C0(U) is not empty

Theorem. If U is a non-empty open set in n, then the set of smooth functions with compact support C0(U) is non-trivial (that is, it contains functionsMathworldPlanetmath 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 functionsMathworldPlanetmath 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:, whose support (http://planetmath.org/SupportOfFunction) is the compact set [a,b].

To see this, let ϕ: be the function defined on this page (http://planetmath.org/InfinitelyDifferentiableFunctionThatIsNotAnalytic), and let


Since ϕ is smooth, it follows that f is smooth. Also, from the definition of ϕ, we see that ϕ(x-a)=0 precisely when xa, and ϕ(b-x)=0 precisely when xb. Thus the support of f is indeed [a,b].

Since U is non-empty and open there exists an xU and ε>0 such that Bε(x)U. Let f be smooth functionMathworldPlanetmath such that suppf=[-ε/2,ε/2], and let


Since 2 (Euclidean norm) is smooth, the claim follows.

Title C0(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