Continuous involutions over $\mathbb{R}$

Today I attended a seminar about classifying every real-valued continuous function on $\mathbb{R}$ such that $f \circ f = id$ (involutions).

Motivation. To classify all continuous involutions over $\mathbb{R}$.

Discussion

We begin by considering the set of all involutions over some set $X$, $$\mathfrak{F}_X = \{f:X\to X\ |\ f\circ f = id_X\}.$$

Claim: For every $f \in \mathfrak{F}_X$ we can say that $f$ is bijective.

Proof. First, assume towards a contradiction that $f$ isn't injective. Thus, let $x, y \in X$ such that $x \neq y$ and $f(x) = f(y)$. However, this implies $$f(f(x)) = f(f(y)) \implies x = y$$ by definition of involution. Absurdity. Therefore, $f$ is injective.

Now, observe that there is an $f(x)$ for every $x\in X$ by definition, and again $f$ maps $f(x)$ to $x$. So $f$ maps $f(x)$ to $x$ for every $x\in X$. Therefore, $f$ is surjective as well as injective i.e. bijective. $\blacksquare$

Now, let $\mathfrak{F} = \{f\in \mathbb{R} | f \text{ is continuous}\}$ be the set of all continuous involutions over the real line. Since every continuous bijection of the real line is strictly monotone, every continuous involution over the real line is either strictly increasing or strictly decreasing. First let us consider the case when it is increasing. Suppose $f$ is strictly increasing, and suppose $f(x) \neq x$ ($x\in \mathbb{R}$). Then we see that,

$$ f(x) > x\implies f(f(x)) = x > f(x) > x\implies x > x.$$

$$ f(x) < x\implies f(f(x)) = x < f(x) < x\implies x < x.$$

An absurdity. Thus, the identity map $f(x) = x \equiv id$ is the only strictly increasing continuous involution over the real line.

Now let us consider the case when it is strictly decreasing. Then we have more examples, such as

$ f(x) = -x. $
$ f(x) = 1-x. $
there's nothing special about either 0 or 1, so in general for $\lambda \in \mathbb{R}$
$$ f(x) = \lambda-x $$
for $c \in (-\infty, 0)$
$$ f(x) = \begin{cases} cx, x > 0\\ x/c, x \leq 0\end{cases} $$

In general if $f$ is a strictly decreasing continuous function from $[0,\infty)$ onto $(-\infty, 0]$,

then we can extend the domain of $f$ to $(-\infty, 0)$ by setting $f(x)=t$ where $x = f(t)$.

As there are uncountably many continuous functions over the real line, the cardinality of $\mathfrak{F}$ is $\mathfrak{c}$.