TITLE: Suppose that $f(x)$ is continuous on $(0, \infty)$ such that for all $x > 0$,$f(x^2) = f(x)$. Prove that $f$ is a constant function.
QUESTION [9 upvotes]: Suppose that $f(x)$ is continuous on $(0, +\infty)$ such that for all $x > 0$,$f(x^2) = f(x)$. Prove that $f$ is a constant function. My attempt is to show that for any point $a \neq b$ , we have $f(a)=f(b)$. But I have no idea on how to get this. Anyone can help?

REPLY [6 votes]: As I understand it:
$$f(x) = f(x^2)\quad  \Rightarrow$$
$$ f(x^\frac{1}{2}) = f(x)\quad \Rightarrow$$
$$f(x) = f(x^\frac{1}{2^n})_{n \in N}$$
take $a,b \in (0,+\infty)$, then:
$$f(a) =\lim_{n\rightarrow\infty} f(a^\frac{1}{2^n})=f(1)$$
$$f(b) =\lim_{n\rightarrow\infty} f(b^\frac{1}{2^n})=f(1)$$
$$f(a)=f(b)\Rightarrow_{def}\mathtt {f\quad is\quad constant}$$
Is it correct? Is it rigorous? My thought process was: continuity has limit in its definition, so I probably have to use lim in writing my proof.