Proof of Sublevel Sets of the Wall-Confined Free Energy are Closed for the Wasserstein Distance

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

Step 1 (Claim 1). Let $c$, $(\lambda_{m})_{m\in\mathbb{N}}$ and $\lambda$ be as in claim 1. Every $\lambda_{m}$ lies in $\mathcal{D}\subseteq\Sigma_{d,R}$ by The Wall-Confined Free Energy: Norm Bound, Lower Bound, Weak-Star Compact Sublevel Sets and Displacement Monotonicity of the Score §bounds, and $\lambda\in\Sigma_{d,R}$; since $(W_{2}(\lambda_{m},\lambda))_{m}$ converges to $0$, Moments of Noncommutative Laws are Lipschitz in the Wasserstein Distance §convergence shows that $\lambda_{m}\to\lambda$ weak-star. By The Wall-Confined Free Energy: Norm Bound, Lower Bound, Weak-Star Compact Sublevel Sets and Displacement Monotonicity of the Score §compact, applied with the bound $c$, there are a strictly increasing sequence $(m_{i})_{i\in\mathbb{N}}$ of natural numbers and $\lambda'\in\mathcal{D}$ such that $\lambda_{m_{i}}\to\lambda'$ weak-star, the real sequence $(\mathcal{E}(\lambda_{m_{i}}))_{i}$ converges, and $\mathcal{E}(\lambda')\le\lim_{i\to\infty}\mathcal{E}(\lambda_{m_{i}})$. The subsequence $(\lambda_{m_{i}})_{i}$ also converges weak-star to $\lambda$, and hence $\lambda=\lambda'$, both by Sequential Weak-Star Compactness of the Noncommutative Laws with a Given Norm Bound §unique. Thus $\lambda\in\mathcal{D}$. Since $\mathcal{E}(\lambda_{m_{i}})\le c$ for every $i$, claim 1 of Order Properties of Limits of Real Sequences, compared with the constant sequence $c$, gives $\lim_{i\to\infty}\mathcal{E}(\lambda_{m_{i}})\le c$, and therefore $\mathcal{E}(\lambda)\le c$.

Step 2 (Claim 2). Let $\lambda\in\mathcal{D}$ and suppose that $\mathcal{E}$ is not lower semicontinuous at $\lambda$ relative to $\mathcal{D}$ (Lower Semicontinuous Function on a Subset of a Metric Space). Then there is a real $\varepsilon>0$ such that for every real $\delta>0$ some $y\in\mathcal{D}$ satisfies $W_{2}(\lambda,y)<\delta$ and not $\mathcal{E}(\lambda)-\varepsilon<\mathcal{E}(y)$, that is, $\mathcal{E}(y)\le\mathcal{E}(\lambda)-\varepsilon$, the order of $\mathbb{R}$ being total. Let $c=\mathcal{E}(\lambda)-\varepsilon$ and $A=\{y\in\mathcal{D}:\mathcal{E}(y)\le c\}\subseteq\Sigma_{d,R}$. By the implication from claim 3 to claim 1 of Characterization of the Closure in a Metric Space by Open Balls in the metric space $(\Sigma_{d,R},W_{2})$, $\lambda$ lies in the closure of $A$, so by Sequential Characterization of the Closure in a Metric Space there is a sequence $(a_{m})_{m\in\mathbb{N}}$ in $A$ that converges to $\lambda$ in $(\Sigma_{d,R},W_{2})$; as $|W_{2}(a_{m},\lambda)-0|=W_{2}(a_{m},\lambda)$, the values of $W_{2}$ being nonnegative (The Noncommutative Quadratic Wasserstein Distance and Optimal Couplings §distance), the real sequence $(W_{2}(a_{m},\lambda))_{m}$ converges to $0$ in the sense of Limit of a Sequence of Real Numbers. Step 1, applied to $c$, $(a_{m})_{m}$ and $\lambda$, gives $\mathcal{E}(\lambda)\le\mathcal{E}(\lambda)-\varepsilon$. But $\mathcal{E}(\lambda)-\varepsilon<\mathcal{E}(\lambda)$ by claim 1 of Elementary Order Arithmetic in an Ordered Field, since $-\varepsilon<0$ by claim 4 there; this contradicts the total order. Hence $\mathcal{E}$ is lower semicontinuous at every $\lambda\in\mathcal{D}$ relative to $\mathcal{D}$, which is claim 2.