Edward Charles Titchmarsh, The theory of the Riemann zeta-function, IX. The general distribution of the zeros pages 191 to 193.
Oxford at the Clarendon press 1951.

**9.12.** We shall now obtain a more precise result of the same kind. $\dagger$

THEOREM 9.12. *For every large positive T,* $\zeta(s)$ *has a zero* $\beta+i \gamma$ *satisfying*

$$|\gamma-T|<\frac{A}{\log\log\log\;T}.$$

This was first proved by Littlewood by a detailed study of the conformal representation used in the previous proof. This involves rather complicated calculations with elliptic functions. We shall give here two proofs which avoid these calculations.

In the first, we replace the rectangles by a succession of circles. Let $T$ be a large positive number, and suppose that $\zeta(s)$ has no zero $\beta+i \gamma $ such that $T-\delta \leq \gamma \leq T+\delta$, where $\delta < \frac{1}{2}$. Then the function

$$f(s)=\log\zeta(s),$$
where the logarithm has its principal value for $\sigma > 2$, is regular in the rectangle

$$-2 \leq \sigma \leq 3, \;\;\;\;\;\;\; T - \delta \leq t \leq T + \delta$$

$\dagger$ Littlewood (3); proofs given here by Titchmarsh (13), Kramaschke (1).

Let $c_v, C_v, \textbf{C}_v, \Gamma_v$ be four concentric circles, with centre $2-\frac{1}{4}v\delta+iT,$ and radii $\frac{1}{4}\delta,\frac{1}{2}\delta,\frac{3}{4}\delta,\delta$ respectively. Consider these sets of circles for $v=0,1,...,n,$ where $n=[12/\delta]+1,$ so that $2-\frac{1}{4}n\delta \leq -1,$ i.e. the centre of the last circle, or to the left of, $\sigma = -1$. Let $m_v, M_v,$ and $\textbf{M}_v$ denote the maxima of $|f(s)|$ on $c_v, C_v,$ and $\textbf{C}_v$ respectively.

Let $A_1,A_2,...$ denote the constants (it is convenient to preserve their identity throughout the proof). We have $\textbf{R} \{ f(s) \} < A_1\log T$ on all the circles, and $|f(2+iT)|<A_2$. Hence the Borel-Carathéodory theorem for the circles $\textbf{C}_0$ and $\Gamma_0$ gives

$$\textbf{M}_0< \frac{\delta+\frac{3}{4}\delta}{\delta-\frac{3}{4}\delta}(A_1\log T+A_2)=7(A_1 \log T + A_2),$$
and in particular

$$|f(2-\frac{1}{4}\delta+iT)|<7(A_1 \log T + A_2).$$

Hence, applying the Borel-Carathéodory theorem to $\textbf{C}_1$ and $\Gamma_1,$

$$\textbf{M}_1<7\{A_1\log T + |f(2-\frac{1}{4}\delta+iT)|\}<(7-7^2)A_1\log T +7^2 A_2.$$

So generally $$\textbf{M}_v < (7+7^{v+1})A_1 \log T + 7^{v+1}A_2$$
or, say, $$\textbf{M}_v < 7^{v}A_3 \log T. \;\;\;\;\;\;\;\;\;\;\;\;\;(9.12.1)$$
Now by Hadamard's three-circles theorem
$$M_v \leq m_{v}^{a}\textbf{M}_{v}^{b},$$

where $a$ and $b$ are positive constants such that $a+b=1;$ in fact $a= \log \frac{3}{2}/\log 3,$ $b=\log 2 / \log 3.$ Also, since the circle $C_{v-1}$ includes the circle $c_v,$ $m_v \leq M_{v-1}.$ Hence

$$M_v \leq M_{v-1}^{a} \textbf{M}_{v}^{b} \;\;\;\;\;\;\;\;(v=1, 2,...,n).$$

Thus $$M_1 \leq M_{0}^{a}\textbf{M}_{1}^{b}, \;\;\;\;\;\;\; M_2 \leq M_{1}^{a}\textbf{M}_{2}^{b} \leq M_{0}^{a^2}\textbf{M}_{1}^{ab}\textbf{M}_{2}^{b},$$
and so on, giving finally
$$M_n \leq M_{0}^{a^n}\textbf{M}_{1}^{a^{n-1}b}\textbf{M}_{2}^{a^{n-2}b}...\textbf{M}_{n}^{b}.$$

Hence, by $(9.12.1),$

$$M_n \leq M_{0}^{a^n}7^{a^{n-1}b+2a^{n-2}b+...+nb}(A_3 \log T)^{a^{n-1}b+a^{n-2}b+...+b}.$$

Now
$$a^{n-1}+2a^{n-2}b+...+nb<n^2$$
$$a^{n-1}b + a^{n-2}b + ... + b = b(1-a^n)/(1-a)=1-a^n.$$
Hence $$M_n \leq M_{0}^{a^n}7^{n^2}(A_3 \log T)^{1-a^n} < A_4 7^{n^2}(\log T)^{1-a^n},$$
since $M_0$ is bounded as $T \rightarrow \infty.$

But $|\zeta(s)|>t^{A_5}$ for $\sigma \leq -1, \;\; t>t_0,$ so that $M_n>A_5 \log T.$ Hence
$$A_5<A_47^{n^2}(\log T)^{-a^n},$$
$$\log \log T < \left(\frac{1}{a}\right)^{n} \left(n^2 \log 7 -\log \frac{A_5}{A_4}\right),$$
$$\log \log \log T < n \log \frac{1}{a} + A_6 \log n,$$
so that
$$\delta < \frac{12}{n-1} < \frac{A}{\log \log \log T},$$
and the result follows.