Let $\chi$ be a Dirichlet character and $L(s, \chi)$ be the corresponding Lfunction. Is $$\frac{1}{L(1+it, \chi)}$$ unbounded for $t \in \mathbb{R}$? I'm aware that this is true if $L=\zeta$, but I'm not sure of general $L$.
1 Answer
Yes, this is known to hold, and for more general families of $L$functions. See in particular Theorem 3.1 and the discussion following it in "Extremal values of Dirichlet $L$functions in the halfplane of absolute convergence " by Jörn Steuding.

$\begingroup$ I found no Theorem 0.8 in Lamzouri's paper (journal or arXiv). Also, unless I'm missing something, Littlewood's result only bounds $L(1+it,\chi)$ and does not ensure the existence of $t$ such that $L(1+it,\chi)$ is small. $\endgroup$ Apr 6, 2022 at 5:32

$\begingroup$ Apologies; I misunderstood and thought you were interested in large values, not small values. I'll give another reference. $\endgroup$ Apr 6, 2022 at 15:28

$\begingroup$ Littlewood's idea still works, but you need some information about the value distribution of twisted characters. Example: Large values of $L(1,\chi)$ can be found along a sequence of $\chi$ with $\chi(p)=1$ for all $p\leq O(\log q_{\chi})$ (such $\chi$ exist via Pell equation arguments; see Montgomery and Weinberger link.springer.com/article/10.1007/BF01351721). When $\chi$ is fixed and $t$ is large, one must know something about the value distribution of $\chi(p)p^{it}$ to ensure small (or large) values of $L(1+it,\chi)$. Plug this into Littlewood's truncated Euler product. $\endgroup$ Apr 7, 2022 at 0:11