Theorem. If $U$ is a non-empty open set in $\sR^n$ , then the set of smooth functions with compact support $C^\infty_0(U)$ is non-trivial (that is, it contains functions other than the zero function).

Remark. This theorem may seem to be obvious at first sight. A way to notice that it is not so obvious, is to formulate it for analytic functions with compact support: in that case, the result does not hold; in fact, there are no nonconstant analytic functions with compact support at all. One important consequence of this theorem is the existence of partitions of unity.

Proof of the theorem: Let us first prove this for $n=1$ : If $a<b$ be real numbers, then there exists a smooth non-negative function $f:\sR \to \sR$ , whose support is the compact set $[a,b]$ .

To see this, let $\phi\colon \sR \to \sR$ be the function defined on this page, and let $$ f(x) = \phi(x-a) \phi(b-x). $$ Since $\phi$ is smooth, it follows that $f$ is smooth. Also, from the definition of $\phi$ , we see that $\phi(x-a)=0$ precisely when $x\le a$ , and $\phi(b-x)=0$ precisely when $x\ge b$ . Thus the support of $f$ is indeed $[a,b]$ .

Since $U$ is non-empty and open there exists an $x\in U$ and $\varepsilon>0$ such that $B_\varepsilon(x)\subseteq U$ . Let $f$ be smooth function such that $\operatorname{supp} f =[-\varepsilon/2,\varepsilon/2]$ , and let $$ h(z) = f(\Vert x-z \Vert^2). $$ Since $\lVert\cdot \rVert^2$ (Euclidean norm) is smooth, the claim follows.