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Prove that there is no continuous function $f:\mathbb{R}\to\mathbb{R}$ such that for all $x\in \mathbb{R},f(f(x))= \cos x$.

I would so appreciate if someone can give me a hint!

user26857
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    See http://mathoverflow.net/questions/17605/how-to-solve-ffx-cosx (and also http://math.stackexchange.com/questions/65876/thoughts-about-ffx-ex, http://math.stackexchange.com/questions/312385/continuous-function-f-mathbbr-to-mathbbr-such-that-ffx-x) – Watson Aug 25 '16 at 12:59
  • can you @Watson please write this excellent proof :) – Hamza Aug 25 '16 at 13:08

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I'm following Sergei Ivanov's argument from this answer.

Suppose that $f$ is such a continuous function. Using real analysis, the equation $\cos(x) -x =0$ has a unique real solution $x_0$. The number $y_0=f(x_0)$ satisfies $$\cos(y_0)=f(f(y_0))=f(f(f(x_0))) = f(\cos(x_0))=f(x_0)=y_0,$$ so that it is a fixed point of $\cos$, which yields $y_0=x_0$.

Therefore, $x_0$ being a fixed point of $\cos = f \circ f$, you could find a neighborhood $V$ of $x_0$ such that $f$ is injective in $V$. In particular, $f\vert_V$ is monotone. In all cases, $f \circ f$ is increasing on $V$. But $\cos$ is decreasing around $x_0$, as you can see for instance here:

$\qquad\qquad\qquad$enter image description here

Community
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Watson
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    The reason $x_0$ must have a neighborhood $V$ on which $f$ is injective is not that $x_0$ is a fixed point of $f$. Indeed, $h(x)=x^2$ has a fixed point at $0$ and there is no neighborhood of $0$ on which $h$ is injective. Rather, the point is that $f \circ f = \cos$, and $\cos$ is injective on a small neighborhood of $x_0$, so $f$ must be too. –  Aug 25 '16 at 13:39
  • @mathguy: yes, obviously. I swapped $f$ and $\cos$… Thank you! – Watson Aug 25 '16 at 13:54