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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!

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  • $\begingroup$ Is there a reason that you label this general functional equation as differential equation (ODE)? Where are the derivatives? $\endgroup$ – Lutz Lehmann Nov 15 '18 at 12:05
  • $\begingroup$ 1) Monotonic functions are a tiny subset of continuous functions, so I can't see how this will help. 2) Continuity is a property that happens at specific points, so this can't possibly hope to work. (Also, there's some very weird circular/backwards argument stuff going on). $\endgroup$ – user3482749 Nov 15 '18 at 12:05
  • $\begingroup$ @LutzL sorry for that. I'll modify the post $\endgroup$ – Zero Nov 15 '18 at 12:12
  • $\begingroup$ @user3482749 I thought if $f\in C^{0}(\mathbb{R})$, $\lim_{x \to a} f(x) = \pm \infty$ for some $a \in \mathbb{R}$ cannot occur. That's where I use the continuity's property $\endgroup$ – Zero Nov 15 '18 at 13:00
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    $\begingroup$ Note that for $x\approx 0$ you have $f(f(x))=|x|+O(x^2)$. Thus $f$ has to be locally invertible for positive and negative $x$ separately and discontinuity might follow by arguments similar to Find a real function $f:\mathbb{R}\to\mathbb{R}$ such that $f(f(x)) = -x$? and linked posts. $\endgroup$ – Lutz Lehmann Nov 15 '18 at 14:17
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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.

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    $\begingroup$ I tried to express my idea my more precisely, and I chose to write it as an answer instead of modifying the question. $\endgroup$ – Zero Nov 15 '18 at 17:00
  • $\begingroup$ Yes. The expression for $f^2$ has been cooked up with a couple of "features" and you need to decide which is the important one. Is it (i) the behaviour at the origin is $|x|$, or (ii) the behaviour at $\pm\infty$ is as you've said? In this case it's (ii). The sin and log are distractions to a large extent. $\endgroup$ – Richard Martin Nov 16 '18 at 9:17
  • $\begingroup$ @RichardMartin Yes, and finally I realized I'm supposed to focus my attention to just some of the properties. Thanks for your directions $\endgroup$ – Zero Nov 16 '18 at 13:37

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