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AN INTRODUCTION TO SOBOLEV SPACES

© Steve Taylor, Montana State University.

Preface: These notes were written to supplement the graduate level PDE course at Montana State University. Sobolev Spaces have become an indispensable tool in the theory of partial differential equations and all graduate-level courses on PDE's ought to devote some time to the study of the more important properties of these spaces. The object of these notes is to give a self-contained and brief treatment of the important properties of Sobolev spaces. The main aim is to give clear proofs of all of the main results without writing an entire book on the subject! Why did I write these notes? Much of the existing literature on the subject seems to fall into two categories, either long treatises on the subject with the most general assumptions possible (and thus unsuitable for part of a PDE course), or very sketchy discussions confined to a chapter of a PDE text.

CONTENTS:

1. The Spaces Wj , p(Ω) and W0j , p(Ω)... 1

2. Extension Theorems... 10

3. Sobolev Inequalities and Imbedding Theorems... 13

4. Compactness Theorems... 20

5. Interpolation Results... 25

6. The Spaces Hk(Ω) and H0k(Ω)... 27

7. Trace Theorems... 29

Appendix: Some Spaces of Continuous Functions... 34

References... 35 In these notes, Ω is a domain (i.e. an open, connected set) in Rn.

1. The Spaces Wj , p(Ω) and W0j , p(Ω) Definitions: Suppose 1p< ∞. Then

(i) Llocp (Ω)={u: uLp(K) for every compact subset K of Ω} (ii) u is locally integrable in Ω if uL1loc(Ω) .

(iii) Let u and v be locally integrable functions defined in . We say that v is the αth weak derivative of u if for every φ ∈C0(Ω)

(2)

uDαφ dx

=(−1)|α|

### ∫

dx,

and we say that Dαu=v in the weak sense.

(iv) Let u and v be in Llocp (Ω) . We say that v is the αth strong derivative of u if for each compact subset K of Ω there exists a sequence {φj} in C|α|(K) such that φju in Lp(K) and Dαφjv in Lp(K).

THEOREM 1 If Dαu=v and Dβv =w in the weak sense then Dα+ βu=w in the weak sense.

PROOF Let ψ ∈C0

(Ω) and φ = Dβψ . Then

uDα +βψ dx

=(−1)|α|

φv dx=(−1)|

### ∫

vDβψ dx =(−1)|+|β|

### ∫

ψw dx.

Definition (mollifiers): Let ρ ∈C0(Rn) be such that

(i) Supp ρ ⊂B1(0) , (recall that "supp" denotes the support of a function, and Br(p) denotes an open ball of radius r and center p).

(ii)

### ∫

ρ(x) dx=1, (iii) ρ(x)≥0 .

If ε > 0 then we set (provided that the integral exists) Jεu(x )= 1

εn ρ(xy

ε )u(y) dy

### ∫

.

Jεu is called a mollifier of u. Note that if u is locally integrable in Ω and if K is a compact subset of Ω then Jεu∈C(K) provided that ε < dist(K,∂Ω). Suppose now that

uLlocp (Ω) . Clearly

Jεu(x )= ρ(y)u(x − εy) dy

B1( 0)

### ∫

,

so for p>1 we have (if 1 / p+1 / q=1)

| Jεu(x)|≤ {ρ(y)}1/ q{ρ(y)}1/ p|u(x− εy)| dy

B1( 0)

### ∫

(3)

≤( ({ρ(y)}1/ q)qdx)1/ q( ({ρ(y)}1/ p| u(x− εy)|)p dy)1/ p

B1( 0)

B1( 0)

### ∫

.

Hence | Jεu(x)|p≤ ρ(y)|u(x− εy)|p dy

B1( 0)

### ∫

, and this trivially holds if p=1 too. Integrating this, we see that

| Jεu(x)|p

K dx

B1( 0)ρ(y)

### ∫

K|u(x− εy)|p dx dy

≤ ρ(y) |u(x)|pdx dy

K0

B1( 0)

= |u(x)|p dx

K0

### ∫

,

where K0 is a compact subset of Ω, KInterior(K0) and ε <dist(K,∂K0) . i.e. we have

|| Jεu||Lp( K)≤||u||Lp( K

0). (1)

LEMMA 2 If uLlocp (Ω) and K is a compact subset of Ω then || Jεuu||Lp( K )0 as ε →0.

PROOF Let K0 be a compact subset of Ω where KInterior(K0) and let ε <dist(K,∂K0) . Let δ >0 and let w∈C(K0) be such that || uw||Lp( K0)< δ. Then applying (1) to uw , we obtain

|| JεuJεw||Lp( K )< δ. (2)

But Jεw(x)w(x)= ρ(y){w(x− εy)w(x)} dy

B1( 0 )

### ∫

, and this goes to zero uniformly on K

as ε →0. Hence, if ε is sufficiently small, we have

|| Jεww||Lp( K )< δ. (3)

Hence, by (2) and (3)

|| Jεuu||Lp( K )≤||w−u||Lp( K)+|| JεuJεw||Lp( K )+|| Jεww||Lp( K)<3δ. (4) Since δ is arbitrary, || Jεuu||Lp( K )→0 as ε →0. 

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The proof of the following theorem contains some other important approximating properties of mollifiers.

THEOREM 3 Suppose that u and v are in Llocp (Ω) . Then Dαu=v in the weak sense if and only if Dαu=v in the strong Lp sense.

PROOF Suppose that Dαu=v in the strong Lp sense. Let φ ∈C0(Ω) and let K = supp φ. Let ε >0 and take ψ ∈C|α|(K) so that ||ψ −u||Lp( K)< ε and

|| Dαψ −v||Lp( K )< ε. Then

| uDαφ dx

K (−1)|α|

K dx|≤|

KψDαφ dx(−1)|α|

### ∫

KφDαψ dx|

+ | (u− ψ)Dαφ dx|

K +|

### ∫

K(vDαψ)φ dx|

≤||u− ψ||Lp( K)|| Dαφ||Lq( K )+||v−Dαψ||Lp( K)||φ||Lq( K )

≤ ε(||Dαφ||Lq( K )+||φ||Lq( K )),

where q is the conjugate exponent of p (if p=1 then q= ∞ and if p>1 then 1 / p+1 / q=1). But ε is arbitrary, so the LHS must be zero. So Dαu=v in the weak sense.

Conversely, suppose that Dαu=v in the weak sense and let K be a compact subset of Ω. Then Jεu∈C(K) if ε < dist(K,∂Ω) and we have for all x in K

DαJεu(x)= ε−n Dxαρ(xy

ε )u(y) dy

### ∫

= εn(−1)|α| Dyαρ(xy

ε )u(y) dy

= ε−n ρ(xy

ε )v(y) dy

### ∫

= Jεv(x).

But by Lemma 2, || Jεuu||Lp( K )0 and || DαJεuv||Lp( K)=|| Jεvv||Lp( K)→0 as ε →0.

Thus Dαu=v in the strong sense.

Definitions (i) |u|j, p=( | Dαu(x)|p dx

|α|j

### ∑

)1/ p.

(ii) C ˆ j, p(Ω)={u∈Cj(Ω): | u|j, p < ∞} .

(iii) Hj , p(Ω) = completion of ˆ C j, p(Ω) with respect to the norm | |j, p .

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Hj , p(Ω) is called a Sobolev space. We will encounter other such spaces as well.

Recall that the completion of a normed linear space is a larger space in which all Cauchy sequences converge (i.e. it is a Banach space). It is constructed by first defining a space of equivalence classes of Cauchy sequences. Two Cauchy sequences {xm} , {ym} are said to be in the same equivalence class if lim

m→ ∞|| xmym||=0 . A member x of the old space is identified with the equivalence class of the sequence {x, x,x, . . .} of the new space and in this sense the new space contains the old space. Further, the old space is dense in its completion. Moreover, if a normed linear space X is dense in a Banach space Y, then Y is the completion of X.

Recall that for 1≤ p< ∞, Lp(Ω) is the completion of C0(Ω) with respect to the usual "p norm". This knowledge allows us to see what members of Hj , p(Ω) "look like".

Members of Lp(Ω) are equivalence classes of measurable functions with finite p norms, two functions being in the same equivalence class if they differ only on a set of measure zero.

Suppose that {um} is a Cauchy sequence in ˆ C j, p(Ω). Then for |α|≤ j , {Dαum} is a Cauchy sequence in Lp(Ω). Hence, there are members uα of Lp(Ω) such that Dαumuα in Lp(Ω). Hence, according to our definition of strong derivatives, u0 is in Lp(Ω) and uα is the α strong derivative of u0. Hence we see that

Hj , p(Ω)={uLp(Ω): u has strong Lp derivatives of order ≤ j in Lp(Ω) and there exists a sequence {um} in ˆ C j, p(Ω) such that DαumDαu in Lp(Ω)}.

Definition Wj , p(Ω)={uLp(Ω): the weak derivatives of order ≤ j of u are in Lp(Ω)}

Note that by Theorem 3, an equivalent definition of Wj , p(Ω) is obtained by writing "strong derivatives" instead of "weak derivatives". Because of this, we see easily that

Hj , p(Ω)⊂Wj , p(Ω) . In fact, Hj , p(Ω)=Wj, p(Ω). This is not obvious because for

members of Hj , p(Ω) we can find sequences of nice functions such that DαumDαu in the topology of Lp(Ω), while according to our definition of strong derivatives, such limits exist only in the topology of Llocp (Ω) for members of Wj , p(Ω) . Before proving that

Hj , p(Ω)=Wj, p(Ω), we need the concept of a partition of unity.

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LEMMA 4 Let ERn and let G be a collection of open sets U such that E⊂{∪U: U∈G} . Then there exists a family F of non-negative functions f ∈C0(Rn) such that 0f (x)1 and

(i) for each f ∈F, there exists U∈G such that supp fU ,

(ii) if KE is compact then supp fK is non-empty for only finitely many f ∈F ,

(iii)

f (x)

f∈F

### ∑

=1 for each xE (because of (ii), this sum is finite),

(iv) if G ={Ω1,Ω2, . ..} where each i is bounded and iE then the family F of such functions can be constructed so that F ={f1, f2, ..} and supp fj ⊂ Ωj.

The family of functions F is called a partition of unity subordinate to the cover G.

PROOF Suppose first that E is compact, so there exists a positive integer N such that E⊂ ∪i=1

N Ui, where each Ui ∈G. Pick compact sets EiUi such that E⊂ ∪i=1

N Ei. Let gi = JεiχEi , where εi is chosen to be so small that supp giUi. Then gi ∈C0

(Ui) and gi >0 on a neighborhood of Ei. Let g=

### ∑

i=1N gi, and let S= supp g⊂ ∪iN=1Ui. If ε <dist(E,∂S) then k = JεχS is zero on E and h=g+kC(Rn). Further, h>0 on Rn and h=g on E. Thus F ={fi: fi =gi / h} does the job.

If E is open, let

Ei =EB i(0)∩{x: dist(x,∂E)≥1 i}.

Thus Ei is compact and E= ∪i=1

N Ei. Let Gi be the collection of all open sets of the form U[Interior(Ei+1)−Ei2], where U∈G and E0 = E1= ∅. The members of Gi provide an open cover for the compact set EiInterior(Ei1), so they possess a partition of unity Fi with finitely many elements. We let

s(x)= g∈F g(x)

i

i=1

### ∑

and observe that only finitely many terms are represented and that s>0 on E. Now we let F be the collection of all functions of the form

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f (x)= g(x)

s(x), xE 0, xE

 

 This F does the job.

If E is not open, note that any partition of unity for ∪U is a partition of unity for E.

For the proof of (iv), let H be the partition of unity obtained above and let fi = sum of functions h in H such that supp h⊂ Ωi, but supp h⊄ Ωj, j<i . Note that each h is represented in one and only one of these sums and that the sums are finite since each Ω i is a compact subset of E. Thus the functions fi provide the required partition of unity.  THEOREM 5 (Meyers and Serrin, 1964) Hj , p(Ω)=Wj, p(Ω).

PROOF We already know that Hj , p(Ω)⊂Wj , p(Ω) . The opposite inclusion follows if we can show that for every u∈Wj ,p and for every ε >0 we can find wC ˆ j, p such that for |α|≤ j , || DαwDαu||Lp(Ω)< ε.

For m ≥1 let

m={x∈Ω: || x||<m, dist(x,∂Ω)> 1 m}

and let Ω0= Ω−1 = ∅. Let {ψm} be the partition of unity of part (iv), Theorem 4, subordinate to the cover {Ωm+2− Ω m} . Each uψm is j times weakly differentiable and has support in Ωm+2 − Ω m. As in the "conversely" part of the proof of Theorem 3, we can pick

εm >0 so small that wm =Jεm(uψm) has support in Ωm+3− Ω m1 and |wmm|j, p< ε 2m. Let w = Σm=1

wm. This is a C function because on each set Ωm+2− Ω m we have w =wm2+wm1+wm+wm+1+wm+2. Further,

|| DαwDαu||Lp(Ω)=||Σm=1

Dα(wmm)||Lp(Ω)

≤ Σm=1

|| Dα(wmm)||Lp(Ω)

≤ Σm=1

ε / 2m = ε.

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Remarks

(i) The proof shows that in fact C(Ω)∩C ˆ j, p(Ω) is dense in Wj , p(Ω) .

(ii) Clearly members of C(Ω)∩C ˆ j, p(Ω) are not necessarily continuous on ∂Ω or even bounded near ∂Ω. It would be very useful to have the knowledge that C(Ω )∩C ˆ j , p(Ω) or Cj(Ω )∩C ˆ j, p(Ω) is also dense in Wj , p(Ω) . But the following example shows that this cannot always be expected.

Problem 1 Let Ω ={(x, y) : 1<x2+y2 <2, y0 if x>0}, i.e. an annulus minus the positive x-axis. Let w(x, y)= θ, the angular polar coordinate of (x,y). Clearly w is in W1,1(Ω) because it is a bounded continuously differentiable function. Show that we cannot find a φ ∈C1(Ω ) such that |u− ϕ|1,1<2π. (Note that Ω is the whole annulus).

The reason for the failure of the domain in Problem 1 is that the domain is on each side of part of its boundary. The following definition expresses the idea of a domain lying on only one side of its boundary.

Definition A domain Ω has the segment property if for each x∈∂Ω there exists an open ball U centered at x and a vector y such that if z∈Ω ∩U then z+ty∈Ω for

0<t<1.

We will not need the following theorem, so we don't prove it. For a proof, see Adam's book. However, see Lemma 9 for the simpler version of the result that we will need.

THEOREM 6 If has the segment property then the set of restrictions to of functions in C0(Rn) is dense in Wm, p(Ω) .

THEOREM 7 Change of Variables and the Chain Rule. Let V, Ω be domains in Rn and let T: V → Ω be invertible. Suppose that T and T−1 have continuous, bounded derivatives of order j . Then if u∈Wj ,p(Ω) we have v =uoT∈Wj , p(V) and the derivatives of v are given by the chain rule.

PROOF Let y denote coordinates in Ω and let x denote coordinates in V ( y=T(x) ). If fLp(Ω) then foTLp(V ) because

| foT|p dx

V =

| f |pJ dyconst.

### ∫

| f |p dy (5)

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(Here J is the Jacobian of T1).

If u∈Wj ,p(Ω), let {um} be a sequence in ˆ C j, p(Ω) converging to u in Wj , p(Ω) and set vm=um oT . By the chain rule, if |α|≤ j

Dxαvm =

### ∑

β ≤ α(Dyβum)oT Rα,β

where the Rα,β are bounded terms involving T and its derivatives. But for |β|≤ j DyβuLp(Ω) ⇒(Dyβu)oT ∈Lp(V)(Dyβu)oTRαLp(V) since the Rα are bounded.

Further,

|| Dxαvm − Σβ ≤α(Dyβu)oTRα,β||Lp( V )=||Σβ ≤α(DyβumDyβu)oTRα,β||Lp( V )

≤ Σβ ≤α||(DyβumDyβu)oTRα,β||Lp( V )

const.Σβ ≤α||(DyβumDyβu)oT||Lp( V )

const.Σβ ≤α|| DyβumDyβu||Lp(Ω)

by (5). So (α =0 case), vmv=uoT in Lp(V ) and

Dxαvm→ Σβ ≤α(Dyβu)oT Rα,β in Lp(V ). This shows that v∈Wj, p(V) and

Dxαv = Σβ ≤ α(Dyβu)oT Rα.  Definition W0j , p(Ω) =completion of C0

(Ω) with respect to the norm | |j, p .

Remarks (i) Clearly W0j , p(Ω)⊂Wj , p(Ω) because C0

(Ω)⊂C ˆ j , p(Ω).

(ii) Saying that f ∈W0

j, p(Ω) is a generalized way of saying that f and its derivatives of order ≤ j−1 vanish on ∂Ω. e.g. W01, p(Ω)∩W2, p(Ω) is a useful space for studying solutions of the Dirichlet problem for second order elliptic PDE's.

(iii) C0j(Ω)⊂W0j, p(Ω) because if f ∈C0

j(Ω), we know that if ε is sufficiently small then Jε f ∈C0

(Ω) and Jε ff in | |j, p norm.

Problem 2 Show that Wj , p(Rn)=W0j, p(Rn) . Hint: Why is it enough to show that C ˆ j, p(Rn)⊂W0j , p(Rn) ?

Problem 3 Show that if Ω is a domain in Rn, f ∈W0

j, p(Ω) and if f is extended to be zero outside Ω then the new function is in Wj , p(Rn).

Problem 4 Show that if y∈C1[0,1] and y(0)=y(1)=0 then y∈W0

1, p(0,1) . Use this fact to show that for any fLp(0,1) there is a unique y∈W0

1, p(0,1)∩W2, p(0,1) such that y"y= f . Hint: Solve the problem first with f ∈C0

(0,1) and then take limits.

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2. Extension Theorems

Most of the important Sobolev inequalities and imbedding theorems that we will derive in the next section are most easily derived for the space W0j , p(Ω) which (see Problem 3) can be viewed as being a subspace of Wj , p(Rn). Direct derivations of these results for the spaces Wj , p(Ω) are tedious and difficult because of the boundary behavior of the functions (Adams uses the direct derivation approach in his book). In this section we investigate the existence of extension operators that allow us to extend functions in

Wj , p(Ω) to be functions in Wj , p(Rn). This will allow us to easily deduce the Sobolev

imbedding theorems for the spaces Wj , p(Ω) from the corresponding results for Wj , p(Rn).

LEMMA 8 Let uRn and fLp(Rn). Set fδ(x)= f (x +δu) . Then lim

δ →0 fδ = f in Lp(Rn).

PROOF Given ε >0, let φ ∈C0(Rn) be such that || f − φ||Lp< ε. Since φδ → φ uniformly on a sufficiently large ball containing the supports of all φδ (say, for δ ≤1), we can pick δ so small that ||φ − φδ||Lp< ε. Then

|| ffδ||Lp≤|| f − φ||Lp+||φ − φδ||Lp+||φδfδ||Lp<3ε.  LEMMA 9 Let R+n ={x ∈Rn: xn >0} . C(R +n)∩C ˆ j, p(R+n) is dense in Wj , p(R+n).

PROOF Suppose f is in Wj , p(R+n) let ε > 0 and pick φ ∈C(R+n)∩C ˆ j, p(R+n) so that

|| Dαφ − Dαf ||Lp( R+n)< ε for all |α|≤ j . We take the vector of Lemma 8 to be u=(0,0,0, . . ,1) and define functions ψαLp(Rn) as

ψα(x )= Dαφ(x) , xn>0 0 , xn≤0

 

Observe that for each δ >0, φδ ∈C(R +n)∩C ˆ j, p(R+n) . By Lemma 8, we can pick δ >0 so that, for all |α|≤ j , ||ψδα− ψα||Lp( Rn)< ε. But this implies that || DαφδDαφ||Lp( R

+n)< ε. Hence

|| DαφδDαf ||Lp( R+n)≤|| DαφδDαφ||Lp( R+n)+||Dαφ − Dαf ||Lp( R+n)<2ε. 

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LEMMA 10 There exists a linear mapping E0: Wj, p(R+n)→Wj, p(Rn) such that E0f = f in R+n and |E0f |j , pRnC| f |Rj, p+n, where C depends on only n and p.

PROOF If f ∈C(R +n), define

E0f (x)= f (x) , xn ≥0

ckf (x1, x2, . . , xn1,−kxn)

k=1

### ∑

j+1 , xn <0

 

where the constants ck are chosen so that E0f (x)∈Cj(Rn) , i.e.

(−k)mck

k=1 j+1

### ∑

=1, m=0,1,2, . . , j .

It is easy to check that there is a constant C depending on only n and p such that

|| DαE0f ||Lp( Rn)C|| Dαf ||Lp( R+n). (6) If now f ∈Wj, p(R+n) , take a sequence fm ∈C(R +n)∩C ˆ j, p(R+n) converging to f in

Wj , p(R+n) (we can do this by Lemma 9). Then fm is a Cauchy sequence and (6) implies

that E0fm is a Cauchy sequence in Wj , p(Rn). We denote the limit by E0f . Since

|| DαE0fm||Lp( Rn)C|| Dαfm||Lp( R+n), taking limits shows that f satisfies (6).

Definition A domain Ω is of class Cm if ∂Ω can be covered by bounded open sets Ωj

such that there are mappings ψj:Ω jB , where B is the unit ball centered at the origin and

(i) ψj(Ωj∩ Ω)=BR+n (ii) ψj(Ωj∩ ∂Ω)=B∩ ∂R+n (iii) ψj ∈Cm(Ω j) and ψj

1 ∈Cm(B ).

(Because of (iii), all derivatives of order ≤m of ψj and its inverse are bounded).

THEOREM 11 If is a bounded domain of class Cm then there exists a bounded linear extension operator E:Wm , p(Ω)→Wm , p(Rn).

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PROOF Since ∂Ω is compact (boundaries are always closed), we might as well assume that the number of sets Ωj covering ∂Ω is a finite number N. Let U= ∪Nj=1j and let d =dist(∂Ω,∂U ). Setting 0 ={x ∈Ω: dist(x,∂Ω)>d / 2} , we see that0,Ω1,Ω2, . .,ΩN cover Ω. These sets also cover Ω , which is compact, so by the first part of the proof of Lemma 4, there exists a finite partition of unity θ012, . . ,θN for Ω and supp θj ⊂ Ωj. Recall that the support of a function is the closure of the set on which that function is non-zero. Hence, supp θj is even bounded away from ∂Ωj.

Let f ∈Wm, p(Ω) . Then fθj ∈Wm, p(Ω ∩ Ωj), so by our chain rule theorem (Theorem 7)

wj =( fθj)oψj1Wm, p(R+nB). Clearly supp wj is bounded away from

∂B , so we can extend wj to be a member of Wm, p(R+n) by letting it be zero in R+nB . We can further extend wj to all of Rn by use of the extension operator E0 of Lemma 10. Let

w ˜ j =E0w . If ρ <1 is chosen so that supp θj oψj

−1B ρ(0) , then we observe from the way that E0 was constructed that supp ˜ w jB ρ(0). Consequently, supp

w ˜ j oψj ⊂ ψj(B ρ(0)) is bounded away from ∂Ωj. Further, again by Theorem 7, this function is in Wm, p(Ωj). We extend it to be in Wm, p(Rn) by defining it to be zero outside Ωj. If we call the extended function gj, it is clear from our construction that gj = j on Ω ∩ Ωj and that

|gj|m, pRnC| f |m , p , where C is independent of f. Finally, we let g0 denote the function obtained by extending fθ0 to be zero outside Ω and define Ef = ΣjN=0gj. 

Remarks The theorem can be improved in a number of ways:

(i) We can allow Ω to be unbounded if ∂Ω is bounded (e.g. Ω is the exterior of a bounded domain).

(ii) We can allow Ω to be of class Cm−1,1 instead of Cm (i.e. the derivatives of order m−1 of the functions ψj are Lipschitz continuous. The proof of this requires a better version of Theorem 7 which we don't have time to prove here. Note that for the case m =1, the boundary could have corners.

(iii) Calderón has proved an extension theorem for domains satisfying the cone property (see the definition below) and a few other minor assumptions. The proof is much too time-consuming for us and it relies on the Calderón-Zygmund inequality, which also has a very lengthy proof. (See [Ad] for this).

Definition A domain Ω is said to satisfy the cone property if there exist positive constants α, h such that for each x∈Ω there exists a right spherical cone Vx⊂ Ω with height h and opening α.

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3. Sobolev Inequalities and Imbedding Theorems

THEOREM 12 If Ω ⊂Rn satisfies the cone condition (with height h and opening α) and if p>1, mp>n then Wm, p(Ω)⊂CB(Ω) and there is a constant C depending on only α, h, n and p such that for all u∈Wm , p(Ω), sup|u|≤C|u|m , p.

Note: Ω does not have to be bounded as Friedman suggests in his Theorem 9.1!

PROOF Initially, suppose that u is in ˆ C m, p(Ω). Let g∈C(R) be such that g(t)=1 if t1 / 2 and g(t)=0 if t1. Let x∈Ω and let (r, θ) denote polar coordinates centered at x. Here, θ =(θ12, . . ,θn1) denotes the angular coordinates and we can describe the cone with vertex x in polar coordinates as Vx ={(r,θ) : 0≤rh,θ ∈A}. Clearly, we have

u(x)= − ∂

∂r{g(r / h)

0

h u(r,θ)} dr

= (−1)m

(m−1)! rm1m

∂rm{g(r / h)

0

### ∫

h u(r,θ)} dr ,

after m-1 integrations by parts. Next, we integrate with respect to the angular measure dSθ, noting that the left-hand-side becomes a constant times u(x).

u(x)=c rm−1m

∂rm{g(r / h)

0

h u(r,θ)} drdSθ

A

=c rmnm

∂rm{g(r / h)

0

### ∫

h u(r,θ)} rn1drdSθ

### ∫

A

=c rm−nm

∂rm{g(r / h)u(r,θ)} dV

Vx

### ∫

.

Applying Hölder's inequality to this, we obtain

|u(x)|const.||rm−n||Lq( Vx)||∂rmm{g(r / h)u(r,θ)}||Lp( V

x)

const.|| rmn||Lq( Vx)|u|m, p .

But rm−n is in Lq(Vx) if n−1+(mn)q> −1, which is the case because q= p−1p and mp>n . Thus, we obtain sup|u|C|u|m , p. To extend this result to arbitrary u∈Wm , p(Ω), take a sequence {uk} of functions in ˆ C m, p(Ω) converging to u in the | |m, p norm. Then sup|ujuk|≤C|ujuk|m , p, showing that the sequence is a Cauchy sequence in CB(Ω).

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Thus u is in CB(Ω) and taking the limit of sup|uj|≤C|uj|m , p shows that u satisfies the same

inequality. 

Problem 5. Modify the proof to show that the theorem also applies to the case p=1, m=n.

Problem 6. Show that a similar theorem holds for W0m, p(Ω) and the cone condition is not required. Note that here we can even conclude that W0m, p(Ω)⊂{uCB(Ω ) : u=0 on

∂Ω} .

COROLLARY 13 If Ω ⊂Rn satisfies the cone condition (with height h and opening α) and if p>1, (mk)p>n then Wm, p(Ω)⊂CBk(Ω) and there is a constant C depending on only α, h, n, k and p such that for all u∈Wm , p(Ω) sup

|α|k| Dαu|C|u|m , p.

PROOF Apply the previous theorem to the derivatives Dαu for |α|≤k .Problem 7 What can you conclude if p=1 and mk=n ? See Problem 5.

Problem 8 What is the corresponding theorem for W0m, p(Ω) ? See Problem 6.

THEOREM 14 If Ω ⊂Rn is any domain and p>n then W01, p(Ω)⊂C0,α(Ω ), where α =1−np and there exists a constant C depending on only p and n such that for all u∈W0

1, p(Ω)

|u(x)u(y)|

|| xy||αC || Diu||Lp() i=1

### ∑

n .

PROOF Let u∈C0(Ω) . We might as well assume that u∈C0(Rn) . Let d =|| x −y|| , Sx =Bd(x), Sy =Bd(y) and S =SxSy. Then

|u(x)u(y)| vol(S)= |u(x)u(y)| dz

### ∫

S

|u(x)u(z)|+|u(z)u(y)| dz

S

|u(x)u(z)| dz

Sx

+

### ∫

Sy|u(z)u(y)| dz

But if (r,θ) are the polar coordinates of z in a coordinate system centered at x, we get

|u(x)u(z)|≤ ∫0r |∂ρ∂u| dρ, which implies

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|u(x)u(z)| dz

Sx

0d

### ∫

0r|∂ρ∂u| dρrn1drdSθ

≤ |∂u

∂ρ| dρrn1drdSθ

0

d 0

d

= dn

n |∂u

∂ρ| dρ

0

d dSθ

= dn

n ρ1n|∂u

∂ρ|ρn1

0

d dSθ

= dn

n ρ1−n|∂u

∂ρ| dz

Sx

### ∫

dn

n ||ρ1n||Lq( Sx)||∂u

∂ρ||Lp( Sx)

where q= p

p−1. A simple calculation shows that

||ρ1n||Lq( Sx)=const. d

1−n p

and it is easy to see that

||∂u

∂ρ||Lp( Sx)const. || Diu||Lp() i=1

### ∑

n .

Also, vol(S)=const. dn and the integral over Sy can be estimated in a similar fashion.

Putting this together yields

|u(x)u(y)|Cd

1−n

p || Diu||Lp() i=1

### ∑

n

which is precisely the inequality that we wanted. Further, we know from Theorem 12 applied to Rn that sup|u|C|u|1, p. Combining this with the previous inequality shows that for u∈C0

(Ω) we have || u||C0,α(Ω )C|u|1, p. Thus, if we now let u∈W0

1, p(Ω) and take a sequence {um} of functions in C0(Ω) converging to u in | |1,p norm, it follows that {um} converges in C0,α(Ω ) . Thus u∈C0,α(Ω ) , and taking limits shows that u satisfies the

inequality in the statement of the theorem. 

References

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