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With a graduate student, I'm going through the paper (Proc. London Math. Soc. (3) 47 (1983), no. 2, 193–224.)

Here's the background and notation.

We have a quadratic character $\chi$ modulo $q$, with Siegel zero $\beta_0$. $\eta=((1-\beta_0)\log q)^{-1}$, so $3\le \eta\ll q$ is known. Let $L=\log q$. We take $$ q^{250}\le x\le q^{500}. $$

Lemma 3 (stated on p. 198) says $$ \sum_{p<x,\chi(p)=1}\frac{\log(p)}{p}\ll L\log(\eta)^{-1/2} $$ Heath-Brown says "This is not necessarily the best bound of its type, but suffices for our purposes." I think the proof he gives is flawed. I know results like this are found elsewhere. My question is

Can this proof be fixed? If not, can you provide a reference for a more robust proof?

(Side note: In a recent blog post on this same paper, Tao proves his Lemma 5 which is similar, summing instead $1/p$, up to a $o(1)$ error. "For more precise estimates on the $o(1)$ error, see the paper of Heath-Brown (particularly Lemma 3).")

The proof starts in Section 4 on p. 206. The overall structure is to write $$ \frac{L^\prime}{L}(s,\chi)-\frac{L^\prime}{L}(s^\prime,\chi) $$ as a both sum over zeros and as a sum over primes. Here $s=1+L^{-1}$ and $s^\prime=1+aL^{-1}$, where $a$ is to be chosen later. The zeros side comes down to estimating $$ aL^{-1}\sum_{\rho\ne\beta_0}|\rho-1|^{-2}. $$ The zeros $\rho$ at height $\ge 1$ give no trouble. Heath-Brown quotes Prachar to estimate the number of zeros in the disk $|s-1|\le r$ as $$ \ll 1+r\log q $$ for $r\le 2$. To count zeros below height $1$ it would make sense to take $r=\sqrt{2}$ here.

With $r_0$ the closest non-Siegel zero to $1$, he quotes the Deuring-Heilbronn phenomenon to say $r_0\gg L^{-1}\log\eta$.

One would expect then the bound to involve the $aL^{-1}$ term, the number of zeros at height below $1$, and the worst case for the term being summed, namely $r_0^{-2}\ll L^2\log(\eta)^{-2}$, i.e. a bound of $$ aL(1+\sqrt{2}L)\log(\eta)^{-2}\ll aL^2\log(\eta)^{-2}. $$ Heath-Brown uses instead the count of zeros inside the circle of radius $r_0$ (which makes no sense, by definition of $r_0$) and gets a better estimate $\ll aL/\log\eta$. The correct (I think) estimate does not suffice for the error bound the lemma claims.

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"Those oft are stratagems which errors seem, Nor is it Homer nods, but we that Dream."

Heath-Brown's proof is fine. It needs just one more line of explanation.
Note that $$ \sum_{\substack{\rho \neq \beta \\ |\gamma| \le 1}} \frac{1}{|\rho -1|^2} \ll \int_{r_0}^{2} \# \{ \rho \neq \beta: |\rho -1| \le x \} \frac{dx}{x^3}, $$
and by the quoted line from Prachar this is $$ \ll \int_{r_0}^2 (1+x \log q) \frac{dx}{x^3} \ll r_0^{-2} + (\log q) r_0^{-1}, $$ which is what Heath-Brown writes.

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    $\begingroup$ (And now someone can use this answer for that epigram question.) $\endgroup$ – Lucia Oct 9 '15 at 23:34
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    $\begingroup$ "Good-Nature and Good-Sense must ever join; To err is Humane; to Forgive, Divine." $\endgroup$ – Stopple Oct 10 '15 at 4:25
  • $\begingroup$ Fair enough! I didn't mean any offense with my epigram, and I hope none was taken. It was just that I couldn't resist it, in view of that epigram question running around recently. $\endgroup$ – Lucia Oct 10 '15 at 4:31
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    $\begingroup$ None taken! And I enjoyed reading Pope. $\endgroup$ – Stopple Oct 10 '15 at 19:44

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