Estimating the height required to find a given small value of $|\zeta(s)|$ near the line $\sigma=1$

Question

There are some qualitative theorems of Bohr, Jessen and Titchmarsh (e.g. The Theory of the Riemann zeta function, E.C. Titchmarsh, pages 306-308) proving that there is a $K=K(a,\alpha,\beta)$ such that the number of solutions to $\zeta(s)=a$, $a\neq 0$, in the rectangle $1/2<\alpha<\beta<1$, $0<t<T$, is greater than $KT$. However, there doesn't seem to be an obvious way of using the method described in Titchmarsh to determine the size of $K$ as a function of the variable $a$.

I am interested in a similar quantitative question:

At what height $T(\alpha,\delta)$ are you guaranteed there exists an $s$ such that $|\zeta(s)|=\delta$ in the region $\sigma>1-\alpha$, $0<t<T$?

Obviously, if there is a zero in the region then you're done. For sufficiently small $\delta$ it follows from Euler's product formula that such points cannot be significantly beyond the line $\sigma=1$.

Assuming that $$\frac{1}{|\rho_a\zeta'(\rho_a)|}\leq C$$ at the points $\rho_a$ where $\zeta(\rho_a)=a$ and $0<\alpha<7/4$, I find that

$$\log T(\alpha,\delta)= o \left(\delta^{-7/\alpha}\right),$$ but I think this bound is bad.

Answer 1

It suffices to construct points in the rectangle with $|\zeta(s)|\leq\delta$. This can be done, even to the right of 1, by diophantine approximation: Pick some $t$, such that for the first $k$ primes $p_1, \ldots, p_k$ we have $t\log p_i\bmod 2\pi\in[\frac{4\pi}{5}, \frac{6p}{5}]$. Then $\prod_{i=1}^k\left(1-\frac{1}{p^{1+it}}\right)^{-1}$ is small. Pick a small $\sigma>1$, depending on $k$, such that $\prod_{i=k+1}^\infty\left(1-\frac{1}{p^{s}}\right)^{-1}$ is bounded by e.g. 2 in the half plane $\Re\;s>\sigma$. If $k$ is large, then $\sigma$ will be close to 1, so $\prod_{i=1}^k\left(1-\frac{1}{p^{\sigma+it}}\right)^{-1}$ is still small.

This procedure can be made completely explicit, however, when using the pigeon hole principle for constructing $t$, you get an exponential dependency on $k$, and since $\prod_{i=1}^k\left(1+\frac{1}{p}\right)^{-1}\asymp\frac{1}{\log k}$, the final result is double exponential.

If we want to stay to the right of 1, we cannot do much better, since under RH we have $\frac{1}{\zeta(1+it)}\ll\log\log t$. If we do not want to stay there, things become more complicated, since the product expansion is no longer valid.