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

Throughout, $\alpha$, $\beta$, $\varepsilon$ and $\delta$ denote real numbers, semicontinuity of $\Phi$ and $\Psi$ is relative to $K$ in the metric space $(X,d)$, and we use freely that $2=1+1$ satisfies $0<2$: indeed $0\le1$ by claim 1 of Elementary Arithmetic in an Ordered Field and $0\ne1$ in a field, so $0<1$, and adding $1$ gives $1<2$ by claim 1 of Elementary Order Arithmetic in an Ordered Field. Hence $2^{-1}$ exists and $0<2^{-1}$ by claim 7 of Elementary Order Arithmetic in an Ordered Field, and $2^{-1}+2^{-1}=2\cdot2^{-1}=1$.

Proof of claim 1. Fix $\alpha$ with $0<\alpha$. Since $\Psi$ is lower semicontinuous on $K$ and $0\le\alpha$, claim 3 of Sums and Nonnegative Multiples of Semicontinuous Functions shows that $\alpha\Psi$ is lower semicontinuous on $K$. By claim 1 of Semicontinuity Under Negation and Characterization of Continuity the function $-(\alpha\Psi)$ is upper semicontinuous on $K$. As $\Phi$ is upper semicontinuous on $K$, claim 1 of Sums and Nonnegative Multiples of Semicontinuous Functions shows that $\Phi+\bigl(-(\alpha\Psi)\bigr)$ is upper semicontinuous on $K$; its value at $x\in K$ is $\Phi(x)-\alpha\Psi(x)$, so it is exactly $\Phi-\alpha\Psi$.

Since $K$ is nonempty and compact in $X$, claim 1 of Semicontinuous Functions Attain Their Extrema on a Compact Set provides $x\in K$ with $\Phi(y)-\alpha\Psi(y)\le\Phi(x)-\alpha\Psi(x)$ for every $y\in K$. If $x'\in K$ has the same property, then taking $y=x'$ in the inequality for $x$ and $y=x$ in the inequality for $x'$ gives both $\Phi(x')-\alpha\Psi(x')\le\Phi(x)-\alpha\Psi(x)$ and the reverse inequality, so the two values coincide by antisymmetry of the total order of $\mathbb{R}$. Thus $M_{\alpha}$ is well defined.

Proof of claim 2. Let $0<\beta\le\alpha$ and let $x$ be a maximiser at level $\alpha$. From $\beta\le\alpha$ and $0\le\Psi(x)$, claim 5 of Elementary Arithmetic in an Ordered Field gives $\beta\Psi(x)\le\alpha\Psi(x)$; claim 4 of Elementary Order Arithmetic in an Ordered Field turns this into $-\alpha\Psi(x)\le-\beta\Psi(x)$, and adding $\Phi(x)$ (claim 3 of Elementary Arithmetic in an Ordered Field) yields

$$M_{\alpha}=\Phi(x)-\alpha\Psi(x)\;\le\;\Phi(x)-\beta\Psi(x)\;\le\;M_{\beta},$$

the last inequality because $x\in K$ and $M_{\beta}$ is the maximum value of $\Phi-\beta\Psi$ on $K$.

Next let $z\in Z$ and $0<\alpha$. Then $\Psi(z)=0$, so $\alpha\Psi(z)=0$ by claim 1 of Zero Products and Elementary Identities in a Field; since the additive inverse of $0$ is $0$ and $0$ is the additive identity of the field, $\Phi(z)-\alpha\Psi(z)=\Phi(z)$. As $z\in K$, this gives $\Phi(z)\le M_{\alpha}$.

Put $S=\{\,M_{\alpha}:\alpha\in\mathbb{R},\ 0<\alpha\,\}$. It is nonempty, containing $M_{1}$. Fix $z_{0}\in Z$, which exists because $Z$ is nonempty; the previous paragraph shows that $\Phi(z_{0})\le M_{\alpha}$ for every $\alpha$ with $0<\alpha$, so $\Phi(z_{0})$ is a lower bound of $S$. By Existence of the Infimum of a Nonempty Subset of $\mathbb{R}$ Bounded Below the greatest lower bound $M=\inf S$ exists in $\mathbb{R}$. For an arbitrary $z\in Z$ the number $\Phi(z)$ is likewise a lower bound of $S$, so $\Phi(z)\le M$ because $M$ is the greatest lower bound.

Proof of claim 3. Let $0<\varepsilon$. By claim 4 of Approximation Property of the Supremum and the Infimum in $\mathbb{R}$ applied to $S$ there is an element of $S$ strictly below $M+\varepsilon$; that is, there is $\alpha_{0}$ with $0<\alpha_{0}$ and $M_{\alpha_{0}}<M+\varepsilon$. Let $\alpha_{0}\le\alpha$. Claim 2 gives $M_{\alpha}\le M_{\alpha_{0}}$, and claim 2 of Elementary Order Arithmetic in an Ordered Field then gives $M_{\alpha}<M+\varepsilon$. Finally $M\le M_{\alpha}$ because $M$ is a lower bound of $S$.

Proof of claim 4. We first prove the final assertion. Let $0<\alpha$ and let $x$ be a maximiser at level $\alpha$. From $0<\alpha$ and $0\le\Psi(x)$, claim 5 of Elementary Arithmetic in an Ordered Field gives $0\le\alpha\Psi(x)$; claim 4 of Elementary Order Arithmetic in an Ordered Field gives $-\alpha\Psi(x)\le0$, and adding $\Phi(x)$ gives $\Phi(x)-\alpha\Psi(x)\le\Phi(x)$. Since $M\le M_{\alpha}=\Phi(x)-\alpha\Psi(x)$, transitivity gives $M\le\Phi(x)$.

Now let $0<\varepsilon$ and put $\varepsilon'=\varepsilon\cdot2^{-1}$, so that $0<\varepsilon'\le\varepsilon$ by claims 8 and 5 of Elementary Order Arithmetic in an Ordered Field. By claim 3 there is $\alpha_{1}$ with $0<\alpha_{1}$ such that

$$M\le M_{\beta}<M+\varepsilon'\qquad\text{for every }\beta\text{ with }\alpha_{1}\le\beta .$$

Put $\alpha_{0}=2\alpha_{1}+2$. From $0<\alpha_{1}$ and $0<2$ we get $0<2\alpha_{1}$ by claim 5 of Elementary Order Arithmetic in an Ordered Field, hence $2\le2\alpha_{1}+2$ by claim 3 of Elementary Arithmetic in an Ordered Field, and $1<2$ gives $1\le\alpha_{0}$.

Let $\alpha_{0}\le\alpha$ and let $x$ be a maximiser at level $\alpha$. Put $\beta=\alpha\cdot2^{-1}$. Multiplying $\alpha_{0}\le\alpha$ by the nonnegative number $2^{-1}$ (claim 5 of Elementary Arithmetic in an Ordered Field) gives $\alpha_{1}+1=\alpha_{0}\cdot2^{-1}\le\beta$, and $\alpha_{1}\le\alpha_{1}+1$, so $\alpha_{1}\le\beta$ and therefore $M\le M_{\beta}<M+\varepsilon'$. Also $M\le M_{\alpha}$ by claim 3 applied with the same $\alpha_{1}$, or directly because $M$ is a lower bound of $S$.

Because $2^{-1}+2^{-1}=1$ we have $\beta+\beta=\alpha$, so

$$M_{\alpha}=\Phi(x)-\alpha\Psi(x)=\bigl(\Phi(x)-\beta\Psi(x)\bigr)-\beta\Psi(x)\;\le\;M_{\beta}-\beta\Psi(x),$$

using $x\in K$ for the last step. Adding $\beta\Psi(x)-M_{\alpha}$ to both sides gives $\beta\Psi(x)\le M_{\beta}-M_{\alpha}$. From $M\le M_{\alpha}$ we get $-M_{\alpha}\le-M$, hence $M_{\beta}-M_{\alpha}\le M_{\beta}-M<\varepsilon'$. Therefore $\beta\Psi(x)<\varepsilon'$, and multiplying by the positive number $2$ (claim 10 of Elementary Order Arithmetic in an Ordered Field) gives

$$\alpha\Psi(x)=2\beta\Psi(x)<2\varepsilon'=\varepsilon .$$

That $0\le\alpha\Psi(x)$ was shown above. Since $1\le\alpha_{0}\le\alpha$ and $0\le\Psi(x)$, claim 5 of Elementary Arithmetic in an Ordered Field gives $\Psi(x)=1\cdot\Psi(x)\le\alpha\Psi(x)<\varepsilon$, and $0\le\Psi(x)$ holds by hypothesis.

Finally, adding $\alpha\Psi(x)$ to $M_{\alpha}=\Phi(x)-\alpha\Psi(x)$ gives $\Phi(x)=M_{\alpha}+\alpha\Psi(x)$. Combining $M_{\alpha}<M+\varepsilon'$ with $\alpha\Psi(x)<\varepsilon$ and $\varepsilon'\le\varepsilon$ gives $\Phi(x)<M+\varepsilon'+\varepsilon\le M+2\varepsilon$, while $M\le\Phi(x)$ was proved above.

Proof of claim 5. The terms of $(x_{k})_{k\in\mathbb{N}}$ lie in $K$, which is nonempty and compact in $X$, so by Every Sequence in a Compact Subset of a Metric Space Has a Cluster Point There the sequence has a cluster point in $(X,d)$ lying in $K$.

Let $\hat{x}\in K$ be any cluster point of $(x_{k})_{k\in\mathbb{N}}$.

Step (a): $\Psi(\hat{x})=0$. Let $0<\varepsilon$. Since $\Psi$ is lower semicontinuous at $\hat{x}$ relative to $K$, there is $\delta$ with $0<\delta$ such that every $y\in K$ with $d(\hat{x},y)<\delta$ satisfies $\Psi(\hat{x})-\varepsilon<\Psi(y)$. By claim 4 there is $\alpha_{0}$ with $1\le\alpha_{0}$ such that $\Psi(x)<\varepsilon$ for every $\alpha$ with $\alpha_{0}\le\alpha$ and every maximiser $x$ at level $\alpha$. By the hypothesis on $(\alpha_{k})_{k\in\mathbb{N}}$, applied with $R=\alpha_{0}$, there is $N\in\mathbb{N}$ with $\alpha_{0}<\alpha_{k}$ for every $k$ with $N\le k$. Applying Cluster Point of a Sequence in a Metric Space with the radius $\delta$ and this $N$ produces $k\in\mathbb{N}$ with $N\le k$ and $d(x_{k},\hat{x})<\delta$; by the symmetry axiom of Metric Space also $d(\hat{x},x_{k})<\delta$. For this $k$ we have $\alpha_{0}\le\alpha_{k}$, so $\Psi(x_{k})<\varepsilon$, while $x_{k}\in K$ gives $\Psi(\hat{x})-\varepsilon<\Psi(x_{k})$. Hence $\Psi(\hat{x})-\varepsilon<\varepsilon$, that is $\Psi(\hat{x})<2\varepsilon$.

Suppose $0<\Psi(\hat{x})$. Applying the previous paragraph with $\varepsilon=\Psi(\hat{x})\cdot4^{-1}$, which is positive by claims 7 and 5 of Elementary Order Arithmetic in an Ordered Field, gives $\Psi(\hat{x})<2\varepsilon=\Psi(\hat{x})\cdot2^{-1}$, contradicting claim 8 of Elementary Order Arithmetic in an Ordered Field. Hence $\Psi(\hat{x})\le0$, and with the standing hypothesis $0\le\Psi(\hat{x})$ we conclude $\Psi(\hat{x})=0$, so $\hat{x}\in Z$.

Step (b): $\Phi(\hat{x})=M$. Since $\hat{x}\in Z$, claim 2 gives $\Phi(\hat{x})\le M$. For the reverse inequality, let $0<\varepsilon$. Since $\Phi$ is upper semicontinuous at $\hat{x}$ relative to $K$, there is $\delta$ with $0<\delta$ such that every $y\in K$ with $d(\hat{x},y)<\delta$ satisfies $\Phi(y)<\Phi(\hat{x})+\varepsilon$. Applying Cluster Point of a Sequence in a Metric Space with the radius $\delta$ and with $N$ any element of $\mathbb{N}$ produces $k\in\mathbb{N}$ with $d(x_{k},\hat{x})<\delta$, hence $d(\hat{x},x_{k})<\delta$. By the final assertion of claim 4 applied at level $\alpha_{k}$ we have $M\le\Phi(x_{k})$, and therefore

$$M\le\Phi(x_{k})<\Phi(\hat{x})+\varepsilon .$$

Suppose $\Phi(\hat{x})<M$. Applying the previous sentence with $\varepsilon=\bigl(M-\Phi(\hat{x})\bigr)\cdot2^{-1}$, which is positive, gives $M<\Phi(\hat{x})+\bigl(M-\Phi(\hat{x})\bigr)\cdot2^{-1}$; but adding $-\Phi(\hat{x})$ to both sides turns this into $M-\Phi(\hat{x})<\bigl(M-\Phi(\hat{x})\bigr)\cdot2^{-1}$, contradicting claim 8 of Elementary Order Arithmetic in an Ordered Field. Hence $M\le\Phi(\hat{x})$, and antisymmetry gives $\Phi(\hat{x})=M$.

Finally, $\hat{x}\in Z$ by step (a), and claim 2 gives $\Phi(z)\le M=\Phi(\hat{x})$ for every $z\in Z$, so $\Phi$ attains a maximum over $Z$ at $\hat{x}$ with value $M$. $\blacksquare$