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Matrix-integral bound from a convex mixture of row isometries

Proved
RybinAI2026.P01.matrix_integral_inequality_row_isometry_mix

by miao · Sep 8, 2026 · Mathlib c5ea003 (Lean v4.30.0)

integral-inequalitymatrix-analysispositive-definite-matrices

Let nnn be a natural number and let A,B,C,DA,B,C,DA,B,C,D be real symmetric positive-definite n×nn\times nn×n matrices. Write βM(u,v)=uTMv\beta_M(u,v)=u^{\mathsf T}MvβM​(u,v)=uTMv and let ddd be the original unnormalized double spherical integral. Let TTT be a real linear isometric equivalence of Euclidean space and let t∈[0,1]t\in[0,1]t∈[0,1]. Suppose TTT preserves the quadratic form of BBB:

βB(Tu,Tu)=βB(u,u)for every u.\beta_B(Tu,Tu)=\beta_B(u,u)\qquad\text{for every }u.βB​(Tu,Tu)=βB​(u,u)for every u.

Suppose also that the output numerator is the following convex mixture:

β(A+B)−(C+D)(u,v)=t βB−D(u,v)+(1−t) βB−D(Tu,v)for every u,v.\beta_{(A+B)-(C+D)}(u,v) =t\,\beta_{B-D}(u,v)+(1-t)\,\beta_{B-D}(Tu,v) \qquad\text{for every }u,v.β(A+B)−(C+D)​(u,v)=tβB−D​(u,v)+(1−t)βB−D​(Tu,v)for every u,v.

Then

d(A+B,C+D)≤d(B,D).d(A+B,C+D)\leq d(B,D).d(A+B,C+D)≤d(B,D).

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.

Preamble
import Definitions.Def_rybin2026_p01_matrix_integral

open Matrix RybinAI2026.P01
Formal statement
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
Source
https://rybindmitry.github.io/problems/1.html, Problem 1 and its defining integral. Derived sufficient condition using convexity of absolute value and denominator-preserving orthogonal changes of variables; not a separately stated theorem in the source.

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