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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.

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2 Answers

up vote 3 down vote accepted

$f$ 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 $2$.

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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.

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