Return to Revisions

Based on your normalization, $L(s,f)$ is defined as an Euler product for $\Re(s)>\frac{k+1}{2}$, so $L(s,f)$ is non-zero in that right-half plane. Now Jacquet–Shalika MR0432596 showed that that non-zero region extends to the line $\Re(s)=\frac{k+1}{2}$ (for $\mathrm{GL}(n)$, in fact) (this is analogous to the proof of non-vanishing on the line $s=1$ for $\zeta(s)$). Furthermore, $L(s,f)$ satisfies a functional equation $\Lambda(s,f)=i^k\Lambda(k-s,W_Nf)$ where $N$ is the level of $f$, $W_N$ is the Atkin-Lehner operator, so $W_Nf$ is also weight $k$ and level $N$, and $\Lambda(s,f)=N^{s/2}(2\pi)^{-s}\Gamma(s)L(s,f)$. So, $s=\frac{k}{2}$ is the central point (which is indeed $\neq1$ if $k>2$) (stuff like Beilinson–Deligne–Bloch–Kato basically says that vanishing away from the central point should be easier to understand). Now the values of $L(s,f)$ at $0\lt s\leq\frac{k-1}{2}$ are related to values of a possibly different modular form ($W_Nf$) at $k-s$, i.e. in the non-vanishing range $\Re(s)\geq\frac{k+1}{2}$. The only complicating part of this relation is the Gamma factors, but at these $s$, both $s$ and $k-s$ are $>0$, so they are not poles of $\Gamma(s)$. Hence, $L(s,f)\neq0$ at all positive $s$ except possibly in the strip $\frac{k-1}{2}\lt\Re(s)\lt\frac{k+1}{2}$.