11 Zero density theorems

Definition 11.1 Zero density exponents

For $σ \in R$ and $T > 0$ , let $N (σ, T)$ denote the number of zeroes $ρ$ of the Riemann zeta function with $Re (ρ) \geq σ$ and $| Im (ρ) | \leq T$ .

If $1 / 2 \leq σ < 1$ is fixed, we define the zero density exponent $A (σ) \in [- \infty, \infty)$ to be the infimum of all (fixed) exponents $A$ for which one has

N (σ - δ, T) ≪ T^{A (1 - σ) + o (1)}

whenever $T$ is unbounded and $δ > 0$ is infinitesimal.

The shift by $δ$ is for technical convenience, it allows for $A (σ)$ to control (very slightly) the zeroes to the left of $Re s = σ$ . In non-asymptotic terms: $A (σ)$ is the infimum of all $A$ such that for every $ε > 0$ there exists $C, δ > 0$ such that

N (σ - δ, T) \leq C T^{A (1 - σ) + ε}

whenever $T \geq C$ .

Lemma 11.2 Basic properties of

A

$σ \mapsto (1 - σ) A (σ)$ is non-increasing and left-continuous, with $A (1 / 2) = 2$ .
If the Riemann hypothesis holds, then $A (σ) = - \infty$ for all $1 / 2 < σ \leq 1$ .

Proof ▶

Remark 11.3

One can ask what happens if one omits the $δ$ shift. Thus, define $A_{0} (σ)$ to be the infimum of all fixed exponents $A$ for which $N (σ, T) ≪ T^{A (1 - σ) + o (1)}$ for unbounded $T$ . Then it is not difficult to see that

lim_{σ^{'} \to σ^{+}} A (σ) \leq A_{0} (σ) \leq A (σ)

for any fixed $1 / 2 < σ < 1$ ; thus $A_{0}$ is basically the same exponent at $A$ , except possibly at jump discontinuities of the left-continuous function $A$ , in which case it could theoretically take on a different value. (But we do not expect such discontinuities to actually exist.) Thus there is not a major difference between $A (σ)$ and $A_{0} (σ)$ , but the former has some very slight technical advantages (such as the aforementioned left continuity).

The quantity $‖ A ‖_{\infty} := sup_{1 / 2 \leq σ < 1} A (σ)$ is of particular importance to the theory of primes in short intervals; see Section 13. From Lemma 11.2 we have $‖ A ‖_{\infty} \geq 2$ . It is conjectured that this is an equality.

Heuristic 11.4 Density hypothesis

One has $‖ A ‖_{\infty} = 2$ . Equivalently, $A (σ) \leq 2$ for all $1 / 2 \leq σ < 1$ .

Indeed, the Riemann hypothesis implies the stronger assertion that $A (σ) = - \infty$ for all $12 < σ < 1$ . However, for many applications to the prime numbers in short intervals, the density hypothesis is almost as powerful; see Section 13.

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

Lemma 11.5 Zero density from large values

Let $1 / 2 < σ < 1$ . Then

A (σ) (1 - σ) \leq max (sup_{τ \geq 2} {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 7.4(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 $N (σ - o (1), T) ≪ T^{B + O (ε) + o (1)}$ as $T \to \infty$ .

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

Suppose that one has a zero $σ_{r} + i t_{r}$ of this form. Then by a standard approximation to the zeta function [ 88 , Theorem 1.8 ] , one has

\sum_{n \leq T} \frac{1}{n^{σ_{r} + i t_{r}}} ≪ T^{- 1 / 2} .

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 $t_{r}$ by a factor of at most $2$ , we either have

| \sum_{T^{δ_{1}} \leq n \leq T} \frac{1}{n^{σ_{r} + i t_{r}}} | ≫ T^{- δ_{2}}

for all $r$ , or

\sum_{n \leq T^{δ_{1}}} \frac{1}{n^{σ_{r} + i t_{r}}} ≪ T^{- δ_{2}}

for all $r$ .

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

| \sum_{T^{δ_{1}} \leq n \leq T} \frac{ψ (n / N)}{n^{σ_{r} + i t_{r}}} | ≫ T^{- δ_{2} - o (1)}

for some fixed bump function $ψ$ supported on $[1 / 2, 1]$ , and some $T^{δ_{1}} ≪ N ≪ T$ .

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

| \sum_{n \in I} ψ (n / N) (n / N)^{- σ_{r}} n^{- i t_{r}} | ≫ N^{σ} T^{- δ_{2} - o (1)} .

Performing a Fourier expansion of $ψ (n / N) (n / N)^{- σ_{r}}$ in $\log n$ and using the triangle inequality, we can bound

\sum_{n \in I} ψ (n / N) (n / N)^{- σ_{r}} n^{- i t_{r}} ≪_{A} \int_{R} | \sum_{n \in I} \frac{1}{n^{i t}} | (1 + | t - t_{r} |)^{- A} d t

for any $A > 0$ , so by the triangle inequality we conclude that

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

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 $T = N^{τ + o (1)}$ for some $2 \leq τ \leq 1 / δ_{1}$ , then we conclude that

| \sum_{n \in I} \frac{1}{n^{i t_{r}^{'}}} | ≫ N^{σ - δ_{2} / δ_{1} + o (1)}

for all remaining $r$ . By Definition 8.1 we then have (for $δ_{2}$ small enough)

R ≪ N^{{LV}_{ζ} (σ, τ) + ε + o (1)} ≪ T^{{LV}_{ζ} (σ, τ) / τ + ε + o (1)}

and the claim follows in this case from 1.

In the case $N ≍ T$ , a standard application of the Euler–Maclaurin formula (see e.g., [ 168 , (2.1.2) ] ) yields

\sum_{T^{δ_{1}} \leq n \leq T} \frac{ψ (n / N)}{n^{σ_{r} + i t_{r}}} ≪ T^{- σ_{r}}

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

| \sum_{n} ψ (n / N) (n / N)^{- σ_{r}} n^{- i t_{r}} | ≫ N^{σ} T^{- δ_{2} - o (1)} .

Applying the van der Corput $B$ -process (see, e.g., [ 91 , § 8.3 ] ) or the approximate functional equation we have

\sum_{n} ψ (n / N) (n / N)^{- σ_{r}} n^{- i t_{r}} = e (\frac{t_{r}}{2 π} \log \frac{t_{r}}{2 π} - \frac{t_{r}}{2 π} + \frac{1}{8}) \sum_{m} ψ (2 π t_{r} / m N) (2 π t_{r} / m N)^{- σ_{r}} m^{i t_{r}} (2 π m^{2} / t_{r})^{- 1 / 2} + O (T^{o (1)})

and thus

| \sum_{m} ψ (2 π t_{r} / m N) (2 π t_{r} / m N)^{1 - σ_{r}} m^{- i t_{r}} | ≫ M^{1 / 2} N^{σ - 1 / 2} T^{- δ_{2} - o (1)};

where $M := 2 π T / N ≪ N^{1 / 2}$ . In particular

\sum_{m \in [M / 10, 10 M]} ψ (2 π t_{r} / m N) (2 π t_{r} / m N)^{1 - σ_{r}} m^{- i t_{r}} ≫ M^{σ} T^{- δ_{2} - o (1)};

since $N ≫ T^{1 / 2}$ and $σ \geq 1 / 2$ . Performing a Fourier expansion as before, we conclude that

\sum_{m \in [M / 10, 10 M]} m^{- i t_{r}^{'}} ≪ M^{σ} T^{- δ_{2} - o (1)}

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

Now suppose instead we are in the latter (“Type II”) case 3. We multiply both sides of 3 by the mollifier $\sum_{m \leq T^{δ_{2} / 2}} \frac{1}{m^{σ_{r} + i t_{r}}}$ to obtain

| 1 + \sum_{T^{δ_{2} / 2} \leq n \leq T^{δ_{1} + δ_{2} / 2}} \frac{a_{n}}{n^{σ_{r} + i t_{r}}} | = o (1)

where $a_{n}$ is some sequence with $a_{n} ≪ T^{o (1)}$ . 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 $[N, 2 N]$ with $T^{δ_{2} / 2} ≪ N ≪ T^{δ_{1} + δ_{2} / 2}$ such that

| \sum_{n \in I} \frac{a_{n}}{n^{σ_{r} + i t_{r}}} | ≫ T^{- o (1)}

and hence by Fourier expansion of $\frac{1}{n^{σ_{r}}}$ in $\log n$

| \sum_{n \in I} \frac{a_{n}}{n^{i t_{r}^{'}}} | ≫ N^{σ_{r}} T^{- o (1)}

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 $T = N^{τ + o (1)}$ for some $\frac{1}{δ_{1} + δ_{2} / 2} \leq τ \leq \frac{1}{δ_{2} / 2}$ . Applying Lemma 7.3, we conclude that

R ≪ N^{LV (σ, τ) + o (1)} = T^{LV (σ, τ) / τ + o (1)}

and the claim follows in this case from 2.

Recently, a partial converse to the above lemma was established:

Lemma 11.6 Large values from zero density

[ 125 , Theorem 1.2 ] If $τ > 0$ and $1 / 2 \leq σ \leq 1$ are fixed, then

{LV}_{ζ} (σ, τ) / τ \leq max (\frac{1}{2}, sup_{σ \leq σ^{'} \leq 1} A (σ^{'}) (1 - σ^{'}) + \frac{σ^{'} - σ}{2}) .

Proof ▶

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

| \sum_{n \in I} \frac{1}{n^{i t_{r}}} | ≫ N^{σ - o (1)}

uniformly for all $r$ . By [ 125 , Theorem 1.2 ] , we have for any fixed $δ > 0$ that

R ≪ T^{δ} sup_{σ - δ \leq σ^{'} \leq 1} T^{\frac{σ^{'} - σ}{2}} N (σ^{'}, O (T)) + T^{\frac{1 - σ}{2} + δ} .

Using Definition 11.1, we conclude that

R ≪ T^{max (\frac{1}{2}, sup_{σ - δ \leq σ^{'} \leq 1} A (σ^{'}) (1 - σ^{'}) + \frac{σ^{'} - σ}{2}) + O (δ)}

and thus

{LV}_{ζ} (σ, τ) \leq τ max (\frac{1}{2}, sup_{σ - ε \leq σ^{'} \leq 1} A (σ^{'}) (1 - σ^{'}) + \frac{σ^{'} - σ}{2}) + O (δ) .

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

The suprema in Lemma 11.5 require unbounded values of $τ$ , but thanks to the ability to raise to a power, we can reduce to a bounded range of $τ$ . Here is a basic such reduction, suited for machine-assisted proofs:

Corollary 11.7

Let $1 / 2 < σ < 1$ and $τ_{0} > 0$ . 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)

with the convention that the first supremum is $- \infty$ if it is vacuous (i.e., if $τ_{0} < 2$ ).

Implemented at zero_density_estimate.py as:
lv_zlv_to_zd(hypotheses, sigma_interval, tau0)

Proof ▶

Denote the right-hand side by $B$ , thus

LV (σ, τ) \leq B τ

for all $τ_{0} \leq τ \leq 2 τ_{0}$ , and

{LV}_{ζ} (σ, τ) \leq B τ

whenever $2 \leq τ < τ_{0}$ . From Lemma 7.8 we then have

LV (σ, τ) \leq B τ

for all $k τ_{0} \leq τ \leq 2 k τ_{0}$ and natural numbers $k$ . Note that the intervals $[k τ_{0}, 2 k τ_{0}]$ cover all of $[τ_{0}, \infty)$ , hence we have

LV (σ, τ) \leq B τ

for all $τ \geq τ_{0}$ . In particular

\underset{τ \to \infty}{lim sup} LV (σ, τ) / τ \leq B .

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

{LV}_{ζ} (σ, τ) \leq B τ

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

{LV}_{ζ} (\frac{1}{2} + \frac{1}{τ - 1} (σ - \frac{1}{2}), \frac{τ}{τ - 1}) \leq B \frac{τ}{τ - 1}

for $τ \geq 2$ . Thus

sup_{τ \geq 2} \frac{{LV}_{ζ} (σ, τ)}{τ} \leq B .

The claim now follows from Lemma 11.5.

For machine assisted proofs, one can simply take $τ_{0}$ to be a sufficiently large quantity, e.g., $τ_{0} = 3$ for $σ$ not too close to $1$ , and larger for $σ$ approaching $1$ , to recover the full power of Lemma 11.5. However, the amount of case analysis required increases with $τ_{0}$ . For human written proofs, the following version of Corollary 11.7 is more convenient:

Corollary 11.8

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

A (σ) (1 - σ) \leq max (sup_{2 \leq τ < 4 τ_{0} / 3} {LV}_{ζ} (σ, τ) / τ, sup_{2 τ_{0} / 3 \leq τ \leq τ_{0}} LV (σ, τ) / τ) .

Implemented at zero_density_estimate.py as:
lv_zlv_to_zd(hypotheses, sigma_interval, tau0)

Proof ▶

Applying Corollary 11.7 with $τ$ replaced by $4 τ_{0} / 3$ , it suffices to show that

sup_{4 τ_{0} / 3 \leq τ \leq 8 τ_{0} / 3} LV (σ, τ) / τ \leq sup_{2 τ_{0} / 3 \leq τ \leq τ_{0}} LV (σ, τ) / τ .

But this follows from Lemma 7.8, since the intervals $[2 k τ_{0} / 3, k τ_{0}]$ for $k = 2, 3$ cover all of $[4 τ_{0} / 3, 8 τ_{0} / 3]$ .

The following special case of the above corollary is frequently used in practice to assist with the human readability of zero density proofs:

Corollary 11.9

Let $1 / 2 < σ < 1$ and $τ_{0} > 0$ . Suppose that one has the bounds

LV (σ, τ) \leq (3 - 3 σ) \frac{τ}{τ_{0}}

for $2 τ_{0} / 3 \leq τ \leq τ_{0}$ , and

{LV}_{ζ} (σ, τ) \leq (3 - 3 σ) \frac{τ}{τ_{0}}

for $2 \leq τ < 4 τ_{0} / 3$ . Then $A (σ) \leq \frac{3}{τ_{0}}$ .

The reason why this particular special case is convenient is because the inequality

2 - 2 σ \leq (3 - 3 σ) \frac{τ}{τ_{0}}

obviously holds for $τ \geq 2 τ_{0} / 3$ . That is to say, we automatically verify 6 in regimes where the Montgomery conjecture holds. In fact, we can do a bit better, thanks to subdivision:

Corollary 11.10

Let $1 / 2 < σ < 1$ and $τ_{0} > 0$ . Suppose that one has the bound 7 for $2 \leq τ < 4 τ_{0} / 3$ , and the Montgomery conjecture $LV (σ, τ) \leq 2 - 2 σ$ whenever $0 \leq τ \leq τ_{0} + σ - 1$ . Then $A (σ) \leq \frac{3}{τ_{0}}$ .

Proof ▶

We may assume that $τ_{0} \geq 3 - 3 σ$ , since otherwise the claim follows from the Riemann–von Mangoldt bound

A (σ) (1 - σ) \leq A (1 / 2) (1 - 1 / 2) = 1.

By Lemma 7.4(ii) we have

LV (σ, τ) \leq max (2 - 2 σ, 3 - 3 σ + τ - τ_{0})

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

11.1 Known zero density bounds

Let us see some examples of these corollaries in action.

Theorem 11.11

The Montgomery conjecture implies the density hypothesis.

Proof ▶

Apply Corollary 11.9 with $τ_{0} = 3 / 2$ (so that 7 is vacuously true).

Theorem 11.12

The Lindelof hypothesis implies the density hypothesis, and also that $A (σ) \leq 0$ for $3 / 4 < σ \leq 1$ .

Proof ▶

The first result is proved in [ 83 ] , and the second result is due to [ 45 ] . We will apply Corollary 11.8. From Corollary 8.8 we see that ${LV}_{ζ} (σ, τ) = - \infty$ whenever $σ > 1 / 2$ and $τ \geq 1$ , so for any choice of $τ_{0}$ we have

sup_{2 \leq τ < 4 τ_{0} / 3} {LV}_{ζ} (σ, τ) / τ = - \infty .

From Theorem 7.9 and Lemma 7.8 we have

LV (σ, τ) \leq max ((2 - 2 σ) k, τ + (1 - 2 σ) k)

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

LV (σ, τ) \leq (2 - 2 σ) τ + O (1),

and hence by taking $τ_{0}$ large enough, we can make $sup_{2 τ_{0} / 3 \leq τ \leq τ_{0}} LV (σ, τ) / τ$ bounded by $2 - 2 σ + ε$ for any $ε > 0$ . This already gives the density hypothesis bound $A (σ) \leq 2$ . For $σ > 3 / 4$ , we may additionally apply Theorem 8.14 to make $sup_{2 τ_{0} / 3 \leq τ \leq τ_{0}} LV (σ, τ) / τ$ arbitrarily small, giving the bound $A (σ) \leq 0$ .

There are similar results assuming weaker versions of the Lindelof hypothesis. For instance, we have

Theorem 11.13 Ingham’s first bound

[ 82 ] (See also [ 168 ] ) For any $1 / 2 < σ < 1$ , we have

A (σ) \leq 2 + 4 μ (1 / 2) .

Proof ▶

We give here a proof (somewhat different from the original proof) that passes through Corollary 11.7. We apply Corollary 11.7 with $τ_{0}$ chosen so that $μ (1 / 2) τ_{0} < σ - 1 / 2$ . From Corollary 8.6 we then have

A (σ) (1 - σ) \leq sup_{τ_{0} \leq τ \leq 2 τ_{0}} LV (σ, τ) / τ .

For any integer $k \geq 0$ and $k \leq τ \leq k + 1$ , we see from 9 that

LV (σ, τ) \leq (2 - 2 σ) (k + 1)

and

LV (σ, τ) \leq τ + (1 - 2 σ) k;

multiplying the first inequality by $2 σ - 1$ , the second by $2 - 2 σ$ , and summing, we conclude that

LV (σ, τ) \leq (τ + 2 σ - 1) (2 - 2 σ);

inserting this bound we have

A (σ) \leq 2 + \frac{2 σ - 1}{τ_{0}} .

Optimizing in $τ_{0}$ , we obtain the claim.

Theorem 11.14 Ingham’s second bound

[ 83 ] For any $1 / 2 < σ < 1$ , one has $A (σ) \leq \frac{3}{2 - σ}$ .

Recorded in literature.py as:
add_zero_density_ingham_1940()
Derived in derived.py as:
prove_zero_density_ingham_1940()
prove_zero_density_ingham_1940_v2()

Proof ▶

We apply Corollary 11.10 with $τ_{0} := 2 - σ$ . Here we have $4 τ_{0} / 3 < 2$ since $σ > 1 / 2$ , so the claim 7 is automatic; and the Montgomery conjecture hypothesis follows from Theorem 7.9.

Either of Theorem 11.13 or Theorem 11.14 implies an older result of Carlson [ 18 ] that $A (σ) \leq 4 σ$ for $1 / 2 < σ < 1$ . Recorded in literature.py as:
add_zero_density_carlson_1921()

Theorem 11.15 Huxley bound

[ 68 ] For any $1 / 2 < σ < 1$ , one has $A (σ) \leq \frac{3}{3 σ - 1}$ . (In particular, the density hypothesis holds for $σ \geq 5 / 6$ .)

Recorded in literature.py as:
add_zero_density_huxley_1972()
Derived in derived.py as:
prove_zero_density_huxley_1972()
prove_zero_density_huxley_1972_v2()

Proof ▶

We apply Corollary 11.10 with $τ_{0} := 3 σ - 1$ . The Montgomery conjecture hypothesis follows from Theorem 7.12. So it remains to show that 7 holds for $2 \leq τ < 4 τ_{0} / 3$ . For $σ \leq 5 / 6$ we have $4 τ_{0} / 3 \leq 2$ , so the claim is vacuously true in this case. For $σ > 5 / 6$ we use Corollary 8.6 and the bound $μ (1 / 2) \leq 1 / 6$ from Table 6.1 to conclude that ${LV}_{ζ} (σ, τ) = - \infty$ whenever $σ > 1 / 2 + τ / 6$ , but this is precisely $τ < 6 σ - 3$ . Since $6 σ - 3 > 4 τ_{0} / 3$ when $σ > 5 / 6$ , we obtain the claim.

Theorem 11.16 Guth–Maynard bound

For any $1 / 2 < σ < 1$ , one has $A (σ) \leq \frac{15}{3 + 5 σ}$ .

Recorded in literature.py as:
add_zero_density_guth_maynard_2024()
Derived in derived.py as:
prove_zero_density_guth_maynard_2024()
prove_zero_density_guth_maynard_2024_v2()

Proof ▶

We may assume that $7 / 10 < σ < 8 / 10$ , since the bound follows from the Ingham and Huxley bounds otherwise. We apply Corollary 11.9 with $τ_{0} := \frac{3 + 5 σ}{5}$ . We have $4 τ_{0} / 3 < 2$ , so the claim 7 is vacuous and we only need to establish 6. We split into the subranges $13 / 5 - 2 σ \leq τ < τ_{0}$ and $2 τ_{0} / 3 \leq τ \leq 13 / 5 - 2 σ$ . In the former range we use Theorem 10.27 (and 8), and reduce to showing that

18 / 5 - 4 σ \leq (3 - 3 σ) \frac{τ}{τ_{0}},

and

τ + 12 / 5 - 4 σ \leq (3 - 3 σ) \frac{τ}{τ_{0}}

for $13 / 5 - 2 σ \leq τ < τ_{0}$ . The first inequality follows from

18 / 5 - 4 σ \leq (3 - 3 σ) \frac{13 / 5 - 2 σ}{τ_{0}}

which one can numerically check holds in the range $7 / 10 < σ < 8 / 10$ . Finally, the third inequality is obeyed with equality when $τ = τ_{0}$ and the right-hand side has a larger slope in $τ$ than the left (since $τ_{0} \geq 3 - 3 σ$ ), so the claim follows as well.

In the remaining region $2 τ_{0} / 3 \leq τ \leq 13 / 5 - 2 σ$ , we use Theorem 7.9 and 8 to reduce to showing that

τ + 1 - 2 σ \leq (3 - 3 σ) \frac{τ}{τ_{0}}

in this range. This follows again from 10 which guarantees the inequality at the right endpoint $τ = 13 / 5 - 2 σ$ .

Theorem 11.17 Jutila zero density theorem

[ 102 ] The zero density hypothesis is true for $σ \geq 11 / 14$ .

Derived in derived.py as:
prove_zero_density_jutila_1977()
prove_zero_density_jutila_1977_v2()

Proof ▶

We apply Corollary 11.8 with $τ_{0} := 3 / 2$ , then it suffices to show that

LV (σ, τ) \leq (2 - 2 σ) τ

for all $1 \leq τ \leq 3 / 2$ .

From the $k = 3$ case of Theorem 7.16 we have

LV (σ, τ) \leq max (2 - 2 σ, τ + \frac{10 - 16 σ}{3}, τ + 18 - 24 σ) .

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

In fact, we can do better:

Theorem 11.18 Heath-Brown zero density theorem

[ 59 ] For $11 / 14 \leq σ < 1$ , one has $A (σ) \leq \frac{9}{7 σ - 1}$ (in particular, this recovers Theorem 11.17). For any $3 / 4 \leq σ \leq 1$ , one has $A (σ) \leq max (\frac{3}{10 σ - 7}, \frac{4}{4 σ - 1})$ (which is a superior bound when $σ \geq 20 / 23$ ).

Recorded in literature.py as:
add_zero_density_heathbrown_1979()
Derived in derived.py as:
prove_zero_density_heathbrown_1979a()
prove_zero_density_heathbrown_1979b()
prove_zero_density_heathbrown_1979a_v2()
prove_zero_density_heathbrown_1979b_v2()

Proof ▶

For the first estimate, we apply Corollary 11.9 with $τ_{0} := \frac{7 σ - 1}{3}$ . To verify 6, we apply the $k = 3$ version of Theorem 7.16, which gives

LV (σ, τ) \leq max (2 - 2 σ, τ + \frac{10 - 16 σ}{3}, τ + 18 - 24 σ) .

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

τ + \frac{10 - 16 σ}{3} \leq (3 - 3 σ) \frac{τ}{τ_{0}}

for $2 τ_{0} / 3 \leq τ \leq τ_{0}$ . This holds with equality at $τ = τ_{0}$ , hence holds for $τ \leq τ_{0}$ as well since $τ_{0} \geq 3 - 3 σ$ . As for 7, we invoke Theorem 9.6 and reduce to showing that

2 τ + 6 - 12 σ \leq (3 - 3 σ) \frac{τ}{τ_{0}}

for $2 \leq τ \leq 4 τ_{0} / 3$ . Since $6 - 12 σ$ is negative, the ratio of the left-hand side and right-hand side is increasing in $τ$ , so it suffices to verify this claim at the endpoint $τ = 4 τ_{0} / 3$ . The claim then simplifies to $τ_{0} \leq \frac{3}{4} (4 σ - 1)$ , which one can verify from the choice of $τ_{0}$ and the hypothesis $σ \geq 11 / 14$ .

For the second estimate, we take $τ_{0} := min (10 σ - 7, \frac{3}{4} (4 σ - 1))$ . To verify 6, we now use Theorem 7.14 and 8, and reduce to showing that

τ + 10 - 13 σ \leq (3 - 3 σ) \frac{τ}{τ_{0}}

for $2 τ_{0} / 3 \leq τ \leq τ_{0}$ . The inequality holds at $τ = τ_{0}$ since $τ_{0} \leq 10 σ - 7$ , and hence for all smaller $τ$ since $τ_{0} \geq 3 - 3 σ$ . As for 7, we can repeat the previous arguments since $τ_{0} \leq \frac{3}{4} (4 σ - 1)$ .

With the aid of computer assistance, we were able to strengthen the second claim here. We first need a lemma:

Lemma 11.19

$(3 / 40, 31 / 40)$ is an exponent pair. In particular, by Corollary 6.8, $μ (7 / 10) \leq 3 / 40$ .

Derived in derived.py as:
best_proof_of_exponent_pair(frac(3,40), frac(31,40))

Proof ▶

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

(3 / 40, 31 / 40) = x y A W + (1 - x) y A B A W + (1 - y) W

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 $β (α) \leq 3 / 40 + 7 α / 10$ can be derived from Theorem 4.16 (as well as the classical bounds in Corollary 4.8). We also note that the corollary $μ (7 / 10) \leq 3 / 40 = 0.075$ is immediate from [ 169 , Theorem 2.4 ] , which in fact gives the slightly stronger bound $μ (7 / 10) \leq 218 / 3005 = 0.07254 \dots$ .

Theorem 11.20 Improved Heath-Brown zero density theorem

For any $7 / 10 < σ \leq 1$ , one has $A (σ) \leq \frac{3}{10 σ - 7}$ .

Derived in derived.py as:
prove_zero_density_heathbrown_extended()

Proof ▶

We apply Corollary 11.10 with $τ_{0} := 10 σ - 7$ . 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 ${LV}_{ζ} (σ, τ) = - \infty$ whenever $σ > \frac{3}{40} τ + \frac{7}{10}$ , or equivalently $τ < \frac{4}{3} (10 σ - 7)$ , which then immediately gives 7.

Theorem 11.21 Bourgain result on density hypothesis

The density hypothesis holds for $σ > 25 / 32$ .

Recorded in literature.py as:
add_zero_density_bourgain_2000()

Proof ▶

The arguments below are a translation of the original arguments of Bourgain [ 16 ] to our notational framework.

In view of Theorem 11.17 (or Theorem 11.18), we may assume that $25 / 32 < σ < 11 / 14$ . Set $ρ := LV (σ, τ)$ . As in the proof of Theorem 11.17, it suffices to show that

ρ \leq (2 - 2 σ) τ

for all $1 \leq τ \leq 3 / 2$ .

From the $k = 3$ case of Theorem 7.16 we have

ρ \leq max (2 - 2 σ, τ + \frac{10 - 16 σ}{3}, τ + 18 - 24 σ)

which in the $σ < 11 / 14$ regime simplifies to

ρ \leq max (2 - 2 σ, τ + 18 - 24 σ)

and this already suffices unless

τ \geq \frac{24 σ - 18}{2 σ - 1} .

In the regime $σ > 25 / 32$ and $τ \leq 3 / 2$ , the bound 13 certainly implies

ρ \leq min (1, 4 - 2 τ)

and also

max (1, 2 τ - 2) = 1

so we may invoke Corollary 10.30 to conclude that

ρ \leq max (α_{2} + 2 - 2 σ, α_{1} + α_{2} / 2 + 2 - 2 σ, - α_{2} + 2 τ + 4 - 8 σ, - 2 α_{1} + τ + 12 - 16 σ, 4 α_{1} + 3 - 4 σ)

for any $α_{1}, α_{2} \geq 0$ .

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

Now suppose that $τ > \frac{4 (1 + σ)}{5}$ . In this case we choose $α_{1} = \frac{τ}{8} - \frac{9 σ - 7}{2}$ and $α_{2} = \frac{5 τ}{4} - (1 + σ)$ , which can be checked to be nonnegative using the hypotheses on $σ, τ$ . In this case one can again check that 15 implies 11.

We can improve this bound as follows:

Theorem 11.22 Improved Bourgain density hypothesis bound

For $17 / 22 \leq σ \leq 4 / 5$ , one has $A (σ) \leq max (\frac{2}{9 σ - 6}, \frac{9}{8 (2 σ - 1)})$ . Thus one has $A (σ) \leq \frac{9}{8 (2 σ - 1)}$ for $38 / 49 \leq σ \leq 4 / 5$ and $A (σ) \leq \frac{2}{9 σ - 6}$ for $17 / 22 \leq σ \leq 38 / 49$ .

Derived in derived.py as:
prove_zero_density_bourgain_improved()

The arguments can be pushed to some $σ$ below $17 / 22$ , but in that range the estimate in Corollary 11.31 becomes superior, so we do not pursue this further.

Proof ▶

We apply Corollary 11.10 with $τ_{0} := min (\frac{9 (3 σ - 2)}{2}, \frac{8 (2 σ - 1)}{3})$ . For future reference we note that

\frac{1}{3} < \frac{3 - 3 σ}{τ_{0}} < \frac{1}{2} .

It suffices to show that

LV (σ, τ) \leq \frac{τ}{τ_{0}} (3 - 3 σ)

for $2 τ_{0} / 3 \leq τ \leq τ_{0}$ , as well as

{LV}_{ζ} (σ, τ) \leq \frac{τ}{τ_{0}} (3 - 3 σ)

for $2 \leq τ < 4 τ_{0} / 3$ . For 18 we use the twelfth moment bound in Theorem 9.6. Since the slope of $2 τ - 12 (σ - 1 / 2)$ in $τ$ exceeds that of $\frac{τ}{τ_{0}} (3 - 3 σ)$ by 16, it suffices to check the bound at the endpoint, i.e., to show that

8 τ_{0} / 3 - 12 (σ - 1 / 2) \leq 4 - 4 σ

or equivalently $τ_{0} \leq \frac{3 (4 σ - 1)}{4}$ , which one can easily check to be the case.

Now we prove 17. Set $ρ := LV (σ, τ)$ . From the $k = 3$ case of Theorem 7.16 we again have 12, which implies the required bound $ρ \leq \frac{τ}{τ_{0}} (3 - 3 σ)$ unless one has

τ \geq - max (\frac{10 - 16 σ}{3}, 18 - 24 σ) / (1 - \frac{3 - 3 σ}{τ_{0}})

In this regime, one can also check from 12 that

ρ \leq min (1, 4 - 2 τ)

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

\begin{aligned} ρ & \leq max (α_{2} + 2 - 2 σ, α_{1} + α_{2} / 2 + 2 - 2 σ, - α_{2} + 2 τ + 4 - 8 σ, \\ - 2 α_{1} + τ + 12 - 16 σ, 4 α_{1} + 2 + max (1, 2 τ - 2) - 4 σ) \end{aligned}

for any $α_{1}, α_{2} \geq 0$ .

We first consider the case when $38 / 49 \leq σ \leq 4 / 5$ , so that $τ_{0} = 8 (2 σ - 1) / 3$ . As in the proof of Theorem 11.21, we set

α_{2} := max (\frac{5 τ}{4} - (1 + σ), 0)

and

α_{1} := \frac{τ}{3} - \frac{2}{3} (7 σ - 5) - α_{2} / 6.

With this choice, the expressions $α_{1} + α_{2} / 2 + 2 - 2 σ$ and $- 2 α_{1} + τ + 12 - 16 σ$ are both equal to $\frac{τ}{3} + \frac{16 - 20 σ}{3} + \frac{α_{2}}{3}$ , while $- α_{2} + 2 τ + 4 - 8 σ$ is equal to

\frac{τ}{3} + \frac{16 - 20 σ}{3} + \frac{α_{2}}{3} - \frac{4}{3} (α_{2} - \frac{5 τ}{4} - (1 + σ))

which is less than or equal to the previous expression by definition of $α_{2}$ . Finally, the expression $4 α_{1} + 1 + max (2, 2 τ - 2) - 4 σ$ is equal to

\frac{4 τ}{3} + \frac{46 - 68 σ}{3} + max (1, 2 τ - 2) - \frac{2 α_{2}}{3} .

Thus it remains to show the bounds

α_{2} + 2 - 2 σ \leq \frac{τ}{τ_{0}} (3 - 3 σ)

\frac{τ}{3} + \frac{16 - 20 σ}{3} + \frac{α_{2}}{3} \leq \frac{τ}{τ_{0}} (3 - 3 σ)

and

\frac{4 τ}{3} + \frac{46 - 68 σ}{3} + max (1, 2 τ - 2) - \frac{2 α_{2}}{3} \leq \frac{τ}{τ_{0}} (3 - 3 σ) .

For 23, we trivially have $2 - 2 σ \leq \frac{τ}{τ_{0}} (3 - 3 σ)$ since $τ \geq 2 τ_{0} / 3$ , and the slope of $\frac{5 τ}{4} - (1 + σ) + 2 - 2 σ$ in $τ$ certainly exceeds $\frac{3 - 3 σ}{τ_{0}}$ by 16, so it suffices to check the endpoint

\frac{5 τ_{0}}{4} - (1 + σ) + 2 - 2 σ \leq 3 - 3 σ

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

\frac{τ_{0}}{3} + \frac{16 - 20 σ}{3} + \frac{\frac{5 τ_{0}}{4} - (1 + σ)}{3} \leq 3 - 3 σ

and

\frac{τ}{3} + \frac{16 - 20 σ}{3} + \frac{0}{3} \leq 2 - 2 σ .

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

\frac{4 τ}{3} + \frac{46 - 68 σ}{3} + max (1, 2 τ_{0} - 2) - \frac{2 (\frac{5 τ}{4} - (1 + σ))}{3} \leq \frac{τ}{τ_{0}} (3 - 3 σ)

which by 16 would follow from

\frac{4 τ_{0}}{3} + \frac{46 - 68 σ}{3} + max (1, 2 τ_{0} - 2) - \frac{2 (\frac{5 τ_{0}}{4} - (1 + σ))}{3} \leq 3 - 3 σ

and one can check that this applies for $σ \geq 38 / 49$ .

Now suppose that we are in the case $16 / 21 \leq σ \leq 38 / 49$ , so that $τ_{0} = \frac{9 (3 σ - 2)}{2} \leq \frac{72}{49} < \frac{3}{2}$ (so in particular $max (1, 2 τ - 2) = 1$ for $τ \leq τ_{0}$ ). We set

α_{2} := max (11 - 16 σ + τ, 0)

and

α_{1} := \frac{τ}{3} - \frac{2}{3} (7 σ - 5) - α_{2} / 6.

Note that for $σ \leq 38 / 49$ , one has

(5 τ / 4 - (1 + σ)) - (11 - 16 σ + τ) \leq 15 σ - 12 - τ_{0} / 4 \leq 0

and hence

α_{2} \geq 5 τ / 4 - (1 + σ) .

As before, we conclude that the quantities $α_{1} + α_{2} / 2 + 2 - 2 σ$ and $- 2 α_{1} + τ + 12 - 16 σ$ are both equal to $\frac{τ}{3} + \frac{16 - 20 σ}{3} + \frac{α_{2}}{3}$ , while $- α_{2} + 2 τ + 4 - 8 σ$ 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

11 - 16 σ + τ_{0} + 2 - 2 σ \leq 3 - 3 σ

which one can check to be true (with room to spare) for $σ \geq 16 / 21$ . For 24, we use 16 as before to reduce to showing that

\frac{τ_{0}}{3} + \frac{16 - 20 σ}{3} + \frac{11 - 16 σ + τ}{3} \leq 3 - 3 σ

and

\frac{τ}{3} + \frac{16 - 20 σ}{3} + \frac{0}{3} \leq 2 - 2 σ .

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

\frac{4 τ}{3} + \frac{46 - 68 σ}{3} + 1 - \frac{2 (11 - 16 σ + τ)}{3} \leq \frac{τ}{τ_{0}} (3 - 3 σ)

which by 16 follows from

\frac{4 τ_{0}}{3} + \frac{46 - 68 σ}{3} + 1 - \frac{2 (11 - 16 σ + τ_{0})}{3} \leq 3 - 3 σ,

but this is an identity.

Theorem 11.23 Bourgain zero density theorem

[ 15 , Proposition 3 ] Let $(k, ℓ)$ be an exponent pair with $k < 1 / 5$ , $ℓ > 3 / 5$ , and $15 ℓ + 20 k > 13$ . Then, for any $σ > \frac{ℓ + 1}{2 (k + 1)}$ , one has

A (σ) \leq \frac{4 k}{2 (1 + k) σ - 1 - ℓ}

assuming either that $k < \frac{11}{85}$ , or that $\frac{11}{85} < k < \frac{1}{5}$ and $σ > \frac{144 k - 11 ℓ - 11}{170 k - 22}$ .

Corollary 11.24 Special case of Bourgain’s zero density theorem

[ 15 , Corollary 4 ] One has

A (σ) \leq \frac{4}{30 σ - 25}

for $\frac{15}{16} \leq σ \leq 1$ and

A (σ) \leq \frac{2}{7 σ - 5}

for $\frac{17}{19} \leq σ \leq \frac{15}{16}$ .

Recorded in literature.py as:
add_zero_density_bourgain_1995()

Proof ▶

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.

It was remarked in [ 15 ] that further zero density estimates could be obtained by using additional exponent pairs. This we do here:

Corollary 11.25 Optimized Bourgain zero density bound

One has

A (σ) \leq {\begin{cases} \frac{11}{12 (4 σ - 3)} & \frac{3}{4} < σ \leq \frac{14}{15}, \\ \frac{391}{2493 σ - 2014} & \frac{14}{15} < σ \leq \frac{2841}{3016}, \\ \frac{22232}{163248 σ - 134765} & \frac{2841}{3016} < σ \leq \frac{859}{908}, \\ \frac{356}{2742 σ - 2279} & \frac{859}{908} < σ \leq \frac{1625}{1692}, \\ \frac{2609588}{20732766 σ - 17313767} & \frac{1625}{1692} < σ \leq \frac{3334585}{3447984}, \\ \frac{75872}{9 (81024 σ - 69517)} & \frac{3334585}{3447984} < σ \leq \frac{974605}{1005296}, \\ \frac{288}{3616 σ - 3197} & \frac{974605}{1005296} < σ \leq \frac{5857}{6032}, \\ \frac{86152}{1447460 σ - 1311509} & \frac{5857}{6032} < σ < 1. \end{cases}

Implemented at zero_density_estimate.py as:
bourgain_ep_to_zd()

Proof ▶

Let $S (σ)$ denote the closure of the region

\begin{array}{r} {(k, ℓ) : 0 < k < \frac{1}{5}, \frac{3}{5} < ℓ < 1, 15 ℓ + 20 k > 13, \frac{ℓ + 1}{2 (k + 1)} < σ, \\ k < \frac{11}{85} or (k > \frac{11}{85} and \frac{144 k - 11 ℓ - 11}{170 k - 22} < σ)} \end{array}

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

min_{(k, ℓ) \in S (σ) \cap H} \frac{4 k}{2 (1 + k) σ - 1 - ℓ}

is a convex optimisation problem for each $3 / 4 < σ < 1$ . We take the following choices of $(k, ℓ)$ (found with the aid of computer assistance).

(\frac{11}{85} - ε, \frac{59}{85} + 2 ε), (\frac{391}{4595} + ε, \frac{3461}{4595}), (\frac{2779}{38033} + ε, \frac{58699}{76066}), (\frac{89}{1282} + ε, \frac{997}{1282}),

(\frac{652397}{9713986} + ε, \frac{7599781}{9713986}), (\frac{2371}{43205} + ε, \frac{280013}{345640}), (\frac{9}{217} + ε, \frac{1461}{1736}), (\frac{10769}{351096} + ε, \frac{609317}{702192}) .

Of these exponent pairs:

$(\frac{11}{85}, \frac{59}{85})$ is the intersection of the lines $k = 1 / 5$ and $15 ℓ + 20 k = 13$ ;
$(\frac{391}{4595}, \frac{3461}{4595})$ is an intersection of the line $15 ℓ + 20 k = 13$ 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 $σ = 1$ by choosing other exponent pairs, however it turns out such results are superseded by other zero density estimates.

Lemma 11.26 1980 Ivic zero density bound

[ 87 ] , [ 88 , Theorem 11.2 ] We have

A (σ) \leq \frac{4}{2 σ + 1}

for $17 / 18 \leq σ \leq 1$ , and

A (σ) \leq \frac{24}{30 σ - 11}

for $155 / 174 \leq σ \leq 17 / 18$ .

Recorded in literature.py as:
add_zero_density_ivic_1980()

Proof ▶

From Lemma 8.16 we have

LV (σ, τ) \leq max (2 - 2 σ, τ + 9 - 12 σ, τ - \frac{84 σ - 65}{6})

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

{LV}_{ζ} (σ, τ) \leq max (τ + (3 - 6 σ), 3 τ + 19 (1 / 2 - σ)) .

We apply Corollary 11.9 with $τ_{0} := max (\frac{30 σ - 11}{8}, \frac{6 σ + 3}{4})$ , and reduce to showing that 6 for $2 τ_{0} / 3 \leq τ \leq τ_{0}$ and 7 for $2 \leq τ < 4 τ_{0} / 3$ . But this follows from the preceding estimates after routine calculations.

One can also use bounds on $μ$ to obtain zero density theorems:

Lemma 11.27 Zero density from

μ

bound

[ 130 , Theorem 12.3 ] If $1 / 2 \leq α \leq 1$ and $\frac{α + 1}{2} \leq σ \leq 1$ , then

A (σ) \leq μ (α) \frac{2 (3 σ - 1 - 2 α)}{(2 σ - 1 - α) (σ - α)} .

Corollary 11.28 1971 Montgomery zero density bound

[ 130 ] , [ 88 , Theorem 11.3 ] For any $9 / 10 \leq σ \leq 1$ and $1 / 2 \leq α \leq 1$ one has

A (σ) (1 - σ) \leq \frac{7}{6} μ (5 σ - 4) .

In particular, for $152 / 155 \leq σ \leq 1$ , one has

A (σ) \leq min (35 / 36, 1600 (1 - σ)^{1 / 2}) .

Recorded in literature.py as:
add_zero_density_montgomery_1971()

Proof ▶

Apply the previous lemma with $α = 5 σ - 4$ .

Lemma 11.29 Preliminary large values estimate

If $m \geq 2$ is an integer, $3 / 4 < σ \leq 1$ , and $(k, ℓ)$ is an exponent pair, then

LV (σ, τ) \leq max (2 - 2 σ, m (2 - 4 σ) + m τ, min (X, Y))

where

X := 2 τ / 3 + 4 m (3 - 4 σ) / 3

and

Y := max (τ + 3 m (3 - 4 σ), (k + ℓ) τ / k + k (1 + 2 k + 2 ℓ) (3 - 4 σ) / k) .

Proof ▶

Lemma 11.30 General zero density estimate

[ 88 , (11.76), (11.77) ] If $(k, ℓ)$ is an exponent pair, and $m \geq 2$ an integer, then

A (σ) \leq \frac{3 m}{(3 m - 2) σ + 2 - m}

whenever

\begin{aligned} σ & \geq min (\frac{6 m^{2} - 5 m + 2}{8 m^{2} - 7 m + 2}, \\ max (\frac{9 m^{2} - 4 m + 2}{12 m^{2} - 6 m + 2}, \frac{3 m^{2} (1 + 2 k + 2 ℓ) - (4 k + 2 ℓ) m + 2 k + 2 ℓ}{4 m^{2} (1 + 2 k + 2 ℓ) - (6 k + 4 ℓ) m + 2 k + 2 ℓ})) . \end{aligned}

Implemented at zero_density_estimate.py as:
ivic_ep_to_zd(exp_pairs, m=2)

Proof ▶

With the hypothesis on $σ$ , one sees from Lemma 11.29 that

LV (σ, τ) \leq max (2 - 2 σ, τ - \frac{(4 m - 2) σ + 2 - 2 m}{m} + 2 - 2 σ)

for $0 \leq τ < \frac{(4 m - 2) σ + 2 - 2 m}{m}$ , and hence for all $τ \geq 0$ by by Lemma 7.4(ii). Meanwhile, from Theorem 9.6 one has

{LV}_{ζ} (σ, τ) \leq 2 τ - 12 (σ - 1 / 2)

for all $τ \geq 2$ . The claim then follows from Corollary 11.9 with $τ_{0} := \frac{(3 m - 2) σ + 2 - m}{m}$ after a routine calculation.

Corollary 11.31 1980-1984 Ivic zero density bound

[ 87 ] , [ 88 , Theorem 11.4 ] One can bound $A (σ$ ) by

\begin{aligned} \frac{3}{2 σ} & for \frac{3831}{4791} \leq σ \leq 1; \\ \frac{9}{7 σ - 1} & for \frac{41}{53} \leq σ \leq 1; \\ \frac{6}{5 σ - 1} & for \frac{13}{17} \leq σ \leq 1; \\ \frac{15}{13 σ - 3} & for \frac{127}{167} \leq σ \leq 1; \\ \frac{9}{8 σ - 2} & for \frac{47}{62} \leq σ \leq 1; \end{aligned}

Recorded in literature.py as:
add_zero_density_ivic_1980()
add_zero_density_ivic_1984()
Derived in derived.py as:
prove_zero_density_ivic_1984()

Proof ▶

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

The first bound has been improved:

Theorem 11.32 2000 Bourgain zero density theorem

[ 13 ] One has $A (σ) \leq 3 / 2 σ$ for $3734 / 4694 \leq σ \leq 1$ .

Recorded in literature.py as:
add_zero_density_bourgain_2002()

Lemma 11.33 Preliminary large values theorem

If $1 / 2 \leq σ \leq 1$ and $τ < 8 σ - 5$ , then

LV (σ, τ) \leq max (2 - 2 σ, 6 τ / 5 + (20 - 32 σ) / 5) .

Proof ▶

Corollary 11.34 Zero density estimates for

σ

close to

3 / 4

[ 88 , Theorem 11.5 ] One has $A (σ) \leq \frac{3}{7 σ - 4}$ for $3 / 4 \leq σ \leq 10 / 13$ , and $A (σ) \leq \frac{9}{8 σ - 2}$ for $10 / 13 \leq σ \leq 1$ .

Proof ▶

For $3 / 4 \leq σ \leq 10 / 13$ , we see from Lemma 11.33 that the bound

LV (σ, τ) \leq max (2 - 2 σ, τ + 7 - 10 σ)

holds for $0 \leq τ < 8 σ - 5$ , and hence for all $τ \geq 0$ by Lemma 7.7. Meanwhile, from Lemma 9.5 we have

{LV}_{ζ} (σ, τ) \leq τ - 4 (σ - 1 / 2)

for all $1 / 2 \leq σ \leq 1$ and $τ \geq 2$ . The claim then follows from Corollary 11.9 with $τ_{0} := 7 σ - 4$ after a routine calculation. Similarly, for $10 / 13 \leq σ \leq 1$ , we have

LV (σ, τ) \leq max (2 - 2 σ, τ + \frac{11 - 17 σ}{3})

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

Theorem 11.35 Pintz zero density theorem

[ 141 , Theorem 1 ] If $k \geq 4$ , $ℓ \geq 3$ are integers and $σ = 1 - η$ is such that

\frac{1}{k (k + 1)} \leq η < \frac{1}{k (k - 1)}

and

\frac{1}{2 ℓ (ℓ + 1)} \leq η < \frac{1}{2 ℓ (ℓ - 1)}

then

A (σ) \leq max (\frac{3}{ℓ (1 - 2 (ℓ - 1) η)}, \frac{4}{k (1 - (k - 1) η)}) .

Recorded in literature.py as:
add_zero_density_pintz_2023()

Proof ▶

We apply Corollary 11.9 with

τ_{0} := min (ℓ (1 - 2 (ℓ - 1) η), \frac{3}{4} (k (1 - (k - 1) η))) - ε

for an arbitrarily small $ε$ . It then suffices to show that 6 holds for $2 τ_{0} / 3 \leq τ \leq τ_{0}$ and 7 holds for $2 \leq τ < 4 τ_{0} / 3$ .

To prove 7, it suffices by Lemma 8.5 to show that $σ > τ β (1 / τ)$ for all $2 \leq τ < 4 τ_{0} / 3$ . By 36 one has $2 \leq τ < k (1 - (k - 1) η)$ . Meanwhile, from Lemma 4.23 one has

τ β (1 / τ) \leq 1 + max (\frac{τ - r}{r (r - 1)}, - \frac{1}{r (r - 1)}, - \frac{2 τ}{r^{2} (r - 1)})

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

\frac{r - τ}{r (r - 1)}, \frac{2 τ}{r^{2} (r - 1)} \geq η

or equivalently

τ \in [\frac{r^{2} (r - 1) η}{2}, r (1 - (r - 1) η)] .

But one can check that these intervals for $3 \leq r \leq k$ cover the entire range $2 \leq τ < 4 τ_{0} / 3$ , as required.

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

sup_{1 \leq τ \leq τ_{0}} β (1 / τ) τ < 2 σ - 1 = 1 - 2 η .

Using 37, 35 we obtain the claim whenever

τ \in [r^{2} (r - 1) η + ε, r (1 - 2 (r - 1) η) - ε]

for some $3 \leq r \leq ℓ$ . These cover the range $[18 η + ε, τ_{0}]$ . For the remaining range $[1, 18 η + ε]$ we use the van der Corput bound

τ β (1 / τ) \leq \frac{τ}{2} \leq 9 η

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

The range of the second bound in Lemma 11.26 was recently extended:

Theorem 11.36 Chen-Debruyne-Vidas density theorem

[ 19 ] For any $279 / 314 \leq σ \leq 17 / 18$ , one has $A (σ) \leq \frac{24}{30 σ - 11}$ .

Recorded in literature.py as:
add_zero_density_chen_debruyne_vindas_2024()

The following result appears in an unpublished preprint of Kerr, and is based on the large values theorems in Theorem 10.32:

Proposition 11.37

[ 103 , Theorems 6, 7 ] One has $A (σ) \leq \frac{3}{2 σ}$ for $σ \geq 23 / 29$ , and

A (σ) \leq max (\frac{36}{138 σ - 89}, \frac{114 σ - 79}{(1 - σ) (138 σ - 89)})

for $127 / 168 \leq σ \leq 107 / 138$ .

The current best known zero density estimates (excepting the unpublished result in Proposition 11.37) are summarized in Table 11.1.

Derived in derived.py as:
compute_best_zero_density()

Table 11.1 Current best upper bound on

A (σ)

$A (σ)$ bound

Range

Reference

$\frac{3}{2 - σ}$

$\frac{1}{2} \leq σ \leq \frac{7}{10} = 0.7$

Theorem 11.14

$\frac{15}{3 + 5 σ}$

$\frac{7}{10} \leq σ < \frac{19}{25} = 0.76$

Theorem 11.16

$\frac{9}{8 σ - 2}$

$\frac{19}{25} \leq σ < \frac{127}{167} = 0.7604 \dots$

Corollary 11.31

$\frac{15}{13 σ - 3}$

$\frac{127}{167} \leq σ < \frac{13}{17} = 0.7647 \dots$

Corollary 11.31

$\frac{6}{5 σ - 1}$

$\frac{13}{17} \leq σ < \frac{17}{22} = 0.7727 \dots$

Corollary 11.31

$\frac{2}{9 σ - 6}$

$\frac{17}{22} \leq σ < \frac{41}{53} = 0.7735 \dots$

Theorem 11.22

$\frac{9}{7 σ - 1}$

$\frac{41}{53} \leq σ < \frac{7}{9} = 0.7777 \dots$

Corollary 11.31

$\frac{9}{8 (2 σ - 1)}$

$\frac{7}{9} \leq σ < \frac{1867}{2347} = 0.7954 \dots$

Theorem 11.22

$\frac{3}{2 σ}$

$\frac{1867}{2347} \leq σ < \frac{4}{5} = 0.8$

Theorem 11.32

$\frac{3}{2 σ}$

$\frac{4}{5} \leq σ < \frac{7}{8} = 0.875$

Corollary 11.31

$\frac{3}{10 σ - 7}$

$\frac{7}{8} \leq σ < \frac{279}{314} = 0.8885 \dots$

Theorem 11.18

$\frac{24}{30 σ - 11}$

$\frac{279}{314} \leq σ < \frac{155}{174} = 0.8908 \dots$

Theorem 11.36

$\frac{24}{30 σ - 11}$

$\frac{155}{174} \leq σ \leq \frac{9}{10} = 0.9$

Theorem 11.26

$\frac{3}{10 σ - 7}$

$\frac{9}{10} < σ \leq \frac{31}{34} = 0.9117 \dots$

Theorem 11.20

$\frac{11}{48 σ - 36}$

$\frac{31}{34} < σ < \frac{14}{15} = 0.9333 \dots$

Corollary 11.25

$\frac{391}{2493 σ - 2014}$

$\frac{14}{15} \leq σ < \frac{2841}{3016} = 0.9419 \dots$

Corollary 11.25

$\frac{22232}{163248 σ - 134765}$

$\frac{2841}{3016} \leq σ < \frac{859}{908} = 0.9460 \dots$

Corollary 11.25

$\frac{356}{2742 σ - 2279}$

$\frac{859}{908} \leq σ < \frac{23}{24} = 0.9583 \dots$

Corollary 11.25

$\frac{3}{24 σ - 20}$

$\frac{23}{24} \leq σ < \frac{2211487}{2274732} = 0.9721 \dots$

Theorem 11.35

$\frac{86152}{1447460 σ - 1311509}$

$\frac{2211487}{2274732} \leq σ < \frac{39}{40} = 0.975$

Corollary 11.25

$\frac{2}{15 σ - 12}$

$\frac{39}{40} \leq σ < \frac{41}{42} = 0.9761 \dots$

Theorem 11.35

$\frac{3}{40 σ - 35}$

$\frac{41}{42} \leq σ < \frac{59}{60} = 0.9833 \dots$

Theorem 11.35

$\frac{3}{n (1 - 2 (n - 1) (1 - σ))}$

$1 - \frac{1}{2 n (n - 1)} \leq σ < 1 - \frac{1}{2 n (n + 1)}$

(for integer $n \geq 6$ )

Theorem 11.35

\includegraphics[width=0.5\linewidth ]{chapter/zero_density_estimate_plot.png} — Figure 11.1 The bounds in Table 11.1, compared against the existing literature bounds on $A (σ)$ .

For completeness, we list in Table 11.2 some historical zero density theorems not already covered, which have now been superseded by more recent estimates.

Table 11.2 Historical upper bounds on

A (σ)

$A (σ)$ bound	Range	Reference
$4 σ$	$\frac{1}{2} \leq σ \leq 1$	Carlson (1921) [ 18 ]
$2$	$4 / 5 \leq σ \leq 1$	Montgomery (1969) [ 129 ]
$2$	$0.8080 \leq σ \leq 1$	Forti–Viola (1972) [ 40 ]
$\frac{39}{115 σ - 75}$	$55 / 67 \leq σ \leq 189 / 230$	Huxley (1973) [ 78 ]
$2$	$189 / 230 \leq σ \leq 78 / 89$	Huxley (1973) [ 78 ]
$\frac{48}{37 (2 σ - 1)}$	$78 / 89 \leq σ \leq 61 / 74$	Huxley (1973) [ 78 ]
$\frac{3}{2 σ}$	$37 / 42 \leq σ \leq 1$	Huxley (1975) [ 79 ]
$\frac{48}{37 (2 σ - 1)}$	$61 / 74 \leq σ \leq 37 / 42$	Huxley (1975) [ 79 ]
$2$	$0.80119 \leq σ \leq 1$	Huxley (1975) [ 79 ]
$2$	$4 / 5 \leq σ \leq 1$	Huxley (1975) [ 80 ]
$\frac{6}{5 σ - 1}$	$67 / 87 \leq σ \leq 1$	Ivić (1979) [ 90 ]
$\frac{3}{34 σ - 25}$	$28 / 37 \leq σ \leq 74 / 95$	Ivić (1979) [ 90 ]
$\frac{9}{7 σ - 1}$	$74 / 95 \leq σ \leq 1$	Ivić (1979) [ 90 ]
$\frac{3}{2 σ}$	$4 / 5 \leq σ \leq 1$	Ivić (1979) [ 90 ]
$\frac{68}{98 σ - 47}$	$115 / 166 \leq σ \leq 1$	Ivić (1979) [ 90 ]
$\frac{3}{2 σ}$	$3831 / 4791 \leq σ \leq 1$	Ivić (1980) [ 87 ]
$\frac{9}{7 σ - 1}$	$41 / 53 \leq σ \leq 1$	Ivić (1980) [ 87 ]
$\frac{6}{5 σ - 1}$	$13 / 17 \leq σ \leq 1$	Ivić (1980) [ 87 ]
$\frac{4}{2 σ + 1}$	$17 / 18 \leq σ \leq 1$	Ivić (1980) [ 87 ]
$\frac{24}{30 σ - 11}$	$155 / 174 \leq σ \leq 17 / 18$	Ivić (1980) [ 87 ]
$\frac{3}{7 σ - 4}$	$3 / 4 \leq σ \leq 10 / 13$	Ivić (1983) [ 84 ]
$\frac{9}{8 σ - 2}$	$10 / 13 \leq σ \leq 1$	Ivić (1983) [ 84 ]
$\frac{15}{22 σ - 10}$	$10 / 13 \leq σ \leq 5 / 6$	Ivić (1984) [ 85 ]
$\frac{3 k}{(3 k - 2) σ + 2 - k}$	$\frac{9 k^{2} - 3 k + 2}{12 k^{2} - 5 k + 2} \leq σ \leq 1$ ; $k \geq 2$	Ivić (1984) [ 85 ]
$58.05 (1 - σ)^{1 / 2}$	$1 / 2 \leq σ \leq 1$	Ford (2002) [ 39 ]
$6.42 (1 - σ)^{1 / 2}$	$9 / 10 \leq σ \leq 1$	Heath-Brown (2017) [ 63 ]
$3 \sqrt{2} (1 - σ)^{1 / 2} + 18 (1 - σ)$	$17 / 18 \leq σ \leq 1$	Pintz (2023) [ 141 ]

TODO: enter this table into literature.py

11.2 Estimates for $σ$ very close to $1 / 2$ or $1$

Some additional estimates were established for $σ$ sufficiently close to $1 / 2$ or $1$ .

Turán [ 170 ] introduced the power sum method to establish

A (1 - η) \leq 2 + η^{0.14}

for $η$ small enough. Halász and Turán [ 45 ] combined this method with the large values approach of Halász [ 44 ] to improve the bound to

A (1 - η) \leq C η^{1 / 2}

with $C = 12, 000$ for sufficiently small $η$ . See [ 140 ] for an alternate proof of these results.

The constant $C$ in 38 was improved to $1304.37$ by Montgomery [ 130 , Theorem 12.3 ] (see also the remark after [ 88 , (11.97) ] for a correction), to $58.05$ by Ford [ 39 ] , to $5.03$ by Heath-Brown [ 63 ] (the latter exploiting the resolution of the Vinogradov mean value conjecture [ 17 ] ), and to any $C > 3 \sqrt{2} = 4.242 \dots$ in [ 141 ] . See also an explicit version at [ 7 ] .

“Log-free” zero density estimates of the form

N (1 - η, T) ≪ T^{B η}

for various $B$ were established starting with the work of Linnik [ 118 , 119 ] and developed further in [ 170 ] , [ 38 ] , [ 11 ] , [ 101 ] , [ 41 ] , [ 42 ] , [ 62 ] . An explicit version of such estimates may be found in [ 8 ] .

There is some work establishing bounds on $N (σ, T)$ for $σ$ very close to $1 / 2$ (and not necessarily fixed), although these bounds do not make further improvements on $A (σ)$ . Specifically, bounds of the form

N (σ, T) ≪ T^{1 - θ (2 σ - 1)} \log T

for $T \geq 2$ (say) were established for $θ = 1 / 8$ by Selberg [ 159 ] (see [ 160 ] for an explicit version), any $0 < θ < 1 / 2$ by Jutila [ 100 ] , and any $0 < θ < 4 / 7$ by Conrey (claimed in [ 23 ] , with a full proof given in [ 6 ] ). Note that the density hypothesis would follow if we could establish the claim for all $0 < θ < 1$ , but an improvement to Ingham’s bound (Theorem 11.14) would only occur once $θ$ exceeded $2 / 3$ .

11.3 A heuristic for zero density estimates

We can now state a rough heuristic as to what zero density estimates to expect from a given large value theorem:

[Predicting a zero density estimate from a large value theorem] Suppose that $1 / 2 \leq σ \leq 1$ and $τ_{0} \geq 1$ are such that one can prove $LV (σ, τ_{0}) \leq 3 - 3 σ$ (i.e., the Montgomery conjecture holds here with a multiplicative loss of $3 / 2$ ). Then in principle, one can hope to prove $A (σ) \leq 3 / τ_{0}$ . Conversely, if one cannot prove $LV (σ, τ_{0}) \leq 3 - 3 σ$ , then the bound $A (σ) \leq 3 / τ_{0}$ is likely out of reach.

We justify this heuristic as follows, though we stress that the arguments that follow are not fully rigorous. In the first part, we simply apply Corollary 11.9. In practice, the 6 is often more delicate than 7 and ends up being the limiting factor for the bounds; furthermore, within 6, it is the right endpoint $τ = τ_{0}$ of the range $2 τ_{0} / 3 \leq τ \leq τ_{0}$ that ends up being the bottleneck; but this is precisely the claimed criterion $LV (σ, τ_{0}) \leq 3 - 3 σ$ . We remark that in some cases (particularly for $σ$ close to one), the estimate 7 ends up being more of the bottleneck than 6, and so one should view $3 / τ_{0}$ here as a theoretical upper limit of methods rather than as a guaranteed bound. (In particular, the need to also establish the bound ${LV}_{ζ} (σ, \frac{4}{3} τ_{0} - ε) < 4 - 4 σ$ for $ε > 0$ small can sometimes be a more limiting factor.)

Conversely, suppose that

LV (σ, τ_{0}) > 3 - 3 σ,

but that one still wants to prove the bound $A (σ) \leq 3 / τ_{0}$ . Heuristically, Theorem 11.6 suggests that in order to do this, it is necessary to establish the bound ${LV}_{ζ} (σ, τ) / τ \leq \frac{3}{τ_{0}} (1 - σ)$ for all $τ \geq 2$ . In particular, one should show that

{LV}_{ζ} (σ, 2 τ_{0}) \leq 6 - 6 σ .

Let us consider the various options one has to do this. There are ways to control zeta large values that do not apply to general large value estimates, such as moment estimates of the zeta function, exponent pairs, or control of $β$ and $μ$ . However, at our current level of understanding, these techniques only control ${LV}_{ζ} (σ, τ)$ for relatively small values of $τ$ , and in practice $2 τ_{0}$ is too large for these methods to apply; this exponent also tends to be too large for direct application of standard large value theorems to be useful. Hence, the most viable option in practice is raising to a power (Lemma 7.8), using

{LV}_{ζ} (σ, 2 τ_{0}) \leq k {LV}_{ζ} (σ, 2 τ_{0} / k)

for some natural number $k \geq 2$ . However, the most natural choice $k = 2$ is blocked due to our hypothesis 39, while in practice the $k \geq 3$ choice is blocked because of Lemma 7.5. Hence it appears heuristically quite difficult to establish $A (σ) \leq 3 / τ_{0}$ with current technology, in the event that 39 occurs.

In Table 11.3 we list some examples in which the heuristic can actually be attained. Note that this only covers some, but not all, of the best known zero density estimates in Table 11.1, as there are often other bounds that need to be established that prevent the heuristic limit of $3 / τ_{0}$ from actually being attained; so one should take the heuristic with a certain grain of salt.

Table 11.3 Examples of large value theorems, the values of

τ_{0}

and

A (σ)

they suggest, and rigorous zero density theorems that attain the predicted value for at least some ranges of

σ

Large value theorem	Predicted choice of $τ_{0}$	Predicted bound $\frac{3}{τ_{0}}$ on $A (σ)$	Matching zero density theorem(s)
Theorem 7.9	$2 - σ$	$\frac{3}{2 - σ}$	Theorem 11.14
Theorem 7.12	$3 σ - 1$	$\frac{3}{3 σ - 1}$	Theorem 11.15
Theorem 7.14	$10 σ - 7$	$\frac{3}{10 σ - 7}$	Theorems 11.18, 11.20
Theorem 7.16, $k = 3$	$\frac{7 σ - 1}{3}$	$\frac{9}{7 σ - 1}$	Theorems 11.18, 11.31
Lemma 11.29, $m = 2$	$\frac{4 σ}{2}$	$\frac{3}{2 σ}$	Corollary 11.31, Theorem 11.32
Lemma 11.29, $m = 3$	$\frac{7 σ - 1}{3}$	$\frac{9}{7 σ - 1}$	Theorems 11.18, Corollary 11.31
Lemma 11.29, $m = 4$	$\frac{10 σ - 2}{4}$	$\frac{6}{5 σ - 1}$	Corollary 11.31
Lemma 11.33	$7 σ - 4$	$\frac{3}{7 σ - 4}$	Corollary 11.34
Lemma 11.33	$\frac{8 σ - 2}{3}$	$\frac{9}{8 σ - 2}$	Corollary 11.34
Theorem 10.27	$\frac{5 σ - 3}{3}$	$\frac{15}{5 σ - 3}$	Theorem 11.16

One consequence of Heuristic 11.3 is that, in the regimes where the heuristic is accurate, combining multiple large values theorems together are unlikely to achieve new zero density theorems that could not be accomplished with each large value theorem separately.

11 Zero density theorems

11.1 Known zero density bounds

11.2 Estimates for σ very close to 1/2 or 1

11.3 A heuristic for zero density estimates

11.2 Estimates for $σ$ very close to $1 / 2$ or $1$