Arzelà-Ascoli Applied

Problem 1. Let $F : [0, 1]^2 \to \mathbb R$ be continuous. For each $x \in [0, 1]$ , define $f_x := F(x, \cdot) : [0, 1] \to \mathbb R$ . Prove that the set

$\mathcal F := \{ f_x \in \mathcal C([0, 1], \mathbb R) : x \in [0, 1] \}$

is compact.

(Click for Solution)

Solution. For boundedness, we note that for any $f_x$ ,

$\displaystyle \| f_x \|_\infty = \sup_{y \in K} | F(x,y) | \leq \sup_{(x,y) \in K \times K} |F(x,y)| = \|F\|_\infty,$

as required.

For closedness, fix any sequence $\{f_{x_n}\} \subseteq \mathcal F$ and suppose $f_{x_n} \to f$ . Since $\{x_n\} \subseteq [0, 1]$ is bounded, by the Bolzano-Weierstrass theorem, it contains a convergent subsequence $x_{n_k} \to x$ . Hence, for any $y \in [0, 1]$ , since $F(\cdot, y)$ is continuous,

$\begin{aligned} f(y) = \lim_{n \to \infty} f_{x_n}(y) &= \lim_{n \to \infty} F(x_{n}, y) \\ &= \lim_{k \to \infty} F(x_{n_k}, y) = F(x,y) = f_x(y), \end{aligned}$

so that $f = f_x$ , as required.

For equicontinuity, fix $\epsilon > 0$ . We will consider $K \times K$ equipped with the supremum metric (thus inducing the box topology) for computational simplicity. Fix $k > 0$ to be tuned. For each $(x, y) \in K \times K$ , there exists $\delta_{x,y} > 0$ such that

$\| (x,y) - (u,v) \|_\infty < \delta_{x,y} \quad \Rightarrow \quad |F(x,y) - F(u,v)| < k \cdot \epsilon.$

Let $d$ denote the supremum metric. Since $\{B_{d}((x,y), \delta_{x,y}/2) : (x,y) \in K \times K\}$ forms an open cover for the compact space $K \times K$ , we can extract a finite sub cover $\{B_d((x_i,y_i), \delta_i) : i = 1,\dots, n\}$ , where we denoted $\delta_i = \delta_{x_i,y_i}$ for brevity.