Proof of Adding a Constant to a Viscosity Subsolution or Supersolution on the Wasserstein Space

Each result cited is universally quantified over the data in its own statement, and is applied to the data named here. Throughout, $\mathcal{D}$ is regarded as a subset of the metric space $(\mathcal{P}_{2}(\mathbb{R}^{d}),W_{2})$ of Square-Integrable Vector Fields Against a Probability Measure on Euclidean Space, and Test Functions: Standing Notation §measures, as in The Delta-Envelopes of a Function on the Penalty Domain Relative to a Penalty Pair; it is nonempty, since it contains $\mathcal{D}_{\Sigma}$, which is nonempty (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). For $a,b,c\in\mathbb{R}$ we use that $a\le b$ holds if and only if $a+c\le b+c$, by the compatibility of the order with addition in the ordered field $\mathbb{R}$; and for a real function $w$ on $\mathcal{D}$, $w+s$ denotes the function with value $w(\mu)+s$ at $\mu$.

Step 1 (Semicontinuous envelopes of a translate). Let $g:\mathcal{D}\to\mathbb{R}$ and let $x\in\mathcal{D}$; let $A_{g}(x)$, $B_{g}(x)$ be the sets of Upper and Lower Semicontinuous Envelopes of a Real-Valued Function for the nonempty subset $\mathcal{D}$. For $c\in\mathbb{R}$ and a positive radius $\rho$, the condition $g(y)\le c$ for every $y\in\mathcal{D}$ with $W_{2}(y,x)\le\rho$ is equivalent to $g(y)+s\le c+s$ for the same $y$; hence $c\in A_{g}(x)$ if and only if $c+s\in A_{g+s}(x)$, and in the same way $c\in B_{g}(x)$ if and only if $c+s\in B_{g+s}(x)$. Consequently $g+s$ is bounded above near each point of $\mathcal{D}$ if and only if $g$ is, and likewise for bounded below.

Suppose $g$ is bounded above near each point, and let $g^{*}$ and $(g+s)^{*}$ be the upper semicontinuous envelopes, the infima of the sets $A$. If $c'\in A_{g+s}(x)$, then $c'-s\in A_{g}(x)$, so $g^{*}(x)\le c'-s$, i.e. $g^{*}(x)+s\le c'$; thus $g^{*}(x)+s$ is a lower bound of $A_{g+s}(x)$, and, the infimum being the greatest lower bound (Lower Bound and Greatest Lower Bound in a Totally Ordered Set), $g^{*}(x)+s\le(g+s)^{*}(x)$. If $c\in A_{g}(x)$, then $c+s\in A_{g+s}(x)$, so $(g+s)^{*}(x)\le c+s$, i.e. $(g+s)^{*}(x)-s\le c$; thus $(g+s)^{*}(x)-s$ is a lower bound of $A_{g}(x)$ and $(g+s)^{*}(x)-s\le g^{*}(x)$. Hence

If $g$ is bounded below near each point, the same argument with the sets $B$, upper bounds in place of lower bounds and the lower semicontinuous envelopes, the suprema of these sets, being least upper bounds (Upper Bound and Least Upper Bound), gives

Step 2 (Claim 1). By Viscosity Subsolution, Supersolution and Solution of a Second-Order Equation on the Wasserstein Space §subsolution, $u$ has penalty-subordinate growth from above. Let $\delta>0$; by Penalty-Subordinate Growth of a Function on the Penalty Domain §above there is $C\in\mathbb{R}$ with $u(\mu)\le C+\delta\,\mathcal{E}(\mu)$ for every $\mu\in\mathcal{D}$, hence $u(\mu)+s\le(C+s)+\delta\,\mathcal{E}(\mu)$; as $\delta$ was arbitrary, $u+s$ has penalty-subordinate growth from above. By The Delta-Envelopes of a Function on the Penalty Domain Relative to a Penalty Pair §minus, $u-\delta\mathcal{E}$ is bounded above near each point of $\mathcal{D}$, $u^{-}_{\delta}=(u-\delta\mathcal{E})^{*}$ and $(u+s)^{-}_{\delta}=((u+s)-\delta\mathcal{E})^{*}$; as $(u+s)-\delta\mathcal{E}=(u-\delta\mathcal{E})+s$ pointwise, (1a) with $g=u-\delta\mathcal{E}$ gives

Now let $\delta>0$, let $\varphi$ be an intrinsic test function on $\mathcal{D}$, let $\hat{\mu}\in\mathcal{D}$ be a point at which the function with value $(u+s)^{-}_{\delta}(\mu)-\varphi(\mu)$ has a local maximum relative to $\mathcal{D}$, and let $\varepsilon>0$. By (2a) this function is $\mu\mapsto\bigl(u^{-}_{\delta}(\mu)-\varphi(\mu)\bigr)+s$; so there is a positive $\rho$ with $\bigl(u^{-}_{\delta}(\mu)-\varphi(\mu)\bigr)+s\le\bigl(u^{-}_{\delta}(\hat{\mu})-\varphi(\hat{\mu})\bigr)+s$ for every $\mu\in\mathcal{D}$ with $W_{2}(\hat{\mu},\mu)<\rho$, and subtracting $s$ shows that $u^{-}_{\delta}-\varphi$ has a local maximum relative to $\mathcal{D}$ at $\hat{\mu}$ (Local Maximum of a Function Relative to a Subset of a Metric Space). As $u$ is a viscosity subsolution of $F$, Viscosity Subsolution, Supersolution and Solution of a Second-Order Equation on the Wasserstein Space §subsolution with $\delta$, $\varphi$, $\hat{\mu}$ and $\varepsilon$ gives $\nu\in\mathcal{D}_{\Sigma}$, $\pi\in\Pi(\nu,\hat{\mu})$, $t\in\mathbb{R}$, $q\in L^{2}(\nu;\mathbb{R}^{d})$ and $Y\in\mathcal{S}(d)$ with

Put $t'=t+s$. By (2a), $(u+s)^{-}_{\delta}(\nu)-(u+s)^{-}_{\delta}(\hat{\mu})=u^{-}_{\delta}(\nu)-u^{-}_{\delta}(\hat{\mu})$ and $t'-(u+s)^{-}_{\delta}(\hat{\mu})=t-u^{-}_{\delta}(\hat{\mu})$, so the second and third conditions hold for $u+s$ with $t'$ in place of $t$; the first, fourth and fifth do not involve the function. Finally, by The Bundle of Vector Fields over a Set of Measures, Second-Order Equation Operators on the Wasserstein Space, and Their Delta-Shifts §shifted, with $r=t+\delta\,\mathcal{E}(\nu)$, $q'=q+\delta\,\Sigma(\nu)\in L^{2}(\nu;\mathbb{R}^{d})$ (so $(\nu,q')\in\mathcal{V}(\mathcal{D}_{\Sigma})$) and $Y'=Y+\delta\,H_{\mathcal{E}}(\nu)\in\mathcal{S}(d)$, and by the hypothesis of claim 1 at $(\nu,r,q',Y')$,

So $\nu$, $\pi$, $t'$, $q$, $Y$ satisfy the six conditions of Viscosity Subsolution, Supersolution and Solution of a Second-Order Equation on the Wasserstein Space §subsolution for $u+s$, $F'$, $\delta$, $\varphi$, $\hat{\mu}$ and $\varepsilon$. As these were arbitrary, $u+s$ is a viscosity subsolution of $F'$ relative to the penalty pair.

Step 3 (Claim 2). By Viscosity Subsolution, Supersolution and Solution of a Second-Order Equation on the Wasserstein Space §supersolution, $u$ has penalty-subordinate growth from below; given $\delta>0$ and $C$ with $-C-\delta\,\mathcal{E}(\mu)\le u(\mu)$ on $\mathcal{D}$ (Penalty-Subordinate Growth of a Function on the Penalty Domain §below), we get $-(C-s)-\delta\,\mathcal{E}(\mu)\le u(\mu)+s$, so $u+s$ has penalty-subordinate growth from below. By The Delta-Envelopes of a Function on the Penalty Domain Relative to a Penalty Pair §plus, $u^{+}_{\delta}=(u+\delta\mathcal{E})_{*}$ and $(u+s)^{+}_{\delta}=((u+\delta\mathcal{E})+s)_{*}$, so (1b) gives $(u+s)^{+}_{\delta}(\mu)=u^{+}_{\delta}(\mu)+s$ for $\mu\in\mathcal{D}$. The rest is Step 2 with local minima (Local Minimum of a Function Relative to a Subset of a Metric Space) in place of local maxima and Viscosity Subsolution, Supersolution and Solution of a Second-Order Equation on the Wasserstein Space §supersolution in place of the subsolution clause: a local minimum of $(u+s)^{+}_{\delta}-\varphi$ at $\hat{\mu}$ is one of $u^{+}_{\delta}-\varphi$; the data $\nu,\pi,t,q,Y$ provided for $u$ and $F$, with $-\varepsilon\le F^{+}_{\delta}(\nu,t,q,Y)$, give for $u+s$ the data $\nu,\pi,t+s,q,Y$; and with $r=t-\delta\,\mathcal{E}(\nu)$, $q'=q-\delta\,\Sigma(\nu)$, $Y'=Y-\delta\,H_{\mathcal{E}}(\nu)$, the hypothesis of claim 2 gives

Hence $u+s$ is a viscosity supersolution of $F'$ relative to the penalty pair.