Limits of Penalized Maxima on a Compact Set

Let $(X,d)$ be a metric space, equipped with the collection of its subsets that are open in $(X,d)$, a topology by Metric Open Sets Form a Topology, and let $K\subseteq X$ be nonempty and compact in $X$. Let $\mathbb{R}$ be the set of real numbers with the addition, multiplication and order $\le$ of its ordered field structure; for $s,t\in\mathbb{R}$ write $s<t$ to mean that $s\le t$ and $s\ne t$, and write $s-t$ for $s+(-t)$. Equip $X\times X$ with the product metric $d_{X\times X}$ obtained from $d$ and $d$, a metric by claim 1 of The Product Metric is a Metric; by A Product of Compact Subsets is Compact in the Product Metric the set $K\times K$ is compact in $X\times X$.

Let $u:K\to\mathbb{R}$ be upper semicontinuous on $K$, let $v:K\to\mathbb{R}$ be lower semicontinuous on $K$, and let $\psi:K\times K\to\mathbb{R}$ be lower semicontinuous on $K\times K$, with respect to the metric $d_{X\times X}$, and such that $0\le\psi(x,y)$ for every $(x,y)\in K\times K$, with $\psi(x,y)=0$ if and only if $x=y$.

For $\alpha\in\mathbb{R}$ with $0<\alpha$ let $\Phi_{\alpha}:K\times K\to\mathbb{R}$ be the function

Then the following hold.

1. (Attainment) The function $u-v:K\to\mathbb{R}$ with $(u-v)(x)=u(x)-v(x)$ is upper semicontinuous on $K$, and there is exactly one $M\in\mathbb{R}$ such that $M=(u-v)(x_{M})$ for some $x_{M}\in K$ and $(u-v)(x)\le M$ for every $x\in K$. For each $\alpha$ with $0<\alpha$, the function $\Phi_{\alpha}$ is upper semicontinuous on $K\times K$, and there is exactly one $M_{\alpha}\in\mathbb{R}$ such that $M_{\alpha}=\Phi_{\alpha}(p)$ for some $p\in K\times K$ and $\Phi_{\alpha}(q)\le M_{\alpha}$ for every $q\in K\times K$.

In claims 2 to 6, $M$ and $M_{\alpha}$ denote these values, and for each $\alpha$ with $0<\alpha$ we let $(x_{\alpha},y_{\alpha})$ be any point of $K\times K$ with $\Phi_{\alpha}(x_{\alpha},y_{\alpha})=M_{\alpha}$; the claims hold for every such choice.

2. (Ordering) $M\le M_{\alpha}$ for every $\alpha$ with $0<\alpha$, and $M_{\beta}\le M_{\alpha}$ whenever $0<\alpha$ and $\alpha\le\beta$.

3. (Convergence) For every $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$ there is $\alpha_{0}\in\mathbb{R}$ with $0<\alpha_{0}$ such that every $\alpha$ with $\alpha_{0}\le\alpha$ satisfies $M_{\alpha}\le M+\varepsilon$.

4. (Vanishing penalty) $M\le u(x_{\alpha})-v(y_{\alpha})$ for every $\alpha$ with $0<\alpha$; and for every $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$ there is $\alpha_{1}\in\mathbb{R}$ with $0<\alpha_{1}$ such that every $\alpha$ with $\alpha_{1}\le\alpha$ satisfies

5. (Vanishing distance) For every $\eta\in\mathbb{R}$ with $0<\eta$ there is $\alpha_{2}\in\mathbb{R}$ with $0<\alpha_{2}$ such that every $\alpha$ with $\alpha_{2}\le\alpha$ satisfies $d(x_{\alpha},y_{\alpha})<\eta$.

6. (Cluster points) Let $\hat{x}\in K$ be a point with the following property: for all $\delta,\beta\in\mathbb{R}$ with $0<\delta$ and $0<\beta$ there is $\alpha$ with $\beta\le\alpha$ and $d(\hat{x},x_{\alpha})<\delta$. Then $u(\hat{x})-v(\hat{x})=M$.