# 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

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?

• Suppose $0<k<1$. Note that $f(x)=f(kx)=f(k^2x)= f(k^3x)= ...$. These converge to $0$, so you have (by continuity) that $f(x)=f(0)$. Commented Feb 25, 2018 at 10:06
• Not the same but similar and related: math.stackexchange.com/questions/1111361/… Commented Feb 25, 2018 at 10:48
• $|k|\ne 1$..... Commented Feb 25, 2018 at 12:21
• You are reminded to use tag that describes the questions. Your question is not about "functions", although your question has a function $f$ in it.
– user99914
Commented Feb 26, 2018 at 1:16

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 $0$.

• How would we achieve 2)? How could we deduce we may assume $|k| < 1$ and what do you exactly mean by that? What is the relevance? Thank you! Commented Dec 11, 2019 at 15:07