I am very interested in functions $\gamma:\mathbb R\to\mathbb R$ with the following property: $$\gamma^2(x)=x$$ One form of a function satisfying this is $$f(x)=\frac{a-x}{1+bx}$$ Which has the property $f^2(x)=x$. Infinitely many more functions with this property can be obtained by finding some other injective function $g$, its inverse $g^{-1}$, and then composing $g, g^{-1},$ and $f$ as follows: $$g^{-1}\circ f\circ g$$ However, I am not very interested in involutory functions of this form, since they seem to all be ripoffs of the general form that I already stated.

In fact, it seems that all involutory functions can be put in the form $$g^{-1}\circ f\circ g$$ for some $g$, and for some $a,b$. I can't find any counterexamples, but I don't know how to prove it either. It seems to me that the best way to approach this would be to set up some kind of differential equation like $$(f'\circ f)(x)=\frac{1}{f'(x)}$$ But I have absolutely no idea how I might show that any involutory function can be put in the aforementioned form.

Any ideas?

NOTE: I'm sure there are some elaborate piecewise-defined answers that can destroy my conjecture. However, I can't expect people to know what I mean when I ask to prove this for all "reasonable" functions - so I will establish some stricter restrictions on $\gamma$. The function must be expressible using some finite composition of these functions and their inverses: $$\phi_1(x,a)=x+a$$ $$\phi_2(x,a)=ax$$ $$\phi_3(x,a)=x^a$$ $$\phi_4(x,a)=a^x$$ For example, $x^2+x+1$ can be expressed as $$\phi_1(\phi_3(x,2),\phi_1(x,1))$$

  • $\begingroup$ what is the role played by $n$? $\endgroup$ – Francesco Polizzi Aug 12 '17 at 16:45
  • $\begingroup$ Whoops, sorry. I had $\gamma^n$ but decided to change it to $2$. I'll fix it! $\endgroup$ – Frpzzd Aug 12 '17 at 16:46
  • $\begingroup$ The answer will very likely depend on your setting. Just sets and functions? Spaces and continuous maps? Groups and homomorphisms? $\endgroup$ – Randall Aug 12 '17 at 16:46
  • $\begingroup$ @Randall I just changed my question to specify that $\gamma$ maps reals to reals. $\endgroup$ – Frpzzd Aug 12 '17 at 16:48
  • 1
    $\begingroup$ @Nilknarf If you don't require continuity any involutive bijection can do e.g. the transposition $t_{a,b}$ which exchanges only $a\not=b$. $\endgroup$ – Duchamp Gérard H. E. Aug 12 '17 at 16:51

There is a general approach using power series. Suppose that $g(x) = b_1x+b_2x^2+b_3x^3+\dots\quad$ and we require that $x+f(x)=g(-xf(x))\;$ for some power series $f(x)$. The connection with $g(x)$ implies $f(f(x))=x.\quad$ Solving for the coefficients of $f(x)$ term by term gives the expansion $$f(x)=-x+b_1x^2-b_1^2x^3+(b_1^3+b_2)x^4-(b_1^4+3b_1b_2)x^5+\dots$$ which is the general form of involution with fixed point $0$ unlike your $f(x)=(a-x)/(1+bx)$ where $f(0)=a$ and $f(a)=0$ with $a\neq 0$. However, If $a=0$ then $$f(x)=-x/(1+bx)=-x+bx^2-b^2x^3+b^3x^4+\dots$$ which is the case where $g(x)=bx$.

You might ask for involutions $f(x)=x+a_2x^2+\dots\;$ but the only example is $f(x)=x$.


You want an involution $h$ for which no invertible $g$ satisfies $hg=gf$. In fact if $h$ is the identity function this is equivalent to $g=gf$, which by invertibility implies $f$ is the identity function. This fails for any $a,\,b$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.