# All solutions of $f\left(\frac{1}{x}\right)=\sqrt{x}f(x)$

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)?

• $f(x) =x^{-1/4}$ satisfies the functional equation. Also any constant multiple of a function that satisfies the equation also satisfies it. Mar 16, 2020 at 17:29
• If $f$ and $g$ are positive functions satisfying the equation, for any $t \in \mathbb{R}$, $f^tg^{1-t}$ satisfies the equation too. Mar 16, 2020 at 17:52
• @Yves Daoust $\sum_{n=-\infty}^\infty e^{-n^2\pi x}=\vartheta _{3}(0,e^{-\pi x})$ where $\vartheta$ is the Jacobi theta function. It converges for $x\gt 0$. mathworld.wolfram.com/JacobiThetaFunctions.html Mar 16, 2020 at 17:54
• If by a quadratic function you mean $f(x)=ax^2+bx+c$ then I believe no such solutions exist Mar 16, 2020 at 17:57
• @მამუკა ჯიბლაძე I may have erred, then. Mar 16, 2020 at 17:58

## 1 Answer

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 $$g$$.