The claim (i) follows from the Phragmen-Lindelöf principle. The claim (ii) follows from Hölder. The claim (iii) follows from standard upper and lower bounds on \(\zeta (\sigma +it)\) for \(\sigma \geq 1\). The claim (iv) follows from (i)-(iii), and (v) is trivial. The claim (vi) follows easily from the functional equation.

For (vii), the bound \(M(\sigma ,A) \leq A \mu (\sigma ) + 1\) is trivial, which implies that

Suppose for contradiction that

thus there is \(\delta {\gt}0\) such that

for all \(A\geq 0\). By convexity, this gives

for all sufficiently small \(\varepsilon \), and then by the Cauchy integral formula and Hölder’s inequality we can conclude that

for unbounded \(|t|\), leading to

Sending \(A\) to infinity and \(\varepsilon \) to zero, we obtain a contradiction.

Follows from Hölder’s inequality and the standard estimates

and

for any unbounded \(T{\gt}1\), due to [ 95 ] and [ 96 ] respectively.

We first show the lower bound, or equivalently that

whenever \(\tau \geq 2\) and \(\sigma \geq 1/2\). Accordingly, let \(N\) be unbounded, \(T = N^{\tau +o(1)}\), \(I \subset [N,2N]\), and \(W\) be a \(1\)-separated subset of \([T,2T]\) such that

for \(t \in W\). By standard Fourier analysis (or by Perron’s formula and contour shifting), this gives

and hence by Hölder

so on summing in \(r\)

By Definition 9.1, the left-hand side is \(\ll T^{M(\sigma _0,A)+o(1)}\). Since \(T = N^{\alpha +o(1)}\), we obtain

giving the claim.

For the converse bound, let \(M\) be the right-hand side of 1. From Lemma 8.9 we have \(M \geq 1\). By [ 144 , § 8.1 ] it will suffice to show that for any \(V{\gt}0\) and any \(1\)-separated \(W \subset [T,2T]\) with

for all \(t \in W\), one has

The claim is clear if \(V \geq T^C\) or \(V \leq T^C\) for some sufficiently large \(C\), so we may assume that \(V = T^{O(1)}\). We also clearly can assume \(|W|\geq 1\). Using the Riemann–Siegel formula [ 144 , Theorem 4.1 ] and dyadic decomposition, we have either

or

for some \(I \subset [N,2N]\) and \(1 \leq N \ll T^{1/2}\), and all \(t \in W\). In either case, we can perform summation by parts and conclude that

or

for some \(I'\) in \([N,2N]\) and all \(t \in W\). As \(\sigma _0 \geq 1/2\), the letter hypothesis is stronger than the former, so we may assume the former. If \(N = T^{o(1)}\) then this would imply that \(V \ll T^{o(1)}\), and we would be done from the trivial bound \(R \ll T\) since \(M \geq 1\). Hence, after passing to a subsequence, we can assume that \(N = T^{1/\tau +o(1)}\) for some \(2 {\lt} \tau {\lt} \infty \). We can also assume that \(V = N^{\sigma -\sigma _0+o(1)}\) for some \(\sigma \in \mathbf{R}\). If \(\sigma \leq \sigma _0\) then \(V \ll T^{o(1)}\) and we are done as before, so we may assume \(\sigma {\gt} \sigma _0\); in particular, \(\sigma \geq 1/2\). From Lemma 8.2 we have

and hence by definition of \(M\)

as required.

Apply Lemma 9.5 with \(\sigma _0 = 1/2\) and \(A=4\), using Lemma 9.2(iv).

From Lemma 8.11 with the exponent pair \((1/2,1/2)\) from Lemma 5.10 we have

If \(2\tau - 12 (\sigma -1/2) \geq 0\), the claim is immediate; if instead \(2\tau - 12 (\sigma -1/2) {\lt} 0\), use Lemma 8.4.

Let \((N,T,V,(a_n)_{n \in [N,2N]},J,W)\) be a zeta large value pattern with \(N\), \(V = N^{\sigma +o(1)}\), \(T = N^{\tau +o(1)}\) and \(W = N^{\mathrm{LV}_\zeta (\sigma ,\tau )+o(1)}\). Our task is to show that

Write \(a(n) = 1_I(n)\). By a Fourier analytic expansion we can bound

for some fixed \(\varepsilon {\gt}0\) and all \(t \in W\), hence

In particular, we can truncate to large values of \(\zeta (1/2+it_1)\), in the sense that

Summing in \(t\) and using the \(1\)-separation to bound the sum of \(1/(1+|t_1-t|)\) by \(T^{o(1)}\), we conclude that

Hence by dyadic pigeonholing we have

for some \(V' \geq T^{-o(1)} N^{-1/2} V\), and thus

for all \(t'\) in some \(1\)-separated subset \(W'\) of \([T/4, 3T]\) with

Applying [ 103 , Theorem 2 ] (treating different cases using the bounds [ 103 , (7), (8), (9) ] ), we have the bound

and thus

and the claim now follows from the lower bound on \(V'\).

This is a computation using Lemma 8.11, Theorem 8.15, and Lemma 9.5; see [ 144 ] for details.

This is a routine modification of the proof of Lemma 9.5.

From Lemma 8.11 with the exponent pair \((k,\ell )\) we have

In particular, for \(\sigma -1/2 \leq h\tau \) one has

and for \(\sigma -1/2 \geq h\tau \) one has

The claim then follows from Lemma 9.12.

Apply Corollary 9.13 with the exponent pair \((4/18, 11/18) = BABA(1/6, 2/3)\) from Corollary 5.11.