Local Compactness and Compact Support
Continuous functions used to test measures are often required to vanish outside a compact set. This requirement is expressed by the support of a function, which includes not only the points where the function is nonzero but also their limit points. The space of continuous functions with compact support is denoted by \(C_c(X)\). To define it, we first specify the topological setting.
A topological space \(X\) is Hausdorff if any two distinct points have disjoint open neighborhoods. A subset \(K\subseteq X\) is compact if every open cover of \(K\) has a finite subcover. A neighborhood of \(x\in X\) means a set containing an open set that contains \(x\).
Here closure is always taken in the space \(X\), unless a different space is stated. In a Hausdorff space, compact subsets are closed. This fact makes the two forms of local compactness equivalent.
Proof. Suppose first that \(x\) has a compact neighborhood \(K\). By the definition of neighborhood, there is an open set \(U\) with \(x\in U\subseteq K\). Since \(X\) is Hausdorff and \(K\) is compact, \(K\) is closed. Therefore \(\overline U\subseteq K\). The set \(\overline U\) is closed in \(X\), hence closed in \(K\); a closed subset of a compact space is compact. Thus \(U\) is an open neighborhood of \(x\) with compact closure. Conversely, if \(U\) is open, \(x\in U\), and \(\overline U\) is compact, then \(\overline U\) is a compact neighborhood of \(x\), because it contains the open neighborhood \(U\). This proves both implications. \(\square\)
The real line \(\mathbb{R}\) with its usual topology is locally compact Hausdorff: for each \(x\in\mathbb{R}\), the open interval \((x-1,x+1)\) has compact closure \([x-1,x+1]\). Any compact Hausdorff space is also locally compact: the whole space is a compact neighborhood of each of its points. In particular, a closed bounded interval with its relative topology is compact Hausdorff and locally compact.
Support and the Space \(C_c(X)\)
The set \(\{x:f(x)\ne0\}\) is open when \(f\) is continuous, because it is the inverse image of the open set \(\mathbb{R}\setminus\{0\}\) or \(\mathbb{C}\setminus\{0\}\). Its closure, the support, is closed. The support is not usually the same as the nonzero set: it also contains points that can be approached by nonzero points.
Proof. Put \(U=\{x:f(x)\ne0\}\). By definition, \(U\subseteq\overline U=\operatorname{supp}f\), so \(f=0\) on \(X\setminus\operatorname{supp}f\). If \(F\) is any closed set such that \(f=0\) on \(X\setminus F\), then every point of \(U\) belongs to \(F\); since \(F\) is closed, it also contains \(\overline U\). Hence \(\operatorname{supp}f\subseteq F\), proving the smallest-set assertion. Now assume \(X\) is Hausdorff. If \(\operatorname{supp}f\) is compact, take \(K=\operatorname{supp}f\); the first part shows that \(f\) vanishes outside \(K\). Conversely, suppose \(f\) vanishes outside a compact \(K\). Then \(U\subseteq K\), so \(\operatorname{supp}f=\overline U\subseteq K\), since compact sets in a Hausdorff space are closed. The support is closed in \(X\), hence is closed in \(K\); it is therefore compact. \(\square\)
This criterion explains why compact support is a condition on the closure of the nonzero set, not just on where the function happens to be nonzero. It also gives a practical test: find a compact set outside which the function is zero, or calculate the closure of the nonzero set directly.
Worked Supports
Worked Example: A Clipped Function on the Real Line
Define \(f:\mathbb{R}\to\mathbb{R}\) by \(f(x)=\max(1-x^2,0)\). Determine its support and decide whether \(f\in C_c(\mathbb{R})\).
The function is continuous because it is the maximum of the two continuous functions \(1-x^2\) and \(0\). It is nonzero exactly when \(1-x^2>0\), which is equivalent to \(x^2<1\), or \(-1<x<1\). Thus its nonzero set is \((-1,1)\). The closure of that set in \(\mathbb{R}\) is \([-1,1]\), so $$ \operatorname{supp}f=[-1,1]. $$ The interval \([-1,1]\) is compact, and therefore \(f\in C_c(\mathbb{R})\). Check: \(f(-1)=f(1)=0\), but every open neighborhood of either endpoint contains points in \((-1,1)\) where \(f\ne0\). Both endpoints must therefore be in the support.
Worked Example: A Continuous Function Without Compact Support
Let \(g(x)=e^{-x^2}\) on \(\mathbb{R}\). Find its support and test whether it has compact support.
For every real \(x\), \(e^{-x^2}>0\). Consequently the nonzero set is all of \(\mathbb{R}\), its closure is \(\mathbb{R}\), and \(\operatorname{supp}g=\mathbb{R}\). The real line is not compact: the open intervals \((-n,n)\), for positive integers \(n\), cover \(\mathbb{R}\), and no finite collection of them covers it. Hence \(g\notin C_c(\mathbb{R})\). Check: although \(g(x)\) tends to zero as \(|x|\) tends to infinity, it is never exactly zero; its support is still the entire real line.
Worked Example: A Bump-Shaped Function on the Real Line
Define \(h:\mathbb{R}\to\mathbb{R}\) by $$ h(x)= \begin{cases} \exp\!\left(-\dfrac{1}{1-x^2}\right),& |x|<1,\\ 0,& |x|\ge1. \end{cases} $$ Determine its support and verify continuity at the two joining points.
For \(|x|<1\), the exponent is a finite real number, so \(h(x)>0\). For \(|x|\ge1\), \(h(x)=0\). The nonzero set is therefore \((-1,1)\), and its closure is \([-1,1]\). Thus \(\operatorname{supp}h=[-1,1]\), a compact set.
Inside \((-1,1)\), the formula is a composition of continuous functions. To check continuity at \(1\), let \(x\) approach \(1\) from below. Then \(1-x^2\) is positive and tends to \(0\), so \(-1/(1-x^2)\) tends to \(-\infty\), and its exponential tends to \(0=h(1)\). As \(x\) approaches \(1\) from above, \(h(x)=0\). The two one-sided limits equal the defined value. At \(-1\), approaching from above again gives \(1-x^2\to0\) through positive values and hence \(h(x)\to0=h(-1)\); approaching from below gives the constant value \(0\). This verifies continuity at both endpoints. Check: the function is positive at every point strictly between \(-1\) and \(1\), so neither endpoint can be omitted from the closure of its nonzero set.
Worked Example: Relative Support on a Compact Interval
Take \(X=[0,1]\) with its relative topology and define \(q(x)=\max(0,1-4|x-\tfrac12|)\). Find \(\operatorname{supp}q\), and compare it with the support of \(r(x)=x\) on the same space.
The function \(q\) is continuous. Its nonzero condition is $$ 1-4\left|x-\tfrac12\right|>0 \quad\Longleftrightarrow\quad \left|x-\tfrac12\right|<\tfrac14 \quad\Longleftrightarrow\quad \tfrac14<x<\tfrac34. $$ The closure in \(X=[0,1]\) is \([\tfrac14,\tfrac34]\), so \(\operatorname{supp}q=[\tfrac14,\tfrac34]\). In particular, \(q\) is zero at both endpoints of its support, but those points are limits of points where it is positive.
For \(r(x)=x\), the nonzero set in \(X\) is \((0,1]\). Its closure in the relative topology of \(X\) is \([0,1]\), so \(\operatorname{supp}r=[0,1]\). Since \(X\) is compact, every closed subset of \(X\) is compact; both functions belong to \(C_c(X)\). Check: \(q\) is nonzero at \(x=\tfrac12\), but zero at \(x=\tfrac14\) and \(x=\tfrac34\), confirming that the computed support is the closure, not the nonzero set itself.
Basic Operations and a Common Pitfall
Proof. The sum, scalar multiple, and product of continuous functions are continuous. For the sum, a point outside \(\operatorname{supp}f\cup\operatorname{supp}g\) has an open neighborhood on which \(f=0\) and an open neighborhood on which \(g=0\), by the support characterization above. The intersection of those neighborhoods is open and both functions vanish there, so \(f+g\) vanishes there too. The point is not in \(\operatorname{supp}(f+g)\), proving the first inclusion. For the product, if a point is outside \(\operatorname{supp}f\), there is an open neighborhood on which \(f=0\), and hence \(fg=0\) there. The same conclusion holds if the point is outside \(\operatorname{supp}g\). Thus the product support is contained in both supports, proving the second inclusion. Finally, each support is compact. A finite union of compact sets is compact, and a closed subset of a compact set is compact. The support inclusions therefore show that \(f+g\) and \(fg\) have compact support. For \(af\), if \(a=0\), its support is empty; if \(a\ne0\), its support equals \(\operatorname{supp}f\). In either case it has compact support. \(\square\)
A common mistake is to report the nonzero set as the support. For instance, for the function \(q\) above, one might write \(\operatorname{supp}q=(\tfrac14,\tfrac34)\), because that is exactly where \(q\ne0\). That set is not the support: it is not closed in \(X\). The definition requires taking its closure in \(X\), giving \([\tfrac14,\tfrac34]\). The endpoint check confirms the correction: every relative neighborhood of either endpoint contains points strictly between the endpoints where \(q\) is nonzero.
On \(\mathbb{R}\), compactness can often be checked using the fact that a closed bounded subset is compact. In a general locally compact Hausdorff space, there is no universal notion of “bounded,” so one must use compactness itself. On a compact space \(X\), every continuous function belongs to \(C_c(X)\), because its support is closed in \(X\) and therefore compact. On a noncompact space, continuity alone does not guarantee compact support, as the function \(e^{-x^2}\) demonstrates.
Check Your Understanding
For each question, use closure in the stated space when determining a support.
- On \(\mathbb{R}\), find the support of \(f(x)=\max(0,1-|x|)\). Is \(f\in C_c(\mathbb{R})\)?
- On \(\mathbb{R}\), determine the support of \(\sin x\). Does it have compact support?
- On \(X=[-2,2]\) with its relative topology, find the support of \(f(x)=x^2\). Does it belong to \(C_c(X)\)?
- Give a reason that a continuous function on a compact Hausdorff space always has compact support.
- If \(f,g\in C_c(X)\), explain why the support of \(f+g\) is contained in the union of their supports, even when cancellation makes the sum zero at some points inside that union.