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The Neumann function and the L p Neumann problem in chord-arc domains

  • Steve Hofmann EMAIL logo and Derek Sparrius
Published/Copyright: February 25, 2025

Abstract

We construct the Neumann function in a 1-sided chord-arc domain (i.e., a uniform domain with an Ahlfors regular boundary), and establish size and Hölder continuity estimates up to the boundary. We then obtain a Kenig-Pipher type theorem, in which L p solvability of the Neumann problem is shown to yield solvability in L q for 1 < q < p, and in the Hardy space H 1, in 2-sided chord-arc domains, under suitable background hypotheses.

2020 Mathematics Subject Classification: 35J25; 42B37; 42B30

1 Introduction, history and discussion of main results

We consider solutions of the divergence form elliptic equation Lu = 0, in a domain Ω R n , where

L = d i v A i , j = 1 n x i A i , j x j

and A = A(x) is a matrix of real, measurable coefficients, satisfying the uniform ellipticity condition

(1.1) λ | ξ | 2 A ( x ) ξ , ξ i , j = 1 n A i j ( x ) ξ j ξ i , A L ( R n ) Λ ,

for some λ > 0, Λ < ∞, and for all ξ R n , and a.e. x R n . We do not assume symmetry of the coefficients.

This paper includes two main sets of results (terminology will be defined in Section 2). First, in Section 4, in the setting of a 1-sided chord arc domain, we establish a Harnack inequality and obtain Hölder continuity, up to the boundary, for solutions with (at least locally) vanishing Neumann condition. Using these results, we construct, and prove estimates for, the Neumann function. Then in Section 5, in the setting of a 2-sided chord arc domain, we prove a result of “Calderón-Zygmund type” for solvability of the Neumann problem: we show that if (N) p (the Neumann problem with data in L p ) is solvable for L, for some p > 1, then it is also solvable with data in L q , 1 < q < p, as well as in the Hardy space H 1(∂Ω). As background conditions, we require solvability of the Regularity problem (R) p for L, and the Dirichlet problem[1] (D) p, for the transpose operator L*, with coefficients A i , j * A j . i . We remark that in the case that L is the Laplacian, these background hypotheses are known to be verified [1], [2]; see also Ref. [3] for a partial result concerning the Neumann problem.

These results follow the program of Kenig and Pipher[2] [4], who proved analogous results in the setting of a starlike Lipschitz domain. Many of our arguments follow those of Ref. [4] rather closely, but there is one essential difference in our work, compared to that of Ref. [4]: in the latter paper, the authors obtain the aforementioned Hölder continuity up to the boundary via a reflection argument, a method that is completely unavailable in our setting, in which the boundary need not be given locally as a graph, and in which there is no distinguished direction. Instead, we adapt the program of Moser to establish a Harnack inequality up to the boundary, which then leads, as usual, to Hölder continuity.

While the present manuscript was in preparation, we learned of two different projects that overlap with (different parts of) ours, and that have been conducted independently. First, our results concerning the Harnack inequality and Hölder continuity up to the boundary, for solutions with vanishing Neumann condition, have been obtained independently in work of David, Decio, Engelstein, Mayboroda, and Michetti [5], as part of their study of the Robin boundary value problem. Second, our results concerning the “Calderón-Zygmund” theorem for solvability of the L p Neumann problem, have been obtained independently by Feneuil and Li [6]. In fact, the results in Ref. [6] are somewhat more general than ours: they are able to relax the 2-sided chord arc condition, and they are able to dispense with the background hypothesis that the L p Regularity problem is solvable (assuming merely that the Dirichlet problem for L* is solvable with data in L p).

We remark that our results concerning existence of, and estimates for, the Neumann function have found application in recent very interesting work of Mourgoglou and Tolsa [3], who have obtained L p solvability results for the Neumann problem in chord-arc domains with very big pieces of (boundaries of) super-domains in which the Neumann problem is solvable in L q with q > p, with uniform estimates.

2 Notation, definitions, and some useful known results

  1. Unless otherwise stated, we use the letters c, C to denote harmless positive constants, not necessarily the same at each occurrence, which depend only on dimension and the constants appearing in the hypotheses of the theorems (which we refer to as the “allowable parameters”). We shall also sometimes write ab, ab, and ab to mean, respectively, that aCb, acb, and 0 < ca/bC, where the constants c, C are as above, unless explicitly noted to the contrary. At times, we shall designate by M a particular constant whose value will remain unchanged throughout the proof of a given lemma or proposition, but which may have a different value during the proof of a different lemma or proposition. On the other hand, when we attach a subscript, e.g., M 1, we mean a particular constant that has been fixed henceforth.

  2. Our ambient space is R n , n ≥ 3.

  3. Ω will always denote an open set in R n .

  4. The open n-dimensional Euclidean ball of radius r and center x will be denoted B(x, r), or sometimes B r (x) (or simply B r when the center is left implicit, or just B when both center and radius are implicit). For x ∈ ∂Ω, a surface ball is denoted Δ(x, r) ≔ B(x, r) ∩ ∂Ω (sometimes Δ r (x) or simply Δ r , or just Δ).

  5. Given an open subset Ω R n , if B = B r (x) with x Ω ̄ , we shall write Ω B = Ω r (x) ≔ B(x, r) ∩Ω (we use both notations interchangeably), and we simply write Ω r when the center is implicit.

  6. Given a Euclidean ball B or surface ball Δ, its radius will be denoted r B or r Δ, respectively.

  7. Given a Euclidean or surface ball B = B(x, r) or Δ = Δ(x, r), its concentric dilate by a factor of κ > 0 will be denoted κBB(x, κr) or κΔ ≔ Δ(x, κr).

  8. Given an open set Ω R n , for x ∈ Ω, we set δ(x) ≔ dist(x, ∂Ω).

  9. We let H n 1 denote (n − 1)-dimensional Hausdorff measure, and let σ H n 1 Ω denote the surface measure on ∂Ω.

  10. For a Borel set E R n , we let 1 E denote the usual indicator function of E, i.e., 1 E (x) = 1 if xE, and 1 E (x) = 0 if xE.

  11. Given a Borel subset E R n , we denote the average value of a function f, with respect to a Borel measure m , by

    f E , m E f d m 1 m ( E ) E f d m

    (we shall use the simplified notation f E when the measure m is implicitly understood).

  12. We shall use the letter I (and sometimes J) to denote a closed n-dimensional Euclidean dyadic cube with sides parallel to the co-ordinate axes, and we let (I) denote the side length of I.

Definition 2.1.

(ADR) (aka Ahlfors-David regular). We say that a set E R n , of Hausdorff dimension n − 1, is ADR if it is closed, and if there is some uniform constant C such that

(2.2) 1 C r n 1 σ Δ ( x , r ) C r n 1 , r ( 0 , d i a m ( E ) ) , x E ,

where diam(E) may be infinite. Here, Δ(x, r) ≔ EB(x, r) is the surface ball of radius r, and as above, σ H n 1 E is the “surface measure” on E.

Definition 2.3.

(Corkscrew condition). Following [7], we say that an open set Ω R n satisfies the corkscrew condition if for some uniform constant c > 0 and for every ball BB(x, r), with x ∈ ∂Ω and 0 < r < diam(∂Ω), there is a ball B(x B , cr) ⊂ B(x, r) ∩Ω. The point x B ⊂ Ω is called a corkscrew point relative to B (or relative to the corresponding surface ball Δ = B ∩ ∂Ω). We note that we may allow r < Cdiam(∂Ω) for any fixed C, simply by adjusting the constant c. In order to emphasize that B(x B , cr) ⊂ Ω, we shall sometimes refer to this property as the interior corkscrew condition. If Ω ext R n \ Ω ̄ satisfies the corkscrew condition, then we say that Ω satisfies an exterior corkscrew condition.

Definition 2.4.

(Harnack Chains, and the Harnack Chain condition [7]). Given two points x, x′ ∈ Ω, and a pair of numbers M, N ≥ 1, an (M, N)-Harnack Chain connecting x to x′, is a chain of open balls B 1, …, B N ⊂ Ω, with xB 1, x′ ∈ B N , B k B k+1 ≠ ∅ and M −1diam(B k ) ≤ dist(B k , ∂Ω) ≤ Mdiam(B k ). We say that Ω satisfies the Harnack Chain condition if there is a uniform constant M such that for any two points x, x′ ∈ Ω, there is an (M, N)-Harnack Chain connecting them, with N depending only on M and the ratio | x x | / min δ Ω ( x ) , δ Ω ( x ) .

Definition 2.5.

(1-sided NTA (aka uniform), NTA, and 2-sided NTA domains). If Ω R n satisfies the interior corkscrew and Harnack Chain conditions we say that it is a 1-sided NTA (1-sided nontangentially accessible, aka “uniform”) domain. If, in addition, it satisfies the exterior corkscrew condition, then we say that Ω is NTA. If both Ω and Ω ext R n \ Ω ̄ are NTA domains, then we say that Ω is 2-sided NTA.

Definition 2.6.

(1-sided CAD, CAD, and 2-sided CAD). We say that a connected open set Ω R n is a 1-sided chord-arc domain (abbreviated 1-sided “CAD”), if it is 1-sided NTA, and if ∂Ω is (n − 1)-dimensional ADR. If, in addition, Ω satisfies the exterior corkscrew condition, then we say that it is a CAD. If both Ω and Ω ext R n \ Ω ̄ are CAD, then we say that Ω is a 2-sided CAD.

For future reference, we now recall a fundamental fact that holds in a 1-sided chord arc domain Ω. Let ω L y and G L (y, ⋅) denote, respectively, the elliptic-harmonic measure and Green function for L in Ω, with pole at y ∈ Ω.

Lemma 2.7

[8]. Let Ω R n be a 1-sided CAD. Let BB(x, r), with x ∈ ∂Ω, 0 < r < diam(∂Ω). Then for y ∈ Ω\2B we have

(2.8) ω L y ( Δ ) r n 2 G L ( y , x Δ ) ω L y ( Δ ) .

The implicit constants in (2.8) depend only on ellipticity, dimension, and on the constants in the 1-sided CAD character.

The result is proved in Ref. [8] in the case that Ω is a Lipschitz domain, but the proof carries over routinely mutatis mutandis to the 1-sided chord arc setting.

3 Preliminary results

In the sequel, we shall assume that Ω is unbounded, with unbounded boundary, and at minimum satisfies the 1-sided CAD condition. In parts of Section 5, we shall further assume that Ω is a 2-sided CAD. Our arguments may be used to treat either bounded domains, or unbounded domains with unbounded boundaries, but in order to simplify the exposition, we will focus on the latter case (otherwise, we leave it to the interested reader to verify that our methods may be adapted, mutatis mutandis, to handle the case of bounded domains).

For an unbounded open set Ω R n , we define Y 1,2 ( Ω ) { f L 2 * ( Ω ) : f L 2 ( Ω ) } , with norm

f Y 1,2 ( Ω ) = f L 2 * ( Ω ) + f L 2 ( Ω ) ,

where 2* ≔ 2n/(n − 2). We note that, when Ω is an unbounded 1-sided chord-arc domain, Y 1,2(Ω) is a homogeneous Sobolev space. Indeed, in that setting, assuming a priori that f L 2 * ( Ω ) , we have f L 2 * ( Ω ) f L 2 ( Ω ) (as follows from the standard embedding results in R n , combined with the extension theorem of Jones [9], which applies in our setting; we shall return to this point momentarily). Hence,

f Y 1,2 ( Ω ) f L 2 ( Ω ) , f Y 1,2 ( Ω ) ,

so in particular, Y 1,2 is a Hilbert Space with the inner product f , g Ω f g ̄ .

We let Y 0 1,2 ( Ω ) denote the closure of C 0 ( Ω ) in the Y 1,2 norm.

Closely related to Y 1,2(Ω) is the homogeneous Sobolev space W ̇ 1,2 ( Ω ) , defined as the collection of equivalence classes (modulo constants) of functions f L loc 2 ( Ω ) , whose weak gradients ∇f belong to L 2(Ω), endowed with the norm

f W ̇ 1,2 ( Ω ) = f L 2 ( Ω ) .

Similarly, we define W ̇ 0 1,2 ( Ω ) to be the closure of C 0 ( Ω ) under the norm  W ̇ 1,2 ( Ω ) , and we let W ̇ 1,2 ( Ω ) = ( W ̇ 0 1,2 ( Ω ) ) * denote the dual space of W ̇ 0 1,2 ( Ω ) .

Remark 3.1.

We observe that W ̇ 1,2 ( Ω ) is isomorphic to Y 1,2(Ω): each element of the latter is a realization of an equivalence class in the former, namely, it is the particular realization belonging to L 2 * ( Ω ) ; moreover, each equivalence class in W ̇ 1,2 ( Ω ) has such a realization. To simplify the notation when there is no chance of confusion, we shall often make no notational distinction between an equivalence class, and any particular representative of that class (e.g., when considering ∇f). On the other hand, at times in the sequel, it will be useful to make a distinction between the equivalence class in W ̇ 1,2 , and some particular representative of that class. When we wish to do so, we shall write [f] to denote the equivalence class, and f to denote a particular realization; in this case, by convention we shall always take f to be the particular realization in Y 1,2(Ω).

As usual, W 1,2(Ω) ≔ {fL 2(Ω): ∇fL 2(Ω)} is the inhomogeneous Sobolev space with norm

f W 1,2 ( Ω ) = f L 2 ( Ω ) + f L 2 ( Ω ) ,

and W 0 1,2 ( Ω ) is the closure of C 0 ( Ω ) in the W 1,2 norm.

We also define H ̇ 1 / 2 ( Ω ) = B ̇ 1 / 2 2,2 ( Ω ) , the Besov space defined as the space of equivalence classes of functions f L loc 2 ( Ω ) , modulo constants, with

(3.2) f H ̇ 1 / 2 ( Ω ) = Ω Ω | f ( x ) f ( y ) | 2 | x y | n d σ ( y ) d σ ( x ) 1 / 2 .

Having defined our space modulo constants, we equip H ̇ 1 / 2 ( Ω ) with the norm defined in (3.2), which in turn will make H ̇ 1 / 2 ( Ω ) a Hilbert space. We define H ̇ 1 / 2 ( Ω ) ( H ̇ 1 / 2 ( Ω ) ) * , the dual of H ̇ 1 / 2 ( Ω ) .

Remark 3.3.

As was the case the for W ̇ 1,2 (see Remark 3.1 above), we often shall make no notational distinction between an equivalence class in H ̇ 1,2 ( Ω ) and any particular representative of that class (e.g., when considering differences f(x) − f(y)). On the other hand, at times in the sequel, it will be useful to make a distinction between the equivalence class in H ̇ 1,2 , and some particular representative of that class. When we wish to do so, we shall write [f] to denote the equivalence class, and f to denote a particular realization.

We note for future reference that standard trace-extension theory holds in our domains. In particular, there exist extension operators E Ω : W ̇ 1,2 ( Ω ) W ̇ 1,2 ( R n ) and Y 1,2 ( Ω ) Y 1,2 ( R n ) , and E Ω : H ̇ 1 / 2 ( Ω ) W ̇ 1,2 ( R n ) , and a trace operator Tr : W ̇ 1,2 ( Ω ) H ̇ 1 / 2 ( Ω ) , satisfying

(3.4) E Ω ( Φ ) W ̇ 1,2 ( R n ) Φ W ̇ 1 / 2 ( Ω ) , E Ω ( Φ ) Y 1,2 ( R n ) Φ Y 1 / 2 ( Ω ) ,

(3.5) E Ω ( φ ) W ̇ 1,2 ( R n ) φ H ̇ 1 / 2 ( Ω ) ,

and

(3.6) Tr ( Φ ) H ̇ 1 / 2 ( Ω ) Tr E Ω ( Φ ) H ̇ 1 / 2 ( Ω ) E Ω ( Φ ) W ̇ 1,2 ( R n ) Φ W ̇ 1,2 ( Ω ) .

where in each case, the implicit constants depend only on dimension and the ADR, CS, and HC constants. A few words of explanation are in order. Since our domains are unbounded 1-sided chord arc domains with unbounded boundaries, they are in particular (ɛ, ∞) domains in the sense of Jones [9], and we may deduce the extension property (3.4) by using the extension operator constructed in Ref. [9], and following the proof of Ref. [9], Theorem 1]. The latter result applies directly to the inhomogeneous Sobolev spaces W 1,2, but the proof carries over to the homogenous Sobolev spaces W ̇ 1,2 and Y 1,2 used here. We leave the details to the interested reader. Similarly, (3.5) and (3.6) are homogeneous versions of Theorems 2 and 1, respectively, in Ref. [10]. Again the proofs carry over to our setting, and again we omit the details. In the case of (3.6), the first step is by definition, the last step is simply (3.4), and the middle step is the homogeneous version of Ref. [10], Theorem 1]. For a more general result (which can also be adapted to the setting of homogeneous smoothness spaces), we refer the reader to Ref. [[11], Theorem 5.1].[3] In particular, in Ref. [11], it is shown that the trace operator is independent of the choice of extension of Φ to W 1,2 ( R n ) (or in our case, to W ̇ 1,2 ( R n ) .)

Recall that given a Borel subset E R n , we denote the average value of a function f, with respect to an implicitly understood Borel measure m , by

f E E f d m 1 m ( E ) E f d m .

We note for future reference the following well known estimate, for any ball B (in fact, even for convex domains, see, e.g., [[12], Lemma 7.16]) and for uW 1,2(B):

(3.7) | u ( x ) u B | B | x z | 1 n | u ( z ) | d z , a.e.  x B .

Remark 3.8.

We further note for future reference that there exists a constant K 0 = K 0(CS) ≥ 2 such that, for R > 0, and for every x Ω ̄ , with δ(x) < R, setting Ω R = Ω R (x): = Ω ∩ B(x, R), we can find a point x 0 Ω K 0 R \ Ω R Ψ R , with δ(x 0) > R/5, such that B 0B R/10(x 0) ⊂ Ψ R .

Proposition 3.9.

Suppose that Ω R n satisfies the Corkscrew (CS) and Harnack Chain (HC) conditions. Let K 0 be the constant in Remark 3.8. Let R > 0 and let x Ω ̄ , with δ(x) < R, and as usual define Ω R = Ω ∩ B(x, R). Let Ψ R Ω K 0 R \ Ω R , and x 0 ∈ Ψ R be as in Remark 3.8, so that B 0B R/10(x 0) ⊂ Ψ R . Then there is a constant K sufficiently large such that for all uW 1,2 KR ),

(3.10) | u ( y ) u B 0 | Ω K R | y z | 1 n | u ( z ) | d z , a.e.  y Ω R ,

where the implicit constant and K depend only on dimension and the CS and HC constants.

Proof.

By HC, given y ∈ Ω R , and bearing in mind that K 0 = K 0(CS), we see that there exists a chain of balls B 0, B 1, …, B N with radii r 0, r 1, …, r N , and with B 0B R/10(x 0) as above, such that

  1. yB N ;

  2. B k B k−1 ≠ ∅, 1 ≤ kN;

  3. r k ≈ dist(B k , ∂Ω);

  4. For each 1 ≤ kN, there is a ball B k * of radius r k * , such that B k B k 1 B k * and

    r k * r k r k 1 δ ( z ) y z

    for all z B k * ;

  5. The balls B k * satisfy the bounded overlap property k 1 B k * ( z ) C , for all z ∈ Ω;

  6. k B k * Ω K R for some K large enough, depending only on the CS and HC constants.

By a density argument, we may assume that uC 1 KR ). Since yB N , we have

(3.11) u ( y ) u B 0 = u ( y ) u B N + u B N u B N 1 + u B 1 + u B 1 u B 0 u ( y ) u B N + k = 1 N r k B k * u B N y z 1 n u d z + k = 1 N r k B k * u ,

where we have applied Poincaré’s inequality to the telescoping sum, and then used (3.7) in B N , in the last two steps.

By property (iv), we can estimate the last term in (3.11) as

k = 1 N r k B k * u d z k = 1 N B k * y z 1 n u d z .

Plugging the latter estimate into (3.11), and using properties (v) and (vi), we obtain (3.10). □

Corollary 3.12.

Suppose that Ω R n satisfies the Corkscrew and Harnack Chain conditions. Let K be as in Proposition 3.9. Fix x 1 Ω ̄ , and define Ω R = Ω R (x 1) as above. Then for uW 1,2 KR ),

(3.13) | u ( x ) u ( y ) | Ω K R | x z | 1 n + | y z | 1 n | u ( z ) | d z , a.e.  x , y Ω R .

In addition,

(3.14) | u ( x ) u Ω R | Ω K R | x z | 1 n | u ( z ) | d z , a.e.  x Ω R .

In particular, for 2* ≔ 2n/(n − 2),

(3.15) R 1 u u Ω R L 2 ( Ω R ) u u Ω R L 2 * ( Ω R ) u L 2 ( Ω K R )

Proof.

If δ(x 1) ≥ R (so that Ω R = B R ), the result is standard (with K = 1). We therefore suppose that δ(x 1) < R. By a density argument, we may assume that uC 1 KR ). To prove (3.13), we let B 0 denote the ball in Remark 3.8 and Proposition 3.9, and apply (3.10) to each summand in the inequality

| u ( x ) u ( y ) | | u ( x ) u B 0 | + | u ( y ) u B 0 | .

To prove (3.14), we write

| u ( x ) u Ω R | = Ω R u ( x ) u ( y ) d y ,

and then apply (3.13) to the integrand to obtain the bound

Ω R Ω K R | x z | 1 n + | y z | 1 n | u ( z ) | d z d y Ω K R | x z | 1 n | u ( z ) | d z + R 1 n Ω K R | u ( z ) | d z Ω K R | x z | 1 n | u ( z ) | d z .

Finally, (3.15) follows immediately from Hölder’s inequality, and then (3.14) and the Hardy-Littlewood-Sobolev theorem. □

The next result is an extension of Ref. [[12], Theorem 7.21] beyond the setting of convex domains. The proof follows that in Ref. [12], using the estimates established above.

Theorem 3.16.

Suppose that Ω R n satisfies the CS and HC conditions and let K 0 and K be as in Remark 3.8 and Proposition 3.9, respectively. Fix x 1 ∈ ∂Ω, and 2R < diam(∂Ω)/K 0, and let Ω R = Ω ∩ B(x 1, R). Let hW 1,23KR ) and suppose that there exists a constant M such that

(3.17) Ω r h d x M r n 1

for all balls B r B 3KR = B(x 1, 3KR). Then there exists positive constants c 0 and C depending on the allowable parameters such that

(3.18) Ω 2 R exp c 0 M h h Ω 2 R d x C R n

Proof.

We first observe that since (3.17) holds for all balls B r B 3KR , we in fact have that

(3.19) Ω 2 KR B r h d x M r n 1

for every B r . Indeed, if B r B 3KR , then (3.17) applies directly; otherwise if B r meets both Ω2KR and also R n \ Ω 3 KR , then rKR, and we may simply apply (3.17) to Ω2KR .

Next, we use [12], Lemma 7.20] (the proof of which does not require any extra assumptions on Ω) in the domain Ω2KR , to deduce that if f satisfies

Ω 2 KR B r f d x M r n 1 ,

for all balls B r , then there exist constants C 1 and C 2 depending only on n such that

Ω 2 KR exp 1 C 1 M I 1 ( f 1 Ω 2 KR ) d x C 2 R n

where I α f = α n f . Applying the latter estimate with f = ∇h, and using (3.14) with u = h, and with R replaced by 2R, we obtain (3.18). □

4 The Neumann function

In this section we assume Ω R n is a 1-sided CAD, i.e., Ω has an ADR boundary and satisfies (CS) and (HC) conditions, unless otherwise stated. Moreover, we continue to assume that Ω and ∂Ω are unbounded (we leave it to the interested reader to verify that the construction may be adapted to the case of bounded domains). Much of our construction will follow that of Ref. [4], except for the proof of Hölder continuity at the boundary for the Neumann function, where we must use a different approach since the reflection method in Ref. [4] is not available in our setting.

Let L = −divA∇, where A = (a ij (x)) is a matrix of bounded measurable coefficients satisfying the uniform ellipticity condition

λ ξ 2 a i j ξ i ξ j , A ξ Λ ξ ξ R n ,

where 0 < λ ≤ Λ < ∞. In the sequel, all implicit constants will depend only on ellipticity, dimension, and the ADR, CS, and HC constants.

We do not assume that A is a symmetric matrix. However, we then run into the issue that there is not necessarily a canonical choice of the matrix A representing the operator L = −divA∇, and hence there is not necessarily a canonical representation of the conormal derivative. The following example illustrates the idea.

Example 4.1.

Let R + n + 1 = { ( x , t ) : x R n , t 0 } and R + n + 1 = R + n + 1 | t = 0 = R n . Consider the following Neumann problem

(N L,A ) L u = 0  in  R + n + 1 u ν A = f H ̇ 1 / 2 ( R n ) ,

where L = −divA∇, the coefficient matrix A = A(x) is independent of the transverse variable t, and the conormal derivative is, at least formally, given by the formula νAu, where ν is the outer unit normal vector on ∂Ω. By Lax-Milgram, there is a unique u W ̇ R + n + 1 (thus, unique modulo constants) which is a weak solution to (N L,A ), which by definition means that for any Φ W ̇ ( R n + 1 ) , we have

(4.2) f , Tr ( Φ ) = R + n + 1 A u Φ d x d t .

We now modify the matrix A in the following way. Let b ( x ) = [ b 1 ( x ) , , b n ( x ) ] L ( R n ; R n ) be a (weakly) divergence free vector-valued function on R n , i.e., denoting the gradient for R n + 1 as ∇ = [∇, ∂ t ] where ∇ is the gradient vector on R n , we suppose that

(4.3) R n b f d x = 0 , f W 1,1 ( R n ) .

We write the matrix A in the block formwhere A is an n × n matrix, B is an n × 1 column vector, C is a 1 × n row vector, and D is a scalar. We now define the matrix A ̃ , by

Observe that A ̃ and A have precisely the same ellipticity. Moreover, for Φ C 0 R + n + 1 , using (4.3) and the fact that b is t-independent, we see that

R + n + 1 t u b Φ t Φ b u d x d t = 0 ,

and this identity may be extended by density to Φ Y 0 1,2 R + n + 1 . Thus,

L u , Φ = R + n + 1 A u Φ d x d t = R + n + 1 A ̃ u Φ d x d t , Φ Y 0 1,2 R + n + 1 ,

i.e., there may be more than one coefficient matrix for which the associated form yields the same operator. On the other hand, if u is a weak solution of Lu = 0 in R + n + 1 , we have

0 = R + n + 1 A u Φ d x d t = R + n + 1 A ̃ u Φ d x d t , Φ Y 0 1,2 R + n + 1 .

i.e., u is also a weak solution of L ̃ u d i v ( A ̃ u ) = 0 in R + n + 1 . Furthermore, one can use Lax-Milgram to construct a weak solution u ̃ of N L ̃ , A ̃ , in the sense of (4.2) but with A replaced by A ̃ . In general, the weak solution u ̃ of N L ̃ , A ̃ need not agree with the solution u of N L,A , thus, the Neumann problem itself depends on the particular choice of coefficient matrix A.

To resolve the possibly ambiguity in the matrix A, we assume for the remainder of this paper that A has been fixed.

Let us introduce some further notation that we will use in the sequel.

We let A* denote the transpose (hence also the adjoint, since our coefficients are real) of the coefficient matrix A.

We reiterate that for now, we will continue to follow the presentation in Ref. [[4], Section 2], with some technical adjustments owing to the unbounded scale-invariant setting in which we work.

We recall the notational convention in Remark 3.3, namely that when it is useful to distinguish an equivalence class in H ̇ 1 / 2 ( Ω ) from a particular representative in that class, we shall denote the former by [φ], and the latter by φ. Let Lip c (∂Ω) denote the space of Lipschitz functions with compact support on ∂Ω, and set

(4.4) H 1 / 2 ( Ω ) [ φ ] H ̇ 1 / 2 ( Ω ) : [ φ ]  has a realization φ Lip c ( Ω ) s .

We note that H 1 / 2 ( Ω ) is dense in H ̇ 1 / 2 ( Ω ) .

Remark 4.5.

In the sequel, when distinguishing between equivalence classes in H ̇ 1,2 ( Ω ) and their particular realizations, we shall use the convention that when [ φ ] H 1 / 2 ( Ω ) , we shall always let φ denote the particular realization of the equivalence class belonging to Lip c (∂Ω).

Lemma 4.6.

For each x ∈ Ω, elliptic measure ω x may be identified with an element of H ̇ 1 / 2 ( Ω ) as follows: there is a bounded linear functional ϒ x defined on H ̇ 1 / 2 ( Ω ) , which is associated to ω x in the sense that for all [ φ ] H 1 / 2 ( Ω ) , with realization φ ∈ Lip c (∂Ω) as in the preceeding remark,

(4.7) ϒ x , [ φ ] = Ω φ d ω x .

Here, ⟨⋅, ⋅⟩ denotes the duality pairing of H ̇ 1 / 2 ( Ω ) and its dual H ̇ 1 / 2 ( Ω ) . More generally,

(4.8) ϒ x , [ φ ] = u ( x ) , [ φ ] H ̇ 1 / 2 ( Ω ) ,

where u is the Y 1,2(Ω) realization of the W ̇ 1,2 ( Ω ) weak solution to the Dirichlet problem with data [φ]. Moreover,

(4.9) ϒ x H ̇ 1 / 2 ( Ω ) δ ( x ) 2 n 2 .

In addition, there is a unique v x (⋅) ∈ Y 1,2(Ω) such that (recalling the notation in Remark 3.1)

(4.10) Ω A * v x Φ d y = ϒ x , Tr ( [ Φ ] ) , [ Φ ] W ̇ 1,2 ( Ω ) .

In particular, for [ Φ ] W ̇ 1,2 ( Ω ) such that Tr ( [ Φ ] ) = [ φ ] H 1 / 2 , we have

(4.11) Ω A * ( y ) v x ( y ) Φ ( y ) d y = Ω φ d ω x .

Furthermore,

(4.12) v x Y 1,2 ( Ω ) v x 2 δ ( x ) 2 n 2

where the implicit constant depends only on the allowable parameters.

Remark 4.13.

Recall that, in order to simplify the exposition, we have assumed that Ω and ∂Ω are unbounded; if Ω were bounded, we would have had to modify the right hand side of (4.11), so that it would vanish when φ ≡ 1, and to work with a subspace of inhomogeneous W 1,2, whose elements are normalized to have mean value zero in Ω, rather than with the homogeneous Sobolev spaces W ̇ 1,2 and Y 1,2. We refer the reader to Ref. [[4], Section 2] for details.

Remark 4.14.

Taking Φ C 0 ( Ω ) (or even in Y 0 1,2 ( Ω ) ), we see that v x (⋅) is a weak solution of the adjoint equation L*v ≔ divA*∇v = 0 in Ω.

Proof of Lemma 4.6.

We shall use the equivalence class notation in Remarks 3.1 and 3.3.

We first consider the Dirichlet problem

(4.15) L u = 0  in  Ω [ u ] | Ω = [ φ ] H ̇ 1 / 2 ( Ω ) [ u ] W ̇ 1,2 ( Ω ) .

Here, we interpret [u]|∂Ω = [φ] in the trace sense. We construct the unique solution via the standard Lax-Milgram argument. Let [ φ ] H ̇ 1,2 ( Ω ) , and let [ Φ ] W ̇ 1,2 ( Ω ) be the extension as in (3.5), i.e., such that Tr([Φ]) = [φ], with

(4.16) [ Φ ] W ̇ 1,2 ( Ω ) [ φ ] H ̇ 1 / 2 ( Ω ) ,

where the implicit constant depends only on n, ADR, (CS) and (HC).

Then d i v A Φ W ̇ 1,2 ( Ω ) , with

d i v A Φ W ̇ 1,2 ( Ω ) [ φ ] H ̇ 1 / 2 ( Ω ) ,

so by Lax-Milgram, there is a unique [ w ] W ̇ 0 1,2 ( Ω ) such that divAw = divA∇Φ in the weak sense in Ω, satisfying

(4.17) [ w ] W ̇ 0 1,2 ( Ω ) [ φ ] H ̇ 1 / 2 ( Ω ) .

Set u ≔ Φ − w, so that [u] is a weak solution of (4.15), and by (4.16) and (4.17) satisfies

(4.18) [ u ] W ̇ 1,2 ( Ω ) φ H ̇ 1 / 2 ( Ω ) .

Note that if [Φ1] is any other W ̇ 1,2 ( Ω ) extension of [φ], with corresponding [ w 1 ] W ̇ 0 1,2 ( Ω ) such that divAw 1 = divA∇Φ1, and [u 1] ≔ [Φ1w 1] another solution to (4.15), then [ u u 1 ] W ̇ 0 1,2 ( Ω ) is a weak solution, hence u = u 1 in the sense of W ̇ 1,2 . Thus, the problem (4.15) is well-posed,

We now fix the particular realization of [ u ] W ̇ 1,2 ( Ω ) such that uY 1,2(Ω), so that

u Y 1,2 ( Ω ) u 2 [ φ ] H ̇ 1 / 2 ( Ω ) .

By Moser’s local boundedness estimate, for each fixed x ∈ Ω, we therefore have

(4.19) u ( x ) B ( x , δ ( x ) / 2 ) u 2 * 1 / 2 * δ ( x ) 2 n 2 u Y 1,2 ( Ω ) δ ( x ) 2 n 2 [ φ ] H ̇ 1 / 2 ( Ω ) .

Since the solution map [φ] ↦ [u] is linear, we also have that, for each x ∈ Ω, the mapping φu(x) defines a bounded linear functional on H ̇ 1 / 2 ( Ω ) , hence an element of H ̇ 1 / 2 , with norm at most C δ ( x ) 2 n 2 . We denote this linear functional by ϒ x , so that (4.8) and (4.9) hold. On the other hand, ∂Ω is regular in the sense of Wiener at every point, so for [ φ ] H 1 / 2 ( Ω ) , we also have

ϒ x , [ φ ] = u ( x ) = Ω φ d ω x ,

where we are using the convention in Remark 4.5. Thus, (4.7) holds.

We now turn to the construction of v x . It is enough to find [ v x ] W ̇ 1,2 ( Ω ) satisfying the right-hand inequality in (4.12): we may then fix the particular realization of [v x ] belonging to Y 1,2(Ω). The existence and uniqueness (modulo constants) of such a [ v x ] W ̇ 1,2 ( Ω ) comes from finding the variational solution to the Neumann problem

(4.20) L * v x = 0  in  Ω v x ν A * = ϒ x H ̇ 1 / 2 ( Ω ) [ v x ] W ̇ 1,2 ( Ω ) ,

where / ν A * denotes the outer conormal derivative associated to L* = −divA*∇, the adjoint operator. We observe that formally, (4.20) can be thought of as solving the (adjoint) Neumann problem with data G x ν A , where G x G(x, ⋅) is the Green’s function associated with Ω, with pole at x. Solving the Neumann problem (4.20) in the variational sense means by definition that we seek a unique [ v x ] W ̇ 1,2 ( Ω ) satisfying (4.10). Using the trace estimate (3.6), we see that such a [v x ] exists by virtue of the Lax-Milgram lemma, and satisfies

v x L 2 ( Ω ) ϒ x H ̇ 1 / 2 ( Ω ) δ ( x ) 2 n 2 ,

where in the last step we have used (4.9). We omit the routine details. Hence, we have solved the problem (4.20) with the estimate (4.12), as desired. Finally, (4.11) follows immediately from (4.10) and (4.7). □

Note that the adjoint (i.e., transpose) matrix A* = A T satisfies the same properties as A. Hence, Lemma 4.6 also holds with the roles of L = −divA∇ and L* = −divA*∇ reversed, with adjoint elliptic measure ω * x and a linear functional ϒ x in place of ω x and ϒ x , and with a unique solution v x * to the Neumann problem for L. Continuing to follow the strategy of Ref. [4], we now obtain results relating v x v(x, ⋅) and v x * v * ( x , ) .

Corollary 4.21.

For x, y ∈ Ω, v(x, y) = v*(y, x).

Proof.

Note that (4.10) holds for all [ Φ ] W ̇ 1,2 ( Ω ) . We now take Φ = v z * , which then yields

ϒ x , Tr v z * = Ω A * v x v z * d y = Ω v x A v z * d y = ϒ z , Tr ( [ v x ] ) ,

where in the last step we have used the adjoint version of (4.10). Since v x (⋅) and v z * ( ) are Y 1,2(Ω) solutions to L*v = 0 and Lv = 0, respectively, it follows from (4.8) (and its adjoint version) that v x ( z ) = v z * ( x ) . □

Corollary 4.22.

For x, y ∈ Ω,

(4.23) v ( x , y ) δ ( y ) 2 n 2 δ ( x ) 2 n 2 .

Proof.

Applying Moser’s local boundedness estimate to the adjoint solution v x (⋅) = v(x, ⋅), we see that

v ( x , y ) B ( y , δ ( y ) / 2 ) v x 2 * 1 / 2 * δ ( y ) 2 n 2 v x Y 1,2 ( Ω ) δ ( y ) 2 n 2 δ ( x ) 2 n 2 ,

where the last inequality follows from (4.12). □

Definition 4.24.

(Neumann Function). Let G(x, y) be the Green’s function for L in Ω. The Neumann function N(x, y), x, y in Ω is defined to be N(x, y) = G(x, y) − v(x, y).

We shall sometimes use the notation N x (⋅) ≔ N(x, ⋅), and G x (⋅) ≔ G(x, ⋅), thus N x = G x v x .

Lemma 4.25.

Let [ Φ ] W ̇ 1,2 ( Ω ) , with realization Φ ∈ Y 1,2(Ω), and suppose that x ∈ Ω(Φ). Then

Ω A * N x Φ = 0 ,

i.e., N x ( ) / ν A * = 0 in the variational sense.

Proof.

We may assume that Tr ( [ Φ ] ) H 1 / 2 ( Ω ) (see (4.4)), and that Φ Y 1,2 ( Ω ) C ( Ω ̄ ) ; indeed the set of all such [Φ] is dense in W ̇ 1,2 ( Ω ) . The conclusion of the lemma then follows immediately from (4.11) and the standard Riesz formula

(4.26) Ω A * G x Φ = Φ ( x ) Ω φ d ω x , Φ Y 1,2 ( Ω ) C ( Ω ̄ ) , Tr ( Φ ) C c ( Ω ) .

Remark 4.27.

In the sequel, we shall often work with bounded sub-domains of Ω. Given y 0 Ω ̄ , and for R > 0, set Ω R = Ω R (y 0) ≔ Ω ∩ B(y 0, R). Our results are new only when δ(y 0) < 2R, and in that case, matters may be reduced to the case that y 0 ∈ ∂Ω, by an appropriate fattening of the ball. On the other hand, if δ(y 0) ≥ 2R, then our estimates are simply the classical interior estimates of De Giorgi-Nash-Moser. In the sequel, we shall sometimes apply these results in contexts where the estimates may be required to hold both at the boundary and in the interior, but we shall only give explicit references for the former.

We recall the following fact (see [13], Lemma 3.61]).

Proposition 4.28.

For each Ω R as above, with y 0 ∈ ∂Ω, there is a 1-sided chord-arc subdomain T R ⊂ Ω, and a constant κ ≥ 8, such that Ω4KR T R ⊂ Ω κKR , where K > 1 is the constant in Proposition 3.9 and Corollary 3.12, and where κ and the 1-sided chord-arc constants (i.e., the (CS), (HC) and ADR constants) for T R depend only on n, and the ADR, (CS) and (HC) constants for Ω.

For the reader’s convenience, we mention that in the notation of Ref. [13], T R is the “Carleson tent” T Δ, with Δ = Δ(y 0, 4KR) ≔ ∂Ω ∩ B(y 0, 4KR).

Our next goal is to show that the Neumann function is Hölder continuous up to the boundary. At this point, we shall depart from the treatment in Ref. [4], as the reflection argument used there is unavailable in our setting. Instead, we shall use the methods of Moser. The following local boundedness result is known: see Ref. [14], Theorem 1.6 and its proof, and Remark 1.8].

Lemma 4.29.

Let y 0 ∈ ∂Ω, R > 0 and set Ω R = Ω R (y 0) as above. Let uW 1,22R ). Suppose that u satisfies either

(4.30) Ω A * u Φ d y = 0 , Φ W 0 1,2 B ( y 0 , 2 R ) ,

or, that u ≥ 0 in Ω2R and that

(4.31) Ω A * u Φ d y 0 , Φ W 0 1,2 B ( y 0 , 2 R ) , Φ 0 .

Then

(4.32) sup Ω R | u | Ω 2 R u 2 1 / 2 ,

where the implicit constants depend only on n, ADR, (CS), (HC) and ellipticity.

Remark 4.33.

Observe that by Lemma 4.25, we may apply Lemma 4.29 to the Neumann function u = N x , provided that x ∈ Ω\Ω2R .

Remark 4.34.

In the sequel, we let T R be the 1-sided chord-arc subdomain in Proposition 4.28. In particular, T R is a trace-extension domain, depending only on n, ADR, (CS) and (HC).

Lemma 4.35.

Fix Ω R as above. Let K be the constant in Proposition 3.9. Suppose that uW 1,2(T R ), u ≥ 0 is a local supersolution to the adjoint Neumann problem in Ω4KR , in the sense that

(4.36) Ω A * u Φ d y 0 , Φ W 0 1,2 B ( y 0 , 4 K R ) , Φ 0 .

Then

(4.37) Ω u 2 u 2 η 2 d y Ω η 2 d y , η C 0 B ( y 0 , 4 K R ) , η 0 .

where the implicit constants depend only on n, ADR, (CS), (HC) and ellipticity.

Proof.

Let η C 0 B ( y 0 , 4 K R ) , and define u ɛ = u + ɛ, for ɛ > 0. Then u ε 1 η 2 W 0 1,2 ( B ( y 0 , 4 K R ) ) (here we have used the extension property to define u outside of T R ; see Remark 4.34), and hence u ε 1 η 2 is a valid test function in (4.36). Note that

u ε 1 η 2 = u ε 2 η 2 u ε + 2 u ε 1 η η .

Since ∇u ɛ = ∇u, by ellipticity and (4.36) we therefore have

Ω u 2 u ε 2 η 2 d y Ω u ε 2 η 2 A * u u ε d y = Ω A * u u ε 1 η 2 d y + 2 Ω u ε 1 η A * u η d y = 0 + O Ω u ε 1 η | u | | η | d y γ Ω u 2 u ε 2 η 2 d y + γ 1 Ω η 2 d y ,

where γ ∈ (0, 1) is at our disposal. Choosing γ small enough depending only on ellipticity, we may hide the small term, to obtain

Ω u 2 u ε 2 η 2 d y Ω η 2 d y ,

uniformly in ɛ. Letting ɛ → 0, we obtain (4.37). □

Our next goal is to establish a Harnack inequality up to the boundary for non-negative solutions with vanishing Neumann condition, and then to use that result to show that the Neumann function is locally Hölder continuous, again up to the boundary. To this end, we will follow Moser’s proof of the interior Harnack inequality. We fix R > 0 and a point y 0 ∈ ∂Ω, and let Ω R ≔ Ω ∩ B(y 0, R) as above.

Theorem 4.38.

(Weak Harnack Inequality). Let uW 1,2(T R ), u ≥ 0, be a local supersolution to the adjoint Neumann problem in Ω4KR , in the sense of Lemma 4.35, i.e., (4.36) holds. Then for some s > 0 sufficiently small,

(4.39) Ω 2 R u s 1 / s inf Ω R u ,

where s and the implicit constants depend only on dimension, ADR, (CS), (HC), and ellipticity.

Proof.

We use the standard argument of Moser, following the treatment in Ref. [12], Chapter 8], adapted to our setting in which we allow estimates up to the boundary. We momentarily fix ɛ > 0, and set u ɛ u + ɛ. We claim that, in particular, for all β > 0, u ε β is a W 1,2 local subsolution in the sense of (4.31), i.e.,

(4.40) Ω A * u ε β Φ d y 0 , Φ W 0 1,2 B ( y 0 , 2 R ) , Φ 0 .

Indeed, first note that for all β > 0, u ε β is bounded (depending on ɛ and β), and hence it readily follows that u ε β W 1,2 ( T r ) , and that u ε β 1 Φ W 0 1,2 ( B ( y 0 , 2 R ) ) (here we have used the extension property to define u outside of T R ; see Remark 4.34). We then have

Ω A * u ε β Φ d y = β Ω A * u ε β 1 u Φ d y = β Ω A * u Φ u ε β 1 d y β ( β + 1 ) Ω A * u u Φ u ε β 2 d y I + I I .

Then I ≤ 0, since u ε β 1 Φ is a valid test function in (4.36), and II ≤ 0 trivially, by ellipticity and the fact that u ɛ > 0 and Φ ≥ 0. Hence, the claim (4.40) holds.

We may therefore apply Lemma 4.29 to the subsolution u ε β , to obtain

sup Ω R u ε β Ω 2 R u ε 2 β 1 / 2 ,

i.e.,

Ω 2 R u ε 2 β 1 / ( 2 β ) inf Ω R u ε inf Ω R u + ε .

Choosing β = s/2, and letting ɛ → 0, we obtain

(4.41) Ω 2 R u s 1 / s inf Ω R u , s > 0 .

To complete the proof of Theorem 4.38, it therefore suffices to show that for some s > 0 small enough,

(4.42) Ω 2 R u s 1 / s Ω 2 R u s 1 / s 1 .

To this end, we first note that Lemma 4.35 applies to u. In particular, setting h ≔  log u, by (4.37), we see that hW 1,23KR ).

Let B r B 3KR , and let η be a smooth bump function adapted to B r , i.e., η C 0 ( B 4 r / 3 ) , with 0 ≤ η ≤ 1 and η ≡ 1 on B r , satisfying ∇ηr −1. By (4.37), we then have

Ω r | h | d x r n / 2 Ω r | h | 2 d x 1 / 2 M r n 1 ,

where M depends only on the allowable parameters; i.e., (3.17) holds for all B r B 3KR , so we may apply Theorem 3.16 to h to obtain

Ω 2 R exp c M h h Ω 2 R d x C R n ,

for some uniform constant c > 0 small enough. Taking s = c/M, we therefore have

Ω 2 R u s Ω 2 R u s = Ω 2 R exp s ( h h Ω 2 R ) Ω 2 R exp s ( h h Ω 2 R ) 1 ,

which proves (4.42), and thus also Theorem 4.38. □

Theorem 4.43.

(Harnack Inequality). Fix Ω R = Ω R (y 0) as above, with y 0 ∈ ∂Ω. Let K be the constant in Proposition 3.9. Suppose that uW 1,2(T R ), u ≥ 0 is a local weak solution to the Neumann problem with vanishing data in Ω4KR , in the sense that

(4.44) Ω A * u Φ d y = 0 , Φ W 0 1,2 B ( y 0 , 4 K R ) .

Then for all B r = B r (z) ⊂ B R/2,

(4.45) sup Ω r u inf Ω r u ,

where the implicit constants depend only on n, ADR, (CS), (HC) and ellipticity.

Proof.

First note that (4.32) holds for Ω r = Ω r (z), i.e.,

(4.46) sup Ω r u Ω 2 r u 2 1 / 2 , B r B R .

We now claim that

(4.47) Ω 2 r u 2 1 / 2 Ω 4 r u s 1 / s B r B R / 2 ,

where s > 0 is the exponent in Theorem 4.38, and where the implicit constants depend only on the allowable parameters. Let us take the claim for granted momentarily, and suppose that B r B R/2. Then B 2r B R , so we may apply Theorem 4.38 with 2r in place of R, to obtain

(4.48) Ω 4 r u s 1 / s inf Ω 2 r u inf Ω r u .

Combining (4.46), (4.47), and (4.48), we obtain (4.45), as desired.

It remains only to verify (4.47). To this end, observe that (4.46) is a weak reverse Hölder inequality, holding for all B r B R , including the case that B r is an interior ball, as well as the case that B r is centered on ∂Ω. Since weak reverse Hölder inequalities enjoy a well-known self-improvement property that allows the exponent on the right hand side to be taken arbitrarily close to zero (see, e.g., [15], Appendix B]), we find in particular that

Ω r u 2 1 / 2 Ω 2 r u s 1 / s , B r B R .

Suppose now that B r B R/2. Then B 2r B R , and we may replace r by 2r in the last estimate to obtain (4.47). □

As a corollary of the Harnack inequality, we obtain local Holder continuity up to the boundary.

Corollary 4.49.

Fix Ω R = Ω R (y 0) as above, with y 0 ∈ ∂Ω. Let K be the constant in Proposition 3.9. Suppose that uW 1,2(T R ) is a local solution to the Neumann problem with vanishing data in Ω4KR , in the sense that (4.44) holds. Then there exist C < ∞ and α > 0 depending only on dimension, ellipticity, ADR, (CS) and (HC), such that for all B r B R/2,

(4.50) u ( z ) u ( y ) C z y r α Ω 2 r u 2 d x 1 / 2 , y , z Ω r .

Proof.

We follow the standard argument, using the Harnack inequality. As in Theorem 4.43, we allow B r to be an interior ball, or to be centered on ∂Ω. Set

M ( r ) sup Ω r u , m ( r ) inf Ω r u .

and set

osc ( r ) M ( r ) m ( r ) .

Let B r B R/2, and let κ be the constant in Proposition 4.28. Note that in particular, in the case that B r is centered on ∂Ω, we then have that T r/κK ⊂ Ω r . Hence, we may then apply Theorem 4.43 with Ω r/κK in place of Ω R , to obtain that for some C 0 depending only on allowable parameters,

(4.51) M ( r / κ K ) m ( r ) = sup Ω r / κ K u m ( r ) C 0 inf Ω r / κ K u m ( r ) = C 0 m ( r / κ K ) m ( r ) ,

and also

(4.52) M ( r ) m ( r / κ K ) = sup Ω r / κ K M ( r ) u C 0 inf Ω r / κ K M ( r ) u = C 0 M ( r ) M ( r / κ K ) ,

Consequently, adding (4.51) and (4.52), we obtain

osc ( r ) + osc ( r / κ K ) C 0 osc ( r ) osc ( r / κ K ) ,

i.e., with θ = (C 0 − 1)/(C 0 + 1) < 1, we have

osc ( r / κ K ) θ osc ( r ) .

We now obtain (4.50) by a standard iteration argument, using Lemma 4.29 to bound osc(r) in terms of the L 2 average over Ω2r . □

As an immediate consequence of Corollary 4.49, we obtain local Hölder continuity of the Neumann function, up to the boundary.

Theorem 4.53.

Fix Ω R = Ω R (y 0) as above, with y 0 ∈ ∂Ω, and suppose that x ∈ Ω\Ω κKR , where κ is the constant in Proposition 4.28, and K is the constant in Proposition 3.9. Then the Neumann function N(x, ⋅) is Hölder continuous in Ω R/2, i.e., setting N x N(x, ⋅), we have

(4.54) N x ( z ) N x ( y ) z y R α Ω R N x 2 1 / 2 , y , z Ω R / 2 .

Remark 4.55.

Note that, after possible correction of the boundary trace of v x on a set of σ-meaure zero on ∂Ω, we may take N x to be Hölder continuous all the way to the boundary, i.e., (4.54) holds for all y , z B R / 2 Ω ̄ , and with x Ω ̄ \ B κ K R . The analogous statement holds for u in Corollary 4.49.

Our next goal is to prove the following Theorem:

Theorem 4.56.

Let Ω R n be an NTA domain with ADR boundary. Then for all x , y Ω ̄ , xy, we have

(4.57) N ( x , y ) x y 2 n .

Proof.

By Remark 4.55 (and its analogue with the roles of x and y reversed), N(x, y) extends continuously to the boundary in each variable separately, thus it is enough to prove (4.57) for x, y ∈ Ω. To this end, we recall (see Definition 4.24) that N(x, y) ≔ G(x, y) − v(x, y). By Corollary 4.22, we already know that

(4.58) | v ( x , y ) | δ ( x ) ( 2 n ) / 2 δ ( y ) ( 2 n ) / 2 .

Moreover, it is of course well known that

(4.59) G ( x , y ) | x y | 2 n .

Consequently, if |xy| < min(δ(x), δ(y)), then (4.57) holds. We may therefore suppose that

| x y | min δ ( x ) , δ ( y ) .

We claim that the latter estimate actually self improves, to give

(4.60) | x y | max δ ( x ) , δ ( y ) .

Indeed, suppose that, say, |xy| ≥ δ(x). Fix x ̂ Ω such that | x x ̂ | = δ ( x ) . Then

δ ( y ) | y x ̂ | | y x | + | x x ̂ | = | x y | + δ ( x ) 2 | x y | ,

which proves (4.60).

Consequently, we may assume that (4.60) holds, otherwise there is nothing to prove. But if (4.60) holds, then

| G ( x , y ) | | x y | 2 n δ ( x ) ( 2 n ) / 2 δ ( y ) ( 2 n ) / 2 ,

which, combined with (4.58), yields the following:

(4.61) | N ( x , y ) | δ ( x ) ( 2 n ) / 2 δ ( y ) ( 2 n ) / 2 .

Next, we claim that

(4.62) | N ( x , y ) | | x y | ( 2 n ) / 2 min δ ( x ) ( 2 n ) / 2 , δ ( y ) ( 2 n ) / 2 .

Let us now establish this claim. Again, we may suppose that (4.60) holds, otherwise we have nothing to prove. Set R ≔ |xy|. By Lemma 4.29 and Remark 4.33, Moser’s local boundedness estimate holds up to the boundary for N(x, ⋅), hence,

| N ( x , y ) | B ( y , R / 2 ) Ω | N ( x , z ) | 2 * 1 / 2 * B ( y , R / 2 ) Ω | G ( x , z ) | 2 * 1 / 2 * + B ( y , R / 2 ) Ω | v x ( z ) | 2 * 1 / 2 * R 2 n + R n / 2 * v x Y 1,2 ( Ω ) R 2 n + R ( 2 n ) / 2 δ ( x ) ( 2 n ) / 2 ,

since G satisfies (4.59), and v x satisfies (4.12), where we have used that −n/2* = (2 − n)/2, that

(4.63) | x z | | x y | = R , z B ( y , R / 2 ) ,

and that, by the corkscrew condition,

| B ( y , R / 2 ) Ω | | B ( y , R / 2 ) | R n .

On the other hand, by Corollary 4.21, N(x, y) = N*(y, x) ≔ G*(y, x) − v*(y, x), where G*(y, x) = G(x, y) is the adjoint Green function. Clearly, all of our estimates for N(x, y) also apply to the adjoint Neumann function N*(y, x), with the roles of x and y reversed, and therefore, since condition (4.60) is symmetric with respect to x and y, we may reverse the roles of x and y in the previous argument to get

| N ( x , y ) | | x y | ( 2 n ) / 2 δ ( y ) ( 2 n ) / 2 ,

which proves (4.62), as claimed.

With (4.62) in hand, for x ∈ Ω fixed, and for γ > 0 (small), set

K γ ( x ) sup y Ω : y x | N ( x , y ) | | x y | 2 n + γ | x y | ( 2 n ) / 2 δ ( x ) ( 2 n ) / 2 .

Note that

K γ ( x ) γ 1 < ,

for each γ > 0, by virtue of (4.62).

Thus, our goal is to show that K γ (x) ≤ C 0, uniformly for all x ∈ Ω, and for all small γ > 0; in that case, we then obtain (4.57) by letting γ → 0.

In fact, we shall prove that for all suitably small ɛ > 0, for all small γ > 0, and for arbitrary x ∈ Ω,

(4.64) 1 2 K γ ( x ) C ε + C ε K γ ( x ) ,

with C depending only on the allowable parameters, and with C ɛ depending on these and on ɛ. Choosing ɛ ≤ 1/(4C), we may hide the small term on the left hand side of the inequality, and we are done.

Let us now prove (4.64). Fix x ∈ Ω, and γ > 0. We may then choose y ∈ Ω\{x} such that

(4.65) 1 2 K γ ( x ) | N ( x , y ) | R 2 n + γ R ( 2 n ) / 2 δ ( x ) ( 2 n ) / 2 ,

where we have set R ≔ |xy|. We may suppose that δ(y) < R/10, otherwise (4.62) trivially implies (4.57).

Set

Ω R / 4 Ω B ( y , R / 4 ) .

Define also

Ω R / 4 + Ω R / 4 { z Ω : δ ( z ) ε 2 R } , Ω R / 4 Ω R / 4 \ Ω R / 4 + .

Observe that by the Corkscrew condition, for ɛ small enough we have

(4.66) | Ω R / 4 | R n .

Note also that by [13], Lemma 5.3],

(4.67) | Ω R / 4 | ε 2 R n .

Now, by Lemma 4.29 and Remark 4.33,

| N ( x , y ) | Ω R / 4 | N ( x , z ) | 2 d z 1 / 2 R n Ω R / 4 | N ( x , z ) | 2 d z 1 / 2 R n Ω R / 4 + | N ( x , z ) | 2 d z 1 / 2 + R n Ω R / 4 | N ( x , z ) | 2 d z 1 / 2 I 1 + I 2 .

Since Ω R / 4 + Ω R / 4 , by (4.62) and (4.66) we have

I 1 C ε R n | Ω R / 4 | 1 / 2 R 2 n C ε R 2 n C ε R 2 n + γ R ( 2 n ) / 2 δ ( x ) ( 2 n ) / 2 .

Furthermore, by (4.67)(4.63), and the definition of K γ (x),

I 2 ε K γ ( x ) R 2 n + γ R ( 2 n ) / 2 δ ( x ) ( 2 n ) / 2 .

Dividing through by R 2−n + γR (2−n)/2 δ(x)(2−n)/2, and using (4.65), we obtain (4.64), as desired. □

Corollary 4.68.

Let x , y , y 0 Ω ̄ , with |yy 0| < |xy 0|/2. Then

(4.69) N ( x , y ) N ( x , y 0 ) y y 0 α min x y 2 n α , x y 0 2 n α .

Proof.

As above, let κ and K be the constants in Propositions 4.28 and 3.9, respectively. Observe that it is enough to prove the result when 2κK|yy 0| < |xy 0|, because in the range 2|yy 0| < |xy 0| ≤ 2κK|yy 0|, we have |yy 0| ≈ |xy 0| ≈ |xy|, and the desired bound follows trivially from the pointwise estimate in Theorem 4.56.

Thus, we now suppose that 2κK|yy 0| < |xy 0|. Define R so that |xy 0| = κKR, hence x ∈ Ω\Ω κKR (y 0), and |yy 0| < R/2. Then by Theorem 4.53, Remark 4.55, and Theorem 4.56,

N ( x , y ) N ( x , y 0 ) y y 0 R α Ω R ( y 0 ) N 2 ( x , z ) d z 1 / 2 y y 0 R α Ω R ( y 0 ) x z 2 ( 2 n ) d z 1 / 2 y y 0 R α x y 0 2 n y y 0 α x y 0 2 n α y y 0 α min x y 2 n α , x y 0 2 n α

where we used that | x y | x y 0 x z for all zB(y 0, R), since in particular, κK > 2. □

5 The Neumann problem

In this section, we show that L p solvability of the Neumann problem, for some p > 1, paired with L p solvability of the regularity problem, and L p ′ solvability of the Dirichlet problem[4] for the adjoint operator L*, implies L q solvability of the Neumann problem for 1 < q < p. As before, we let L = −divA∇, and we interpret Lu = 0 in the weak sense. As above, for the sake of lightening the exposition, we shall continue to focus on the case that Ω and ∂Ω are unbounded. We leave it to the interested reader to make the appropriate adjustments needed to treat the case of a bounded domain.

We first discuss the variational Neumann problem.

5.1 W ̇ 1,2 weak solutions of the Neumann problem

We begin by constructing the weak W ̇ 1,2 solution to the Neumann problem with data f H ̇ 1 / 2 ( Ω ) , and we then establish several properties of the solution.

In the sequel, for f H ̇ 1 / 2 ( Ω ) , we let ⟨f, −⟩ denote the action of f as an element of H ̇ 1 / 2 ( Ω ) , acting on H ̇ 1 / 2 ( Ω ) . We remind the reader of the equivalence class notation in Remarks 3.1 and 3.3. Recall also that for [ Φ ] W ̇ 1,2 ( Ω ) , we have Tr [ Φ ] H ̇ 1 / 2 ( Ω ) .

Theorem 5.1.

Let Ω R n be an unbounded 1-sided CAD, with unbounded boundary. Then for each f H ̇ 1 / 2 ( Ω ) , there exists a unique (modulo constants) weak solution [ u ] W ̇ 1,2 ( Ω ) of the Neumann problem

L u = 0  in  Ω u ν A = f  on  Ω [ u ] W ̇ 1,2 ( Ω ) ,

i.e., there is a unique [ u ] W ̇ 1,2 ( Ω ) satisfying

(5.2) Ω A u Φ d x = f , Tr [ Φ ] , [ Φ ] W ̇ 1,2 ( Ω ) .

Conversely, if [ u ] W ̇ 1,2 ( Ω ) is a weak solution of the equation Lu = 0 in Ω, then f = ∂u/∂ν A exists as an element of H ̇ 1 / 2 ( Ω ) , and satisfies (5.2).

Proof.

By the trace estimate (3.6), the mapping [ Φ ] f , Tr ( [ Φ ] ) defines a bounded linear functional on W ̇ 1 / 2 ( Ω ) . The existence and uniqueness of such a [u] now follows directly from Lax-Milgram.

To establish the converse, following [4], we let E Ω : H ̇ 1 / 2 ( Ω ) W ̇ 1,2 ( R n ) be the extension operator (see (3.5)), and note that the mapping [ φ ] Ω A u E Ω φ d x , with [ φ ] H ̇ 1 / 2 ( Ω ) , defines a bounded linear functional on H ̇ 1 / 2 ( Ω ) , by virtue of the extension estimate (3.5). We denote this linear functional by [φ] ↦ ⟨f, [φ]⟩, with f H ̇ 1 / 2 ( Ω ) . Given [ φ ] H ̇ 1 / 2 ( Ω ) , we suppose now that [Φ] is any other extension of [φ] to W ̇ 1,2 ( Ω ) . Then Ψ Φ E Ω φ is a realization of an element of W ̇ 0 1,2 ( Ω ) , whence it follows that ∫Ω Au ⋅ ∇Ψ dx = 0, since Lu = 0 in the weak sense. Hence, (5.2) holds for all [ Φ ] W ̇ 1,2 ( Ω ) , so that f = ∂u/∂ν A in the variational sense. □

We shall continue to assume, as above, that Ω is a 1-sided CAD, until noted otherwise (i.e., until Section 5.2, at which point we shall impose the stronger 2-sided CAD condition).

Lemma 5.3.

Let uY 1,2(Ω) ∩ C(Ω) (observe that we do not assume that u is a solution). Then

(5.4) Ω A * ( y ) y N ( x , y ) u ( y ) d y = u ( x ) , a.e. x Ω .

Proof. Note that by (4.11) and the standard Riesz formula (4.26), we see that (5.4) holds under the somewhat more restrictive hypothesis that u Y 1,2 ( Ω ) C ( Ω ̄ ) , Tr ( u ) Lip c ( Ω ) . In particular,

(5.5) Ω A * ( y ) y N ( x , y ) Φ ( y ) d y = Φ ( x ) , Φ C c ( Ω ) .

Temporarily fix x ∈ Ω, and let 0 < ɛ < δ(x)/4. Fix ψ C c ( B ( 0,1 ) ) , with 0 ≤ ψ ≤ 1, ψ ≡ 1 on B(0, 1/2), and ψ 1 , and set ψ ε ( y ) ψ ( x y ε ) . For future reference we also set ψ γ ( y ) ψ ( x y γ ) , with 0 < ɛ < γ/2 < δ(x)/4. We then view the LHS of (5.4) as

(5.6) Ω u ( y ) A * ( y ) y N ( x , y ) d y lim ε 0 Ω u ( y ) A * ( y ) y N ( x , y ) 1 ψ ε ( y ) d y ,

We seek to show that the limit in (5.6) exists, and is equal to u(x) for a.e. x. To this end, momentarily fix γ > 2ɛ, and define ψ γ ( y ) ψ ( x y γ ) as above. Note that for ɛ < γ/2, we have that ψ ɛ = 0 in the support of 1 − ψ γ . Hence,

lim ε 0 Ω u ( y ) 1 ψ γ ( y ) A * ( y ) y N ( x , y ) 1 ψ ε ( y ) d y = Ω u ( y ) 1 ψ γ ( y ) A * ( y ) y N ( x , y ) d y = 0 , 0 < γ < δ ( x ) / 2 ,

by Lemma 4.25. It therefore remains to consider

lim ε 0 I ε , γ ( x ) lim ε 0 Ω u ψ γ A * N ( x , ) ( 1 ψ ε ) d y .

Fix a non-negative ρ C c B ( 0,1 ) , with ∫ρ = 1, let ρ γ (y) ≔ γ n ρ(y/γ), and set

u γ ρ γ u .

Note that u γ ψ γ C c ( Ω ) , since we assume that γ < δ(x)/2. Then for any such γ, by (5.5), we have

lim ε 0 I I ε , γ ( x ) lim ε 0 Ω u γ ψ γ A * N ( x , ) ( 1 ψ ε ) d y = Ω u γ ψ γ A * N ( x , ) d y = u γ ( x ) ,

since ψ γ (x) = 1. Then lim γ→0 lim ɛ→0 II ɛ,γ (x) = u(x) for all x ∈ Ω, since u γ u pointwise. It therefore remains only to show that

(5.7) lim γ 0 lim sup ε 0 | I ε , γ ( x ) I I ε , γ ( x ) | = 0 , a.e. x Ω .

To this end, set

E ε , γ ( x ) Ω ( u u γ ) ψ γ ψ ε A * N ( x , ) d y Ω ( u ψ γ u γ ψ γ ) A * ψ ε N ( x , ) d y = : E ε , γ ( x ) E ε , γ ( x ) ,

and observe that

I ε , γ ( x ) I I ε , γ ( x ) = Ω ( u u γ ) ψ γ A * N ( x , ) ( 1 ψ ε ) d y = Ω ( u u γ ) ψ γ ( 1 ψ ε ) A * N ( x , ) d y + E ε , γ ( x ) = 0 + E ε , γ ( x ) ,

since N(x, ⋅) is a weak adjoint solution in the support of the test function (uu γ )ψ γ  (1 − ψ ɛ ).

To handle the first error term E ε , γ , we observe that for γ < δ(x)/2, by definition of ψ γ ,

| E ε , γ ( x ) | u u γ L B x , δ ( x ) / 2 Ω | ψ ε ( y ) | | N ( x , y ) | d y ,

and in turn, by definition of ψ ɛ , and then Caccioppoli’s inequality and Theorem 4.56 (the size estimate for the Neumann function), we see that

Ω | ψ ε ( y ) | | N ( x , y ) | d y ε 1 ε / 2 | x y | ε | y N ( x , y ) | d y ε n 1 ε / 2 | x y | < ε | y N ( x , y ) | 2 d y 1 / 2 ε n 2 ε / 4 | x y | < 2 ε | N ( x , y ) | 2 d y 1 / 2 1 ,

Consequently, since u γ u uniformly on compacta in Ω (recall that u is continuous in Ω)

lim γ 0 lim sup ε 0 | E ε , γ ( x ) | lim γ 0 u u γ L B x , δ ( x ) / 2 = 0 , x Ω .

We now turn to the second error term E ε , γ . Since ɛ < γ/2, we see that ψ γ ≡ 1 on the support of ∇ψ ɛ . Hence, using the definition of ψ ɛ , and the size estimates for the Neumann function, we have

| E ε , γ | | ( u u γ ) | | ψ ε | | N ( x , ) | d y ε 1 n ε / 2 | x y | < ε | u ( y ) u γ ( y ) | d y ε M ( 1 Ω u ) ( x ) + M ( 1 Ω u γ ) ( x ) ,

where M denotes the Hardy-Littlewood maximal operator. Thus, lim sup ε 0 | E ε , γ ( x ) | = 0 , for a.e. x ∈ Ω, and for each fixed γ < δ(x)/2. This establishes (5.7), and hence also (5.4). □

Let L c , 0 2 ( Ω ) denote the collection of fL 2(∂Ω) with compact support, and mean value zero:

(5.8) L c , 0 2 ( Ω ) f L 2 ( Ω ) : f  has compact support on  Ω ,  and  Ω f d σ = 0 .

We note that L c , 0 2 ( Ω ) is dense in L p (∂Ω), for 1 < p < ∞ (as may be seen by approximating the identity by a sum of dyadic martingale difference operators; recall that we have assumed that ∂Ω is unbounded), and in the usual atomic Hardy space H 1(∂Ω). Furthermore, L c , 0 2 ( Ω ) H ̇ 1 / 2 ( Ω ) , and f L c , 0 2 ( Ω ) acts on [ φ ] H ̇ 1 / 2 ( Ω ) via integration:

(5.9) f , [ φ ] = Ω f φ d σ , f L c , 0 2 ( Ω ) , [ φ ] H ̇ 1 / 2 ( Ω ) .

Recall that N(x, y) ≔ G(x, y) − v x (y), where v x (y) was constructed in Lemma 4.6. In particular, there is an equivalence class [ Tr ( N ( x , ) ) ] H ̇ 1 / 2 ( Ω ) C ̇ α ( Ω ) , by Lemma 4.6 and Corollary 4.68.

Theorem 5.10.

Let f H ̇ 1 / 2 ( Ω ) , and let u be the particular Y 1,2(Ω) realization of the W ̇ 1,2 ( Ω ) weak solution to the problem (N) in Theorem 5.1. Then

(5.11) u ( x ) = f , [ Tr ( N ( x , ) ) ] , a.e. x Ω

(5.12) u ( x ) = Ω f ( y ) N ( x , y ) d σ ( y ) x Ω , f L c , 0 2 ( Ω ) .

Proof.

It is enough to prove (5.11) for f H ̇ 1 / 2 ( Ω ) , since (5.12) then follows by (5.9), and the continuity of u and N. In turn, by Lemma 5.3, it suffices to show that

(5.13) Ω u ( y ) A * ( y ) y N ( x , y ) d y = f , [ Tr ( N ( x , ) ) ] .

Fix x ∈ Ω, and let ψ C c ( B ( 0,1 ) ) , with 0 ≤ ψ ≤ 1, ψ ≡ 1 on B(0, 1/2), and ψ 1 . For 0 < ɛ < δ(x)/4, set ψ ε ( y ) ψ ( x y ε ) . As above, the LHS of (5.13) is defined as the limit in (5.6), and we need to show that this limit equals the RHS of (5.13). To this end, note that N(x, ⋅)(1 − ψ ɛ ) is a valid Y 1,2 test function, which equals N(x, ⋅) outside of B(x, ɛ) (in particular on the boundary). Hence,

(5.14) Ω A ( y ) u ( y ) y N ( x , y ) 1 ψ ε ( y ) d y = f , [ Tr ( N ( x , ) ) ] ,

uniformly for all sufficiently small ɛ, since u is the weak solution of (N). This proves (5.13), in light of (5.6). □

We next prove a localized version of the preceding result.

Lemma 5.15.

Let f L c , 0 2 ( Ω ) , with u as in Theorem 5.10. Then for Φ C c ( R n ) and a.e. x ∈ Ω,

(5.16) Φ ( x ) u ( x ) = Ω u A Φ N ( x , ) d y Ω N ( x , ) A u Φ d y + Ω N ( x , y ) Φ ( y ) f ( y ) d σ ( y ) .

Proof.

By Lemma 5.3, we have

Φ ( x ) u ( x ) = Ω A ( Φ u ) N ( x , ) d y , a.e. x Ω .

Consequently,

(5.17) Φ ( x ) u ( x ) = Ω u A Φ N ( x , ) d y Ω N ( x , y ) A u Φ d y + Ω A u ( N ( x , ) Φ ) d y .

Given x ∈ Ω, define ψ ε ( y ) ψ ( x y ε ) exactly as in the proof of Theorem 5.10, and note that (5.13) and (5.14) may be generalized as follows, by the same argument: since u is the weak solution of (N) with data f L c , 0 2 ( Ω ) , and N(x, ⋅) Φ (1 − ψ ɛ ) ∈ Y 1,2(Ω) is a valid test function that equals N(x, ⋅)Φ on the boundary, we have

(5.18) Ω A u N ( x , ) Φ d y = lim ε 0 Ω A ( y ) u ( y ) y N ( x , y ) Φ ( y ) 1 ψ ε ( y ) d y = f , [ Tr ( N ( x , ) Φ ) ] = Ω N ( x , y ) Φ ( y ) f ( y ) d σ ( y ) ,

where in the last step we have used (5.9). Plugging (5.18) into the last term in (5.17), we obtain (5.16) for a.e. x ∈ Ω. □

For future reference, we now record standard boundary Caccioppoli inequalities, with either vanishing Dirichlet data, or vanishing Neumann data. Recall that by Theorem 5.1, if [ u ] W ̇ 1 / 2 ( Ω ) solves Lu = 0 in Ω, in the weak sense, then ∂u/∂ν A exists in H ̇ 1 / 2 ( Ω ) .

Lemma 5.19.

Let r > 0, z ∈ ∂Ω, and as usual, set Ω r = Ω r (z) ≔ Ω ∩ B(z, r), and Δ r = ∂Ω ∩ B(z, r); similarly for Δ2r , Ω2r . Suppose that u is the Y 1,2(Ω) realization of a weak solution [ u ] W ̇ 1 / 2 ( Ω ) , with either (i) Tr(u) ≡ 0 on Δ2r ; or (ii) ∂u/∂ν A ≡ 0 on Δ2r ; or (iii) suppose that u = N(⋅, y), with y Ω \ Ω 2 r ̄ , or that u = N(⋅, y) − N(⋅, y′), with y , y Ω \ Ω 2 r ̄ . Then

(5.20) Ω r u 2 r 2 Ω 2 r u 2 d x

In addition, if K is the constant in Proposition 3.9, and 2* = 2n/(n − 2), then

(5.21) Ω r / K | u | 2 * 1 / 2 * Ω 2 r | u | 2 1 / 2

Proof.

Inequality (5.20) is standard in scenario (i), so we consider only scenarios (ii) and (iii). Let φ C 0 ( B ( z , 2 r ) ) with 0 ≤ φ ≤ 1, φ ≡ 1 on B(z, r), φ r 1 . Then

(5.22) Ω r u 2 d x Ω u 2 φ 2 d x Ω A u u φ 2 d x = Ω A u ( u φ 2 ) d x 2 Ω A u φ u φ d x I + I I .

In either case (ii) or (iii), 2Y 1,2(Ω), hence is the realization of [ u φ 2 ] W ̇ 1,2 ( Ω ) , with Tr[ 2] vanishing in the sense of H ̇ 1 / 2 ( Ω ) , outside of Δ2r . Consequently, I = 0, as follows in case (ii) by Theorem 5.1 (specifically (5.2), with f = ∂u/∂ν A ), and in case (iii) by Lemma 4.25 with the roles of x and y (and of A and A*) reversed.

As usual, for any ɛ > 0, we have

| I I | ε Ω | u | 2 φ 2 d x + ε 1 r 2 Ω 2 r u 2 d x .

Choosing ɛ small enough, depending only on ellipticity, we may hide the small term to obtain (5.20).

To prove (5.21), we apply first (3.15) with R = r/K, and then (5.20), to obtain

Ω r / K | u u Ω r / K | 2 * 1 / 2 * r 2 Ω r | u | 2 1 / 2 Ω 2 r | u | 2 1 / 2 .

Combining this estimate with the trivial bound | u Ω r / K | Ω 2 r | u | 2 1 / 2 , we obtain (5.21). □

Remark 5.23.

In the case that Ω2r = B 2r ⊂ Ω, estimate (5.21) is well known, with K = 1, and follows from (3.7) and the usual interior Caccioppoli inequality.

5.2 L p theory

We now turn to the issue of L p solvability. We begin by defining nontangential “cones” and maximal functions, as follows. For a sufficiently large constant K 1 = K 1(HC, CS), and given x ∈ ∂Ω, set

(5.24) Γ ( x ) y Ω : | x y | < K 1 δ ( y ) .

For K 1 large enough, Γ(x) defines a nontangential approach region such that each CS point relative to any ball B(x, r) is connected to the boundary point x via a Harnack path lying within Γ(x). Given R > 0, we also define truncated versions of the cones (from above, and from below, respectively):

(5.25) Γ R ( x ) Γ ( x ) y Ω : δ ( y ) < R , Γ R ( x ) Γ ( x ) y Ω : δ ( y ) R .

Following [4], we also define a variant of the nontangential maximal function as follows:

N ̃ ( u ) ( x ) = sup y Γ ( x ) B ( y , δ ( y ) / 2 ) u 2 1 / 2 .

as well as its truncated versions

(5.26) N ̃ R ( u ) ( x ) = sup y Γ R ( x ) B ( y , δ ( y ) / 2 ) u 2 1 / 2 , N ̃ R ( u ) ( x ) = sup y Γ R ( x ) B ( y , δ ( y ) / 2 ) u 2 1 / 2 .

We are now able to define L p solvability for the Neumann and regularity problems for L. We recall that the space L c , 0 2 ( Ω ) is defined in (5.8), and is dense in L p (∂Ω), 1 < p < ∞, and in Hardy space H 1(∂Ω). Moreover, L c , 0 2 ( Ω ) H ̇ 1 / 2 ( Ω ) (see (5.9)).

Definition 5.27.

Given f L c , 0 2 ( Ω ) , let [ u ] W ̇ 1,2 ( Ω ) be the weak solution to the Neumann problem given by Theorem 5.1, with data f. Let 1 < p < ∞. We say that (N) p is solvable for L if there is a uniform constant C such that for each f L c , 0 2 ( Ω ) (or each f L c , 0 2 L p when p > 2),

N ̃ ( u ) L p ( Ω ) C f L p ( Ω ) .

We say that ( N ) H 1 (H 1 denotes Hardy space) is solvable for L if there is a uniform constant C such that for each f L c , 0 2 ( Ω ) ,

N ̃ ( u ) L 1 ( Ω ) C f H 1 ( Ω ) .

Henceforth, we shall assume that Ω is a 2-sided CAD, unless noted explicitly to the contrary. In the latter setting, we recall, for future reference, some constructions and estimates concerning boundary Sobolev spaces presented in Ref. [16].

5.2.1 Boundary Sobolev spaces L ̇ 1 p ( Ω )

In Ref. [16], Section 3.6], it was shown that one can make sense of tangential derivatives on ∂Ω provided that ∂Ω is ADR, and that the measure theoretic boundary ∂Ω coincides σ-a.e. with the topological boundary. In particular, a CAD satisfies these conditions. The tangential derivatives are defined as follows: let ν = (ν 1, …, ν n ) be the measure theoretic outward unit normal to ∂Ω. Then for φ C 0 1 ( R n ) supported in a neighborhood of ∂Ω, set

τ i j φ = ν i ( j φ ) | Ω ν j ( i φ ) | Ω , i , j = 1 , , n .

Let g L loc 1 ( Ω , d σ ) , and suppose that for some p ∈ (1, ∞), and some positive M < ∞,

Ω g ( τ i j φ ) d σ M φ L p ( Ω ) , φ C 0 1 ( R n ) .

Since the restrictions of all such φ form a dense class in L p ′(∂Ω), it follows that there is a function hL p (∂Ω), with h L p ( Ω ) M , such that

Ω g ( τ i j φ ) d σ = Ω h φ d σ , φ C 0 1 ( R n ) .

In this case, we say that g has a weak tangential derivative τ i j g , and we set τ i j g h . Set

| D tan g | i , j = 1 n τ i j g 2 1 / 2 ,

provided that the tangential derivatives τ i j g , 1 i , j n exist in the weak sense. Using our usual notation to denote an equivalence class modulo constants, we define

L ̇ 1 p ( Ω , d σ ) [ g ]  with  g L loc p ( Ω ) : | D tan g | L p ( Ω ) ,

and we set

(5.28) [ g ] L ̇ 1 p ( Ω ) D tan g L p ( Ω ) .

We next define the Calderón-Scott [17] maximal operator

(5.29) Λ * g ( x ) sup r > 0 1 r Δ ( x , r ) g g Δ ( x , r ) d σ ,

for x ∈ ∂Ω, where Δ(x, r) ≔ B(x, r) ∩ ∂Ω, and g Δ(x,r) ≔ ⨏Δ(x,r) g. In Ref. [16], it is shown (see [16], (3.6.11), (4.3.19)]) that in a 2-sided CAD,

(5.30) D tan g L p ( Ω ) Λ * g L p ( Ω ) , 1 < p < ,

where of course the implicit constants depend only on p, n, ADR and the CS and HC constants for Ω and Ωext. We observe that (5.30) is an extension of a result of [17], first proved in the flat setting. We caution the reader that the precise statements of [16], (3.6.11)] and [16], (4.3.19)] are inhomogeneous versions of (5.28) and (5.30), respectively, but the proof in Ref. [16] actually yields the version of (5.30) stated here.

Recall that δ(x) ≔ dist(x, ∂Ω), for x ∈ Ω. We continue to follow the approach in Ref. [4].

Theorem 5.31.

Assume u W loc 1,2 ( Ω ) solves Lu = 0 in Ω, and N ̃ ( u ) L p ( d σ ) for some 1 < p < ∞. Then u converges nontangentially a.e. (dσ) to a limit g, with [ g ] L ̇ 1 p ( Ω ) .

Proof.

Fix z ∈ ∂Ω with N ̃ ( u ) ( z ) < . Let x, y ∈ Γ(z), each within a distance ɛ > 0 of z. To establish a.e. (dσ) existence of the nontangential limit, it is enough to show that for all such x, y,

(5.32) u ( x ) u ( y ) C ε N ̃ ( u ) ( z ) .

To prove (5.32), we may assume y z x z = ε . For K 1 in (5.24) large enough, by HC and CS, there exists a set of points { z k } k = 0 N Γ ( z ) , with z 0 = x, z N = y, N ≈ 1 + log(ɛ/|yz), z k+1B k B(z k , δ(z k )/8), and δ(z k ) ≈ δ(z k+1) ≈ θ k ɛ, for some uniform θ = θ(CS, HC) < 1.

Note that by construction, B k B k + 1 B k B ( z k , δ ( z k ) / 2 ) . Hence, with u B k B k u ,

u ( z k ) u ( z k + 1 ) | u ( z k ) u B k | + | u ( z k + 1 ) u B k | B k u u B k 2 1 / 2 + B k + 1 u u B k 2 1 / 2 B k u u B k 2 1 / 2 δ ( z k ) B k u 2 1 / 2 θ k ε N ̃ ( u ) ( z ) ,

where we have used Moser’s local boundedness estimate, then Poincaré’s inequality, and finally that by construction, δ(z k ) ≈ δ(z k+1) ≈ θ k ɛ. Since θ < 1, it then follows that

u ( x ) u ( y ) k = 0 N 1 u ( z k ) u ( z k + 1 ) k = 0 N 1 θ k ε N ̃ ( u ) ( z ) ε N ̃ ( u ) ( z ) .

Thus, (5.32) holds. Let g denote the nontangential limit of u. Letting yz nontangentially in (5.32), we see that in particular, for a.e. z ∈ ∂Ω,

(5.33) u ( x ) g ( z ) C ε N ̃ ( u ) ( z ) , x Γ ( z ) , | x z | ε .

It remains to establish that g is a representative of an equivalence class modulo constants with [ g ] L ̇ 1 p ( d σ ) . We observe first that g L loc p ( d σ ) . Indeed, given x ∈ ∂Ω, we let x Δ ∈ Ω be a corkscrew point relative to Δ ≔ Δ(x, 1), so that x Δ z 1 , and x Δ ∈ Γ(z), for all z ∈ Δ. Then

Δ g ( z ) p d σ ( z ) 1 / p Δ u ( x Δ ) p d σ ( z ) 1 / p + Δ u ( x Δ ) g ( z ) p d σ ( z ) 1 / p | u ( x Δ ) | + N ̃ ( u ) L p ( Ω ) ,

where we have used (5.33) with ɛ ≈ 1 and x = x Δ. By Moser’s local boundedness, and our hypothesis, the last expression is finite, thus g L loc p ( d σ ) .

Finally, we need to show that | D tan g | L p ( Ω ) . By (5.30), it is enough to verify that

(5.34) Λ g L p ( Ω ) N ̃ ( u ) L p ( Ω ) ,

where Λ g is defined in (5.29). To this end, fix x ∈ ∂Ω, let r > 0, set Δ r ≔ Δ(x, r), and let x Δ r Ω be a CS point relative to Δ r , so that x Δ r Γ ( z ) , and | x Δ r z | r , for all z ∈ Δ r . Then

1 r Δ r g g Δ r d σ = 1 r Δ r g ( z ) Δ r g ( y ) d σ ( y ) d σ ( z ) 2 r Δ r g ( z ) u ( x Δ r ) d σ ( z ) Δ r N ̃ ( u ) d σ ,

where in the last step we have used (5.33) with ɛ = r. Taking a supremum in r, and letting M denote the usual Hardy-Littlewood maximal operator, we obtain the pointwise bound

Λ g M N ̃ ( u ) .

The L p bound (5.34) now follows. □

5.2.2 The regularity problem (R) p

Definition 5.35.

Fix 1 < p < ∞. The regularity problem (R) p , with data in L ̇ 1 p ( d σ ) , is solvable for L if for each [ g ] L ̇ 1 p ( Ω ) H ̇ 1 / 2 ( Ω ) , the variational solution [ u ] W ̇ 1,2 ( Ω ) with data [g], with realization uY 1,2(Ω), satisfies

N ̃ ( u ) L p ( Ω ) g L ̇ 1 p ( Ω ) .

Moreover, [g] has a realization g such that u converges to g nontangentially a.e. on ∂Ω.

We shall also need the definition of the Dirichlet problem. Set

(5.36) N ( u ) ( x ) sup y Γ ( x ) | u ( y ) | , x Ω .

Definition 5.37.

Fix 1 < p < ∞. The Dirichlet problem (D) p, with data in L p ′(dσ) is solvable for L* if for each fL p ′(∂Ω), there is a unique solution u satisfying

L * u = 0  in  Ω u | Ω = f L p ( Ω ) N ( u ) L p ( Ω ) .

with u converging to f nontangentially a.e. on ∂Ω. Moreover, the solution u satisfies

N ( u ) L p ( Ω ) f L p ( Ω ) .

Remark 5.38.

It is well known that in a chord arc domain Ω R n (thus, ∂Ω is (n − 1)-dimensional ADR), solvability of (D) p for L* implies (in fact is equivalent to) the following local, scale invariant L p estimate for the adjoint Poisson kernel k L * :

(5.39) Δ ( z 0 , R ) k L * X p d σ 1 / p R 1 n ,

where X is any Corkscrew point with R/2 ≤ |Xz 0|≲ Rδ(X), and where the various implicit constants are uniform for all z 0 ∈ ∂Ω, and all R > 0 (with R < diam(Ω) in the case of a bounded domain). We refer the reader to Ref. [18], Chapter 2]. In fact, the chord arc assumption is not needed (provided that (5.39) is suitably interpreted), and ADR alone suffices (see [19]).

It was shown in Ref. [4] that L p solvability of the Neumann problem paired with L p solvability of the regularity problem on the ball (and more generally bounded starlike Lipschitz domains) implies L q solvability of the Neumann problem for 1 < q < p. We will follow rather closely the arguments used in Ref. [4], with appropriate adjustments to allow us to treat the class of 2-sided chord arc domains.

In the next lemma, we shall use the following simple geometric fact: if M 1 is chosen sufficiently large, depending only on allowable parameters, then given a surface ball Δ R / M 1 ( z 0 ) , for each truncated cone Γ R / M 1 ( z ) with z Δ R / M 1 ( z 0 ) , and for every point x Γ R / M 1 ( z ) , we have

(5.40) B ( x , 3 δ ( x ) / 4 ) B ( z 0 , R / 2 ) , x Γ R / M 1 ( z ) , z Δ R / M 1 ( z 0 ) .

Theorem 5.41.

(Localization of (R) p ). Suppose that Ω is an unbounded, 2-sided chord-arc domain, with unbounded boundary. Let 1 < p < ∞ and suppose that (R) p is solvable for L in Ω, and (D) p is solvable for L*. Let R > 0, and let ϒ R (z 0) = Ω2R (z 0)\Ω R (z 0) be an annulus contained in Ω centered at z 0 ∈ ∂Ω. Let u Y 1,2 ( Ω ) C ( Ω 2 R ( z 0 ) ̄ ) be a solution to Lu = 0 in Ω, with data g such that [ g ] L ̇ 1 p ( Ω ) H ̇ 1 / 2 ( Ω ) . Then for M 1 chosen as above,

(5.42) Δ R / M 1 ( z 0 ) | N ̃ R / M 1 ( u ) | p d σ Δ 2 R ( z 0 ) | D tan g | p d σ + R p Δ 2 R ( z 0 ) | g | p d σ + ϒ R ( z 0 ) | u | 2 d x p / 2 + R p ϒ R ( z 0 ) | u | 2 d x p / 2 .

Proof.

We follow the proof of [4], Theorem 5.19]. Let φ C 0 ( B 2 R ( z 0 ) ) be such that 0 ≤ φ ≤ 1, φ ≡ 1 on B R (z 0), and |∇φ| ≲ R −1. Then by the Riesz formula (4.26), for all ξ ∈ Ω,

( φ u ) ( ξ ) = Ω φ g d ω ξ + Ω A ( φ u ) G ( ξ , ) d y w ( ξ ) + F ( ξ ) .

Note that by (5.40), we have N ̃ R / M 1 ( u ) = N ̃ R / M 1 ( ( φ u ) ) in Δ R / M 1 ( z 0 ) , so it is enough to prove (5.42) with u replaced by w and by F on the left hand side of the inequality.

By the solvability of (R) p , we have

Δ R / M 1 ( z 0 ) | N ̃ R / M 1 ( w ) | p d σ 1 / p R ( 1 n ) / p φ g L ̇ 1 p ( Ω ) R 1 Δ 2 R ( z 0 ) | g | p d σ 1 / p + Δ 2 R ( z 0 ) | D tan g | p d σ 1 / p .

It remains to consider the contribution of ∇F. One may write F as

F ( ξ ) = Ω u A φ G ( ξ , ) d y + Ω φ A u G ( ξ , ) d y = Ω u A φ G ( ξ , ) d y Ω G ( ξ , ) A u φ d y F 1 ( ξ ) + F 2 ( ξ )

where we used the fact that u is a weak solution. From our choice of φ, we see that ∇φ vanishes in Ω\ϒ R (z 0). In particular, for all x as in (5.40), F 1 is a solution in B(x, 3δ(x)/4). It then follows that for such x, we can use Caccioppoli’s inequality to obtain

(5.43) B ( x , δ ( x ) / 2 ) | F 1 ( ξ ) | 2 d ξ 1 / 2 1 δ ( x ) B ( x , 3 δ ( x ) / 4 ) | F 1 ( ξ ) | 2 d ξ 1 / 2 .

Let ξB(x, 3δ(x)/4). Using (5.40), Caccioppoli’s inequality, Harnack’s inequality, and Carleson’s estimate, we obtain

| F 1 ( ξ ) | R 1 ϒ R ( z 0 ) | G ( ξ , ) | 2 d y 1 / 2 ϒ R ( z 0 ) | u | 2 d y 1 / 2 R n / 2 2 G ( ξ , x R ) ϒ R ( z 0 ) | u | 2 d y 1 / 2

where x R is a corkscrew point chosen so that R ≤ |x R z 0| ≲ Rδ(x R ). For z Δ R / M 1 ( z 0 ) , and x Γ R / M 1 ( z ) , we may apply (5.40) and Harnack’s inequality to obtain

B ( x , 3 δ ( x ) / 4 ) | G ( ξ , x x R ) | 2 d ξ 1 / 2 G ( x , x R ) .

Hence, for z Δ R / M 1 ( z 0 ) , and x Γ R / M 1 ( z ) , the CFMS estimate (Lemma 2.7) then implies that

1 δ ( x ) B ( x , δ ( x ) / 2 ) | F 1 ( ξ ) | 2 d ξ 1 / 2 R n 2 ϒ R ( z 0 ) | u | 2 d y 1 / 2 G ( x , x R ) δ ( x ) R n 2 ϒ R ( z 0 ) | u | 2 d y 1 / 2 ω x R Δ z , δ ( x ) δ ( x ) n 1 R n 2 ϒ R ( z 0 ) | u | 2 d y 1 / 2 M ( k L * x R 1 Δ 2 R ( z 0 ) ) ( z ) ,

where M denotes the usual Hardy-Littlewood maximal operator on ∂Ω. Combining the latter estimate with (5.43), and then using the definition of N ̃ R / M 1 ( F 1 ) and (5.39) (with 2R in place of R), we see that

Δ R / M 1 ( z 0 ) | N ̃ R / M 1 ( F 1 ) | p d σ 1 / p R n 2 ϒ R ( z 0 ) | u | 2 d y 1 / 2 Δ 2 R ( z 0 ) k L * x R p d σ 1 / p R 1 ϒ R ( z 0 ) | u | 2 d y 1 / 2 .

Similarly,

| F 2 ( ξ ) | R 1 ϒ R ( z 0 ) | G ( ξ , y ) | 2 d y 1 / 2 ϒ R ( z 0 ) | u | 2 d y 1 / 2 R n / 2 1 G ( ξ , x R ) ϒ R ( z 0 ) | u | 2 d y 1 / 2 .

Consequently, a slight variation of the previous argument yields the estimate

Δ R / M 1 ( z 0 ) | N ̃ R / M 1 ( F 2 ) | p d σ 1 / p ϒ R ( z 0 ) | u | 2 d y 1 / 2 .

We recall the definition of the truncated cones N ̃ R (see (5.25) and (5.26)).

Theorem 5.44.

(Localization of (N) p ). Let M 1 be the constant in Theorem 5.41. Let 1 < p < ∞ and suppose that (N) p and (R) p are solvable for L in Ω, and that (D) p is solvable for L*. Let R > 0 and define an annulus Ψ ̃ R = Ψ ̃ R ( z ) Ω 4 K 0 R ( z ) \ Ω R / 4 ( z ) , contained in Ω and centered at z ∈ ∂Ω. Let u be the particular Y 1,2(Ω) realization of the W ̇ 1,2 ( Ω ) weak solution to the problem (N) in Theorem 5.1, with data f L c , 0 2 (or in L c , 0 2 L p ( Ω ) , if p > 2). Suppose further that supp(f) ∩ Δ(z, 10K 0 R) = ∅. Then, setting gu|∂Ω, we have

(5.45) Δ R / ( 4 M 1 ) ( z ) | N ̃ R / ( 4 M 1 ) ( u ) | p d σ 1 / p R 1 Δ R / 2 ( z ) | g | p d σ 1 / p + Ω R | u | 2 d x 1 / 2 + R 1 Ψ ̃ R | u | 2 d x 1 / 2 ,

where Ω R = Ω R (z).

Proof.

We follow closely the proof of [4], Theorem 6.10]. By the solvability of (N) p and Theorem 5.31, g = u|∂Ω exists as a nontangential limit almost everywhere on ∂Ω, and [ g ] L ̇ 1 p ( Ω ) . Observe that since f L c , 0 2 , the solution u and its boundary data g satisfy the hypotheses of Theorem 5.41, so applying the latter with R/4 in place of R, we need only show that g verifies

(5.46) Δ ( z , R / 2 ) τ i j g p d σ 1 / p 1 R Ψ ̃ R u 2 d x 1 / 2 , 1 i , j n .

Let Φ C c ( B ( z , 2 R ) ) with Φ ≡ 1 on B(z, R), and Φ R 1 . By Lemma 5.15, if x ∈ Ω R/2(z), then u = I + II + III, the three terms occurring on the right hand side of (5.16) in Lemma 5.15. Note that III = 0, since f and Φ|∂Ω have disjoint supports.

Next, note that II = II(x) = −∫Ω N(x, ⋅) Au ⋅ ∇Φ dy is continuous in Ω R / 2 ( z ) ̄ , by Corollary 4.68 (with the roles of x and y reversed), since ∇Φ is supported in B(z, 2R)\B(z, R). We let h be a Lipschitz function supported in Δ(z, R/2), with h L p 1 , and fixing i, j ∈ {1, 2, …, n}, we test II against τ i j h . Note that the latter function belongs in particular to L c , 0 2 ( Ω ) . Let v denote the Y 1,2 solution to the adjoint Neumann problem with data τ i j h . By the adjoint version of (5.12), we have that

v ( y ) = Ω N ( x , y ) τ i j h ( x ) d σ ( x ) .

It then follows that

(5.47) Ω I I ( x ) τ i j h ( x ) d σ ( x ) = Ω A u ( y ) Φ ( y ) Ω N ( x , y ) τ i j h ( x ) d σ ( x ) d y = Ω A u ( y ) Φ ( y ) v ( y ) d y .

Since τ i j h L with compact support, we have v C ( Ω ̄ ) , and we claim that

(5.48) v | Ω L p ( Ω ) 1 .

Indeed, let θ L c , 0 2 ( Ω ) , with θ L p ( Ω ) = 1 , and let w be the Y 1,2 solution of the Neumann problem with data θ, so that w(x) = ∫∂Ω N(x, y)θ(y)dσ(y), by (5.12). Then,

Ω θ ( y ) v ( y ) d σ ( y ) = Ω τ i j h ( x ) Ω N ( x , y ) θ ( y ) d σ ( y ) d σ ( x ) = Ω h ( x ) τ i j w ( x ) d σ ( x ) .

The solvability of (N) p implies that N ̃ ( w ) L p θ L Ω p , hence τ i j w L p ( Ω ) 1 , by Theorem 5.31. Consequently, since h L p 1 , we have

Ω θ ( y ) v ( y ) d σ ( y ) h L p ( Ω ) τ i j w L p ( Ω ) 1 .

Since L c , 0 2 ( Ω ) is dense in L p (∂Ω), we may take the supremum over all such θ to establish the claimed estimate (5.48).

Since (D) p is solvable for L*, we have

(5.49) N v L p ( Ω ) v L p ( Ω ) 1 ,

where as usual, N is the standard nontangential maximal operator defined in (5.36).

Note that the (adjoint) conormal derivative of v is τ i j h , hence the support of N A * v is contained in Δ(z, R/2). As usual, we set Ψ R = Ψ R ( z ) = Ω K 0 R ( z ) \ Ω R ( z ) . We claim that

(5.50) Ψ R v 2 1 / 2 Ψ ̃ R v p 1 / p ,

and (for future reference)

(5.51) Ψ R v 2 1 / 2 R 1 Ψ R * v 2 1 / 2 R 1 Ψ ̃ R v p 1 / p ,

where Ψ R * Ω 2 K 0 R ( z ) \ Ω R / 2 ( z ) is a fattened version of Ψ R , with Ψ R Ψ R * Ψ ̃ R . Indeed, if p < 2 (hence p′ > 2), (5.50) is simply Hölder’s inequality, with no fattening necessary, and (5.51) follows from (5.50) and Caccioppoli’s inequality (including the boundary version in Lemma 5.19), after first covering Ψ R by a uniformly bounded number of subdomains of the form Ω cR = Ω ∩ B cR , where the balls B cR are centered in Ω ̄ , and c is a sufficiently small, uniform positive constant. We omit the routine details. In the case p > 2, we again cover Ψ R as above, and use the fact that v already enjoys an L 2 L 2 * weak reverse Hölder inequality at scales rR, as follows from the adjoint version of (5.21) and Remark 5.23; the bound (5.50) then follows from the well known fact that weak reverse Hölder inequalities satisfy a self-improvement property that allows any positive exponent on the right hand side (see, e.g., [15], Appendix B]). To obtain (5.51), we combine Caccioppoli’s inequality with the same self-improvement property. Again we omit the details.

Further, by (5.49),

(5.52) Ψ ̃ R v p 1 / p R ( n 1 ) / p .

Since supp(∇Φ) ⊂ Ω2R  \Ω R ⊂ Ψ R , and Φ R 1 , by (5.47), we have

Ω I I τ i j h d σ R 1 R n Ψ R u 2 1 / 2 Ψ R v 2 1 / 2 R n 1 R ( n 1 ) / p Ψ R u 2 1 / 2 ,

where in the last step we have used (5.50) and (5.52). In turn, taking a supremum over all Lipschitz h as above (i.e., with support in Δ(z, R/2), and h L p ( Ω ) 1 ), we see that

Δ ( z , R / 2 ) | τ i j I I | p d σ 1 / p Ψ R u 2 1 / 2 R 1 Ψ ̃ R u 2 1 / 2 ,

where the last estimate follows from Caccioppoli’s inequality (up to the boundary), exactly like the first inequality in (5.51).

It remains to consider term I = I(x) = ∫Ω uA∇Φ ⋅∇N(x, ⋅)dy (the first term on the right hand side of (5.16) in Lemma 5.15). We observe that I(x) is continuous in Ω R / 2 ( z ) ̄ , as may be seen by applying Cauchy-Schwarz to the integral defining I(x) − I(x′), with x , x Ω R / 2 ( z ) ̄ , and then using first the Caccioppoli inequality of Lemma 5.19, and then Corollary 4.68 (each with the roles of x and y reversed). Thus, as for term II, with h as above, we test against τ i j h , to obtain

Ω I ( x ) τ i j h ( x ) d σ ( x ) = Ω u A Φ v d y ,

where, exactly as before, v is the solution of the adjoint Neumann problem with data τ i j h . Proceeding as we have done for term II, we have

Ω I τ i j h d σ R 1 R n Ψ R u 2 1 / 2 Ψ R v 2 1 / 2 R n 1 R ( n 1 ) / p R 1 Ψ R u 2 1 / 2 ,

where in the last setp we have used (5.51) and (5.52). Consequently,

Δ ( z , R / 2 ) | τ i j I | p d σ 1 / p R 1 Ψ ̃ R u 2 1 / 2 .

Since I + II = u = g in Δ(z, R/2), we obtain (5.46), as desired. □

Before proceeding to the main result in this section, let us recall that for 1 < s ≤ ∞, an H 1 s-atom is a function a satisfying

  1. supp(a) ⊂ Δ(y 0, r), for some surface ball Δ(y 0, r) ⊂ ∂Ω.

  2. aL s (∂Ω), with a L s ( Ω ) σ ( Δ ( x , r ) ) 1 / s r ( n 1 ) / s , and s′ = s/(s − 1).

  3. ∂Ω a dσ = 0.

It is of course well known that for each s > 1, and each fH 1, there is a decomposition f = ∑ k λ k a k , where a k is an s-atom for each k, and f H 1 inf k | λ k | , where the infimum runs over all such decompositions (with s fixed). Note that in particular, an s-atom belongs to L c , 0 2 ( Ω ) , if s ≥ 2, and conversely, any f L c , 0 2 ( Ω ) belongs to H 1(∂Ω). Note also that if a is an s-atom adapted to a surface ball Δ of radius r, then

(5.53) a L p ( Ω ) σ ( Δ ) 1 / p r ( n 1 ) / p , 1 p s .

Theorem 5.54.

Suppose that (N) p and (R) p are solvable for L, and (D) p is solvable for L*, for some 1 < p < ∞. Then (N) q is solvable for L, for all 1 < qp, and so is ( N ) H 1 (Hardy space solvability).

Proof.

We follow the strategy previously used in Ref. [4] (and to some extent, [20]). It suffices to establish the Hardy space estimate

(5.55) N ̃ ( u ) L 1 ( Ω ) f H 1 ( Ω ) .

for f L c , 0 2 ( Ω ) , where [ u ] W ̇ 1,2 ( Ω ) is the weak solution to the Neumann problem given by Theorem 5.1, with data f. In that case, the map f N ̃ ( u ) will then be bounded from L p to L p , and from H 1 to L 1, so we may obtain L q solvability, 1 < q < p, by interpolating.

In turn, to prove (5.55), it suffices to show that if a is an H 1 2-atom (or an H 1 p-atom, if p > 2), then

(5.56) N ̃ ( u ) L 1 ( Ω ) 1 ,

where u is the solution to (N) p with data a, and the implicit constant is uniform. To this end, we fix a 2-atom a (or a p-atom, if p > 2), adapted to a surface ball Δ(y 0, r), and let u be the solution to the Neumann problem with data a. We now make the standard decomposition into dyadic annuli:

N ̃ ( u ) L 1 ( Ω ) = Δ ( y 0 , 4 r ) N ̃ ( u ) d σ + k = 2 S k N ̃ ( u ) d σ I 0 + k = 2 I k ,

where S k = S k ( y 0 ) Δ y 0 , 2 k + 1 r \ Δ y 0 , 2 k r .

We first dispose of the “local term”:

I 0 r ( n 1 ) / p N ̃ ( u ) L p ( Ω ) r ( n 1 ) / p a L p ( Ω ) 1 ,

where in the last two steps we have used the assumed L p solvability, and then (5.53).

We now treat the “far away terms” ∑I k . Let M 1 be the constant in Theorem 5.41, and in Theorem 5.44. Let k ≥ 2, and choose R = 2 k r/(100K 0) ≈ 2 k r, so that S k may be covered by a uniformly bounded number of surface balls of the form Δ j ≔ Δ(z j , R/(4M 1)), with z j S k , and with the further property that Δ(y 0, r) ∩ Δ(z j , 10K 0 R) = ∅ for each j. We can then bound N ̃ ( u ) by a sum of its truncated versions (see (5.25) and (5.26), and write

I k j = 1 M Δ j N ̃ R / ( 4 M 1 ) ( u ) d σ + Δ j N ̃ R / ( 4 M 1 ) ( u ) d σ I k + I k ,

where M = M(n, ADR, M 1, K 0). Thus, it suffices to establish the desired estimate for each individual summand in I k and I k . We fix z = z j , and Δ = Δ(z, R/(4M 1)) = Δ j as above. We first treat I k . Since a is supported in Δ(y 0, r), we can apply Theorem 5.44 to obtain

(5.57) Δ ( z , R / ( 4 M 1 ) ) N ̃ R / ( 4 M 1 ) ( u ) p d σ 1 / p R 1 Δ R / 2 ( z ) | g | p d σ 1 / p + Ω R | u | 2 d x 1 / 2 + R 1 Ψ ̃ R | u | 2 d x 1 / 2 R 1 Δ R / 2 ( z ) | g | p d σ 1 / p + R 1 Ω R * | u | 2 d x 1 / 2 ,

where Ψ ̃ R = Ψ ̃ R ( z ) Ω 4 K 0 R ( z ) \ Ω R / 4 ( z ) , and Ω R * Ω 4 K 0 R ( z ) , and where in the last step, we have used the boundary Caccioppoli inequality of Lemma 5.19. By our choice of R, we have

| x y | > 2 k 1 r 2 r > 2 | y y 0 | , x Ω R * ̄ , y Δ ( y 0 , r ) .

Consequently, for all such x and y, Corollary 4.68 implies that

(5.58) N ( x , y ) N ( x , y 0 ) y y 0 α x y 2 n α 2 k α ( 2 k r ) 2 n 2 k α R 2 n .

Therefore, since a has mean value zero and is supported in Δ(y 0, r),

(5.59) | u ( x ) | = Ω N ( x , y ) N ( x , y 0 ) a ( y ) d σ ( y ) 2 k α R 2 n , x Ω R * ̄ ,

by the case p = 1 of (5.53). Here, of course, u(x) = g(x), when x Δ R / 2 ( z ) Ω Ω R * ̄ . Hence, I k is the sum of a uniformly bounded number of terms, each satisfying an estimate of the form

Δ ( z , R / ( 4 M 1 ) ) N ̃ R / ( 4 M 1 ) ( u ) d σ R n 1 Δ ( z , R / ( 4 M 1 ) ) N ̃ R / ( 4 M 1 ) ( u ) d σ R n 2 Δ R / 2 ( z ) | g | p d σ 1 / p + R n 2 Ω R * u 2 d x / 2 2 k α ,

where we have used (5.57) and then (5.59) (including its boundary version) in the last two steps. Summing in k, we obtain k = 2 I k 1 .

It remains to treat the terms I k . Let k ≥ 2. We note that given any y 1 ∈ Ω with δ(y 1) ≥ R/(4M 1), and any xB(y 1, 3δ(y 1)/4), we have

| x y 0 | | x y | δ ( x ) R / ( 16 M 1 ) 2 k r r > | y y 0 | , y Δ ( y 0 , r ) .

For such x, we therefore deduce from Corollary 4.68 (and Theorem 4.56) that (5.58), and hence also (5.59), continue to hold. It follows that for any y 1 ∈ Ω, with δ(y 1) ≥ R/(4M 1), by Caccioppoli’s inequality we have

B ( y 1 , δ ( y 1 ) / 2 ) u 2 1 / 2 δ ( y 1 ) 1 B ( y 1 , 3 δ ( y 1 ) / 4 ) u 2 1 / 2 2 k α R 1 n .

By definition (see (5.25) and (5.26)), the latter bound then holds for N ̃ R / ( 4 M 1 ) ( u ) ( x ) , for all x ∈ ∂Ω. Therefore,

Δ ( z , R / ( 4 M 1 ) ) N ̃ R / ( 4 M 1 ) ( u ) d σ 2 k α R 1 n R n 1 = 2 k α .

Since I k is the sum of a uniformly bounded number of such terms, we see that k = 2 I k 1 . □


Corresponding author: Steve Hofmann, Department of Mathematics, University of Missouri, Columbia, MO 65211, USA, E-mail: 

Dedicated to Prof. Robert Fefferman.


Acknowledgments

The first author is supported by NSF grant DMS-2349846.

  1. Research ethics: Not applicable.

  2. Informed consent: Not applicable.

  3. Author contributions: Both authors contributed to this work.

  4. Use of Large Language Models, AI and Machine Learning Tools: None declared.

  5. Conflict of interest: The authors state no conflict of interest.

  6. Research funding: This work was supported by NSF under grant DMS-2349846.

  7. Data availability: Not applicable.

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Received: 2024-09-03
Accepted: 2025-01-30
Published Online: 2025-02-25

© 2025 the author(s), published by De Gruyter, Berlin/Boston

This work is licensed under the Creative Commons Attribution 4.0 International License.

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