# If $f(kx)=f(x),\space\forall x\in\Bbb R$ and $k>0,k\ne 1$, is $f$ bounded?

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?

Thoughts:

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?

Motivation:

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.

• Why should $f([-1,1])$ be bounded? – Daniel Fischer Oct 24 at 14:40
• This is true if $f$ is continuous. – Kenta S Oct 24 at 15:12
• @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. – Invisible Oct 24 at 15:22
• – 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.

Let

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

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 $$f(x)=(1-\text{frac}(1-\frac1M))^{-1}=M$$.

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