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

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!

• Is there a reason that you label this general functional equation as differential equation (ODE)? Where are the derivatives? – Lutz Lehmann Nov 15 '18 at 12:05
• 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). – user3482749 Nov 15 '18 at 12:05
• @LutzL sorry for that. I'll modify the post – Zero Nov 15 '18 at 12:12
• @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 – Zero Nov 15 '18 at 13:00
• 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. – Lutz Lehmann Nov 15 '18 at 14:17

## 1 Answer

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.

• I tried to express my idea my more precisely, and I chose to write it as an answer instead of modifying the question. – Zero Nov 15 '18 at 17:00
• 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. – Richard Martin Nov 16 '18 at 9:17
• @RichardMartin Yes, and finally I realized I'm supposed to focus my attention to just some of the properties. Thanks for your directions – Zero Nov 16 '18 at 13:37