Proof of Properties of the Upper Semicontinuous Envelope

Conventions. The order $\le$ and the arithmetic of $\mathbb{R}$ are those of the ordered field of real numbers; $\le$ is in particular a total order, hence antisymmetric, and of any two real numbers one is at least the other. Multiplication by a nonnegative real number preserves $\le$: if $a\le b$ and $0\le\lambda$ then either $a=b$, and the products are equal, or $a<b$, and then claim 10 of Elementary Order Arithmetic in an Ordered Field applies when $0<\lambda$, while $\lambda=0$ makes both products $0$. Among the ordered field axioms is the compatibility of $\le$ with addition: if $a\le b$ then $a+c\le b+c$ for every $c\in\mathbb{R}$. That axiom is what we use for non-strict inequalities; claim 1 of Elementary Order Arithmetic in an Ordered Field is its strict counterpart and is cited only where the inequality is strict. Throughout, $A_{u}(x)$ and the witnessing radii are as described in Upper and Lower Semicontinuous Envelopes of a Real-Valued Function, and $d$ is symmetric and satisfies the triangle inequality by the metric axioms. The claims are established below in the order 1, 3, 2, 4, 6, 5; each argument uses only claims established before it.

Claim 1. As recorded in Upper and Lower Semicontinuous Envelopes of a Real-Valued Function §upper, $u(x)$ is a lower bound for $A_{u}(x)$. Since $u^{*}(x)$ is the greatest lower bound of $A_{u}(x)$, we get $u(x)\le u^{*}(x)$.

Claim 3. Let $x\in S$ and let $\varepsilon\in\mathbb{R}$ be positive. Since $v$ is upper semicontinuous at $x$ relative to $S$, there is a positive $\delta\in\mathbb{R}$ such that every $y\in S$ with $d(x,y)<\delta$ satisfies $v(y)<v(x)+\varepsilon$. Put $r=\delta\cdot2^{-1}$, which is positive and satisfies $r<\delta$ by claim 8 of Elementary Order Arithmetic in an Ordered Field. If $y\in S$ satisfies $d(y,x)\le r$, then $d(x,y)=d(y,x)\le r<\delta$, so $d(x,y)<\delta$ by claim 2 of Elementary Order Arithmetic in an Ordered Field, whence $u(y)\le v(y)<v(x)+\varepsilon$ and in particular $u(y)\le v(x)+\varepsilon$. Therefore $v(x)+\varepsilon\in A_{u}(x)$, and since $u^{*}(x)$ is a lower bound for $A_{u}(x)$ we get $u^{*}(x)\le v(x)+\varepsilon$. As $\varepsilon$ was an arbitrary positive number, Comparison of Real Numbers with Arbitrary Positive Slack §slack-above gives $u^{*}(x)\le v(x)$.

Claim 2. Let $x\in S$ and let $\varepsilon\in\mathbb{R}$ be positive; put $\varepsilon'=\varepsilon\cdot2^{-1}$, positive with $\varepsilon'<\varepsilon$ by claim 8 of Elementary Order Arithmetic in an Ordered Field. The set $A_{u}(x)$ is nonempty and bounded below, so claim 4 of Approximation Property of the Supremum and the Infimum in $\mathbb{R}$ provides $c\in A_{u}(x)$ with $c<u^{*}(x)+\varepsilon'$. Let $r$ be a positive real number witnessing $c\in A_{u}(x)$, so that $u(z)\le c$ for every $z\in S$ with $d(z,x)\le r$, and put $\delta=r\cdot2^{-1}$, positive with $\delta+\delta=r$ by claim 8.

Let $y\in S$ satisfy $d(x,y)<\delta$, and let $z\in S$ satisfy $d(z,y)\le\delta$. Then $d(z,x)\le d(z,y)+d(y,x)$, and $d(z,y)+d(y,x)<\delta+\delta=r$ by claim 3 of Elementary Order Arithmetic in an Ordered Field, since $d(z,y)\le\delta$ and $d(y,x)=d(x,y)<\delta$; hence $d(z,x)<r$ by claim 2 there, and in particular $d(z,x)\le r$, so $u(z)\le c$. Thus $\delta$ witnesses $c\in A_{u}(y)$, and therefore

the last step by claim 1 of Elementary Order Arithmetic in an Ordered Field. Claim 2 there gives $u^{*}(y)<u^{*}(x)+\varepsilon$. As $\varepsilon$ was arbitrary, $u^{*}$ is upper semicontinuous at $x$ relative to $S$, and as $x$ was arbitrary, $u^{*}$ is upper semicontinuous on $S$.

For the second assertion of the claim, fix $x\in S$ and run the previous paragraph with $\varepsilon=1$, which is positive by claim 6 of Elementary Order Arithmetic in an Ordered Field; it produces $c$ and $\delta$ as above. Put $\rho=\delta\cdot2^{-1}$, positive with $\rho<\delta$ by claim 8. Every $y\in S$ with $d(y,x)\le\rho$ satisfies $d(x,y)=d(y,x)\le\rho<\delta$, hence $u^{*}(y)\le c$ by the previous paragraph. So $\rho$ witnesses $c\in A_{u^{*}}(x)$, and $u^{*}$ is bounded above near each point of $S$.

Claim 4. If $u^{*}(x)=u(x)$ for every $x\in S$, then $u$ is upper semicontinuous on $S$ by claim 2. Conversely, if $u$ is upper semicontinuous on $S$, then claim 3 applied with $v=u$ gives $u^{*}(x)\le u(x)$ for every $x\in S$; with claim 1 and the antisymmetry of $\le$ this gives $u^{*}(x)=u(x)$.

Claim 6. Let $x\in S$ and let $c\in A_{v}(x)$, with witnessing radius $r$. For $y\in S$ with $d(y,x)\le r$ we have $u(y)\le v(y)\le c$, so $r$ also witnesses $c\in A_{u}(x)$. Hence $A_{v}(x)\subseteq A_{u}(x)$. Since $u^{*}(x)$ is a lower bound for $A_{u}(x)$ it is a lower bound for $A_{v}(x)$, and since $v^{*}(x)$ is the greatest lower bound of $A_{v}(x)$ we conclude $u^{*}(x)\le v^{*}(x)$.

Claim 5. Let $x\in S$ and let $\varepsilon\in\mathbb{R}$ be positive. By claim 4 of Approximation Property of the Supremum and the Infimum in $\mathbb{R}$ there is $c\in A_{u}(x)$ with $c<u^{*}(x)+\varepsilon$; let $r$ be a positive witnessing radius for $c$. Let $\varepsilon'$ be the least of $\varepsilon$ and $r$, which exists and is positive by claim 9 of Elementary Order Arithmetic in an Ordered Field, being equal to one of them.

We first find $z\in S$ with $d(z,x)\le\varepsilon'$ and $u^{*}(x)-\varepsilon'<u(z)$. Suppose no such $z$ exists. Then, by the totality of $\le$, every $z\in S$ with $d(z,x)\le\varepsilon'$ satisfies $u(z)\le u^{*}(x)-\varepsilon'$, so $\varepsilon'$ witnesses $u^{*}(x)-\varepsilon'\in A_{u}(x)$ and therefore $u^{*}(x)\le u^{*}(x)-\varepsilon'$. Adding $-u^{*}(x)+\varepsilon'$ to both sides gives $\varepsilon'\le0$ by the compatibility of $\le$ with addition, contradicting $0<\varepsilon'$.

Fix such a $z$. Since $d(z,x)\le\varepsilon'\le r$, we get $u(z)\le c<u^{*}(x)+\varepsilon$, hence $u(z)<u^{*}(x)+\varepsilon$ by claim 2 of Elementary Order Arithmetic in an Ordered Field and so $u(z)-u^{*}(x)<\varepsilon$ by claim 1 there. Since $\varepsilon'\le\varepsilon$, claim 4 of that lemma gives $-\varepsilon\le-\varepsilon'$, so $u^{*}(x)-\varepsilon\le u^{*}(x)-\varepsilon'$ by the compatibility of $\le$ with addition; combined with $u^{*}(x)-\varepsilon'<u(z)$ and claim 2 of that lemma this gives $u^{*}(x)-\varepsilon<u(z)$, whence $-\varepsilon<u(z)-u^{*}(x)$ by claim 1 there. Claim 9 of Properties of the Absolute Value in an Ordered Field now gives $|u(z)-u^{*}(x)|<\varepsilon$, and $d(z,x)\le\varepsilon'\le\varepsilon$. This proves the first assertion of claim 5.

For the sequence, write $k$ also for the image in $\mathbb{R}$ of $k\in\mathbb{N}$ under the canonical map of Properties of the Canonical Map from the Natural Numbers to an Ordered Field. By claim 2 there, $1\le k$; since $0<1$ by claim 6 of Elementary Order Arithmetic in an Ordered Field, claim 2 there gives $0<k$, so $k^{-1}$ exists and is positive by claim 7 there. Applying the first assertion with $\varepsilon=k^{-1}$ and choosing, for each $k\in\mathbb{N}$, one of the points it provides — an appeal to countable choice — yields a point $x_{k}\in S$ with

and these points form a sequence $(x_{k})_{k\in\mathbb{N}}$ in $S$.

Let $\eta\in\mathbb{R}$ be positive. By claim 3 of The Archimedean Property of the Real Numbers there is $N\in\mathbb{N}$ with $N^{-1}<\eta$. Let $k\in\mathbb{N}$ satisfy $k\ge N$. Then $N\le k$ in $\mathbb{R}$: this is an equality if $k=N$, and follows from claim 6 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field otherwise. Multiplying $N\le k$ by the positive number $N^{-1}k^{-1}$ gives $k^{-1}\le N^{-1}$. Hence

using claim 2 of Elementary Order Arithmetic in an Ordered Field to chain the comparisons. By Convergent Sequence in a Metric Space, $(x_{k})$ converges to $x$ in $(M,d)$ and $(u(x_{k}))$ converges to $u^{*}(x)$ in $(\mathbb{R},d_{\mathbb{R}})$.