Proof of Completeness of the Space of Continuous Vector-Valued Functions under the Supremum Metric

Throughout, $d$ is the Euclidean distance on $\mathbb{R}^{k}$, a metric by Euclidean Distance is a Metric on $\mathbb{R}^n$, and we use the componentwise inequalities $|x^{i}-y^{i}|\le d(x,y)\le\sum_{l=1}^{k}|x^{l}-y^{l}|$ of claim 1 of Componentwise Estimates, Transpose Identities, and Indefinite Riemann Integrals (with $d(x,y)=|x-y|$ as recorded there).

Claim 1. Fix $h,h'\in\mathcal{C}$. Each component $h^{i}-h'^{i}$ is continuous on $[a,b]$ (differences of continuous functions are continuous, Sum and Product Rules for One-Dimensional Derivatives and Continuity), hence attains a maximum and a minimum by Extreme Value Theorem on a Compact Interval; with $M_i$ denoting the larger of the absolute values of that maximum and minimum, $|h^{i}(t)-h'^{i}(t)|\le M_i$ for all $t\in[a,b]$. Then $d(h(t),h'(t))\le\sum_iM_i$ for all $t$, so the set of values is nonempty and bounded above, and the supremum exists by the least upper bound property; $d_{\infty}(h,h')$ is finite.

$\mathcal{C}$ is nonempty (constant functions). The metric axioms: $d_{\infty}\ge0$ is clear; $d_{\infty}(h,h')=0$ forces $d(h(t),h'(t))=0$ for every $t$, hence $h(t)=h'(t)$ for every $t$ since $d$ is a metric; symmetry is inherited from $d$; and for $h,h',h''\in\mathcal{C}$ and every $t$,

so the right-hand side is an upper bound of the values and dominates their supremum: $d_{\infty}(h,h'')\le d_{\infty}(h,h')+d_{\infty}(h',h'')$.

Claim 2. Let $(h_n)_{n}$ be a Cauchy sequence in $(\mathcal{C},d_{\infty})$, and let $\varepsilon>0$ be given. For every $t\in[a,b]$ and every component $i$, $|h_n^{i}(t)-h_m^{i}(t)|\le d(h_n(t),h_m(t))\le d_{\infty}(h_n,h_m)$, so $(h_n^{i}(t))_n$ is a Cauchy sequence of real numbers and converges by Every Cauchy Sequence of Real Numbers Converges; define $h(t)$ componentwise as the limit.

Uniform estimate. Choose $N$ with $d_{\infty}(h_n,h_m)<\varepsilon/2$ for $n,m\ge N$. Fix $t\in[a,b]$ and $n\ge N$. For every $m\ge N$,

and the last sum tends to $0$ as $m\to\infty$ by componentwise convergence; hence $d(h_n(t),h(t))\le\varepsilon/2$. Since $t$ was arbitrary, the set of values $d(h_n(t),h(t))$ ($t\in[a,b]$) is bounded above by $\varepsilon/2$, for every $n\ge N$.

Continuity of the limit. Fix a component $i$, a point $t\in[a,b]$, and $\eta>0$. Apply the uniform estimate with $\varepsilon$ replaced by a value making the bound at most $\eta/3$, obtaining $n$ with $d(h_n(s),h(s))\le\eta/3$ for all $s\in[a,b]$; then choose $\delta>0$ from continuity of $h_n^{i}$ at $t$ with $|h_n^{i}(s)-h_n^{i}(t)|<\eta/3$ for $|s-t|<\delta$, $s\in[a,b]$. Then for such $s$,

Hence $h\in\mathcal{C}$, so $d_{\infty}(h_n,h)$ is defined, and the uniform estimate gives $d_{\infty}(h_n,h)\le\varepsilon/2<\varepsilon$ for all $n\ge N$: the sequence converges to $h$ in $(\mathcal{C},d_{\infty})$. Every Cauchy sequence converges, so $(\mathcal{C},d_{\infty})$ is complete. $\blacksquare$