Proof of The Continuous Functions with Compact Support Form a Real Normed Space under the Uniform Norm

Throughout, $C_{c}(X)\subseteq\mathbb{R}^{X}$, two functions on $X$ are equal when their values agree at every point, and a sequence in $C_{c}(X)$ is a sequence in $\mathbb{R}^{X}$ (routine facts of the setting); identities in $C_{c}(X)$ are therefore checked pointwise.

Step 1: the vector space. By The Space of Continuous Functions with Compact Support and the Uniform Norm (ii) the sum is a binary operation on $C_{c}(X)$ and the scalar multiplication a function $\mathbb{R}\times C_{c}(X)\to C_{c}(X)$, and by Continuous Functions with Compact Support: the Zero Function, Closure under Operations, Vanishing off the Support, Existence of the Maximum Modulus, and Uniform Continuity (i) the constant function $0$ lies in $C_{c}(X)$. Each law of a real vector space that does not involve the zero or negatives, and the identity $f+0=f$, reduces at every $x\in X$ to the corresponding law of the field $\mathbb{R}$; for instance $\bigl(\lambda(f+g)\bigr)(x)=\lambda\bigl(f(x)+g(x)\bigr)=\lambda f(x)+\lambda g(x)=(\lambda f+\lambda g)(x)$. In particular Vector Space over a Field (iii) holds with the zero $0$.

For Vector Space over a Field (iv), let $v\in C_{c}(X)$ and let $z\in C_{c}(X)$ satisfy $h+z=h$ for every $h\in C_{c}(X)$. Taking $h=0$ gives $z=0+z=0$. The function $y=(-1)v$ lies in $C_{c}(X)$, and $(v+y)(x)=v(x)-v(x)=0$ for every $x$, so $v+y=0=z$. Hence $C_{c}(X)$ is a real vector space. The argument shows that $0$ is the only $z$ with $h+z=h$ for all $h$, so $0$ is the zero vector, and since $f+(-1)f=0$, the negative of $f$ is $(-1)f$. This proves (i); in particular the difference satisfies $(f-g)(x)=f(x)-g(x)$ for $f,g\in C_{c}(X)$ and $x\in X$.

Step 2: the bounds. Let $f\in C_{c}(X)$ and $S=\{0\}\cup\{|f(x)|:x\in X\}$, so that $\|f\|_{u}$ is the greatest element of $S$ by The Space of Continuous Functions with Compact Support and the Uniform Norm (iii). Each $|f(x)|$ lies in $S$, hence $|f(x)|\le\|f\|_{u}$. If $C\ge0$ and $|f(x)|\le C$ for every $x\in X$, then every element of $S$ is at most $C$; in particular $\|f\|_{u}\le C$, since $\|f\|_{u}\in S$. This proves (iii).

Step 3: the norm. Let $f,g\in C_{c}(X)$ and $\lambda\in\mathbb{R}$; we check the conditions of Norms and Real Normed Spaces (v). Nonnegativity: $0\in S$, so $\|f\|_{u}\ge0$. Definiteness: if $\|f\|_{u}=0$, then $|f(x)|\le0$, so $f(x)=0$, for every $x$ by Step 2; thus $f=0$, the zero vector by Step 1. Absolute homogeneity: by Step 2, $|(\lambda f)(x)|=|\lambda|\,|f(x)|\le|\lambda|\,\|f\|_{u}$ for every $x$, so $\|\lambda f\|_{u}\le|\lambda|\,\|f\|_{u}$ by Step 2 again. If $\lambda\neq0$, the same inequality applied to $\lambda^{-1}$ and $\lambda f$ gives $\|f\|_{u}=\|\lambda^{-1}(\lambda f)\|_{u}\le|\lambda|^{-1}\|\lambda f\|_{u}$, hence equality; if $\lambda=0$, then $\lambda f=0$ and both sides vanish, since $\|0\|_{u}=\max\{0\}=0$. Triangle inequality: for every $x$,

$$|(f+g)(x)|\le|f(x)|+|g(x)|\le\|f\|_{u}+\|g\|_{u},$$

so $\|f+g\|_{u}\le\|f\|_{u}+\|g\|_{u}$ by Step 2. Hence $\|\cdot\|_{u}$ is a norm on the real vector space $C_{c}(X)$, and (ii) holds.

Step 4: uniform convergence. Suppose first that $\|f_{m}-f\|_{u}\to0$ in the sense of Convergent Sequences of Real Numbers (i), and let $\varepsilon>0$. There is $N\in\mathbb{N}$ with $\|f_{m}-f\|_{u}<\varepsilon$ for every $m\ge N$, and then, by Steps 1 and 2,

$$|f_{m}(x)-f(x)|=|(f_{m}-f)(x)|\le\|f_{m}-f\|_{u}<\varepsilon\qquad\text{for every }m\ge N\text{ and every }x\in X,$$

which is uniform convergence of $(f_{m})$ to $f$ on $X$ as in Pointwise and Uniform Convergence of a Sequence of Real-Valued Maps (ii). Conversely, suppose $(f_{m})$ converges uniformly to $f$ on $X$, and let $\varepsilon>0$. There is $N\in\mathbb{N}$ with $|(f_{m}-f)(x)|=|f_{m}(x)-f(x)|<\varepsilon/2$ for every $m\ge N$ and every $x\in X$, so Step 2 with $C=\varepsilon/2$ gives $0\le\|f_{m}-f\|_{u}\le\varepsilon/2<\varepsilon$ for every $m\ge N$. Hence $\|f_{m}-f\|_{u}\to0$, which proves (iv).