The claim (i) is clear using the Riemann-von Mangoldt formula [ 144 , Theorem 1.7 ] and the functional equation. The claim (ii) is also clear.

Write the right-hand side as \(B\), then \(B \geq 0\) (from Lemma 7.4(iii)) and we have

for all \(\tau \geq 1\), and

whenever \(\varepsilon {\gt}0\) and \(\tau \) is sufficiently large depending on \(\varepsilon \) (and \(\sigma \)). It would suffice to show, for any \(\varepsilon {\gt}0\), that \(N(\sigma -o(1),T) \ll T^{B+O(\varepsilon )+o(1)}\) as \(T \to \infty \).

By dyadic decomposition, it suffices to show for large \(T\) that the number of zeroes with real part at least \(\sigma -o(1)\) and imaginary part in \([T,2T]\) is \(\ll T^{B+O(\varepsilon )+o(1)}\). From the Riemann-von Mangoldt theorem, there are only \(O(\log T)\) zeroes whose imaginary part is within \(O(1)\) of a specified ordinate \(t \in [T,2T]\), so it suffices to show that given some zeroes \(\sigma _r + i t_r\), \(r=1,\dots ,R\) with \(\sigma -o(1) \leq \sigma _r {\lt} 1\) and \(t_r \in [T,2T]\) \(1\)-separated, that \(R \ll T^{B+O(\varepsilon )+o(1)}\).

Suppose that one has a zero \(\sigma _r+i t_r\) of this form. Then by a standard approximation to the zeta function [ 144 , Theorem 1.8 ] , one has

Let \(0 {\lt} \delta _1 {\lt} \varepsilon \) be a small quantity (independent of \(T\)) to be chosen later, and let \(0 {\lt} \delta _2 {\lt} \delta _1\) be sufficiently small depending on \(\delta _1,\delta _2\). By the triangle inequality, and refining the sequence \(t_r\) by a factor of at most \(2\), we either have

for all \(r\), or

for all \(r\).

Suppose we are in the former (“Type I”) case, we perform a smooth partition of unity, and conclude that

for some fixed bump function \(\psi \) supported on \([1/2,1]\), and some \(T^{\delta _1} \ll N \ll T\).

We divide into several cases depending on the size of \(N\). First suppose that \(N \ll T^{1/2}\). The variable \(n\) is restricted to the interval \(I := [\max (N/2, T^{\delta _1}), N]\). We have

Performing a Fourier expansion of \(\psi (n/N) (n/N)^{-\sigma _r}\) in \(\log n\) and using the triangle inequality, we can bound

for any \(A{\gt}0\), so by the triangle inequality we conclude that

for some \(t'_r = t_r + O(T^{o(1)})\). By refining the \(t_r\) by a factor of \(T^{o(1)}\) if necessary, we may assume that the \(t'_r\) are \(1\)-separated, and by passing to a subsequence we may assume that \(T = N^{\tau +o(1)}\) for some \(2 \leq \tau \leq 1/\delta _1\), then we conclude that

for all remaining \(r\). By Definition 8.1 we then have (for \(\delta _2\) small enough)

and the claim follows in this case from 1.

In the case \(N \asymp T\), a standard application of the Euler–Maclaurin formula (see e.g., [ 277 , (2.1.2) ] ) yields

which leads to a contradiction. So the only remaining case is when \(T^{1/2} \ll N \ll o(T)\). Here we can ignore the cutoffs on \(n\) and obtain

Applying the van der Corput \(B\)-process (see, e.g., [ 149 , § 8.3 ] ) or the approximate functional equation we have

and thus

where \(M := 2\pi T/N \ll N^{1/2}\). In particular

since \(N \gg T^{1/2}\) and \(\sigma \geq 1/2\). Performing a Fourier expansion as before, we conclude that

for some \(t'_r = t_r + O(T^{o(1)})\), and one can argue as in the \(N \ll T^{1/2}\) case (partitioning \([M/10, 10M]\) into \(O(1)\) intervals each contained in some \([M',2M']\) with \(M' \ll T^{1/2}\)).

Now suppose instead we are in the latter (“Type II”) case 3. We multiply both sides of 3 by the mollifier \(\sum _{m \leq T^{\delta _2/2}} \frac{1}{m^{\sigma _r+it_r}}\) to obtain

where \(a_n\) is some sequence with \(a_n \ll T^{o(1)}\). By dyadic decomposition and the pigeonhole principle, and refining the \(t_r\) by a factor of \(O(T^{o(1)})\) as needed, we can then find an interval \(I\) in \([N,2N]\) with \(T^{\delta _2/2} \ll N \ll T^{\delta _1+\delta _2/2}\) such that

and hence by Fourier expansion of \(\frac{1}{n^{\sigma _r}}\) in \(\log n\)

for some \(t'_r = t_r + O(T^{o(1)})\); by refining the \(t_r\) by a further factor of \(T^{o(1)}\) we may assume that the \(t'_r\) are also \(1\)-separated; we can also pigeonhole so that \(T = N^{\tau +o(1)}\) for some \(\frac{1}{\delta _1+\delta _2/2} \leq \tau \leq \frac{1}{\delta _2/2}\). Applying Lemma 7.3, we conclude that

and the claim follows in this case from 2.

Let \(N \geq 1\) be unbounded, \(T = N^{\tau +o(1)}\), and \(I \subset [N,2N]\) be an interval, and \(t_1,\dots ,t_R \in [T,2T]\) be \(1\)-separated with

uniformly for all \(r\). By [ 206 , Theorem 1.2 ] , we have for any fixed \(\delta {\gt}0\) that

Using Definition 11.1, we conclude that

and thus

Here the implied constant in the \(O()\) notation is understood to be uniform in \(\delta \). Letting \(\delta \) go to zero, and using left-continuity of \(\mathrm{A}\), we obtain the claim.

Denote the right-hand side by \(B\), thus

for all \(\tau _0 \leq \tau \leq 2\tau _0\), and

whenever \(2 \leq \tau {\lt} \tau _0\). From Lemma 7.8 we then have

for all \(k\tau _0 \leq \tau \leq 2k\tau _0\) and natural numbers \(k\). Note that the intervals \([k\tau _0, 2k\tau _0]\) cover all of \([\tau _0,\infty )\), hence we have

for all \(\tau \geq \tau _0\). In particular

Also, combining the previous estimate with 4 using Lemma 8.3(iii) we have

for all \(\tau \geq 2\). By Lemma 8.3(iv), this implies that

for \(\tau \geq 2\). Thus

The claim now follows from Lemma 11.5.

Applying Corollary 11.7 with \(\tau \) replaced by \(4\tau _0/3\), it suffices to show that

But this follows from Lemma 7.8, since the intervals \([2k\tau _0/3, k\tau _0]\) for \(k=2,3\) cover all of \([4\tau _0/3,8\tau _0/3]\).

We may assume that \(\tau _0 \geq 3-3\sigma \), since otherwise the claim follows from the Riemann–von Mangoldt bound

By Lemma 7.4(ii) we have

for all \(\tau \geq 0\). But both expressions on the right-hand side are bounded by \((3-3\sigma ) \frac{\tau }{\tau _0}\) for \(2\tau _3 \leq \tau \leq \tau _0\) and \(\tau _0 \geq 3-3\sigma \), so the claim follows from Corollary 11.9.

Apply Corollary 11.9 with \(\tau _0=3/2\) (so that 7 is vacuously true).

The first result is proved in [ 138 ] , and the second result is due to [ 91 ] . We will apply Corollary 11.8. From Corollary 8.8 we see that \(\mathrm{LV}_\zeta (\sigma ,\tau ) = -\infty \) whenever \(\sigma {\gt} 1/2\) and \(\tau \geq 1\), so for any choice of \(\tau _0\) we have

From Theorem 7.9 and Lemma 7.8 we have

for any natural number \(k\) and \(\tau \geq 1\); setting \(k\) to be the integer part of \(\tau \) we conclude in particular that

and hence by taking \(\tau _0\) large enough, we can make \(\sup _{2\tau _0/3 \leq \tau \leq \tau _0} \mathrm{LV}(\sigma ,\tau )/\tau \) bounded by \(2-2\sigma + \varepsilon \) for any \(\varepsilon {\gt}0\). This already gives the density hypothesis bound \(\mathrm{A}(\sigma ) \leq 2\). For \(\sigma {\gt} 3/4\), we may additionally apply Theorem 8.14 to make \(\sup _{2\tau _0/3 \leq \tau \leq \tau _0} \mathrm{LV}(\sigma ,\tau )/\tau \) arbitrarily small, giving the bound \(\mathrm{A}(\sigma ) \leq 0\).

We give here a proof (somewhat different from the original proof) that passes through Corollary 11.7. We apply Corollary 11.7 with \(\tau _0\) chosen so that \(\mu (1/2) \tau _0 {\lt} \sigma -1/2\). From Corollary 8.6 we then have

For any integer \(k \geq 0\) and \(k \leq \tau \leq k+1\), we see from 9 that

and

multiplying the first inequality by \(2\sigma -1\), the second by \(2-2\sigma \), and summing, we conclude that

inserting this bound we have

Optimizing in \(\tau _0\), we obtain the claim.

We apply Corollary 11.10 with \(\tau _0 := 2-\sigma \). Here we have \(4\tau _0/3 {\lt} 2\) since \(\sigma {\gt}1/2\), so the claim 7 is automatic; and the Montgomery conjecture hypothesis follows from Theorem 7.9.

We apply Corollary 11.10 with \(\tau _0 := 3\sigma -1\). The Montgomery conjecture hypothesis follows from Theorem 7.12. So it remains to show that 7 holds for \(2 \leq \tau {\lt} 4\tau _0/3\). For \(\sigma \leq 5/6\) we have \(4\tau _0/3 \leq 2\), so the claim is vacuously true in this case. For \(\sigma {\gt} 5/6\) we use Corollary 8.6 and the bound \(\mu (1/2) \leq 1/6\) from Table 6.1 to conclude that \(\mathrm{LV}_\zeta (\sigma ,\tau ) = -\infty \) whenever \(\sigma {\gt} 1/2 + \tau /6\), but this is precisely \(\tau {\lt} 6\sigma - 3\). Since \(6\sigma -3 {\gt} 4\tau _0/3\) when \(\sigma {\gt} 5/6\), we obtain the claim.

We may assume that \(7/10 {\lt} \sigma {\lt} 8/10\), since the bound follows from the Ingham and Huxley bounds otherwise. We apply Corollary 11.9 with \(\tau _0 := \frac{3+5\sigma }{5}\). We have \(4\tau _0/3 {\lt} 2\), so the claim 7 is vacuous and we only need to establish 6. We split into the subranges \(13/5-2\sigma \leq \tau {\lt} \tau _0\) and \(2\tau _0/3 \leq \tau \leq 13/5-2\sigma \). In the former range we use Theorem 10.27 (and 8), and reduce to showing that

and

for \(13/5-2\sigma \leq \tau {\lt} \tau _0\). The first inequality follows from

which one can numerically check holds in the range \(7/10 {\lt} \sigma {\lt} 8/10\). Finally, the third inequality is obeyed with equality when \(\tau =\tau _0\) and the right-hand side has a larger slope in \(\tau \) than the left (since \(\tau _0 \geq 3-3\sigma \)), so the claim follows as well.

In the remaining region \(2\tau _0/3 \leq \tau \leq 13/5-2\sigma \), we use Theorem 7.9 and 8 to reduce to showing that

in this range. This follows again from 10 which guarantees the inequality at the right endpoint \(\tau = 13/5-2\sigma \).

We apply Corollary 11.8 with \(\tau _0 := 3/2\), then it suffices to show that

for all \(1 \leq \tau \leq 3/2\).

From the \(k=3\) case of Theorem 7.16 we have

But all terms on the right-hand side can be verified to be less than or equal to \((2-2\sigma )\tau \) when \(1 \leq \tau \leq 3/2\) and \(\sigma \geq 11/14\), giving the claim.

For the first estimate, we apply Corollary 11.9 with \(\tau _0 := \frac{7\sigma -1}{3}\). To verify 6, we apply the \(k=3\) version of Theorem 7.16, which gives

When \(\sigma \geq 11/14\) one has \(18-24\sigma \leq \frac{10-16\sigma }{3}\), so by 8 we need to show that

for \(2\tau _0/3 \leq \tau \leq \tau _0\). This holds with equality at \(\tau =\tau _0\), hence holds for \(\tau \leq \tau _0\) as well since \(\tau _0 \geq 3-3\sigma \). As for 7, we invoke Theorem 9.7 and reduce to showing that

for \(2 \leq \tau \leq 4\tau _0/3\). Since \(6-12\sigma \) is negative, the ratio of the left-hand side and right-hand side is increasing in \(\tau \), so it suffices to verify this claim at the endpoint \(\tau = 4\tau _0/3\). The claim then simplifies to \(\tau _0 \leq \frac{3}{4} (4\sigma -1)\), which one can verify from the choice of \(\tau _0\) and the hypothesis \(\sigma \geq 11/14\).

For the second estimate, we take \(\tau _0 := \min ( 10\sigma -7, \frac{3}{4} (4\sigma -1))\). To verify 6, we now use Theorem 7.14 and 8, and reduce to showing that

for \(2\tau _0/3 \leq \tau \leq \tau _0\). The inequality holds at \(\tau =\tau _0\) since \(\tau _0 \leq 10\sigma -7\), and hence for all smaller \(\tau \) since \(\tau _0 \geq 3-3\sigma \). As for 7, we can repeat the previous arguments since \(\tau _0 \leq \frac{3}{4} (4\sigma -1)\).

This can be derived from the Watt exponent pair \(W := (89/560, 1/2 + 89/560)\) from Theorem 5.12 as well as the \(A\) and \(B\) transforms and convexity (Lemmas 5.4, 5.8, 5.9) after observing that

with \(x=37081/40415\) and \(y=476897/493711\). (One could of course also use more recent exponent pairs that are stronger, such as the Bourgain exponent pair \((13/84, 1/2+13/84)\).) We remark that one could also obtain this result from Lemma 5.3, after observing that the required bound \(\beta (\alpha ) \leq 3/40 + 7\alpha /10\) can be derived from Theorem 4.16 (as well as the classical bounds in Corollary 4.8). We also note that the corollary \(\mu (7/10) \leq 3/40 = 0.075\) is immediate from [ 279 , Theorem 2.4 ] , which in fact gives the slightly stronger bound \(\mu (7/10) \leq 218/3005 = 0.07254\dots \).

We apply Corollary 11.10 with \(\tau _0 := 10\sigma -7\). The claim 6 again follows from Theorem 7.14 and 8 as in the proof of Theorem 11.18. Meanwhile, from Lemma 11.19 and Corollary 8.7 we have \(\mathrm{LV}_\zeta (\sigma ,\tau ) = -\infty \) whenever \(\sigma {\gt} \frac{3}{40} \tau + \frac{7}{10}\), or equivalently \(\tau {\lt} \frac{4}{3} (10\sigma -7)\), which then immediately gives 7.

The arguments below are a translation of the original arguments of Bourgain [ 21 ] to our notational framework.

In view of Theorem 11.17 (or Theorem 11.18), we may assume that \(25/32 {\lt} \sigma {\lt} 11/14\). Set \(\rho := \mathrm{LV}(\sigma ,\tau )\). As in the proof of Theorem 11.17, it suffices to show that

for all \(1 \leq \tau \leq 3/2\).

From the \(k=3\) case of Theorem 7.16 we have

which in the \(\sigma {\lt} 11/14\) regime simplifies to

and this already suffices unless

In the regime \(\sigma {\gt} 25/32\) and \(\tau \leq 3/2\), the bound 13 certainly implies

and also

so we may invoke Corollary 10.30 to conclude that

for any \(\alpha _1, \alpha _2 \geq 0\).

We now divide into cases. First suppose that \(\tau \leq \frac{4(1+\sigma )}{5}\). In this case we set \(\alpha _1 := \frac{\tau }{3} - \frac{2}{3} (7\sigma -5)\) (which can be checked to be nonnegative using 14 and \(\sigma \geq 25/32\)) and \(\alpha _2=0\), and one can check that 15 implies 11 in this case (with some room to spare).

Now suppose that \(\tau {\gt} \frac{4(1+\sigma )}{5}\). In this case we choose \(\alpha _1 = \frac{\tau }{8} - \frac{9\sigma -7}{2}\) and \(\alpha _2 = \frac{5\tau }{4} - (1+\sigma )\), which can be checked to be nonnegative using the hypotheses on \(\sigma ,\tau \). In this case one can again check that 15 implies 11.

We apply Corollary 11.10 with \(\tau _0 := \min ( \frac{ 9(3\sigma -2)}{2}, \frac{8(2\sigma -1)}{3} )\). For future reference we note that

It suffices to show that

for \(2\tau _0/3 \leq \tau \leq \tau _0\), as well as

for \(2 \leq \tau {\lt} 4\tau _0/3\). For 18 we use the twelfth moment bound in Theorem 9.7. Since the slope of \(2\tau - 12 (\sigma -1/2)\) in \(\tau \) exceeds that of \(\frac{\tau }{\tau _0} (3-3\sigma )\) by 16, it suffices to check the bound at the endpoint, i.e., to show that

or equivalently \(\tau _0 \leq \frac{3(4\sigma -1)}{4}\), which one can easily check to be the case.

Now we prove 17. Set \(\rho := \mathrm{LV}(\sigma ,\tau )\). From the \(k=3\) case of Theorem 7.16 we again have 12, which implies the required bound \(\rho \leq \frac{\tau }{\tau _0}(3-3\sigma )\) unless one has

In this regime, one can also check from 12 that

(with room to spare) so we may apply Corollary 10.30 to obtain

for any \(\alpha _1, \alpha _2 \geq 0\).

We first consider the case when \(38/49 \leq \sigma \leq 4/5\), so that \(\tau _0 = 8(2\sigma -1)/3\). As in the proof of Theorem 11.21, we set

and

With this choice, the expressions \(\alpha _1+\alpha _2/2 + 2-2\sigma \) and \(-2\alpha _1 + \tau + 12 - 16 \sigma \) are both equal to \(\frac{\tau }{3} + \frac{16-20\sigma }{3} + \frac{\alpha _2}{3}\), while \(-\alpha _2 + 2\tau +4-8\sigma \) is equal to

which is less than or equal to the previous expression by definition of \(\alpha _2\). Finally, the expression \(4\alpha _1 + 1+\max (2,2\tau -2)-4\sigma \) is equal to

Thus it remains to show the bounds

and

For 23, we trivially have \(2-2\sigma \leq \frac{\tau }{\tau _0}(3-3\sigma )\) since \(\tau \geq 2\tau _0/3\), and the slope of \(\frac{5\tau }{4} - (1+\sigma ) + 2-2\sigma \) in \(\tau \) certainly exceeds \(\frac{3-3\sigma }{\tau _0}\) by 16, so it suffices to check the endpoint

which one can check to be valid for \(\sigma \leq 4/5\). Now we turn to 24. From 16 it suffices to show that

and

The former is an identity, and the latter simplifies to \(\tau \geq 14\sigma -10\), which one can check follows from 19 (with some room to spare) in the regime \(38/49 \leq \sigma \leq 4/5\), giving the claim. Finally, for 25 it suffices to show that

which by 16 would follow from

and one can check that this applies for \(\sigma \geq 38/49\).

Now suppose that we are in the case \(17/22 \leq \sigma \leq 38/49\), so that \(\tau _0 = \frac{9(3\sigma -2)}{2} \leq \frac{72}{49} {\lt} \frac{3}{2}\) (so in particular \(\max (1,2\tau -2) = 1\) for \(\tau \leq \tau _0\)). We set

and

Note that for \(\sigma \leq 38/49\), one has

and hence

As before, we conclude that the quantities \(\alpha _1+\alpha _2/2 + 2-2\sigma \) and \(-2\alpha _1 + \tau + 12 - 16 \sigma \) are both equal to \(\frac{\tau }{3} + \frac{16-20\sigma }{3} + \frac{\alpha _2}{3}\), while \(-\alpha _2 + 2\tau +4-8\sigma \) is less than or equal to this quantity. Thus it suffices to show 23, 24, 25 as before.

For 23 we argue as before to reduce to showing that

which one can check to be true (with room to spare) for \(\sigma \geq 17/22\). For 24, we use 16 as before to reduce to showing that

and

The first inequality is an identity, and the latter again reduces to \(\tau \geq 14\sigma -10\) which one can check follows from 19. For 25 it suffices to show that

which by 16 follows from

but this is an identity.

Apply Theorem 11.23 with the classical pairs \((\frac{1}{14},\frac{11}{14})\) and \((\frac{1}{6}, \frac{2}{3})\) respectively from Proposition 5.10.

Let \(\mathcal{S}(\sigma )\) denote the closure of the region

One may verify that \(\mathcal{S}(\sigma )\) is a convex polygon for all \(3/4 {\lt} \sigma {\lt} 1\), and thus so is \(\mathcal{S}(\sigma ) \cap H\), where \(H\) is the convex hull of exponent pairs. Thus

is a convex optimisation problem for each \(3/4 {\lt} \sigma {\lt} 1\). We take the following choices of \((k, \ell )\) (found with the aid of computer assistance).

Of these exponent pairs:

• \((\frac{11}{85}, \frac{59}{85})\) is the intersection of the lines \(k = 1/5\) and \(15\ell + 20k = 13\);

• \((\frac{391}{4595}, \frac{3461}{4595})\) is an intersection of the line \(15\ell + 20k = 13\) and the boundary of \(H\);

• all other exponent pairs are vertices of \(H\).

The desired result follows from taking a minimum over the implied bounds. Sharper bounds are possible close to \(\sigma = 1\) by choosing other exponent pairs, however it turns out such results are superseded by other zero density estimates.

From Lemma 8.16 we have

for all \(\tau \geq 0\). Meanwhile, applying Lemma 8.11 with the exponent pair \((2/7,4/7)\) we have

We apply Corollary 11.9 with \(\tau _0 := \max ( \frac{30\sigma -11}{8}, \frac{6\sigma +3}{4} )\), and reduce to showing that 6 for \(2\tau _0/3 \leq \tau \leq \tau _0\) and 7 for \(2 \leq \tau {\lt} 4\tau _0/3\). But this follows from the preceding estimates after routine calculations.

Apply the previous lemma with \(\alpha = 5\sigma -4\).

See [ 144 , (11.74) ] .

With the hypothesis on \(\sigma \), one sees from Lemma 11.29 that

for \(0 \leq \tau {\lt} \frac{(4m-2)\sigma + 2-2m}{m}\), and hence for all \(\tau \geq 0\) by by Lemma 7.4(ii). Meanwhile, from Theorem 9.7 one has

for all \(\tau \geq 2\). The claim then follows from Corollary 11.9 with \(\tau _0 := \frac{(3m-2)\sigma +2-m}{m}\) after a routine calculation.

Apply Lemma 11.30 with \(m=2\) and \((k,\ell ) = (\frac{97}{251}, \frac{132}{251})\) for the first claim; the remaining claims follow from taking \(m=3, 4, 5, 6\) and the trivial exponent pair \((0, 1)\).

See [ 144 , (11.95) ] .

For \(3/4 \leq \sigma \leq 10/13\), we see from Lemma 11.33 that the bound

holds for \(0 \leq \tau {\lt} 8\sigma -5\), and hence for all \(\tau \geq 0\) by Lemma 7.7. Meanwhile, from Lemma 9.6 we have

for all \(1/2 \leq \sigma \leq 1\) and \(\tau \geq 2\). The claim then follows from Corollary 11.9 with \(\tau _0 := 7\sigma -4\) after a routine calculation. Similarly, for \(10/13 \leq \sigma \leq 1\), we have

for \(0 \leq \tau {\lt} \frac{11\sigma -5}{3}\), hence for all \(\tau \geq 0\) by Lemma 7.4(ii); the claim then follows from Corollary 11.9 with \(\tau _0 := \frac{8\sigma -2}{3}\) after a routine calculation.

We apply Corollary 11.9 with

for an arbitrarily small \(\varepsilon \). It then suffices to show that 6 holds for \(2\tau _0/3 \leq \tau \leq \tau _0\) and 7 holds for \(2 \leq \tau {\lt} 4\tau _0/3\).

To prove 7, it suffices by Lemma 8.5 to show that \(\sigma {\gt} \tau \beta (1/\tau )\) for all \(2 \leq \tau {\lt} 4 \tau _0/3\). By 36 one has \(2 \leq \tau {\lt} k(1-(k-1)\eta )\). Meanwhile, from Lemma 4.23 one has

for any \(r \geq 3\). So by 34 it suffices to find \(3 \leq r \leq k\) such that

or equivalently

But one can check that these intervals for \(3 \leq r \leq k\) cover the entire range \(2 \leq \tau {\lt} 4\tau _0/3\), as required.

To prove 6, it suffices by Lemma 7.10 and 8 to show that

Using 37, 35 we obtain the claim whenever

for some \(3 \leq r \leq \ell \). These cover the range \([18\eta +\varepsilon , \tau _0]\). For the remaining range \([1, 18\eta +\varepsilon ]\) we use the van der Corput bound

from Corollary 4.8, which suffices since \(\eta \leq \frac{1}{k(k-1)} \leq \frac{1}{12}\).