How to analyze $f(f(x))=-x^3+\sin(x^2+\ln(1+\left|x\right|))$?

Question

Define $f\in C^{0}\left(\mathbb{R}\right)$ satisfying $f(f(x))=-x^3+\sin(x^2+\ln(1+\left|x\right| ))$. Prove that this equation has no continuous solution.

To figure out the proof, I thought like this:
If $f$ is monotonic we can conclude that $f(f(x))$ is monotonically increasing, which is contradictory to that $-x^3+\sin(x^2+\ln(1+\left|x\right|))$ can be strictly decreasing for sufficiently large or sufficiently small $x$.
So $f$ isn't monotonic. But I can't get more in this way.

Then I tried another way to solve this problem, that is to find contradiction when $x\rightarrow \infty$.
Given $f\in C^{0}\left(\mathbb{R}\right)$. I thought if $f(x)\rightarrow\infty$, we must have $x\rightarrow \infty$.
From the equation $\displaystyle\lim_{x\to +\infty}f(f(x))=-\infty$ and $\displaystyle\lim_{x\to -\infty}f(f(x))=+\infty$, we get $\displaystyle\lim_{x\to \infty}f(x)=\infty$. Then I thought we can claim that $\displaystyle\lim_{x\to -\infty}f(x)$ exists, and by discussing whether it equals $+\infty$ or $-\infty$ we can find contradiction.

Is there something wrong in my analysis?
And any other ideas to solve this problem?
I would appreciate it if you share your thoughts on this problem!

Answer 1

In this proof, we aim to prove a general conclusion：

For any $f\in C^{0}(\mathbb{R})$, $\displaystyle\lim_{x\to +\infty}f(f(x))=-\infty$ and $\displaystyle\lim_{x\to -\infty}f(f(x))=+\infty$ cannot be simultaneously true.

Lemma. If $f\in C^{0}(\mathbb{R})$ satisfying $\displaystyle\lim_{x\to \infty}f(f(x))=\infty$, we have $\displaystyle\lim_{x\to \infty}f(x)=\infty$.
Proof. If it's not true, we can find a sequence $(x_n)_{n\in \mathbb{N}}$ satisfying $\displaystyle \lim_{n\to \infty}x_n=\infty$ but $(f(x_n))_{n\in \mathbb{N}}$ is bounded.
Applying $\displaystyle\lim_{x\to \infty}f(f(x))=\infty$ we have $\displaystyle\lim_{n\to \infty}f(f(x_n))=\infty$. It generates contradiciton because it means $(f(x_n))_{n\in \mathbb{N}}$ cannot be bounded.

If $\displaystyle\lim_{x\to +\infty}f(f(x))=-\infty$ and $\displaystyle\lim_{x\to -\infty}f(f(x))=+\infty$ can be simultaneously true, from the lemma we get $\displaystyle\lim_{x\to \infty}f(x)=\infty$. Certainly, we have,$\displaystyle\lim_{x\to -\infty}f(x)=+\infty$ or $-\infty$ and $\displaystyle\lim_{x\to +\infty}f(x)=+\infty$ or $-\infty$.
If $\displaystyle\lim_{x\to +\infty}f(x)=+\infty$, then we'll get $\displaystyle\lim_{x\to +\infty}f(f(x))=+\infty$ which generates contradiction. So we must have $\displaystyle\lim_{x\to +\infty}f(x)=-\infty$. Similarly we must have $\displaystyle\lim_{x\to -\infty}f(x)=+\infty$.
But using $\displaystyle\lim_{x\to +\infty}f(x)=-\infty$ and $\displaystyle\lim_{x\to -\infty}f(x)=+\infty$ we have $\displaystyle\lim_{x\to +\infty}f(f(x))=+\infty$, which also generates contradiction.
So we arrive at that conclusion, which also works out the original problem.