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Smooth locally defined functions can be glued by a partition of unity

Statement

Let (Ui)i∈I be an open cover of a smooth manifold M, let fi:Ui→R be smooth for each i, and let (ϕi)i∈I be a smooth partition of unity subordinate to (Ui). Then the pointwise formula F(p):=∑iϕi(p)fi(p) defines a smooth function F:M→R.

Facts & Assumptions

Given: An open cover (Ui)i∈I of M, smooth functions fi:Ui→R, and a smooth partition of unity (ϕi)i∈I subordinate to (Ui).

[F1]

In a smooth partition of unity subordinate to (Ui), the support family (supp⁡(ϕi)) is locally finite and each support lies in Ui (Smooth partitions of unity subordinate to an open cover).

[L1]

Smooth maps paste over an open cover (Smooth maps paste over an open cover).

[L2]

A locally finite sum of smooth functions is smooth (A locally finite sum of smooth functions is smooth).

[A1]

For each i, the product ϕifi is smooth on Ui.

Proof

technique · direct
1.1F1L1A1givenconstruct

Fix i∈I. On the open set Ui let Gi:=ϕifi, and on the open set M∖supp⁡(ϕi) let Hi:=0. By [A1], the map Gi is smooth on Ui; by [F1], the two open sets cover M and on the overlap Ui∖supp⁡(ϕi) one has ϕi=0, so Gi=Hi. Therefore [L1] pastes them to a smooth global function Fi:M→R with Fi=ϕifi on Ui and supp⁡(Fi)⊆supp⁡(ϕi).

2.1F1L2step 1.1

By [F1] and step 1.1, the family (supp⁡(Fi))i∈I is locally finite. Hence the sum F:=∑iFi is well defined and smooth by [L2].

3.1F1step 1.1step 2.1∎

Let p∈M. If p∈Ui, step 1.1 gives Fi(p)=ϕi(p)fi(p), including when ϕi(p)=0. If p∉Ui, then [F1] implies p∉supp⁡(ϕi), so the zero extension in step 1.1 gives Fi(p)=0; the summand ϕi(p)fi(p) is interpreted as zero there. Local finiteness in [F1] ensures that only finitely many supports contain p, and all other summands vanish. Thus the stated pointwise sum equals ∑iFi(p)=F(p) and is smooth by step 2.1.

Depends on

Used by

Dependency tree · two levels

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Sources