Let $\chi$ be a Dirichlet character and $L(s, \chi)$ be the corresponding L-function. 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 1


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 half-plane 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. $\endgroup$
    – 2734364041
    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. $\endgroup$
    – 2734364041
    Apr 7, 2022 at 0:11

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