# Involutions (of real functions)

1. Feb 27, 2009

### mnb96

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 $$f:\Re\rightarrow\Re$$
A function $$f$$, with the property $$f=f^{-1}$$ 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?

$$f(x)=a-x$$

$$f(x)=\frac{a}{x}$$

$$f(x) = \frac{1}{x-a}+a$$

2. Feb 27, 2009

### yyat

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.

Last edited: Feb 27, 2009
3. Feb 27, 2009

### mnb96

...you are actually right.
I'll try to state my problem in a better way:

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

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

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