Showing that $f\colon (-1,1)\rightarrow \mathbb R$ is constant

Question

Let $f\colon(-1,1)\to \mathbb{R}$ be a real valued function. $f$ is continuous at $0$ with $f(0)=k$. If $f(x)=f(x^2)$ for all $x\in (-1,1)$ then $f(x)=k$ for all $x$ in domain.

I am trying to prove it rigorously. It is given to me that $f$ is continuous at $0$. By definition, it means

for every $\epsilon>0$, there exists a $\delta>0$ such that $|x-0|<\delta\implies |f(x)-k|<\epsilon$

Now $f(x)=f(x^2)$ so that means we also have $|f(x^2)-k|<\epsilon$

But I am not able to proceed further here. I request you to give me some hints so that I can solve it on my own.

Edit (after positing the question magically I got some idea:)):

I take the sequence $s_n=x^{2n}$.Since $|x|<1$, then $\lim_{n\to\infty} f(x^{2n})=f(0)$. Therefore $f(x)=k$ as $f(x^{2n})=f(x)$

Edit 2: The above sequence is incorrect. We have to take $s_n=x^{2^n}$. I thank @Kavi for pointing it out.

Answer 1

You have made a mistake. You get $f(x)=f(x^{2^{n}})$, not $f(x)=f(x^{2n})$. Since $f(x)=f(x^{2^{n}})$ for any positive integer $n$ (by a simple induction argument), and for $|x|<1$ we have $x^{2^{n}}\to 0$ we get $f(x^{2^{n}}) \to f(0)=k$. Hence $f(x)=k$ for any $x$.

Answer 2

Assume there are $x_1,x_2 \in (-1,1)$, $x_1 \not =x_2$ s.t.

$f(x_1)\not = f(x_2).$

$a_n:= (x_1)^{2n}$ , $n=1,2,3...$

$b_n:=(x_2)^{2n} , n=1,2,3,..$.

Then

$\lim_{ n \rightarrow \infty} a_n= \lim_{n \rightarrow \infty } b_n=0.$

Since $f$ continuos

$\lim_{n \rightarrow \infty} f(a_n)=f(0)=0$, and

$\lim_{n \rightarrow \infty} f(b_n)=f(0)=0.$

On the other hand:

$f(x_1)=f(a_n)$ , $n=1,2,3....$

$f(x_2)= f(b_n)$ , $n =1,2,3...$

Hence

$\lim_{n \rightarrow \infty}f(a_n)=f(x_1),$

$\lim_{n \rightarrow \infty}f(b_n)= f(x_2),$

Since $f(x_1) \not = f(x_2)$, a contradiction.