Find all the bijective functions $f:[0,1]\to[0,1]$ such that $x=\frac{1}{2}\big(f(x)+f^{-1}(x)\big)$ for all $x\in[0,1]$.

Question

Find all bijective functions $ f : [0,1] \to [0,1]$ that satisfy the equation $$x=\frac{1}{2} \big(f(x) +f^{-1} (x)\big)\,\forall x \in[0,1]\,.$$

I honestly don't know how to approach this. I tried to plug in different values, but this doesn't work.

Answer 1

Suppose $x \in I = [0, 1]$ and $f(x) = x + a$. The only way this can work is if $f^{-1}(x) = x - a$ (and in particular $x - a$ is also in $I$), so we also get $f(x - a) = x$.

Looking at this as $f(x - a) = (x - a) + a$, we can apply the result above with $x - a$ instead of $x$, which tell us that $f(x - 2a) = x - a$. Repeating this process, $f(x - na) = x - (n - 1)a$ for any integer $n \ge 0$.

In particular, $x - na \in I$ for all $n$. But $I$ is bounded above and below, which means that $a = 0$: otherwise the value $x - na$ will eventually wander out of $I$ for some large $n$ and we will get a contradiction.

So we have shown that $a = 0$ for any choice of $x$, which means that $f(x) = x$ always, and the identity function is the only solution.

(Note that we did not need to assume $f$ is continuous.)