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Lemma A.3 (Bisection interval), p. 13 — the root of h lies in [λ₁(1 + √(v₁ᵀΣv₁)/ρ), λ₁(1 + √Tr[Σ]/ρ)]

Proved
WassKF.Bisect.lemma_A_3

by mikedeng1 · Oct 7, 2026 · Mathlib 0df444a (Lean v4.33.1)

bisectiondistributionally-robust-optimizationp2o-batch-pfp2bp2o-gran-per-chapterp2o-plan-paperp2o-v1semidefinite-programming

Let Σ∈S++d\Sigma \in \mathbb{S}^d_{++}Σ∈S++d​, let D∈S+dD \in \mathbb{S}^d_+D∈S+d​ with D≠0D \neq 0D=0, and let ρ>0\rho > 0ρ>0. Let λ1\lambda_1λ1​ be the largest eigenvalue of DDD and v1v_1v1​ a corresponding eigenvector with ∥v1∥=1\|v_1\| = 1∥v1​∥=1. If γ⋆\gamma^\starγ⋆ satisfies γ⋆Id≻D\gamma^\star I_d \succ Dγ⋆Id​≻D and solves

h(γ⋆)=ρ2−⟨Σ,(Id−γ⋆(γ⋆Id−D)−1)2⟩=0,h(\gamma^\star) = \rho^2 - \big\langle \Sigma, \big(I_d - \gamma^\star(\gamma^\star I_d - D)^{-1}\big)^2\big\rangle = 0,h(γ⋆)=ρ2−⟨Σ,(Id​−γ⋆(γ⋆Id​−D)−1)2⟩=0,

then γmin⁡≤γ⋆≤γmax⁡\gamma_{\min} \le \gamma^\star \le \gamma_{\max}γmin​≤γ⋆≤γmax​, where

γmin⁡=λ1(1+v1⊤Σv1/ρ),γmax⁡=λ1(1+Tr[Σ]/ρ).\gamma_{\min} = \lambda_1\Big(1 + \sqrt{v_1^\top \Sigma v_1}/\rho\Big),\qquad \gamma_{\max} = \lambda_1\Big(1 + \sqrt{\mathrm{Tr}[\Sigma]}/\rho\Big).γmin​=λ1​(1+v1⊤​Σv1​​/ρ),γmax​=λ1​(1+Tr[Σ]​/ρ).

These are the initial bounds LBLBLB and UBUBUB of Algorithm 1.

Formalization Note The page says only that v1v_1v1​ is "a corresponding eigenvector"; the normalisation v1⊤v1=1v_1^\top v_1 = 1v1⊤​v1​=1 is added, because γmin⁡\gamma_{\min}γmin​ scales with ∥v1∥\|v_1\|∥v1​∥ and the lemma is false for a rescaled eigenvector. "The largest eigenvalue" is encoded as λ1Id−D⪰0\lambda_1 I_d - D \succeq 0λ1​Id​−D⪰0 together with Dv1=λ1v1Dv_1 = \lambda_1 v_1Dv1​=λ1​v1​. The lemma is stated for every nonzero D⪰0D \succeq 0D⪰0 (the class of Lemmas A.1–A.2), which contains D=∇f(S)D = \nabla f(S)D=∇f(S); the page's "the solution" is read as "every solution with γ⋆Id≻D\gamma^\star I_d \succ Dγ⋆Id​≻D".

Preamble
import Mathlib
import Definitions.Def_WassKF_Bisect_hFun

open Matrix
Formal statement
namespace WassKF.Bisect

/-- Lemma A.3 (Bisection interval), p. 13: for `ρ > 0`, every solution `γ⋆` with `γ⋆I_d ≻ D` of
`h(γ⋆) = 0` lies in `[γ_min, γ_max]`, `γ_min = λ₁(1 + √(v₁ᵀΣv₁)/ρ)`, `γ_max = λ₁(1 + √Tr[Σ]/ρ)`
(A.7), where `λ₁` is the largest eigenvalue of `D` (`λ₁I − D ⪰ 0` and `Dv₁ = λ₁v₁`) and `v₁` is a
corresponding eigenvector, normalised to `v₁ᵀv₁ = 1` (added: the page says only "an eigenvector").
`D` is any nonzero positive semidefinite matrix, as in Lemmas A.1–A.2. -/
theorem lemma_A_3 {n m : ℕ} (Sigma D : Matrix (Fin n ⊕ Fin m) (Fin n ⊕ Fin m) ℝ)
    (ρ : ℝ) (hρ : 0 < ρ) (hSigma : Sigma.PosDef) (hD : D.PosSemidef) (hD0 : D ≠ 0)
    (lam1 : ℝ) (v₁ : Fin n ⊕ Fin m → ℝ)
    (hlam : (lam1 • (1 : Matrix (Fin n ⊕ Fin m) (Fin n ⊕ Fin m) ℝ) - D).PosSemidef)
    (hv : D *ᵥ v₁ = lam1 • v₁) (hv1 : v₁ ⬝ᵥ v₁ = 1)
    (γstar : ℝ)
    (hγstar : (γstar • (1 : Matrix (Fin n ⊕ Fin m) (Fin n ⊕ Fin m) ℝ) - D).PosDef)
    (hroot : hFun Sigma D ρ γstar = 0) :
    lam1 * (1 + Real.sqrt (v₁ ⬝ᵥ (Sigma *ᵥ v₁)) / ρ) ≤ γstar ∧
      γstar ≤ lam1 * (1 + Real.sqrt Sigma.trace / ρ) := by sorry

end WassKF.Bisect
Source
Shafieezadeh-Abadeh, Nguyen, Kuhn, Mohajerin Esfahani, Wasserstein Distributionally Robust Kalman Filtering, arXiv:1809.08830v3, p. 13, Lemma A.3, (A.7)
Human review
  • Endorsed by Shuze Chen · Oct 8, 2026

    Confirmed by the moderator at approval.

  • Endorsed by mikedeng1 · Oct 8, 2026

    Confirmed by the mission captain (proposal self-audit).

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