Proof of Negation, Restriction, and Separated Differences of Semicontinuous Functions

Throughout, order and addition arithmetic in $\mathbb{R}$ is taken from Elementary Order Arithmetic in an Ordered Field, whose claim 1 is strict compatibility of $<$ with addition, claim 2 is mixed transitivity of $<$ and $\le$, and claim 3 is addition of inequalities; and $-(-t)=t$ for $t\in\mathbb{R}$ by claim 5 of Additive Cancellation and Elementary Additive Identities in a Field.

Claim 1. Suppose first that $u$ is upper semicontinuous at $x$ relative to $A$, and let $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$. Choose $\delta\in\mathbb{R}$ with $0<\delta$ such that every $y\in A$ with $d_X(x,y)<\delta$ satisfies $u(y)<u(x)+\varepsilon$. Fix such a $y$. Adding $-u(x)-u(y)-\varepsilon$ to both sides of $u(y)<u(x)+\varepsilon$ and using claim 1 of the order arithmetic lemma gives

$$-u(x)-\varepsilon<-u(y),$$

that is, $(-u)(x)-\varepsilon<(-u)(y)$. As $\varepsilon$ was arbitrary, $-u$ is lower semicontinuous at $x$ relative to $A$.

Conversely, suppose $-u$ is lower semicontinuous at $x$ relative to $A$, let $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$, and choose $\delta$ with $0<\delta$ such that every $y\in A$ with $d_X(x,y)<\delta$ satisfies $-u(x)-\varepsilon<-u(y)$. Adding $u(x)+u(y)+\varepsilon$ to both sides and using claim 1 of the order arithmetic lemma gives $u(y)<u(x)+\varepsilon$. Hence $u$ is upper semicontinuous at $x$ relative to $A$. This proves the first equivalence.

Applying the first equivalence to the function $-u$ in place of $u$ gives: $-u$ is upper semicontinuous at $x$ relative to $A$ if and only if $-(-u)$ is lower semicontinuous at $x$ relative to $A$. Since $-(-u)$ and $u$ take the same value at every point of $A$, they are the same function, and the second equivalence follows.

Claim 2. Suppose $u$ is upper semicontinuous at $x$ relative to $A$ and let $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$. Choose $\delta$ with $0<\delta$ such that every $y\in A$ with $d_X(x,y)<\delta$ satisfies $u(y)<u(x)+\varepsilon$. If $z\in A'$ satisfies $d_X(x,z)<\delta$, then $z\in A$ because $A'\subseteq A$, so $u(z)<u(x)+\varepsilon$; since $u|_{A'}(z)=u(z)$ and $u|_{A'}(x)=u(x)$, this reads $u|_{A'}(z)<u|_{A'}(x)+\varepsilon$. Hence $u|_{A'}$ is upper semicontinuous at $x$ relative to $A'$. The lower semicontinuous case is identical, with the inequality $u(x)-\varepsilon<u(y)$ in place of $u(y)<u(x)+\varepsilon$.

Claim 3. Suppose $u$ is upper semicontinuous at $x$ relative to $A$ and $w$ is lower semicontinuous at $x$ relative to $A$. By claim 1, $-w$ is upper semicontinuous at $x$ relative to $A$. For every $z\in A$ we have $u(z)+(-w)(z)=u(z)+(-w(z))=u(z)-w(z)$, so $u-w$ is the pointwise sum of $u$ and $-w$ in the sense of Sums and Nonnegative Multiples of Semicontinuous Functions; by claim 1 of that lemma, $u-w$ is upper semicontinuous at $x$ relative to $A$.

Suppose instead that $u$ is lower semicontinuous at $x$ relative to $A$ and $w$ is upper semicontinuous at $x$ relative to $A$. By claim 1, $-w$ is lower semicontinuous at $x$ relative to $A$, and by claim 3 of Sums and Nonnegative Multiples of Semicontinuous Functions the pointwise sum $u-w$ of $u$ and $-w$ is lower semicontinuous at $x$ relative to $A$.

Claim 4. Let $(x_0,y_0)\in A\times B$ and let $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$. By claim 8 of Elementary Order Arithmetic in an Ordered Field there is $\eta\in\mathbb{R}$ with $0<\eta$ and $\eta+\eta=\varepsilon$.

Since $f$ is upper semicontinuous at $x_0$ relative to $A$, there is $\delta_1\in\mathbb{R}$ with $0<\delta_1$ such that every $x\in A$ with $d_X(x_0,x)<\delta_1$ satisfies $f(x)<f(x_0)+\eta$. Since $g$ is lower semicontinuous at $y_0$ relative to $B$, there is $\delta_2\in\mathbb{R}$ with $0<\delta_2$ such that every $y\in B$ with $d_Y(y_0,y)<\delta_2$ satisfies $g(y_0)-\eta<g(y)$.

Let $\delta$ be the minimum of $\delta_1$ and $\delta_2$. By claim 2 of Elementary Properties of the Minimum of Two Elements, $\delta$ equals $\delta_1$ or $\delta_2$, so $0<\delta$; and by claim 1 of that lemma, $\delta\le\delta_1$ and $\delta\le\delta_2$.

Let $(x,y)\in A\times B$ satisfy $d_{X\times Y}((x_0,y_0),(x,y))<\delta$. By the definition of the product metric, $d_{X\times Y}((x_0,y_0),(x,y))$ is the maximum of $d_X(x_0,x)$ and $d_Y(y_0,y)$, so by claim 1 of Elementary Properties of the Maximum of Two Elements,

$$d_X(x_0,x)\le d_{X\times Y}((x_0,y_0),(x,y)),\qquad d_Y(y_0,y)\le d_{X\times Y}((x_0,y_0),(x,y)).$$

By claim 2 of the order arithmetic lemma these give $d_X(x_0,x)<\delta$ and $d_Y(y_0,y)<\delta$, and then, again by claim 2, $d_X(x_0,x)<\delta_1$ and $d_Y(y_0,y)<\delta_2$. Consequently

$$f(x)<f(x_0)+\eta,\qquad g(y_0)-\eta<g(y).$$

Adding $-g(y_0)-g(y)+\eta$ to both sides of the second inequality and using claim 1 of the order arithmetic lemma gives $-g(y)<-g(y_0)+\eta$. Adding this to the first inequality by claim 3 of the order arithmetic lemma gives

$$f(x)-g(y)<f(x_0)+\eta+(-g(y_0)+\eta)=f(x_0)-g(y_0)+\varepsilon,$$

where the last equality uses $\eta+\eta=\varepsilon$ and commutativity and associativity of addition. That is, $h(x,y)<h(x_0,y_0)+\varepsilon$.

Since $\varepsilon$ was arbitrary, $h$ is upper semicontinuous at $(x_0,y_0)$ relative to $A\times B$, and since $(x_0,y_0)$ was an arbitrary point of $A\times B$, $h$ is upper semicontinuous on $A\times B$.