TheoremBase

Noise Penalty Pairs with Closed Score Along Noise Couplings

A noise penalty pair has closed score along noise couplings if, whenever measures in the score domain with uniformly bounded scores are joined to a measure of the penalty domain by couplings of vanishing noise cost, the limit lies in the score domain and the scores converge weakly to its score along the couplings.

Statement

In the setting of First-Order Equations on the Noise Wasserstein Space Relative to a Noise Penalty Pair: Standing Notation, let (D,DΣ,E,Σ)(\mathcal{D},\mathcal{D}_{\Sigma},\mathcal{E},\Sigma) be a noise penalty pair on Pρa\mathcal{P}^{a}_{\rho}. For ν∈DΣ\nu\in\mathcal{D}_{\Sigma} the score Σ(ν)\Sigma(\nu) lies in TνaT^{a}_{\nu}, hence in L2(ν;Xa)L^{2}(\nu;X^{a}), by Noise Penalty Pairs on the Noise Wasserstein Space: the Penalty, Its Score, and Their Domains §pair. Sequences of couplings of vanishing noise cost and weak convergence along them are those of First-Order Equations on the Noise Wasserstein Space Relative to a Noise Penalty Pair: Standing Notation §couplings, the measures involved lying in D⊆Pρa\mathcal{D}\subseteq\mathcal{P}^{a}_{\rho}.

(Closed score along noise couplings) The noise penalty pair (D,DΣ,E,Σ)(\mathcal{D},\mathcal{D}_{\Sigma},\mathcal{E},\Sigma) has closed score along noise couplings if the following holds for every nonnegative R∈RR\in\mathbb{R}. Let (νn)n∈N(\nu_{n})_{n\in\mathbb{N}} be a sequence in DΣ\mathcal{D}_{\Sigma} with

∥Σ(νn)∥νn≤R(n∈N),\lVert\Sigma(\nu_{n})\rVert_{\nu_{n}}\le R\qquad(n\in\mathbb{N}),

let ν∈D\nu\in\mathcal{D}, and let (πn)n∈N(\pi_{n})_{n\in\mathbb{N}} be a sequence of couplings of vanishing noise cost from (νn)n∈N(\nu_{n})_{n\in\mathbb{N}} to ν\nu. Then ν∈DΣ\nu\in\mathcal{D}_{\Sigma}, so that Σ(ν)∈Tνa⊆L2(ν;Xa)\Sigma(\nu)\in T^{a}_{\nu}\subseteq L^{2}(\nu;X^{a}) by Noise Penalty Pairs on the Noise Wasserstein Space: the Penalty, Its Score, and Their Domains §pair, and the sequence (Σ(νn))n∈N(\Sigma(\nu_{n}))_{n\in\mathbb{N}} converges weakly to Σ(ν)\Sigma(\nu) along (πn)n∈N(\pi_{n})_{n\in\mathbb{N}}.

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