The claim (i) follows from Lemma 10.2(iv), and the remaining claims then follow from Lemma 11.2.

Write the right-hand side as \(B\), then \(B \geq 0\) (from Lemma 10.11(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 \(N^*(\sigma ,T) \ll T^{B+O(\varepsilon )+o(1)}\) for unbounded \(T\).

By dyadic decomposition, it suffices to show for unbounded \(T\) that the additive energy of imaginary parts of zeroes in \([T,2T]\) is \(\ll T^{B+O(\varepsilon )+o(1)}\). As in the proof of Lemma 11.5, we can assume the imaginary parts are \(1\)-separated (here we take advantage of the triangle inequality in Lemma 10.2(iii)).

Suppose that one has a zero \(\sigma '+i t\) of this form. Then by standard approximations to the zeta function, 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 \(t'\) by a factor of at most \(2\), we either have

for all zeroes, or 3 for all zeroes.

Suppose we are in the former (“Type I”) case, we can dyadically partition and conclude from the pigeonhole principle that

for some interval \(I\) in some \([N,2N]\) with \(T^{\delta _1} \ll N \ll T\), with at most \(O(\log T)\) different choices for \(I\). Performing a Fourier expansion of \(n^{\sigma '}\) in \(\log n\) and using the triangle inequality one can then deduce that

for some \(t' = t + O(T^{o(1)})\); refining the \(t\) by a factor of \(T^{o(1)}\) if necessary, we may assume that the \(t'\) are \(1\)-separated and that the interval \(I\) is independent of \(t'\), and by passing to a subsequence we may assume that \(T = N^{\tau +o(1)}\) for some \(1 \leq \tau \leq 1/\delta _1\), then

for all \(t'\). If we let \(\Sigma '\) denote the set of such \(t'\), then by Definition 10.9 we then have (for \(\delta _2\) small enough) we have

By Lemma 10.2(i) this implies that the set \(\Sigma \) of imaginary parts of zeroes under consideration also obeys the bound

and the claim follows in this case from 1.

The Type II case similarly follows from 2 exactly as in the proof of Lemma 11.5.

Repeat the proof of Corollary 11.7.

See [ 104 , Lemma 4 ] .

We first suppose that \(\sigma \leq 3/4\). Here we apply Corollary 12.4 with \(\tau _0 = 2\). The \(\mathrm{LV}^*_\zeta \) supremum is now trivial, so it suffices to show that

whenever \((\sigma ,\tau ,\rho ,\rho ^*,s) \in \mathcal{E}\) with \(2 \leq \tau \leq 3\). Let \(k\) be the first integer for which \(1 \leq \tau /k \leq 3/2\), thus \(k=2,3\) and also \(\tau /(k+1) \leq 1\). By Lemma 10.12, there exist tuples

for some \(\rho '\), \(s'\), \(\rho ''\) and \(s''\) satisfying

Applying Corollary 10.21 to the former tuple of 4 and using \(\rho ' \le \rho /k\), we have

Write \(\tau ' := \tau /k\). Applying Lemma 7.9 to the first tuple of 4 one has

while applying Lemma 7.9 to the second tuple of 4 (recalling that \(\tau /(k+1) \le 1\)) gives

and thus

and

A tedious calculation shows that for \(1 \leq \tau ' \leq 3/2\), we have

and

Since

we obtain the claim.

Now suppose that \(\sigma {\gt} 3/4\). From Theorem 11.18 and Lemma 12.2(i) we are already done when \(\sigma \geq 25/28\), so we may assume \(\sigma {\lt} 25/28\).

Here we apply Corollary 12.4 with \(\tau _0 = 4\sigma -1\). To control the \(\mathrm{LV}^*_\zeta \) term, we need to establish

whenever \((\sigma ,\tau ,\rho ,\rho ^*,s) \in \mathcal{E}_\zeta \) and \(2 \leq \tau {\lt} 4\sigma -1\). We use Lemma 8.3(ii) followed by Lemma 9.7 to give

so the claim reduces to verifying

This holds with equality when \(\tau = 4\sigma -1\), and the slope in \(\tau \) is higher on the left-hand side for \(\sigma {\gt}1/2\), so the claim 9 follows.

It remains to establish

whenever \((\sigma ,\tau ,\rho ,\rho ^*,s) \in \mathcal{E}\) and \(4\sigma -1 \leq \tau \leq 2(4\sigma -1)\). Let \(k\) be the first integer for which \((4\sigma -1)/2 \leq \tau /k \leq 3(4\sigma -1)/4\), thus \(k=2,3\) and also \(\tau /(k+1) \leq 4\sigma -1\). By Lemma 10.12, we have 4. From Theorem 7.12 we have

and also

and hence

Among other things, this implies that \(\rho /k \leq 1\).

From Theorem 10.20 and \(\rho ' \le \rho /k\), we have

where \(\tau ' := \tau /k\). This expression is complicated, so we divide into cases. First suppose that \(\rho /k+1 \geq 5\rho /4k + \tau '/2\). In this case the first maximum in the above expression is \(\rho /k+1\), and we simplify to

which after solving for \(\rho ^*/k\) gives

Inserting 11, one can verify after a tedious analysis (using the hypothesis \(3/4 \leq \sigma {\lt} 25/28\)) that

as required.

It remains to treat the case where \(\rho /k+1 {\gt} 5\rho /4k + \tau '/2\). Using 11 one can check that this forces

so that 11 now becomes

The bound 12 becomes

which simplifies to

Inserting 17 and 16, one can eventually show (again using the hypothesis \(3/4 \leq \sigma {\lt} 25/28\)) that 15 holds as required.

Throughout assume that \(3/4 \le \sigma \le 5/6\). Choose

We will show that

for all \((\sigma , \tau , \rho , \rho ^*, s) \in \mathcal{E}\) for which \(\tau _0 \le \tau \le 2\tau _0\), and that

for all \((\sigma , \tau , \rho , \rho ^*, s) \in \mathcal{E}_\zeta \) such that \(2 \le \tau \le \tau _0\). The desired result (i) then follows from Corollary 12.4 and computing the piecewise maximum of 18 and 21.

First, consider 21. Suppose that \((\sigma , \tau , \rho , \rho ^*, s)\in \mathcal{E}_\zeta \) with \(3/4 \le \sigma \le 5/6\) and \(2 \le \tau \le \tau _0\). Then, from Theorem 9.7 we have

Furthermore, by Theorem 7.12 and Lemma 7.8 with \(k = 2\), one has \(\rho \le 2\max (2 - 2\sigma , 4 - 6\sigma + \tau /2)\). However since \(\tau \le \tau _0 = 8\sigma - 4\), this simplifies to

Since \(\sigma \ge 3/4\), this also implies that \(\rho \le 1\). For future reference we also note that

By Theorem 10.20, one has

Since \(\rho \le 1\), one has \(\rho + 1 \ge 2\rho \). Thus the middle term in the first maximum may be omitted, and we are left with two cases to consider.

Case 1: If \(\rho + 1 \ge 5\rho /4 + \tau /2\) then

Solving for \(\rho ^*\) gives

Applying 23 to each term,

i.e. 21 holds. The last inequality may be verified by inspecting the growth rates with respect to \(\tau \) of each term (using 24), and checking that the desired inequality holds at \(\tau = 2\).

Case 2: If \(\rho + 1 {\lt} 5\rho /4 + \tau /2\), then

Solving for \(\rho \) gives

If \(\tau \ge 4\sigma - 1\), then apply 23 termwise to get

since the RHS is increasing faster in \(\tau \) and at \(\tau = 4\sigma - 1\) we have equality.

On the other hand if \(\tau {\lt} 4\sigma - 1\) then we apply 22 termwise to get

by inspecting growth rates in \(\tau \) and noting that at \(\tau = 4\sigma - 1\) one has equality.

Thus we have shown that if \((\sigma , \tau , \rho , \rho ^*, s) \in \mathcal{E}_\zeta \) with \(3/4 \le \sigma \le 5/6\) and \(2 \le \tau \le 8\sigma - 4\), then

which is 21.

Now consider 18. Suppose that \(\tau _0 \le \tau \le 2\tau _0\) and \((\sigma , \tau , \rho , \rho ^*, s) \in \mathcal{E}\). Note that the interval \([\tau _0, 2\tau _0]\) is covered by intervals \(I_k := [(4\sigma - 2)k, (4\sigma - 2)(k + 1)]\) with \(k = 2, 3\). Suppose that \(\tau \in I_k\), and write

Then, by Theorem 7.12 and \(\tau ' \ge 4\sigma - 2\) one has

Also, from Theorem 7.12 and \(\tau \le (4\sigma - 2)(k + 1)\) one has

so that for \(k = 2,3\) one has \(\rho /k \le (2 - 2\sigma )(k + 1)/k \le 3 - 3\sigma \). In summary,

Next, by Lemma 10.12,

for \(\tau ' := \tau /k\) and some \(\rho ' \le \rho /k\) and \(s' \le s/k\).

Applying Theorem 10.20 to this tuple, then applying \(\rho ' \le \rho /k\), one has

By 25 one has \(\rho /k \le 1\) (since \(\sigma \ge 3/4\)) so there are only two cases to consider:

Case 1: \(\rho /k + 1 \ge 5\rho /(4k) + \tau '/2\) then

Solving for \(\rho ^*/k\), we get

If \(\tau ' \ge 3\sigma - 1\) then 25 reduces to \(\rho /k \le 3-3\sigma \). Substituting this bound gives

However the RHS is bounded by

for all \(\tau ' \ge 3\sigma - 1\) since the desired bound holds at \(\tau ' = 3\sigma - 1\) (where one has equality).

On the other hand, if \(4\sigma - 2 \le \tau ' \le 3\sigma - 1\) then 25 reduces to \(\rho /k \le 4 - 6\sigma + \tau '\). Substituting this bound gives

where the last inequality may be established by checking that it holds at both \(\tau ' = 4\sigma - 2\) and \(\tau ' = 3\sigma - 1\) (where one has equality). To summarise, by taking \(k = 2, 3\) in the 26 and 29, one has

in this case.

Case 2: \(\rho /k + 1 {\lt} 5\rho /(4k) + \tau '/2\) then

Solving for \(\rho ^*/k\) gives

Proceeding as before, if \(\tau ' \ge 3\sigma - 1\) then 25 becomes \(\rho /k \le 3 - 3\sigma \), and substituting gives

One may verify that the RHS is bounded by

(with some room to spare) by checking at the endpoint \(\tau ' = 3\sigma - 1\).

Similarly, if \(4\sigma - 2 \le \tau ' \le 3\sigma - 1\) then using \(\rho /k \le 4 - 6\sigma + \tau '\) from 25 one has

One can check that the RHS is bounded by

by checking the required inequalities hold at \(\tau ' = 4\sigma - 2\) and \(\tau ' = 3\sigma - 1\) (in each case, with some room to spare).

Combining all the cases, by taking \(k= 2,3\) we have shown that for \(3/4 \le \sigma \le 4/5\) and \(\tau _0 \le \tau \le 2\tau _0\) one has

which is the first part of 18.

The case for \(4/5 \le \sigma \le 5/6\) may be treated similarly. Here one needs to verify that

for \(\tau ' \ge 3\sigma - 1\), and that

for \(4\sigma - 2 \le \tau ' \le 3\sigma - 1\). The treatment is analogous to before, so we omit the proof.

Throughout assume \(7/10 \le \sigma \le 3/4\) and take \(\tau _0 = 2\) in Corollary 12.4. It suffices to show that if \((\sigma , \tau , \rho , \rho ^*, s) \in \mathcal{E}\) with \(2 \le \tau \le 4\), then either

or

Note for future reference the crude bounds

Let

Via Theorem 7.9 and Lemma 7.8, one has

and via Theorem 10.27 and Lemma 7.8, one has

Since \(\sigma \le 4/5\), we may drop the first term, i.e.

Combining 35 and 36,

One can verify that all intervals are proper for \(7/10 \le \sigma \le 3/4\) and \(k = 2,3\).

Since \((\sigma , \tau , \rho , \rho ^*, s) \in \mathcal{E}\), by Lemma 10.12 one has \((\sigma , \tau ', \rho ', \rho ^*/k, s') \in \mathcal{E}\) for some \(\rho ' \le \rho /k\) and \(s'\le s/k\). Applying Theorem 10.20to the first tuple followed by \(\rho ' \le \rho /k\), and noting that \(\rho /k + 1 \ge 2\rho /k\) since \(\rho /k \le 1\) by 37, one obtains

First, suppose that \(\rho /k + 1 {\lt} 5\rho /(4k) + \tau '/2\) so that

Solving for \(\rho ^*/k\) gives

One may verify that the RHS is bounded by

by substituting each case of 37. This involves the tedious verification of the following four inequalities:

for \(1 \le \tau ' \le 13/5 - 2\sigma \);

for \(13/5 - 2\sigma \le \tau ' \le 6/5\);

for \(k = 2, 3\) and \(6/5 \le \tau ' \le 2(\sigma - 1/5) + 2(1-\sigma )/k\);

for \(k = 2, 3\) and \(2(\sigma - 1/5) + 2(1-\sigma )/k \le \tau ' \le 1 + 1/k\). For instance, in the case of 43, the LHS is increasing faster with respect to \(\tau '\) than the RHS in view of 32, and the inequality holds at the upper limit \(\tau ' = 13/5 - 2\sigma \) (with some room to spare). The other inequalities may be verified similarly, with the exception of

which is equivalent to \(6 - 9\sigma \le 3(53 - 75\sigma )/(4(5\sigma + 3))\tau '\). For \(\sigma {\lt} 53/75\) the LHS is negative while the RHS is positive so the inequality holds. For \(\sigma \ge 53/75\) one may verify that the inequality holds at the lower limit \(\tau ' = 6/5\).

In the remainder of the proof we assume \(\rho /k + 1 \ge 5\rho /(4k) + \tau '/2\) so that 42 becomes

Solving for \(\rho ^*/k\) gives

The case where \(\tau ' \ge 6/5\) is simpler so we handle it first. Applying the last two cases of 37 it suffices to verify that

for \(k = 2, 3\) and \(6/5 \le \tau ' \le 2(\sigma - 1/5) + 2(1 - \sigma )/k\);

for \(k = 2, 3\) and \(2(\sigma - 1/5) + 2(1 - \sigma )/k \le \tau ' \le 1 + 1/k\). Note also that one has equality when \(\tau ' = 2(\sigma - 1/5) + 2(1 - \sigma )/k\) and \(k = 2\).

Lastly, consider the case where \(1 \le \tau ' \le 6/5\). Applying \(\rho /k \le \min (1 - 2\sigma + \tau ', 18/5 - 4\sigma )\) from the first two cases of 37, one obtains

However one may verify that the RHS of both of the above inequalities are bounded by \(5(18 - 19\sigma )/(2(5\sigma + 3))\tau '\) for \(1 \le \tau ' \le 6/5\) by checking at \(\tau ' = 13/5 - 2\sigma \). Thus

Meanwhile, by Lemma 10.12,

for some \(\rho ''\le \rho /(k-1)\) and \(s''\le s/(k-1)\). Applying Theorem 10.20 to this tuple (and applying \(\rho ''\le \rho /(k-1)\)) gives

By expanding the first maximum and simplifying, one of the following inequalities must hold:

If 51 holds, then solving for \(\rho ^*/k\) gives

For \(\tau ' \le 13/5 - 2\sigma \), we apply \(\rho /k \le 1 - 2\sigma + \tau '\) from 37 (along with the inequalities \(4 - 4\sigma \ge 0\), \(3 - 4\sigma \ge 0\), \(6 - 8\sigma \ge 0\) and \((k - 1)/k \le 2/3\)),

for all \(1 \le \tau ' \le 13/5 - 2\sigma \). The last inequality is verified using 32 and checking at both \(\tau ' = 1\) and \(\tau ' = 13/5-2\sigma \). Similarly, for \(13/5 - 2\sigma \le \tau ' \le 6/5\) we use \(\rho /k \le 18/5-4\sigma \) and verify that

where the last inequality is verified using 32 and checking at the lower limit \(\tau ' = 13/5 - 2\sigma \).

Suppose now that 52 holds. Solving for \(\rho ^*/k\) gives

Similarly to before, applying the first two cases of 37 allows one to verify that for \(1 \le \tau ' \le 6/5\),

and

each with room to spare. Therefore, for \(1 \le \tau ' \le 6/5\), one has

However, if \(\tau ' \le \sigma + 1/2 - (2\sigma - 1)/(2k)\) we may apply \(\rho /k \le 1 - 2\sigma + \tau '\) to get

where by 32, the last inequality is verified by checking that it holds at the upper limit \(\tau ' = \sigma + 1/2 - (2\sigma - 1)/(2k)\) for \(k = 2,3\). For \(\tau ' {\gt} \sigma + 1/2 - (2\sigma - 1)/(2k)\), we once again apply \(\rho /k \le 1 - 2\sigma + \tau '\) to get

where now the last inequality is verified at the lower limit \(\tau ' = \sigma + 1/2 - (2\sigma - 1)/(2k)\). Therefore, in view of 47 and 54, one has

in this case, as required.

Lastly, suppose that 53 holds. Then solving for \(\rho ^*/k\) gives

Proceeding as before, we use \(\rho /k \le 1 - 2\sigma + \tau '\) from 37 together with \((k - 1)/k \le 2/3\) to get

where the last inequality is verified at \(\tau ' = 6/5\). Furthermore, using \(\rho /k \le \min (1 -2\sigma +\tau ', 18/5-4\sigma )\) and \((k - 1)/k \ge 1/2\) one has

for \(1 \le \tau ' \le 6/5\). Therefore,

If \(1 \le \tau ' \le 1 + 2\sigma /15\), we use the bound \(\rho /k \le 1 - 2\sigma + \tau '\) to get

where by 32 it suffices to check the inequality at the upper limit \(\tau ' = 1 + 2\sigma /15\) (where we have equality if \(k = 2\)). On the other hand if \(1 + 2\sigma /15 \le \tau ' \le 6/5\), we use \(\rho /k \le 1 - 2\sigma + \tau '\) to get

where by 32 it suffices to check the last inequality at \(\tau ' = 1 + 2\sigma /15\) (where we have equality). Therefore, one has

in this case too.

Throughout assume that \(3/4 \le \sigma \le 4/5\) and take \(\tau _0 := 8\sigma - 4\) in Corollary 12.4. It suffices to show that

for all \((\sigma , \tau , \rho , \rho ^*, s) \in \mathcal{E}\) satisfying \(\tau _0 \le \tau \le 2\tau _0\), and

for all \((\sigma , \tau , \rho , \rho ^*, s) \in \mathcal{E}_\zeta \) such that \(2 \le \tau \le \tau _0\). In the proof of Theorem 12.7 we have already shown that 56 holds in the large range \(65/86 \le \sigma \le 5/6\), so it remains to prove 55. Given \(\sigma , \tau \), let \(k \ge 2\) be the integer for which

so that \(k = 2, 3\) for \(\tau _0 \le \tau {\lt} 2\tau _0\), and as before write \(\tau ' := \tau /k\).

By Theorem 10.27 and Lemma 7.8 one has

and from Theorem 7.12 and Lemma 7.8, for any integer \(\ell \),

so that in particular, taking \(\ell = k + 1\) and noting that \(\tau /(k + 1) \le 4\sigma - 2\) by 57,

Combining everything, one obtains (for \(k \ge 2\))

First, suppose that \(6/5 \le \tau ' \le (4\sigma - 2)(k + 1)/k\). By Lemma 10.12, \((\sigma , \tau ', \rho ', \rho ^*/k, s') \in \mathcal{E}\) for some \(\rho ' \le \rho /k\) and \(s' \le s/k\). Applying Theorem 10.20, and noting that \(\rho /k + 1 \ge 2\rho /k\) since \(\rho /k \le 1\) by 62,

If \(\rho /k + 1 \ge 5\rho /(4k) + \tau '/2\), then

Solving for \(\rho ^*/k\) gives

Applying \(\rho /k \le \min (12/5 - 4\sigma + \tau ', 3 - 3\sigma )\) to the RHS, one may ultimately verify that

If \(\rho /k + 1 \le 5\rho /(4k) + \tau '/2\) one has

and solving for \(\rho ^*/k\) gives

Once again applying \(\rho /k \le \min (12/5 - 4\sigma + \tau ', 3 - 3\sigma )\), one again ultimately obtains

Now suppose that \(4\sigma - 2 \le \tau ' \le 6/5\). By Lemma 10.12, for any integer \(k \ge 2\) one has \((\sigma , \tau /(k - 1), \rho ', \rho ^*/(k - 1), s') \in \mathcal{E}\) for some \(\rho ' \le \rho /(k - 1)\) and \(s' \le s/(k - 1)\). Applying Theorem 10.20 to this tuple, followed by \(\rho ' \le \rho /(k - 1)\) and rearranging, one obtains

Consider the first maximum of 67. If \(\rho + k - 1\) is maximal, then

Solving for \(\rho ^*\) and dividing by \(k\) gives

Since \(3/4 \le \sigma \le 1\) and \(1/2 \le (k - 1)/k \le 2/3\), we have

Bounding the RHS with \(\rho /k \le \min (4 - 6\sigma + \tau ', 18/5 - 4\sigma )\), one ultimately obtains

Now suppose \(5\rho /4 + \tau '/2\) is maximal in 67. Then

Solving for \(\rho ^*\) and dividing by \(k\) gives

As before, since \(3/4 \le \sigma \le 1\) and \(1/2 \le (k - 1)/k \le 2/3\), one has

Once again we apply \(\rho /k \le \min (4 - 6\sigma + \tau ', 18/5 - 4\sigma )\) to ultimately obtain

Lastly, if \(2\rho \) is maximal in 67, then

Solving for \(\rho ^*\) and dividing by \(k\) gives

Once again we apply \(\rho /k \le \min (4 - 6\sigma + \tau ', 18/5 - 4\sigma )\) to ultimately obtain

in this case too.

Fix \(664/877 \le \sigma \le 31/40\) and take \(\tau _0 = 2\). It suffices to show that

for all \((\sigma , \tau , \rho , \rho ^*, s) \in \mathcal{E}\) satisfying \(2 \le \tau \le 4\).

Let \(k = 2\) if \(2 \le \tau {\lt} 3\) and \(k = 3\) otherwise, and as usual let \(\tau ' = \tau /k\). By Lemma 10.12, \((\sigma , \tau ', \rho ', \rho ^*/k, s') \in \mathcal{E}\) for some \(\rho ' \le \rho /k\) and \(s' \le s/k\). Applying Theorem 10.20, and noting that \(\rho /k + 1 \ge 2\rho /k\) since \(\rho /k \le 1\) by 62,

Rearranging the inequality and solving for \(\rho ^*/k\), one must either have

or

Here we will divide our argument into several cases. Suppose first that

By Theorem 10.27, one has

and by Theorem 7.9, one has

Combining the two inequalities, we have

Substituting 72 into 70 and 71, one may verify that

for all \((9\sigma - 1)/5 \le \tau ' \le 1 + 1/k\) (with equality occurring at \(\tau ' = 2\sigma - 2/5 + 2(1 - \sigma )/k\) and \(k = 2\)).

Now consider the case where

Taking \(k = 10\) in Theorem 7.16, one has

Suppose first that \(\sigma \ge 281/371\). Then, this reduces to

By Lemma 10.12, for any integer \(k \ge 2\) one has \((\sigma , \tau /(k - 1), \rho ', \rho ^*/(k - 1), s') \in \mathcal{E}\) for some \(\rho ' \le \rho /(k - 1)\) and \(s' \le s/(k - 1)\). Applying Theorem 10.20 to this tuple, followed by \(\rho ' \le \rho /(k - 1)\) and rearranging, one obtains

By considering each case of the two maximums individually, and solving for \(\rho ^*/k\), one obtains

Bounding each term on the RHS using 77, one may verify that in each case

for \(1 \le \tau ' \le (9\sigma - 1)/5\) and \(k = 2,3\).

Suppose now that \(\sigma {\lt} 281/371\) so that 76 reduces to

If \(1 \le \tau ' {\lt} 77\sigma /2 - 28\) then substituting \(\rho /k \le \max (2 - 2\sigma , 60 - 80\sigma + \tau ')\) into 82, one may ultimately verify in each case that

for \(k = 2,3\), with equality when \(\tau ' = 77\sigma /2 - 28\) and \(k = 2\) (here we make use of the assumption \(\sigma \ge 664/877 = 0.7571\ldots \)).

Lastly, if \(77\sigma /2 - 28 \le \tau ' \le (9\sigma - 1)/5\) then 88 simplifies to \(\rho /k \le 60-80\sigma +\tau '\). Substituting this into 70 and 71, one obtains in either case that

with equality when \(\tau ' = 77\sigma /2 - 28\) and \(k = 2\).

Fix \(42/55 \le \sigma \le 79/103\) and take \(\tau _0 = 2\). It suffices to show that

for all \((\sigma , \tau , \rho , \rho ^*, s) \in \mathcal{E}\) for which \(2 \le \tau \le 4\). As before let

so that in particular \(1 \le \tau ' \le (k + 1)/k\).

By Lemma 10.12, for \(k = 2, 3\),

for some \(\rho ' \le \rho /k\), \(s' \le s/k\), \(\rho '' \le \rho /(k + 1)\) and \(s'' \le s/(k + 1)\). Applying Theorem 7.16 with \(k = 6\) to each tuple, one has

First suppose \(\sigma \ge 97/127\), in which case the above simplifies to

This combines to give

First suppose that \(\tau ' \ge (11\sigma - 5)/3\). Applying Theorem 10.20 to the first tuple of 91, and noting that \(\rho /k + 1 \ge 2\rho /k\) since \(\rho /k \le 1\),

Rearranging the inequality and solving for \(\rho ^*/k\), one must either have

or

In either case, upon substituting the last two cases of 95 one obtains

Now suppose that \(\tau ' {\lt} (11\sigma - 5)/3\). Then, note that for \(k = 2, 3\) one has

for some \(\rho ''' \le \rho /(k-1)\) and \(s''' \le s/(k-1)\). Applying Theorem 10.20 to this tuple, followed by \(\rho ''' \le \rho /(k - 1)\) and rearranging, one obtains

By considering each case of the two maximums individually, and solving for \(\rho ^*/k\), one obtains

Note that for all \(1 \le \tau ' {\lt} (11\sigma - 5)/3\), one has by 95 that \(\rho /k \le 2 - 2\sigma \). Substituting this into 105, one obtains

in this case too. Combined with 101, we have shown that

for all \((\sigma , \tau , \rho , \rho ^*, s) \in \mathcal{E}\) for \(97/127 \le \sigma \le 79/103\) and \(2 \le \tau \le 4\), as required.

The proof in the range \(42/55 \le \sigma \le 97/127\) is similar. Here 92 reduces to

so that

If \(\tau ' \ge 46\sigma - 34\), we substitute the last two cases of this bound into 99 and 100, one may verify that

with equality when \(\tau ' = 46\sigma - 34 + 2(1 - \sigma )/k\) and \(k = 2\).

On the other hand if \(1 \le \tau ' {\lt} 46\sigma - 34\) then substituting \(\rho /k \le 2 - 2\sigma \) into 105 gives

in each case. Combined with 115, the desired result follows for \(42/55 \le \sigma \le 97/127\).

Fix \(79/103 \le \sigma \le 84/109\) and take

Let

Suppose that \((\sigma , \tau , \rho , \rho ^*, s) \in \mathcal{E}\). We will first show

By Lemma 10.12, one has that

for some \(\rho ' \le \rho /k\), \(s' \le s/k\), \(\rho '' \le \rho /(k + 1)\) and \(s'' \le s/(k + 1)\). First suppose that \(\sigma \ge 33/43\). In this range, Theorem 7.16 with \(k = 5\) gives

For \(\tau \ge \tau _0\), the first inequality reduces to \(\rho /k \le 18/5 - 28\sigma /5 + \tau '\), while for \(\tau \le 2\tau _0\) the second inequality reduces to

Combining these two inequalities gives

In particular this implies \(\rho /k \le 1\). Applying Theorem 10.20 to the first tuple of 121,

By considering each case of the first maximum and solving for \(\rho ^*/k\), one must either have

or

However one may verify in both cases that

for \(33/43\le \sigma \le 84/109\) and \(\tau _0 \le \tau \le 2\tau _0\) (with equality at \(\tau = 2(13\sigma - 3)/5\)).

Now consider \(\sigma \le 33/43\). In this range of \(\sigma \), Theorem 7.16 with \(k = 5\) gives

which gives

Substituting this bound into 122 and 123, one obtains

for \(79/103 \le \sigma \le 33/43\) and \(\tau _0 \le \tau \le 2\tau _0\) (with equality at \(\tau = 74\sigma - 54\)). Combined with 124, one obtains 120.

It remains to show that

for all \((\sigma , \tau , \rho , \rho ^*, s) \in \mathcal{E}_\zeta \) for which \(79/103 \le \sigma \le 84/109\) and \(2 \le \tau \le \tau _0\). This follows from substituting \(\rho /k \le 6 - 12\sigma + 2\tau '\) (Theorem 9.7) into 122 and 123.