How Many Involutions Can Be Defined in the Set of Real Functions?

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  • #1
mnb96
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Hello,
the following problem popped in a different thread but the original one went off-topic, and I thought this question deserved a thread itself:

Let's consider the entire set of the real functions [tex]f:\Re\rightarrow\Re[/tex]
A function [tex]f[/tex], with the property [tex]f=f^{-1}[/tex] is called involution.

How many involutions is it possible to find in the set of real functions?
I know the following three forms: are there more?

[tex]f(x)=a-x[/tex]

[tex]f(x)=\frac{a}{x}[/tex]

[tex]f(x) = \frac{1}{x-a}+a[/tex]
 
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  • #2
Your second and third examples are undefined for x=0, x=a respectively, so they do not give involutions R->R.

Edit: If you define f(0)=0, f(a)=a respectively, this does give involutions, although discontinuous ones.

As another example, the non-continuous function that swaps the intervals [0,1] and [2,3] is an involution. You may want to consider only continuous, differentiable or analytic functions.
 
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  • #3
...you are actually right.
I'll try to state my problem in a better way:

Let's consider a subset of the real numbers [tex]A \subseteq \Re[/tex], and the family of continous functions [tex]f:A \rightarrow A[/tex]

In this way, all the functions I listed should be involutions. The second and the third one are involutions by simply letting [tex]A = \Re - \{0\}[/tex] and [tex]A = \Re - \{a\}[/tex]

My question remains the same: what/how many are the involution which we can define?
 
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