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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$$f(x)=f(x^2)=f(x^4)=\ldots = f(x^{2^n})$$ 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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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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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 '13 at 19:55
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$\begingroup$ @Mathematics There's no reason to separate into cases. $\endgroup$ – user61527 Dec 19 '13 at 19:57
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$\begingroup$ @Mathematics Think about the hint for a while. $\endgroup$ – user61527 Dec 19 '13 at 19:58
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$. $\endgroup$ – Did Jun 19 '14 at 12:23