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)?

  • 1
    $\begingroup$ $f(x) =x^{-1/4}$ satisfies the functional equation. Also any constant multiple of a function that satisfies the equation also satisfies it. $\endgroup$ Mar 16, 2020 at 17:29
  • 2
    $\begingroup$ If $f$ and $g$ are positive functions satisfying the equation, for any $t \in \mathbb{R}$, $f^tg^{1-t}$ satisfies the equation too. $\endgroup$
    – Aphelli
    Mar 16, 2020 at 17:52
  • $\begingroup$ @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 $\endgroup$
    – Poder Rac
    Mar 16, 2020 at 17:54
  • $\begingroup$ If by a quadratic function you mean $f(x)=ax^2+bx+c$ then I believe no such solutions exist $\endgroup$ Mar 16, 2020 at 17:57
  • $\begingroup$ @მამუკა ჯიბლაძე I may have erred, then. $\endgroup$
    – Poder Rac
    Mar 16, 2020 at 17:58

1 Answer 1


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 $g$.


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