# Continuous involutions on $\mathbb R$ with at least two fixed points

Thanks to the posts 65876 and 66141, I was thinking about the functional square roots of the identity function on $\mathbb R$, namely involutions. An involution on $\mathbb R$ is a function $f:\mathbb R \to \mathbb R$ such that $f(f(x))=x$ for all $x \in \mathbb R$. I will restrict myself to functions that are continuous everywhere.

It is an easy exercise that every continuous involution on $\mathbb R$ has at least one fixed point. After a little effort, I have a complete characterization of continuous involutions on $\mathbb R$ with precisely one fixed point. Namely, these are precisely the functions that of the form:

$$f(x) = \begin{cases} a + \varphi(x-a) & x \geq a, \\ a - \varphi^{-1}(a-x) & x \leq a. \end{cases}$$ where $a \in \mathbb R$ and $\varphi: [0,\infty) \to [0, \infty)$ is a continuous strictly increasing bijection with $\varphi(0)=0$. Here $a$ is the fixed point of $f$. (A special example is the function $x \mapsto 2a-x$ for a fixed $a \in \mathbb R$.)

I want to know if there are any nontrivial examples other than the above family of functions.

Are there continuous involutions on $\mathbb R$, other than the identity, that have at least two fixed points?

Note. Searching on Math.SE, I found this question on involutions of $\mathbb R$. Gerry's answer gives a reference to a paper by J F Ritt, On certain real solutions of Babbage's functional equation, Annals of Math 17 (1916) 113-122. The paper touches upon fixed points, but does not address or answer my question.

-
A continuous involution of $\mathbb{R}$ is either order-preserving or order-reversing, since it's bijective. If it's order-preserving, it's not hard to show that it's the identity, and if it's order-reversing it has exactly one fixed point (since it flips the half line above a fixed point with the half line below it). – user83827 Sep 22 '11 at 3:09
@ccc Please write that comment as an answer. Thanks, it looks so obvious now ;) – Srivatsan Sep 22 '11 at 3:14
Both the answers were very helpful, but @ccc's answer made me "see" the reason a bit more clearly, so I accepted that one. – Srivatsan Sep 26 '11 at 20:04

A continuous involution of $\mathbb{R}$ is either order-preserving or order-reversing, since it's bijective. If it's order-preserving it's not hard to show that it's the identity, and if it's order-reversing it has exactly one fixed point (since it flips the half line above any fixed point with the half line below it).

-
+1 Nice answer. – Srivatsan Sep 22 '11 at 3:33
Thanks! It's hard to tell the difference between an answer and a comment sometimes. – user83827 Sep 22 '11 at 3:45

None other than the identity.

If $f(f(x))=x$, then $f$ is surjective. Also, $f$ is injective since $f(a)=f(b)\Rightarrow f(f(a))=f(f(b))\Rightarrow a=b$. So $f$ is invertible.

Now you have that $f(x)=f^{-1}(x)$. The graph of $f$ must be symmetric about the line $y=x$.

Suppose that $a$ and $b$ are two fixed points discretely separated with $a<b$ and no fixed points in between. The graph of $f$ either lies entirely below $y=x$ or above it on the interval $(a,b)$. (Otherwise, the intermediate value theorem gives another fixed point inbetween.) This contradicts the symmetry of the graph. (If say this section of the graph lies below the diagonal, then on the same interval, the graph of $f^{-1}$ lies above the diagonal.)

So any example with more than one fixed point will have an entire interval $[a,b]$ where $f(x)=x$. If the graph ever deviates from the identity function (say after $x=b$) then the graph either moves above or below $y=x$, again contradicting the symmetry.

-
I am not sure this is fully right. What if for some $x \in (a,b)$, $f(x)$ takes you outside the interval? Perhaps $f(x) > b$. I don't see how your argument works then... – Srivatsan Sep 22 '11 at 3:04
@Srivatsan Continuity and the intermediate value theorem imply that that cannot happen. Continuous invertible functions are monotonic. For example, if $c$ is in the interval, and $f(c)>b$, then somehow we would get from $a$ up to $f(c)$ and then back down to $b$ with a continuous invertible function. – alex.jordan Sep 22 '11 at 3:10
Sorry, I was citing the mean value theorem, when I meant to cite the intermediate value theorem. – alex.jordan Sep 22 '11 at 3:15
I am not very clear with the answer. Please give me some time before I read it properly. – Srivatsan Sep 22 '11 at 3:16
@Srivatsan: Try this minor variant. $C=\{x\in\mathbb{R}:f(x)=x\}$ is closed. If $C$ is not connected, let $(a,b)$ be an order component of $\mathbb{R}\setminus C$; then $a,b\in C$, and alex’s first argument yields a contradiction, so $C$ is connected and hence an interval. If $C$ is neither a singleton nor $\mathbb{R}$, it has an endpoint, and alex’s last paragraph applies. – Brian M. Scott Sep 22 '11 at 3:55