Approximating C_0(X) Functions by Compactly Supported Ones

Vanishing at Infinity and Compact Support

On a noncompact space, a continuous function need not have compact support: it may remain nonzero arbitrarily far out. The space \(C_0(X)\) consists of continuous functions that become uniformly small outside a compact set, while \(C_c(X)\) consists of continuous functions that vanish outside a compact set. The key to approximating a function in \(C_0(X)\) is to preserve it where its magnitude is appreciable and taper it continuously to zero outside a compact region.

Throughout, \(X\) is a locally compact Hausdorff space. No measure is being used: all convergence and norms below are pointwise or uniform on \(X\), not almost everywhere. Functions may be real-valued or complex-valued, and \(|f(x)|\) denotes the absolute value or complex modulus.

The level-set definition expresses “vanishing at infinity” without requiring a distance or a notion of going to infinity along a sequence. It also immediately identifies the compact set on which an approximation should match \(f\). The norm in this definition is finite: for \(f\in C_0(X)\), the set \(K=\{|f|\ge1\}\) is compact, and \(|f|<1\) outside \(K\). Continuity makes \(f\) bounded on \(K\), so it is bounded on all of \(X\).

Proof. Let \(f\in C_c(X)\), and put \(K=\operatorname{supp}f\), which is compact. The continuous function \(f\) is bounded on \(K\), and it is zero on \(X\setminus K\), so it is bounded on \(X\). For any \(\varepsilon>0\), the set \(F_\varepsilon=\{x:|f(x)|\ge\varepsilon\}\) is closed because \(x\mapsto |f(x)|\) is continuous and \([\varepsilon,\infty)\) is closed. Also \(F_\varepsilon\subseteq K\), since \(f(x)\ne0\) for every \(x\in F_\varepsilon\). Thus \(F_\varepsilon\) is a closed subset of the compact set \(K\), hence compact. This is exactly the defining condition for \(f\in C_0(X)\). \(\square\)

The Cutoff Principle

We use the locally compact Hausdorff cutoff result: if \(K\subseteq U\subseteq X\), where \(K\) is compact and \(U\) is open, there is an \(\eta\in C_c(X)\) with \(0\le\eta\le1\), \(\eta=1\) on \(K\), and \(\operatorname{supp}\eta\subseteq U\). This is the compactly supported Urysohn cutoff construction. In particular, taking \(U=X\) gives a compactly supported function equal to one on any prescribed compact set.

Proof. If \(X\) is empty, the only function on \(X\) is the zero function and the claim holds with \(g=0\). Otherwise, fix \(f\in C_0(X)\) and \(\delta>0\), and set \(\varepsilon=\delta/2\). The set \(K=\{x:|f(x)|\ge\varepsilon\}\) is compact by the definition of \(C_0(X)\). Apply the cutoff result with \(U=X\) to obtain \(\eta\in C_c(X)\) such that \(0\le\eta\le1\) and \(\eta=1\) on \(K\). Define \(g=\eta f\). The product of continuous functions is continuous, and \(\operatorname{supp}g\subseteq\operatorname{supp}\eta\), so \(g\in C_c(X)\). For \(x\in K\), \(g(x)=f(x)\). For \(x\notin K\), \(|f(x)|<\varepsilon\), and therefore \[ |f(x)-g(x)|=|1-\eta(x)|\,|f(x)|\le |f(x)|<\varepsilon, \] because \(0\le\eta(x)\le1\). Hence \(\|f-g\|_\infty\le\varepsilon=\delta/2<\delta\). If \(K\) is empty, the same cutoff construction may use \(\eta=0\); the estimate \(|f(x)|<\varepsilon\) holds everywhere. Thus the conclusion includes that case as well. \(\square\)

The proof gives a useful error estimate: the approximation agrees exactly with \(f\) wherever \(|f|\ge\varepsilon\), and its error is at most \(\varepsilon\) everywhere else. The cutoff may be chosen independently of the values of \(f\); it only needs to equal one on the compact level set.

Explicit Cutoffs on the Real Line

On \(\mathbb{R}\), with its usual topology, a convenient cutoff for any \(R>0\) is \[ \eta_R(x)= \begin{cases} 1,& |x|\le R,\\ R+1-|x|,& R<|x|<R+1,\\ 0,& |x|\ge R+1. \end{cases} \]

The formulas agree at \(|x|=R\) and \(|x|=R+1\), so \(\eta_R\) is continuous. It takes values in \([0,1]\), equals one on \([-R,R]\), and has support \([-R-1,R+1]\), a compact interval. For a continuous function \(f\) vanishing at infinity, \(g_R=\eta_R f\) is therefore in \(C_c(\mathbb{R})\). The error is zero on \([-R,R]\); outside that interval it is \(|1-\eta_R(x)|\,|f(x)|\). The following examples compute that error explicitly.

Why Sharp Truncation Is Not Enough

A tempting construction is to keep \(f\) on a compact interval and set it to zero immediately outside. That generally produces a discontinuity at the boundary, so it need not belong to \(C_c(X)\). For example, define \(q(x)=e^{-|x|}\chi_{[-2,2]}(x)\), where \(\chi_{[-2,2]}\) is one on \([-2,2]\) and zero elsewhere. At \(x=2\), \(q(2)=e^{-2}\), but for every \(x>2\), \(q(x)=0\). The right-hand limit at \(2\) is zero, not \(q(2)\); hence \(q\) is not continuous and is not in \(C_c(\mathbb{R})\).

The tapered cutoff fixes precisely this boundary problem: \(\eta_2\) decreases continuously from one to zero on \(2<|x|<3\). Multiplication by it preserves continuity while making the support compact. This is the practical content of the density theorem: a function in \(C_0(X)\) need not itself have compact support, but its small tail can be discarded with arbitrarily small uniform error.

When \(X\) is compact, every continuous function has compact support, since its support is a closed subset of \(X\). In that case \(C_c(X)=C_0(X)=C(X)\), and the density statement is immediate. The theorem is most useful on noncompact spaces, where it allows continuous functions vanishing at infinity to be handled through compactly supported test functions without changing them by much in the uniform norm.