# Functional Equation: Squeeze it

Determine whether there exists a function $$f:\mathbb{R}\rightarrow\mathbb{R}$$ such that $$f\big(x^3+x\big)\le x\le\big(f(x)\big)^3+f(x)$$ for all $$x\in\mathbb{R}$$.

Let $$g(x) := x^3+x$$. Then the problem asks for $$f$$ such that $$f(g(x)) \leq x \leq g(f(x))$$.
Since $$g$$ is bijective, choose $$f = g^{-1}$$, and the inequality becomes $$x \leq x \leq x$$, which is satisfied for all $$x$$.
Let $$x$$ be arbitrary, let $$y := g^{-1}(x)$$. We have $$g(f(x)) \geq x$$, but we also have $$f(g(y)) \leq y$$, so $$g(f(x)) = g(f(g(y)) \leq g(y) = x$$ by monotony of $$g$$. Hence, $$g(f(x)) = x$$, so $$f(x) = g^{-1}(x)$$.
• Correct! And the only possible solution is the inverse of $x^3+x$! May 27 '20 at 4:12