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

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

To see this, let $\varphi :ℝ\to ℝ$ be the function defined 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 the definition 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 and open there exists an $x\in U$ and $\epsilon >0$ such that $B_{\epsilon }(x)\subseteq U$. Let $f$ be smooth function such that $\mathrm{supp}f=[-\epsilon /2,\epsilon /2]$, and let

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

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

Title	$C_0^{\mathrm{\infty }}(U)$ is not empty
Canonical name	Cinfty0UIsNotEmpty
Date of creation	2013-03-22 13:43:57
Last modified on	2013-03-22 13:43:57
Owner	matte (1858)
Last modified by	matte (1858)
Numerical id	17
Author	matte (1858)
Entry type	Theorem
Classification	msc 26B05