Proof of Arzela-Ascoli Theorem for Vector-Valued Functions on a Compact Interval

Throughout, $|x-y|$ is the Euclidean distance on $\mathbb{R}^{k}$, by claim 2 of the elementary properties of the Euclidean norm, and we use the triangle inequality of claim 6 of that lemma without further comment. Let $\mathcal{C}$ denote the set of maps $[a,b]\to\mathbb{R}^{k}$ with continuous components, equipped with the supremum metric $d_{\infty}$, as in the completeness lemma for continuous vector-valued functions; by claim 1 of that lemma $d_{\infty}$ is finite and is a metric on $\mathcal{C}$, and by claim 2 the metric space $(\mathcal{C},d_{\infty})$ is complete. Note first that $M\ge0$, since $|f_{1}(a)|\le M$ and norms are nonnegative.

Step 1: every $f_{n}$ lies in $\mathcal{C}$. Fix $n$ and $t_{0}\in[a,b]$, and let $\varepsilon>0$. Hypothesis (ii) supplies $\delta>0$ such that $|f_{m}(s)-f_{m}(t)|<\varepsilon$ for every $m$ and all $s,t\in[a,b]$ with $|s-t|<\delta$; in particular $|f_{n}(t)-f_{n}(t_{0})|<\varepsilon$ whenever $|t-t_{0}|<\delta$. By claim 4 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$, $|f_{n}^{i}(t)-f_{n}^{i}(t_{0})|\le|f_{n}(t)-f_{n}(t_{0})|<\varepsilon$ for each component index $i$, so every component $f_{n}^{i}$ is continuous on $[a,b]$. Hence $f_{n}\in\mathcal{C}$.

Step 2: a countable set of dyadic points. For a natural number $m\ge1$ put

a finite set, and let $Q=\bigcup_{m\ge1}Q_{m}$. Since $p\,2^{-m}=(2p)2^{-(m+1)}$ we have $Q_{m}\subseteq Q_{m+1}$. Consecutive points of $Q_{m}$ differ by $(b-a)2^{-m}$, so

because $t$ lies in one of the $2^{m}$ intervals $[a+(b-a)p2^{-m},\,a+(b-a)(p+1)2^{-m}]$ that cover $[a,b]$, and one may take $q$ to be that interval's left endpoint. Listing the elements of $Q_{1}$ in increasing order, then those of $Q_{2}\setminus Q_{1}$ in increasing order, then those of $Q_{3}\setminus Q_{2}$, and so on, produces a sequence $(q_{r})_{r\in\mathbb{N}}$ whose set of terms is $Q$; each $Q_{m}$ being finite, for every $m$ there is a natural number $R_{m}$ with $Q_{m}\subseteq\{q_{1},\dots,q_{R_{m}}\}$.

Step 3: diagonal extraction. Let $\bar{B}$ be the closed ball in $\mathbb{R}^{k}$ of centre the origin and radius $M$; by hypothesis (i), $f_{n}(t)\in\bar{B}$ for all $n$ and $t$. By the compactness of closed Euclidean balls together with the fact that a compact subset of a metric space is sequentially compact, $\bar{B}$ is a sequentially compact subset of $\mathbb{R}^{k}$: every sequence of points of $\bar{B}$ has a subsequence converging to a point of $\bar{B}$.

We construct strictly increasing maps $\sigma_{r}:\mathbb{N}\to\mathbb{N}$ for $r\in\mathbb{N}$, each a subsequence selector refining the previous one. Applying sequential compactness of $\bar{B}$ to the sequence $\bigl(f_{n}(q_{1})\bigr)_{n\in\mathbb{N}}$ yields a strictly increasing $\sigma_{1}$ such that $\bigl(f_{\sigma_{1}(n)}(q_{1})\bigr)_{n}$ converges. If $\sigma_{r}$ has been constructed, apply sequential compactness to $\bigl(f_{\sigma_{r}(n)}(q_{r+1})\bigr)_{n}$ to obtain a strictly increasing $\tau:\mathbb{N}\to\mathbb{N}$ with $\bigl(f_{\sigma_{r}(\tau(n))}(q_{r+1})\bigr)_{n}$ convergent, and set $\sigma_{r+1}=\sigma_{r}\circ\tau$, again strictly increasing. We record the nesting explicitly. For all $s\le r$ there is a strictly increasing $\theta_{s,r}:\mathbb{N}\to\mathbb{N}$ with $\sigma_{r}=\sigma_{s}\circ\theta_{s,r}$. This follows by induction on $r$ for fixed $s$: for $r=s$ take $\theta_{s,s}$ to be the identity, and if $\sigma_{r}=\sigma_{s}\circ\theta_{s,r}$ then $\sigma_{r+1}=\sigma_{r}\circ\tau=\sigma_{s}\circ(\theta_{s,r}\circ\tau)$, where $\tau$ is the selector used at that construction step and $\theta_{s,r}\circ\tau$ is strictly increasing as a composition of strictly increasing maps. Consequently, for $s\le r$, the sequence $\bigl(f_{\sigma_{r}(n)}(q_{s})\bigr)_{n}$ is a subsequence of a subsequence, hence a subsequence, of $\bigl(f_{\sigma_{s}(n)}(q_{s})\bigr)_{n}$, which converges by construction; so it converges too, by the fact that a subsequence of a convergent sequence has the same limit.

Define $n_{j}=\sigma_{j}(j)$. Since $\sigma_{j+1}$ selects a subsequence of the one selected by $\sigma_{j}$, there is a strictly increasing $\rho$ with $\sigma_{j+1}=\sigma_{j}\circ\rho$, and $\rho(j+1)\ge j+1$ by the growth bound for strictly increasing sequences of natural numbers; as $\sigma_{j}$ is strictly increasing this gives $n_{j+1}=\sigma_{j}(\rho(j+1))\ge\sigma_{j}(j+1)>\sigma_{j}(j)=n_{j}$. So $(f_{n_{j}})_{j\in\mathbb{N}}$ is a subsequence of $(f_{n})_{n\in\mathbb{N}}$. Fix $r$. For $j\ge r$ we have $n_{j}=\sigma_{j}(j)=\sigma_{r}\bigl(\theta_{r,j}(j)\bigr)$, so $n_{j}$ is the $\theta_{r,j}(j)$-th term of the sequence selected by $\sigma_{r}$. The index map $j\mapsto\theta_{r,j}(j)$ is strictly increasing on $\{j\in\mathbb{N}:j\ge r\}$: if $j'>j\ge r$ then $\sigma_{r}(\theta_{r,j'}(j'))=n_{j'}>n_{j}=\sigma_{r}(\theta_{r,j}(j))$ by the strict increase of $(n_{j})_{j}$ just established, and $\sigma_{r}$ is strictly increasing, hence injective and order-reflecting, so $\theta_{r,j'}(j')>\theta_{r,j}(j)$. Therefore $\bigl(f_{n_{j}}(q_{r})\bigr)_{j\ge r}$ is a subsequence of the convergent sequence $\bigl(f_{\sigma_{r}(n)}(q_{r})\bigr)_{n}$ and therefore converges; adjoining the finitely many terms with $j<r$ does not affect this, so $\bigl(f_{n_{j}}(q_{r})\bigr)_{j\in\mathbb{N}}$ converges for every $r$.

Step 4: the subsequence is Cauchy for $d_{\infty}$. Let $\varepsilon>0$. By hypothesis (ii) choose $\delta>0$ such that $|f_{n}(s)-f_{n}(t)|<\varepsilon/3$ for every $n$ and all $s,t\in[a,b]$ with $|s-t|<\delta$. Choose a natural number $m\ge1$ with $(b-a)2^{-m}<\delta$. Every sequence $\bigl(f_{n_{j}}(q)\bigr)_{j}$ with $q\in Q_{m}$ converges by Step 3 and is therefore Cauchy; as $Q_{m}$ is finite, there is $J$ such that

Now let $t\in[a,b]$ and $j,l\ge J$. By $(*)$ pick $q\in Q_{m}$ with $|t-q|\le(b-a)2^{-m}<\delta$. Then

Since $t$ was arbitrary, $d_{\infty}(f_{n_{j}},f_{n_{l}})\le\varepsilon$ for all $j,l\ge J$. As $\varepsilon>0$ was arbitrary, $(f_{n_{j}})_{j}$ is a Cauchy sequence in $(\mathcal{C},d_{\infty})$.

Step 5: the limit and its properties. By completeness of $(\mathcal{C},d_{\infty})$ there is $f\in\mathcal{C}$ with $d_{\infty}(f_{n_{j}},f)\to0$; that is, $f$ has continuous components and $\sup_{t\in[a,b]}|f_{n_{j}}(t)-f(t)|\to0$, which is the asserted convergence, the supremum being finite by claim 1 of Completeness of the Space of Continuous Vector-Valued Functions under the Supremum Metric.

For the bound, fix $t\in[a,b]$ and let $\eta>0$. Choosing $j$ with $d_{\infty}(f_{n_{j}},f)<\eta$ gives

Since $\eta>0$ was arbitrary, $|f(t)|\le M$.

Finally let $\varepsilon>0$ and let $\delta>0$ be as in the uniform equicontinuity of the family for this $\varepsilon$, and let $s,t\in[a,b]$ satisfy $|s-t|<\delta$. For every $j$,

Given $\eta>0$, choosing $j$ with $d_{\infty}(f_{n_{j}},f)<\eta/2$ gives $|f(s)-f(t)|<\varepsilon+\eta$; as $\eta>0$ was arbitrary, $|f(s)-f(t)|\le\varepsilon$.