On the system of functional equations $f(x^y)=\sqrt[y]{f(x)}$ and $ f(x^y)f(x^{-y})=1 $

Question

Find all functions $f:\Bbb R^+\to\Bbb R^+$ s.t. for all $x,y \in \Bbb R^+$ the following are simultaneously valid:

$$ f(x^y)=\sqrt[y]{f(x)}$$

$$ f(x^y)f(x^{-y})=1 $$

First I added the equations together:

$$ f(x^y)+f(x^y)f(x^{-y})=1+ \sqrt[y]{f(x)}$$

Then I factored the LHS:

$$ f(x^y)(1+f(x^{-y})=1+ \sqrt[y]{f(x)}$$

Then I isolated $f(x^y):$

$$ f(x^y)=\frac{1+\sqrt[y]{f(x)}}{1+f(x^{-y})} $$

Am I on the right track?

Answer 1

Here is a partial solution: assuming differentiability of $f$, if we take the derivative of both sides of the first equation with respect to $y$ we obtain $$ f'(x^y)\,x^y\ln x = -\frac{1}{y^2}\ln f(x)\,(f(x))^{1/y}, \tag{1} $$ which, for $y=1$, yields $$ f'(x)\,x\ln x = -f(x)\ln f(x). \tag{2} $$ Integrating $(2)$ we obtain $$ \ln(\ln f(x))=-\ln(\ln x)+C \Rightarrow f(x)=\exp\left(\frac{C}{\ln x}\right)\quad(x\neq 1). \tag{3} $$ One can easily check that $(3)$ satisfies both functional equations if $x\neq 1$. In order to satisfy the functional equations also at $x=1$ we must have $f(1)=1$.

Answer 2

I know Gonçalo's answer has been accepted already, but here is another approach on how to solve this problem, perhaps in a more systematic way.

First things first, it is to be noted that the second equation can be ignored, because it is derived from the first one; indeed, one has : $f(x^y)f(x^{-y}) = f(x)^{1/y}f(x)^{-1/y} = 1$.

Secondly, the main equation can be rewritten as $g(x^y) = \frac{1}{y}g(x)$, where $g = \ln \circ f$. This equation can be viewed as an eigenvalue problem for the composition operator $C_\phi$, where $\phi(x) = x^y$, so that $C_\phi(g) = g \circ \phi$ $-$ note that it is linear. The goal is then to determine the eigenfunctions of $C_\phi$ associated to the eigenvalue $\frac{1}{y}$.

It is known from Lie theory that this precise operator can be represented by the formula $C_\phi = e^{tA}$, with $t = \ln y$ and $A = x\ln x \,\partial_x$, where $\partial_x$ denotes the derivative with respect to $x$. As $C_\phi$ and $A$ will have the same eigenvectors (since $C_\phi$ is a function of $A$), let's determine the eigenfunctions $h_\lambda$ of $A$ : $$ A(h_\lambda)(x) = x\ln x \,h_\lambda'(x) = \lambda h_\lambda(x) $$ hence, by separation of the variables, $$ \frac{\mathrm{d}h_\lambda}{h_\lambda} = \lambda\frac{\mathrm{d}x}{x\ln x} $$ and, after integration, $$ h_\lambda(x) = \alpha\ln^\lambda(x) $$ with $\alpha$ a constant. In consequence, one has $C_\phi(h_\lambda) = e^{\lambda t}h_\lambda = y^\lambda h_\lambda$, hence $g = h_{-1}$ and finally $f(x) = e^{g(x)} = e^{\alpha/\ln(x)}$.