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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.

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    $\begingroup$ Hint first huh? $\endgroup$
    – Kaster
    Dec 19, 2013 at 19:51
  • $\begingroup$ what did u say? @Kaster $\endgroup$
    – Topology
    Dec 19, 2013 at 19:53
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    $\begingroup$ Hint: Suppose not. $\endgroup$
    – Aeryk
    Dec 19, 2013 at 19:53
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    $\begingroup$ Hint: $f(0.1) = f(0.01) = f(0.0001) = \dots =f(0)$. $\endgroup$
    – Yury
    Dec 19, 2013 at 19:54
  • $\begingroup$ Cancelled roughly twenty misplaced $. $\endgroup$
    – Did
    Jun 19, 2014 at 12:23

3 Answers 3

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By induction we have $$ f(x)=f(x^{2^n}) \quad \forall x \in (-1,1), \quad \forall n \in \mathbb{N}. $$ Thus $$ f(x)=\lim_nf(x^{2^n})=f(0) \quad \forall x \in (-1,1). $$

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$$f(x)=f(x^2)=f(x^4)=\ldots = f(x^{2^n})$$ 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, $$f(x)=f(0)$$

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Hint: This can be done by contradiction, with a sequential characterization of continuity.

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  • $\begingroup$ so in contrary there r two cases. right? first $f(x)$>$f(0)$? like this? $\endgroup$
    – Topology
    Dec 19, 2013 at 19:55
  • $\begingroup$ @Mathematics There's no reason to separate into cases. $\endgroup$
    – user61527
    Dec 19, 2013 at 19:57
  • $\begingroup$ So how shall we proceed? @T.Bongers $\endgroup$
    – Topology
    Dec 19, 2013 at 19:57
  • $\begingroup$ @Mathematics Think about the hint for a while. $\endgroup$
    – user61527
    Dec 19, 2013 at 19:58
  • $\begingroup$ Ok..Ok @T.Bongers $\endgroup$
    – Topology
    Dec 19, 2013 at 19:58

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