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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.

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    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).2011-09-22
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    @ccc Please write that comment as an answer. Thanks, it looks so obvious now ;)2011-09-22
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    Both the answers were very helpful, but @ccc's answer made me "see" the reason a bit more clearly, so I accepted that one.2011-09-26

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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).

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    Thanks! It's hard to tell the difference between an answer and a comment sometimes.2011-09-22
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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

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.

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    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...2011-09-22
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    @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.2011-09-22
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    Sorry, I was citing the mean value theorem, when I meant to cite the intermediate value theorem.2011-09-22
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    I am not very clear with the answer. Please give me some time before I read it properly.2011-09-22
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    @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.2011-09-22