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PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck pass
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Verified Lyapunov residual, descent, and stability checks make an inexact Newton-Kleinman step safe

Statement

Consider a trial gain Kt in the safeguarded-inexact branch of Finite-budget exact or safeguarded inexact Newton-Kleinman policy iteration, with At=A−BKt. Suppose outward certificates give W≻0, μ>0, At∗W+WAt⪯−μI, an inverse bound κ≥∥LAt−1∥F→F, a residual bound r for a Hermitian approximation P~ to At∗P+PAt+Q+Kt∗RKt=0, and an upper bound η≥λmax⁡(P~−Xj) with η+κr≤0. Then At is Hurwitz, its exact cost matrix Pt exists and is positive semidefinite, and Pt⪯Xj. Thus every accepted line-search step preserves stability and does not increase the exact closed-loop cost matrix.

Facts & Assumptions

Given: The current exact cost matrix Xj, trial gain, residual data, and verified bounds in the statement.

[F1]

Finite-budget exact or safeguarded inexact Newton-Kleinman policy iteration: a trial is accepted only if its Lyapunov margin, inverse bound, residual bound, and descent upper bound satisfy the stated tests.

[F2]

Standard CARE and DARE data, equations, and conventions: Q=C∗C⪰0 and R≻0.

[F3]

The Hurwitz Lyapunov integral is the unique solution and has the stated positivity: for Hurwitz At, the solution of At∗P+PAt+S=0 is the unique integral ∫0∞etAt∗SetAt dt, and is positive semidefinite for S⪰0.

[F4]

A Sylvester equation is uniquely solvable exactly when the spectra are disjoint: when At is Hurwitz, the Lyapunov operator E↦At∗E+EAt is invertible.

[F5]

Verified point and uniform linear residual certificate data specifies bounds for an approximate-inverse defect and a preconditioned linear residual, with a usable certificate requiring q<1. The raw Lyapunov residual bound r and inverse-norm bound κ in this proposition are separate hypotheses of the Statement.

Proof

1.1F1givenalgebra

If Atv=λv with v≠0, the certificate gives 2Re⁡(λ)v∗Wv=v∗(At∗W+WAt)v≤−μ∥v∥22<0. Because W≻0, every eigenvalue has negative real part, so At is Hurwitz.

2.1givenF1F2F3F4F5step 1.1algebra∎

Let Pt be the exact solution of the Lyapunov equation; it exists uniquely by [F3] and [F4], and Pt⪰0 by [F2]–[F3]. If Rt denotes the exact residual of P~, then LAt(P~−Pt)=Rt. The Statement's certified inverse and raw residual bounds give ∥P~−Pt∥F≤κr; these are distinct from the certificate quantities described in [F5]. For each unit vector v, v∗(Pt−Xj)v≤v∗(P~−Xj)v+∥Pt−P~∥2≤η+κr≤0. Thus Pt⪯Xj.

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