The question is :
Let $f : \mathbb {R} \longrightarrow \mathbb {R}$ be a continuous function such that $f(x) = f(x^2)$.Then show that $f(x) = f(0)$ , $\forall x \in \mathbb {R}$.
How can I solve it?Please help me.Thank you in advance.
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4Similar: http://math.stackexchange.com/q/338802/321264, http://math.stackexchange.com/q/980236/321264 – StubbornAtom Jul 29 '16 at 17:42
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1I say duplicate. – Jul 29 '16 at 17:43
2 Answers
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Big hints:
To begin with, $f(x)=f(\lvert x\rvert)$.
Let $0\le x<1$. $$f(x)-f(0)=f(x^2)-f(0)=f(x^4)-f(0)=f(x^8)-f(0)=\cdots$$
How much can $f(1)$ be, in light of (2) ?
Let $x>1$. $$f(x)-f(1)=f(x^{1/2})-f(1)=f(x^{1/4})-f(1)=\cdots$$
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Hint: you can show that for all $x$ and all integers $n$, $f(x^{2^n})=f(x)$.
What happens letting $n \to \infty$ if $|x|<1$ ? What if $|x| \geq 1$ ?
Dark
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Yeah @Dark I have got it.Since $|x| < 1$ , $x^{2n}$ tends to $0$ as $n$ is tending to infinity. – Jul 29 '16 at 17:44