1270 CHAPTER 37. BASIC THEORY OF SOBOLEV SPACES
Proof: Let {ψ i}∞
i=1 be a partition of unity subordinate to the given open cover withψ i ∈C∞
c (Ui) and let u ∈ Xm,p (U) . Thus
u =∞
∑k=1
ψku.
Consider Uk for some k. Let ak be the special vector associated with Uk such that
tak +U ∩Uk ⊆U (37.0.7)
for all t ∈ (0,1) and consider only t small enough that
spt(ψk)− tak ⊆Uk (37.0.8)
Pick l (t)> 1/t which is also large enough that
tak +U ∩Uk +B(
0,1
l (t)
)⊆U, spt(ψk)+B
(0,
1l (tk)
)− tak ⊆Uk. (37.0.9)
This can be done because tak+U∩Uk is a compact subset of U and so has positive distanceto UC and spt(ψk)− tak is a compact subset of Uk having positive distance to UC
k . Let tkbe such a value for t and for φ l a mollifier, define
vtk (x)≡∫Rn
ũ(x+ tkak−y)ψk (x+ tkak−y)φ l(tk) (y)dy (37.0.10)
where as usual, ũ is the zero extention of u off U. For vtk (x) ̸= 0, it is necessary that
x+ tkak−y ∈ spt(ψk) for some y ∈ B(
0, 1l(tk)
). Therefore, using 37.0.9, for vtk (x) ̸= 0, it
is necessary that
x ∈ y− tkak +U ∩ spt(ψk)⊆ B(
0,1
l (tk)
)+ spt(ψk)− tkak
⊆ B(
0,1
l (tk)
)+ spt(ψk)− tkak ⊆Uk
showing that vtk has compact support in Uk. Now change variables in 37.0.10 to obtain
vtk (x)≡∫Rn
ũ(y)ψk (y)φ l(tk) (x+ tkak−y)dy. (37.0.11)
For x ∈U ∩Uk, the above equals zero unless
y− tkak−x ∈ B(
0,1
l (tk)
)which implies by 37.0.9 that
y ∈ tkak +U ∩Uk +B(
0,1
l (tk)
)⊆U