Subconvexity bounds and zero-free regions When I see results in analytic number theory, I often have trouble seeing how they relate and their relative strength. Here is a specific question that should help me (and hopefully others too).
Here is my understanding, please correct anything that is wrong here.
An $L$ function $L(s)$ comes with an analytic conductor $q$. I will also write $s = \sigma + it$ ($\sigma,t\in\mathbb{R}$). We believe that it satisfies a Riemann hypothesis:
$L(s)\neq 0$ whenever $\sigma>\frac12$
and a Lindelöf hypothesis:
$|L(1/2+it)|\ll |qt|^\epsilon$.
Ignoring real zeroes, we have bounds towards the Riemann hypothesis, i.e. zero-free regions of the form:
$L(s)\neq 0$ whenever $\sigma > 1-\frac{c}{\log(q(1+|t|))}$ and $t\neq 0$,
and sometimes improved zero-free regions
$L(s)\neq 0$ whenever $\sigma > 1-\frac{c}{\log(q(1+|t|))^a}$ and $t\neq 0$, for some $a<1$.
We also have bounds towards the Lindelöf hypothesis, i.e. convexity bounds
$L(1/2+it) \ll |qt|^{1/4}$,
and sometimes subconvexity bounds, i.e.
$L(1/2+it) \ll |qt|^{1/4-\delta}$ for some $\delta>0$.
Both zero-free regions and subconvexity bounds are related to bounding the values of the $L$-function in the critical strip and to bounding certain sums (character/exponential/...). However, I don't see very well how they relate: for instance, I found out that in the $q$-aspect for Dirichlet $L$-functions, subconvexity bounds are known (Burgess) but improved zero-free regions are not (according to the answer to this question).

Can you clarify the relations between:

*

*bounds of various sums

*subconvexity bounds

*zero-free regions?


Feel free to separate the $t$-aspect from the $q$-aspect if you think that it will clarify the answer.
 A: Let's just discuss the $t$-aspect, i.e. bounds for the zeta function and its zeroes.
Let $T$ be a large ordinate, and let $H$ be a medium-sized quantity (much larger than $1$, but much less than $T$).  Roughly speaking, subconvexity bounds at ordinates $t = T + O(H)$ measure the average failure of the Riemann hypothesis in this region, whilst zero-free regions at these ordinates measure the worst-case failure of the Riemann hypothesis here.
Indeed, consider a rectangular contour with corners $\frac{1}{2}+\varepsilon+iT$, $\frac{1}{2}+\varepsilon+i(T+H)$, $2 + i(T+H)$, and $2+iT$ for some very small $\varepsilon>0$.  Suppose first that there are no zeroes of $\zeta$ in this rectangle, so we have a holomorphic branch of $\log \zeta$ in this rectangle.  By Cauchy's theorem, the integral of $\log \zeta$ on this rectangle is zero.  For medium-sized $H$ (and perturbing $T$ and $H$ slightly if needed), the contribution of the short horizontal edges of the rectangle are small, and from elementary bounds the contribution of the right vertical edge is $O(H)$.  Sending $\varepsilon \to 0$, and taking real parts, we conclude (heuristically at least) that
$$ \frac{1}{H} \int_T^{T+H} \log |\zeta(\frac{1}{2}+it)|\ dt = o( \log T )$$
(in fact we can get a better error term here than $o(\log T)$).  This is a cousin of the well known fact that the Riemann hypothesis implies the Lindelof hypothesis.
Now suppose we have some zeroes $\sigma_j + i t_j$ in the rectangle.  To maintain a holomorphic branch of $\log \zeta$ in the rectangle, we need to cut out some slits connecting $\frac{1}{2}+it_j$ to $\sigma_j + it_j$.  If one does so and performs the usual contour integration calculations, one eventually arrives (heuristically at least) at the Jensen-type formula
$$ \frac{1}{H} \int_T^{T+H} \log |\zeta(\frac{1}{2}+it)|\ dt = \frac{2\pi }{H} \sum_{T \leq t_j \leq T+H} (\sigma_j-\frac{1}{2})_+ + o( \log T ).$$
Recall from the Riemann-von Mangoldt formula that there are about $\frac{H}{2\pi} \log T$ zeroes $\sigma_j+it_j$ with $T \leq t_j \leq T+H$; by the functional equation, they are symmetric around the critical line.  The expression $ \frac{2\pi }{H} \sum_{T \leq t_j \leq T+H} (\sigma_j-\frac{1}{2})_+$ is thus at worst about $\frac{1}{4} \log T$, which is attained when about half of the zeroes have abscissas $\sigma_j$ close to 1, and the other half have abscissas close to zero.  This is basically a slightly averaged form of the convexity bound $\zeta(\frac{1}{2}+iT) \ll T^{1/4+o(1)}$.
More generally, a subconvexity bound of the form $\zeta(\frac{1}{2}+iT) \ll T^{\alpha+o(1)}$ is basically asserting that near the ordinate $T$, the quantities $(\sigma_j-\frac{1}{2})_+$ have average value at most $\alpha+o(1)$, thus as mentioned previously subconvexity controls the average failure of the Riemann hypothesis.  In particular, if the Lindelof conjecture is true, $\alpha=0$ and most of the zeroes at ordinates near $T$ are within $o(1)$ of the critical line (this observation is I believe due to Backlund).
Meanwhile, zero-free regions are upper bounds on $(\sigma_j-\frac{1}{2})_+$, that is to say worst-case failure of the Riemann hypothesis near a given ordinate $T$.  The worst-case bounds the average-case, so a really good zero-free region does give some subconvexity bound, but this is extremely inefficient (in particular, the best zero-free region currently available unconditionally does not come anywhere close to implying the best subconvexity bounds currently known).
I like to formalise the above discussion using limiting profiles of the log-zeta function: https://terrytao.wordpress.com/2015/03/01/254a-supplement-7-normalised-limit-profiles-of-the-log-magnitude-of-the-riemann-zeta-function-optional/ .  I think Titchmarsh's book also has some discussion of these topics.
Finally, $\zeta(\frac{1}{2}+it)$ can be related to partial sums $\sum_{n \leq N} \frac{1}{n^{1/2+it}}$ by elementary arguments (or the approximate functional equation), and by dyadic decomposition and summation by parts these partial sums can in turn be related to exponential sums such as $\frac{1}{N^{1/2}} \sum_{N \leq n \leq 2N} \frac{1}{n^{it}} = \frac{1}{N^{1/2}} \sum_{N \leq n \leq 2N} e(  - \frac{t}{2\pi} \log n )$.  A subconvexity bound $\zeta(\frac{1}{2}+iT) \ll T^{\alpha+o(1)}$ is basically the same thing as an exponential sum bound $\sum_{N \leq n \leq 2N} e(  - \frac{t}{2\pi} \log n ) \ll N^{1/2} T^{\alpha+o(1)}$ (see e.g. these notes of mine for some more rigorous statements of this form).
