TITLE: For which functions is the (generalized) Riemann hypothesis known?
QUESTION [6 upvotes]: In [1], Lin Weng shows that the Riemann hypothesis (RH) holds for certain linear combinations of shifted completed Riemann zetas. Further, Deligne's result on the Riemann hypothesis for function fields gives a RH for the latter. I would like to know whether these are all examples for which the RH is known.
To be precise, we say that a meromorphic function on $\mathbb C$ 
satisfies the generalized Riemann hyporthesis (genRH), if all of its poles and zeroes lie in the union of a finite number of vertical lines $c+i\mathbb R$ and the real line. Are there any known examples, other than those in [1], or those derived from Deligne's work, for which the genRH has been proven?
Of particular interest would be a function which can be written as a Dirichlet series for $\Re(s)>>0$.
[1] Symmetries and the Riemann hypothesis. Algebraic and arithmetic structures of moduli spaces (Sapporo 2007), 173–223, 
Adv. Stud. Pure Math., 58, Math. Soc. Japan, Tokyo, 2010.

REPLY [2 votes]: The answer is that there's trivially many, even if you want it to be defined by a Dirichlet series.  Dirichlet polynomials (such as $1+2^{-s}$) will work.  But that's cheap because these are too small (bounded on vertical lines).   Starting with any Dirichlet series $f(s)=\sum a_n n^{-s}$ that analytically continues to an entire function, you can take the function $\exp f(s)$, which can also be defined as a Dirichlet series and is nowhere zero.  But that's cheap because these are too big (not order 1 functions) and hence cannot be defined as a product over their zeros.
If you wanted functions that are neither too big nor too small then conjecturally all L-functions satisfy this, but there isn't a single function that you can prove this for, and for all we know no such function exists.  There is an earlier mathoverflow question that defines this precisely and asks for a function satisfying the weaker property of having a "zero-free region", which is provided in the answer.