12 Zero density energy theorems

Definition 12.1 Zero density exponents

For $1 / 2 \leq σ \leq 1$ and $T > 0$ , let $N^{*} (σ, T)$ denote the additive energy $E_{1} (Σ)$ of the imaginary parts of the zeroes $ρ$ of the Riemann zeta function with $Re (ρ) \geq σ$ and $| Im (ρ) | \leq T$ . For fixed $1 / 2 \leq σ \leq 1$ , the zero density exponent $A^{*} (σ) \in [- \infty, \infty)$ is the infimum of all exponents $A^{*}$ for which one has

N^{*} (σ - δ, T) ≪ T^{A^{*} (1 - σ) + o (1)}

for all unbounded $T$ and infinitesimal $δ > 0$ .

The exponent $A^{*} (σ)$ is also essentially referred to as $B (σ)$ in [ 77 ] (though without the technical shift by $δ$ in that reference).

Implemented at zero_density_energy_estimate.py as:
Zero_Density_Energy_Estimate

Lemma 12.2 Basic properties of

A^{*}

We have the trivial bounds

$2 A (σ), 4 A (σ) - \frac{1}{1 - σ} \leq A^{*} (σ) \leq 3 A (σ)$

for any $1 / 2 \leq σ \leq 1$ .
$σ \mapsto (1 - σ) A^{*} (σ)$ is non-increasing, with $A^{*} (1 / 2) = 6$ and $A^{*} (1) = - \infty$ .
If the Riemann hypothesis holds, then $A^{*} (σ) = - \infty$ for all $1 / 2 < σ \leq 1$ .

Implemented at zero_density_energy_estimate.py as:
add_trivial_zero_density_energy_estimates(hypotheses)

Proof ▶

Upper bounds on $A^{*} (σ)$ can be obtained from large value energy theorems via the following relation.

Lemma 12.3 Zero density energy from large values energy

Let $1 / 2 < σ < 1$ . Then

A^{*} (σ) (1 - σ) \leq max (sup_{τ \geq 1} {LV}_{ζ}^{*} (σ, τ) / τ, \underset{τ \to \infty}{lim sup} {LV}^{*} (σ, τ) / τ) .

Proof ▶

Write the right-hand side as $B$ , then $B \geq 0$ (from Lemma 10.11(iii)) and we have

{LV}_{ζ}^{*} (σ, τ) \leq B τ

for all $τ \geq 1$ , and

{LV}^{*} (σ, τ) \leq (B + ε) τ

whenever $ε > 0$ and $τ$ is sufficiently large depending on $ε$ (and $σ$ ). It would suffice to show, for any $ε > 0$ , that $N^{*} (σ, T) ≪ T^{B + O (ε) + o (1)}$ for unbounded $T$ .

By dyadic decomposition, it suffices to show for unbounded $T$ that the additive energy of imaginary parts of zeroes in $[T, 2 T]$ is $≪ T^{B + O (ε) + o (1)}$ . 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 $σ^{'} + i t$ of this form. Then by standard approximations to the zeta function, one has

\sum_{n \leq T} \frac{1}{n^{σ^{'} + i t}} ≪ T^{- 1} .

Let $0 < δ_{1} < ε$ be a small quantity (independent of $T$ ) to be chosen later, and let $0 < δ_{2} < δ_{1}$ be sufficiently small depending on $δ_{1}, δ_{2}$ . By the triangle inequality, and refining the sequence $t^{'}$ by a factor of at most $2$ , we either have

| \sum_{T^{δ_{1}} \leq n \leq T} \frac{1}{n^{σ^{'} + i t}} | ≫ T^{- δ_{2}}

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

| \sum_{n \in I} \frac{1}{n^{σ^{'} + i t}} | ≫ T^{- δ_{2} - o (1)}

for some interval $I$ in some $[N, 2 N]$ with $T^{δ_{1}} ≪ N ≪ T$ , with at most $O (\log T)$ different choices for $I$ . Performing a Fourier expansion of $n^{σ^{'}}$ in $\log n$ and using the triangle inequality one can then deduce that

| \sum_{n \in I} \frac{1}{n^{i t^{'}}} | ≫ N^{σ^{'}} T^{- δ_{2} - o (1)}

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 $T = N^{τ + o (1)}$ for some $1 \leq τ \leq 1 / δ_{1}$ , then

| \sum_{n \in I} \frac{1}{n^{i t^{'}}} | ≫ N^{σ - δ_{2} / δ_{1} + o (1)}

for all $t^{'}$ . If we let $Σ^{'}$ denote the set of such $t^{'}$ , then by Definition 10.9 we then have (for $δ_{2}$ small enough) we have

E_{1} (Σ^{'}) ≪ N^{{LV}_{ζ}^{*} (σ, τ) + ε + o (1)} ≪ T^{{LV}_{ζ}^{*} (σ, τ) / τ + ε + o (1)} .

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

E_{1} (Σ) ≪ T^{{LV}_{ζ}^{*} (σ, τ) / τ + ε + o (1)} .

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.

Corollary 12.4

Let $1 / 2 < σ < 1$ and $τ_{0} > 0$ be fixed. Then

A^{*} (σ) (1 - σ) \leq max (sup_{2 \leq τ < τ_{0}} {LV}_{ζ}^{*} (σ, τ) / τ, sup_{τ_{0} \leq τ \leq 2 τ_{0}} {LV}^{*} (σ, τ) / τ)

Implemented at zero_density_energy_estimate.py as:
lver_to_energy_bound(LVER, LVER_zeta, sigma_interval)

Proof ▶

12.1 Known additive energy bounds

Proposition 12.5 Additive energy under the Lindelof hypothesis

Let $1 / 2 \leq σ \leq 1$ be fixed. Then one has

A^{*} (σ) \leq 8 - 4 σ

and $A^{*} (σ) \leq 0$ if $σ > 3 / 4$ .

Proof ▶

Theorem 12.6 Heath-Brown’s additive energy bound

[ 80 , Theorem 2 ] Let $1 / 2 \leq σ \leq 1$ be fixed. Then one can bound $A^{*} (σ)$ by

\begin{aligned} \frac{10 - 11 σ}{(2 - σ) (1 - σ)} & for 1 / 2 \leq σ \leq 2 / 3; \\ \frac{18 - 19 σ}{(4 - 2 σ) (1 - σ)} & for 2 / 3 \leq σ \leq 3 / 4; \\ \frac{12}{4 σ - 1} & for 3 / 4 \leq σ \leq 1. \end{aligned}

Recorded in literature.py as:
add_zero_density_energy_heath_brown_1979()
Derived in derived.py as:
prove_heath_brown_energy_estimate()

Proof ▶

We first suppose that $σ \leq 3 / 4$ . Here we apply Corollary 12.4 with $τ_{0} = 2$ . The ${LV}_{ζ}^{*}$ supremum is now trivial, so it suffices to show that

ρ^{*} \leq max (\frac{10 - 11 σ}{2 - σ}, \frac{18 - 19 σ}{4 - 2 σ}) τ

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

(σ, \frac{τ}{k}, ρ^{'}, \frac{ρ^{*}}{k}, s^{'}), (σ, \frac{τ}{k + 1}, ρ^{″}, \frac{ρ^{*}}{k + 1}, s^{″}) \in E .

for some $ρ^{'}$ , $s^{'}$ , $ρ^{″}$ and $s^{″}$ satisfying

ρ^{'} \leq \frac{ρ}{k}, s^{'} \leq \frac{s}{k}, ρ^{″} \leq \frac{ρ}{k + 1}, s^{″} \leq \frac{s}{k + 1} .

Applying Corollary 10.21 to the former tuple of 4 and using $ρ^{'} \leq ρ / k$ , we have

\frac{ρ^{*}}{k} \leq max (\frac{3 ρ}{k} + 1 - 2 σ, \frac{ρ}{k} + 4 - 4 σ, \frac{5 ρ}{2 k} + \frac{3 - 4 σ}{2}) .

Write $τ^{'} := τ / k$ . Applying Lemma 7.9 to the first tuple of 4 one has

ρ / k \leq τ^{'} + 1 - 2 σ

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

ρ / k = \frac{k + 1}{k} \frac{ρ}{k + 1} \leq \frac{k + 1}{k} (2 - 2 σ) \leq 3 - 3 σ

and thus

ρ / k \leq min (τ^{'} + 1 - 2 σ, 3 - 3 σ)

and

\begin{aligned} ρ^{*} / k \leq max ( & 3 min (τ^{'} + 1 - 2 σ, 3 - 3 σ) + 1 - 2 σ, \\ min (τ^{'} + 1 - 2 σ, 3 - 3 σ) + 4 - 4 σ, \\ 5 min (τ^{'} + 1 - 2 σ, 3 - 3 σ) / 2 + (3 - 4 σ) / 2) . \end{aligned}

A tedious calculation shows that for $1 \leq τ^{'} \leq 3 / 2$ , we have

3 min (τ^{'} + 1 - 2 σ, 3 - 3 σ) + 1 - 2 σ \leq \frac{10 - 11 σ}{2 - σ} τ^{'},

min (τ^{'} + 1 - 2 σ, 3 - 3 σ) + 4 - 4 σ \leq max (\frac{7 - 7 σ}{2 - σ}, 6 - 6 σ) τ^{'}

and

5 min (τ^{'} + 1 - 2 σ, 3 - 3 σ) / 2 + (3 - 4 σ) / 2 \leq \frac{18 - 19 σ}{4 - 2 σ} τ^{'} .

Since

max (\frac{7 - 7 σ}{2 - σ}, 6 - 6 σ) \leq max (\frac{10 - 11 σ}{2 - σ}, \frac{18 - 19 σ}{4 - 2 σ})

we obtain the claim.

Now suppose that $σ > 3 / 4$ . From Theorem 11.18 and Lemma 12.2(i) we are already done when $σ \geq 25 / 28$ , so we may assume $σ < 25 / 28$ .

Here we apply Corollary 12.4 with $τ_{0} = 4 σ - 1$ . To control the ${LV}_{ζ}^{*}$ term, we need to establish

ρ^{*} \leq \frac{12 (1 - σ)}{4 σ - 1} τ

whenever $(σ, τ, ρ, ρ^{*}, s) \in E_{ζ}$ and $2 \leq τ < 4 σ - 1$ . We use Lemma 8.3(ii) followed by Lemma 9.7 to give

ρ^{*} \leq 3 ρ \leq 3 (2 τ - 12 (σ - 1 / 2))

so the claim reduces to verifying

3 (2 τ - 12 (σ - 1 / 2)) \leq \frac{12 (1 - σ)}{4 σ - 1} τ .

This holds with equality when $τ = 4 σ - 1$ , and the slope in $τ$ is higher on the left-hand side for $σ > 1 / 2$ , so the claim 9 follows.

It remains to establish

ρ^{*} \leq \frac{12 (1 - σ)}{4 σ - 1} τ

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

ρ / k \leq max (2 - 2 σ, τ^{'} + 4 - 6 σ)

and also

ρ / k = \frac{k + 1}{k} \frac{ρ}{k + 1} \leq \frac{k + 1}{k} (2 - 2 σ) \leq 3 - 3 σ

and hence

ρ / k \leq min (max (2 - 2 σ, τ^{'} + 4 - 6 σ), τ^{'} + 4 - 6 σ, 3 - 3 σ) .

Among other things, this implies that $ρ / k \leq 1$ .

From Theorem 10.20 and $ρ^{'} \leq ρ / k$ , we have

\begin{aligned} ρ^{*} / k \leq 1 - 2 σ & + \frac{1}{2} max (\frac{ρ}{k} + 1, \frac{2 ρ}{k}, \frac{5 ρ}{4 k} + \frac{τ^{'}}{2}) \\ + \frac{1}{2} max (\frac{ρ^{*}}{k} + 1, \frac{4 ρ}{k}, \frac{3 ρ^{*}}{4 k} + \frac{ρ}{k} + \frac{τ^{'}}{2}) \end{aligned}

where $τ^{'} := τ / k$ . This expression is complicated, so we divide into cases. First suppose that $ρ / k + 1 \geq 5 ρ / 4 k + τ^{'} / 2$ . In this case the first maximum in the above expression is $ρ / k + 1$ , and we simplify to

ρ^{*} / k \leq 3 / 2 - 2 σ + ρ / 2 k + max (ρ^{*} / k + 1, 4 ρ / k, 3 ρ^{*} / 4 k + ρ / k + τ^{'} / 2) / 2,

which after solving for $ρ^{*} / k$ gives

ρ^{*} / k \leq max (ρ / 2 k + 4 - 4 σ, 5 ρ / 2 k + (3 - 4 σ) / 2, 8 ρ / 5 k + 2 τ^{'} / 5 + (12 - 16 σ) / 5) .

Inserting 11, one can verify after a tedious analysis (using the hypothesis $3 / 4 \leq σ < 25 / 28$ ) that

ρ^{*} / k \leq \frac{12 (1 - σ)}{4 σ - 1} τ^{'}

as required.

It remains to treat the case where $ρ / k + 1 > 5 ρ / 4 k + τ^{'} / 2$ . Using 11 one can check that this forces

4 σ - 2 \leq τ^{'} \leq \frac{3}{4} (4 σ - 1),

so that 11 now becomes

ρ / k \leq 3 - 3 σ .

The bound 12 becomes

ρ^{*} / k \leq 1 - 2 σ + (5 ρ / 4 k + τ^{'} / 2) / 2 + max (ρ^{*} / k + 1, 4 ρ / k, 3 ρ^{*} / 4 k + ρ / k + τ^{'} / 2) / 2

which simplifies to

ρ^{*} / k \leq max (5 ρ / 4 k + τ^{'} / 2 + 3 - 4 σ, 21 ρ / 8 k + τ^{'} / 4 + 1 - 2 σ, 9 ρ / 5 k + 4 τ^{'} / 5 + (8 - 16 σ) / 5) .

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

We found the following estimates with the use of computer-aided proof discovery, which improve on Theorem 12.6 in various ranges of $σ$ . First, by using Theorem 10.20 in place of Corollary 10.21 in the proof of the previous theorem, it is possible to obtain an improved additive energy estimate for $σ \geq 3 / 4$ . A human-readable proof is contained in the following theorem.

Theorem 12.7

For $3 / 4 \leq σ \leq 5 / 6$ one has

A^{*} (σ) \leq max (\frac{18 - 19 σ}{2 (3 σ - 1) (1 - σ)}, \frac{4 (10 - 9 σ)}{5 (4 σ - 1) (1 - σ)}) .

Derived in derived.py as:
prove_improved_heath_brown_energy_estimate()

Proof ▶

Throughout assume that $3 / 4 \leq σ \leq 5 / 6$ . Choose

τ_{0} = 8 σ - 4.

We will show that

ρ^{*} / τ \leq {\begin{cases} \frac{18 - 19 σ}{2 (3 σ - 1)}, & 3 / 4 \leq σ < 4 / 5, \\ \frac{7 (1 - σ)}{3 σ - 1}, & 4 / 5 \leq σ \leq 5 / 6, \end{cases}

for all $(σ, τ, ρ, ρ^{*}, s) \in E$ for which $τ_{0} \leq τ \leq 2 τ_{0}$ , and that

ρ^{*} / τ \leq max (\frac{45 - 46 σ}{4 (4 σ - 1)}, \frac{4 (10 - 9 σ)}{5 (4 σ - 1)}),

for all $(σ, τ, ρ, ρ^{*}, s) \in E_{ζ}$ such that $2 \leq τ \leq τ_{0}$ . The desired result (i) then follows from Corollary 12.4 and computing the piecewise maximum of 18 and 21.

First, consider 21. Suppose that $(σ, τ, ρ, ρ^{*}, s) \in E_{ζ}$ with $3 / 4 \leq σ \leq 5 / 6$ and $2 \leq τ \leq τ_{0}$ . Then, from Theorem 9.7 we have

ρ \leq 2 τ - 12 (σ - 1 / 2) .

Furthermore, by Theorem 7.12 and Lemma 7.8 with $k = 2$ , one has $ρ \leq 2 max (2 - 2 σ, 4 - 6 σ + τ / 2)$ . However since $τ \leq τ_{0} = 8 σ - 4$ , this simplifies to

ρ \leq 4 - 4 σ .

Since $σ \geq 3 / 4$ , this also implies that $ρ \leq 1$ . For future reference we also note that

\frac{4}{5} < max (\frac{45 - 46 σ}{4 (4 σ - 1)}, \frac{4 (10 - 9 σ)}{5 (4 σ - 1)}) < 2, (3 / 4 \leq σ \leq 5 / 6) .

By Theorem 10.20, one has

ρ^{*} \leq 1 - 2 σ + \frac{1}{2} max (ρ + 1, 2 ρ, \frac{5}{4} ρ + \frac{τ}{2}) + \frac{1}{2} max (ρ^{*} + 1, 4 ρ, \frac{3}{4} ρ^{*} + ρ + \frac{τ}{2}) .

Since $ρ \leq 1$ , one has $ρ + 1 \geq 2 ρ$ . Thus the middle term in the first maximum may be omitted, and we are left with two cases to consider.

Case 1: If $ρ + 1 \geq 5 ρ / 4 + τ / 2$ then

ρ^{*} \leq 1 - 2 σ + \frac{ρ + 1}{2} + \frac{1}{2} max (ρ^{*} + 1, 4 ρ, \frac{3}{4} ρ^{*} + ρ + \frac{τ}{2}) .

Solving for $ρ^{*}$ gives

ρ^{*} \leq max (4 - 4 σ + ρ, \frac{3}{2} - 2 σ + \frac{5}{2} ρ, \frac{2}{5} (6 - 8 σ + τ + 4 ρ)) .

Applying 23 to each term,

\begin{aligned} ρ^{*} & \leq max (8 - 8 σ, \frac{23}{2} - 12 σ, \frac{2}{5} (22 - 24 σ + τ)) \\ \leq max (\frac{45 - 46 σ}{4 (4 σ - 1)} τ, \frac{4 (10 - 9 σ)}{5 (4 σ - 1)} τ), \end{aligned}

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

Case 2: If $ρ + 1 < 5 ρ / 4 + τ / 2$ , then

ρ^{*} \leq 1 - 2 σ + \frac{5}{8} ρ + \frac{τ}{4} + \frac{1}{2} max (ρ^{*} + 1, 4 ρ, \frac{3}{4} ρ^{*} + ρ + \frac{τ}{2}) .

Solving for $ρ$ gives

ρ^{*} \leq max (3 - 4 σ + \frac{τ}{2} + \frac{5}{4} ρ, 1 - 2 σ + \frac{τ}{4} + \frac{21}{8} ρ, \frac{8 - 16 σ + 4 τ + 9 ρ}{5}) .

If $τ \geq 4 σ - 1$ , then apply 23 termwise to get

\begin{aligned} ρ^{*} & \leq max (8 - 9 σ + \frac{τ}{2}, \frac{23}{2} - \frac{25}{2} + \frac{τ}{4}, \frac{4}{5} (11 - 13 σ + τ)) \\ \leq max (\frac{45 - 46 σ}{4 (4 σ - 1)} τ, \frac{4 (10 - 9 σ)}{5 (4 σ - 1)} τ), \end{aligned}

since the RHS is increasing faster in $τ$ and at $τ = 4 σ - 1$ we have equality.

On the other hand if $τ < 4 σ - 1$ then we apply 22 termwise to get

\begin{aligned} ρ^{*} & \leq max (\frac{21}{2} - 19 σ + 3 τ, \frac{67 - 134 σ + 22 τ}{4}, \frac{2}{5} (31 - 62 σ + 11 τ)) \\ \leq max (\frac{45 - 46 σ}{4 (4 σ - 1)} τ, \frac{4 (10 - 9 σ)}{5 (4 σ - 1)} τ), \end{aligned}

by inspecting growth rates in $τ$ and noting that at $τ = 4 σ - 1$ one has equality.

Thus we have shown that if $(σ, τ, ρ, ρ^{*}, s) \in E_{ζ}$ with $3 / 4 \leq σ \leq 5 / 6$ and $2 \leq τ \leq 8 σ - 4$ , then

ρ^{*} / τ \leq min (\frac{45 - 46 σ}{4 (4 σ - 1)}, \frac{4 (10 - 9 σ)}{5 (4 σ - 1)}),

which is 21.

Now consider 18. Suppose that $τ_{0} \leq τ \leq 2 τ_{0}$ and $(σ, τ, ρ, ρ^{*}, s) \in E$ . Note that the interval $[τ_{0}, 2 τ_{0}]$ is covered by intervals $I_{k} := [(4 σ - 2) k, (4 σ - 2) (k + 1)]$ with $k = 2, 3$ . Suppose that $τ \in I_{k}$ , and write

τ^{'} := τ / k .

Then, by Theorem 7.12 and $τ^{'} \geq 4 σ - 2$ one has

ρ / k \leq max (2 - 2 σ, 4 - 6 σ + τ^{'}) = 4 - 6 σ + τ^{'} .

Also, from Theorem 7.12 and $τ \leq (4 σ - 2) (k + 1)$ one has

ρ / (k + 1) \leq max (2 - 2 σ, 4 - 6 σ + τ / (k + 1)) \leq 2 - 2 σ

so that for $k = 2, 3$ one has $ρ / k \leq (2 - 2 σ) (k + 1) / k \leq 3 - 3 σ$ . In summary,

ρ / k \leq min (3 - 3 σ, 4 - 6 σ + τ^{'}) .

Next, by Lemma 10.12,

(σ, τ^{'}, ρ^{'}, ρ^{*} / k, s^{'}) \in E

for $τ^{'} := τ / k$ and some $ρ^{'} \leq ρ / k$ and $s^{'} \leq s / k$ .

Applying Theorem 10.20 to this tuple, then applying $ρ^{'} \leq ρ / k$ , one has

\begin{aligned} \frac{ρ^{*}}{k} \leq 1 - 2 σ & + \frac{1}{2} max (\frac{ρ}{k} + 1, \frac{2 ρ}{k}, \frac{5 ρ}{4 k} + \frac{τ^{'}}{2}) \\ + \frac{1}{2} max (\frac{ρ^{*}}{k} + 1, \frac{4 ρ}{k}, \frac{3}{4} \frac{ρ^{*}}{k} + \frac{ρ}{k} + \frac{τ^{'}}{2}) . \end{aligned}

By 25 one has $ρ / k \leq 1$ (since $σ \geq 3 / 4$ ) so there are only two cases to consider:

Case 1: $ρ / k + 1 \geq 5 ρ / (4 k) + τ^{'} / 2$ then

\frac{ρ^{*}}{k} \leq \frac{3}{2} - 2 σ + \frac{ρ}{2 k} + \frac{1}{2} max (\frac{ρ^{*}}{k} + 1, \frac{4 ρ}{k}, \frac{3}{4} \frac{ρ^{*}}{k} + \frac{ρ}{k} + \frac{τ^{'}}{2}) .

Solving for $ρ^{*} / k$ , we get

\frac{ρ^{*}}{k} \leq max (4 (1 - σ) + \frac{ρ}{k}, \frac{(3 - 4 σ) + 5 ρ / k}{2}, \frac{2}{5} ((6 - 8 σ) + τ^{'} + \frac{4 ρ}{k})) .

If $τ^{'} \geq 3 σ - 1$ then 25 reduces to $ρ / k \leq 3 - 3 σ$ . Substituting this bound gives

ρ^{*} / k \leq max (7 - 7 σ, 9 - 19 σ / 2, 32 (1 - σ) / 5) .

However the RHS is bounded by

\frac{18 - 19 σ}{6 σ - 2} τ^{'}

for all $τ^{'} \geq 3 σ - 1$ since the desired bound holds at $τ^{'} = 3 σ - 1$ (where one has equality).

On the other hand, if $4 σ - 2 \leq τ^{'} \leq 3 σ - 1$ then 25 reduces to $ρ / k \leq 4 - 6 σ + τ^{'}$ . Substituting this bound gives

\begin{aligned} ρ^{*} / k & \leq max (8 - 10 σ + τ^{'}, \frac{23 - 34 σ + 5 τ^{'}}{2}, \frac{2}{5} (22 - 32 σ + 5 τ^{'})) \\ \leq \frac{18 - 19 σ}{6 σ - 2} τ^{'} \end{aligned}

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

ρ^{*} \leq \frac{18 - 19 σ}{6 σ - 2} τ, (τ_{0} \leq τ \leq 2 τ_{0})

in this case.

Case 2: $ρ / k + 1 < 5 ρ / (4 k) + τ^{'} / 2$ then

ρ^{*} / k \leq 1 - 2 σ + \frac{5 ρ}{8 k} + \frac{τ^{'}}{4} + \frac{1}{2} max (\frac{ρ^{*}}{k} + 1, \frac{4 ρ}{k}, \frac{3}{4} \frac{ρ^{*}}{k} + \frac{ρ}{k} + \frac{τ^{'}}{2}) .

Solving for $ρ^{*} / k$ gives

ρ^{*} / k \leq max (3 - 4 σ + \frac{τ^{'}}{2} + \frac{5}{4} \frac{ρ}{k}, 1 - 2 σ + \frac{τ^{'}}{4} + \frac{21}{8} \frac{ρ}{k}, \frac{1}{5} (8 - 16 σ + 4 τ^{'} + 9 \frac{ρ}{k})) .

Proceeding as before, if $τ^{'} \geq 3 σ - 1$ then 25 becomes $ρ / k \leq 3 - 3 σ$ , and substituting gives

\begin{array}{r} ρ^{*} / k \leq max (\frac{27 - 31 σ}{4} + \frac{τ^{'}}{2}, \frac{71 - 79 σ}{8} + \frac{τ^{'}}{4}, 7 - \frac{43}{5} σ + \frac{4}{5} τ^{'}) . \end{array}

One may verify that the RHS is bounded by

\frac{18 - 19 σ}{6 σ - 2} τ^{'}

(with some room to spare) by checking at the endpoint $τ^{'} = 3 σ - 1$ .

Similarly, if $4 σ - 2 \leq τ^{'} \leq 3 σ - 1$ then using $ρ / k \leq 4 - 6 σ + τ^{'}$ from 25 one has

ρ^{*} / k \leq max (8 - \frac{23 σ}{2} + \frac{7 τ^{'}}{4}, \frac{23}{2} - \frac{71 σ}{4} + \frac{23 τ^{'}}{8}, \frac{44 - 70 σ + 13 τ^{'}}{5}) .

One can check that the RHS is bounded by

\frac{18 - 19 σ}{6 σ - 2} τ^{'}

by checking the required inequalities hold at $τ^{'} = 4 σ - 2$ and $τ^{'} = 3 σ - 1$ (in each case, with some room to spare).

Combining all the cases, by taking $k = 2, 3$ we have shown that for $3 / 4 \leq σ \leq 4 / 5$ and $τ_{0} \leq τ \leq 2 τ_{0}$ one has

ρ^{*} \leq \frac{18 - 19 σ}{6 σ - 2} τ

which is the first part of 18.

The case for $4 / 5 \leq σ \leq 5 / 6$ may be treated similarly. Here one needs to verify that

max (7 - 7 σ, 9 - 19 σ / 2, 32 (1 - σ) / 5) \leq \frac{7 (1 - σ)}{3 σ - 1} τ^{'},

max (\frac{27 - 31 σ}{4} + \frac{τ^{'}}{2}, \frac{71 - 79 σ}{8} + \frac{τ^{'}}{4}, 7 - \frac{43}{5} σ + \frac{4}{5} τ^{'}) \leq \frac{7 (1 - σ)}{3 σ - 1} τ^{'}

for $τ^{'} \geq 3 σ - 1$ , and that

max (8 - 10 σ + τ^{'}, \frac{23 - 34 σ + 5 τ^{'}}{2}, \frac{2}{5} (22 - 32 σ + 5 τ^{'})) \leq \frac{7 (1 - σ)}{3 σ - 1} τ^{'},

max (8 - \frac{23 σ}{2} + \frac{7 τ^{'}}{4}, \frac{23}{2} - \frac{71 σ}{4} + \frac{23 τ^{'}}{8}, \frac{44 - 70 σ + 13 τ^{'}}{5}) \leq \frac{7 (1 - σ)}{3 σ - 1} τ^{'}

for $4 σ - 2 \leq τ^{'} \leq 3 σ - 1$ . The treatment is analogous to before, so we omit the proof.

Using Theorem 10.27, it is possible to obtain improved energy estimates near $σ = 3 / 4$ , which are given by the next two theorems.

Theorem 12.8

For $7 / 10 \leq σ \leq 3 / 4$ , one has

A^{*} (σ) \leq max (\frac{5 (18 - 19 σ)}{2 (5 σ + 3) (1 - σ)}, \frac{2 (45 - 44 σ)}{(2 σ + 15) (1 - σ)}) .

Derived in derived.py as:
prove_zero_density_energy_2()

Proof ▶

Throughout assume $7 / 10 \leq σ \leq 3 / 4$ and take $τ_{0} = 2$ in Corollary 12.4. It suffices to show that if $(σ, τ, ρ, ρ^{*}, s) \in E$ with $2 \leq τ \leq 4$ , then either

ρ^{*} \leq \frac{5 (18 - 19 σ)}{2 (5 σ + 3)} τ

ρ^{*} \leq \frac{2 (45 - 44 σ)}{2 σ + 15} τ .

Note for future reference the crude bounds

1 < \frac{5 (18 - 19 σ)}{2 (5 σ + 3)} < 2, 1 < \frac{2 (45 - 44 σ)}{2 σ + 15} < \frac{7}{4} .

Let

k := {\begin{cases} 2, & 2 \leq τ < 3, \\ 3, & 3 \leq τ \leq 4, \end{cases}, τ^{'} := τ / k .

Via Theorem 7.9 and Lemma 7.8, one has

ρ / k \leq max (2 - 2 σ, 1 - 2 σ + τ^{'})

and via Theorem 10.27 and Lemma 7.8, one has

ρ / k \leq max (2 - 2 σ, 18 / 5 - 4 σ, 12 / 5 - 4 σ + τ^{'}) .

Since $σ \leq 4 / 5$ , we may drop the first term, i.e.

ρ / k \leq max (18 / 5 - 4 σ, 12 / 5 - 4 σ + τ^{'}) .

Combining 35 and 36,

ρ / k \leq {\begin{cases} 1 - 2 σ + τ^{'}, & 1 \leq τ^{'} \leq 13 / 5 - 2 σ, \\ 18 / 5 - 4 σ, & 13 / 5 - 2 σ \leq τ^{'} \leq 6 / 5, \\ 12 / 5 - 4 σ + τ^{'}, & 6 / 5 \leq τ^{'} \leq 2 (σ - 1 / 5) + 2 (1 - σ) / k, \\ (2 - 2 σ) (k + 1) / k, & 2 (σ - 1 / 5) + 2 (1 - σ) / k \leq τ^{'} \leq 1 + 1 / k . \end{cases}

One can verify that all intervals are proper for $7 / 10 \leq σ \leq 3 / 4$ and $k = 2, 3$ .

Since $(σ, τ, ρ, ρ^{*}, s) \in E$ , by Lemma 10.12 one has $(σ, τ^{'}, ρ^{'}, ρ^{*} / k, s^{'}) \in E$ for some $ρ^{'} \leq ρ / k$ and $s^{'} \leq s / k$ . Applying Theorem 10.20to the first tuple followed by $ρ^{'} \leq ρ / k$ , and noting that $ρ / k + 1 \geq 2 ρ / k$ since $ρ / k \leq 1$ by 37, one obtains

\frac{ρ^{*}}{k} \leq 1 - 2 σ + \frac{1}{2} max (\frac{ρ}{k} + 1, \frac{5 ρ}{4 k} + \frac{τ^{'}}{2}) + \frac{1}{2} max (\frac{ρ^{*}}{k} + 1, \frac{4 ρ}{k}, \frac{3 ρ^{*}}{4 k} + \frac{ρ}{k} + \frac{τ^{'}}{2}) .

First, suppose that $ρ / k + 1 < 5 ρ / (4 k) + τ^{'} / 2$ so that

\frac{ρ^{*}}{k} \leq 1 - 2 σ + \frac{5 ρ}{8 k} + \frac{τ^{'}}{4} + \frac{1}{2} max (\frac{ρ^{*}}{k} + 1, \frac{4 ρ}{k}, \frac{3}{4} \frac{ρ^{*}}{k} + \frac{ρ}{k} + \frac{τ^{'}}{2}) .

Solving for $ρ^{*} / k$ gives

\frac{ρ^{*}}{k} \leq max (3 - 4 σ + \frac{τ^{'}}{2} + \frac{5}{4} \frac{ρ}{k}, 1 - 2 σ + \frac{τ^{'}}{4} + \frac{21}{8} \frac{ρ}{k}, \frac{1}{5} (8 - 16 σ + 4 τ^{'} + 9 \frac{ρ}{k})) .

One may verify that the RHS is bounded by

\frac{5 (18 - 19 σ)}{2 (5 σ + 3)} τ^{'}

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

max (\frac{17}{4} - \frac{13}{2} σ + \frac{7}{4} τ^{'}, \frac{29}{8} - \frac{29}{4} σ + \frac{23}{8} τ^{'}, \frac{17}{5} - \frac{34}{5} σ + \frac{13}{5} τ^{'}) \leq \frac{5 (18 - 19 σ)}{2 (5 σ + 3)} τ^{'}

for $1 \leq τ^{'} \leq 13 / 5 - 2 σ$ ;

max (\frac{15}{2} - 9 σ + \frac{τ^{'}}{2}, \frac{209}{20} - \frac{25}{2} σ + \frac{τ^{'}}{4}, \frac{202}{25} - \frac{52}{5} σ + \frac{4}{5} τ^{'}) \leq \frac{5 (18 - 19 σ)}{2 (5 σ + 3)} τ^{'}

for $13 / 5 - 2 σ \leq τ^{'} \leq 6 / 5$ ;

max (6 - 9 σ + \frac{7}{4} τ^{'}, \frac{73}{10} - \frac{25}{2} σ + \frac{23}{8} τ^{'}, \frac{148}{25} - \frac{52}{5} σ + \frac{13}{5} τ^{'}) \leq \frac{5 (18 - 19 σ)}{2 (5 σ + 3)} τ^{'}

for $k = 2, 3$ and $6 / 5 \leq τ^{'} \leq 2 (σ - 1 / 5) + 2 (1 - σ) / k$ ;

\begin{aligned} max ( & 3 - 4 σ + \frac{τ^{'}}{2} + \frac{5 (2 - 2 σ)}{4} \frac{k + 1}{k}, 1 - 2 σ + \frac{τ^{'}}{4} + \frac{21 (2 - 2 σ)}{8} \frac{k + 1}{k}, \\ \frac{1}{5} (8 - 16 σ + 4 τ^{'} + 9 (2 - 2 σ) \frac{k + 1}{k})) \leq \frac{5 (18 - 19 σ)}{2 (5 σ + 3)} τ^{'} \end{aligned}

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

6 - 9 σ + \frac{7}{4} τ^{'} \leq \frac{5 (18 - 19 σ)}{2 (5 σ + 3)} τ^{'}

which is equivalent to $6 - 9 σ \leq 3 (53 - 75 σ) / (4 (5 σ + 3)) τ^{'}$ . For $σ < 53 / 75$ the LHS is negative while the RHS is positive so the inequality holds. For $σ \geq 53 / 75$ one may verify that the inequality holds at the lower limit $τ^{'} = 6 / 5$ .

In the remainder of the proof we assume $ρ / k + 1 \geq 5 ρ / (4 k) + τ^{'} / 2$ so that 42 becomes

\frac{ρ^{*}}{k} \leq \frac{3}{2} - 2 σ + \frac{ρ}{2 k} + \frac{1}{2} max (\frac{ρ^{*}}{k} + 1, \frac{4 ρ}{k}, \frac{3}{4} \frac{ρ^{*}}{k} + \frac{ρ}{k} + \frac{τ^{'}}{2}) .

Solving for $ρ^{*} / k$ gives

\frac{ρ^{*}}{k} \leq max (4 - 4 σ + \frac{ρ}{k}, \frac{3 - 4 σ + 5 ρ / k}{2}, \frac{2}{5} (6 - 8 σ + τ^{'} + \frac{4 ρ}{k})) .

The case where $τ^{'} \geq 6 / 5$ is simpler so we handle it first. Applying the last two cases of 37 it suffices to verify that

max (\frac{32}{5} - 8 σ + τ^{'}, \frac{15}{2} - 12 σ + \frac{5}{2} τ^{'}, \frac{156}{25} - \frac{48}{5} σ + 2 τ^{'}) \leq \frac{5 (18 - 19 σ)}{2 (5 σ + 3)} τ^{'}

for $k = 2, 3$ and $6 / 5 \leq τ^{'} \leq 2 (σ - 1 / 5) + 2 (1 - σ) / k$ ;

\begin{aligned} max (4 - 4 σ + (2 - 2 σ) \frac{k + 1}{k}, \frac{3 - 4 σ}{2} + (5 - 5 σ) \frac{k + 1}{k}, \\ \frac{2}{5} (6 - 8 σ + τ^{'} + (8 - 8 σ) \frac{k + 1}{k})) \leq \frac{5 (18 - 19 σ)}{2 (5 σ + 3)} τ^{'} \end{aligned}

for $k = 2, 3$ and $2 (σ - 1 / 5) + 2 (1 - σ) / k \leq τ^{'} \leq 1 + 1 / k$ . Note also that one has equality when $τ^{'} = 2 (σ - 1 / 5) + 2 (1 - σ) / k$ and $k = 2$ .

Lastly, consider the case where $1 \leq τ^{'} \leq 6 / 5$ . Applying $ρ / k \leq min (1 - 2 σ + τ^{'}, 18 / 5 - 4 σ)$ from the first two cases of 37, one obtains

\frac{3 - 4 σ + 5 ρ / k}{2} \leq min (4 - 7 σ + \frac{5}{2} τ^{'}, \frac{21}{2} - 12 σ),

\frac{2}{5} (6 - 8 σ + τ^{'} + \frac{4 ρ}{k}) \leq min (4 - \frac{32}{5} σ + 2 τ^{'}, \frac{204}{25} - \frac{48}{5} σ + \frac{2}{5} τ^{'}) .

However one may verify that the RHS of both of the above inequalities are bounded by $5 (18 - 19 σ) / (2 (5 σ + 3)) τ^{'}$ for $1 \leq τ^{'} \leq 6 / 5$ by checking at $τ^{'} = 13 / 5 - 2 σ$ . Thus

\frac{ρ^{*}}{k} \leq max (\frac{5 (18 - 19 σ)}{2 (5 σ + 3)} τ^{'}, 4 - 4 σ + \frac{ρ}{k}) .

Meanwhile, by Lemma 10.12,

(σ, τ / (k - 1), ρ^{″}, ρ^{*} / (k - 1), s^{″}) \in E

for some $ρ^{″} \leq ρ / (k - 1)$ and $s^{″} \leq s / (k - 1)$ . Applying Theorem 10.20 to this tuple (and applying $ρ^{″} \leq ρ / (k - 1)$ ) gives

\begin{aligned} \frac{ρ^{*}}{k - 1} \leq 1 - 2 σ & + \frac{1}{2} max (\frac{ρ}{k - 1} + 1, \frac{2 ρ}{k - 1}, \frac{5 ρ}{4 (k - 1)} + \frac{τ}{2 (k - 1)}) \\ + \frac{1}{2} max (\frac{ρ^{*}}{k - 1} + 1, \frac{4 ρ}{k - 1}, \frac{3 ρ^{*}}{4 (k - 1)} + \frac{ρ}{k - 1} + \frac{τ}{2 (k - 1)}) . \end{aligned}

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

ρ^{*} \leq (\frac{3}{2} - 2 σ) (k - 1) + \frac{ρ}{2} + \frac{1}{2} max (ρ^{*} + k - 1, 4 ρ, \frac{3 ρ^{*}}{4} + ρ + \frac{τ}{2}),

ρ^{*} \leq (1 - 2 σ) (k - 1) + ρ + \frac{1}{2} max (ρ^{*} + k - 1, 4 ρ, \frac{3 ρ^{*}}{4} + ρ + \frac{τ}{2}),

ρ^{*} \leq (1 - 2 σ) (k - 1) + \frac{5}{8} ρ + \frac{τ}{4} + \frac{1}{2} max (ρ^{*} + k - 1, 4 ρ, \frac{3 ρ^{*}}{4} + ρ + \frac{τ}{2}) .

If 51 holds, then solving for $ρ^{*} / k$ gives

ρ^{*} / k \leq max ((4 - 4 σ) \frac{k - 1}{k} + \frac{ρ}{k}, \frac{(3 - 4 σ) (k - 1) / k + 5 ρ / k}{2}, \frac{2}{5} ((6 - 8 σ) \frac{k - 1}{k} + τ^{'} + \frac{4 ρ}{k})) .

For $τ^{'} \leq 13 / 5 - 2 σ$ , we apply $ρ / k \leq 1 - 2 σ + τ^{'}$ from 37 (along with the inequalities $4 - 4 σ \geq 0$ , $3 - 4 σ \geq 0$ , $6 - 8 σ \geq 0$ and $(k - 1) / k \leq 2 / 3$ ),

\frac{ρ^{*}}{k} \leq max (\frac{11}{3} - \frac{14}{3} σ + τ^{'}, \frac{7}{2} - \frac{19}{3} σ + \frac{5}{2} τ^{'}, \frac{16}{5} - \frac{16}{3} σ + 2 τ^{'}) < \frac{5 (18 - 19 σ)}{2 (5 σ + 3)} τ^{'}

for all $1 \leq τ^{'} \leq 13 / 5 - 2 σ$ . The last inequality is verified using 32 and checking at both $τ^{'} = 1$ and $τ^{'} = 13 / 5 - 2 σ$ . Similarly, for $13 / 5 - 2 σ \leq τ^{'} \leq 6 / 5$ we use $ρ / k \leq 18 / 5 - 4 σ$ and verify that

\frac{ρ^{*}}{k} \leq max (\frac{94}{15} - \frac{20}{3} σ, 10 - \frac{34}{3} σ, \frac{184}{25} - \frac{128}{15} σ + \frac{2}{5} τ^{'}) < \frac{5 (18 - 19 σ)}{2 (5 σ + 3)} τ^{'}

where the last inequality is verified using 32 and checking at the lower limit $τ^{'} = 13 / 5 - 2 σ$ .

Suppose now that 52 holds. Solving for $ρ^{*} / k$ gives

\frac{ρ^{*}}{k} \leq max ((3 - 4 σ) \frac{k - 1}{k} + 2 \frac{ρ}{k}, (1 - 2 σ) \frac{k - 1}{k} + 3 \frac{ρ}{k}, \frac{2}{5} ((4 - 8 σ) \frac{k - 1}{k} + τ^{'} + 6 \frac{ρ}{k}))

Similarly to before, applying the first two cases of 37 allows one to verify that for $1 \leq τ^{'} \leq 6 / 5$ ,

(3 - 4 σ) \frac{k - 1}{k} + 2 \frac{ρ}{k} \leq \frac{2}{3} (3 - 4 σ) + 2 min (1 - 2 σ + τ^{'}, 18 / 5 - 4 σ) \leq \frac{5 (18 - 19 σ)}{2 (5 σ + 3)} τ^{'}

and

\frac{2}{5} ((4 - 8 σ) \frac{k - 1}{k} + τ^{'} + 6 \frac{ρ}{k}) \leq \frac{2}{5} (\frac{1}{2} (4 - 8 σ) + τ^{'} + 6 min (1 - 2 σ + τ^{'}, 18 / 5 - 4 σ)) \leq \frac{5 (18 - 19 σ)}{2 (5 σ + 3)} τ^{'},

each with room to spare. Therefore, for $1 \leq τ^{'} \leq 6 / 5$ , one has

\frac{ρ^{*}}{k} \leq max (\frac{5 (18 - 19 σ)}{2 (5 σ + 3)} τ^{'}, (1 - 2 σ) \frac{k - 1}{k} + \frac{3 ρ}{k})

However, if $τ^{'} \leq σ + 1 / 2 - (2 σ - 1) / (2 k)$ we may apply $ρ / k \leq 1 - 2 σ + τ^{'}$ to get

(1 - 2 σ) \frac{k - 1}{k} + \frac{3 ρ}{k} \leq (1 - 2 σ) \frac{k - 1}{k} + 3 (1 - 2 σ + τ^{'}) \leq max (\frac{5 (18 - 19 σ)}{2 (5 σ + 3)}, \frac{2 (45 - 44 σ)}{2 σ + 15}) τ^{'},

where by 32, the last inequality is verified by checking that it holds at the upper limit $τ^{'} = σ + 1 / 2 - (2 σ - 1) / (2 k)$ for $k = 2, 3$ . For $τ^{'} > σ + 1 / 2 - (2 σ - 1) / (2 k)$ , we once again apply $ρ / k \leq 1 - 2 σ + τ^{'}$ to get

4 - 4 σ + \frac{ρ}{k} \leq 5 - 6 σ + τ^{'} \leq max (\frac{5 (18 - 19 σ)}{2 (5 σ + 3)}, \frac{2 (45 - 44 σ)}{2 σ + 15}) τ^{'},

where now the last inequality is verified at the lower limit $τ^{'} = σ + 1 / 2 - (2 σ - 1) / (2 k)$ . Therefore, in view of 47 and 54, one has

\frac{ρ^{*}}{k} \leq max (\frac{5 (18 - 19 σ)}{2 (5 σ + 3)}, \frac{2 (45 - 44 σ)}{2 σ + 15}) τ^{'}

in this case, as required.

Lastly, suppose that 53 holds. Then solving for $ρ^{*} / k$ gives

\frac{ρ^{*}}{k} \leq max ((3 - 4 σ) \frac{k - 1}{k} + \frac{τ^{'}}{2} + \frac{5}{4} \frac{ρ}{k}, (1 - 2 σ) \frac{k - 1}{k} + \frac{τ^{'}}{4} + \frac{21}{8} \frac{ρ}{k}, \frac{(8 - 16 σ) (k - 1) / k + 4 τ^{'} + 9 ρ / k}{5}) .

Proceeding as before, we use $ρ / k \leq 1 - 2 σ + τ^{'}$ from 37 together with $(k - 1) / k \leq 2 / 3$ to get

(3 - 4 σ) \frac{k - 1}{k} + \frac{τ^{'}}{2} + \frac{5}{4} \frac{ρ}{k} \leq \frac{13}{4} - \frac{31}{6} σ + \frac{7}{4} τ^{'} \leq \frac{2 (45 - 44 σ)}{2 σ + 15} τ^{'},

where the last inequality is verified at $τ^{'} = 6 / 5$ . Furthermore, using $ρ / k \leq min (1 - 2 σ + τ^{'}, 18 / 5 - 4 σ)$ and $(k - 1) / k \geq 1 / 2$ one has

\begin{aligned} \frac{(8 - 16 σ) (k - 1) / k + 4 τ^{'} + 9 ρ / k}{5} & \leq min (\frac{13}{5} - \frac{26}{5} σ + \frac{13}{5} τ^{'}, \frac{182}{25} - \frac{44}{5} σ + \frac{4}{5} τ^{'}) \\ \leq max (\frac{5 (18 - 19 σ)}{2 (5 σ + 3)}, \frac{2 (45 - 44 σ)}{2 σ + 15}) τ^{'} \end{aligned}

for $1 \leq τ^{'} \leq 6 / 5$ . Therefore,

\frac{ρ^{*}}{k} \leq max (\frac{5 (18 - 19 σ)}{2 (5 σ + 3)} τ^{'}, \frac{2 (45 - 44 σ)}{2 σ + 15} τ^{'}, (1 - 2 σ) \frac{k - 1}{k} + \frac{τ^{'}}{4} + \frac{21}{8} \frac{ρ}{k}) .

If $1 \leq τ^{'} \leq 1 + 2 σ / 15$ , we use the bound $ρ / k \leq 1 - 2 σ + τ^{'}$ to get

(1 - 2 σ) \frac{k - 1}{k} + \frac{τ^{'}}{4} + \frac{21}{8} \frac{ρ}{k} \leq (1 - 2 σ) \frac{k - 1}{k} + \frac{τ^{'}}{4} + \frac{21}{8} (1 - 2 σ + τ^{'}) \leq \frac{2 (45 - 44 σ)}{2 σ + 15} τ^{'}

where by 32 it suffices to check the inequality at the upper limit $τ^{'} = 1 + 2 σ / 15$ (where we have equality if $k = 2$ ). On the other hand if $1 + 2 σ / 15 \leq τ^{'} \leq 6 / 5$ , we use $ρ / k \leq 1 - 2 σ + τ^{'}$ to get

4 - 4 σ + \frac{ρ}{k} \leq 5 - 6 σ + τ^{'} \leq \frac{2 (45 - 44 σ)}{2 σ + 15} τ^{'}

where by 32 it suffices to check the last inequality at $τ^{'} = 1 + 2 σ / 15$ (where we have equality). Therefore, one has

\frac{ρ^{*}}{k} \leq max (\frac{5 (18 - 19 σ)}{2 (5 σ + 3)}, \frac{2 (45 - 44 σ)}{2 σ + 15}) τ^{'}

in this case too.

Theorem 12.9

For $3 / 4 \leq σ \leq 4 / 5$ , one has

A^{*} (σ) \leq max (\frac{197 - 220 σ}{8 (5 σ - 1) (1 - σ)}, \frac{3 (29 - 30 σ)}{5 (5 σ - 1) (1 - σ)}, \frac{4 (10 - 9 σ)}{5 (4 σ - 1) (1 - σ)})

Derived in derived.py as:
prove_zero_density_energy_3()

Proof ▶

Throughout assume that $3 / 4 \leq σ \leq 4 / 5$ and take $τ_{0} := 8 σ - 4$ in Corollary 12.4. It suffices to show that

ρ^{*} \leq max (\frac{197 - 220 σ}{8 (5 σ - 1)}, \frac{3 (29 - 30 σ)}{5 (5 σ - 1)}) τ

for all $(σ, τ, ρ, ρ^{*}, s) \in E$ satisfying $τ_{0} \leq τ \leq 2 τ_{0}$ , and

ρ^{*} \leq \frac{4 (10 - 9 σ)}{5 (4 σ - 1)} τ

for all $(σ, τ, ρ, ρ^{*}, s) \in E_{ζ}$ such that $2 \leq τ \leq τ_{0}$ . In the proof of Theorem 12.7 we have already shown that 56 holds in the large range $65 / 86 \leq σ \leq 5 / 6$ , so it remains to prove 55. Given $σ, τ$ , let $k \geq 2$ be the integer for which

k \leq \frac{τ}{4 σ - 2} < k + 1

so that $k = 2, 3$ for $τ_{0} \leq τ < 2 τ_{0}$ , and as before write $τ^{'} := τ / k$ .

By Theorem 10.27 and Lemma 7.8 one has

ρ / k \leq max (18 / 5 - 4 σ, 12 / 5 - 4 σ + τ^{'}) = {\begin{cases} 18 / 5 - 4 σ, & τ^{'} \leq 6 / 5 \\ 12 / 5 - 4 σ + τ^{'}, & τ^{'} > 6 / 5 \end{cases}

and from Theorem 7.12 and Lemma 7.8, for any integer $ℓ$ ,

ρ / ℓ \leq max (2 - 2 σ, 4 - 6 σ + τ / ℓ) = {\begin{cases} 2 - 2 σ, & τ / ℓ \leq 4 σ - 2, \\ 4 - 6 σ + τ / ℓ, & τ / ℓ > 4 σ - 2. \end{cases}

so that in particular, taking $ℓ = k + 1$ and noting that $τ / (k + 1) \leq 4 σ - 2$ by 57,

ρ / k = \frac{k + 1}{k} \frac{ρ}{k + 1} \leq \frac{k + 1}{k} max (2 - 2 σ, 4 - 6 σ + \frac{τ}{k + 1}) = (2 - 2 σ) \frac{k + 1}{k} \leq 3 - 3 σ .

Combining everything, one obtains (for $k \geq 2$ )

ρ / k \leq {\begin{cases} 4 - 6 σ + τ^{'}, & 4 σ - 2 \leq τ^{'} \leq 2 σ - 2 / 5, \\ 18 / 5 - 4 σ, & 2 σ - 2 / 5 \leq τ^{'} \leq 6 / 5, \\ 12 / 5 - 4 σ + τ^{'}, & 6 / 5 \leq τ^{'} \leq σ + 3 / 5, \\ 3 - 3 σ, & σ + 3 / 5 \leq τ^{'} \leq (4 σ - 2) (k + 1) / k . \end{cases}

First, suppose that $6 / 5 \leq τ^{'} \leq (4 σ - 2) (k + 1) / k$ . By Lemma 10.12, $(σ, τ^{'}, ρ^{'}, ρ^{*} / k, s^{'}) \in E$ for some $ρ^{'} \leq ρ / k$ and $s^{'} \leq s / k$ . Applying Theorem 10.20, and noting that $ρ / k + 1 \geq 2 ρ / k$ since $ρ / k \leq 1$ by 62,

\frac{ρ^{*}}{k} \leq 1 - 2 σ + \frac{1}{2} max (\frac{ρ}{k} + 1, \frac{5 ρ}{4 k} + \frac{τ^{'}}{2}) + \frac{1}{2} max (\frac{ρ^{*}}{k} + 1, \frac{4 ρ}{k}, \frac{3 ρ^{*}}{4 k} + \frac{ρ}{k} + \frac{τ^{'}}{2}) .

If $ρ / k + 1 \geq 5 ρ / (4 k) + τ^{'} / 2$ , then

\frac{ρ^{*}}{k} \leq \frac{3}{2} - 2 σ + \frac{ρ}{2 k} + \frac{1}{2} max (\frac{ρ^{*}}{k} + 1, \frac{4 ρ}{k}, \frac{3 ρ^{*}}{4 k} + \frac{ρ}{k} + \frac{τ^{'}}{2}) .

Solving for $ρ^{*} / k$ gives

\frac{ρ^{*}}{k} \leq max (4 - 4 σ + \frac{ρ}{k}, \frac{3 - 4 σ + 5 ρ / k}{2}, \frac{2}{5} (6 - 8 σ + τ^{'} + \frac{4 ρ}{k})) .

Applying $ρ / k \leq min (12 / 5 - 4 σ + τ^{'}, 3 - 3 σ)$ to the RHS, one may ultimately verify that

\frac{ρ^{*}}{k} \leq max (\frac{197 - 220 σ}{8 (5 σ - 1)}, \frac{3 (29 - 30 σ)}{5 (5 σ - 1)}) τ^{'} .

If $ρ / k + 1 \leq 5 ρ / (4 k) + τ^{'} / 2$ one has

ρ^{*} / k \leq 1 - 2 σ + \frac{5 ρ}{8 k} + \frac{τ^{'}}{4} + \frac{1}{2} max (\frac{ρ^{*}}{k} + 1, \frac{4 ρ}{k}, \frac{3}{4} \frac{ρ^{*}}{k} + \frac{ρ}{k} + \frac{τ^{'}}{2})

and solving for $ρ^{*} / k$ gives

ρ^{*} / k \leq max (3 - 4 σ + \frac{τ^{'}}{2} + \frac{5}{4} \frac{ρ}{k}, 1 - 2 σ + \frac{τ^{'}}{4} + \frac{21}{8} \frac{ρ}{k}, \frac{1}{5} (8 - 16 σ + 4 τ^{'} + 9 \frac{ρ}{k})) .

Once again applying $ρ / k \leq min (12 / 5 - 4 σ + τ^{'}, 3 - 3 σ)$ , one again ultimately obtains

ρ^{*} / k \leq \frac{197 - 220 σ}{8 (5 σ - 1)} τ^{'} .

Now suppose that $4 σ - 2 \leq τ^{'} \leq 6 / 5$ . By Lemma 10.12, for any integer $k \geq 2$ one has $(σ, τ / (k - 1), ρ^{'}, ρ^{*} / (k - 1), s^{'}) \in E$ for some $ρ^{'} \leq ρ / (k - 1)$ and $s^{'} \leq s / (k - 1)$ . Applying Theorem 10.20 to this tuple, followed by $ρ^{'} \leq ρ / (k - 1)$ and rearranging, one obtains

\begin{aligned} ρ^{*} \leq (1 - 2 σ) (k - 1) & + \frac{1}{2} max (ρ + k - 1, 2 ρ, \frac{5 ρ}{4} + \frac{τ}{2}) \\ + \frac{1}{2} max (ρ^{*} + k - 1, 4 ρ, \frac{3 ρ^{*}}{4} + ρ + \frac{τ}{2}) . \end{aligned}

Consider the first maximum of 67. If $ρ + k - 1$ is maximal, then

ρ^{*} \leq (\frac{3}{2} - 2 σ) (k - 1) + \frac{ρ}{2} + \frac{1}{2} max (ρ^{*} + k - 1, 4 ρ, \frac{3 ρ^{*}}{4} + ρ + \frac{τ}{2}) .

Solving for $ρ^{*}$ and dividing by $k$ gives

\frac{ρ^{*}}{k} \leq max ((4 - 4 σ) \frac{k - 1}{k} + \frac{ρ}{k}, \frac{(3 - 4 σ) (k - 1) / k + 5 ρ / k}{2}, \frac{2}{5} ((6 - 8 σ) (k - 1) / k + τ^{'} + \frac{4 ρ}{k})) .

Since $3 / 4 \leq σ \leq 1$ and $1 / 2 \leq (k - 1) / k \leq 2 / 3$ , we have

\frac{ρ^{*}}{k} \leq max (\frac{8}{3} (1 - σ) + \frac{ρ}{k}, \frac{3}{4} - σ + \frac{5}{2} \frac{ρ}{k}, \frac{2}{5} (3 - 4 σ + τ^{'} + \frac{4 ρ}{k})) .

Bounding the RHS with $ρ / k \leq min (4 - 6 σ + τ^{'}, 18 / 5 - 4 σ)$ , one ultimately obtains

ρ^{*} / k \leq max (\frac{197 - 220 σ}{8 (5 σ - 1)}, \frac{3 (29 - 30 σ)}{5 (5 σ - 1)}) τ^{'} .

Now suppose $5 ρ / 4 + τ^{'} / 2$ is maximal in 67. Then

ρ^{*} \leq (1 - 2 σ) (k - 1) + \frac{5}{8} ρ + \frac{τ}{4} + \frac{1}{2} max (ρ^{*} + k - 1, 4 ρ, \frac{3 ρ^{*}}{4} + ρ + \frac{τ}{2}) .

Solving for $ρ^{*}$ and dividing by $k$ gives

\frac{ρ^{*}}{k} \leq max ((3 - 4 σ) \frac{k - 1}{k} + \frac{τ^{'}}{2} + \frac{5}{4} \frac{ρ}{k}, (1 - 2 σ) \frac{k - 1}{k} + \frac{τ^{'}}{4} + \frac{21}{8} \frac{ρ}{k}, \frac{8}{5} ((1 - 2 σ) \frac{k - 1}{k} + \frac{τ^{'}}{2} + \frac{9}{8} \frac{ρ}{k})) .

As before, since $3 / 4 \leq σ \leq 1$ and $1 / 2 \leq (k - 1) / k \leq 2 / 3$ , one has

\frac{ρ^{*}}{k} \leq max (\frac{3 - 4 σ}{2} + \frac{τ^{'}}{2} + \frac{5}{4} \frac{ρ}{k}, \frac{1}{2} - σ + \frac{τ^{'}}{4} + \frac{21}{8} \frac{ρ}{k}, \frac{8}{5} (\frac{1}{2} - σ + \frac{τ^{'}}{2} + \frac{9}{8} \frac{ρ}{k})) .

Once again we apply $ρ / k \leq min (4 - 6 σ + τ^{'}, 18 / 5 - 4 σ)$ to ultimately obtain

ρ^{*} / k \leq max (\frac{197 - 220 σ}{8 (5 σ - 1)}, \frac{3 (29 - 30 σ)}{5 (5 σ - 1)}) τ^{'} .

Lastly, if $2 ρ$ is maximal in 67, then

ρ^{*} \leq (1 - 2 σ) (k - 1) + ρ + \frac{1}{2} max (ρ^{*} + k - 1, 4 ρ, \frac{3 ρ^{*}}{4} + ρ + \frac{τ}{2}) .

Solving for $ρ^{*}$ and dividing by $k$ gives

ρ^{*} / k \leq max ((3 - 4 σ) \frac{k - 1}{k} + \frac{2 ρ}{k}, (1 - 2 σ) \frac{k - 1}{k} + \frac{3 ρ}{k}, \frac{8}{5} ((1 - 2 σ) \frac{k - 1}{k} + \frac{τ^{'}}{4} + \frac{3}{2} \frac{ρ}{k})) .

Once again we apply $ρ / k \leq min (4 - 6 σ + τ^{'}, 18 / 5 - 4 σ)$ to ultimately obtain

ρ^{*} / k \leq max (\frac{197 - 220 σ}{8 (5 σ - 1)}, \frac{3 (29 - 30 σ)}{5 (5 σ - 1)}) τ^{'}

in this case too.

Modest improvements are possible by incorporating more large value estimates; these are recorded in the next few theorems.

Theorem 12.10

For $664 / 877 \leq σ \leq 31 / 40$ , one has

A^{*} (σ) \leq max (\frac{72 - 91 σ}{7 (11 σ - 8) (1 - σ)}, \frac{5 (18 - 19 σ)}{2 (5 σ + 3) (1 - σ)}) .

Derived in derived.py as:
prove_zero_density_energy_4()

Proof ▶

Fix $664 / 877 \leq σ \leq 31 / 40$ and take $τ_{0} = 2$ . It suffices to show that

ρ^{*} \leq max (\frac{72 - 91 σ}{7 (11 σ - 8)}, \frac{5 (18 - 19 σ)}{2 (5 σ + 3)}) τ

for all $(σ, τ, ρ, ρ^{*}, s) \in E$ satisfying $2 \leq τ \leq 4$ .

Let $k = 2$ if $2 \leq τ < 3$ and $k = 3$ otherwise, and as usual let $τ^{'} = τ / k$ . By Lemma 10.12, $(σ, τ^{'}, ρ^{'}, ρ^{*} / k, s^{'}) \in E$ for some $ρ^{'} \leq ρ / k$ and $s^{'} \leq s / k$ . Applying Theorem 10.20, and noting that $ρ / k + 1 \geq 2 ρ / k$ since $ρ / k \leq 1$ by 62,

\frac{ρ^{*}}{k} \leq 1 - 2 σ + \frac{1}{2} max (\frac{ρ}{k} + 1, \frac{5 ρ}{4 k} + \frac{τ^{'}}{2}) + \frac{1}{2} max (\frac{ρ^{*}}{k} + 1, \frac{4 ρ}{k}, \frac{3 ρ^{*}}{4 k} + \frac{ρ}{k} + \frac{τ^{'}}{2}) .

Rearranging the inequality and solving for $ρ^{*} / k$ , one must either have

\frac{ρ^{*}}{k} \leq max (4 - 4 σ + \frac{ρ}{k}, \frac{3 - 4 σ + 5 ρ / k}{2}, \frac{2}{5} (6 - 8 σ + τ^{'} + \frac{4 ρ}{k})) .

\frac{ρ^{*}}{k} \leq max (3 - 4 σ + \frac{τ^{'}}{2} + \frac{5}{4} \frac{ρ}{k}, 1 - 2 σ + \frac{τ^{'}}{4} + \frac{21}{8} \frac{ρ}{k}, \frac{1}{5} (8 - 16 σ + 4 τ^{'} + 9 \frac{ρ}{k})) .

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

τ^{'} \geq \frac{9 σ - 1}{5} .

By Theorem 10.27, one has

ρ / k \leq max (18 / 5 - 4 σ, 12 / 5 - 4 σ + τ^{'})

and by Theorem 7.9, one has

ρ / (k + 1) \leq max (2 - 2 σ, 1 - 2 σ + τ / (k + 1)) .

Combining the two inequalities, we have

ρ / k \leq {\begin{cases} 18 / 5 - 4 σ, & (9 σ - 1) / 5 \leq τ^{'} < 6 / 5, \\ 12 / 5 - 4 σ + τ^{'}, & 6 / 5 \leq τ^{'} < 2 σ - 2 / 5 + 2 (1 - σ) / k, \\ (2 - 2 σ) (k + 1) / k, & 2 σ - 2 / 5 + 2 (1 - σ) / k \leq τ^{'} \leq (k + 1) / k . \end{cases}

Substituting 72 into 70 and 71, one may verify that

\frac{ρ^{*}}{k} \leq \frac{5 (18 - 19 σ)}{2 (5 σ + 3)} τ^{'}

for all $(9 σ - 1) / 5 \leq τ^{'} \leq 1 + 1 / k$ (with equality occurring at $τ^{'} = 2 σ - 2 / 5 + 2 (1 - σ) / k$ and $k = 2$ ).

Now consider the case where

1 \leq τ^{'} \leq \frac{9 σ - 1}{5} .

Taking $k = 10$ in Theorem 7.16, one has

ρ / k \leq max (2 - 2 σ, 19 / 5 - 29 σ / 5 + τ^{'}, 60 - 80 σ + τ^{'}) .

Suppose first that $σ \geq 281 / 371$ . Then, this reduces to

\begin{aligned} ρ / k & \leq max (2 - 2 σ, 19 / 5 - 29 σ / 5 + τ^{'}) . \end{aligned}

By Lemma 10.12, for any integer $k \geq 2$ one has $(σ, τ / (k - 1), ρ^{'}, ρ^{*} / (k - 1), s^{'}) \in E$ for some $ρ^{'} \leq ρ / (k - 1)$ and $s^{'} \leq s / (k - 1)$ . Applying Theorem 10.20 to this tuple, followed by $ρ^{'} \leq ρ / (k - 1)$ and rearranging, one obtains

\begin{aligned} ρ^{*} \leq (1 - 2 σ) (k - 1) & + \frac{1}{2} max (ρ + k - 1, 2 ρ, \frac{5 ρ}{4} + \frac{τ}{2}) \\ + \frac{1}{2} max (ρ^{*} + k - 1, 4 ρ, \frac{3 ρ^{*}}{4} + ρ + \frac{τ}{2}) . \end{aligned}

By considering each case of the two maximums individually, and solving for $ρ^{*} / k$ , one obtains

\begin{aligned} \frac{ρ^{*}}{k} \leq max ( & (4 - 4 σ) \frac{k - 1}{k} + \frac{ρ}{k}, \frac{(3 - 4 σ) (k - 1) / k + 5 ρ / k}{2}, \\ \frac{2}{5} ((6 - 8 σ) \frac{k - 1}{k} + τ^{'} + \frac{4 ρ}{k}), (3 - 4 σ) \frac{k - 1}{k} + 2 \frac{ρ}{k}, \\ (1 - 2 σ) \frac{k - 1}{k} + 3 \frac{ρ}{k}, \frac{2}{5} ((4 - 8 σ) \frac{k - 1}{k} + τ^{'} + 6 \frac{ρ}{k}), \\ (3 - 4 σ) \frac{k - 1}{k} + \frac{τ^{'}}{2} + \frac{5}{4} \frac{ρ}{k}, (1 - 2 σ) \frac{k - 1}{k} + \frac{τ^{'}}{4} + \frac{21}{8} \frac{ρ}{k}, \\ \frac{(8 - 16 σ) (k - 1) / k + 4 τ^{'} + 9 ρ / k}{5}) . \end{aligned}

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

\frac{ρ^{*}}{k} < \frac{5 (18 - 19 σ)}{2 (5 σ + 3)} τ^{'}

for $1 \leq τ^{'} \leq (9 σ - 1) / 5$ and $k = 2, 3$ .

Suppose now that $σ < 281 / 371$ so that 76 reduces to

\frac{ρ}{k} \leq max (2 - 2 σ, 60 - 80 σ + τ^{'}) .

If $1 \leq τ^{'} < 77 σ / 2 - 28$ then substituting $ρ / k \leq max (2 - 2 σ, 60 - 80 σ + τ^{'})$ into 82, one may ultimately verify in each case that

\frac{ρ^{*}}{k} \leq max (\frac{72 - 91 σ}{7 (11 σ - 8)}, \frac{5 (18 - 19 σ)}{2 (5 σ + 3)}) τ^{'}

for $k = 2, 3$ , with equality when $τ^{'} = 77 σ / 2 - 28$ and $k = 2$ (here we make use of the assumption $σ \geq 664 / 877 = 0.7571 \dots$ ).

Lastly, if $77 σ / 2 - 28 \leq τ^{'} \leq (9 σ - 1) / 5$ then 88 simplifies to $ρ / k \leq 60 - 80 σ + τ^{'}$ . Substituting this into 70 and 71, one obtains in either case that

\frac{ρ^{*}}{k} \leq max (\frac{72 - 91 σ}{7 (11 σ - 8)}, \frac{5 (18 - 19 σ)}{2 (5 σ + 3)}) τ^{'}

with equality when $τ^{'} = 77 σ / 2 - 28$ and $k = 2$ .

Theorem 12.11

For $42 / 55 \leq σ \leq 79 / 103$ , one has

A^{*} (σ) \leq max (\frac{18 - 19 σ}{6 (15 σ - 11) (1 - σ)}, \frac{3 (18 - 19 σ)}{4 (4 σ - 1) (1 - σ)}) .

Derived in derived.py as:
prove_zero_density_energy_5()

Proof ▶

Fix $42 / 55 \leq σ \leq 79 / 103$ and take $τ_{0} = 2$ . It suffices to show that

ρ^{*} \leq max (\frac{18 - 19 σ}{6 (15 σ - 11)}, \frac{3 (18 - 19 σ)}{4 (4 σ - 1)}) τ

for all $(σ, τ, ρ, ρ^{*}, s) \in E$ for which $2 \leq τ \leq 4$ . As before let

k := {\begin{cases} 2, & 2 \leq τ < 3, \\ 3, & 3 \leq τ \leq 4, \end{cases} τ^{'} := τ / k,

so that in particular $1 \leq τ^{'} \leq (k + 1) / k$ .

By Lemma 10.12, for $k = 2, 3$ ,

(σ, τ^{'}, ρ^{'}, ρ^{*} / k, s^{'}), (σ, τ / (k + 1), ρ^{″}, ρ^{*} / (k + 1), s^{″}) \in E

for some $ρ^{'} \leq ρ / k$ , $s^{'} \leq s / k$ , $ρ^{″} \leq ρ / (k + 1)$ and $s^{″} \leq s / (k + 1)$ . Applying Theorem 7.16 with $k = 6$ to each tuple, one has

\begin{aligned} ρ / k & \leq max (2 - 2 σ, 11 / 3 - 17 σ / 3 + τ^{'}, 36 - 48 σ + τ^{'}), \\ ρ / (k + 1) & \leq max (2 - 2 σ, 11 / 3 - 17 σ / 3 + τ / (k + 1), 36 - 48 σ + τ / (k + 1)) . \end{aligned}

First suppose $σ \geq 97 / 127$ , in which case the above simplifies to

\begin{aligned} ρ / k & \leq max (2 - 2 σ, 11 / 3 - 17 σ / 3 + τ^{'}), \\ ρ / (k + 1) & \leq max (2 - 2 σ, 11 / 3 - 17 σ / 3 + τ / (k + 1)) . \end{aligned}

This combines to give

ρ / k \leq {\begin{cases} 2 - 2 σ, & 1 \leq τ^{'} < (11 σ - 5) / 3, \\ 11 / 3 - 17 σ / 3 + τ^{'}, & (11 σ - 5) / 3 \leq τ^{'} < (11 σ - 5) / 3 + 2 (1 - σ) / k, \\ (2 - 2 σ) (k + 1) / k, & (11 σ - 5) / 3 + 2 (1 - σ) / k \leq τ^{'} \leq 1 + 1 / k . \end{cases}

First suppose that $τ^{'} \geq (11 σ - 5) / 3$ . Applying Theorem 10.20 to the first tuple of 91, and noting that $ρ / k + 1 \geq 2 ρ / k$ since $ρ / k \leq 1$ ,

\frac{ρ^{*}}{k} \leq 1 - 2 σ + \frac{1}{2} max (\frac{ρ}{k} + 1, \frac{5 ρ}{4 k} + \frac{τ^{'}}{2}) + \frac{1}{2} max (\frac{ρ^{*}}{k} + 1, \frac{4 ρ}{k}, \frac{3 ρ^{*}}{4 k} + \frac{ρ}{k} + \frac{τ^{'}}{2}) .

Rearranging the inequality and solving for $ρ^{*} / k$ , one must either have

\frac{ρ^{*}}{k} \leq max (4 - 4 σ + \frac{ρ}{k}, \frac{3 - 4 σ + 5 ρ / k}{2}, \frac{2}{5} (6 - 8 σ + τ^{'} + \frac{4 ρ}{k}))

\frac{ρ^{*}}{k} \leq max (3 - 4 σ + \frac{τ^{'}}{2} + \frac{5}{4} \frac{ρ}{k}, 1 - 2 σ + \frac{τ^{'}}{4} + \frac{21}{8} \frac{ρ}{k}, \frac{1}{5} (8 - 16 σ + 4 τ^{'} + 9 \frac{ρ}{k})) .

100

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

ρ^{*} / k \leq \frac{3 (18 - 19 σ)}{4 (4 σ - 1)} τ^{'}, (\frac{11 σ - 5}{3} \leq τ^{'} \leq 1 + \frac{1}{k}) .

101

Now suppose that $τ^{'} < (11 σ - 5) / 3$ . Then, note that for $k = 2, 3$ one has

(σ, τ / (k - 1), ρ^{‴}, ρ^{*} / (k - 1), s^{‴}) \in E

for some $ρ^{‴} \leq ρ / (k - 1)$ and $s^{‴} \leq s / (k - 1)$ . Applying Theorem 10.20 to this tuple, followed by $ρ^{‴} \leq ρ / (k - 1)$ and rearranging, one obtains

\begin{aligned} ρ^{*} \leq (1 - 2 σ) (k - 1) & + \frac{1}{2} max (ρ + k - 1, 2 ρ, \frac{5 ρ}{4} + \frac{τ}{2}) \\ + \frac{1}{2} max (ρ^{*} + k - 1, 4 ρ, \frac{3 ρ^{*}}{4} + ρ + \frac{τ}{2}) . \end{aligned}

102

By considering each case of the two maximums individually, and solving for $ρ^{*} / k$ , one obtains

\begin{aligned} \frac{ρ^{*}}{k} \leq max ( & (4 - 4 σ) \frac{k - 1}{k} + \frac{ρ}{k}, \frac{(3 - 4 σ) (k - 1) / k + 5 ρ / k}{2}, \\ \frac{2}{5} ((6 - 8 σ) \frac{k - 1}{k} + τ^{'} + \frac{4 ρ}{k}), (3 - 4 σ) \frac{k - 1}{k} + 2 \frac{ρ}{k}, \\ (1 - 2 σ) \frac{k - 1}{k} + 3 \frac{ρ}{k}, \frac{2}{5} ((4 - 8 σ) \frac{k - 1}{k} + τ^{'} + 6 \frac{ρ}{k}), \\ (3 - 4 σ) \frac{k - 1}{k} + \frac{τ^{'}}{2} + \frac{5}{4} \frac{ρ}{k}, (1 - 2 σ) \frac{k - 1}{k} + \frac{τ^{'}}{4} + \frac{21}{8} \frac{ρ}{k}, \\ \frac{(8 - 16 σ) (k - 1) / k + 4 τ^{'} + 9 ρ / k}{5}) . \end{aligned}

105

Note that for all $1 \leq τ^{'} < (11 σ - 5) / 3$ , one has by 95 that $ρ / k \leq 2 - 2 σ$ . Substituting this into 105, one obtains

ρ^{*} / k \leq \frac{3 (18 - 19 σ)}{4 (4 σ - 1)} τ^{'}

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

ρ^{*} \leq \frac{3 (18 - 19 σ)}{4 (4 σ - 1)} τ

for all $(σ, τ, ρ, ρ^{*}, s) \in E$ for $97 / 127 \leq σ \leq 79 / 103$ and $2 \leq τ \leq 4$ , as required.

The proof in the range $42 / 55 \leq σ \leq 97 / 127$ is similar. Here 92 reduces to

\begin{aligned} ρ / k & \leq max (2 - 2 σ, 36 - 48 σ + τ^{'}), \\ ρ / (k + 1) & \leq max (2 - 2 σ, 36 - 48 σ + τ / (k + 1)) \end{aligned}

so that

ρ / k \leq {\begin{cases} 2 - 2 σ, & 1 \leq τ^{'} < 46 σ - 34, \\ 36 - 48 σ + τ^{'}, & 46 σ - 34 \leq τ^{'} < 46 σ - 34 + 2 (1 - σ) / k, \\ (2 - 2 σ) (k + 1) / k, & 46 σ - 34 + 2 (1 - σ) / k \leq τ^{'} \leq 1 + 1 / k . \end{cases}

111

If $τ^{'} \geq 46 σ - 34$ , we substitute the last two cases of this bound into 99 and 100, one may verify that

\frac{ρ^{*}}{k} \leq \frac{18 - 19 σ}{6 (15 σ - 11)} τ^{'}

115

with equality when $τ^{'} = 46 σ - 34 + 2 (1 - σ) / k$ and $k = 2$ .

On the other hand if $1 \leq τ^{'} < 46 σ - 34$ then substituting $ρ / k \leq 2 - 2 σ$ into 105 gives

\frac{ρ^{*}}{k} \leq \frac{18 - 19 σ}{6 (15 σ - 11)} τ^{'}, (1 \leq τ^{'} \leq 46 σ - 34)

in each case. Combined with 115, the desired result follows for $42 / 55 \leq σ \leq 97 / 127$ .

Theorem 12.12

For $79 / 103 \leq σ \leq 84 / 109$ , one has

A^{*} (σ) \leq max (\frac{18 - 19 σ}{2 (37 σ - 27) (1 - σ)}, \frac{5 (18 - 19 σ)}{2 (13 σ - 3) (1 - σ)}) .

Derived in derived.py as:
prove_zero_density_energy_6()

Proof ▶

Fix $79 / 103 \leq σ \leq 84 / 109$ and take

τ_{0} = {\begin{cases} (36 σ - 16) / 5, & 79 / 103 \leq σ < 33 / 43, \\ 38 σ - 28, & 33 / 43 \leq σ \leq 84 / 109. \end{cases}

Let

k := {\begin{cases} 2, & τ_{0} \leq τ < 3 τ_{0} / 2 \\ 3, & 3 τ_{0} / 2 \leq τ \leq 2 τ_{0} \end{cases}, τ^{'} := τ / k .

Suppose that $(σ, τ, ρ, ρ^{*}, s) \in E$ . We will first show

ρ^{*} \leq max (\frac{18 - 19 σ}{2 (37 σ - 27)}, \frac{5 (18 - 19 σ)}{2 (13 σ - 3)}) τ, (τ_{0} \leq τ \leq 2 τ_{0}) .

120

By Lemma 10.12, one has that

(σ, τ^{'}, ρ^{'}, ρ^{*} / k, s^{'}), (σ, τ / (k + 1), ρ^{″}, ρ^{*} / (k + 1), s^{″}) \in E

121

for some $ρ^{'} \leq ρ / k$ , $s^{'} \leq s / k$ , $ρ^{″} \leq ρ / (k + 1)$ and $s^{″} \leq s / (k + 1)$ . First suppose that $σ \geq 33 / 43$ . In this range, Theorem 7.16 with $k = 5$ gives

ρ / k \leq max (2 - 2 σ, 18 / 5 - 28 σ / 5 + τ^{'}),

ρ / (k + 1) \leq max (2 - 2 σ, 18 / 5 - 28 σ / 5 + τ / (k + 1)) .

For $τ \geq τ_{0}$ , the first inequality reduces to $ρ / k \leq 18 / 5 - 28 σ / 5 + τ^{'}$ , while for $τ \leq 2 τ_{0}$ the second inequality reduces to

ρ / k = \frac{k + 1}{k} \frac{ρ}{k + 1} \leq \frac{k + 1}{k} (2 - 2 σ) \leq 3 - 3 σ .

Combining these two inequalities gives

ρ / k \leq min (18 / 5 - 28 σ / 5 + τ^{'}, 3 - 3 σ), (τ_{0} \leq τ \leq 2 τ_{0}) .

In particular this implies $ρ / k \leq 1$ . Applying Theorem 10.20 to the first tuple of 121,

\frac{ρ^{*}}{k} \leq 1 - 2 σ + \frac{1}{2} max (\frac{ρ}{k} + 1, \frac{5 ρ}{4 k} + \frac{τ^{'}}{2}) + \frac{1}{2} max (\frac{ρ^{*}}{k} + 1, \frac{4 ρ}{k}, \frac{3 ρ^{*}}{4 k} + \frac{ρ}{k} + \frac{τ^{'}}{2}) .

By considering each case of the first maximum and solving for $ρ^{*} / k$ , one must either have

\frac{ρ^{*}}{k} \leq max (4 - 4 σ + \frac{ρ}{k}, \frac{3 - 4 σ + 5 ρ / k}{2}, \frac{2}{5} (6 - 8 σ + τ^{'} + \frac{4 ρ}{k}))

122

\frac{ρ^{*}}{k} \leq max (3 - 4 σ + \frac{τ^{'}}{2} + \frac{5}{4} \frac{ρ}{k}, 1 - 2 σ + \frac{τ^{'}}{4} + \frac{21}{8} \frac{ρ}{k}, \frac{1}{5} (8 - 16 σ + 4 τ^{'} + 9 \frac{ρ}{k})) .

123

However one may verify in both cases that

\frac{ρ^{*}}{k} \leq \frac{5 (18 - 19 σ)}{2 (13 σ - 3)} τ^{'}

124

for $33 / 43 \leq σ \leq 84 / 109$ and $τ_{0} \leq τ \leq 2 τ_{0}$ (with equality at $τ = 2 (13 σ - 3) / 5$ ).

Now consider $σ \leq 33 / 43$ . In this range of $σ$ , Theorem 7.16 with $k = 5$ gives

ρ / k \leq max (2 - 2 σ, 30 - 40 σ + τ^{'}),

ρ / (k + 1) \leq max (2 - 2 σ, 30 - 40 σ + τ / (k + 1))

which gives

ρ / k \leq min (30 - 40 σ + τ^{'}, 3 - 3 σ), (τ_{0} \leq τ \leq 2 τ_{0}) .

Substituting this bound into 122 and 123, one obtains

\frac{ρ^{*}}{k} \leq \frac{18 - 19 σ}{2 (37 σ - 27)} τ^{'}

for $79 / 103 \leq σ \leq 33 / 43$ and $τ_{0} \leq τ \leq 2 τ_{0}$ (with equality at $τ = 74 σ - 54$ ). Combined with 124, one obtains 120.

It remains to show that

ρ^{*} \leq max (\frac{18 - 19 σ}{2 (37 σ - 27)}, \frac{5 (18 - 19 σ)}{2 (13 σ - 3)}) τ

125

for all $(σ, τ, ρ, ρ^{*}, s) \in E_{ζ}$ for which $79 / 103 \leq σ \leq 84 / 109$ and $2 \leq τ \leq τ_{0}$ . This follows from substituting $ρ / k \leq 6 - 12 σ + 2 τ^{'}$ (Theorem 9.7) into 122 and 123. □

Theorem 12.13

For $84 / 109 \leq σ \leq 5 / 6$ , one has

A^{*} (σ) \leq max (\frac{18 - 19 σ}{9 (3 σ - 2) (1 - σ)}, \frac{4 (10 - 9 σ)}{5 (4 σ - 1) (1 - σ)}) .

Derived in derived.py as:
prove_zero_density_energy_7()

Theorem 12.14

For $165 / 226 \leq σ \leq 42 / 55$ one has

A^{*} (σ) \leq max (\frac{457 - 546 σ}{2 (61 - 58 σ) (1 - σ)}, \frac{5 (18 - 19 σ)}{2 (5 σ + 3) (1 - σ)}) .

Derived in derived.py as:
prove_zero_density_energy_12()

Table 12.1 records the sharpest known unconditional upper bounds on $A^{*} (σ)$ for $1 / 2 \leq σ \leq 1$ (except when close to $σ = 1$ , when sharper estimates are available by applying Lemma 12.2 with known zero-density bounds).

Table 12.1 Current best upper bound on

A^{*} (σ)

$A^{*} (σ)$ bound	Range	Reference
$\frac{10 - 11 σ}{(2 - σ) (1 - σ)}$	$\frac{1}{2} \leq σ \leq \frac{2}{3} = 0.6666 \dots$	Theorem 12.6
$\frac{18 - 19 σ}{(4 - 2 σ) (1 - σ)}$	$\frac{2}{3} \leq σ \leq \frac{7}{10} = 0.7$	Theorem 12.6
$\frac{5 (18 - 19 σ)}{2 (5 σ + 3) (1 - σ)}$	$\frac{7}{10} \leq σ \leq \frac{539 - \sqrt{42121}}{460} = 0.7255 \dots$	Theorem 12.8
$\frac{2 (45 - 44 σ)}{(2 σ + 15) (1 - σ)}$	$\frac{539 - \sqrt{42121}}{460} \leq σ \leq \frac{165}{226} = 0.7300 \dots$	Theorem 12.8
$\frac{457 - 546 σ}{2 (61 - 58 σ) (1 - σ)}$	$\frac{165}{226} \leq σ \leq \frac{5831 + \sqrt{60001}}{8240} = 0.7373 \dots$	Theorem 12.14
$\frac{5 (18 - 19 σ)}{2 (5 σ + 3) (1 - σ)}$	$\frac{5831 + \sqrt{60001}}{8240} \leq σ \leq \frac{42}{55} = 0.7636 \dots$	Theorem 12.14, 12.10
$\frac{18 - 19 σ}{6 (15 σ - 11) (1 - σ)}$	$\frac{42}{55} \leq σ \leq \frac{97}{127} = 0.7637 \dots$	Theorem 12.11
$\frac{3 (18 - 19 σ)}{4 (4 σ - 1) (1 - σ)}$	$\frac{97}{127} \leq σ \leq \frac{79}{103} = 0.7669 \dots$	Theorem 12.11
$\frac{18 - 19 σ}{2 (37 σ - 27) (1 - σ)}$	$\frac{79}{103} \leq σ \leq \frac{33}{43} = 0.7674 \dots$	Theorem 12.12
$\frac{5 (18 - 19 σ)}{2 (13 σ - 3) (1 - σ)}$	$\frac{33}{43} \leq σ \leq \frac{84}{109} = 0.7706 \dots$	Theorem 12.12
$\frac{18 - 19 σ}{9 (3 σ - 2) (1 - σ)}$	$\frac{84}{109} \leq σ \leq \frac{1273 - \sqrt{128689}}{1184} = 0.7721 \dots$	Theorem 12.13
$\frac{4 (10 - 9 σ)}{5 (4 σ - 1) (1 - σ)}$	$\frac{1273 - \sqrt{128689}}{1184} \leq σ \leq \frac{5}{6} = 0.8333 \dots$	Theorem 12.7, 12.9, 12.13
$\frac{12}{4 σ - 1}$	$\frac{5}{6} \leq σ \leq 1$	Theorem 12.6

\includegraphics[width=0.5\textwidth ]{chapter/zero_density_energy_estimate.png} — Figure 12.1 Comparison of bounds on $A^{*} (σ)$ under various assumptions.