Notes on polytope decomposition

The notes here are derived from these notes of Pace Nielsen.

Let [math]\displaystyle{ 1/4 \leq \varepsilon \leq 1/3 }[/math]. The problem here is to optimise the ratio J/I, where

[math]\displaystyle{ J := 3 \int\int_{x+y \leq 1-\varepsilon} (\int_0^{3/2-x-y} F(x,y,z)\ dz)^2\ dx dy }[/math]

[math]\displaystyle{ I := \int\int\int_{x+y+z \leq 3/2} F(x,y,z)^2\ dx dy dz }[/math]

for symmetric F supported on [math]\displaystyle{ R := \{ (x,y,z): x+y+z \leq 3/2 \} }[/math] subject to the vanishing marginal condition

[math]\displaystyle{ \int_0^{3/2-x-y} F(x,y,z)\ dz = 0 }[/math] when [math]\displaystyle{ x+y \geq 1+\varepsilon }[/math].

(Throughout these notes, [math]\displaystyle{ x,y,z }[/math] are understood to be non-negative.

We partition R six identical pieces, one of which is

[math]\displaystyle{ R_{xyz} := \{ (x,y,z) \in R: x+y \leq y+z \leq z+x \}. }[/math]

We then partition [math]\displaystyle{ R_{xyz} }[/math] further into 10 pieces:

Name	Inequalities	[math]\displaystyle{ x,y }[/math] inequalities	[math]\displaystyle{ z }[/math] inequalities
[math]\displaystyle{ R_{xyz} }[/math]	[math]\displaystyle{ x,y,z \geq 0; x+y+z \leq 3/2; x+y \leq y+z \leq z+x }[/math]	[math]\displaystyle{ x \geq y \geq 0; 2x+y \leq 3/2 }[/math]	[math]\displaystyle{ x \leq z \leq 3/2-x-y }[/math]
[math]\displaystyle{ A_{xyz} }[/math]	[math]\displaystyle{ z+x \leq 1-\varepsilon }[/math]	[math]\displaystyle{ 2x \leq 1-\varepsilon }[/math]	[math]\displaystyle{ z \leq 1-\varepsilon-x }[/math]
[math]\displaystyle{ B_{xyz} }[/math]	[math]\displaystyle{ y+z \leq 1-\varepsilon \leq z+x \leq 1+\varepsilon }[/math]	[math]\displaystyle{ 2x \leq 1+\varepsilon; x+y \leq 1-\varepsilon; y \leq 1/2+\varepsilon }[/math]	[math]\displaystyle{ 1-\varepsilon-x \leq z \leq \min(1+\varepsilon-x, 1-\varepsilon-y) }[/math]
[math]\displaystyle{ C_{xyz} }[/math]	[math]\displaystyle{ x+y \leq 1-\varepsilon \leq y+z; z+x \leq 1+\varepsilon }[/math]	[math]\displaystyle{ x+y \leq 1-\varepsilon; x \leq 1/2+\varepsilon }[/math]	[math]\displaystyle{ 1-\varepsilon-y \leq z \leq \min(1+\varepsilon-x }[/math]
[math]\displaystyle{ D_{xyz} }[/math]	[math]\displaystyle{ 1-\varepsilon \leq x+y; z+x \leq 1+\varepsilon }[/math]	[math]\displaystyle{ x+y \geq 1-\varepsilon; 2x \leq 1+\varepsilon }[/math]	[math]\displaystyle{ z \leq 1+\varepsilon-x }[/math]
[math]\displaystyle{ E_{xyz} }[/math]	[math]\displaystyle{ y+z \leq 1-\varepsilon; 1+\varepsilon \leq z+x }[/math]	[math]\displaystyle{ x+y \leq 1-\varepsilon; y \leq 1/2-\varepsilon }[/math]	[math]\displaystyle{ 1+\varepsilon-x \leq z \leq 1-\varepsilon-y }[/math]
[math]\displaystyle{ S_{xyz} }[/math]	[math]\displaystyle{ x+y \leq 1-\varepsilon \leq y+z \leq 1+\varepsilon \leq z+x }[/math] [math]\displaystyle{ z \leq 1/2+\varepsilon }[/math]	[math]\displaystyle{ x+y \leq 1-\varepsilon; x \geq 1/2; y \geq 1/2-2\varepsilon }[/math]	[math]\displaystyle{ \max(1-\varepsilon-y,1+\varepsilon-x) \leq z \leq \min(1/2+\varepsilon,1+\varepsilon-y) |- | \lt math\gt T_{xyz} }[/math]	[math]\displaystyle{ x+y \leq 1-\varepsilon \leq y+z \leq 1+\varepsilon \leq z+x }[/math] [math]\displaystyle{ z \geq 1/2+\varepsilon; x \geq 1/2-\varepsilon }[/math]	[math]\displaystyle{ x+y \leq 1-\varepsilon; x \geq 1/2-\varepsilon }[/math]	[math]\displaystyle{ \max(1-\varepsilon-y,1+\varepsilon-x,1/2-\varepsilon) \leq z \leq 1+\varepsilon-y }[/math]
[math]\displaystyle{ U_{xyz} }[/math]	[math]\displaystyle{ x+y \leq 1-\varepsilon \leq y+z \leq 1+\varepsilon \leq z+x }[/math] [math]\displaystyle{ x \leq 1/2-\varepsilon }[/math]	[math]\displaystyle{ x \leq 1/2-\varepsilon }[/math]	[math]\displaystyle{ \max(1-\varepsilon-y,1+\varepsilon-x) \leq z \leq 1+\varepsilon-y }[/math]
[math]\displaystyle{ G_{xyz} }[/math]	[math]\displaystyle{ x+y \leq 1-\varepsilon; 1+\varepsilon \leq y+z }[/math]	[math]\displaystyle{ x+y \leq 1-\varepsilon; x \leq 1/2-\varepsilon }[/math]	[math]\displaystyle{ 1+\varepsilon-y \leq z }[/math]
[math]\displaystyle{ H_{xyz} }[/math]	[math]\displaystyle{ 1-\varepsilon \leq x+y; y+z \leq 1+\varepsilon \leq z+x }[/math]	[math]\displaystyle{ x+y \geq 1-\varepsilon }[/math]	[math]\displaystyle{ 1+\varepsilon-x \leq z \leq 1+\varepsilon-y }[/math]