This is a very naive question which really does little more than highlight my ignorance of how converse theorems really work.

Take an algebraic gadget which should be conjecturally associated to an automorphic representation. For example, take a finite image continuous complex representation of the absolute Galois group of a number field. Or take an elliptic curve over a totally real field. There are converse theorems of the form "if the $L$-function of this gadget and the $L$-function of sufficiently many twists of this gadget have analytic continuation, are bounded in vertical strips, and satisfy the expected functional equations, then the gadget is indeed associated to an automorphic representation". For 2-dimensional gadgets (for example 2-dimensional complex Galois representations, or elliptic curves) one only needs to consider abelian twists---this is essentially proved in Jacquet--Langlands.

However it is nowadays pretty standard that these $L$-functions have meromorphic continuation and the expected functional equation. In the finite image case this follows from Brauer's theorem in finite group theory, and in the elliptic curve case this follows from the fact that these curves are now known to be potentially modular, following work of Kisin, Taylor and others.

So one might ask whether *meromorphic* continuation + functional equation, plus converse theorem techniques, is enough to prove *something* about the algebraic gadget (other than "it potentially comes from an automorphic representation" -- something which we already know and are using to get the meromorphic continuation). But presumably there is some serious theoretical obstruction to proving anything interesting here, or else it would all have been done for Artin $L$-functions in the 70s.

What is the obstruction?

provedin the 2-d Artin case over Q that analytic continuation of the function alone is enough. See Booker's Annals paper---MR2031863. The same result is true for modular forms of some small levels I think: e.g. Hecke's work (also cited by Emmanuel) didn't need any twists. Of course you may well know all this already so apologies. $\endgroup$ – Kevin Buzzard Apr 30 '11 at 7:48