Continuous $f$ such that $f(x)=f(x^2)$ is constant?

Question

Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous function such that $f(x)=f(x^2)$ for all $x\in\mathbb{R}$. I've proven that also $f\left(y^{(2^{-n})}\right)=f(y)$ for all $n\in\mathbb{N}, y\geq0$. How can I deduce that $f$ is constant?

Answer 1

For $x>0$, as $x^{1/2^{n}}\rightarrow 1$, then $f(x)=f(x^{1/2^{n}})\rightarrow f(1)$, so $f(x)=f(1)$. Now $f(0)=\lim_{x\rightarrow 0^{+}}f(x)=f(1)$. And we know that $f$ is an even function.

Answer 2

If $f$ is continous, for all sequence $(x_n)_{n \in \mathbb{N}}$ such as $x_n \underset{n \rightarrow +\infty}{\rightarrow}x$ then $$ f\left(x_n\right) \underset{n \rightarrow +\infty}{\rightarrow}f\left(x\right) $$ It should help you conclude.