bounded, by the compactness of $[0,1]$.
You can replace $2$ by any $a>1$, if that is done in your hypothesis.