Let $f:\Bbb R\to\Bbb R$ be a function satisfying $$f(kx)=f(x),\space\forall x\in\Bbb R\space\text{ and some } k>0,k\ne 1$$

Does $f$ have to be bounded?


I think $f$ has to be bounded and would like to verify it and improve what I have so far.

$$f(kx)=f(x),\space\forall x\in\Bbb R \implies f\left(k^nx\right)=f\left(\frac{x}{k^m}\right),\forall m,n\in\Bbb N$$

Since $f$ is defined on the whole $\Bbb R,(\forall x\in\Bbb R)(\exists ! y\in\Bbb R)$ s. t. $y=f(x)$.

Let's look at the closed interval $[-1,1]$. $$f\left([-1,1]\right)=f\left([-k,k]\right)=f\left(\left[-\frac1k,\frac1k\right]\right)$$ so, I think, we have actually covered all the possible outputs of $f(x)$. No matter how big $|x|$ is, there is always some $y\in\Bbb R$ with $|y|<|x|$ s. t.$x=ky$ and, hence $f(x)=f(y)$, which we have already found in $[-1,1]$.

May I ask if my arguments are valid?


I was thinking about the periodic composition $(f\circ g)(x)$, where $f$ has the above properties and $g$ isn't periodic. We could take $g(x)=k^{\frac{x}n}$. Then: $$f(g(x))=f\left(k^{\frac{x}n}\right)=f\left(k\cdot k^{\frac{x}n}\right)=f\left(k^{\frac{x+n}n}\right)$$

and $f(g(x))$ has a prime period $\tau_0=n$.

If $g$ were periodic, $f(g(x))$ would definitely be periodic, no matter what $f$ were. The examples I thought of were periodic functions like constants or everywhere discontinuous functions like the Dirichlet function. However, I realized $f$ doesn't necessarily need to be periodic, so I focused just on the weaker property of boundedness.

Thank you in advance!

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    $\begingroup$ Why should $f([-1,1])$ be bounded? $\endgroup$ – Daniel Fischer Oct 24 at 14:40
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    $\begingroup$ This is true if $f$ is continuous. $\endgroup$ – Kenta S Oct 24 at 15:12
  • $\begingroup$ @DanielFischer, I made a cardinal error and totally forgot about other cases when $f$ isn't continuous, unlike when it is and is bounded by the Bolzano-Weierstrass theorem. $\endgroup$ – Invisible Oct 24 at 15:22
  • $\begingroup$ Related question. $\endgroup$ – Invisible Nov 8 at 7:19

(Suppose $k=2$)

Assuming continuity:

$\sup_{x\in\mathbb R} \left| f(x)\right|=\sup_{x\in[-1,1] }\left| f(x)\right|<\infty$

Without continuity:

Let $f((2n+1)2^m)=2n+1$ for $n,m\in \mathbb Z$ and $f(x)=0$ for others.

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$$ f(x)=\begin{cases} (1-\text{frac}(\log_k|x|))^{-1} & x\ne 0\\ 0 & x=0 \end{cases} $$

Here, $\text{frac}(x):=x-\lfloor x\rfloor$ denotes the fractional part of $x$. Then $f(x)$ is unbounded.

P.S. Proof that $f(x)$ is unbounded.

Let $M>1$ be arbitrary, and let $x=k^{1-1/M}$. We then have $f(x)=(1-\text{frac}(1-\frac1M))^{-1}=M$.

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  • $\begingroup$ That is quite amazing! Is there any analytic way to prove it is not bounded but not just giving an example? $\endgroup$ – AlexanderGrey Oct 24 at 15:55

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