On page 266 of Fourier Analysis by Stein and Shakarchi, it is shown that
If $s > 1$, then $$\prod_{\chi} L(s,\chi)\geq 1,$$ where the product is taken over all Dirichlet characters. In particular the product is real-valued.
The sketch of the proof is to rewrite $L(s,\chi)$ using Dirichlet's product formula. $$\begin{aligned} \prod_{\chi} L(s, \chi)&= \exp\left(\sum_{\chi}\sum_p \log_1\left(\frac{1}{1 - \chi(p)p^{-s}}\right)\right)\\ &=\exp\left(\sum_\chi \sum_p \sum_{k=1}^\infty \frac{1}{k}\frac{\chi(p^k)}{p^{ks}}\right)\\ &=\exp\left(\sum_p\sum_{k=1}^\infty \sum_\chi \frac{1}{k}\frac{\chi(p^k)}{p^{ks}}\right)\\ &=\exp\left(\varphi(q)\sum_p\sum_{k=1}^\infty \frac{1}{k}\frac{\delta_0(p^k)}{p^{ks}}\right) \geq 1 \end{aligned}$$ since the terms inside the exponential are non-negative.
A lemma that is used:
The Dirichlet characters are multiplicative. Moreover, $$\delta_l(m) = \frac{1}{\varphi(q)}\sum_\chi \overline{\chi(l)}\chi(m),$$ where the sum is over all Dirichlet characters.
Some clarification for notations:
- $\log_1$ is a function defined as $$\log_1\left(\frac{1}{1-z}\right) = \sum_{k=1}^\infty \frac{z^k}{k}\quad\text{for } |z|<1.$$
- $p$ takes value from all prime numbers.
- $\chi(\cdot)$ is the Dirichlet character.
- $\varphi(q)$ is the order of group $\mathbb{Z}^*(q)$.
- $\delta_l(n) = \mathbb{I}(n\equiv l\pmod q)$.
My question is regarding the last equation in the proof, where the book says with $l = 0$, the lemma gives $\sum_\chi \chi(p^k) = \varphi(q)\delta_0(p^k)$. To get this from the lemma, I would assume that $\chi(0) = 1$. However, should not $\chi(0) = 0$ since $0$ is never relatively prime to $q$? Am I missing something?