Matrix-integral bound from a convex mixture of row isometries
ProvedRybinAI2026.P01.matrix_integral_inequality_row_isometry_mixLet be a natural number and let be real symmetric positive-definite matrices. Write and let be the original unnormalized double spherical integral. Let be a real linear isometric equivalence of Euclidean space and let . Suppose preserves the quadratic form of :
Suppose also that the output numerator is the following convex mixture:
Then
This is a sufficient condition for the maximum inequality in Problem 1, including equality endpoints t=0 and t=1. It uses a symmetry of the denominator quadratic form to compare integrated numerators, and need not give a pointwise bound by the original input kernel. It imposes neither a commutativity assumption on all four matrices nor an arbitrary congruence invariance rule. Dimension zero is allowed by the formal statement.
import Definitions.Def_rybin2026_p01_matrix_integral open Matrix RybinAI2026.P01
theorem RybinAI2026.P01.matrix_integral_inequality_row_isometry_mix {n : ℕ}
(A B C D : Matrix (Fin n) (Fin n) ℝ)
(hA : A.PosDef) (hB : B.PosDef) (hC : C.PosDef) (hD : D.PosDef)
(e : Euclidean n ≃ₗᵢ[ℝ] Euclidean n) (t : ℝ) (ht : 0 ≤ t) (ht1 : t ≤ 1)
(hBsym : ∀ u : Euclidean n, bilinear B (e u) (e u) = bilinear B u u)
(hnum : ∀ u v : Euclidean n,
bilinear ((A+B)-(C+D)) u v =
t * bilinear (B-D) u v + (1-t) * bilinear (B-D) (e u) v) :
distance (A+B) (C+D) ≤ distance B D := by
sorry