This is the sixth thread for the Polymath8b project to obtain new bounds for the quantity

$\displaystyle H_m := \liminf_{n \rightarrow\infty} (p_{n+m} - p_n),$

either for small values of ${m}$ (in particular ${m=1,2}$) or asymptotically as ${m \rightarrow \infty}$. The previous thread may be found here. The currently best known bounds on ${H_m}$ can be found at the wiki page (which has recently returned to full functionality, after a partial outage).

The current focus is on improving the upper bound on ${H_1}$ under the assumption of the generalised Elliott-Halberstam conjecture (GEH) from ${H_1 \leq 8}$ to ${H_1 \leq 6}$, which looks to be the limit of the method (see this previous comment for a semi-rigorous reason as to why ${H_1 \leq 4}$ is not possible with this method). With the most general Selberg sieve available, the problem reduces to the following three-dimensional variational one:

Problem 1 Does there exist a (not necessarily convex) polytope ${R \subset [0,1]^3}$ with quantities ${0 \leq \varepsilon_1,\varepsilon_2,\varepsilon_3 \leq 1}$, and a non-trivial square-integrable function ${F: {\bf R}^3 \rightarrow {\bf R}}$ supported on ${R}$ such that

• ${R + R \subset \{ (x,y,z) \in [0,2]^3: \min(x+y,y+z,z+x) \leq 2 \},}$
• ${\int_0^\infty F(x,y,z)\ dx = 0}$ when ${y+z \geq 1+\varepsilon_1}$;
• ${\int_0^\infty F(x,y,z)\ dy = 0}$ when ${x+z \geq 1+\varepsilon_2}$;
• ${\int_0^\infty F(x,y,z)\ dz = 0}$ when ${x+y \geq 1+\varepsilon_3}$;

and such that we have the inequality

$\displaystyle \int_{y+z \leq 1-\varepsilon_1} (\int_{\bf R} F(x,y,z)\ dx)^2\ dy dz$

$\displaystyle + \int_{z+x \leq 1-\varepsilon_2} (\int_{\bf R} F(x,y,z)\ dy)^2\ dz dx$

$\displaystyle + \int_{x+y \leq 1-\varepsilon_3} (\int_{\bf R} F(x,y,z)\ dz)^2\ dx dy$

$\displaystyle > 2 \int_R F(x,y,z)^2\ dx dy dz?$

(Initially it was assumed that ${R}$ was convex, but we have now realised that this is not necessary.)

An affirmative answer to this question will imply ${H_1 \leq 6}$ on GEH. We are “within almost two percent” of this claim; we cannot quite reach ${2}$ yet, but have got as far as ${1.959633}$. However, we have not yet fully optimised ${F}$ in the above problem.

The most promising route so far is to take the symmetric polytope

$\displaystyle R = \{ (x,y,z) \in [0,1]^3: x+y+z \leq 3/2 \}$

with ${F}$ symmetric as well, and ${\varepsilon_1=\varepsilon_2=\varepsilon_3=\varepsilon}$ (we suspect that the optimal ${\varepsilon}$ will be roughly ${1/6}$). (However, it is certainly worth also taking a look at easier model problems, such as the polytope ${{\cal R}'_3 := \{ (x,y,z) \in [0,1]^3: x+y,y+z,z+x \leq 1\}}$, which has no vanishing marginal conditions to contend with; more recently we have been looking at the non-convex polytope ${R = \{x+y,x+z \leq 1 \} \cup \{ x+y,y+z \leq 1 \} \cup \{ x+z,y+z \leq 1\}}$.) Some further details of this particular case are given below the fold.

There should still be some progress to be made in the other regimes of interest – the unconditional bound on ${H_1}$ (currently at ${270}$), and on any further progress in asymptotic bounds for ${H_m}$ for larger ${m}$ – but the current focus is certainly on the bound on ${H_1}$ on GEH, as we seem to be tantalisingly close to an optimal result here.

— 1. Analysis of the symmetric polytope —

In the symmetric situation described above, we have a single vanishing marginal condition

• ${\int_0^{3/2-x-y} F(x,y,z)\ dz = 0}$ when ${(x,y) \in [0,1]^2}$ is such that ${1+\varepsilon_3 \leq x+y \leq 3/2}$

and the inequality we wish to prove is that

$\displaystyle 3\int_{x+y \leq 1-\varepsilon} (\int_0^{\min(1,3/2-x-y)} F(x,y,z)\ dz)^2\ dx dy$

$\displaystyle > 2 \int_R F(x,y,z)^2\ dx dy dz.$

A routine calculus of variations computation, using Lagrange multipliers, tells us that the extremal ${F}$ should take the form

$\displaystyle F(x,y,z) = G(x,y) + G(y,z) + G(z,x) \ \ \ \ \ (1)$

where ${G}$ is symmetric and supported the disconnected set ${T_0 \cup T_1}$, where ${T_0}$ is the triangle

$\displaystyle T_0 := \{ (x,y) \in [0,1]^2: x+y \leq 1-\varepsilon\}$

and ${T_1}$ is the trapezoid

$\displaystyle T_1 := \{ (x,y) \in [0,1]^2: 1+\varepsilon \leq x+y \leq 3/2 \}.$

(Strictly speaking, we have not established that an extremiser actually exists, but for the purposes of numerics it seems to be reasonable to proceed as if an extremiser does exist.) One also has the eigenfunction equation

$\displaystyle (M''_{3,\varepsilon,1})^{-1} \int_0^{\min(1,3/2-x-y)} F(x,y,z)\ dz = G(x,y)$

whenever ${(x,y) \in [0,1]^2}$ is such that ${x+y \leq 1-\varepsilon}$, where ${M''_{3,\varepsilon,1}}$ is the quantity that we are trying to push above ${2}$. It seems that this eigenfunction equation should indicate where exactly we expect the singularities of ${F}$ and ${G}$ to lie, and how bad they will be (our best guess right now is that the extremiser is continuous but not differentiable at certain points), but this has not yet been fully analysed.

If we use the ansatz (1), we arrive at a moderately complicated variational problem involving one unknown symmetric function ${G(x,y)}$, which is constrained to be supported in ${T_0 \cup T_1}$ and obeys the vanishing marginal condition

$\displaystyle \int_0^{3/2-x-y} (G(x,y) + G(y,z) + G(z,x))\ dz = 0$

whenever ${(x,y)}$ lies in ${T_1}$. If we let ${G_0, G_1}$ be the restrictions of ${G}$ to ${T_0,T_1}$ respectively (so that ${G = G_0 + G_1}$), this constraint becomes

$\displaystyle (3/2-x-y) G_1(x,y) + H_0(x,y) + H_1(x,y) + H_0(y,x) + H_1(y,x) = 0$

where

$\displaystyle H_0(x,y) := \int_{0 \leq z \leq \min(3/2-x-y,1-\varepsilon-y)} G_0(y,z)\ dz$

and

$\displaystyle H_1(x,y) := \int_{1+\varepsilon-y \leq z \leq 3/2-x-y} G_1(y,z)\ dz.$

This equation can be used to solve for ${G_1}$ in terms of ${G_0}$, thus reducing to a variational problem for an unconstrained function ${G_0}$ on the triangle ${T_0}$. In the hexagonal region

$\displaystyle \{ (x,y) \in T_1: x, y \geq 1/2 - \varepsilon \}$

the functions ${H_1(x,y), H_1(y,x)}$ vanish, and we simply have

$\displaystyle G_1(x,y) = -\frac{H_0(x,y) + H_0(y,x)}{3/2-x-y}$

(note that ${H_0}$ vanishes at the boundary ${x+y=3/2}$, so there is no singularity here). If ${\varepsilon \geq 1/4}$, this covers all of ${T_1}$, and completes the reduction to a variational problem for ${G_0}$. We do not believe this range to be the optimal value for ${\varepsilon}$, but might be worth considering as a warmup.

Now suppose ${\varepsilon < 1/4}$. In the triangular region

$\displaystyle \{ (x,y) \in T_1: x \leq 1/2-\varepsilon\} \ \ \ \ \ (2)$

(bounded by the vertices ${(\varepsilon,1)}$, ${(1/2-\varepsilon,1)}$, ${(1/2-\varepsilon,1/2+2\varepsilon)}$) we have vanishing of ${H_1(y,x)}$ but not ${H_1(x,y)}$, leading to the equation

$\displaystyle (3/2-x-y) G_1(x,y) + H_0(x,y) + H_0(y,x) + \int_{1+\varepsilon-y}^{3/2-x-y} G_1(z,y)\ dz = 0. \ \ \ \ \ (3)$

To solve this equation, we first differentiate it in ${x}$ to obtain

$\displaystyle (3/2-x-y) \partial_x G_1(x,y) - G_1(x,y) + \partial_x H_0(x,y) + \partial_x H_0(y,x)$

$\displaystyle - G_1(3/2-x-y,y) = 0,$

so if we define

$\displaystyle K_1(x,y) := G_1(x,y) + G_1(3/2-x-y,y)$

and

$\displaystyle L_0(x,y) := \partial_x H_0(x,y) + \partial_x H_0(y,x)$

then

$\displaystyle \partial_x G_1(x,y) = \frac{K_1(x,y)}{3/2-x-y} - \frac{L_0(x,y)}{3/2-x-y} \ \ \ \ \ (4)$

and thus by the substitution ${x \mapsto 3/2-x-y}$ (which preserves the triangle (2), as well as ${K}$)

$\displaystyle -\partial_x G_1(3/2-x-y,y) = \frac{K_1(x,y)}{x} - \frac{L_0(3/2-x-y,y)}{x}$

and so on subtracting

$\displaystyle \partial_x K_1(x,y) = K_1(x,y) (\frac{1}{3/2-x-y} - \frac{1}{x}) - \frac{L_0(x,y)}{3/2-x-y}$

$\displaystyle + \frac{L_0(3/2-x-y,y)}{x};$

multiplying by ${x (x+y-3/2)}$ and rearranging, this becomes

$\displaystyle \partial_x ( x (x+y-3/2) K_1(x,y) ) = L_0(x,y) x$

$\displaystyle + L_0(3/2-x-y,y) (x+y-3/2),$

so we have

$\displaystyle K_1(x,y) = N_0(x,y) + C_1(y) (3/2-y) ( \frac{1}{x} + \frac{1}{3/2-x-y} )$

for some function ${C_1(y)}$, where

$\displaystyle N_0(x,y) := \frac{1}{x(x+y-3/2)} \int_{(3/2-y)/2}^x L_0(x',y) x'$

$\displaystyle + L_0(3/2-x'-y,y) (x'+y-3/2)\ dx',$

where we are using the signed integral, thus ${\int_b^a = -\int_a^b}$ when ${a>b}$. Note that when ${x = (3/2-y)/2}$, ${N_0(x,y)}$ vanishes and ${K_1(x,y) = 2 G_1(x,y)}$, and so

$\displaystyle G_1( (3/2-y)/2, y ) = 2 C_1(y).$

Meanwhile, from (4) we have

$\displaystyle \partial_x G_1(x,y) = \frac{N_0(x,y)-L_0(x,y)}{3/2-x-y} + C_1(y) \frac{(3/2-y)}{(3/2-x-y)^2}$

$\displaystyle + C_1(y) (\frac{1}{x} + \frac{1}{3/2-x-y})$

and so

$\displaystyle G_1(x,y) = P_0(x,y) + C_1(y) ( \frac{3/2-y}{3/2-x-y} + \log \frac{x}{3/2-x-y} )$

where

$\displaystyle P_0(x,y) := \int_{(3/2-y)/2}^x \frac{N_0(x',y)-L_0(x',y)}{3/2-x'-y}\ dx'.$

On the other hand, from (3) with ${x=1/2-\varepsilon}$ one has

$\displaystyle (1+\varepsilon-y) G_1(1/2-\varepsilon,y) + H_0(1/2-\varepsilon,y) + H_0(y,1/2-\varepsilon) = 0$

and so we may solve for ${C_1}$ as

$\displaystyle C_1(y) = - \frac{ \frac{1}{1+\varepsilon-y} (H_0(1/2-\varepsilon,y) + H_0(y,1/2-\varepsilon)) + P_0(1/2-\varepsilon,y)}{ \frac{3/2-y}{1+\varepsilon-y} + \log \frac{1/2-\varepsilon}{1+\varepsilon-y} }$

which gives an explicit (but messy) formula for ${G_1}$ in terms of ${G_0}$ in the triangle (2); by symmetry one also describes ${G_1}$ in the triangle

$\displaystyle \{ (x,y) \in T_1: y \leq 1/2-\varepsilon \},$

and this completes the description of ${G_1}$ in terms of ${G_0}$. (The exact description is, however, rather messy, in particular requiring a certain amount of integration; it is not clear whether we should actually use it for the purposes of efficient numerics.)