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I am confused with what seems to be a standard notation in analytic number theory and I'd appreciate any clarification. I am interested in the zero density estimates, for example link.springer.com/article/10.1007/BF01403187 . In this paper and in many other sources I have seen, $N_{\chi}(\alpha, T)$ is defined to be the number of zeros of $L(s, \chi)$ in the rectangle $\alpha \leq \sigma \leq 1$, $|t| \leq T$ where $\sigma$ is the real part of $s$.

What I was confused with is, is this the number of zeros counting with multiplicity (as in if $\rho$ is a zero of order $2$, then is $\rho$ counted twice) or not? All except one reference I have seen so far has no mention of multiplicity, so I thought it was counting without multiplicity. but it's not clear to me and I was hoping to get a clarification from an expert who knows this material.

This made me also wonder about in the notation $$ \sum_{\rho} $$ where the sum is over the non-trivial zeroes of the L function, is this also sum taking into account the multiplicity (so $\sum_{\rho} f(\rho)$ is actually $$ \sum_{\rho, distinct} (multiplicity \ of \ \rho) \ f(\rho) ? $$ ) or not? For this one what I had in mind in particular was the explicit formula for $\sum_{n \leq X} \chi(n) \Lambda(n)$. I have thought that it meant a sum without taking into account the multiplicity, but maybe this was incorrect.

It's not clear to me looking at the sources. Thanks in advance for the clarification. very appreciated.

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Zeros are always counted with multiplicity, both in $N_\chi(\alpha,T)$ and in sums over zeros. This becomes clear when you look at how this quantity is estimated. Note also that the multiplicity of each zero $s$ of $L(s,\chi)$ is small, namely $O(\log q(2+|s|))$ by Jensen's formula.

For example, in Gallagher's paper the crucial step to look at is on p.336: "Since there are $\ll r\mathcal{L}$ zeros in the disc $|s-w|\leq r$ etc." This estimate comes from Jensen's formula, which counts the zeros with multiplicity. So in the next display, when $N_\chi(\alpha,T)$ is bounded from above, zeros are taken into account with their multiplicities.

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    $\begingroup$ That's good to know, thank you! $\endgroup$
    – Johnny T.
    Oct 3, 2020 at 16:28

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