# $C^{\infty}_{0}(U)$ is not empty

Theorem. If $U$ is a non-empty open set in $\mathbb{R}^{n}$, 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 be real numbers, then there exists a smooth non-negative function $f:\mathbb{R}\to\mathbb{R}$, whose support (http://planetmath.org/SupportOfFunction) is the compact set $[a,b]$.

To see this, let $\phi\colon\mathbb{R}\to\mathbb{R}$ be the function defined on this page (http://planetmath.org/InfinitelyDifferentiableFunctionThatIsNotAnalytic), 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 $x\leq a$, and $\phi(b-x)=0$ precisely when $x\geq b$. 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

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

Since $\lVert\cdot\rVert^{2}$ (Euclidean norm) is smooth, the claim follows. $\Box$

Title $C^{\infty}_{0}(U)$ is not empty Cinfty0UIsNotEmpty 2013-03-22 13:43:57 2013-03-22 13:43:57 matte (1858) matte (1858) 17 matte (1858) Theorem msc 26B05