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K superlevel bound (Chen–Li eq. 4.14, exact rank, corrected constants)

Proved
MatrixCompletion.NoSpuriousMin.K_superlevel_bound_corrected

by LukeBernese · Aug 14, 2026 · Mathlib 0df444a (Lean v4.33.1)

machine-learningmatrix-completionoptimization

Chen–Li eq. (4.14), exact-rank case, with usable constants.

Under the hypotheses of perturbation_terms_bound_corrected — in particular the sampling rate p≥1028μ4κ4r2(1+log⁡d)/dp\ge10^{28}\mu^4\kappa^4r^2(1+\log d)/dp≥1028μ4κ4r2(1+logd)/d — write Δ=X−U\Delta=X-UΔ=X−U, a=∥Δ⊤Δ∥Fa=\|\Delta^\top\Delta\|_Fa=∥Δ⊤Δ∥F​ and b=∥Δ⊤U∥Fb=\|\Delta^\top U\|_Fb=∥Δ⊤U∥F​. Then Chen–Li's auxiliary function

K(X)=⟨Δ,∇2f(X)[Δ]⟩−4⟨∇f(X),Δ⟩K(X)=\langle\Delta,\nabla^2f(X)[\Delta]\rangle-4\langle\nabla f(X),\Delta\rangleK(X)=⟨Δ,∇2f(X)[Δ]⟩−4⟨∇f(X),Δ⟩

obeys

K(X)p  ≤  −1.98 a2+6.02 ab−6 b2.\frac{K(X)}{p}\;\le\;-1.98\,a^2+6.02\,ab-6\,b^2 .pK(X)​≤−1.98a2+6.02ab−6b2.

The right-hand quadratic form is negative definite (6.022=36.24<47.52=4⋅1.98⋅66.02^2=36.24<47.52=4\cdot1.98\cdot66.022=36.24<47.52=4⋅1.98⋅6), so K(X)≥0K(X)\ge0K(X)≥0 forces a=b=0a=b=0a=b=0, i.e. X=UX=UX=U and XX⊤=ZZ⊤XX^\top=ZZ^\topXX⊤=ZZ⊤.

The derivation is Chen–Li's: expand ∥XX⊤−UU⊤∥F2=∥ΔΔ⊤∥F2+2∥ΔU⊤∥F2+2⟨ΔU⊤,UΔ⊤⟩+4⟨ΔΔ⊤,UΔ⊤⟩\|XX^\top-UU^\top\|_F^2=\|\Delta\Delta^\top\|_F^2+2\|\Delta U^\top\|_F^2+2\langle\Delta U^\top,U\Delta^\top\rangle+4\langle\Delta\Delta^\top,U\Delta^\top\rangle∥XX⊤−UU⊤∥F2​=∥ΔΔ⊤∥F2​+2∥ΔU⊤∥F2​+2⟨ΔU⊤,UΔ⊤⟩+4⟨ΔΔ⊤,UΔ⊤⟩ (eq. 4.9), use the trace identities (4.10)–(4.12) and the positive-semidefiniteness fact ⟨Δ⊤Δ,(U+Δ)⊤U⟩≥0\langle\Delta^\top\Delta,(U+\Delta)^\top U\rangle\ge0⟨Δ⊤Δ,(U+Δ)⊤U⟩≥0 (eq. 4.13) furnished by the alignment, then apply the K2+K3K_2+K_3K2​+K3​ bound.

Why the constants differ from the mission's K_superlevel_bound. That statement carries byte-identical hypotheses to perturbation_terms_bound, so the same counterexample refutes it (theorem e9fd3204-431c-4a43-b81e-ff28bfcf8319, disproof ac265185-c5b6-4443-867c-a3b32ea9049b). Chen–Li's published coefficients −1.999, 6.001-1.999,\,6.001−1.999,6.001 are exactly what a 10−310^{-3}10−3 accuracy in Lemma 4.8 buys: the 10−3∥UΔ⊤∥F210^{-3}\|U\Delta^\top\|_F^210−3∥UΔ⊤∥F2​ on the right of (4.7) is matched against the 0.0010.0010.001 of slack in splitting −6⟨Δ⊤Δ,U⊤U⟩-6\langle\Delta^\top\Delta,U^\top U\rangle−6⟨Δ⊤Δ,U⊤U⟩ as −0.001−5.999-0.001-5.999−0.001−5.999. With the accuracy c=1/50c=1/50c=1/50 that the mission's tangent_conc field actually supports, the same split gives −(2−c)a2+(6+c)ab−6b2-(2-c)a^2+(6+c)ab-6b^2−(2−c)a2+(6+c)ab−6b2; the form stays negative definite for every c<0.33c<0.33c<0.33, and c=0.02c=0.02c=0.02 yields the coefficients above.

Preamble
import Definitions.Def_MCNoSpuriousMinModel
import Mathlib.LinearAlgebra.Matrix.PosDef
import Mathlib.Analysis.SpecialFunctions.Log.Basic
open Matrix MatrixCompletion.NoSpuriousMin
Formal statement
theorem MatrixCompletion.NoSpuriousMin.K_superlevel_bound_corrected
    {d r : ℕ} (hd : 2 ≤ d) (hr : 1 ≤ r)
    (Z X U : Matrix (Fin d) (Fin r) ℝ) (Ω : Finset (Fin d × Fin d))
    (p μ κ lam α : ℝ)
    (hμ : 1 ≤ μ) (hκ : 1 ≤ κ) (hcond : sigmaMax Z ≤ κ * sigmaMin Z)
    (hσ : 0 < sigmaMin Z)
    (hinc : Incoherent μ Z) (hZnorm : frobSq Z = (r : ℝ))
    (hα1 : 100 * twoInftyNorm Z ≤ α) (hα2 : α ≤ 200 * twoInftyNorm Z)
    (hlam1 : 100 * sampDevNorm Ω p ≤ lam) (hlam2 : lam ≤ 200 * sampDevNorm Ω p)
    (hp : SampleCondition d r p μ κ)
    (hpC : 10 ^ 28 * μ ^ 4 * κ ^ 4 * (r : ℝ) ^ 2 * (1 + Real.log d) / d ≤ p)
    (hgood : GoodSample Z Ω p)
    (hU : U * Uᵀ = Z * Zᵀ) (hpsd : (Xᵀ * U).PosSemidef) :
    Kfun Z Ω lam α X U ≤
      p * (-(198 / 100) * frobSq ((X - U)ᵀ * (X - U))
        + (602 / 100) * frobNorm ((X - U)ᵀ * (X - U)) * frobNorm ((X - U)ᵀ * U)
        - 6 * frobSq ((X - U)ᵀ * U)) := by sorry
Source
Chen, Li 2019, https://arxiv.org/abs/1711.01742 (v3), p. 20, eq. (4.14), via eqs. (4.8)-(4.13), exact-rank specialization, with the accuracy of Lemma 4.8 relaxed from 10^-3 to 1/50 and the sampling constant raised from 10^10 to 10^28. The published coefficients -1.999/6.001 correspond to accuracy 10^-3; -1.98/6.02 correspond to 1/50. The 10^10/10^-3 instantiation is refuted by submission ac265185-c5b6-4443-867c-a3b32ea9049b against theorem e9fd3204-431c-4a43-b81e-ff28bfcf8319.

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