Do odd-weight cusp forms have analytic rank 0? Let $f(z)=\sum_{n\ge 1}a_nq^n$ be a cusp form, where $q=e^{2\pi i z}$. Let $
L(s) = \sum_{n\ge 1} a_nn^{-s}
$ be its corresponding L-function. The completed L-function of $L(s)$, $\Lambda(s)$, should satisfy a functional equation $\Lambda(s)=\bar \Lambda(a-s)$ for some integer $a$.
The analytic rank of $f(s)$ is defined as the order of vanishing of $L$ at $a/2$.
Question: If $f$ is of odd weight, is its analytic rank $0$ (or equivalently, $L(a/2)\neq0$)?
 A: It is conjectured that $L$-functions of motives do not have zeros or poles on the real line, except possibly at integers. In particular, if $f$ is a newform of odd weight $k$, then $L(f,s)$ should not vanish at $s=k/2$.
This conjecture appears in Fontaine, Perrin-Riou, Autour des conjectures de Bloch et Kato: cohomologie galoisienne et valeurs de fonctions L. Motives (Seattle, WA, 1991), 599–706, Proc. Sympos. Pure Math., 55, Part 1, Amer. Math. Soc., Providence, RI, 1994. See Remark 4.5.3.i).
I don't know if this is proven for $L$-functions of modular forms.
A: The statement needs some fixing. You must mean newform, rather than cusp form, as otherwise it could be interpreted that linear combinations can be taken (although they must possess identical sign of the functional equation to be allowed).
This said, I think the answer is expected to be no as discussed by François Brunault, BUT the related question allowing (nonholomorphic) Maass forms of eigenvalue 1/4 has a different answer, as noted by Armitage. That is, contrary to the $L$-function expectations otherwise, there are $L$-functions of motivic weight 1 with $\Lambda(s)=-\Lambda(1-s)$, which vanish at $s=1/2$ due to this symplectic root number.
https://doi.org/10.1007/BF01404125
