I want to show $f(f(x))=-x^{3}+g(x)$ has no continuous solution $f:\mathbb{R}\to\mathbb{R}$.Here $g$ is a continuous periodic function with its period $T>0.$ Any help will be thanked.
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1First show that $f$ is proper, then that $f(f(x)) \sim -x^3$. In particular $xf(f(x)) < 0$ for large $|x|$. Thus there are large $C,A,B > 0$, signs $s_-,s_+$ such that $f(\pm([A,\infty)) \subset s_{\pm}[B,\infty)$, and if $|x| \geq C$ then $|f(x)| \geq A$. Assume that $s_+=1$, then for all $x >C$, $f(x) >A > 0$ so $f(f(x)) > B$ and $xf(f(x)) > 0$, impossible. So $s_+=-1$. Similarly $s_-=-1$, thus if $x > C$, $f(x) < -A$, thus $f(f(x)) > -(-B)=B > 0$ ie $xf(f(x)) > 0$. We have a contradiction in all cases. – Aphelli Jul 31 '19 at 12:06
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@Mindlack I think s-=1? – Tree23 Jul 31 '19 at 12:38
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No: if $s_-=1$, then $f$ maps “large” negative numbers to “large” negative numbers, so $xf(f(x)) > 0$ for $|x|$ large enough, $x < 0$. – Aphelli Jul 31 '19 at 13:05
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The ideas in Is there a continuous $f$ satisfying $f(f(x))=-x^3+x$? and Functional Square Root of $1-x^3$ might help. – Lutz Lehmann Jul 31 '19 at 14:08
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A general discussion of functional square roots with links for deeper theory and numerical experiments. By the general principles there one would expect that $f^{\circ2n}(x)\sim -x^{3^n}$ for large $x$, so that $f(x)\sim -x^{\sqrt{3}}$, which has no real interpretation for negative $x$. – Lutz Lehmann Jul 31 '19 at 14:22