Let \(X\) be a space. A subset \(\mathcal{F}\) of \(\mathcal{C}(X, \mathbb{R})\) is said to vanish uniformly at infinity if given \(\epsilon > 0\), there is a compact subspace \(C\) of \(X\) such that \(\abs{f(x)} < \epsilon\) for \(x \in X - C\) and \(f \in \mathcal{F}\). If \(\mathcal{F}\) consists of a single function \(f\), we say simply that \(f\) vanishes at infinity. Let \(\mathcal{C}_0(X, \mathbb{R})\) denote the set of continuous functions \(f: X \to \mathbb{R}\) that vanish at infinity.
Theorem. Let \(X\) be locally compact Hausdorff; give \(\mathcal{C}_0(X, \mathbb{R})\) the uniform topology. A subset \(\mathcal{F}\) of \(\mathcal{C}_0(X, \mathbb{R})\) has compact closure if and only if it is pointwise bounded, equicontinuous, and vanishes uniformly at infinity.
[Hint: Let \(Y\) denote the one-point compactification of \(X\). Show that \(\mathcal{C}_0(X, \mathbb{R})\) is isometric with a closed subspace of \(\mathcal{C}(Y, \mathbb{R})\) if both are given the sup metric.]
Solution
Let \(Y=X\cup\{\infty\}\) be the one-point compactification. For \(f\in C_0(X,\mathbb R)\) define \[ Tf(x)=f(x)\ (x\in X),\qquad Tf(\infty)=0. \] The condition that \(f\) vanish at infinity is exactly continuity of \(Tf\) at \(\infty\). Hence \(T\) is an isometry from \(C_0(X,\mathbb R)\), with the sup metric, onto \[ E=\{F\in C(Y,\mathbb R):F(\infty)=0\}. \] Evaluation at \(\infty\) is continuous, so \(E\) is closed in \(C(Y,\mathbb R)\).
Apply Ascoli’s theorem on the compact Hausdorff space \(Y\). For a family \(\mathcal F\subset C_0(X,\mathbb R)\), pointwise boundedness of \(T\mathcal F\) on \(Y\) is exactly pointwise boundedness on \(X\) (all values at \(\infty\) are \(0\)). Equicontinuity at points of \(X\) is the stated equicontinuity. Equicontinuity at \(\infty\) says: for every \(\varepsilon>0\) there is a neighborhood \(Y-C\) of \(\infty\), with \(C\subset X\) compact, such that \[ |f(x)|=|Tf(x)-Tf(\infty)|<\varepsilon \] for all \(x\notin C\) and all \(f\in\mathcal F\); this is precisely uniform vanishing at infinity. Therefore Ascoli on \(Y\) says that \(T\mathcal F\) has compact closure iff the three stated conditions hold. Since \(T\) is an isometry onto the closed subspace \(E\), this is equivalent to compactness of \(\overline{\mathcal F}\) in \(C_0(X,\mathbb R)\).