Show that such a function has a fix point

Question

I can't solve the following:

If $f$ is continuous function from $\mathbb{R}$ to $\mathbb{R}$ with $f(f(x))=x$ for every $x \in \mathbb{R}$, then $f$ itself has a fixed point $y$, i.e. $f(y)=y$

I really don't know how to start. Any hint or suggestion is highly appreciated

Answer 1

Consider any $c$ such that $f(c) \neq c$. WLOG, let $f(c) < c$

Now, $f(f(c)) = c > f(c)$.

Now, since $f(x) - x$ is continuous on the domain $[f(c), c]$ so by the intermediate value theorem, it takes on any value in $[f(c) - c, c - f(c)]$ and in particular $0$ since $f(c) - c < 0$ and $c - f(c) > 0$.

i.e. there is some $y \in [f(c), c]$ such that $f(y) = y$

Answer 2

If $f$ is assumed to be continuous, then it will be sufficient to find values $x_1,x_2$ such that $f(x_1)-x_1\leq 0$ and $f(x_2)-x_2\geq 0,$ and then apply IVT to show there must be some value $x^*$ where $f(x^*)-x=0.$