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As stated above, I am trying to prove that the only solutions to the functional relation

$$ g(r)g(1/r)=1, \quad \text{for}\; r>0, $$

are of the form $g(r)=r^a$ for $a \in \mathbb{R}$. Other properties that I assume are that $\lim_{r \rightarrow 0} g(r)=0$, $g(1)=1$ and $g'(r)>0$ for $r\in(0,1)$ (and by reflection also for $r>1$). The function $g(r)$ should be analytic on the interval up to $r=0$. From these properties we also know that $a>0$.

I know that if we do not impose the monotonicity condition, there could in general also be solutions of the form $g(r)=r^{\alpha(r)}$ with $\alpha(r)-\alpha(1/r)=0$ which is solved by any $\alpha(r)=f(\log(r))$ where $f(x)$ is even. So I am not sure if there are still more possible solutions that I am not considering.

Other than that, I do not see how to argue that only monomonials in $r$ solve this equation since if my function is non-analytic in $r=0$ (e.g. $g(r)=r^{1/2}$), I can not take a series expansion there.

thecommutator
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    Take any $g(r)$ satisfying your "other properties" for $0 < r \le 1$, and define $g(r) = 1/g(1/r)$ for $r > 1$. – Robert Israel Aug 01 '23 at 22:57
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    I think $g(r) = \exp((\log r)^3)$ would also satisfy all those conditions. – Daniel Schepler Aug 01 '23 at 23:12
  • Thanks for the comments. @RobertIsrael you are right, but for arbitrary $g(r)$, I am not guaranteed that all my higher left and right derivatives coincide at $r=1$ which is a property I would want to have ($g'(r)$ does indeed coincide) . – thecommutator Aug 02 '23 at 09:40
  • @DanielSchepler you are right, this is a function that I can not argue 'away'. – thecommutator Aug 02 '23 at 09:41

2 Answers2

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I am unsure on how to prove this for general functions, but below is a proof for finite polynomials alone (provided by a friend).

Let $g(r)=a_nr^n+a_{n-1}r^{n-1}...+a_0r^0$, where $a_n$ is nonzero. If $a_n=0$, then $g(r)=0$ and the functional equation is never satisfied. Now, $$g(r)=\frac{1}{g(1/r)}=\frac{1}{a_nr^{-n}+a_{n-1}r^{1-n}...+a_0r^0}=\frac{r^n}{a_nr^0+a_{n-1}r^1+...+a_0r^n}$$ In order for $g(r)$ to be a polynomial and not a rational function, the denominator must evenly divide the numerator. This implies that all the roots of the denominator are also roots of the numerator. However, we know that $0$ is the only root of the numerator, so it can also be the only root of the denominator. This means that the denominator must be of the form $br^a$. However, we also know $a_n\not=0$, so the denominator must be simply $a_nr^0=a_n$, and $a_0=a_1=a_2...=a_{n-1}=0$. This leads us to $g(r) = a_nr^n=\frac{r^n}{a_n}$, so $a_n=\pm1$.

From this, we see that $g(r)=\pm r^n$ for any integer $n>=0$ if $g(r)$ is a polynomial.

Update: if this can be extended from finite to infinite polynomials, then, by the Taylor series, it is proved for all infinitely differentiable $g(x)$. As noted in the comments (@DanielSchepler with $e^{(\text{log}r)^3}$) however, this is not true, so this extension cannot be made. Perhaps some insight can be gained from this though?

Abc Bcd
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The function $g$ in defined on $(0,\infty )$. It may be complex but has no zeros. Consider $f(x)=\log |g(e^x)|.$ Then $f(x)+f(-x)=0$ and $f$ is a completely arbitrary odd function. All the other conditions that you add are easily transposed in terms of $f$; for instance $g(r)\to_{r\to 0} 0$ is $f(x)\to_{x\to -\infty} -\infty.$ Anyway, we are very far from your conjecture, here equivalent to $f(x)=ax.$

Gérard Letac
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