Let $f:\mathbb{R}\to\mathbb{R}$, $f(kx)=f(x)$. Show that if $f$ is continuous at $0$ then $f$ is a Constant function

Question

Let $f:\mathbb{R}\to\mathbb{R}$ , there is $k \in \mathbb{R}$, $|k|\ne 1$, so that for every $x\in \mathbb{R} , f(kx)=f(x).$

Show that if $f$ continuous at $0$ then $f$ is a constant function.

---

I tried to find something to show it from the definition of continuous function but I didn't know how to continue.

We have that $f$ continuous at $0$ if for every $\epsilon>0$ there is $\delta>0$ such that $|x-0|<\delta \implies |f(x)-f(0)|<\epsilon$

And if $f$ not continuous on $0$ is $f$ still a constant function?

Answer 1

Here is some hints:

0) Deal with the case $k=0.$

We now assume that $k\neq 0$.

1) prove that for all $x\in\mathbb{R}$, you have $f(x)=f(x/k)$.

2) Deduce you may assume $\vert k\vert<1$.

Assume for the rest of the proof that $\vert k\vert<1$.

3) Show that for all $x\in\mathbb{R}$ and all integer $n\geq 1$, you have $f(k^nx)=f(x)$.

4) Conclude using continuity at $0$.