Sufficient condition for Riemann Hypothesis?

Is there an L-function ($L_s=\sum_{n =1}^{\infty} \frac{a_n}{n^s}$) having a functional equation coming from a relation of the type [1]: $\sum_{n =1}^{\infty} a_n \; e^{-2\pi nx}= \frac{A}{x^k} \sum_{n =1}^{\infty} a_n'\; e^{-2\pi\frac{n}{Nx}}$ but not satisfying the Riemann Hypothesis?

Notice that in the above sum there is no term for $n=0$.

(I think Dirichlet L-function and some Modular forms L-function with Fricke involution satisfy such relation and RH, but is this possibly a sufficient condition? Any counter example?)

The L-fonction I know having a functional equation and satisfying RH have also an associated relation of the type [1] (coming from Poisson Summation formula or Fricke involution) sometimes they have a term in 0 (like for Zeta) sometimes not (like for Dirichlet L-functions). Now I know example of L-functions not satisfying RH with functional equation but it seems they appear only with combination of L-functions having a term n=0 in the Poisson summation like formula.

By the way do you know a L-function satisfying RH without a relation like [1] (with possibly a term n=0)?

• Are you assuming the $a_n$'s are multiplicative? – Micah Milinovich Aug 15 '13 at 16:31
• If you take two cuspidal eigenforms of the same weight and level, any linear combination of them will satisfy your functional equation, and have no term at zero. But it is easy to construct such a combination with a zero not on the critical strip. Like quid said, the associated L-function won't have an Euler product. I think we really expect only eigenforms to satisfy RH. – Dror Speiser Aug 16 '13 at 7:01