Proof of Attained Maxima on Closed Balls of the Penalty Domain, for a Wasserstein-Coercive Penalty Pair

Each result cited below is universally quantified over the data in its own statement, and is applied to the data named at the point of use. By Basic Properties of a Wasserstein-Coercive Penalty Pair §bounded-below there is $e_{0}\in\mathbb{R}$ with $e_{0}\le\mathcal{E}(\nu)$ for every $\nu\in\mathcal{D}$. The symmetry and the triangle inequality of $W_{2}$ (The Quadratic Wasserstein Distance is a Metric on the Wasserstein Space §symmetry, The Quadratic Wasserstein Distance is a Metric on the Wasserstein Space §triangle) are used without further mention.

For $\nu\in K$ we have $\delta e_{0}\le\delta\,\mathcal{E}(\nu)$ by claim 5 of Elementary Arithmetic in an Ordered Field, hence $g(\nu)\le c-\delta\,\mathcal{E}(\nu)\le c-\delta e_{0}$. The set $\{g(\nu):\nu\in K\}$ is therefore nonempty, since $\hat{\mu}\in K$, and bounded above by $c-\delta e_{0}$; let $s$ be its least upper bound, which exists by The Real Numbers: Standing Notation and Background §bounds. Then $g(\hat{\mu})\le s$.

By Existence of a Sequence of Positive Real Numbers with Limit Zero there is a sequence $(h_{n})_{n\in\mathbb{N}}$ of positive real numbers with limit $0$; replacing $h_{n}$ by $\min\{h_{n},1\}$ we may assume $h_{n}\le1$ for every $n$, the new sequence being positive by claim 9 of Elementary Order Arithmetic in an Ordered Field and still having limit $0$ by claim 2 of Order Properties of Limits of Real Sequences, squeezed between the constant sequence $0$ and $(h_{n})$. For each $n\in\mathbb{N}$, claim 3 of Approximation Property of the Supremum and the Infimum in $\mathbb{R}$, applied with $\varepsilon=h_{n}$, provides $\nu_{n}\in K$ with

Then $g(\hat{\mu})-1\le s-h_{n}<g(\nu_{n})$, so, using $g(\nu_{n})\le c-\delta\,\mathcal{E}(\nu_{n})$ and claim 1 of Elementary Order Arithmetic in an Ordered Field, $\delta\,\mathcal{E}(\nu_{n})<c-g(\hat{\mu})+1$, and multiplying by $\delta^{-1}$ (claims 7 and 10 of that lemma),

By Wasserstein-Coercive Penalty Pairs §coercive the set $\{\mu\in\mathcal{D}:\mathcal{E}(\mu)\le c'\}$ is sequentially compact in $(\mathcal{P}_{2}(\mathbb{R}^{d}),W_{2})$, so by Sequentially Compact Subset of a Metric Space there are $\nu^{*}$ in that set, in particular $\nu^{*}\in\mathcal{D}$, and a strictly increasing sequence $(n_{k})_{k\in\mathbb{N}}$ in $\mathbb{N}$ such that $(\nu_{n_{k}})_{k\in\mathbb{N}}$ converges to $\nu^{*}$.

$\nu^{*}$ lies in $K$. Let $\varepsilon\in\mathbb{R}$ be positive. By Convergent Sequence in a Metric Space there is $k$ with $W_{2}(\nu_{n_{k}},\nu^{*})<\varepsilon$, and then $W_{2}(\nu^{*},\hat{\mu})\le W_{2}(\nu^{*},\nu_{n_{k}})+W_{2}(\nu_{n_{k}},\hat{\mu})\le\varepsilon+r$, since $\nu_{n_{k}}\in K$. As $\varepsilon$ was arbitrary, $W_{2}(\nu^{*},\hat{\mu})\le r$ by Comparison of Real Numbers with Arbitrary Positive Slack §slack-above, so $\nu^{*}\in K$ and $g(\nu^{*})\le s$.

$\nu^{*}$ is a maximiser. The real sequence $(h_{n})$ converges to $0$ in the metric space $(\mathbb{R},d_{\mathbb{R}})$, $d_{\mathbb{R}}(x,y)=|x-y|$ being the metric of The Absolute Value Metric on the Real Line, so that convergence in $(\mathbb{R},d_{\mathbb{R}})$ is exactly the condition of Limit of a Sequence of Real Numbers; by A Subsequence of a Convergent Sequence Has the Same Limit the subsequence $(h_{n_{k}})_{k\in\mathbb{N}}$ converges to $0$ as well. Let $\varepsilon\in\mathbb{R}$ be positive. By Upper Semicontinuous Function on a Subset of a Metric Space there is a positive $\theta$ with $g(\nu)<g(\nu^{*})+\varepsilon$ for every $\nu\in\mathcal{D}$ with $W_{2}(\nu^{*},\nu)<\theta$. Choose $k$ so large that $W_{2}(\nu_{n_{k}},\nu^{*})<\theta$ and $h_{n_{k}}<\varepsilon$, which is possible because both conditions hold from some index on and the larger of the two indices serves. Then

so $s\le g(\nu^{*})+2\varepsilon$. As $\varepsilon$ was arbitrary, $s\le g(\nu^{*})$ by Comparison of Real Numbers with Arbitrary Positive Slack §slack-above (applied with the positive numbers $2\varepsilon$, which exhaust the positive reals by claim 8 of Elementary Order Arithmetic in an Ordered Field). Together with $g(\nu^{*})\le s$ this gives $g(\nu^{*})=s$, and so $g(\nu)\le s=g(\nu^{*})$ for every $\nu\in K$.