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.


"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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