Is there any continuous function $f : (-2 \: \: 2) \to (-2 \: \: 2)$ such that $f(x)=f(x^2)$ other than constant Is there any continous function $f : (-2 \: \: 2) \to  (-2 \: \: 2)$ such that $f(x)=f(x^2)$ other than constant function
The only clue i could get  is $$f(x)=f(x^2)$$ and now replacing $x$ with $-x$ we get
$$f(-x)=f(x^2)$$ so $f(-x)=f(x)$ so $f$ is even function
 A: Not sure I interpret everything correctly, but if we interpret the question to mean that $f(x)=f(x^2)$ is required whenever both $x$ and $x^2$ are in the range $(-2,2)$ then it seems to me that we have elbow room for the following.

$f$ must be constant in the interval $[-\sqrt2,2)$, but can be anything continuous in $(-2,-\sqrt2)$ as long as it has the right limit when $x\to-\sqrt2-$.



*

*When $|x|<1$ we have $x^{2^n}\to0$, and continuity then forces $f(x)=\lim_{n\to\infty}f(x^{2^n}=f(0)$.

*When $1\le x<2$ we have $x^{2^{-n}}\to1$ and continuity forces $$f(x)=f(\sqrt x)=f(x^{1/4})=\cdots=f(1)=\lim_{t\to1-}f(t)=f(0).$$

*When $-\sqrt2 <x\le -1$ we have $f(x)=f(x^2)$, but $x^2\in [1,2)$ so by the previous bullet $f(x)=f(x^2)=f(0)$.


But if $x\le -\sqrt2$ then $x$ is not a square, and $x^2$ is out of the range, so there are no constraints.
A: If
$|x| < 1$
then
$f(x)
=f(x^2)
=f(x^4)
=...
=f(x^{2^n})
$
for any $n \in \mathbb{N}
$.
Since $f$ is continuous,
letting $n \to \infty$,
$f(x) = f(0)$.
Therefore
$f$ is constant in
$(-1, 1)$.
If
$|x| > 1$,
applying the same reasoning,
eventually
$|x^{2^n}|
\gt 2$
so we seem to need
to have $f(x)$
defined for
$|x| > 2$.
I don't know what to make of this.
A: For all $x \in (-2,2)$ we have
$$
f(x)=f(x^2)=f(x^4)=f(x^8)=f(x^{16})=\ldots=f\left(x^{2^n}\right)
$$
with $n$ a positive integer. Since $f$ is continuous, we have
$$
f(x)=\lim_nf\left(x^{2^n}\right)=f(0) \quad \forall |x|<1
$$
Again, by continuity we have 
$$
f(\pm1)=f(1)=\lim_{x\to1^-}f(x)=f(0).
$$
When $1<x<2$, we have
$$
f(x)=f(x^{2^{-1}})=f(x^{2^{-2}})=\ldots=f\left(x^{2^{-n}}\right)
$$
therefore
$$
f(x)=\lim_nf\left(x^{2^{-n}}\right)=f(x^0)=f(1)=f(0).
$$
Since $f(-x)=f(x)$, it follows that $f(x)=f(0)$ for $-2<x<-1$. 
Hence 
$$
f(x)=f(0) \quad \forall x\in (-2,2)
$$
A: Continuing the previous, let $f(x)=f(x^{1/2})...f(x^{1/2^n})$....this converges to 1 so it gets that value of $f$ at 1 which is the same value as $f$ has in $(-1,1)$. You can process thus for $|x|>1$.
