Functional Square Root of $1-x^3$

Question

I'm completely stuck with the problem of finding such a continuous function $f$ that

$ f(f(x)) = 1 - x^3 $

or prove that it doesn't exist.

After some calculations I tend to the second case - continuous function with such property can not exist. Does anyone have some ideas?

Answer 1

Suppose $f:\mathbb{R} \to \mathbb{R}$ is a continuous function such that $f(f(x)) = 1-x^3$, for all $x \in \mathbb{R}$.

Note that $f$ is injective, since \begin{align*} &f(x) = f(y)\\[4pt] \implies\;&f(f(x))=f(f(y))\\[4pt] \implies\;&1-x^3 = 1-y^3\\[4pt] \implies\;&x^3=y^3\\[4pt] \implies\;&x=y\\[4pt] \end{align*} Since $f$ is injective and continuous, it follows that $f$ is strictly monotonic.

If $f$ is strictly increasing, then \begin{align*} &x < y\\[4pt] \implies\;&f(x) < f(y)\\[4pt] \implies\;&f(f(x)) < f(f(y))\\[4pt] \implies\;&1-x^3 < 1-y^3\\[4pt] \implies\;&x^3 > y^3\\[4pt] \implies\;&x > y\\[4pt] \end{align*} contradiction.

If $f$ is strictly decreasing, then \begin{align*} &x < y\\[4pt] \implies\;&f(x) > f(y)\\[4pt] \implies\;&f(f(x)) < f(f(y))\\[4pt] \implies\;&1-x^3 < 1-y^3\\[4pt] \implies\;&x^3 > y^3\\[4pt] \implies\;&x > y\\[4pt] \end{align*} contradiction.

Hence, no such function $f$ exists.