Functional equation in $x,y$: $f(x)f(y)=f\left(\frac{a}{x}\right)f\left(\frac{a}{y}\right)$

Question

Let $f:(0,+\infty)\to \mathbb{R}$ and $a>0$ such that $f(a)=1$. Prove that, if \begin{align*} f(x)f(y)=f\left(\frac{a}{x}\right)f\left(\frac{a}{y}\right),\quad\forall x,y>0 \end{align*} then $f$ is constant.

Answer 1

The claim is false. The function $$f(x)=\left|\ln x\right|$$ has the property with $a=1$, but it is not constant.

Answer 2

$$ f_1(x)=\begin{cases} x & \mathbf{if}\;\;x\in(0;1]\\ 1/x & \mathbf{if}\;\;x\in(1;+\infty) \end{cases}, \qquad f_2(x)=\begin{cases} 1/x & \mathbf{if}\;\;x\in(0;1]\\ x & \mathbf{if}\;\;x\in(1;+\infty) \end{cases} $$

Answer 3

You have $(f(a))^2=(f(1))^2\Rightarrow f(1)=1$ ($f$ is a function so we discard $f(1)=-1$).

It follows for all $y>0$, $$f(y)=f(\frac ay)$$ Making $a=1$ you have $$f(y)=f(\frac 1y)$$

This is verified by the function $$f(y)=\frac 12(y+\frac 1y)$$ which is not constant.

The claim is false.

EDITION.-My son tells me that for any positive a, the function $$f(x)=\frac{1}{|\ln a|}\left(|\ln \frac xa|+|\ln x|\right)$$ is a counterexample. This is much better than $a=1$