A solution of the functional equation $$f\left(\frac{1}{x}\right)=\sqrt{x}f(x),\quad x\gt 0$$ is $$f(x)=\sum_{n=-\infty}^\infty e^{-n^2\pi x}.$$ Another solution should be a certain quadratic function, though I encountered this problem a long time ago and forgot that function (together with the approach).
So there is one or (maybe) two or more solutions of the above functional equation. But how can I prove, given a set of solutions, that these are all the solutions of the equation?
By the way, does anyone know that quadratic function (if it exists at all)?
These are not all solutions to the equation. We can simply define
$$f(x)=\begin{cases}g(x),&x\ge1\\g(1/x)/\sqrt x,&x<1\end{cases}$$
for any function $g$.