Basic Properties of Sobolev Spaces
1.1.9 Domains of the Class C 0,1 and Domains Having the Cone PropertyProperty
Definition 1.We say that a bounded domainΩbelongs tothe class C0,1 if each pointx∈∂Ω has a neighborhoodU such that the setU ∩Ωis repre- sented by the inequality xn < f(x1, . . . , xn−1) in some Cartesian coordinate system with functionf satisfying a Lipschitz condition.
By Lemma1.1.8 any bounded domain starshaped with respect to a ball belongs to the classC0,1.
Definition 2.A domainΩpossessesthe cone property if each point ofΩ is the vertex of a cone contained in Ωalong with its closure, the cone being represented by the inequalitiesx21+· · ·+x2n−1 < b x2n,0 < xn < a in some Cartesian coordinate system,a, b= const.
Remark 1.It is easy to show that bounded domains of the classC0,1have the cone property. The example of a ball with deleted center shows that the converse assertion is not true.
Lemma 1.LetΩbe a bounded domain having the cone property. ThenΩ is the union of a finite number of domains starshaped with respect to a ball.
Since a domain having the cone property is a union of congruent cones and hence it is a union of domains starshaped with respect to balls of a fixed radius, then Lemma 1 follows immediately from the next lemma.
Lemma 2. If a bounded domain Ω is a union of an infinite number of domainsGαstarshaped with respect to ballsBα⊂Gαof a fixed radiusR >0, then for eachr < R there exists a finite number of domains Ωk(1≤k≤N) starshaped with respect to balls of radius r, contained in Ωk, and such that
kΩk =Ω.
Proof. LetG1 be a domain in the collection{Gα}. Consider the domain Ω1 =
βGβ, where the union is taken over all domains Gβ for which the distance between the centers of the ballsBβ andB1is≤R−r. Obviously, any of the balls Bβ contain the ballC1of radiusrconcentric withB1. Since anyGβ is starshaped with respect toC1, then Ω1 is starshaped with respect toC1.
We define G2 to be any of the domains Gα such that Gα ∩Ω1 = ∅. Repeating the preceding construction, we define a domainΩ2starshaped with
respect to the ballC2 of radius r with the center situated at a distanced >
R−rfrom the center of the ballC1. Analogously, we construct a domainΩ3 starshaped with respect to a ballC3of radiusrwith the center situated at a distance d > R−rfrom the centers of the ballsC1 andC2, and so on.
Clearly this process will stop after a finite number of steps since the centers of the balls C1, C2, . . ., are contained in a bounded domain and the distance between centers is more than R−r >0. The result follows.
Remark 2.Domains of the class C0,1 are sometimes called domains hav- ing the strong Lipschitz property, whereas Lipschitz domains are defined as follows.
Definition 3.A bounded domainΩ is called a Lipschitz domain if each point of its boundary has a neighborhoodU ⊂Rnsuch that a quasi-isometric transformation mapsU ∩Ωonto a cube.
Clearly, domains of the class C0,1 are Lipschitz domains. The following example shows that the converse is not true, i.e., a Lipschitz domain may not have a strong Lipschitz property. Moreover, the Lipschitz domain considered in the next example fails to have the cone property (cf. Remark 1).
Example.LetΩ⊂R2be the union of the rectanglesPk ={x:|x1−2−k|<
2−k−2,0 ≤ x2 < 2−k−2}, k = 1,2, . . ., and the square Q = {x: 0 < x1 <
1, −1 < x2 <0}. Obviously, Ω does not have the cone property. We shall show that Ωcan be mapped onto the squareQ by a quasi-isometric map.
We can easily check that the mapping T0 : x→ y = (y1, y2), being the identity onQand defined on Pk by
y1=
x1−2−k
1−2kx2
+ 2−k, y2=x2,
is quasi-isometric. The imageT0Ωis the union of the squareT0Qand the set {y : 0< y1 < 1,0 ≤ y2 < f(y1)} with f satisfying the Lipschitz condition with the constant 4 and 0≤f(y1)≤1/8 (cf. Fig.4).
Letηbe a piecewise linear function,η= 1 fory >0 andη= 0 fory <−1.
The Lipschitz transformationT1:y→z, defined by z1=y1, z2=y2−f(y1)η(y2),
mapsT0Ωonto the square{z: 0< z1<1,−1< z2<0}. The Jacobian ofT1
is greater than 1/2; thereforeT1 is a quasi-isometric mapping.
Thus,Ωis mapped ontoQby the quasi-isometric mappingT0T1. 1.1.10 Sobolev Integral Representation
Theorem 1.Let Ωbe a bounded domain starshaped with respect to a ballBδ, Bδ⊂Ω, and letu∈Llp(Ω). Then for almost allx∈Ω
1.1 The SpacesLp(Ω),Vp(Ω) andWp(Ω) 17
Fig. 4.
u(x) =δ−n
|β|<l
x δ
β
Bδ
ϕβ y
δ
u(y) dy
+
|α|=l
Ω
fα(x;r, θ)
rn−1 Dαu(y) dy, (1.1.8) wherer=|y−x|,θ= (y−x)r−1,ϕβ∈D(B1),fαare infinitely differentiable functions inx,r,θ such that
|fα| ≤c(D/δ)n−1,
where c is a constant that is independent ofΩandD is the diameter ofΩ.
Proof. It suffices to putδ= 1. Letϕ∈D(B1) and
B1
ϕ(z) dz= 1.
First we assume thatu∈C∞( ¯Ω). Ifx∈Ω, z∈B1, the line segment [z, x] is contained inΩand so the following Taylor’s formula applies:
u(x) =
|β|<l
Dβu(z)
β! (x−z)β+l 1
0
(1−t)l−1
|α|=l
1 α!Dαu
z+t(x−z)
(x−z)αdt.
Multiplying this equality byϕ(z) and integrating overz∈B1, we obtain
u(x) =
|β|<l
B1
Dβu(z)
β! (x−z)βϕ(z) dz +l
|α|=l
1 α!
1 0
dt
×
B1
(1−t)l−1Dαu
z+t(x−z)
(x−z)αϕ(z) dz. (1.1.9) Since
B1
Dβu(z)(x−z)βϕ(z) dz
= (−1)|β|
B1
u(z)Dβz
(x−z)βϕ(z)
dz, |β|< l,
the former of the last two sums in (1.1.9) is a polynomial of degree l−1 in Rn, which can be written in the form of the first term on the right-hand side of (1.1.8). We extend Dαuto be zero outside Ω for|α|=l and perform the change of variabley=z+t(x−z) in the last integral overB1in (1.1.9). Then
x−z= (1−t)−1(x−y), dz= (1−t)−ndy, and the general term of the second sum in (1.1.9) is
1 α!
Rn
dy 1
0
Dαu(y)(x−y)αϕ
y−tx 1−t
dt (1−t)n+1. The last expression can be written as
1 α!
Rn
Dαu(y)(x−y)αK(x, y) dy, with
K(x, y) = 1
0
ϕ
y−tx 1−t
dt
(1−t)n+1 =r−n ∞
r
ϕ(x+θ)n−1d, where r=|x−y| andθ= (y−x)/|x−y|. Here we have made the change of variables
y−tx
1−t =x+θ, that is,y−x= (1−t)θand r= (1−t).
Note that the functionRn\ {y} x→K(x, y) is inC∞(Rn\ {y}) for any fixedy∈Rn. Moreover,K(x, y) = 0 if the line segment [x, y] is not contained in the cone Vx=
z∈B1[x, z]. Thus, formula (1.1.8) holds where fα(x, r, θ) =(−1)ll
α! θα ∞
r
ϕ(x+θ)n−1d.
1.1 The SpacesLp(Ω),Vp(Ω) andWp(Ω) 19 To check the estimate|fα| ≤cDn−1, we observe that
fα(x, r, θ)≤ l α!
D r
ϕ(x+θ)n−1d
≤ l α!Dn−1
x+θ∈B1
ϕ(x+θ)d.
The last integral is dominated by 2 sup{|ϕ(z)| : z ∈ B1}, which gives the required estimate.
The representation (1.1.8) has been proved foru∈C∞( ¯Ω). Supposeu∈ Llp(Ω) andp∈[1,∞). By Theorem1.1.6/1, there is a sequence{ui}i≥1 such that
ui ∈C∞( ¯Ω), ui→u inLp(Ω,loc), ∇l(ui−u)Lp(Ω)→0.
Passing to the limit in (1.1.8) for ui and using the continuity of the integral operator with a weak singularity inLp(Ω), we arrive at (1.1.8) in the general
case. This completes the proof of the theorem.
ForΩ=Rn we obtain a simpler integral representation ofu∈D. Theorem 2. Ifu∈D, then
u(x) =(−1)ll ωn
|α|=l
Rn
θα
α!Dαu(y) dy
rn−l, (1.1.10) where ωn is the(n−1)-dimensional measure ofSn−1, i.e.,
ωn= 2πn/2
Γ(n/2), (1.1.11)
and as in Theorem1,r=|y−x|, θ= (y−x)r−1. Proof. For fixedx∈Rn andθ∈Sn−1, we have
u(x) = (−1)l (l−1)!
∞
0
rl−1 ∂l
∂rlu(x+rθ) dr.
Since
∂l
∂rlu(x+rθ) =
|α|=l
l!
α!θα(Dαu)(x+rθ), it follows that
u(x) = (−1)ll ∞
0
rl−1
|α|=l
θα
α!(Dαu)(x+rθ) dr.
Integration with respect toθ implies (1.1.10).