# Conditions for uniqueness of a Mellin transform

Let $f(x)$ and $F(s)$ be a Mellin pair, such that one is the Mellin inversion of the other in the fundamental strip $S_f$.

Let $g(x)$ and $G(s)$ be a Mellin pair, such that one is the Mellin inversion of the other in the fundamental strip $S_g$.

Assume that all the Mellin integrals converge absolutely over respective strips:

$f(x)=\frac{1}{2\pi} \int\limits_{a-i\infty}^{a+i\infty} F(s) x^{-s}ds$

$F(s)=\int\limits_0^\infty f(x)x^{s-1}dx$

$a\in S_f \cap \mathbb{R}$

$g(x)=\frac{1}{2\pi} \int\limits_{b-i\infty}^{b+i\infty} G(s) x^{-s}ds$

$G(s)=\int\limits_0^\infty g(x)x^{s-1}dx$

$b\in S_g \cap \mathbb{R}$

$\bf{Question}$: If $f(x)=g(x)$ in non-empty intersection $S_f \cap S_g$, what are the necessary and sufficient conditions for $F(s)=G(s)$ in $S_f \cap S_g$?

Any explanation as well as any reference is most welcome!

• why downvoted? O.o – anonymous Jun 23 '18 at 17:01

Conditions met as soon as both integrals exist. Uniqueness is almost everywhere, and $f(x)=g(x)$.
But what about the inverse? Does $f(x)=g(x)$ imply $F(s)=G(s)$? Can $f(x)$ have two different transforms $F_1(s)$ and $F_2(s)$?
Is the answer trivial? $f(x)=g(x) \implies F(s)-G(s)=\int (f(x)-g(x))x^{s-1}dx\equiv 0$?