Proof of Continuous Functions with Compact Support: the Zero Function, Closure under Operations, Vanishing off the Support, Existence of the Maximum Modulus, and Uniform Continuity

Throughout, for a real function $h$ on $X$ write $N(h)=\{x\in X:h(x)\neq0\}$, a subset of $X$ by the routine facts of the setting, so that $\operatorname{supp}h=\operatorname{cl}N(h)$ by Support of a Real Function on a Metric Space (i).

The zero function. The constant function $0$ is continuous, constant maps into $\mathbb{R}$ being continuous (a background fact of the base). As $N(0)=\emptyset$ and no open ball meets $\emptyset$, $\operatorname{supp}0=\operatorname{cl}\emptyset=\emptyset$ by Closure of a Subset of a Metric Space (ii); and $\emptyset$ is compact by Compact Subsets of a Metric Space: Covers by Open Sets, Finite Unions and Increasing Covers; Sublevel Sets of the Distance and Closed Balls (i), since the empty subset of any set of open sets is finite (Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice (iii)) and covers $\emptyset$. This proves (i).

Step 1: two facts on closures. Let $A,B\subseteq X$. If $A\subseteq B$, then $\operatorname{cl}A\subseteq\operatorname{cl}B$, since by Closure of a Subset of a Metric Space (ii) a point of $\operatorname{cl}A$ is one every open ball around which meets $A$, hence meets $B$. Moreover $\operatorname{cl}(A\cup B)\subseteq\operatorname{cl}A\cup\operatorname{cl}B$: if $x\in X$ lies in neither $\operatorname{cl}A$ nor $\operatorname{cl}B$, there are real $r,s>0$ with $B(x,r)\cap A=\emptyset$ and $B(x,s)\cap B=\emptyset$; for $t=\min\{r,s\}$ the ball $B(x,t)$ lies in both balls, so it does not meet $A\cup B$, and $x\notin\operatorname{cl}(A\cup B)$.

Step 2: continuity. The functions $f+g$, $cf$ and $fg$ are continuous, since pointwise sums, real multiples and products of continuous real functions are continuous (a background fact of the base). For $|f|$, let $x\in X$ and $\varepsilon>0$, and choose $\delta>0$ with $|f(y)-f(x)|<\varepsilon$ for every $y\in X$ with $d(x,y)<\delta$; by the reverse triangle inequality

$$\bigl||f(y)|-|f(x)|\bigr|\le|f(y)-f(x)|<\varepsilon$$

for these $y$, so $|f|$ is continuous at $x$. For real numbers $a,b$, The Maximum and Minimum of Two Elements of an Ordered Field: Sum, Difference and Absolute Value (i) and The Maximum and Minimum of Two Elements of an Ordered Field: Sum, Difference and Absolute Value (ii) give $\max\{a,b\}+\min\{a,b\}=a+b$ and $\max\{a,b\}-\min\{a,b\}=|a-b|$, hence

$$\max\{a,b\}=\tfrac12\bigl(a+b+|a-b|\bigr),\qquad\min\{a,b\}=\tfrac12\bigl(a+b-|a-b|\bigr).$$

Applied to $a=f(x)$ and $b=g(x)$ at every $x\in X$, this shows $\max(f,g)=\tfrac12\bigl(f+g+|f+(-1)g|\bigr)$ and $\min(f,g)=\tfrac12\bigl(f+g+(-1)|f+(-1)g|\bigr)$, which are continuous by what was just shown.

Step 3: supports. If $f(x)=g(x)=0$, then $(f+g)(x)=0$, and $\max\{f(x),g(x)\}=\min\{f(x),g(x)\}=0$ by The Maximum and Minimum of Two Elements of a Total Order (ii); hence $N(f+g)$, $N(\max(f,g))$ and $N(\min(f,g))$ are contained in $N(f)\cup N(g)$. Further $N(cf)\subseteq N(f)$, $N(fg)=N(f)\cap N(g)$ because a product of two real numbers is nonzero exactly when both factors are, and $N(|f|)=N(f)$. By Step 1,

$$\operatorname{supp}(f+g)=\operatorname{cl}N(f+g)\subseteq\operatorname{cl}\bigl(N(f)\cup N(g)\bigr)\subseteq\operatorname{supp}f\cup\operatorname{supp}g,$$

and the other inclusions of (ii) follow from the inclusions between the sets $N(\cdot)$ by the monotonicity of the closure in Step 1: $\operatorname{supp}\max(f,g)$ and $\operatorname{supp}\min(f,g)$ as for $f+g$, $\operatorname{supp}(cf)\subseteq\operatorname{cl}N(f)$, $\operatorname{supp}(fg)$ lies in both $\operatorname{cl}N(f)$ and $\operatorname{cl}N(g)$, and $\operatorname{supp}|f|=\operatorname{cl}N(f)=\operatorname{supp}f$. Each of these supports is closed, being a closure, and lies in $\operatorname{supp}f\cup\operatorname{supp}g$, which is compact by Compact Subsets of a Metric Space: Covers by Open Sets, Finite Unions and Increasing Covers; Sublevel Sets of the Distance and Closed Balls (ii) (the set $\{\operatorname{supp}f\}\cup\{\operatorname{supp}g\}$ is finite by Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice (iii) and Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice (v)), or in $\operatorname{supp}f$, which is compact. A closed subset of a compact set is compact, so all six functions have compact support. Together with Step 2 this proves (ii).

Step 4: vanishing off the support. If $f(x)\neq0$, then $x\in N(f)\subseteq\operatorname{cl}N(f)=\operatorname{supp}f$, since a set lies in its closure. This is (iii).

Step 5: the maximum. Let $S=\{0\}\cup\{|f(x)|:x\in X\}$ and $K=\operatorname{supp}f$. If $K=\emptyset$, then $f(x)=0$ for every $x\in X$ by Step 4, so $S=\{0\}$ and $0$ is its greatest element. If $K\neq\emptyset$, the restriction of $|f|$ to $K$ is continuous on $K$ by Restricting a Continuous Map to a Smaller Subset Preserves Continuity (ii), since $|f|$ is continuous on $X$ by Step 2. As $K$ is compact and nonempty, the extreme value theorem gives $x^{*}\in K$ with

$$|f(x)|\le|f(x^{*})|\qquad\text{for every }x\in K.$$

For $x\in X\setminus K$ we have $|f(x)|=0\le|f(x^{*})|$ by Step 4, and also $0\le|f(x^{*})|$. Thus $|f(x^{*})|$ is an element of $S$ and an upper bound of $S$, that is, the greatest element of $S$ in the sense of Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (iii). This proves (iv).

Step 6: uniform continuity. Let $\varepsilon>0$; we find $\delta>0$ with $|f(x)-f(y)|<\varepsilon$ for all $x,y\in X$ with $d(x,y)<\delta$, as Uniformly Continuous Map Between Metric Spaces (i) requires. If $K=\operatorname{supp}f$ is empty, $f=0$ by Step 4 and every $\delta>0$ works. Let $K\neq\emptyset$, and let $\mathcal{U}$ be the set of open balls $B(x,r)$ with $x\in K$ and real $r>0$ such that

$$|f(y)-f(x)|<\varepsilon/2\qquad\text{for every }y\in B(x,2r).$$

The set $\mathcal{U}$ covers $K$: for $x\in K$, continuity of $f$ at $x$ (Continuous Map Between Metric Spaces (i)) gives $\delta_{0}>0$ with $|f(y)-f(x)|<\varepsilon/2$ whenever $d(x,y)<\delta_{0}$, and then $x\in B(x,\delta_{0}/2)\in\mathcal{U}$. Open balls are open, so by Compact Subsets of a Metric Space: Covers by Open Sets, Finite Unions and Increasing Covers; Sublevel Sets of the Distance and Closed Balls (i) some finite $\mathcal{V}\subseteq\mathcal{U}$ covers $K$, and $\mathcal{V}\neq\emptyset$ because $K\neq\emptyset$. For each $U\in\mathcal{V}$ the set of pairs $(x,r)$ as in the definition of $\mathcal{U}$ with $U=B(x,r)$ is nonempty, so by Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice (xi) we may fix such a pair $(x_{U},r_{U})$ for every $U\in\mathcal{V}$. Let $\delta>0$ be the least element of the nonempty finite set $\{r_{U}:U\in\mathcal{V}\}$, given by Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice (vi) and Finite Sets: the Pigeonhole Principle, Uniqueness of the Length, Subsets, Unions, Products, Images, Bounded Sets of Natural Numbers, Extreme Elements, Sets of Maps, Finite Unions and Finite Choice (viii).

Now let $x,y\in X$ with $d(x,y)<\delta$. If $x,y\notin K$, then $f(x)=f(y)=0$ by Step 4. If $x\in K$, then $x\in U$ for some $U\in\mathcal{V}$, that is $d(x_{U},x)<r_{U}$, and

$$d(x_{U},y)\le d(x_{U},x)+d(x,y)<r_{U}+\delta\le2r_{U};$$

so $x,y\in B(x_{U},2r_{U})$, and

$$|f(x)-f(y)|\le|f(x)-f(x_{U})|+|f(x_{U})-f(y)|<\varepsilon .$$

The case $y\in K$ is symmetric. This proves (v).