Proof of The Delta-Envelopes of Bounded Functions and Their Monotonicity in the Weight, for a Wasserstein-Coercive Penalty Pair

Each result cited is universally quantified over the data in its own statement, and is applied to the data named here. By Basic Properties of a Wasserstein-Coercive Penalty Pair §bounded-below fix $e_{0}\in\mathbb{R}$ with $e_{0}\le\mathcal{E}(\mu)$ for every $\mu\in\mathcal{D}$; by Basic Properties of a Wasserstein-Coercive Penalty Pair §lsc, $\mathcal{E}$ is lower semicontinuous on $\mathcal{D}$ relative to $\mathcal{D}$ in $(\mathcal{P}_{2}(\mathbb{R}^{d}),W_{2})$. The set $\mathcal{D}$ contains the nonempty set $\mathcal{D}_{\Sigma}$ (Penalty Pairs on the Wasserstein Space: the Penalty, Its Score, and Their Domains §pair, Penalty Pairs on the Wasserstein Space: the Penalty, Its Score, and Their Domains §nonempty), so it is nonempty. For positive $\delta$ and $\mu\in\mathcal{D}$, multiplying $e_{0}\le\mathcal{E}(\mu)$ by the nonnegative $\delta$ gives $\delta e_{0}\le\delta\,\mathcal{E}(\mu)$ (claim 5 of Elementary Arithmetic in an Ordered Field).

Claim 1. Suppose $u(\mu)\le b$ for every $\mu\in\mathcal{D}$, and let $\delta$ be positive. Put $C=b-\delta e_{0}$. For $\mu\in\mathcal{D}$, adding $b-\delta e_{0}$ to both sides of $\delta e_{0}\le\delta\,\mathcal{E}(\mu)$ gives $b\le C+\delta\,\mathcal{E}(\mu)$, so $u(\mu)\le C+\delta\,\mathcal{E}(\mu)$. As $\delta$ was arbitrary, $u$ has penalty-subordinate growth from above (Penalty-Subordinate Growth of a Function on the Penalty Domain §above). Suppose instead $b\le u(\mu)$ for every $\mu\in\mathcal{D}$, and put $C=-b-\delta e_{0}$. For $\mu\in\mathcal{D}$, adding $b-\delta\,\mathcal{E}(\mu)$ to both sides of $\delta e_{0}\le\delta\,\mathcal{E}(\mu)$ gives $-C-\delta\,\mathcal{E}(\mu)=b+\delta e_{0}-\delta\,\mathcal{E}(\mu)\le b\le u(\mu)$, so $u$ has penalty-subordinate growth from below (Penalty-Subordinate Growth of a Function on the Penalty Domain §below).

Claim 2. Suppose $u(\mu)\le b$ for every $\mu\in\mathcal{D}$. By Claim 1, $u$ has penalty-subordinate growth from above; since $0\,\mathcal{E}(\mu)=0$ (claim 1 of Zero Products and Elementary Identities in a Field), $u(\mu)\le b+0\,\mathcal{E}(\mu)$ for every $\mu\in\mathcal{D}$. Apply Basic Properties of the Delta-Envelopes on the Wasserstein Space §bound with $C=b$ and $\eta=0$, for which $0\le\eta\le\delta$ holds, $\mathcal{E}$ being lower semicontinuous on $\mathcal{D}$: it gives $u^{-}_{\delta}(\mu)\le b-(\delta-0)\,\mathcal{E}(\mu)=b-\delta\,\mathcal{E}(\mu)$ for every $\mu\in\mathcal{D}$. If instead $b\le u(\mu)$ for every $\mu\in\mathcal{D}$, then $u$ has penalty-subordinate growth from below by Claim 1 and $-(-b)-0\,\mathcal{E}(\mu)=b\le u(\mu)$; the second part of Basic Properties of the Delta-Envelopes on the Wasserstein Space §bound, with $C=-b$ and $\eta=0$, gives $b+\delta\,\mathcal{E}(\mu)\le u^{+}_{\delta}(\mu)$ for every $\mu\in\mathcal{D}$.

Claim 3, from above. Suppose $u$ has penalty-subordinate growth from above, and put $\kappa=\delta'-\delta$, positive by claim 1 of Elementary Order Arithmetic in an Ordered Field. The function $-\mathcal{E}$ on $\mathcal{D}$, with value $-\mathcal{E}(\nu)$ at $\nu$, is upper semicontinuous on $\mathcal{D}$ relative to $\mathcal{D}$: given $\nu\in\mathcal{D}$ and a positive $\varepsilon$, the lower semicontinuity of $\mathcal{E}$ at $\nu$ provides a positive radius within which $\mathcal{E}(\nu)-\varepsilon<\mathcal{E}(\nu')$, which is $-\mathcal{E}(\nu')<-\mathcal{E}(\nu)+\varepsilon$ by claim 4 of Elementary Order Arithmetic in an Ordered Field. By Basic Properties of the Delta-Envelopes on the Wasserstein Space §semicontinuity, $u^{-}_{\delta}$ is upper semicontinuous on $\mathcal{D}$ and

Let $h:\mathcal{D}\to\mathbb{R}$ have value $h(\nu)=u^{-}_{\delta}(\nu)+\kappa\bigl(-\mathcal{E}(\nu)\bigr)=u^{-}_{\delta}(\nu)-\kappa\,\mathcal{E}(\nu)$. By claims 2 and 1 of Sums and Nonnegative Multiples of Semicontinuous Functions, applied at every point of $\mathcal{D}$ with the nonnegative multiplier $\kappa$, $h$ is upper semicontinuous on $\mathcal{D}$ relative to $\mathcal{D}$. For $\nu\in\mathcal{D}$, adding $-\kappa\,\mathcal{E}(\nu)$ to both sides of the last display gives

The function $u-\delta'\mathcal{E}$ is bounded above near each point of $\mathcal{D}$ and $u^{-}_{\delta'}$ is its upper semicontinuous envelope, by The Delta-Envelopes of a Function on the Penalty Domain Relative to a Penalty Pair §minus read with $\delta'$. So Properties of the Upper Semicontinuous Envelope §least, in the metric space $(\mathcal{P}_{2}(\mathbb{R}^{d}),W_{2})$ with the nonempty subset $\mathcal{D}$, the function $u-\delta'\mathcal{E}$ and the upper semicontinuous majorant $h$, gives $u^{-}_{\delta'}(\mu)\le h(\mu)=u^{-}_{\delta}(\mu)-\kappa\,\mathcal{E}(\mu)$ for every $\mu\in\mathcal{D}$; adding $\kappa\,\mathcal{E}(\mu)$ to both sides is the claim.

Claim 3, from below. Suppose $u$ has penalty-subordinate growth from below. By Basic Properties of the Delta-Envelopes on the Wasserstein Space §duality, $-u$ has penalty-subordinate growth from above, $(-u)^{-}_{\delta}=-u^{+}_{\delta}$ and $(-u)^{-}_{\delta'}=-u^{+}_{\delta'}$ on $\mathcal{D}$. The part from above, applied to $-u$, gives $-u^{+}_{\delta'}(\mu)+\kappa\,\mathcal{E}(\mu)\le-u^{+}_{\delta}(\mu)$ for every $\mu\in\mathcal{D}$; adding $u^{+}_{\delta}(\mu)+u^{+}_{\delta'}(\mu)-\kappa\,\mathcal{E}(\mu)$ to both sides gives $u^{+}_{\delta}(\mu)\le u^{+}_{\delta'}(\mu)-\kappa\,\mathcal{E}(\mu)$.