Find all (at least one) functions $f\colon \mathbb{R}\to \mathbb{R}$ (or show there is none), such that $$ f(x^3+x)≤x≤f(x^3)+f(x), \quad \text{for all $x\in \mathbb{R}$}. $$ This is a problem asked by a friend of mine, and I do not know how to proceed.

Known easy facts:

  • $f(0)=0$.
  • It seems the problem can be split to $x>0$ and $x<0$ separately. Let $f_{-}(x)=-f(-x)$ for $x>0$, then $f_{-}(x^3+x)\ge x\ge f_{-}(x^3)+f_{-}(x)$.

Question: What is the value of $f(-2)$?

If we let $x=-1$, we see that $f(-2)\le -1$.

Next, let $a$ satisfy $a^3+a=-8$, and note that $a\approx -1.8$. Then let $x=-2$. We have:

$$-2\le f(-8)+f(-2)=f(a^3+a)+f(-2)\le a+f(-2)$$

These inequalities give:


a contradiction. No such function exists.

  • $\begingroup$ Nice. Thanks a lot. $\endgroup$ – Ma Ming Jun 6 '13 at 20:11
  • $\begingroup$ So what goes wrong is when $x<0$, I was considering $x>0$ for a long time. $\endgroup$ – Ma Ming Jun 6 '13 at 20:38

There is no such function, I believe. To see this, first let $h:\mathbb R\to\mathbb R$ be defined by $h(x)=x^3+x$. This is a sum of two strictly increasing functions and thus strictly increasing. Since it is also unbounded in both directions and continuous, it is bijective, so its inverse $h^{-1}:\mathbb R\to\mathbb R$ exists. (And is also strictly increasing.)

By the required inequalities, this means that for every $x\in\mathbb R$ we have $$f(x)=f(h(h^{-1}(x)))\leq h^{-1}(x).$$ This and the inequalities further imply that $$x\leq f(x^3)+f(x)\leq h^{-1}(x^3)+h^{-1}(x)\tag{*}$$ must hold for all $x\in\mathbb R$. It turns out that this is impossible.

To see this, we must only find an $x\in\mathbb R$ such that $x>h^{-1}(x^3)+h^{-1}(x)$, so that $(*)$ fails. Note that, since $h(-1)=(-1)^3+(-1)=-2$, it follows that $h^{-1}(-2)=-1$. Therefore, $$h^{-1}((-2)^3)+h^{-1}(-2)<h^{-1}(-2)+h^{-1}(-2)=-2,$$ so $(*)$ indeed fails. Therefore no such $f$ exists.


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