Proof of Sums and Nonnegative Multiples of Semicontinuous Functions

We use the elementary order arithmetic of the ordered field $\mathbb{R}$; all claim numbers cited below are claims of that lemma unless stated otherwise.

Claim 1. Let $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$. By claim 8 the element $\eta=\varepsilon\cdot 2^{-1}$ satisfies $0<\eta$ and $\eta+\eta=\varepsilon$. Since $u$ is upper semicontinuous at $x$ relative to $A$, there is $\delta_1$ with $0<\delta_1$ such that every $y\in A$ with $d(x,y)<\delta_1$ satisfies $u(y)<u(x)+\eta$; since $v$ is upper semicontinuous at $x$ relative to $A$, there is $\delta_2$ with $0<\delta_2$ such that every $y\in A$ with $d(x,y)<\delta_2$ satisfies $v(y)<v(x)+\eta$. By claim 9 there is $\delta\in\mathbb{R}$ with $\delta\le\delta_1$, $\delta\le\delta_2$, and $\delta$ equal to $\delta_1$ or to $\delta_2$; in either case $0<\delta$.

Let $y\in A$ satisfy $d(x,y)<\delta$. Claim 2 gives $d(x,y)<\delta_1$ and $d(x,y)<\delta_2$, so $u(y)<u(x)+\eta$ and $v(y)<v(x)+\eta$. Claim 3, applied with the first inequality strict and the second in particular non-strict, gives

Hence $u+v$ is upper semicontinuous at $x$ relative to $A$.

Claim 2. Suppose first that $\lambda=0$. Then $(\lambda u)(y)=0\cdot u(y)=0$ for every $y\in A$. Given $\varepsilon$ with $0<\varepsilon$, take $\delta=1$, which satisfies $0<\delta$ by claim 6. Every $y\in A$ satisfies $(\lambda u)(y)=0<0+\varepsilon=(\lambda u)(x)+\varepsilon$, the strict inequality holding by claim 1 applied to $0<\varepsilon$. So $\lambda u$ is upper semicontinuous at $x$ relative to $A$.

Now suppose $0<\lambda$, and let $\varepsilon$ with $0<\varepsilon$ be given. Claim 7 gives $0<\lambda^{-1}$, and claim 5 gives $0<\lambda^{-1}\varepsilon$. Upper semicontinuity of $u$ at $x$ relative to $A$, applied with $\lambda^{-1}\varepsilon$ in place of $\varepsilon$, provides $\delta$ with $0<\delta$ such that every $y\in A$ with $d(x,y)<\delta$ satisfies $u(y)<u(x)+\lambda^{-1}\varepsilon$. For such $y$, claim 10 applied with the multiplier $\lambda$ gives

Hence $\lambda u$ is upper semicontinuous at $x$ relative to $A$.

Claim 3. For a function $f:A\to\mathbb{R}$ write $-f$ for the function on $A$ whose value at $y$ is the additive inverse of $f(y)$.

Suppose $u$ and $v$ are lower semicontinuous at $x$ relative to $A$. By claim 1 of Semicontinuity Under Negation and Characterization of Continuity, $-u$ and $-v$ are upper semicontinuous at $x$ relative to $A$, so by claim 1 above their sum $(-u)+(-v)$ is upper semicontinuous at $x$ relative to $A$. In a field $(-u)(y)+(-v)(y)=-(u(y)+v(y))$ for every $y\in A$, so $(-u)+(-v)$ and $-(u+v)$ are the same function. Applying claim 1 of Semicontinuity Under Negation and Characterization of Continuity in the other direction, with $u+v$ in place of $u$, shows that $u+v$ is lower semicontinuous at $x$ relative to $A$.

Suppose $u$ is lower semicontinuous at $x$ relative to $A$. By claim 1 of Semicontinuity Under Negation and Characterization of Continuity, $-u$ is upper semicontinuous at $x$ relative to $A$, so by claim 2 above $\lambda(-u)$ is upper semicontinuous at $x$ relative to $A$. In a field $\lambda\bigl((-u)(y)\bigr)=-\bigl(\lambda\,u(y)\bigr)$ for every $y\in A$, so $\lambda(-u)$ and $-(\lambda u)$ are the same function, and claim 1 of Semicontinuity Under Negation and Characterization of Continuity applied with $\lambda u$ in place of $u$ shows that $\lambda u$ is lower semicontinuous at $x$ relative to $A$.

Final assertion. Each of claims 1, 2 and 3 concerns the single point $x$, and semicontinuity on $A$ is by definition semicontinuity at every point of $A$ relative to $A$. So if the hypothesis of one of the claims holds at every point of $A$, then its conclusion holds at every point of $A$.