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a r X i v : 1 2 0 9 . 4 0 4 4 v 1 [ m a t h . A P ] 1 8 S e p 2 0 1 2

SOBOLEV SPACES AND ELLIPTIC THEORY ON UNBOUNDEDDOMAINS IN Rn

PHILLIP S. HARRINGTON AND ANDREW RAICH

Abstract. In this article, we develop the theory of weighted L 2 Sobolev spaces on un-bounded domains in Rn . As an application, we establish the elliptic theory for ellipticoperators and prove trace and extension results analogous to the bounded, unweighted case.

Contents

1. Preliminaries 22. Main Results 53. Facts for W m,p (,; X ), 1 < p < 94. Sobolev spaces on M 135. Weighted Besov spaces on and M 186. Interior estimates the proof of Theorem 2.13 277. Elliptic regularity at the boundary 328. Traces of L-harmonic functions 39Appendix A. Background on interpolation the real method 43References 46

In this article, we develop weighted L2

-Sobolev spaces and elliptic theory on unboundeddomains in Rn . For spaces of functions with suitable regularity and domains with regularboundaries, we show that traces exist and functions dened on the boundary extend in abounded manner. In the second part of the paper, we show that elliptic equations gain thefull number of derivatives up to the boundary and satisfying an elliptic equation is sufficientfor taking traces for functions as rough as L2.

With this article, we are laying the groundwork to develop the L2-theory for the and b equations on unbounded domains and their boundaries in Cn . Weighted L2 spaces areinstrumental tools in several complex variables, and the analysis cannot proceed withoutthem.

Unlike the bounded case, however, unweighted Sobolev spaces on unbounded domains fail

to have many critical features, such as the Rellich identity, and so we develop spaces withthese properties in mind. One method to solve b-problem involves using extension and traceoperators, so we need fractional Besov and Sobolev spaces. As a result of our several complex

2010 Mathematics Subject Classication. 46E35, 35J15, 35J25, 46B70.Key words and phrases. Sobolev space, Besov space, unbounded domain, elliptic regularity, weighted Sobolevspace, traces, harmonic functions.The rst author is partially supported by NSF grant DMS-1002332 and second author is partially supportedby NSF grant DMS-0855822.

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variables considerations (to be developed in later papers), both the types of weighted Sobolevspaces we study and the types of results we prove are quite different than what appears in theliterature. See, for example [ Kuf85]. The weights that authors typically study involve powersof the distance to the boundary [ MT06, Can03]. Although these weights are quite natural andreect the geometry of the boundary, they are not the only useful weights in several complex

variables (see, e.g., [Har09, Hor65, Sha85, Koh86, HR11, HRa, Rai10, Str10 ]). With respectto the literature on elliptic theory on unbounded domains, authors seem to be less concernedwith proving trace results for solutions to elliptic equations and more interested in solvability,typically for the Dirichlet problem (e.g., see [ BMT08] and the references contained within).Even when the author solves an elliptic equation with a nonzero boundary condition (e.g.,[Kim08]), the derivatives are the standard derivatives, and not the weighted derivatives thatwe consider. Consequently, we must build the theory from the most basic buiding blocks.

Our Sobolev space techniques mainly involve real interpolation, so they dene Besovspaces. The fractional Sobolev spaces and Besov spaces agree (see, for example, [ LM72] or[BL76]), and in the elliptic regularity section of the paper, we use the fractional Sobolevand Besov spaces interchangeably. In fact, even at the integer levels, the main results hold

for the interpolated spaces Bk;2,2

(,; X ) and Sobolev spaces W k, 2

(,; X ) (the former byinterpolation and the latter by direct proof).

1. Preliminaries

Let Rn be an open set and let : R be C . Dene the weighted L p-space

L p(,) = {f : C : |f | pe dV < }where dV is Lebesgue measure on Cn . Let b be the boundary of . We will always assumethat b is at least Lipschitz, so that integration by parts is always justied. For most resultswe will need additional boundary regularity, as indicated below.

1.1. Hypotheses on , , and . Let A Rn .Let A be the distance function from A, i.e., A(x) = inf yA |x y|. Let U A = {x

Rn : there exists a unique point y A such that A (x) = |y x|} . Dene A : U A A byA(x) = y. The following concepts were introduced in [Fed59].

Denition 1.1. If y A, then dene the reach of A at y by

Reach( A, y) = sup {r 0 : B(y, r ) U A}

and the reach of A to be

Reach( A) = inf {Reach( A, y) : y A}.

The majority of our results use a subset of the following hypotheses. Fix m N, m 2.HI. The domain has a C m boundary with positive reach. Moreover, there exists > 0

and a dening function so that on = {y : b (y) < }, C m ( ) < (i.e., isuniformly C m in the sense of [HRb]).

HII. There exists (0, 1) so that

lim| x |

x

| |2 + =

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where

| |2 =n

j =1

x k

2.

HIII. There exists (0, 1) so that

lim| x |x | |2

= HIV. There exists a constant C m > 0 so that

| k| C m (1 + | |)for 1 k m and x .

HV. Hypotheses (HII)-(HIV) can be extended to Rn .HVI. If denotes the outward unit normal to b, we have

inf r> 0

sup|x |>r,x b

| | 1

< 1.

(HII) and (HIII) have their origin in [ Gan, GH10] (who in turn adapt the ideas in [ KM94]).The family of examples par excellence of weight functions is

(x) = t|x|2

for any nonzero t R . Such functions always satisfy (HII)-(HV) ((HI) is examined in detailin [HRb]). It is possible to construct domains for which (HVI) fails for this choice of ,but observe that if satises (HVI) for (x) = t|x|2, then any isometry of Rn will map to another domain which also satises (HVI) (because the composition of |x|2 with anyisometry will equal |x|2 plus lower order terms).

1.2. Weighted Sobolev spaces. Set D j = x j and dene the weighted differential opera-tors

X j = x j x j = e x j e

, 1 j n

andX = ( X 1, . . . , X n ).

Denition 1.2. Let Y j = X j or D j , 1 j n. For a nonnegative k Z , let the weightedSobolev space

W k,p (,; Y ) = {f L p(,) : Y f L p(,) for | | k}where = ( 1, . . . , n ) is an n-tuple of nonnegative integers and Y = Y 11 Y nn . Thespace W k,p (,; Y ) has norm

f p

W k,p (, ;Y ) = | | k Y

f p

L p (,) .

Also, let

W k,p0 (,; Y ) = {g W k,p (,; Y ) :there exists C c () satisfying g W k,p (,;Y ) 0 as } .

In other words, W k,p0 (,; Y ) is the closure of C c () in the W k,p (,; Y )-norm.3

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Remark 1.3. Our analysis focuses on the weighted spaces W k,p (,; X ) and we prove resultson the spaces W k,p (,; D) only where necessary. The choice of which space to focus on isnot central to the theory. We could have written the arguments with the roles of the twospaces reversed.

1.3. Weighted Sobolev spaces on b . Let > 0 and set M = b. Recall that

= {x Rn : dist(x, M ) < }

For discussions involving M , we always assume (HI) and m 2. Therefore, by [HRb], thereexists > 0 and a dening function so that C m ( ) < and |d| = 1 on b. LetZ 1, . . . , Z n 1 T M be an orthonormal basis near a point x M and let Z n = be the unitoutward normal to . Moreover, Z n is also the unit normal to the level curves of (pointingin the direction in which increases). For 1 j n, set

T j = Z j Z j ().

We call a rst order differential operator T tangential if the rst order component of T istangential. For 1 j n 1, Z j is dened locally, and if U is a neighborhood on whichZ 1, . . . , Z n 1 form a basis of T (M U ), we denote Z j by Z U j to emphasize the dependenceon U . In analogy to X , we dene T = ( T 1, . . . , T n ),

tanT = ( T 1, . . . , T n 1) and tanZ = ( Z 1, . . . , Z n 1)

By (HI) , we can construct an open cover {U j } of where U j are of comparable surfacearea and admit local coordinates Z 1, . . . , Z n 1 with coefficients bounded uniformly in C m 1.Let j be a C m partition of unity subordinate to {U j } where j are uniformly bounded inthe C m norm. With j in hand, we set v j = v j , so v = j =1 v j . Observe that we have thefollowing equivalent norms on W k, 2(,; X ) and W k, 2(,; D), respectively:

v W k, 2

(,;X )

j =1 | | k T

U j v j L2

(,) and v W

k, 2(

,;D )

j =1 | | k Z

U j v j L2

(,)

where T U = Z U Z U () and T U = T U 1 T U | | (and similarly for Z

U ).

For the boundary Sobolev space, set

W k,p (M, ; T ) ={f L p(M, ) : T f L p(M, ), | | k and T j is tangential for 1 j k}.

1.4. Notation for differential operators. In the second part of the paper, we establish theelliptic theory for strongly elliptic operators Rn that satisfy (HI) -(HV) (and sometimes(HVI) as well). Much of our development follows the outline in [Fol95]. Let L be a second

order operator of the form

(1) L =n

j,k =1

X j a jk X k +n

j =1

b j X j + X j b j + b

where a jk and b j are functions on a neighborhood of that are bounded in the C 1 norm,and b j and b are bounded functions on a neighborhood of .

Note that the formal adjoint ( X )= ( 1)| | D .4

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The formal adjoint of L is the operator given by the formula

(Lv, u ) = ( v,Lu )so integration by parts yields that

L=n

j,k =1

X ka

jkX

j +

n

j =1

b j

X j + X

jb j

+ b.

We say that the operator L is strongly elliptic on if there exists a constant > 0 sothat

(2) Re n

j,k =1

a jk j k | |2.

Associated to L is a (nonunique) sesquilinear form D called a Dirichlet form given by

(3) D (v, u ) =n

j,k =1

(X j v, a jk X ku) +n

j =1

(v, b j X j u) +n

j =1

(X j v, b j u) + ( v,bu).

D is called a Dirichlet form for the operator L if D (v, u ) = ( v,Lu ) for all u, v C c ().

The Dirichlet form D given by (3) is called strongly elliptic on if (2) holds.

Denition 1.4. The Dirichlet form D on is called coercive over X if W 1,20 (,; X ) X W 1,2(,; X ), X is closed in W 1,2(,; X ), and there exist C > 0 and 0 such that(4) Re D (u, u ) C u 2W 1,2 (, ;X ) u

2L2 (,) for all u X .

D is called strictly coercive if we can take = 0.

If D is coercive, then D (v, u ) = D (v, u ) + (v, u) is strictly coercive.We can also consider the adjoint Dirichlet form

D (v, u ) = D (u, v) for all u, v W 1,2(,; X ).The form D is called self-adjoint if D = D .

2. Main Results

2.1. Sobolev space and trace theorems. Let > 0 and set M = b and

= {x Rn : dist(x, M ) < }.In Section 5.1, we will use interpolation to dene the Besov space Bs ; p,q. The followingtheorem is the analog of the Trace Theorem [AF03, Theorem 7.39]

Theorem 2.1. Let m 2 and assume that Rn satises (HI)-(HVI) . If 1 k m 1,then the following two conditions on a measurable function u on M are equivalent:

(a) There exists U W k, 2(,; X ) supported in so that u = Tr U ;(b) u B k

12 ;2,2(M, ; T ).

The proof of Theorem 2.1 is divided into two results, each of which is more general thanone direction of Theorem 2.1. In Section 5.2 we will show

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Lemma 2.2. Given the hypotheses of Theorem 2.1, if U W k, 2(,; X ), then Tr U B k

12 ;2,2(M, ; T ) and there exists a constant K independent of U so that

Tr U B k

12 ;2 , 2 (M, ;T )

K U W k, 2 ( ,;X ) .

Remark 2.3. The result also holds (by the same proof) if we replace with c.

The second half of the proof of Theorem 2.1 is proven in Section 5.3, as part of the generalresult:

Theorem 2.4. Let , be integers so that 0 + m 2. If u B +12 ;2,2(M, ; T ), then

u = Tr U

for some U W + +1 ,2(,; X ) supported in satisfying

Tr U = = Tr 1U

1

= 0

and U W + +1 , 2 (,;X ) C u B

for some C independent of U and u.

The trace and extension theorems above allow us to prove the following result concerningthe equality of the spaces with weighted and unweighted derivatives. We also prove thefollowing Rellich identity in Section 5.5.

Proposition 2.5. Let satisfy (HI)-(HVI). Then for 0 k m, W k, 2(,; X ) =W k, 2(,; D). Furthermore, if m 2 and 1 k m 1, then W k, 2(,; X ) embeds compactly in W k 1,2(,; X ).

The analog of Proposition 2.5 for M is contained in Corollary 4.6. It is easier in this casesince C c (M ) is dense in W ,2(M, ; ) where is either Z or T . Likewise, for W

1,20 (,; X ),

the result is easier (though not easy) and is contained Proposition 3.3 and its corollaries.A useful application of the trace and extension theorems is the construction of a simple

(k, 2)-extension operator for each k, 1 k m 1. Recall that a simple ( k, 2)-extensionoperator E : W k, 2(,; X ) W k, 2(Rn ,; X ) is one that satises Eu (x) = u(x) for a.e.x and there exists a constant C = C (k) so that Eu W k, 2 (Rn ,;X ) C u W k, 2 (, ;X ) . InSection 5.3 we will show:

Theorem 2.6. Let satisfy (HI)-(HVI). Then for 1 k m 1, there exists a simple (k, 2)-extension operator.

Our nal embedding result is proven in Section 5.5.

Theorem 2.7. Let M, , and satisfy the hypotheses of Theorem 2.4 If s > 1/ 2 and 1 q , then

B s ;2,q(,; X ) B s 1/ 2;2,q(M, ; T ).6

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2.2. Elliptic regularity solvability. The Sobolev space theory that we develop is pow-erful enough that it allows us to adapt the proofs in the bounded, unweighted setting in astraight forward manner and establish the following theorems, see [ Fol95, Chapter 7]. Inparticular, we can establish that strong ellipticity is equivalent to Gardings inequality andsolve the (X , D ) Boundary Value Problem (BVP): namely, for a closed subspace X satisfying

W 1,20 (,; X ) X W

1,2

(,; X ) and f L2

(), nd u X so that D (v, u) = ( v, f ) forall v X . The case X = W 1,20 (,; X ) is the classical Dirichlet problem, but we also wantto include the case X = W 1,2(,; X ).

Note that C c () X , so a solution of the (X , D ) BVP will satisfy (Lv, u) = ( v, f )for all v C c () and hence will be a distributional solution to Lu = f . Furthermore, therequirement that D (v, u ) = ( v, f ) for all v X leads to a free boundary condition, i.e.,integration by parts imposes a boundary condition on u.

Theorem 2.8 (Gardings inequality) . Let

D (v, u) =n

j,k =1

(X j v, a jk X ku) +n

j =1

(v, b j X j u) +n

j =1

(X j v, b j u) + ( v,bu)

be a strongly elliptic Dirichlet form on and suppose that a jk , b j , b j , b are bounded on .Then D is coercive over W 1,2(,; X ) (and hence over any X W 1,2(,; X ) that contains W 1,20 (,; X )).

The converse to Gardings inequality holds as well.

Theorem 2.9. If the Dirichlet form D is coercive over W 1,20 (,; X ) and a jk C (), then D is strongly elliptic.

We can prove existence and uniqueness of weak solutions for operators giving rise to strictlycoercive Dirichlet forms.

Theorem 2.10. Let X be a closed subspace of W 1,2(,; X ) that contains W 1,20

(,; X )and let D be a Dirichlet form that is strictly coercive over X . There is a bounded, injective operator A : L2(,) X that solves the (X , D ) BVP, that is, D (v,Af ) = ( v, f ) for all v X and f L2(,).

Even in the case D is not strictly coercive, we can still gain information regarding weaksolutions.

Theorem 2.11. Let X be a closed subspace of W 1,2(,; X ) that contains W 1,20 (,; X ).Let D be a Dirichlet form that is coercive over X . Dene

V = {u X : D (v, u) = 0 for all v X }and

W = {u X : D (u, v) = 0 for all v X }.Then dim V = dim W < . Moreover, if f L2(,), there exists u X so that D (v, u) =(v, f ) for all v X if and only if f is orthogonal to W in L2(,) in which case the solution is unique modulo V . In particular, if V = W = {0}, the solution always exists and is unique.

In the case that D is self-adjoint, we can prove that L2(,) has a basis of eigenvectors.We will see in Proposition 2.5 that W 1,2(,; X ) embeds compactly in L2(,). Thus,

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Theorem 2.12. Let X be a closed subspace of W 1,2(,; X ) that contains W 1,20 (,; X ).Suppose that D is a Dirichlet form that is coercive over X and satises D = D . There exists an orthonormal basis {u j } of L2(,) consisting of eigenfunctions for the (X , D ) BVP; that is, for each j , there exists u j X and a constant j R so that D (v, u j ) = j (v, u j ) for all v X . Moreover, j > for all j where is the constant in the coercive estimate ( 4),

lim j j = , and u j C

() for all j .2.3. Elliptic regularity estimates. We can prove elliptic regularity in the interior of in Section 6.

Theorem 2.13. Let Rn satisfy (HI)-(HV) for some m 2. Let L be dened by ( 1) and a jk , b j C +1 () W +1 , () and b j , b C () W , () for some 0 m. Assume that f W ,2(,; X ) and L is strongly elliptic. Suppose that u W 1,2(,; X ) is a weak solution (i.e., D (v, u) = ( v, f ) for all v W 1,20 (,; X )) of the elliptic PDE

Lu = f in .

Then u W +2 ,2loc (,; X ) and if V is open and satises dist( V, b) > 0,

(5) u W +2 , 2 (V, ;X ) C ( f W (,;X ) + u L2 (,))where C = C (dist( V, b), C | | , a jk C +1 () , b j C () , b j C +1 () , b C () ,n , ,).

Note that the inequality in Theorem 2.13 is not an a priori inequality. The meaning of (5) is that if the right-hand side is nite, then u W +2 ,2(V, ; X ).

The case that is bounded is not the only case for which we know the hypothesis thatu W 1,2(,; X ) is satised. Indeed, we if combine Theorem 2.10 and Theorem 2.11 withTheorem 2.13 for = 0, we have the following corollary.

Corollary 2.14. Let L, V , and be as in Theorem 2.13 . Let X be a closed subspace of W 1,2(,; X ) that contains W 1,20 (,; X ). Let D be the Dirichlet form corresponding to Lin X . If any of the following conditions hold:

(i) D is strictly coercive,(ii) f W = {w X : D (w, v) = 0 for all v X } and u is the weak solution that is

orthogonal to V ,then u X and hence in W 2,2(V, ; X ).

We can also prove that elliptic regularity holds near the boundary for weak solutions of the partial differential equation Lu = f , in Sections 7.2 and 7.3.

Theorem 2.15. Let Rn satisfy (HI)-(HVI) with m 3. Let 0 m 3 and the operator L be dened by ( 1) where a jk , b j C +1 () W +1 , () and b j , b W , ().Assume that f W ,2(,; X ) and L is strongly elliptic. Let X be a closed subspace of

W 1,2(,; X ) that contains W 1,20 (,; X ). Suppose that u X is a weak solution (i.e.,

D (v, u ) = ( v, f ) for all v X ) of the elliptic PDE Lu = f in .

Then u W +2 ,2(,; X ) and (6) u W +2 ,2 (,;X ) C ( f W (,;X ) + u L2 (,))where C = C (M , a jk C +1 () , b j C +1 () , b j C +1 () , b C +1 () ,n,, C +3 () ).

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2.4. Boundary values of L-harmonic functions. We conclude the paper with a study of the boundary values of L-harmonic functions. The goal is to show that L-harmonic functions(i.e., functions u satisfying Lu = 0) have unique boundary values in W s 1/ 2,2(b,; T ) whenu W s, 2(,; X ) and s 0.

We rst establish a simple but easily applicable uniqueness condition. Let L be a strongly

elliptic second order operator. We would like to understand conditions on L so that if Lu = 0and u|M = 0, then u = 0. Theorem 2.10 present one condition, and we will show the followingin Section 8.

Lemma 2.16. Let Rn be a domain that satises (HII). Let L be a strongly elliptic operator that has a Dirichlet form D so that for all u W 1,20 (,; X ) there exists a constant c satisfying

(7) Re D (u, u ) c X u L 2 (,) .

If Lu = 0 and u W 1,20 (,; X ), then u 0.

Remark 2.17. If the operator L is of the form L = n j,k =1

X j

a jk X k + b where b > 0, then Lsatises (7).

With this restriction on D , we can prove in Sections 8.1 and 8.2:

Theorem 2.18. Let R n be a domain that satises (HI)-(HVI) for m = 2. Let L be a strongly elliptic operator that has a Dirichlet form D which satises ( 7 ). The map sending

u (Lu, Tr u)

is an isomorphism from

W s, 2(,; X ) W s 2,2(,; X ) W s 1/ 2,2(b,; T )

for 1 s m 1.

With an additional restriction on L, we can prove that L-harmonic functions in L2()have boundary values if s 0 in Section 8.3.

Theorem 2.19. Let Rn be a domain that satises (HI)-(HVI) for m 3. Let L be of the form

(8) L =n

j =1

X j X j + b j X j + X

j b j + b,

If f W s, 2

(,; X ) for s 0 and Lf = 0, then Tr f is well-dened and an element of W s 1/ 2,2(b,; T ).

3. Facts for W m,p (,; X ), 1 < p <

3.1. The spaces W k,q (,; X ) and W k,p (,; X ). Let 1 p < and 1 p + 1q = 1. Fix

k N and let N (k) be the number of multiindices where | | k. As k is xed, we suppressthe argument of N . Let 1, . . . , N be an enumeration of such multiindices. For a vector g,

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we write g = ( g1, . . . , gN ) = ( g ) interchangeably. For functions g1, . . . , gN Lq(,), thereexists a bounded linear functional T g1 ,...,g n on W k,p (,; X ) dened by

T g1 ,...,g N f = N

j =1

X jfg j e dV.

We can show that every functional on W k,p (,; X ) arises in this way.

Proposition 3.1. For Rn , let 1 < p < and 1 p + 1q = 1.

(i) The dual space W k,q (,; X ) := W k,p0 (,; X ) is the set

W k,q (,; X ) = u D() : u =| | k

( 1)| |D g , g Lq(,) for all .

Moreover, the norm on W k,q (,; X ) is given by

u W k,q (,;X ) := sup {|u(f )| : f W k,p0 (,; X ), f W k,p (, ;X ) = 1}

= inf F | | k g

qL q (,)

where F is the set of N -tuples (g1, . . . , gN ) Lq(,)N representing the functional u.

(ii) The dual space W k,p (,; X ) consists of u D() for which there exists a vector g = ( g ) (Lq(,))N so that for all f W k,p (,; X ),

u(f ) =| | k

X f, g .

Moreover, the norm on W k,p (,; X )

u W k,p (,;X ) := sup {|u(f )| : f W k,p

(,; X ), f W k,p (, ;X ) = 1}.Proof. The proof is standard. See, for example, [AF03, Sections 3.9, 3.12, 3.13]

3.2. Approximation by W k,p (,; X ).

Proposition 3.2. Let 1 p < and assume that b satises (HI) for some m 1. Let 1 m be an integer. Then C c (Rn ) is dense in both W ,p(,; X ) and W ,p(,; D) in the sense that if > 0 and f W ,p(,; X ), then there exists C c (Rn ) so that

f W ,p (, ;X )

where C is independent of f , , and . Similarly, if > 0 and f W ,p(,; D), then there

exists C c (R

n

) so that f W ,p (,;D ) where C is independent of f , , and .

Proof. Let > 0 and R be a smooth, nonnegative cut-off function so that R 1 onB(0, R), R 0 off B(0, 2R), and |D R | C | | /R | | for | | 0. For R sufficiently large,it follows that

(1 R )f W ,p (,;X ) .10

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Let g = R f , and extend g to be zero outside of . It is enough to prove the result for g.Let be a C m domain satisfying

B(0, 2R) B(0, 3R) .

Since supp R B(0, 2R), g = g1 . Since is bounded, there exists C > 0 so that

h W ,p ( ,;X ) C h W ,p ( )for any h W ,p(). The function is constructed in the following manner. Extend g tog W ,p(Rn ) following the technique of [AF03, Theorem 5.22]. Since g is identically zero ina neighborhood of , this construction can be used to guarantee that g is identicallyzero on \ . Form by mollifying g in such a way that is also identically zero on \ .Since is constructed so that g W ,p ( ) < /C , we have

g W ,p (,;X ) g W ,p ( ,;X ) < .

3.3. Embeddings and compactness for p = 2 .

Proposition 3.3. Let Rn satisfy (HII). Then the embedding of W 1,20 (,; X ) L2(,) is compact.

Proof. We start by making a number of preliminary calculations. Note that the formaladjoint X j = x j . This means

(9) (X j + X j )f = x j

f

and

(10) [X j , X j ]f = 2

x2 j

f.

Note that for f C c (),

(11) [X j , X j ]f, f = X j X

j f, f X

j X j f, f = X

j f 2L 2 (,) X j f

2L 2 (,) .

Next, we see that for > 0, a small constant/large constant argument yields

(X j + X j )f 2L2 (,) 1 +

12

X j f 2L2 (,) + (1 + ) X

j f 2L2 (,) .

Now set(x) = |(x)|2 + (1 + )(x).

Consequently,

f, f

=n

j =1

(X j + X j )f 2L2 (,) (1 + )([X j , X

j ]f, f )

2 + + 12

n

j =1

X j f 2L 2 (,) .(12)

Since C c () is dense in W 1,20 (,; X ), this inequality holds for all f W

1,20 (,; X ).

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Let {f k} W 1,20 (,; X ) be a bounded sequence and set M = max k f k 2W 1, 2 (, ;X ) . SetI (R) = inf x\ B (0,R ) |(x)|. Since I (R) > 0 for R sufficiently large,

f k f j 2L2 (,) B (0,R ) |f k f j (x)|2e dV + \ B (0,R ) (x)I (R) |f k f j (x)|2e dV C ,R f k f j 2L 2 (B (0,R )) + C f k f j

2W 1, 2 (, ;X )

I (R)

C ,R f k f j 2L 2 (B (0,R )) + C M I (R)

.(13)

Fix an increasing sequence R j , so that R j satises M/I (R j ) 1/j . We may inductivelyconstruct a sequence of subsequences f kmj so that

(i) f km +1j is a subsequence of f kmj ,(ii) lim j f kmj = f km in L

2(B(0, Rm ) ,), and(iii) f km |B (0,R k ) = f k if m.

It is now easy to see from (13) that f k jj is a Cauchy sequence in L2

(,) and hence convergesin L2(,). Thus, W 1,20 (,; X ) embeds compactly in L2(,).

Corollary 3.4. The embedding L2(,) W 1,2(,; X ) is compact.

Corollary 3.5. Let Rn satisfy (HII). There exists a constant C > 0 so that f 2L 2 (,) C X f

2L2 (,)

for f W 1,20 (,; X ).

Proof. By (11) and ( 12),

f 2L 2 (,) = X f

2L2 (,)

n

j =1[X j , X

j ]f, f X f 2L2 (,) +

11 + f, f

C X f 2L2 (,) .

Proposition 3.6. Let Rn satisfy (HIII). Then the embedding of W 1,20 (,; D) L2(,) is compact.

Proof. The proof follows the lines of the proof of Proposition 3.3 with

(x) = |(x)|2 (1 + )(x).replacing ( x).

Corollary 3.7. Let Rn satisfy (HIII). Then the embedding L2(,) W 1,2(,; D)is compact.

Corollary 3.8. Let Rn satisfy (HIII). Then there exists a constant C > 0 so that X f 2L 2 (,) C f

2L2 (,) ,

for all f W 1,20 (,; D).12

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Remark 3.9. Proposition 3.2 and Corollaries 3.5 and 3.8 allow us to dene a number of equivalent ways to measure the W k, 20 (,; X ) norm (which we will use later on ) Let W k0 (,; X ) and

Y j = 12

(X j X j ) = x j

12

x j

= X j + 12

x j

and Y = ( Y 1 , . . . , Y n ).

Note that

Y j 2L 2 (,) =x j

12

x j

, x j

12

x j

=x j

2

L2 (,)+

14

x j

2

L 2 (,) Re

x j

, x j

and

Y j 2L 2 (,) = X j + 12

x j

, X j + 12

x j

= X j 2L2 (,) +

14

x j

2

L 2 (,) + Rex j ,

x j .

Thus,

Y j 2L 2 (,) = 12

x j

2

L 2 (,)+ X j

2L 2 (,) +

14

x j

2

L2 (,),

soY 2L 2 (,)

12

2L 2 (,) + X 2L2 (,) .

It then follows that

Y 2L 2 (,) 2L2 (,) X

2L2 (,) .

and consequently (HIV) shows that for any so that 1 m(14) W , 2 (, ;D ) 2W , 2 (,;X )

| |

Y 2L 2 (,)

where the constants in depend on , n, and . The reason that we introduced Y j is thatY j = e

12

x j e 12 , so

Y j 2L 2 (,) = e 12 x j e 12 2e dx = x j e 12 L2 () .4. Sobolev spaces on M

As above with Proposition 3.1, standard arguments yield

Proposition 4.1. Let 1 < p < and 1 p + 1q = 1. Fix a nonnegative integer k m and let

N = N (k) be as in Proposition 3.1. The dual space to W k,p (M, ; T ) consists of u D(M ) for which there exists a vector g = ( g ) (Lq(M, ))N so that for all f W k,p (M, ; T ),

u(f ) =| | k

T f, g

=| | k

( 1)| | f, (T )g

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where T is a tangential operator of order | | . Moreover, the norm on W k,q (M, ; T ) :=W k,p (M, ; T )

u W k,q (M, ;T ) :=sup {|u(f )| : f W k,p (M, ; T ), f W k,p (M, ;T ) = 1}

=inf F

| | k

g qL q (M, )

where F is the set of N -tuples (g1, . . . , gN ) Lq(M, )N representing the functional u.

4.1. Approximations and embeddings for W k,p (M, ; T ). When considering results onthe boundary, we will generally need both (HII) and (HIII). Adding these, it is helpful toobserve that we have

(15) lim| x |

x

| | = .

Conversely, (HIV) and ( 15) imply both (HII) and (HIII), since (HIV) with k = 2 implies

(16) || n | 2| nC 2(1 + | |).By classical results, we know that C mc (M ) is dense in W k, 2(M, ; T ).Let B = ( j ) be the matrix with bounded C m 1 coefficients so that

Z j =n

=1

j

x .

Since T j = Z j Z j, T j = ( BX ) j . Then T = n =1 X implies

T =n

=1

X x

.

Using the formula for T and (9), we observe that

T j + T j = n

=1

jx

+ jx

= (B) j n

=1

jx

.

If H is the Hessian of , then from ( 10) it follows that

[T j , T j ] =n

, =1

j j 2

x x + j

j x

X + j jx

X j 2 j

x x

=n

, =1 j j

2

x x j j x x

j 2

j x x

= B(H)B T jj n

, =1

j j x

x

+ j 2 j

x x .

The key to the proof of Proposition 3.3 was the construction of , an unbounded functionso that ( f, f ) could be written in terms of inner products involving X f L2 (,) and

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([X j , X j ]f, f ). Adapting the heuristic of Proposition 3.3, we compute

n 1

j =1

T j + T j f 2L2 (M, ) (1 + ) [T j , T

j ]f, f M,

=n 1

j =1

B j f n

=1

j x

f 2L 2 (M, )

+ (1 + ) B(H)B T jj f +n

, =1

j j x

x

+ j 2 j

x x f, f

M,.

Therefore, the analog of in Proposition 3.3 is

(17) M (x) =n 1

j =1

B j +n

=1

jx

2

+ (1 + ) [Tr B(H)B T +n

, =1

j j x x

+ j 2 j

x x .

The matrix B plays a critical role here. We observe that

B =

tanZ

/

where = Z n is the unit outward pointing normal. Now, (HVI) and ( 15) tell us

lim| x |

xM

| tanZ | = ,

as well. Using (HIV) and (HVI) to bound H, we have

M (x) | tanZ |2 O(| tanZ | + 1) .

Hence, we have the following analogue of (HII):BI. There exists > 0 so that M dened by (17) satises

lim| x |

xM

M (x) = .

Proposition 4.2. Let R n satisfy (HI)-(HV) and b satisfy (BI). Then the embedding W 1,2(M, ; T ) L2(M, ) is compact.

Proof. The proof follows the argument of Proposition 3.3.

As earlier, we have the following corollary.

Corollary 4.3. Under the assumptions and notation of Proposition 4.2,

tanT f

2L2 (M, ) C

tanT f

2L 2 (M, )

for some constant C independent of f .15

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A similar argument shows the following Rellich identity. Set

(18) M =n 1

j =1

B n

=1

jx

2

(1 + ) Tr B(H)B T +n

, =1 j j

x x + j

2

j

x x .

As before, (HII)-(HIV) and (HVI) can be used to prove an analogue to (HIII):BII. There exists > 0 so that M satises

lim| x |

xM

M (x) = .

Proposition 4.4. Let Rn satisfy (HI)-(HV) and b satisfy (BII) . Then the embedding W 1,2(M, ; L) L2(M, ) is compact.

Corollary 4.5. Under the assumptions and notation of Proposition 3.3,

tanT f

2L 2 (M, ) C f

2W 1, 2 (M, ;L)

for some constant C independent of f .

Our nal comment on the consequences of (HIV) and (HVI) is the following:BIII. There exist constants C k so that

|( tanZ )k| C k(1 + | tanZ |)

for all x M and 1 k m.Since we have shown that (BI)-(BIII) follow from (HI)-(HVI), we will suppress the indi-

vidual boundary hypotheses and assume only (HVI) in the following.Corollary 4.6. Suppose that Rn satises (HI)-(HVI) . Then if 0 < k m,

f W k (M, ;T ) f W k (M, ;Z ) .

Proof. The proof goes by induction. The k = 1 case is the content of Corollary 4.3 andCorollary 4.5. The higher k follow from the k = 1 case, the inductive hypothesis and thefact that [ T j , T j ] is a function bounded by a multiple of (1 + | tan |).

Proposition 4.7. Let Rn satisfy (HI)-(HVI) . Then there exists K, K > 0 depending on n, m so that for any > 0, u W m, 2(M, ; T ), and 0 < j < m ,

| |= jT u 2L 2 (M, ) K

| |= mT u 2L2 (M, ) + j/ (m j ) u L2 (M, )(19)

u W j, 2 (M, ;T ) K u W m, 2 (M, ;T ) + j/ (m j ) u L 2 (M, ;T )(20)

u W j, 2 (M, ;T ) 2K u j/mW m, 2 (M, ;T ) u

(m j )/mL 2 (M, ;T )(21)

Proof. Note that ( 20) follows from repeated applications of ( 19). Equation ( 21) follows from(20) by choosing so that the two terms on the right-hand side are equal.

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We rst prove the result for m = 2, j = 1. In this case,

T j u 2L2 (M, ) = ( T j u, T j u) = ( T

j T j u, u )

T j T j u L 2 (M, ) u L2 (M, ) T j T j u 2L 2 (M, ) + 14

u 2L2 (M, ) .

However, since lim | x |xM | |2 + = , it follows by Corollary 4.3 that

T j T j u2L2 (M, ) C T j u

2W 1, 2 (M, ;T ) .

This proves the result for the case m = 2, j = 1. We can follow the argument of [AF03,Theorem 5.2] to nish proof.

4.2. Approximation. We can also prove a boundary version of the L2 analog to [AF03,Theorem 5.33], the Approximation Theorem for Rn .

Proposition 4.8. Let Rn satisfy (HI)-(HVI) . There exists a constant C = C (m, n ) sothat for 0 < k m, v W k, 2(M, ; T ), and 0 < 1, there exists v C m (M ) so that:

v v L 2 (M, ) C k| |= k

T u L 2 (M, )

and

v W j, 2 (M, ;T ) C v W k, 2 (M, ;T ) if j k 1

k j v W k, 2 (M, ;T ) if k j m.

Proposition 4.8 means that M has the approximation property.

Proof. In this proof, we work locally and use the boundary operators Y b j = Z j 12 Z j (). Itfollows from Corollary 4.6 that

ve12 W k, 2 (M ) =

| | k(Y b) v L 2 (M, ) v W k, 2 (M, ;T ) .

for 0 k m. Then

ve12 W k, 2 (M )

j =1

v j e12 W k, 2 (M )

where v j = U j v and { U j } is a partition of unity subordinate to {U j }. By the classicaltheory, there exists ,j C m +1c (M U j ) so that

v j e12 ,j e

12 L 2 (M, ) C k

| |= k

Z (v j e12 ) L2 (M, )

and

,j e12 W , 2 (M ) C

v j e12 W k, 2 (M U j ) if k 1

k j W k, 2 (M U j ,;X ) if k m.Since the M U j are of comparable surface area, the constant C arising from the classicalApproximation Theorem can be taken independent of j . Thus, the result follows by summingin j and observing that the decomposition v = j =1 v j is locally nite. 17

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5. Weighted Besov spaces on and M

We start with the following proposition. We initially prove a boundary version becausewe need to strengthen Proposition 3.2 before we can prove an analog for . This is an L2adaptation of the Approximation Theorem, [ AF03, Theorem 7.31].

Proposition 5.1. Let Rn

satisfy (HI)-(HVI) . If 0 < k < m , then W k, 2(M, ; T ) H k/m ; L2(M, ), W m, 2(M, ; T )

Proof. The argument is the same as [ AF03, Theorem 7.31] with Proposition 4.7 lling in for[AF03, Theorem 5.2] and Proposition 4.8 with W m, 2(M, ; T ) replacing the ApproximationTheorem in [AF03].

The importance of Proposition 4.8 is that the Reiteration Theorem (see Theorem A.10)holds for interpolation spaces generated from the weighted L2-Sobolev spaces.

5.1. Real interpolation of boundary Sobolev spaces. We are now ready to dene our

weighted Besov spaces.Denition 5.2. Let 0 < s < , 1 p < , 1 q and m be the smallest integer largerthan s. We dene the Besov space Bs ; p,q(M, ; T ) to be the intermediate spaces betweenL p(M, ) and W m,p (M, ; T ) corresponding to = s/m , i.e.,

B s ; p,q(M, ; T ) = L p(M ), W m,p (M, ; T ) s/m,q ;J .

We dene the Besov space Bs ; p,q(M, ; Z ) to be the intermediate spaces between L p(M, )and W m,p (M, ; Z ) corresponding to = s/m , i.e.,

B s ; p,q(M, ; Z ) = L p(M, ), W m,p (M, ; Z ) s/m,q ;J .

We will focus on the case p = 2 since we only proved an L2 Approximation Theorem. ByTheorem A.4, B s ;2,q(M, ; T ) is a Banach space with interpolation norm

u B s ;2 ,q (M, ;T ) = u; L2(M, ), W m, 2(M, ; T ) s/m,q ;J .

Also, B s ;2,q(M, ; T ) inherits density and approximation properties from W m, 2(M, ; T ). Forexample, { C (M ) : W m, 2 (M, ;T ) < } is dense in B s ;2,q(M, ; T ).

Let satisfy (HI) -(HVI) . Proposition 5.1 and the Reiteration Theorem imply that if 0 k < s < m and s = (1 )k + m, then

B s ;2,q(M, ; T ) = W k, 2(M, ; T ), W m, 2(M, ; T ) ,q ;J .

More generally, if 0 k < s < m and s = (1 )s1 + s2 and 1 q 1, q 2 , then(22) Bs ;2,q(M, ; T ) = B s1 ;2,q1 (M, ; T ), B s2 ;2,q2 (M, ; T ) ,q ;J .

The following corollary is an immediate consequence of Proposition 5.1 and Lemma A.8.

Corollary 5.3.

B m ;2,1(M, ; T ) W m, 2(M, ; T ) B m ;2, (M, ; T ).18

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5.2. Proof of Lemma 2.2 .

Proof of Lemma 2.2. We follow the outline of [AF03, Lemma 7.40]. We may apply theReiteration Theorem to obtain

B := B k12 ;2,2(M, ; T ) = W k 1,2(M, ; T ), W k, 2(M, ; T ) ,2;J

where = 1 12 = 12 . From Theorem A.5, we can apply the discrete version of the J -method

and obtain that u B if and only if there exist ui W k 1,2(M, ; T ) W k, 2(M, ; T ) =W k, 2(M, ; T ) for i Z so that

iZ

ui = u

in W k 1,2(M, ; T ) + W k, 2(M, ; T ) = W k 1,2(M, ; T ) and such that

2 i/ 2 u i W k 1, 2 (M, ;T ) , 2i/ 2 u i W k, 2 (M, ;T ) 2.

Let : M be the map that sends x to the unique point (x) M obtained byowing along Z n . That is, there exists t = tx such that x = etZ n ((x)). The constant > 0is small enough so that each point x can be uniquely represented by x = ( (x), tx ). Inthis way, if U C c () and x , then

U (x) = tx ddt U etZ n ((x)) dt = tx Z n U etZ n ((x)) dt.Let C c (R) be so that

(i) (t) = 1 on [ 1, 1],(ii) (t) = 0 if |t | 2,

(iii) 0 (t) 1 for all t R ,(iv) and there exists c j 0 so that |( j )(t)| c j for all j 1 and t R .

Dene i (t) = (t/ 2i) and i = i+1 i . Then i vanishes outside (2 i , 2i+2 )( 2i+2 , 2i)(and at the endpoints in particular). Also, i L (R) = 1 and i L (R) 2 ic1.

Let U C c (). Dene U i(x) by

U i(x) = U i((x), tx ) = e ( x )

2 tx i (t)Z n Ue 2 etZ n ( (x)) dt= e

( x )2 tx i(t) ddt Ue 2 etZ n ( (x)) dt.

Next, for x M , dene ui by ui(x) = U i(x). Thenui(x) = e

( x )2 0 i(t) ddt Ue 2 etZ n ( (x)) dt.

By the support condition on i and the Fundamental Theorem of Calculus,

(23) ui(x)e ( x )

2 = 2i

2i +2i(t)

ddt

Ue

2

etZ n ( (x))dt = 2

i

2i +2 i (t) Ue

2

etZ n ( (x))dt.

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Since U has compact support, U i and consequently ui vanish for all i Z when |x| issufficiently large. Therefore, the support of Tr U is a compact set on which iZ u i convergesuniformly to u = Tr U . Also, if | | k 1, then

Z tan u i(x)e ( x )

2 =

2i

2i +2

i(t)Z tanx

Z n Ue

2

etZ n

(x)

dt

Recall that

Z jx

=n

=1

j (x) x j

.

On , is bounded in the C k+1 norm, so j is bounded in the C k norm. Consequently, byCauchy-Schwarz,

Z tan u i(x)e ( x )2 (2i+2 )1/ 2C ( j ) C k ( ) 2

i

2i +2

| |+1 Ue

2etZ n ( x )

2dt

1/ 2

.

For a xed t > 0, t [2i , 2i+2 ) for exactly two (adjacent) i. Set Y j f = e2 Z j (fe 2 ) =12 (T j + Z j ). By Corollary 4.6, f W j, 2 (M, ;T ) | | j Y tan f M, . Moreover, the paths etZ nfoliate , so multiplying by 2 i/ 2, squaring, summing over i, and integrating yields

iZ

2 i ui 2W k 1, 2 (M, ;T ) C iZ

| | k 1

2 i Y tan ui 2L 2 (M, ) = C iZ

| | k 1

2 i Z tan u ie ( x )

2 2L2 (M )

= C | | k 1 | |+1 U (x)e ( x )2 2 dx C U W k, 2 (, ;X )

where the last inequality follows from ( 14).Using the second equality in ( 23) and Cauchy-Schwarz, we have

Z tan u i(x)e ( x )2 2 i2(i+2) / 2C 2

i

2i +2Z tan

xU etZ n (x) e

( e tZ n ( x ))2

2 dt1/ 2

2 i2(i+2) / 2C ( jk ) C k ( ) 2i

2i +2

| | Ue

2etZ n ( x )

2dt

1/ 2

.

Therefore,

iZ

2i u i 2W k, 2 (M, ;T ) C iZ

| | k

2i Y tan u i 2L 2 (M, ) = C iZ

2i Z tan u ie ( x )

2 2L2 (M, )

= C ( jk ) C k +1 ( )| | k | | U (x)e ( x )2 2 dx C U W k, 2 (,;X ) ,

where the nal inequality follows from ( 14).Together, these inequalities show that u B k 1/ 2;2 , 2 (M, ;T ) C U W m, 2 ( , ;X ) when U

C c (). Since C c () is dense in W k, 20 (,; X ), the proof is complete.

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5.3. Proof of Theorem 2.4 .

Proof of Theorem 2.4. Let = + + 1. Set B = B +12 ;2,2(M, ; T ). By denition,

B = L2(M, ), W ,2(M, ; T )

, where =

+ 12

.

From the discrete J -method (Theorem A.5), u L2

(M, ) belongs to B if and only if thereexist {u j } jZ W ,2(M, ; T ) so that u = jZ u j where the sum converges in L2(M, )and {2 j J (2 j ; u j )} 2. The latter condition means that there exists K > 0 so that

jZ

22+1

u j 2L2 (M, ) K 2 u 2B and

jZ

22 +1

u j 2W , 2 (M, ;T ) K 2 u 2B .

Let C c (R) be the bump function from the proof of Lemma 2.2. Set

j (t) = (t/ j )

for j Z and > 0 to be decided later. Set (t) = (2t/). It follows that |(k) j (t)| ck jk .Also, for k 1,

supp (k) j [ 2 j , j ] [ j , 2 j ].For y , there exists a unique x M and t [ ,] so that y = etZ n (x). Set (y) = x.

Since C m ( ) < , it follows that the projection C m 1 ( ) < . Set

U j (y) = 1!

e12 (y) (y) j (y)) (y)

u j (y) e12 ( (y)) .

Since = Z n , it is immediate that

Tr U j = Tr U j

= = Tr 1U j 1

= 0

and

Tr

U j = u jfor all j Z .

Thus, we only need to show that U W (,; X ) and is supported in . Sincesupp U j for all j , it follows that U is supported in . Note that if f is a smoothfunction on M , then there exist functions c 1 , 2 , 1 1, 2 n 1, that are bounded inC m 2() so that

Z 1 (f )(y) =n 1

2

c 1 , 2 Z 2 f (y)) .

Also, by construction, Z n (f )(y) = 0. Since U j has support in , (14) shows that we

may use the Y k operators (instead of the X ks) for differentiation. Let = ( 1, . . . , ) be amultiindex of length . Set T = { : is tangential } and N = { : = n} .

Setf j (t) = (t) j (t)t

and g j (x) = u j (x)e

12 (x)

for x M . Observe that|f ( 1 ) j (t)| c 1

j ( 1 ) .21

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The function does not affect the estimates derivatives of are supported where |s| [,2]and the support of derivatives of and derivatives of j cause j (or else the particularcombination j is identically zero). By ( 14) and the fact that U is supported in , it isenough to bound jZ Y

U j (y) L 2 ( ,) to show that U W ,2(,; X ).

Since is bounded in the C m norm and c 1 , 2 are bounded in the C m 2 norm, there exists

functions , that are bounded on so that

Y U j (y) e12 (y) = Z U j (y)e

12 (y) = Z f j (y) g j (y)

= N

=0 | | T

, f ( )

j (y) Z g j (y) .

Thus,

Y U j (y) L 2 ( ,) C N

=1 | | T

f ( ) j (y)2 Y u j (y)

2e ( (y)) dy

= C N

=1 | | T b 2j

2jf ( ) j (t)

2 Y u j (x)2e (x) dtd (x)

C N

=1 | | T b j (2 2 +1) Y u j (x) 2e (x) d(x) C

N

=1

j (2 2 +1) u j W T , 2 (b ,;T )

C j (2 +1) + j (2

2 N +1) u j W T , 2 (b ,;T ) .

where C is independent of j . Set = 2 1 . This means

Y U (y) L2 ( ,) jZ

Y U (y) L2 ( ,)

C jZ

j (2 +1) + j (2

2 N +1) u j W T , 2 (b ,;T ) .(24)

To check that the sum on the right hand side of ( 24) is nite, observe that

jZ

j (2 +1) u j 2W T , 2 (b ,;T ) =

jZ

2 j2 +1

u j 2W T , 2 (b ,;T )

jZ

2 j2 +1

u j 2W , 2 (b ,;T ) K u2B .

To bound the remaining term in ( 24), we use (20) to bound

j (2 2 N +1) u j 2W T , 2 (b ,;T ) K

j (2 2 N +1)

2 u j 2W , 2 (b ,;T ) + 2 T T u j 2L2 (b ,) .

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We require j (2 2 N +1) 2 = j (2 +1) . This means 2 = 2 N j . Since T + N = , it followsthat T T =

N N and

j (2 2 N +1) 2 N j

N N = j (2+1)

since + + 1 = . Thus, since = 2 1 ,

j (2 2 N +1) u j 2W T , 2 (b , ;T ) K jZ

2 2 1 u j W , 2 (b , ;T ) + 2 2 1 u j L2 (b ,) .

Thus, U W ,2(,; X ) and the proof is complete.

The proof of Theorem 2.1 is now complete.

Remark 5.4. If (for example) = 0, then the formula for U j is

U j (y) = e12 (y) (y) j (y) u j (y) e

12 ( (y))

Since Z n ((y)) = 0, we can compute

(Y n U j (y))e12 (y) = Z n U j (y)e

12 (y) = Z n (y)( j ) (y) u j (y) e

12 ( (y))

It follows from the support conditions on and that supp( j ) [( j , j ) ( ,)] = .Therefore

Tr( Y n U j ) = 0for all j . Similarly, Tr( Y kn U j ) = 0 for all k for which Y kn is dened. A similar result alsoholds if > 0 since Tr = 0.

Now that Theorem 2.4 is proven, we can apply our trace and extension theorems to provethat a simple ( k, 2)-extension operator exists.

Proof of Theorem 2.6. Let 1 k m 1 and f W k, 2(,; X ). We begin by constructingfunctions u1, . . . , u k recursively. By Theorem 2.1, Tr f B k

12 ;2,2(M, ; T ) and

Tr f B k 12 ;2 , 2 (M, ;T ) K f W k, 2 (, ;X ) .

Next, by Theorem 2.4, there exists a function u1 W k, 2(,; X ) supported in and sothat

Tr u1 = Tr f and u1 W k, 2 (,;X ) K Tr f B k 12 ;2 , 2 (M, ;T ) .Thus,

u1 W k, 2 (,;X ) K f W k, 2 (,;X ) .It now follows that ( f u1) W k, 2(,; X ) W 10 (,; X ).

Next, T n (f u1) W k 1,2(,; X ) so by Theorem 2.1, Tr( T n (f u1)) B k 3/ 2;2,2(M, ; T )and

Tr( T n (f u1)) B k 12 ;2 ,2 (M, ;T ) K T n (f u1) W k 1, 2 (,;X ) .By Theorem 2.4, there exists a function u2 W k, 2(,; X ) supported in and so thatTr u2 = 0 and Tr( u 2 ) = Tr( T n u2) = Tr( T n (f u1)) and

u2 W k, 2 (,;X ) K Tr( T n (f u1)) B k 32 ;2 , 2 (M, ;T ) .

Thus, ( f u1 u2) W k, 2(,; X ) W 20 (,; X ) andf u1 u2 W k, 2 (,;X ) K f W k, 2 (,;X ) .

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Iterating this process, we can show that there exist functionsu1, . . . , u k W k, 2(,; X )

supported in and so that ( f u1 u j ) W k, 2(,; X ) W j0 (,; X ) and

f u1 u j W k, 2 (, ;X ) K f W k, 2 (, ;X ) .

Thus, f (u1 + + uk) W k, 20 (,; X ).Next, we can write

f = ( u1 + + uk) + ( f u1 uk).The function ( f u1 uk) can be extended by 0 to produce a function in W k, 2(Rn ,; X )and (u1 + + uk) W k, 2(Rn ,; X ). Thus, we dene

Ef (x) =f (x) x u1(x) + + uk(x) x c.

5.4. Approximation of functions in W k, 2

(,; X ). Using the Trace Theorem 2.4, weare now in a position to improve Proposition 3.2 for p = 2 and relax the condition thatf W ,20 (,; X ).

Proposition 5.5. Assume that b satises (HI)-(HVI) for some m 2. Let 1 m 1be an integer. Then C c (Rn ) is dense W ,2(,; X ) in the sense that if > 0, then there exist C c (Rn ) so that

f W , 2 (, ;X ) and

W , 2 (Rn ,;X ) C f W , 2 (, ;X )where C is independent of f , , and .

The function that we construct will actually satisfy supp .Proof. The proof is a consequence of Theorem 2.6. Let > 0. Constructing the functionsu1, . . . , u as in the proof of Theorem 2.6. Then for any 1 j , (f u1 u j ) W ,2(,; X ) W j0 (,; X ) and

f u1 u j W , 2 (, ;X ) K f W , 2 (, ;X ) .

Thus, f (u1 + + u) W ,20 (,; X ) so there exists C c () so thatf (u1 + + u + ) W , 2 (, ;X ) < min{,f W , 2 (, ;X )}.

We can now take = u1 + + u j + .


have a C m

boundary and satisfy (HI)-(HVI). There exist K, K > 0 depending on n, m so that for each > 0, u W m, 2(,; X ), and 0 < j < m ,

| |= j

X u 2L 2 (,) K | |= m

X u 2L 2 (,) + j/ (m j ) u 2L2 (,)(25)

u W j, 2 (, ;X ) K u W m, 2 (,;X ) + j/ (m j ) u L2 (,)(26)

u W j, 2 (, ;X ) 2K u j/mW m, 2 (,;X ) u

(m j )/mL2 (,)(27)

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Proof. The proof of Proposition 5.6 is the same as the proof of Proposition 4.7 after usingTheorem 2.6 to reduce the problem to Rn .

We can also prove an L2 analog to [AF03, Theorem 5.33], the Approximation Theoremfor Rn .


satisfy (HI)-(HVI) for some m N. If 0 < k m then there exists a constant C = C (m, n ) so that for v W k, 2(,; X ) and 0 < 1, there exists v C c () so that:

v v L2 (,) Ck| |= k

X v L 2 (,)

and

v W j, 2 (, ;X ) C v W k, 2 (, ;X ) if j k 1

k j v W k, 2 (,;X ) if k j m.

Proof. The proof uses the same argument as the proof of Proposition 4.8.

Proposition 5.7 means that has the approximation property in the sense of [AF03]. Asa consequence of our improved approximation results, following the proof of Proposition 5.1,we can prove

Proposition 5.8. If Rn satises (HI)-(HVI), then

W k, 2(,; X ) H k/m ; L2(,), W m, 2(,; X ) .

5.5. Besov spaces on and Additional Trace Results. We are now ready to deneweighted Besov spaces on .

Denition 5.9. Let 0 < s < , 1 p < , 1 q and be the smallest integer largerthan s. We dene the Besov space Bs ; p,q(,; X ) to be the intermediate space between

L p

() and W ,p

(,; X ) corresponding to = s/ , i.e.,B s ; p,q(,; X ) = L p(,), W ,p(,; X ) s/,q ;J .

We will focus on the case p = 2 since we only proved an L2 Approximation Theorem. ByTheorem A.4, B s ; p,q(,; X ) is a Banach space with interpolation norm

u B s ;p,q (,;X ) = u; L p(,), W ,p(,; X ) s/,q ;J .

Also, Bs ; p,q(,; X ) inherits density and approximation properties from W ,p(,; X ). Forexample, { C () : W ,p (,;X ) < } is dense in B s ; p,q(,; X ).

For that satises (HI) -(HVI) , Proposition 5.7 and the Reiteration Theorem (TheoremA.10), if 0 k < s < and s = (1 )k + , then

B s ;2,q(,; X ) = W k, 2(,; X ), W ,2(,; X ) ,q ;J .

More generally, if 0 k < s < , s = (1 )s1 + s2, and 1 q 1, q 2 , then(28) Bs ;2,q(,; X ) = B s1 ;2,q1 (,; X ), B s 2 ;2,q2 (,; X ) ,q ;J .

We are now in a position to prove the following Trace Lemma.

Lemma 5.10. The trace operator Tr embeds B1/ 2;2,1( ,; X ) into L2(M ).25

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Proof. Let U be an element in B = B1/ 2;2,1( ,; X ). Without loss of generality, wemay assume that U B 1. By the discrete J -interpolation method, there exist functionsU i , i Z , so that U = iZ U i and

iZ

2 i/ 2 U i L 2 ( ,) C andiZ

2i/ 2 U i W 1,2 ( , ;X ) C

for some constant C . As in the proof of Lemma 2.2, we may assume that the functions U iare smooth and at most nitely many are not identically zero. For any of these functions,we have, for 2i t 2i+1 and x M ,

e ( x )

2 U i (x) t0 dds e ( e sZ n ( x ))2 U i esZ n (x) ds + e ( e tZ n ( x ))2 U i etZ n (x)= t0 Z n e ( e sZ n ( x ))2 U i esZ n (x) ds + e ( e tZ n ( x ))2 U i etZ n (x)

2i +1

0Z n e

( esZ n ( x ))2 U i esZ n (x) ds + e

( e tZ n ( x ))2 U i etZ n (x) .

Averaging t over [2i , 2i+1 ], we now have the estimate

e ( x )

2 U i(x) 2i +1

0Z n e

( e sZ n ( x ))2 U i esZ n (x) ds +

12i 2

i +1

2ie

( e tZ n ( x ))2 U i etZ n (x) dt.

By Cauchy-Schwarz,

e ( x )

2 U i(x) 2(i+1) / 2 2i +1

0Y n U i etZ n (x)

2e (etZ n (x)) dt

1/ 2

+ 2 i/ 2 2i +1

2i|U i etZ n (x) |2e (e

tZ n (x)) dt1/ 2

:= ai(x) + bi (x).

Observe that a i L 2 (M ) C 2i/ 2 U i W 1, 2 ( , ;X ) and bi L2 (M ) 2 i/ 2 U i L2 ( ,) . Sum-ming in i, we have

U L2 (M, ) iZ

e

2 U i L2 (M ) C iZ

2i/ 2 U i W 1, 2 ( ,;X ) + 2 i/ 2 U i L 2 ( ,) C.

As a consequence of Theorem 2.1 and Lemma 5.10, we can now prove our Trace Theoremfor L2 Besov spaces, Theorem 2.7. Theorem 2.7 is an L2-analog of [AF03, Theorem 7.43].

Proof of Theorem 2.7. The proof of Theorem 2.7 follows from (22), Theorem 2.1, Lemma5.10 and the Exact Interpolation Theorem.

We conclude our discussion of Sobolev space results with an extension of Proposition 3.3and Corollary 3.5, namely the proof of Proposition 2.5.

Proof of Proposition 2.5 . We will rst show that v W 1, 2 (,;D ) C v W 1, 2 (,;X ) for someC independent of v. By Theorem 2.1, Tr v B1/ 2;2,2(M, ; T ), and there exists v W 1,2(,; X ) with support in so that

Tr v = Tr v26

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Proof of Theorem 2.13 for = 0. We follow the outline of [Eva10, 6.3, Theorem 1]. ChooseW so that V W , dist( V, bW ) > 0, and dist( W, b) > 0. We rst assume that V andW are bounded. Let C () be a smooth cutoff so that V = 1 and supp W . SinceL is elliptic, by the classical theory u W 2,2loc (,; X ).

Since u is a weak solution of Lu = f , we have D (v, u ) = ( v, f ) for all v W 1,20 (,; X ).

Thus

(29)n

j,k =1 X j va jk X kue dx = vge dxwhere

(30) g = f n

j =1

b j X j u + b j X j u + ( X j b j )u bu.

We would like to use (29) substituting v = X ( 2X u). This is problematic as u W 2,2(W,; X ) and not thrice-differentiable. Instead, we let u C c () be so that u uin W 2,2(W,; X ). We set v = X ( 2X u). In this case, the left-hand side of ( 29) becomes

A =n

j,k =1

X j (X 2X u), a jk X ku ,

and the right-hand side becomes

B = vge dx = X 2X u, f n

j =1

b j X j u + b j X

j u b jx j

u bu

.

Equation ( 29) now says that A = B.

Observe thatA =

n

j,k =1

X X j 2X u, a jk X ku +

n

j,k =1

([X j , X ] 2X u, a jk X ku

=n

j,k =1

X j 2X u, a jk X X ku +n

j,k =1

([X j , X ] 2X u, a jk X ku (31)

+ X j 2X u, a jkx

X ku

.(32)

Since no more than two derivatives of u are taken in A, we can let 0 and observe thatA = B where

A =n

j,k =1

X j 2X u, a jk X X ku +n

j,k =1

([X j , X ] 2X u, a jk X ku + X j

2X u, a jkx

X ku

and

(33) B = X 2X u, f

n

j =1

b j X j u + b j X

j u b jx j

u bu

.

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We continue our investigation of A. Observe that

[X j , X ] =

x j x and [X , X k] = [X , X k]

= [X k , X

]= 0 .

We have

A =n

j,k =1

X j 2X u, a jk X kX u +n

j,k =1


+ X j 2X u, a jkx

X ku

+ X j 2X u, a jk [X , X k]

=0u

=n

j,k =1

2X j X u, a jk X kX u +n

j,k =1


+ X j 2X u, a jkx

X ku

+ 2 x j

X u, a jk X kX u .(34)

The strong ellipticity condition implies that

(35)n

j,k =1

2X j X u, a jk X kX u

X X u L 2 (W, ) .

The remaining terms we bound as follows:

X j 2X u, a jkx

X ku

+ 2 x j

X u, a jk X kX u

C 1 X X u L2 (W, ) X u L2 (W, ) ,

where C 1 depends on a jk C 1 () and C 1 () . In particular, C 1 does not depend on | supp | .Next, using (HIV) , Corollary 3.5 and the fact that x j = X

j X j , we have

([X j , X ] 2X u, a jk X ku C 2 (1 + | |) 2|X u | , |a jk || X ku | C 2 X u L2 (W, ) X u L2 (W, ) + X X u L 2 (W, ) ,

where C 2 depends on C 2 and a jk L () . Thus, using ( 35) and the bounds on the errorterms, we can bound (with C = n2(C 1 + C 2))

|A| X X u 2L2 (W, ) C ( X X u L2 (W, ) X u L 2 (W, ) + X u2L2 (W, )

2

X X u 2L2 (W, ) C 3 X u2L2 (W, ) ,(36)

where C 3 = C 3( a jk C 1 () , C 1 () ,n , ). We can bound B with Cauchy-Schwarz and thesmall constant/large constant inequality. In particular, we can use Corollary 3.5 to showthat for some constant C 4 > 0 where C 4 = C 4( b j L () , b j C 1 () , b L () , C 1 () , n), wehave the estimate

|B | C 4 X X u L2 (W, ) f L2 (W, ) + u W 1, 2 (W, ;X )

4

X X u 2L 2 (W, ) + C 5 f 2L2 (W, ) + u

2W 1, 2 (W, ;X ) ,(37)

where C 5 = C 5( b j L () , b j C 1 () , b L () ,n , ).29

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Combining ( 36) and (37), we have shown that

(38) u 2W 2,2 (V, ;X ) C 6 f 2L2 (W, ) + u

2W 1, 2 (W, ;X ) ,

where C 6 = C 6( a jk C 1 () , C 1 () , b j L () , b j C 1 () , b L () ,n , ). We can improve theestimate ( 38) and replace u 2W 1, 2 (W, ;X ) with u

2L2 (W, ) . Let C

c () be a cutoff so that

|W = 1. Using (29) and strong ellipticity, we estimate

X u 2L2 (,) 1

n

j,k =1

2X j u, a jk X ku = 1

2u, g

C 7 u L 2 (,) f L2 (,) + X u L2 (,) + u L2 (,) ,

where C 7 = C 7( b j L () , b j C 1 () , b L () ,n ,). Using a small constant/large constantargument, we have

X u L2 (W, ) X u L 2 (,) C 8 f L2 (,) + u L 2 (,) ,

where C 8 = C 8( b j L () , b j C 1 () , b L () ,n , ). Thus, we can rene ( 38) by

u 2W 2, 2 (V, ;X ) C f L2 (,) + u2L2 (,) ,

where C = C (dist( V, b), a jk C 1 () , b j L () , b j C 1 () , b L () ,n ,). and have disap-peared from C as C 1 () depends only on dist(V, b). Thus, we can relax the boundednesscondition on V and let V be as in the statement of the theorem.

6.2. 1 case.

Proof of Theorem 2.13, 1. As with the = 0 case, that u W +2 ,2loc (,; X ) follows fromthe classical theory. We will establish ( 5) by induction. The = 0 case has already beenestablished.

Let V W so that dist( V, bW ) > 0 and dist( W, b) > 0.Assume that ( 5) holds for a nonnegative integer , for a jk , b j C +2 () W +2 , (),

for b j , b C +1 () W +1 , (), and for f W +1 ,2(,; X ). Assume further that u W 1,2(,; X ) is a weak solution of Lu = f in . By the induction hypothesis, we have theestimate

u W +2 (W, ;X ) C f W (,;X ) + u L 2 (,) ,

where C = C (dist( W, b), a jk C +1 () , b j C () , b j C +1 () , b C () ,n , ,). Let be amultiindex of length | | = + 1. Let v C c (W ) and set

v = ( X )v and u = X u.Since D (v, u) = ( v, f ), we plug in v = ( X )v and compute

D (v, u ) =n

j,k =1

X j (X )v, a jk X ku +

n

j =1

(X )v, b j X j u + X j (X )v, b j u

+ (X )v,bu

.30

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Integrating by parts gives us

D (v, u) =n

j,k =1

X j v, a jk X kX u +n

j =1

v, b j X j X u + v, X

j (b j X

)u

+ v,bX u

+n

j,k =1

v, [X j , (X )]a jk X ku +

=

v, (D a jk ) X kX u

+n

j =1

v, [X j , (X )]b j u

+ =

v, (D b) X u +

n

j =1

v, (D b j ) X X j u + v, X j (D

b j ) X u

,

so

(39) D (v, u) = D (v, u) + v, g,

where

g =n

j,k =1

[X j , (X )]a jk X ku + =

(D a jk ) X kX u +n

j =1

[X j , (X )]b j u

+ =

(D b) X u +n

j =1

(D b j ) X X j u + X j (D b j ) X

u .

To compute [ X j , (X )], observe that

X j (X )= X j X 1 X

+1 = X

1 X j X

2 X

+1 + [X j , X

1 ]X

2 X

+1

= = ( X )X j + [X j , X 1 ]X

2 X

+1 + + X

1 X

[X j , X

+1 ]

= ( X )X j

x j x 1X 2 X

+1 X

1 X

x j x +1

= ( X )X j + =

c (X )D

x j

for some constant c Thus,

[X j , (X )]= =

c D x j

X

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and we can rewrite

g = =

c n

j,k =1

D x j

X a jk X ku +n

j =1

D x j

X b j u

+ =

D b X u +

n

j,k =1D a jk X kX u +

n

j =1D b j X X j u + X j D b j X u

= =

J

c n

j,k =1

D x j

D J a jk X J X ku +n

j =1

D x j

D J b j X J u

+ =

D b X u +n

j,k =1

D a jk X kX u +n

j =1

D b j X X j u + X j D b j X u .

Thus, if f = X f g, then we can express D (v, u ) = ( v, f ) = ( v, X f ) as D (v, u) =

(v, f ). Since v C c () is arbitrary, u is a weak solution for Lu = f . Since

f L2 (W, ) C f W +1 , 2 (,;X ) + g L2 (,)

it follows from the induction hypothesis that f L2(W ) with

f L2 (W, ) C f W +1 , 2 (,;X ) + u L2 (,)

As a consequence of the = 0 case of Theorem 2.13,

u W 2,2 (V, ;X ) C ( f L 2 (W, ) + u L2 (W, )) C ( f W +1 , 2 (, ;X ) + u L 2 (,)).

Since this inequality holds for any of length ( + 1), it follows that u W +3 ,2(V, ; X )

and u W +3 (W, ;X ) C f W +1 (,;X ) + u L2 (,) .

7. Elliptic regularity at the boundary

The standard technique to prove elliptic regularity at the boundary is to work locally,rotate, and atten the domain. Working on a weighted L2 space complicates the matter andwe instead work with tangential and normal derivatives.

7.1. Tangential Operators. Recall that a rst order differential operator T is a tangentialoperator if the rst order component of T annihilates . By the hypotheses on , thereexists > 0 so that on , there exist rst order differential operators T 1, . . . , T n so that

T j =n

=1

j X ,

where ( j ) is an orthogonal matrix, the components j are bounded in C m 1(), T j istangential for 1 j n 1, the rst order part of T n is the unit outward normal to the

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level curve of and

(40)n

j =1

|T j f |2 = | X f |2.

From ( 40), it is clear that if k m, then

(41) u W k, 2 (, ;X ) | | k

T u L 2 (,) + u W k, 2 (\ ,;X ) .

By assumption |d| = 1 on b so that if is the unit (outward) normal to {x Cn :(x) = }, then i0 =

(x)x i . As an immediate consequence of the Divergence Theorem,

(42)

f x

j

dx =

b

f x

j

d

where d is the surface area measure on b.

7.2. Proof of Theorem 2.15 , = 0 case. We are now ready to prove the regularity of solutions of Lu = f near b.

Proof of Theorem 2.15, = 0 case. Given Theorem 2.13, it is enough to show that u W 2,2(,; X ). Let V, W be smooth, bounded domains so that V W and dist( V, bW ) >0. Let C (Rn ) be a smooth cutoff so that |V = 1 and supp W . Since L is el-liptic and W is bounded, the classical theory yields u W 2,2( W,; X ). By (HI) , it isenough to work locally, i.e., we can assume that supp u is small enough that T 1, . . . , T n arewell-dened on supp u.

The function u is a weak solution of Lu = f , so we have D (v, u) = ( v, f ) for all v X .Thus u satises the free boundary condition for X and equations ( 29) and ( 30) hold. As inthe proof of Theorem 2.13, we would like to use (29) substituting v = X k ( 2X ku). This isproblematic as u W 2,2(W,; X ) and not thrice-differentiable. Instead, we use Proposition3.2 which constructs u C c (Rn ) so that u u in W 2,2(W ,; X ). Let 1 k n 1.Then set v = T k ( 2T ku). In this case, the left-hand side of ( 29) becomes

A :=

n

j,j =1X j T k ( 2T ku), a jj X j u ,

and the right-hand side becomes

B := vge dx = T k ( 2T ku), f n

j =1

b j X j u + b j X

j u b jx j

u bu

.

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Equation ( 29) now says that A = B. Since T k is tangential, T k is also tangential and wecompute

A =n

j,j =1

T k X j ( 2T ku), a jj X j u

+n

j,j =1

([X j , T k ]( 2T ku), a jj X j u

=n

j,j =1

X j ( 2T ku), a jj T kX j u

+n

j,j =1

[X j , T k ]( 2T ku), a jj X j u

+n

k =1

X j ( 2T ku), kk a jj x k

X j u

.(43)

Since no more than two derivatives of u are taken in A, we can let 0 and observe thatA = B where

A =n

j,j =1

X j ( 2T ku), a jj T kX j u

+n

j,j =1


+n

k =1

X j ( 2T ku), kk a jj x k

X j u

and

(44) B = T k ( 2T ku), f

n

j =1

b j X j u + b j X

j u b jx j

u bu

.

We continue our investigation of A. Observe that T k =nk =1 kk X

k kk

x k , so

[X j , T

k ] =

n

k =1 kk

2x j x k +

kk x j X k

2 kk x k x j

and

[T k , X j ] = n

k =1

kk x j

X k .

We have

A =n

j,j =1

X j ( 2T ku), a jj X j T ku

+n

j,j =1


+ X j ( 2

T ku), a jj x k X j

u + X j ( 2

T ku), a jj

[T k , X j

]u

=n

j,j =1

2X j T ku, a jj X j T ku +n

j,j =1


+ X j ( 2T ku), a jj x k

X j u

+ X j ( 2T ku), a jj [T k , X j ]u

+ 2 x j

T ku, a jj X j T ku

.

(45)

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The strong ellipticity condition implies that

(46)n

j,j =1

2X j T ku, a jj X j T ku X T ku2L2 (W ,) .

As in the proof of Theorem 2.13, the remaining terms of ( 45) are bounded by

C 2 X T ku L 2 (W ,) X u L2 (W ,) + X u 2L2 (W ,)where C 2 depends on a jj C 1 () , C 3 () and C 1 () . In particular, C does not dependon the size supp . Thus, using ( 46) and the bounds on the error terms, we can bound

|A| X T ku 2L2 (W ,) C 2( X T ku L 2 (W ,) X u L2 (W ,) + X u 2L2 (W ,)

2

X T ku 2L 2 (W ,) C 3 X u2L 2 (W ,)(47)

where C 3 = C 3( a jj C 1 () , C 1 () ,n,, C 3 () ). We can bound B with Cauchy-Schwarzand the small constant/large constant inequality. In particular, for a constant

C 4 = C 4( b j L () , b j C 1 () , b L () , n, C 2 () ),we have the estimate

|B | C 4 X T ku L 2 (W ,) f L2 (W ,) + u W 1, 2 (W ,;X )

4

X T ku 2L2 (W ,) + C 5 f 2L 2 (W ,) + u

2W 1, 2 (W , ;X )(48)

where C 5 = C 5( b j L () , b j C 1 () , b L () ,n,, C 2 () ). Combining ( 47) and (48), itfollows that(49) tanT u 2W 1, 2 (V ,;X ) C 6 f

2L 2 (W ,) + u

2W 1, 2 (W , ;X )

where C 6 = C 6( a jj C 1 () , C 1 () , b j L () , b j C 1 () , b L () ,n,, C 3 () ).We can improve the estimate ( 49) and replace u 2W 1, 2 (W, ;X ) with u

2L 2 (W, ) . Let

C c () be a cutoff so that |W = 1. Using (29) and strong ellipticity, we estimate

X u 2L2 (,) 1

n

j,j =1

2X j u, a jj X j u = 1

2u, g

C 7 u L 2 (,) f L2 (,) + X u L2 (,) + u L2 (,) ,where C 7 = C 7( b j L () , b j C 1 () , b L () ,n , ).

Using a small constant/large constant argument, we have

X u L 2 (W ,) X u L2 (,) C 8 f L 2 (,) + u L 2 (,)where C 8 = C 8( b j L () , b j C 1 () , b L () , n). Thus, we can rene ( 49) by

tanT u

2W 1, 2 (V,) C f L2 (,) + u

2L2 (,)

where C = C ( a jj C 1 () , b j L () , b j C 1 () , b L () ,n,, C 3 () ). and have disap-peared from C as the bound depended on C 1 () but that bound depends on dist( V, W ).Thus, we can relax the boundedness condition on V and let V be as in the statement of thetheorem.

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It remains to bound T 2n u L 2 (,) . To do this, we recall that T j =n j,j =1 jj X j where

( jj ) is an orthogonal matrix. Therefore, if ( jj ) 1 = ( jj ), then X j = nk =1

j k T k .Therefore,

n

j,j =1

(X j )a jj X j =n

j,j ,k,k =1

T k jk a jj j

k T k .

Let atankk =n j,j =1

jk a jj j k . Since ( jk ) is an orthogonal matrix and the smallest eigenvalue

of a jj is , it follows thatn

k,k =1

a tankk k k | |2.

Therefore, if = (0 ,...0, 1), then we see that atannn . Consequently, using the fact thatLu = f , we see

|T n T n u | C 9 | X tanT u| + | X u | +n

j =1

|b j X j u | + |X j (b j u)| + |bu| + |f |

where C 9 = C 9( C 2 () , ). Since the right-hand side is bounded in L2(,), T n T n u L2(,). Finally, since X T n u L 2 (,) C T 2n u L 2 (,) , the proof of Theorem 2.15 for thecase = 0 is complete.

7.3. The 1 case. Before proving the higher order case, we perform a quick computationregarding tangential and nontangential operators.

Lemma 7.1. Let X be a rst order differential operator with coefficients bounded by C k ( ) for k 1 and let T 1 , . . . , T be tangential operators with coefficients bounded by C 1 ( ) .If T = T 1 T , then

i.XT =

T X

for rst order operators X with coefficients bounded by C k + | | ( ) .ii. With X as in i.,

[X, T ] =

T X

Proof. The proof is by induction on . When = 1 this is self-evident for any k 1, sinceXT = [X, T ] + T X . Observe that for 2 j ,

XT 1 T j = T 1 XT 2 T j + [X, T 1 ]T 2 T j .

If X has coefficients bounded by C k ( ) , then the commutator [ X, T 1 ] is a rst orderdifferential operator with coefficients bounded by C k +1 ( ) . If we apply the inductionhypothesis with = j 1 to both terms, then (i) is proved.

The proof of (ii) follows from proof of (i) .

Proof of Theorem 2.15, 1. This proof is loosely based on the proof of [Fol95, Theorem7.29]. By Theorem 2.13 and the classical theory, we know that if f W ,2(,; X ) andLu = f , then u W +2 ,2loc (,; X ). As with the = 0 case, we can restrict ourselves to for

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> 0 suitably small. Let V, W be bounded subsets and satisfy V W , dist( V, bW ) > 0and dist( W, b) > 0. Choose C c (W ) with |V = 1.

We rst induct on the number of tangential derivatives. The base case is already done.The induction hypothesis is that if 1 and | | , then there exists a constant C thatdoes not depend on V , the size of the support of , or W so that

(50) T u W 1, 2 (W , ;X ) C f W | | 1, 2 (W ,;X ) + u L 2 (,) .Let be a multiindex of length + 1. Let v W 1,2(,; X ). We start by showing

(51) |D (v, T u)| C 1 v W 1,2 (W ,;X ) f W , 2 (W , ;X ) + u L2 (W ,)where C 1 = C 1( a jk C +1 () , b j C +1 () , b j C +1 () , b C +1 () , n, C +1 ( ) , C +2 ( )).

By Proposition 3.2, there exist v C c (Rn ) so that v v in W 1,2(,; X ). Therefore,since D involves at most rst order derivatives,

lim0

D (v, T u) = D (v, T u).

We compute

(52) D (v, T u) =n

j,k =1

X j v, a jk X kT (u)

+n

j =1

v, b j X j T (u) + X j v, b j T

(u)

+ v, bT (u) .

We examine each term separately

X j v, a jk X kT (u) = (T )X j v, a jk X k(u) + X j v, [a jk X k , T

](u)

= (T )X j v, a jk X ku + X j (T )v,

x k

a jk u

+ X j v, [a jk X k , T ](u)

= X j (T )v , a jk X ku

+ [ (T ), X j ]v, a jk X ku + X j (T )v, x k

a jk u

(53)

+ X j v, [a jk X k , T ](u) .

Next,

v, b j X j T (u) = (T )v, b j X j (u) + v, [b j X j , T

](u)

= (T )v, b j X j u + (T )v, b j

x j

u

+ v, [b j X j , T ](u) .(54)

Also,

X j v, b j T

(u) = (T

)

X j v, b j u + X j v, [b

j , T

](u) = X j (T )v), b j u

+ [ (T ), X j ]v, b j u + X j v, [b j , T

](u)

(55)

Plugging (53), (54) and ( 55) into ( 52), we seeD (v, T u) = D ( (T )v, u) + E = (T )v, f

+ E

= T 1 v, T 2 T +1 (f ) + E 37

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where

E =n

j,k =1

[ (T ), X j ]v, a jk X ku + X j (T )v,

x k

a jk u

+ X j v, [a jk X k , T ](u)

+

n

j =1 (T

)

v, b j

x j u + v, [b j X j , T

](u) + [ (T

)

, X j ]v, b

j u

+n

j =1

X j v, [b j , T ](u)

+ v, [b, T ](u) .

Since [, (T )] is tangential and [ , (T )]= [T , ] is a differential operator of order , bythe induction hypothesis

(56) |D (v, T u)| C 2 v W 1, 2 (W , ;X ) f W , 2 (W , ;X ) + |E |

where C 2 = C 2( C ( ) , C ( )). We turn our attention to E . Using integration byparts, [( T ), X ]= [X, T ] (formally), and Lemma 7.1 (with k = 1, since the result is not

improved when the coefficients of X are smooth), it follows from the induction hypothesisthat

|E | C 3 v W 1, 2 (W , ;X )| |

T u W 1, 2 (W ,;X )

C 4 v W 1, 2 (W , ;X ) f W 1, 2 (W ,;X ) + u L 2 (W ,)(57)

where

C 4 = C 4( a jk C +1 () , b j C +1 () , b j C +1 () , b C +1 () , n, C +1 ( ) , C +3 ( )).

By plugging (57) into ( 56) and letting 0, we observe that ( 51) has been veried.Since T (u) X , we can set v = T (u) in (51) and use the coercive estimate ( 4) to

obtain

T (u) 2W 1, 2(, ;X ) C 5 |D (T (u), T (u)) | + T (u) 2L 2 (,)

C 6 T (u) W 1, 2 (,;X ) f W , 2 (,;X ) + T (u) L2 (,)

+ T (u) 2L2 (,)where

C 6 = C 6( a jk C +1 () , b j C +1 () , b j C +1 () , b C +1 () ,n,, C ( ) , C +3 ( )).

Applying a small constant/large constant argument and the induction hypothesis ( 50) for| | , we can nish the proof of (50) for | | = + 1.

We now need to lift the restriction that T is tangential. Without loss of generality, wemay assume that | | = + 2 and T = T T n where T is tangential. We will show thatthere exists a constant C 7 so that

(58) T u L 2 (V ,) C 7 f W , 2 (W ,;X ) + u L2 (W ,)where

C 7 = C 7( a jk C +1 () , b j C +1 () , b j C +1 () , b C +1 () , n, C +1 ( ) , C +3 ( )).38

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The = 0 case follows from (50). Similarly, since the commutator [ T j , T n ] is a rst-orderoperator, we can write the = 1 case as

T = T n T + lower order tangential termsand the estimate again follows from ( 50). We prove the 2 case with an inductionargument. Assume now that ( 58) holds for = 0, . . . , J 1 with J 2. Assume that| | = J . Redene so that T = T T 2n . Note that T contains at most ( J 1) occurrencesof T n . Since u W +2loc () and Lu = f in , we have T Lu = T f a.e. in . We can write

T f = T Lu= ann T u + terms involving T n at most J 1 times and of order at most + 2 .

Since ann > 0, by the induction hypothesis and ( 58), it follows thatT u L2 (V ,) C 8 f W , 2 (W , ;X ) + u L2 (W ,) .

where C 8 = C 8( a jk C +1 () , b j C +1 () , b j C +1 () , b C +1 () ,n,, C +3 ( )).Since the constant C 8 does not depend on the size of V , the estimate holds for all V and

hence u W +2 , 2 (,;X ) C 8 f W , 2 (,;X ) + u W 1,2 (,;X ) .

8. Traces of L-harmonic functions

In this section, we wish to show that L-harmonic functions (i.e., functions u so thatLu = 0) have unique boundary values in W s 1/ 2,2(b,; T ) when u W s, 2(,; X ) ands 0.

We rst establish a simple but easily applicable uniqueness condition by proving Lemma2.16.

Proof Lemma 2.16 . Since Lu = 0 and u W 1,2

0 (,; X ), it follows thatRe D (u, u ) = Re( u,Lu ) = 0.

Since Re D (u, u ) c X u L 2 (,) , it follows that X u = 0. By Corollary 3.5, u L2 (,) X u L2 (,) = 0. Therefore, u = 0 and u is constant (on each component of ). Since

u |M = 0, u 0.

8.1. The s 2 case in Theorem 2.18 .

Lemma 8.1. Let Rn be a domain that satises (HI)-(HVI) with m 3. Let Lbe a strongly elliptic operator that has a Dirichlet form D that satises (7) for all u W 1,20 (,; X ). Let 2 k m 1 be an integer. Then there is a one-to-one correspondence

between Bk 1/ 2;2,2

(M, ; T ) and W k, 2

(,; X ) ker L with norm equivalence.Proof. Assume that U W k, 2(,; X ) and LU = 0. Since U W k, 2(,; X ), Theorem 2.1implies that Tr U B k 1/ 2;2,2(M, ; T ). Since L satises the hypotheses of Lemma 2.16, U is the unique function in W k, 2(,; X ) ker L with boundary value Tr U .

Now assume that u Bk 1/ 2;2,2(M, ; T ). By Theorem 2.1, there exists a function U W k, 2(,; X ) with boundary value u and

U W k, 2 (,;X ) C u B k 1/ 2;2 ,2 (M, ;T ) .39

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Since k 2, LU W k 2,2(,; X ). By Theorem 2.11, there exists U 0 W 1,20 (,; X ) sothat D (v, U 0) = ( v, LU ) for all v W 1,20 (,; X ). Since L satises (7), U 0 is unique. ByTheorem 2.15, U 0 W k, 2(,; X ). Moreover, the mapping

L : W k, 2(,; X ) W 1,20 (,; X ) W k 2,2(,; X )

is a bijective linear mapping, so the Open Mapping Theorem (or, more directly, its corollarythe Bounded Inverse Theorem) prove that its inverse is continuous, i.e.,

U 0 W k, 2 (, ;X ) C LU W k 2, 2 (,;X ) C U W k, 2 (, ;X ) C u B k 1/ 2;2 , 2 (M, ;T ) .

Let U = U U 0. Then LU = 0 and Tr U = Tr U = u andU W k, 2 (,;X ) C u B k 1/ 2;2 ,2 (M, ;T ) .

8.2. The case s = 1 in Theorem 2.18 . We use the arguments in [ Tay96 ] for the following.

Theorem 8.2. Let L be a strongly elliptic operator and S be a rst order operator with bounded coefficients. Set

Au = Lu + Su.There exists a constant C > 0 so that for all u W 1,20 (,; X ),

u 2W 1, 2 (, ;X ) C Au2W 1, 2 (,;X ) + C u

2L2 (,) .

Proof. Observe that for any > 0

|(u,Su )| C u L2 (,) u W 1, 2 (, ;X ) C 2

u 2W 1, 2 (, ;X ) + 1

u 2L 2 (,) .

Therefore,

Re(u,Au ) 2

u 2W 1, 2 (, ;X ) C u 2L2 (,) for all u W

1,20 (,; X ),

sou 2W 1, 2 (,;X ) C Re(u,Au ) + C

u 2L2 (,) .Also,

Re(u,Au ) C Au W 1,2 (,;X ) u W 1, 2 (,;X ) C

2 u 2W 1, 2 (,;X ) +

C 2

Au 2W 1, 2 (, ;X ) .

Putting our inequalities together and choosing > 0 small enough so that we can absorbthe C u 2W 1, 2 (,;X ) term, we see that

u 2W 1, 2 (, ;X ) C Au2W 1, 2 (,;X ) + C u

2L2 (,) .

We next show that L : W 1,20 (,; X ) W 1,2(,; X ) is continuous, injective and has a

bounded inverse.We rst assume that L gives rise to a strictly elliptic Dirichlet form over W 1,20 (,; X ).

ThenRe(v,Lu ) = Re D (v, u) C v W 1, 2 (,;X ) u W 1, 2 (, ;X ) .

Consequently, L : W 1,20 (,; X ) W 1,2(,; X ) andLu W 1,2 (, ;X ) C u W 1, 2 (,;X ) .

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Therefore, L : W 1,20 (,; X ) W 1,2(,; X ) has closed range. If L is not surjective, thereexists a nonzero v W 1,2(,; X ) so that v Range(L). By the Riesz RepresentationTheorem, we can therefore choose v W 1,20 (,; X ) so that v(v) = 0 and w(v) = 0 forw Range(L). In this case

0 = ( v,Lu ) for all u W 1,20 (,; X ).Setting u = v forces v = 0 (and hence v= 0 as well). Therefore, L is surjective. We alsoknow that L is injective as a consequence of Lemma 2.16. Consequently, the inverse to Lexists, call it G. Then G : W 1,2(,; X ) W 1,20 (,; X ). As L2(,) W 1,2(,; X )compactly, G : L2(,) W 1,20 (,; X ) compactly.

We now investigate the equationAu = f

where f W 1,2(,; X ), u W 1,20 (,; X ), and A = L + S as in Theorem 8.2. Wecontinue to assume that L has a strictly elliptic Dirichlet form over W 1,20 (,; X ). If u W 1,20 (,; X ), then there exists v W 1,2(,; X ) so that

u = GvIf Au = f , then

f = AGv = ( L + S )Gv = ( I + SG)v.We know that SG : W 1,2(,; X ) L2(,) and L2(,) W 1,2(,; X ) is compact.Therefore I + SG : W 1,2(,; X ) W 1,2(,; X ) is a compact perturbation of theidentity. The Fredholm alternative implies that the map I + SG is therefore surjective if andonly if it is injective. Lemma 2.16 supplies a condition that guarantees injectivity.

Since the difference between a strongly elliptic operator and a strongly elliptic operatorthat gives rise to a strictly elliptic Dirichlet form is the addition of a multiple of the identity,the case of relevance is S = I for some R . If Lu = v = 0, then

(L + I )u = ( L + I )Gv = ( I + G)v = 0since I + G is injective. We have therefore proved the following.

Proposition 8.3. Let L be a strongly elliptic operator that has a Dirichlet form that satises ( 7 ). Then the map

L : W 1,20 (,; X ) W 1,2(,; X )

is an isomorphism with norm equivalence.

With regard to the the norm equivalence, it follows immediately that Lu W 1, 2 (, ;X ) u W 1,2 (,;X ) . The reverse inequality follows from the Bounded Inverse Theorem. We are

now in a position to improve Lemma 8.1.

Lemma 8.4. Let Rn

be a domain that satises (HI)-(HVI) for m = 2. Let L be a strongly elliptic operator that has a Dirichlet form D which satises ( 7 ). There is a one-to-one correspondence between B1/ 2;2,2(M, ; T ) and W 1,2(,; X ) ker L with norm equivalence.

Proof. We already know that Tr : W 1,2(,; X ) B 1/ 2;2,2(M, ; T ) is continuous. Now letf B 1/ 2;2,2(M, ; T ). By Theorem 2.1, there exists F W 1,2(,; X ) so that Tr F = f and

F W 1, 2 (, ;X ) C f B 1/ 2;2 , 2 (,;X ) .41

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Solving Lu = 0 in and Tr u = f is equivalent to nding v W 1,20 (,; X ) where Lv = LF because we could then set u = F + v and it would follow f

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