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Find all continuous functions $f$ on $\mathbb{R}$ satisfying $f(x)=f(\sin x)$ for all $x$.

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I removed the different equations and functional analysis tag. –  Aryabhata Nov 11 '10 at 23:32
    
Is this homework? –  Qiaochu Yuan Nov 11 '10 at 23:57

3 Answers 3

For those of you who wonder why the sequence $x_{n+1} = \sin x_n$ tends to zero, note that after the first application of sine, we get a number in $\pm 1$, and from there on, the numbers keep their sign, and monotonically approach zero (since $|\sin x| \leq |x|$). So the sequence must approach a limit, which is also a fixed point of the sine function. The only fixed point of sine is zero.

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$f(x)$ must be constant.

We must have that if $x_1 = x$ and $x_{n+1} = \sin x_n$, then $f(x_{n+1}) = f(x_n) = f(x_{n-1}) \dots = f(x)$

Now $x_n \to 0$ as $n \to \infty$. Thus $f(x) = f(0)$

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For each $x$, $\sin(\sin(\sin...\sin(x)...))$ (n-times) goes to zero as $n$ goes to infinity, so by continuity f(x)=0 $f(x)=f(0)$.

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I don't think anything restricts f(0), but then f must be constant –  Ross Millikan Nov 11 '10 at 23:30
    
@Ross: Thank you for the correction. –  Jonas Meyer Nov 11 '10 at 23:33

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