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Let $f$ be a real valued function on $(-1,1)$. $f$ is continuous at $0$. $f(x)=f(x^2)$ on $(-1,1)$.Then how can be $f(x)=f(0)$ on $(-1,1)$ ?

MY ATTEMPT: Given $f$ is continous at $0$. Let's choose a sequence $\{c^{2^n}\}$ converging to $0$. $\lim f(c^{2^n})=f(0)$.how can then proceed

IamKnull
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choton choton
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2 Answers2

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Let $x \in (-1,1)$. By induction we see that

$f(x)=f(x^{2^n})$ for all $n$ . Since $x^{2^n} \to 0$ as $ n \to \infty$ and $f$ is continuous, we have $f(x^{2^n}) \to f(0)$ as $ n \to \infty$ .

Fred
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We get $f(c)= f(c^{2^{n}})$ for all $n$ by induction. For $|c|<1$ we have $\lim_{n \to \infty} c^{n}=0$. This also implies $\lim_{n \to \infty} c^{2^{n}}=0$. Hence $f(c)= f(c^{2^{n}})\to f(0)$ by continuity.

Kavi Rama Murthy
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