Let us consider the following functional equation $$ f(f(x))=-x^3+\sin(x^2+\ln(1+|x|)) $$ How to prove that there is NO continuous function $f:\mathbb{R}\to\mathbb{R}$ satisfies the equation above?

First I notice that $$ F(x):=-x^3+\sin(x^2+\ln(1+|x|)) $$ has only one fixed point $0$. From this it follows that $$ f(0)=0. $$ Otherwise, if we assume that $a:=f(0)>0$, then $f(a)=F(0)=0$. It is easy to see that $f(x)$ has a fixed point in $(0,a)$ by intermediate value theorem. Similarly, we can rule out the case $f(0)<0$. Therefore we have $f(0)=0$

I want to show that $f$ has a nonzero fixed point $x^*$, from which we find a contradiction. This is my initial idea. But I don't how to find such non-zero fixed point.

If I find a closed interval $[a,b]$, which does not include 0, such that $f([a,b])\subset [a,b]$. Then we know that $f$ has at least one fixed point in $[a,b]$ by intermediate value theorem.

Maybe, my idea is wrong. Are there some other approaches?

What follows is the plot of function $F$, which may give you some hints: enter image description here


Another approach: look to large values. Let $g(x)$ be that expression on the right, and note that $\lim_{x\to\infty} g(x)=-\infty$ and $\lim_{x\to-\infty} g(x)=\infty$. Also, there is some $M$ such that $g$ is strictly decreasing on both $(-\infty,-M)$ and $(M,\infty)$. Suppose there is a continuous $f$ with $f\circ f = g$. Then, since $g$ is injective on $(-\infty,-M)$, $f$ is also injective on $(-\infty,-M)$, hence monotone. Let $L=\lim_{x\to\infty} f(x)$, in the extended real line. If $L$ is finite, then by continuity $f(L)=\lim_{x\to\infty}f(f(x))=-\infty$. That's impossible, so either $L=\infty$ or $L=-\infty$. If $L=\infty$, then $\lim_{x\to\infty} f(f(x)) =\lim_{y\to\infty} f(y) = \infty$, a contradiction. If $L=-\infty$, then $-\infty = \lim_{x\to\infty} f(f(x))=\lim_{y\to-\infty} f(y)$, and we'll have the same problem on the other side ($\lim_{x\to-\infty} f(f(x))=\lim_{y\to\infty} f(y)=-\infty$). Either way, we can't have both $\lim_{x\to\infty} f(f(x))=-\infty$ and $\lim_{x\to-\infty} f(f(x))=\infty$ on the same globally continuous $f$. There is no such function.


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