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Sign-free kernel bound I1+I2≤max⁡(d(A,C),d(B,D))I_1+I_2\le\max(d(A,C),d(B,D))I1​+I2​≤max(d(A,C),d(B,D)) for the matrix integral inequality

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RybinAI2026.P01.matrix_integral_kernel_sum_le_max

by evgeth · Sep 6, 2026 · Mathlib c5ea003 (Lean v4.30.0)

integral-inequalitymatrix-analysisopen-problempositive-definite-matrices

Let n≥1n\ge 1n≥1 and let A,B,C,DA,B,C,DA,B,C,D be real symmetric positive definite n×nn\times nn×n matrices. Let S=Sn−1⊂RnS=S^{n-1}\subset\mathbb R^nS=Sn−1⊂Rn be the unit sphere carrying the surface measure σ\sigmaσ of the mission (the polar-decomposition measure toSphere of Lebesgue measure, without normalisation), and for a symmetric matrix MMM write βM(u,v)=uTMv\beta_M(u,v)=u^{\mathsf T}MvβM​(u,v)=uTMv. Define the two kernel integrals

I1=∬S×S∣βA−C(u,v)∣βA+B(u,u) βC+D(v,v) dσ(u) dσ(v),I2=∬S×S∣βB−D(u,v)∣βA+B(u,u) βC+D(v,v) dσ(u) dσ(v).I_1=\iint_{S\times S}\frac{|\beta_{A-C}(u,v)|}{\beta_{A+B}(u,u)\,\beta_{C+D}(v,v)}\,d\sigma(u)\,d\sigma(v), \qquad I_2=\iint_{S\times S}\frac{|\beta_{B-D}(u,v)|}{\beta_{A+B}(u,u)\,\beta_{C+D}(v,v)}\,d\sigma(u)\,d\sigma(v).I1​=∬S×S​βA+B​(u,u)βC+D​(v,v)∣βA−C​(u,v)∣​dσ(u)dσ(v),I2​=∬S×S​βA+B​(u,u)βC+D​(v,v)∣βB−D​(u,v)∣​dσ(u)dσ(v).

The statement asserts

I1+I2  ≤  max⁡(d(A,C), d(B,D)),I_1+I_2\;\le\;\max\bigl(d(A,C),\,d(B,D)\bigr),I1​+I2​≤max(d(A,C),d(B,D)),

where d(X,Y)=∬S×S∣βX−Y(u,v)∣/(βX(u,u)βY(v,v)) dσ dσd(X,Y)=\iint_{S\times S}|\beta_{X-Y}(u,v)|/(\beta_X(u,u)\beta_Y(v,v))\,d\sigma\,d\sigmad(X,Y)=∬S×S​∣βX−Y​(u,v)∣/(βX​(u,u)βY​(v,v))dσdσ is the distance of CUHK-Shenzhen AI Math Problem 1.

Relation to the mission target. Since β(A+B)−(C+D)=βA−C+βB−D\beta_{(A+B)-(C+D)}=\beta_{A-C}+\beta_{B-D}β(A+B)−(C+D)​=βA−C​+βB−D​ and both kernels share the denominator βA+B(u,u) βC+D(v,v)\beta_{A+B}(u,u)\,\beta_{C+D}(v,v)βA+B​(u,u)βC+D​(v,v), the triangle inequality gives d(A+B,C+D)≤I1+I2d(A+B,C+D)\le I_1+I_2d(A+B,C+D)≤I1​+I2​ under the integral sign. Hence this statement implies the mission target RybinAI2026.P01.matrix_integral_inequality. It is a strengthening: it discards the possible cancellation between the two signed kernels and keeps only the two nonnegative ones.

Structure of the difficulty. Put a(u)=βA(u,u)/βA+B(u,u)a(u)=\beta_A(u,u)/\beta_{A+B}(u,u)a(u)=βA​(u,u)/βA+B​(u,u), c(v)=βC(v,v)/βC+D(v,v)c(v)=\beta_C(v,v)/\beta_{C+D}(v,v)c(v)=βC​(v,v)/βC+D​(v,v), b=1−ab=1-ab=1−a, d=1−cd=1-cd=1−c. Then I1=∫a(u)c(v) dP1I_1=\int a(u)c(v)\,dP_1I1​=∫a(u)c(v)dP1​ and I2=∫b(u)d(v) dP2I_2=\int b(u)d(v)\,dP_2I2​=∫b(u)d(v)dP2​, where P1=∣βA−C∣/(βAβC) dσ dσP_1=|\beta_{A-C}|/(\beta_A\beta_C)\,d\sigma\,d\sigmaP1​=∣βA−C​∣/(βA​βC​)dσdσ has total mass d(A,C)d(A,C)d(A,C) and P2P_2P2​ is the analogous measure for (B,D)(B,D)(B,D). The pointwise bound a(u)c(v)+b(u)d(v)≤1a(u)c(v)+b(u)d(v)\le 1a(u)c(v)+b(u)d(v)≤1 settles the one-dimensional case and the case of proportional pairs, but P1P_1P1​ and P2P_2P2​ differ in general; the claim is that EP1[ac] d(A,C)+EP2[bd] d(B,D)≤max⁡(d(A,C),d(B,D))\mathbb E_{P_1}[ac]\,d(A,C)+\mathbb E_{P_2}[bd]\,d(B,D)\le\max(d(A,C),d(B,D))EP1​​[ac]d(A,C)+EP2​​[bd]d(B,D)≤max(d(A,C),d(B,D)) nevertheless.

Evidence. Numerical quadrature in dimension 222 (about 9,0009{,}0009,000 random quadruples plus adversarial hill-climbing) and dimension 333 (300300300 quadruples) found no violation; the ratio (I1+I2)/max⁡(d(A,C),d(B,D))(I_1+I_2)/\max(d(A,C),d(B,D))(I1​+I2​)/max(d(A,C),d(B,D)) approaches 111 only in the degenerate regime where one pair dominates the other. Two natural sufficient conditions are false and should not be attempted: the arithmetic-mean variant EP~1[ac]+EP~2[bd]≤1\mathbb E_{\tilde P_1}[ac]+\mathbb E_{\tilde P_2}[bd]\le 1EP~1​​[ac]+EP~2​​[bd]≤1 with P~i∝∣β∣/(βA+BβC+D)\tilde P_i\propto|\beta|/(\beta_{A+B}\beta_{C+D})P~i​∝∣β∣/(βA+B​βC+D​) (violations up to 1.271.271.27), and stochastic dominance of the ratio pair (βB/βA,βD/βC)(\beta_B/\beta_A,\beta_D/\beta_C)(βB​/βA​,βD​/βC​) under P1P_1P1​ versus P2P_2P2​.

Formalization note. The integrals are iterated Bochner integrals against surfaceMeasure n, exactly as in the mission's definition distance; bilinear M u v is uTMvu^{\mathsf T}MvuTMv, and Matrix.PosDef includes symmetry. The hypothesis 0 < n mirrors the mission target.

Preamble
import Definitions.Def_rybin2026_p01_matrix_integral

open Matrix MeasureTheory
Formal statement
namespace RybinAI2026.P01

/-- Sign-free kernel form of the matrix integral inequality (conjectural strengthening). -/
theorem matrix_integral_kernel_sum_le_max
    {n : ℕ} (hn : 0 < n)
    (A B C D : Matrix (Fin n) (Fin n) ℝ)
    (hA : A.PosDef) (hB : B.PosDef) (hC : C.PosDef) (hD : D.PosDef) :
    (∫ u, ∫ v, |bilinear (A - C) u.1 v.1| /
        (bilinear (A + B) u.1 u.1 * bilinear (C + D) v.1 v.1)
        ∂surfaceMeasure n ∂surfaceMeasure n) +
      (∫ u, ∫ v, |bilinear (B - D) u.1 v.1| /
        (bilinear (A + B) u.1 u.1 * bilinear (C + D) v.1 v.1)
        ∂surfaceMeasure n ∂surfaceMeasure n)
      ≤ max (distance A C) (distance B D) := by
  sorry

end RybinAI2026.P01
Source
Conjectural strengthening of CUHK-Shenzhen AI Math Problem 1 (Prof. Cosme Louart), https://rybindmitry.github.io/problems/1.html, obtained by applying the triangle inequality to the numerator of the mission target RybinAI2026.P01.matrix_integral_inequality; no literature source, supported by numerical experiments (2026-09-06).

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