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!

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.

  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.