Proof of Basic Properties of a Wasserstein-Closed Penalty Pair: Lower Bound, Lower Semicontinuity, Complete Sublevel Sets, Bounded Distances, and Coercive Pairs

Each result cited is universally quantified over the data in its own statement.

We work in the setting of The Intrinsic Calculus on the Wasserstein Space: Standing Notation, and elementary order and arithmetic of real numbers (including square roots and squares of nonnegative reals, and the Archimedean property) is carried by The Real Numbers: Standing Notation and Background §background, which we adopt for that purpose. Throughout, $(\mathcal{P}_{2}(\mathbb{R}^{d}),W_{2})$ is a metric space by The Quadratic Wasserstein Distance is a Metric on the Wasserstein Space §metric, and $W_{2}$ is symmetric by The Quadratic Wasserstein Distance is a Metric on the Wasserstein Space §symmetry. By Penalty Pairs on the Wasserstein Space: the Penalty, Its Score, and Their Domains §pair, $\mathcal{D}\subseteq\mathcal{P}_{2}(\mathbb{R}^{d})$ and $\mathcal{E}:\mathcal{D}\to\mathbb{R}$, so $\mathcal{D}_{c}\subseteq\mathcal{P}_{2}(\mathbb{R}^{d})$ for every $c\in\mathbb{R}$.

Claim 1 (bounded below). Step 1. Apply condition 2 of the Wasserstein-closed property with level $0$: there is $B_{0}\in\mathbb{R}$ with $M_{2}(\mu)\le B_{0}$ for every $\mu\in\mathcal{D}$ with $\mathcal{E}(\mu)\le0$. By Penalty Pairs on the Wasserstein Space: the Penalty, Its Score, and Their Domains §bound there is a nonnegative $C\in\mathbb{R}$ with $-C(1+M_{2}(\mu))\le\mathcal{E}(\mu)$ for every $\mu\in\mathcal{D}$. Put $e_{0}=-C(1+|B_{0}|)$; since $C\ge0$ and $1+|B_{0}|>0$, we have $e_{0}\le0$.

Step 2. Let $\mu\in\mathcal{D}$. If $\mathcal{E}(\mu)\le0$, then $M_{2}(\mu)\le B_{0}\le|B_{0}|$ by Step 1, so $1+M_{2}(\mu)\le1+|B_{0}|$ and, $C$ being nonnegative, $e_{0}=-C(1+|B_{0}|)\le-C(1+M_{2}(\mu))\le\mathcal{E}(\mu)$. If $\mathcal{E}(\mu)>0$, then $e_{0}\le0<\mathcal{E}(\mu)$. In both cases $e_{0}\le\mathcal{E}(\mu)$.

Claim 2 (lower semicontinuity). We verify Lower Semicontinuous Function on a Subset of a Metric Space for the metric space $(\mathcal{P}_{2}(\mathbb{R}^{d}),W_{2})$, the subset $A=\mathcal{D}$ and $u=\mathcal{E}$. Let $x\in\mathcal{D}$ and let $\varepsilon\in\mathbb{R}$ be positive; we argue by contradiction and suppose that no positive $\delta$ has the required property.

Step 1. Then for every $n\in\mathbb{N}$, applying the supposition to $\delta=\tfrac1n$ (positive, as $n$ is positive in $\mathbb{R}$ by The Real Numbers: Standing Notation and Background §numbers), there is $y_{n}\in\mathcal{D}$ with $W_{2}(x,y_{n})<\tfrac1n$ and $\mathcal{E}(y_{n})\le\mathcal{E}(x)-\varepsilon$; we choose one such $y_{n}$ for each $n$, obtaining a sequence $(y_{n})_{n\in\mathbb{N}}$ in $\mathcal{D}$.

Step 2. The sequence $(y_{n})$ converges to $x$ in $(\mathcal{P}_{2}(\mathbb{R}^{d}),W_{2})$ in the sense of Convergent Sequence in a Metric Space: given a positive $\varepsilon'$, claim 3 of The Archimedean Property of the Real Numbers gives $N\in\mathbb{N}$ with $0<\tfrac1N<\varepsilon'$, and for every $n\ge N$ we have $\tfrac1n\le\tfrac1N$, hence $W_{2}(y_{n},x)=W_{2}(x,y_{n})<\tfrac1n<\varepsilon'$ by symmetry of $W_{2}$.

Step 3. Put $c=\mathcal{E}(x)-\varepsilon$. Every $y_{n}$ lies in $\mathcal{D}$ with $\mathcal{E}(y_{n})\le c$, and $(y_{n})$ converges to $x\in\mathcal{P}_{2}(\mathbb{R}^{d})$, so condition 1 of the Wasserstein-closed property at level $c$ gives $\mathcal{E}(x)\le c=\mathcal{E}(x)-\varepsilon$, that is $\varepsilon\le0$, contradicting $0<\varepsilon$. Hence some positive $\delta$ has the property that every $y\in\mathcal{D}$ with $W_{2}(x,y)<\delta$ satisfies $\mathcal{E}(x)-\varepsilon<\mathcal{E}(y)$: indeed the negation of this property for $\delta=\tfrac1n$ is exactly the existence of $y_{n}$ in Step 1, the negation of $\mathcal{E}(x)-\varepsilon<\mathcal{E}(y)$ being $\mathcal{E}(y)\le\mathcal{E}(x)-\varepsilon$ by totality of the order. As $x\in\mathcal{D}$ and $\varepsilon$ were arbitrary, $\mathcal{E}$ is lower semicontinuous on $\mathcal{D}$ relative to $\mathcal{D}$.

Claim 3 (complete sublevel sets). Fix $c\in\mathbb{R}$.

Step 1. By claim 1 of The Restriction of a Metric to a Subset Induces the Subspace Topology, applied to the metric space $(\mathcal{P}_{2}(\mathbb{R}^{d}),W_{2})$ and the subset $\mathcal{D}_{c}$, the restriction of $W_{2}$ to $\mathcal{D}_{c}\times\mathcal{D}_{c}$ is a metric on $\mathcal{D}_{c}$, so $(\mathcal{D}_{c},W_{2})$ is a metric space.

Step 2. Let $(\mu_{n})_{n\in\mathbb{N}}$ be a Cauchy sequence in $(\mathcal{D}_{c},W_{2})$. Since the restricted metric takes the same values $W_{2}(\mu_{m},\mu_{\ell})$, the sequence is also a Cauchy sequence in $(\mathcal{P}_{2}(\mathbb{R}^{d}),W_{2})$, so by Completeness of the Quadratic Wasserstein Space over Euclidean Space §complete it converges in $(\mathcal{P}_{2}(\mathbb{R}^{d}),W_{2})$ to some $\mu\in\mathcal{P}_{2}(\mathbb{R}^{d})$.

Step 3. Each $\mu_{n}$ lies in $\mathcal{D}$ with $\mathcal{E}(\mu_{n})\le c$, so condition 1 of the Wasserstein-closed property gives $\mu\in\mathcal{D}$ and $\mathcal{E}(\mu)\le c$, that is $\mu\in\mathcal{D}_{c}$. The distances $W_{2}(\mu_{n},\mu)$ are the same in $(\mathcal{D}_{c},W_{2})$, so by Convergent Sequence in a Metric Space the sequence converges to the point $\mu$ of $\mathcal{D}_{c}$ in $(\mathcal{D}_{c},W_{2})$. Hence $(\mathcal{D}_{c},W_{2})$ is complete.

Claim 4 (bounded distances). Let $c,B$ be as in the claim and let $\mu,\nu\in\mathcal{D}_{c}$. Both lie in $\mathcal{P}_{2}(\mathbb{R}^{d})\subseteq\mathcal{P}(\mathbb{R}^{d})$, so the product coupling $\pi=\mu\boxtimes\nu$ belongs to $\Pi(\mu,\nu)$ by Couplings on Euclidean Space: Product Coupling, Swap, Finiteness of the Cost, Push-Forward Couplings, Modifying One Marginal, Quantisation, Gluing over a Finitely Supported Measure, and the Lipschitz Bound §product, and Couplings on Euclidean Space: Product Coupling, Swap, Finiteness of the Cost, Push-Forward Couplings, Modifying One Marginal, Quantisation, Gluing over a Finitely Supported Measure, and the Lipschitz Bound §cost-finite gives $I(\pi)\le2M_{2}(\mu)+2M_{2}(\nu)\le2B+2B=4B$. By The Quadratic Wasserstein Distance on Euclidean Space §distance, $W_{2}(\mu,\nu)^{2}\le I(\pi)$, hence $W_{2}(\mu,\nu)^{2}\le4B$.

Claim 5 (coercive pairs). Let $(\mathcal{D}',\mathcal{D}'_{\Sigma},\mathcal{E}',\Sigma')$ be a Wasserstein-coercive penalty pair, not assumed to be Wasserstein-closed, and fix $c\in\mathbb{R}$; we verify the two conditions of Wasserstein-Closed Penalty Pairs §w2-closed for this pair at level $c$. Put $\mathcal{D}'_{c}=\{\mu\in\mathcal{D}':\mathcal{E}'(\mu)\le c\}$. By Wasserstein-Coercive Penalty Pairs §coercive, $K=\mathcal{D}'_{c}$ is sequentially compact in $(\mathcal{P}_{2}(\mathbb{R}^{d}),W_{2})$.

Step 1 (closed sublevel sets). Let $(\mu_{n})_{n\in\mathbb{N}}$ be a sequence in $\mathcal{D}'$ with $\mathcal{E}'(\mu_{n})\le c$ for every $n$, converging to some $\mu\in\mathcal{P}_{2}(\mathbb{R}^{d})$. Every $\mu_{n}$ lies in $K$, so by Sequentially Compact Subset of a Metric Space there are $x\in K$ and a strictly increasing sequence $(n_{k})_{k\in\mathbb{N}}$ in $\mathbb{N}$ such that $(\mu_{n_{k}})_{k\in\mathbb{N}}$ converges to $x$. By A Subsequence of a Convergent Sequence Has the Same Limit the subsequence $(\mu_{n_{k}})$ also converges to $\mu$, so $x=\mu$ by Uniqueness of Limits in a Metric Space. Hence $\mu\in K$, that is $\mu\in\mathcal{D}'$ and $\mathcal{E}'(\mu)\le c$.

Step 2 (bounded sublevel sets). If $K$ is empty, $B=0$ satisfies condition 2 vacuously. Otherwise $K$ is nonempty and, by A Sequentially Compact Subset of a Metric Space is Compact, compact in $\mathcal{P}_{2}(\mathbb{R}^{d})$ with the topology of open sets of $(\mathcal{P}_{2}(\mathbb{R}^{d}),W_{2})$. Define $f:K\to\mathbb{R}$ by $f(\mu)=\sqrt{M_{2}(\mu)}$, the nonnegative square root of the nonnegative real number $M_{2}(\mu)$; here $M_{2}(\mu)$ is a nonnegative real number since $\mu\in K\subseteq\mathcal{P}_{2}(\mathbb{R}^{d})$, by The Second Moment of a Probability Measure on Euclidean Space and the Probability Measures with Finite Second Moment §moment. For $x\in K$ and positive $\varepsilon$ take $\delta=\varepsilon$: if $y\in K$ and $W_{2}(x,y)<\delta$, then The Coordinate Fields of a Coupling, and the Second Moment as a Lipschitz Function of the Wasserstein Distance §lipschitz, read with $m=d$ and applied to $y$ and $x$, together with symmetry of $W_{2}$ gives $|f(y)-f(x)|\le W_{2}(y,x)=W_{2}(x,y)<\varepsilon$. Thus $f$ has the continuity property of Extreme Value Theorem on a Compact Subset of a Metric Space, which gives $x_{\max}\in K$ with $f(\mu)\le f(x_{\max})$ for every $\mu\in K$. Put $B=M_{2}(x_{\max})$. For $\mu\in K$, squaring the inequality $0\le f(\mu)\le f(x_{\max})$ between nonnegative reals gives $M_{2}(\mu)=f(\mu)^{2}\le f(x_{\max})^{2}=B$. This is condition 2 at level $c$.

As $c$ was arbitrary, the pair $(\mathcal{D}',\mathcal{D}'_{\Sigma},\mathcal{E}',\Sigma')$ is Wasserstein-closed. $\blacksquare$