Limits of Penalised Maxima on a Compact Subset of a Metric Space

Let $(X,d)$ be a metric space, equipped with the collection of all subsets that are open in $(X,d)$, which is a topology by The Open Subsets of a Metric Space Form a Topology, and let $K\subseteq X$ be nonempty and compact in $X$. Let $\mathbb{R}$ be the real numbers with the addition, multiplication and order $\le$ of their ordered field structure, where $s<t$ means that $s\le t$ and $s\ne t$ and where $s-t$ abbreviates $s+(-t)$. Let $\mathbb{N}$ be the natural numbers with the order $\le$.

Let $\Phi:K\to\mathbb{R}$ be upper semicontinuous on $K$ and let $\Psi:K\to\mathbb{R}$ be lower semicontinuous on $K$, both relative to $K$ in the metric space $(X,d)$, and assume that

$$0\le\Psi(x)\qquad\text{for every }x\in K .$$

Put

$$Z=\{\,x\in K\ :\ \Psi(x)=0\,\}$$

and assume that $Z$ is nonempty. For $\alpha\in\mathbb{R}$ with $0<\alpha$ let $\Phi-\alpha\Psi:K\to\mathbb{R}$ be the function whose value at $x$ is $\Phi(x)-\alpha\Psi(x)$.

Then the following hold.

1. (Attainment)(i) For every $\alpha\in\mathbb{R}$ with $0<\alpha$, the function $\Phi-\alpha\Psi$ is upper semicontinuous on $K$ and there is $x\in K$ with $\Phi(y)-\alpha\Psi(y)\le\Phi(x)-\alpha\Psi(x)$ for every $y\in K$. The value $\Phi(x)-\alpha\Psi(x)$ is the same at every such $x$; it is denoted $M_{\alpha}$, and a point $x\in K$ with $\Phi(x)-\alpha\Psi(x)=M_{\alpha}$ is called a maximiser at level $\alpha$.

2. (Monotonicity, and the infimum $M$)(ii) If $\alpha,\beta\in\mathbb{R}$ satisfy $0<\beta\le\alpha$, then $M_{\alpha}\le M_{\beta}$. Moreover $\Phi(z)\le M_{\alpha}$ for every $z\in Z$ and every $\alpha\in\mathbb{R}$ with $0<\alpha$. Consequently the set $\{\,M_{\alpha}\ :\ \alpha\in\mathbb{R},\ 0<\alpha\,\}$ is nonempty and bounded below, so it has a greatest lower bound

$$M=\inf\{\,M_{\alpha}\ :\ \alpha\in\mathbb{R},\ 0<\alpha\,\}\in\mathbb{R}$$

by Existence of the Infimum of a Nonempty Subset of $\mathbb{R}$ Bounded Below, and $\Phi(z)\le M$ for every $z\in Z$.

3. (Stabilisation of the penalised maxima)(iii) For every $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$ there is $\alpha_{0}\in\mathbb{R}$ with $0<\alpha_{0}$ such that

$$M\le M_{\alpha}<M+\varepsilon\qquad\text{for every }\alpha\in\mathbb{R}\text{ with }\alpha_{0}\le\alpha .$$

4. (The penalty vanishes)(iv) For every $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$ there is $\alpha_{0}\in\mathbb{R}$ with $1\le\alpha_{0}$ such that every $\alpha\in\mathbb{R}$ with $\alpha_{0}\le\alpha$ and every maximiser $x$ at level $\alpha$ satisfy

$$0\le\alpha\Psi(x)<\varepsilon,\qquad 0\le\Psi(x)<\varepsilon,\qquad M\le\Phi(x)<M+2\varepsilon .$$

Moreover $M\le\Phi(x)$ holds for every $\alpha\in\mathbb{R}$ with $0<\alpha$ and every maximiser $x$ at level $\alpha$, with no restriction on $\alpha$.

5. (Cluster points lie in the zero set)(v) Let $(\alpha_{k})_{k\in\mathbb{N}}$ be a sequence in $\mathbb{R}$ with $0<\alpha_{k}$ for every $k\in\mathbb{N}$ and with the property that for every $R\in\mathbb{R}$ there is $N\in\mathbb{N}$ such that $R<\alpha_{k}$ for every $k\in\mathbb{N}$ with $N\le k$, and for each $k\in\mathbb{N}$ let $x_{k}\in K$ be a maximiser at level $\alpha_{k}$. Then the sequence $(x_{k})_{k\in\mathbb{N}}$ has a cluster point in $(X,d)$ lying in $K$, and every cluster point $\hat{x}$ of $(x_{k})_{k\in\mathbb{N}}$ with $\hat{x}\in K$ satisfies

$$\Psi(\hat{x})=0,\qquad \Phi(\hat{x})=M .$$

In particular $\hat{x}\in Z$ and $\Phi(z)\le\Phi(\hat{x})$ for every $z\in Z$, so $\Phi$ attains a maximum over $Z$ at $\hat{x}$, with value $M$.