TITLE: "Sub-logarithmic" zero-free regions from Deuring-Heilbronn/Linnik's repulsion theorem
QUESTION [6 upvotes]: For each $n\in\mathbb{N}$, let:

$\chi_n\pmod{q_n}$ a real non-principal Dirichlet character ($q_1 < q_2 < \cdots$),
$\beta_n$ the largest real zero of $L(s,\chi_n)$,
$\delta_n := (1-\beta_n)\log(q_n)$.

Let $\chi\pmod{q}$ be a Dirichlet character, and consider $s = \sigma + it$ with $|t| < 1$. In p. 206 of Heath-Brown's "Prime twins and Siegel zeros", it is mentioned that the Deuring-Heilbronn phenomenon implies that there is some absolute constant $C>0$ such that, for each $n\in \mathbb{N}$, the region
$$ \bigg\{ s ~\bigg|~ \sigma \geq 1 - \frac{C\log(\delta_n^{-1})}{\log(q)},\ |t| < 1 \bigg\} $$
has no zeros (besides $\beta_1,\ldots,\beta_n$) of $L(s,\chi)$. (At least, that is how I interpreted the assertion "$r_0 \gg L^{-1}\log \eta$" at the mentioned page). Assuming this statement, it follows that:

(Sub-logarithmic zero-free regions (ZFR)) If there exists a sequence of Siegel zeros $\beta_n$ with $\delta_n \to 0$, then all the other zeros $\sigma + i\gamma$ of $L(s,\chi)$ with $|\gamma| < 1$ for Dirichlet characters $\chi\pmod{q}$ satisfy:
  $$ \sigma < 1 - \frac{1}{o(\log(q))}. $$

It appears to me that this (or slight variations of this) statement is often used in the literature [the only example I have in mind at the moment is Remark 1 in p. 515 (p. 6 in the link) of Granville & Stark's $ABC$ implies no "Siegel zeros" for $L$-functions of characters of negative discriminant, where it is mentioned that $\delta_n \to 0$ implies $\frac{L'}{L}(1,\chi_n) = (1-\beta_n)^{-1} + o(\log(q_n))$].

However, I am having trouble following the deduction of these "sub-logarithmic ZFRs" from the Deuring-Heilbronn phenomenon alone. Using the Deuring-Heilbronn (Linnik's repulsion theorem) as in Théorème 16, Sec. 6 of Bombieri's "Le grande crible", we get that there are absolute constants $c_1,c_2 >0$ such that, fixing $n\in\mathbb{N}$, it holds:
$$ \sigma < 1 - c_1\log\left(c_2\frac{\delta_n^{-1}}{\log(q_n q)/\log(q_n)}\right) \cdot \frac{1}{\log(q_n q)} $$
(I am just taking $T = q_n q$ in Théorème 16). Assuming we have an infinite sequence $q_n \to +\infty$ with $\delta_n \to 0$, for a given $q\in\mathbb{N}$ we may take $q_k \leq q < q_{k+1}$, so that $\log(q_{k+1} q)/\log(q_{k+1})< 2$, and hence:
$$ \sigma < 1 - c_1\frac{\log(\frac{c_2}{2}\delta_{k+1}^{-1})}{\log(q_{k+1} q)}. $$
It appears, then, that to derive the "sub-logarithmic" ZFRs, it is necessary to have $\log(q_{k+1}) \ll \log(q_k)$ as $k\to \infty$, i.e.: the gaps between the conductors of consecutive exceptional characters need to be polynomially bounded, even if we assume $\delta_n \to 0$.
I do not think my conclusion is correct (e.g., I believe I misinterpreted some aspect of Heath-Brown's paper), but I have not been able to get rid of this condition on the growth of the $q_n$. In short, my question is the following:

Q. Is it really possible to derive "sub-logarithmic" ZFRs from Linnik's repulsion theorem (i.e., without additional growth conditions on the $q_n$)?

REPLY [4 votes]: Let $\chi$ be a non-principal real Dirichlet character modulo $q$. Let 
$$\beta_0=1-\frac{1}{\eta\log q}$$
be a real zero of $L(s,\chi)$ satisfying $\eta\geq 100$ for convenience (Heath-Brown's condition is $\eta\geq 3$). Let $\rho=\beta+i\gamma$ be any zero of $L(s,\chi)$ such that $\rho\neq\beta_0$ and $|\gamma|\leq 1$. We strengthen Heath-Brown's claim $r_0\gg L^{-1}\log\eta$ to (note that $L=\log q)$
$$1-\beta\gg\frac{\log\eta}{\log q}.$$
We shall deduce this from Theorem 2 of Jutila's 1977 paper "On Linnik's constant", which is also Heath-Brown's reference. Let us write $\delta$ for the left hand side. If $\delta>1/60$, then the statement is trivial by $\eta\ll q$. So we shall assume that $\delta\leq 1/60$. Then, Jutila's theorem yields readily (using $D\leq 2q$) that
$$1-\beta_0\geq\frac{1}{10}\cdot\frac{q^{-3\delta}}{\log q}.$$
In other words, $\eta\leq 10 q^{3\delta}$. By the assumption $\eta\geq 100$, this implies that $\eta\leq q^{6\delta}$, or equivalently that
$$\delta\geq\frac{1}{6}\cdot\frac{\log\eta}{\log q}.$$
We have verified Heath-Brown's claim.