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A manifold bump for a compact set inside an open set
Statement
Let be a smooth manifold, let be compact, and let be open with . Then there exists a smooth function that equals on an open neighbourhood of and satisfies .
Facts & Assumptions
Given: A compact set and an open set with .
Every point of admits a smooth bump supported in and equal to at that point (A chart bump at a point with prescribed support).
The standard smooth step function is smooth, takes values in , is on and on (The standard smooth step function).
An indexed ambient open cover of a compact subset has a finite subcover (A subspace is compact exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it, clause 2).
Smooth maps are continuous, and composites of smooth maps are smooth (Smooth maps are continuous, Identity maps and composites of smooth maps are smooth).
Sums and scalar multiples satisfy the coordinate derivative rules (Sums, scalar multiples, products and quotients: , , , and when ); smoothness is tested by continuous iterated coordinate derivatives in charts ( maps and multi-index derivative notation in Euclidean space).
Support means the closure in of the nonzero locus.
Proof
If , take and the empty open neighbourhood. Otherwise let be the set of all smooth whose support lies in . For each , put . These sets are open by [L3] and cover by [L1]. The family uses all eligible functions, not a simultaneous choice of one function at each point.
By [L2], there are finitely many such that contains . Define . In every chart each iterated coordinate derivative of is the sum of the corresponding continuous derivatives of the , by repeated use of [L4]. Hence and are smooth. Since all summands are nonnegative, throughout the open set .
Define . It is smooth by [F1] and [L3], takes values in , and equals one on by step 2.1.
The finite union is closed by [F2], and lies in . Off every is zero, so and . Taking the closure of the nonzero locus gives . Together with step 3.1 and the empty case in step 1.1, this proves the claim.
Depends on
A chart bump at a point with prescribed support
The standard smooth step function
A subspace is compact exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it
Smooth maps are continuous
Identity maps and composites of smooth maps are smooth
Sums, scalar multiples, products and quotients: , , , and when
maps and multi-index derivative notation in Euclidean space