Proof about Involutory Functions

Question

I would like to prove (or disprove) that involutory functions (functions that are their own inverses) have no real functional square root/half iterate, but I'm not sure where to start with this. This assumption seems "correct", but that isn't really enough. So far all of these functions that I've come across have some functional square root involving complex numbers. For example, if $()=-x$, then the functional square root is $()=ix$. Another example is that if $()=\frac{1}{x}$, then $()=x^i$. One last example is that if $()=1-x$, then $()=ix+\frac{1}{2}-\frac{1}{2}i$.

Can anybody help me out by giving me some idea how I can begin this proof, or give a counterexample?

For $f(x)=1/x,$ you have $f(f(x))=x$ for all $x,$ so that $f$ is its own inverse. — coffeemath, Apr 19 '17 at 23:29

score 2 · Accepted Answer · answered Apr 19 '17 at 23:37

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One counter-example is to find a function $g:\mathbb{R}\to \mathbb{R}$ such that $g^4=\text{id}_{\mathbb{R}}$.

Clearly this function is the functional root of itself squared; $f:\mathbb{R}\to\mathbb{R}$ defined by $f=g^2$.

One such function could be defined as follows:

$$g(x)=\begin{cases} 1/x \text{ when } x\leq -1\\ -x \text{ when } -1<x<0\\ 0 \text{ when } x=0\\ 1/x \text{ when } 0<x\leq1\\ -x \text{ when } 1< x \end{cases}$$

You can imagine this as splitting the real line into four "quarters" and then cycling them around. The numbers below $-1$ land between $-1$ and $0$, the numbers between $-1$ and $0$ land between $0$ and $1$, and so on.

This is a pretty contrived example - if you added restrictions, like $f$ or $g$ being continuous, I think you might be able to force $g$ to be complex-valued.

answered Apr 19 '17 at 23:37

Alexander Heyes

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If I did limit it to continuous functions, how could I begin a proof? – Franklin Pezzuti Dyer Apr 19 '17 at 23:43
I'm not exactly sure - I think the hint in this post may help:
https://math.stackexchange.com/questions/145639/a-continuous-bijection-f-mathbbr-to-mathbbr-is-an-homeomorphism
– Alexander Heyes Apr 20 '17 at 00:05
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@FranklinP.Dyer If $g^4=\text{id}$, then $g$ is invertible. Continuous invertible is monotone. For a monotone $g$, $g^2$ is increasing. The only increasing continuous involution is identity. – A.Γ. Apr 20 '17 at 00:13

Franklin Pezzuti Dyer · Answer 2 · 2017-04-20T21:21:47.997

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Never mind, there is a counterexample $()$ so that $f^4(x)=x$ because the function $()=\frac{x-1}{x+1}$ has that property, so my theory is disproven.

edited Apr 20 '17 at 21:21

answered Apr 20 '17 at 19:10

Franklin Pezzuti Dyer

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Proof about Involutory Functions

2 Answers2