Arzelà–Ascoli Theorem

Subjects: Metric and Normed Spaces
Links: Space of Continuous Functions From Rn to Rm, Compact Sets in Rn, Compactness in Metric Spaces, Complete Metric Spaces, Continuous Function Spaces, Total Boundedness, Relative Compactness in Metric Spaces, Compact Sets

For the rest of this section, let $K=(K,d_K)$ be a compact metric space and $(X,d_X)$ be a metric space, then we will consider the space of continuous functions

Arzelà-Ascoli Theorem

Let $K$ be a compact metric space, and $X$ be a complete metric space. A subset $\mathcal{H}$ of ${\mathcal{C}}^0(K,X)$ is relatively compact on ${\mathcal{C}}^0(K,X)$ iff $\mathcal{H}$ is equicontinuous and the sets

are relatively compact in $X$ for all $z\in K$

Let $X$ and $K$ be compact metric spaces, then a subet $\mathcal{H}$ of ${\mathcal{C}}^0(K,X)$ is relatively compact iff $\mathcal{H}$ is equicontinuous.

Let $K$ be compact metric space, and $X$ be a complete metric space, the sequence $(f_k)$ in the space ${\mathcal{C}}^0(K,X)$ congerves pointwise to the function $f:K\to X$. If $\mathcal{H}:=\{f_k\mid k\in \mathbb{N}\}$ is equicontinuous then $f$ is continuous and $f_k$ converges uniformly to $f$ or $f_k$ converges in ${\mathcal{C}}^0(K,X)$

Arzelà-Ascoli Theorem to ${\mathbb{R}}^n$

Let $K$ be a compact metric space. A subset $\mathcal{H}$ of ${\mathcal{C}}^0(K,{\mathbb{R}}^n)$ is relatively compact on ${\mathcal{C}}^0(K,{\mathbb{R}}^n)$ iff $\mathcal{H}$ is equicontinuous bounded in ${\mathcal{C}}^0(K,{\mathbb{R}}^n)$

Arzelà–Ascoli Theorem from ${\mathbb{R}}^m$ to ${\mathbb{R}}^n$

Let $A\subseteq {\mathbb{R}}^m$ be compact and let $B\subseteq \mathcal{C}(A,{\mathbb{R}}^n)$. If $B$ is bounded and equicontinuous, then any sequence in $B$ has a uniformly convergent subsequence.

Thus we have a characterization of sequential compactness in $\mathcal{C}(A,{\mathbb{R}}^n)$, when $A$ is compact.