$C_0^{\mathrm{\infty }}(U)$ is not empty

Theorem. If $U$ is a non-empty open set in ${ℝ}^n$, then the set ofsmooth functions with compact support $C_0^{\mathrm{\infty }}(U)$ 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 foranalytic functions with compact support: in that case, the resultdoes not hold; in fact, there are no nonconstant analyticfunctions with compact support at all.One important consequence of this theorem is the existence of partitionsof unity.

Proof of the theorem:Let us first prove this for $n=1$:If be real numbers, then there exists asmooth non-negative function $f:ℝ\to ℝ$, whose support (http://planetmath.org/SupportOfFunction) is thecompact set $[a,b]$.

To see this, let $\varphi :ℝ\to ℝ$ be the functiondefined on this page (http://planetmath.org/InfinitelyDifferentiableFunctionThatIsNotAnalytic),and let

$$f(x)=\varphi (x-a)\varphi (b-x).$$

Since $\varphi$ is smooth, it follows that $f$ is smooth. Also, from thedefinition of $\varphi$, we see that$\varphi (x-a)=0$ precisely when $x\le a$, and$\varphi (b-x)=0$ precisely when $x\ge b$.Thus the support of $f$ is indeed $[a,b]$.

Since $U$ is non-empty andopen there exists an $x\in U$ and $\epsilon >0$ such that$B_{\epsilon }(x)\subseteq U$. Let $f$ be smooth functionsuch that $\mathrm{supp}f=[-\epsilon /2,\epsilon /2]$, andlet

$$h(z)=f({\parallel x-z\parallel }^2).$$

Since ${\parallel \cdot \parallel }^2$ (Euclidean norm) is smooth, the claim follows.$\mathrm{\square }$