 Solution

Here is a quantitative local form of the Landau zero-free-region theorem. Let $t\ge 2$, $0<\eta \le 1/4$, $M\ge 2$, and suppose $|\zeta (z)|\le M$ on the two closed discs of radius $\eta$ centred at $1+\eta /4+it$ and $1+\eta /4+2it$. Then every zero $\beta +it$ satisfiesfor an absolute positive $c$. In particular, if upper bounds of this form hold locally for every large height, they give a zero-free region of this width. The discs are away from the pole, and the Euler product excludes zeros to the right of one.

We state precisely the permitted local logarithmic-derivative lemma. If $f$ is holomorphic on a neighborhood of $|z-z_0|\le R$, $f(z_0)\ne 0$, and $|f|\le M$, then for $|z-z_0|\le R/3$ away from zeros,The zeros are counted with multiplicity. This standard disc estimate, which may be assumed here, follows by factoring nearby zeros and applying a Cauchy estimate for derivatives to the remaining logarithm. When every zero has real part at most one and $\Re z>1$, the zero terms have nonnegative real parts, giving the required lower bound. The estimate with fixed radius ratios is also recorded as Lemma 24.17 in Montgomery and Vaughan's general treatment.

The reciprocal Euler product gives $|\zeta (1+\eta /4+ijt)|\ge 1/\zeta (1+\eta /4)\gg \eta$ for $j=1,2$. Put $B=1+\log (M/\eta )$. The lemma's error on both discs is therefore $O(B/\eta )$. Let $d=1-\beta$. If $d\ge \eta /24$, the desired conclusion already holds after reducing $c$. Otherwise $\beta +it$ is among the local zeros and, for $1<\sigma \le 1+\eta /4$,All other zero terms may be discarded because their real parts are nonnegative. The simple pole at one gives $-{\zeta }^{\prime }(\sigma )/\zeta (\sigma )=(\sigma -1{)}^{-1}+O(1)$. Insert these inequalities into part (a):There is no zero on the line one by the argument in Question 2(a), so $d>0$. Choose $\sigma =1+6d$, which lies in the indicated range. The left side is $(4/7-3/6)/d=1/(14d)$. Hence $d\gg \eta /B$, proving the theorem. The logarithm of the upper bound, rather than the upper bound itself, is what enters the zero-free width.