= Solution
Here is a quantitative local form of the <Landau zero-free-region theorem>. Let $t\ge2$, $0<\eta\le1/4$, $M\ge2$, 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 $\beta+it$ satisfies
$$
\boxed{1-\beta\ge\frac{c\eta}{1+\log(M/\eta)}}
$$
for 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 $|z-z_0|\le R$, $f(z_0)\ne0$, and $|f|\le M$, then for $|z-z_0|\le R/3$ away from zeros,
$$
\frac{f'(z)}{f(z)}=\sum_{|\rho-z_0|\le R/2}\frac1{z-\rho}+O\left(\frac{1+\log(M/|f(z_0)|)}R\right).
$$
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 https://personal.science.psu.edu/rcv4/Vol3/Vol3.pdf[Montgomery and Vaughan's general treatment].
The reciprocal <Euler product> gives $|\zeta(1+\eta/4+ij t)|\ge1/\zeta(1+\eta/4)\gg\eta$ 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 $1<\sigma\le1+\eta/4$,
$$
\Re\frac{\zeta'(\sigma+it)}{\zeta(\sigma+it)}\ge\frac1{\sigma-\beta}-C\frac B\eta,\qquad\Re\frac{\zeta'(\sigma+2it)}{\zeta(\sigma+2it)}\ge-C\frac B\eta.
$$
All other zero terms may be discarded because their real parts are nonnegative. The simple <pole> at one gives $-\zeta'(\sigma)/\zeta(\sigma)=(\sigma-1)^{-1}+O(1)$. Insert these inequalities into part (a):
$$
\frac4{\sigma-\beta}-\frac3{\sigma-1}\ll\frac B\eta.
$$
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.
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