# What are the general functions that commute with 1/x?

Given that $f(x)$ is a function that maps real positive number to real positive number, and $f(x)=(f(x^{-1}))^{-1}$, could you find all the possible $f(x)$?

I know that $f(x)=x^a$ satisfies these conditions for any value of $a$. Is this the only function that satisfies the conditions? If not how can find all the other functions that satisfy the conditions?

• Why it says entire in the title?
– user301452
Commented Dec 10, 2017 at 13:41
• Maybe this is a question on the general solution of the functional equation in question. Commented Dec 10, 2017 at 13:43
• Sorry, I don't know there is a specific definition for "entire function". What I meant here is the general solution to the functional equation. Thank you szw for clarifying.
– JNL
Commented Dec 10, 2017 at 14:50

Call $g(x)=\ln\circ f\circ \exp$. You want exactly $g(-x)=-g(x)$. So, the functions that satisfy that condition are precisely the functions $f$ such that $f(x)=e^{g(\ln x)}$ for some odd function $g:\Bbb R\to\Bbb R$. For instance, $f(x)=e^{\sin \ln x}$.

• $e^{\sin \ln x}$ is not entire. Commented Dec 10, 2017 at 14:33
• Oops, turns out the question in not about entire functions after all. Commented Dec 10, 2017 at 14:51
• @DavidC.Ullrich A reasonable misunderstanding.
– user228113
Commented Dec 10, 2017 at 14:52
• Actually no misunderstanding at all -- the original title specified "entire"... (The question is actually more interesting for entire functions; I've got as far as showing that $f$ must be a polynomial...) Commented Dec 10, 2017 at 15:12
• @DavidC.Ullrich Let me rephrase: it was in fact not obvious that the person who asked did not actually mean to add that hypothesis.
– user228113
Commented Dec 10, 2017 at 15:15

The orignal version of the question asked about entire functions. I had an answer to that almost worked out when the word "entire" disappeared. Here's an answer to the original:

If $f$ is an entire function with $f(z)f(1/z)=1$ for all $z\ne0$ then $f=\pm z^n$.

Since $\lim_{z\to0}f(z)f(1/z)=1$ and $f$ has a zero of at most finite order at the origin it follows that $f$ has at worst polynomial growth at infinity. So a standard exercise shows that

$f$ is a polynomial.

Say $$f(z)=\sum_{n=N}^Ma_nz^n,\quad a_N\ne0,a_M\ne0.$$

Now $$1=\sum_{n=N}^Ma_nz^n\sum_{k=N}^Ma_kz^{-k}.$$

The coefficient of $z^{N-M}$ on the right is $a_Na_M$. But the only non-zero term is the constant term; since $a_Na_M\ne0$ this must be the constant term, which says that $N-M=0$. So $f=a_Nz^N$, and then it's clear that $a_N=\pm1$.