# to prove that $f(x)=f(0)$

Let, $f:(-1,1)\rightarrow\mathbb{R}$ be a function continuous at $x=0$ and given that $f(x)=f(x^2)$ for all $x\in(-1,1)$. Prove that, $f(x)=f(0)$ $\forall x\in(-1,1)$. Ok. First give me some hint.

• Hint first huh? – Kaster Dec 19 '13 at 19:51
• what did u say? @Kaster – Topology Dec 19 '13 at 19:53
• Hint: Suppose not. – Aeryk Dec 19 '13 at 19:53
• Hint: $f(0.1) = f(0.01) = f(0.0001) = \dots =f(0)$. – Yury Dec 19 '13 at 19:54
• Cancelled roughly twenty misplaced $. – Did Jun 19 '14 at 12:23 ## 3 Answers $$f(x)=f(x^2)=f(x^4)=\ldots = f(x^{2^n})$$ where$n$is any natural number So if we let$n$be infinitely large then, since$x\in (−1,1)$,$x^{2^n}$tends to$0$Hence, $$f(x)=f(0)$$ By induction we have $$f(x)=f(x^{2^n}) \quad \forall x \in (-1,1), \quad \forall n \in \mathbb{N}.$$ Thus $$f(x)=\lim_nf(x^{2^n})=f(0) \quad \forall x \in (-1,1).$$ Hint: This can be done by contradiction, with a sequential characterization of continuity. • so in contrary there r two cases. right? first$f(x)$>$f(0)\$? like this? – Topology Dec 19 '13 at 19:55
• @Mathematics There's no reason to separate into cases. – user61527 Dec 19 '13 at 19:57
• So how shall we proceed? @T.Bongers – Topology Dec 19 '13 at 19:57
• @Mathematics Think about the hint for a while. – user61527 Dec 19 '13 at 19:58
• Ok..Ok @T.Bongers – Topology Dec 19 '13 at 19:58