About Period 2 Points of Continuous Invertible Functions
It is often presented without proof in high school or junior college level mathematics classes the following ``theorem''.
“Theorem”. If f is a function, then $f(x)=f^{-1}(x)$ if and only if $f(x)=x$.
Taken literally, this is of course complete nonsense, and is trivially seen to be false: take for example $f(x)=1/x$. For an extended period of time I was under the impression that this ``theorem'' could not be rescued regardless of how many additional restrictions we impose on $f$, mostly because no obvious proof has ever jumped out at me. Yet after deliberating some more about this, I have failed to find a counterexample in sufficiently nice functions---to which the ``theorem'' above is usually applied to. So I became more convinced that the theorem might indeed be true under certain reasonable conditions, and eventually it become apparent that there was an easy proof in a simple case.
Theorem. Let $f:[0,1]→[0,1]$ be a continuous bijection satisfying $f(0)=0$ and $f(1)=1$. Then $f(x)=f^{-1}(x)$ if and only if $f(x)=x$.
Proof. The equation is the same as $f(f(x))=x$. Let $G$ be the graph of $f$. If $f(x)=k$ and $f(k)=x$, then both $(x,k)$ and $(k,x)$ are in $G$. We assume $x\neq k$ by way of contradiction, and suppose without loss of generality that $x<k$, then $(x,k)$ lies above the line $y=x$ while $(k,x)$ lies below it. So $f$ is decreasing on a subinterval of $[x,k]$ but increasing on a subinterval of $[0,x]$, contradiction with the fact that it is monotonic (which is implied by the fact that it is a continuous bijection).
Of course, the assumption $f(0)=0,f(1)=1$ and the restriction of the domain and range of $f$ to $[0,1]$ are arbitrary and this theorem applies more generally. All we really need is for the function $f:(a,b)→\mathbb{R}$ to be injective and continuous with $\mathrm{lim}_{x→a}f(x)$ and $\mathrm{lim}_{x\to b}f(x)$ converging to a finite value (which is equivalent to boundedness, since $f$ is monotone).
Interestingly this implies that a function in the conditions of the theorem above cannot admit proper period $2$ points, i.e. points $x$ so that $f(f(x))=x$ but $f(x)\neq x$. For which $n$, then, can such a function $f$ admit proper period $n$ points (i.e. $f^n(x)=x$ but $f^k(x)\neq x$ for any $0<k<n$)? This sounds like something that would be of interest to dynamical systems problems, where fixed points and questions of stability under a certain map are often of interest; most likely this problem has been fully answered already somewhere out there, though I'll have to look further to find out for sure.
