Perron’s Method and Wiener’s Theorem for a Nonlocal Equation

We study the Dirichlet problem for non-homogeneous equations involving the fractional p-Laplacian. We apply Perron’s method and prove Wiener’s resolutivity theorem.


Introduction
The object of our work is to study the Dirichlet problem where s ∈ (0, 1), p ∈ (1, ∞) and This is a non-linear counterpart to the fractional Laplace Equation We have found the Perron method suitable to investigate the boundary behaviour. The celebrated method, introduced in 1923 by O. Perron (cf. [31]) for the Laplace Equation, has a wide range of applications. It requires little, mainly that a comparison principle is valid. It has been adapted to many non-linear equations and appears in the modern theory of viscosity solutions, too. We shall study the Dirichlet boundary value problem in a bounded domain in R n for an equation that arises in the following way. The problem of minimizing the variational integral among all functions v with given boundary values g in the sense that v −g ∈ W s,p 0 ( ) leads to the Euler-Lagrange Equation |y − x| n+sp dx dy = f φ dx, (1.3) for all φ ∈ C ∞ 0 ( ). Here 0 < s < 1, 1 < p < ∞, and f ∈ L ∞ (R n ). All the appearing functions are defined in the whole space R n , although the bounded domain and its boundary ∂ are at the focus. Notice that the boundary values g are prescribed not only on ∂ but in the whole complement R n \ , so that the exterior values count.
The equation can be written as for a. e. x ∈ , provided that the principal value of the integral converges so well that L s p v ∈ L 1 loc ( ). The so defined L s p is a non-linear and non-local "fractional differential operator" often called the fractional p-Laplacian. Operators of this type were first studied in [19], but in a slightly different form and in the viscosity sense. The notation L s p v = −(− p ) s v suggested, for instance, in [5], [20], [15] and [11] stems from the linear case p = 2, when one can use the Fourier transform to define (− ) s v(x) = (2π |ξ |) 2sf (ξ).

In this case
Unfortunately, the simplification offered by the Fourier transform is not available in the non-linear situation p = 2. In principle, the limit as s 1 leads, under suitable normalization to the much studied p-Laplace operator (see for instance Theorem 3.1 in [8], [1], and [2]). Then formulas (1.2), (1.3), and (1.4) reduce to |∇v| p−2 ∇v · ∇φ dx = f φ dx, div |∇v| p−2 ∇v = −f. (1.5) Let us return to the boundary value problem (1.1). We assume that • p ∈ (1, ∞) and s ∈ (0, 1) • is a bounded domain • f ∈ L ∞ ( ) • g ∈ C(R n ) ∩ L ∞ (R n ) and nothing else. We emphasize that is otherwise quite arbitrary. Its boundary ∂ can even have positive n-dimensional volume! The method works well under more general assumptions on the boundary values. 1 First, the boundedness of g can be relaxed a lot. Second, the continuity is not needed for Perron's method itself, but for "arbitrary" functions g the more interesting properties would fail.
The main concepts are the upper class U g , which contains supersolutions, and the lower class L g which contains subsolutions. Perron's method produces two pointwise defined functions: It follows from the construction that these so-called Perron solutions are ordered: They belong to W s,p loc ( ) and are continuous except possibly on the boundary ∂ . Our first theorem assures that they are solutions, indeed they are local weak solutions (Theorem 22). The method is consistent: if the boundary values g also belong to W s,p (R n ), then the Perron solutions coincide with the unique minimizer of the variational integral (1.2), but in our work we include more general g's.
The question of resolutivity is central: do the Perron solutions coincide? By the constructions, any reasonable solution is squeezed between them. We prove the counterpart to Wiener's celebrated resolutivity theorem in [33] (with prescribed continuous boundary values): That the Perron solutions coincide does not mean that they must attain the correct boundary values at all boundary points. (For example, in the range sp < n isolated boundary points become "ignored". The situation is the same for the ordinary Laplace equation.) We can (for continuous g) drop the distinction and write h g = h g = h g . We say that a given boundary point We show that the property of being regular does not depend on the right-hand side of the equation, see the theorem below and Proposition 30. Thus, to discriminate the regular boundary points, it is enough to consider the special case L s p v = 0. It is surprising how simple the proof is, while the corresponding result for the p-Poisson equation in (1.5) requires, as it were, the Wiener criterium, see [29] and [30].

Theorem 2 A boundary point is either regular for all equations
simultaneously or irregular for all of them. Here s and p are fixed.
If sp > n then the Sobolev embedding guarantees the right boundary values, so that all boundary points are regular. We pay attention to the case sp < n. Regularity is equivalent to the existence of a barrier, an auxiliary function, see our Theorem 26. An interesting consequence, pointed out by Lebesgue for the Laplace equation, is that the more complement around a boundary point the domain has, the better the regularity is: Proposition 3 Suppose that we have two domains so that 1 ⊂ 2 and a common boundary point ξ 0 ∈ ∂ 1 ∩ ∂ 2 . If ξ 0 is regular with respect to 2 , so it is with respect to 1 .
We do not treat the boundary regularity any further. A finer analysis would require more elaborate tools. Neither do we introduce concepts like "(s, p)-capacity." The merit of our treaty is that also for arbitrary domains we obtain results, even for right-hand sides f ≡ 0, comparable to corresponding parts in classical Potential Theory. Needless to say, the method itself is not new. We can partly follow the procedure in [17], which was outlined for equations like (1.5), and for the Resolutivity Theorem we use some ideas from [28]. As always, some proofs are straightforward adaptations, but also new, even surprising, difficulties appear. Although, the focus is on the boundary, the exterior values (R n \ ) might interfere in a disturbing way, so that a slight change of g far away from may effect how the "correct" boundary values are attained. On the other hand, sometimes the exterior helps, as in the surprisingly simple proof of Theorem 2. Points in the exterior are always regular, without any requirement on ∂ , as expected.
Finally, we mention that the lack of suitable, explicitly known, strict supersolutions has lead us to include a chapter of calculations; in particular the case sp ≤ 1 is demanding, since the general theory of the Sobolev Spaces W s,p is of little help in this range.
There are several sophisticated problems that are beyond the reach of our present methods.

Preliminaries
In this section we shall present some background. Preliminary results are also included.

Spaces and notation
The fractional Sobolev spaces W s,p (R n ) with 0 < s < 1 and 1 < p < ∞ are assumed to be known by the reader. We refer to "Hitchhiker's guide to the fractional Sobolev spaces" [13] for a good introduction. The norm is defined through The space W s,p 0 (D) is defined as the closure of C ∞ 0 (D) with respect to the norm of W s,p (R n ). We also note that if u belongs to W s,p (R n ), so do u + , u − , and |u|. The same holds with W s,p 0 (D). Remark 4 In general, it is not the same to take the closure in the norm with the double integral taken only over a subset D : However, when sp = 1 and D has Lipschitz boundary, the resulting space is the same, see Proposition B.1 in [6].
Proof From the hypotheses it follows that we can find φ i ∈ C ∞ 0 (B) and 0 ≤ η i ∈ W s,p (R n ) such that φ i → φ and η i → η in W s,p (R n ). Consider the sequence It is clear that ψ i → η strongly in L p and weakly in W s,p (R n  Proof It is enough to prove that f is continuous when 1/2 < |x| < 2. Denote by K(x) the set consisting of those y so that where y x is the closest point to x such that |y x | = |y|. Then

Notion of solutions
To be on the safe side, we define some basic concepts.

Definition 7
We say that a function u ∈ W s,p (R n ) is a weak supersolution of for every nonnegative φ ∈ W s,p 0 ( ). A weak subsolution is defined by reversing the inequality.
A weak solution is a function u ∈ W s,p (R n ) satisfying We shall also need the corresponding local concepts. By a local weak solution of the equation one means a function u ∈ W s,p loc ( ) ∩ L ∞ (R n ) which satisfies the above equation for all test functions 2 φ ∈ C ∞ 0 ( ). The local weak sub-and supersolutions have a similar definition. By regularity theory a (local) weak solution is continuous in , upon a change in a set of measure zero, see Theorem 1.2 in [12], Theorem 3.13 in [7] (basically Theorem 1.5 in [23]), Theorem 1 in [25] or Theorem 5.4 in [18], all in slighly different settings. Here, Theorem 5.4 in [18] will do. We shall always assume this continuity. Furthermore, (local) weak supersolutions can be made lower semicontinuous in by changing them in a set of measure zero. See [22] for this regularity result. In the same way, (local) subsolutions are upper semicontinuous 3 . As a consequence, the functions are defined at every point. In addition, in Corollary 3.6 in [7], the following Caccioppoli inequality for solutions of L s p u = −f in B R , is proved: We also seize the opportunity to mention the recent papers [10] [24] where non-local equations of type (1.1) have been studied. In addition, in the papers [4], [3] and [9] different types of non-local nonlinear equations are introduced.
Existence The existence of solutions comes easily from the variational integral J (v) defined in (1.2).

The minimizer is a weak solution in .
Proof The existence of a minimizer u is standard and follows by the direct method in the calculus of variations. Since the semi-norm is strictly convex, the minimizer is unique. The Euler-Lagrange equation follows from This is the desired equation.

Remark 9
In the literature the minimization is sometimes taken over all those u ∈ W s,p ( ) for which u = g almost everywhere in the complement R n \ . This will not do here. Even if g were smooth, this does not always yield the same solution in irregular domains. For example, the punctured disk 0 < |x| < 1 has the boundary point 0, which cannot be ignored when sp > n. However, if has a Lipschitz boundary, these two problems are equivalent, see Proposition B.1 in [8].

Comparison Principle For two equations
Proof This is Lemma 9 in [27], where it is shown that If sp > 1 we can directly conclude from the Sobolev boundary values that the constant C is zero. If 0 < sp < 1 one can use the fact that ψ + (x)ψ − (y) = 0, where the notation is as in Lemma 9 in [27].

Proposition 11 If a local weak subsolution u and a local weak supersolution v in satisfy
Proof Let ε > 0. By the hypotheses, the set {u > v + ε} is compactly contained in . (It may be empty.) Therefore, the function (u − v − ε) + lies in W s,p loc ( ) and vanishes outside K for some K ⊂⊂ . Plugging in (u − v − ε) + as a test function yields the same type of inequalities as in the proof Theorem 10. We omit the details.
Definition 12 (Comparison) We say that the lower semicontinuous function v satisfies the comparison principle from above (for L s A similar definition goes for the comparison principle from below of upper semicontinuous functions u. Proposition 11 states that local weak supersolutions satisfy the comparison principle from above. In fact, the converse is also true. We are content to prove a special case below, using an idea from [26]. For instance, the weaker assumption that v is lower semicontinuous and bounded would do. The virtue of the proof is that it does not need the interior continuity of the solution to the obstacle problem, which is known only in the case with zero right-hand side (see [21]).

Proposition 13
Suppose v ∈ W s,p (R n ) ∩ C(R n ) ∩ L ∞ (R n ) satisfies the comparison principle from above in . Then v is a local weak supersolution in .
Proof Fix v and let D be an open set such that D ⊂⊂ . We shall show that v is the solution to an obstacle problem, and as such it is a local weak supersolution. Let u be the minimizer of the variational integral J (u) in (1.2) among the class If only continuous admissible functions are considered, one gets the same u. We do not use the knowledge that u is continuous. We can find a minimizing sequence If they are empty for infinitely many k, we have u ≤ v + 2ε. If not, we can find a regular set (satisfying the exterior ball condition) D k,ε such that Since D k,ε satisfies the exterior ball condition, Lemma 17 implies that there is a weak In addition, since h k = u k ≤ v + 2ε outside D k,ε , the comparison principle valid for v implies h k ≤ v + 2ε in R n . We also note that by construction J (h k ) ≤ J (u k ). However, the function h k is not necessarily admissible. The idea is to construct a suitable admissible function, using h k . We note that outside D k,ε , u k ≤ v + 2ε and as u k ∈ A we have u k ≥ v. By Lemma 29 (proved at the end of the paper), one can modify h k far away so that it becomes a super-or a subsolution with zero right-hand side. Then by comparing with constant functions we obtain where we may assume that M > v L ∞ (R n ) + 1.
Take ε < 1/2, and fix λ ∈ (0, ε/(4M)). Consider the convex combination We note that outside D k,ε we have w k = u k and in D k,ε From the inequality we conclude that the sequence w 1 , w 2 , w 3 , . . . is minimizing. Thus it has a subsequence that converges weakly in L p to u. The subsequence is even strongly convergent in L p , as all minimizing sequences are. We extract a further subsequence that converges a.e. to u. Now the defining identity (2.3) forces the convergence h k → u a.e. Since h k ≤ v + 2ε we can conclude that u ≤ v + 2ε. Hence, u ≤ v + 2ε in any case. Since ε can be chosen how small as we wish, u ≤ v. By construction also u ≥ v and the proof is complete.

Remark 9
To proceed to a more general case with a lower semicontinuous v ∈ L ∞ (R n ) satisfying the comparison principle, one can apply the Proposition to the infimal convolutions which obey the comparison principle, if v does. We do not proceed any further here with this matter.

Exterior Sphere Condition and Poisson Modification
When it comes to the continuity of functions in W s,p (R n ), the three cases sp < n, sp = n, sp > n are different. When sp > n the space contains merely continuous functions. For radial functions, which we shall use as auxiliary tools, this continuity comes for free also in the wider range sp > 1, see Lemma 6.
The case sp < 1 is problematic because it contains functions that not at all obey the boundary values. For example, it is straight forward to verify that the characteristic function of the unit ball, belongs to W s,p (R n ) when sp < 1. In fact, it also belongs to W s,p Note that when sp < 1, the W s,p -seminorm of the characteristic function of a set D is exactly the sp-fractional perimeter (cf. [16]), which is finite if for instance ∂D is Lipschitz.
where B R \ B r denotes the ring domain r < |x| < R. In order to use ω as a barrier 4 , we need to assure that it attains its boundary values.

Theorem 15
The function ω is continuous in R n and radial. Furthermore, ω > 0 in B R \B r .
Proof By the comparison principle ω ≥ 0. Since ω is the unique minimizer, it must be a radial function. (Otherwise, a rotation would spoil the uniqueness.) Thus we know that ω is continuous in R n in the case sp > 1, because of the Sobolev inequality in one variable, see Lemma 6. This settles the case sp > 1.
The boundary values require a delicate analysis which relies on Lemma 35. First, note that sup ω < ∞ by comparison with the function in (4.1). The function Multiplying by a large constant we see that the function w obeys the rules By the comparison principle Thus ω is continuous across the sphere ∂B r . The outer boundary values are not essential to us, but we argue as follows. By rotation an arbitrary boundary point can be brought to the position (−R, 0, . . . , 0). Now the comparison is possible, see Lemma 6. This yields the continuity across the outer sphere.
Finally, we have to show that ω has no zeros in the ring domain. It is convenient to consider the function (x 1 − r) from Proposition 32 in the half-space x 1 > r Multiply it by a small constant to achieve . This prevents ω from having zeros in the half-space x 1 > r. By rotational symmetry, this is enough.
Remark 16 Lemma 4.3 in [18] uses a more implicit construction of the auxiliary barrier needed for Theorem 15.
Perron's method presupposes that in some simple domains the equation can be solved with continuous boundary values, for example in all balls. (For the Laplace equation Poisson's formula originally took care of this initial step.) We shall need this possibility to construct a so-called Poisson modifcation of our functions.

Remark 18
The ball B can be replaced by any domain that satisfies the exterior sphere condition. This includes all domains with a boundary of class C 2 .
Proof Fix a boundary point ξ 0 . Let ε > 0. Since g is continuous, there is δ > 0 such that Let the exterior ball B(y 0 , δ) be tangent to B at ξ 0 so that B(y 0 , δ) ∩ B = {ξ 0 }. Let ω be the barrier in Theorem 15, constructed for the ring δ < |x − y 0 | < R, where R is taken so large that B ⊂ B(y 0 , R). We can choose a large constant C so that It follows that By increasing C, if necessary, we can also guarantee that G(x) is a local weak supersolution of (2.1) in B. We see 5 that we get the opposite inequality lim inf Recall that the right-hand member f of our equation is bounded. Let ψ : → R be a bounded, smooth function, say in C 1 (R n ). Select a regular subdomain D ⊂⊂ . By Lemma 17 and Remark 18 the solution of the problem belongs to C(R n ) ∩ W p,s (R n ). We use the notation for this function. Notice that it coincides with ψ outside D. For the lack of a better name, we say that the function P (ψ, D) is the Poisson modification of ψ with respect to the domain D.
Here ψ was smooth. We want to extend the definition to a bounded and lower semicontinuous function v, obeying the comparison principle from above in . In order to modify it in a ball B ⊂⊂ we use an approximation argument. By semicontinuity there are smooth functions ψ j such that ψ j v in R n .
We can define the Poisson modification of v in B as the limit We spell out the details below. We define v j = P (ψ j , B) Since ψ j is smooth, Lemma 17 implies that each v j is continuous across the boundary of the ball B. Thus v j ∈ C(R n ). By Theorem 10, the sequence v 1 , v 2 , . . . is increasing and v j ≤ v. The pointwise limit exists and satisfies V ≤ v. It is lower semicontinuous. We have (3.1) In B the constructed function V is the limit of an increasing sequence of solutions. We claim that V itself is a local weak solution in B. Since ψ j → v and v is bounded, we may assume that the sequence v j is uniformly bounded. If not, we may consider the uniformly bounded sequence of Lipschitz functions which also converges to v. From the Caccioppoli estimate for local weak solutions (2.3) we have for any K ⊂⊂ B. This implies that v j converges weakly in W s,p (K) and strongly in L p (K). Now pick a smooth test function φ ∈ C ∞ 0 (K).
Since v j is a local weak solution in B, By splitting the integrals into integrals over K and R n \ K we have In the integral we may pass to the limit using the weak convergence in W s,p (K). In the integral |x − y| is bounded away from zero, so that we may use the convergence of v j in L p to pass to the limit. Hence, for any φ ∈ C ∞ 0 (K). Since K is an arbitrary open set compactly contained in B, it follows that V is a local weak solution in B.
Next, we prove that V satisfies the comparison principle. Take D ⊂⊂ and let h ∈ C(D) be a local weak solution in D such that h ≤ V in the complement R n \ D and By the comparison principle, In the case B ∩ D is empty, the result follows from (3.3). Therefore, let us study V more closely in the set B ∩ D assuming that it is not empty. In B ∩ D, V is a local weak solution. In R n \ D, h ≤ V by the hypothesis on h and in D ∩

Perron's Method
Consider again the equation in the domain , where f ∈ L ∞ (R n ). We aim at constructing the solution with given boundary values g ∈ L ∞ (R n ). For simplicity we assume also that g ∈ C(R n ). Perron's method produces two ordered solutions, which do or do not assume the boundary data continuously. The upper class U g consists of all functions v such that is lower semicontinuous and obeys the comparison principle from above in .
The lower class L g consists of all functions u such that is upper semicontinuous and obeys the comparison principle from below in .
The upper class contains all lower semicontinuous weak supersolutions satisfying the boundary conditions. If v belongs to the upper class, so does its Poisson modification V = P (v, B). The property v 1 , v 2 , . . . , v k ∈ U g =⇒ min{v 1 , v 2 , . . . , v k } ∈ U g for the pointwise minimum is decisive.
We define pointwise the upper solution: h g (x) = inf v∈U g v(x), the lower solution: h g (x) = sup u∈L g u(x), when x ∈ R n . It follows directly that We shall avoid the cases ±∞. As the name suggests, the Perron solutions are solutions, indeed. See Theorem 22. A general observation is that if there exists a local weak solution, say h g that attains the boundary values at every point in R n \ (in particular on ∂ ), then The reason is that now h g itself is a member of both classes. Thus the method is consistent.

Remark 20
Since g is bounded, we may restrict ourselves to bounded functions in the upper and lower classes. In the case with right-hand side f = 0, we can simply cut the functions with constants. Otherwise we use some radially decreasing smooth function (it only matters that 0 < C(x) < 1 between the spheres). It satisfies and so every v in U g may be replaced by min{v, aC+b} without affecting the upper solution h g . -The same goes for the lower class and h g .

Continuity of the Perron Solutions
We shall prove that the Perron solutions satisfy the differential equation. Our first step is to establish continuity at interior points.

Proposition 21 The upper and the lower Perron solutions are continuous functions in .
Proof By symmetry, it is enough to write the proof for the upper solution. Let x 0 ∈ and take ε > 0. We will then show that if r is small enough, Take x 1 , x 2 ∈ B r (x 0 ). We can then find two decreasing sequences v 1 i and v 2 i of functions in U g so that

Take a larger ball B R (x 0 ) and use the Perron modification
Then Now, from Theorem 5.4 in [18], we have for x 1 , x 2 ∈ B r (x 0 ), since we can assume that the sequence v i is uniformly bounded (cf. Remark 20), so that V i is uniformly bounded as well. By choosing r small enough, we thus have Letting x 1 and x 2 change places, we obtain sup for r small enough, which completes the proof.

A central property.
We are ready to prove:

Theorem 22 (Perron) The upper and the lower Perron solutions are local weak solutions in .
Proof We prove it only for h g . First we construct a monotonically decreasing sequence of functions w i ∈ U g converging to h g at each rational point in R n . Let q 1 , q 2 , . . . be an enumeration of the rational points. At each rational point q k there is a sequence v k

The function
w i = min{v 1 i , . . . , v i i } also belongs to U g , and in particular

Boundary Values and Barriers
Consider some boundary point ξ 0 ∈ ∂ . The boundary point is called regular if for all continuous g. (By Wiener's resolutivity theorem below, h g = h g .) From the proof of Lemma 17 we can read off the following sufficient condition for regularity, the so-called exterior sphere condition:

The same holds for the lower solution.
It is plain that if the exterior sphere condition holds at each boundary point, then the Perron solutions coincide: h g = h g for continuous g.
Continuous "boundary values" can never fail at points away from ∂ .

The same holds for the lower solution.
Proof Let x 0 ∈ R n \ and fix ε > 0. Since g is continuous, there is δ > 0 such that and let ω be the function in Theorem 15 with B r replaced by B δ/2 (x 0 ) and B R replaced by B R (x 0 ). Then ω is a strictly positive function in R n \ B δ (x 0 ). Consequently, we can find a C > 0 such that By choosing C larger if needed, we can also guarantee that Cω is a local weak supersolution of (2.1) in . Therefore, the function g(x 0 ) + ε + Cω(x) ∈ U g so that Similarly, we can prove In particular, at the point x = x 0 , we have since ω(x 0 ) = 0 Since ε is arbitrary, this yields the desired result.
This last result in the exterior is valid without any assumptions on the boundary of .

Definition 25
We say that the function γ is a barrier at the point ξ 0 ∈ ∂ , if In passing, we see that the expedient function in Theorem 15 does not quite satisfy the definition, since it is zero in a small ball. (Actually, we used infinitely many functions there.)

Theorem 26 (Lebesgue) The boundary point ξ 0 is regular if and only if there exists a barrier at ξ 0 .
Proof That the existence of a barrier is sufficient for regularity follows by the same reasoning as in the proof of Lemma 17. To see this, just replace ω there by γ .
For the necessity, we construct a barrier by solving the problem where we select the the boundary values g so that they satisfy To this end we may assume that ⊂ B 1 and that ξ 0 = 0. One may use by the regularity assumption on ξ 0 . By the comparison principle γ (x) ≥ g(x) > 0 when x = ξ 0 . This shows that γ will do as a barrier. Now Proposition 3 in the Introduction follows immediately.

Wiener's Resolutivity Theorem
The resolutivity theorem, originally formulated in 1925 by N. Wiener for harmonic functions (cf. [33]), states that the Perron solutions coincide under fairly general conditions: the domain is arbitrary, the boundary values are continuous, and the right-hand side is bounded.
Let be a bounded domain. (We allow its boundary ∂ to be arbitrary; it may even have positive n-dimensional volume.) Consider the equation The boundary values g are continuous and bounded: g ∈ C ∞ (R n ) ∩ L ∞ (R n ). Perron's method produces two solutions such that We shall prove that h g = h g . This is called resolutivity and the common solution is denoted by h g .
The obstacle problem. We use an obstacle problem as an auxiliary tool. Given a smooth ψ ∈ L ∞ (R n ) we let ψ act as an obstacle in order to obtain a suitable supersolution. Let D be a domain and consider the problem of minimizing the variational integral Here the obstacle ψ also induces the boundary values.
By the direct method in the Calculus of Variations, the existence of a unique minimizer v is established. It is plain to verify that the solution of the obstacle problem is a weak supersolution. Thus it is lower semicontinuous by [22], again by the modification in Lemma 29. It is a solution in the open set {v > ψ} ∩ D where the obstacle does not hinder (under our assumption that ψ is continuous). This obstacle problem has been studied more closely in [21].
The resolutivity. We turn to the resolutivity question. For the proof of Theorem 1 we assume that g ∈ C(R n ) ∩ L ∞ (R n ) and f ∈ L ∞ ( ). We claim that the upper and the lower solutions of (1.1) coincide.
Proof Theorem 1 A reduction to the case when g ∈ C ∞ (R n ) is possible. To see this, let ε > 0. There is a function φ ∈ C ∞ (R n ) such that We deduce that h g ≤ h g . Always, h g ≥ h g and so h g = h g . Therefore we may assume that g ∈ C ∞ (R n ).
The proof is based on the expedient fact that the solution to the equation with the boundary values g taken merely in the sense of the Sobolev norm W s,p 0 ( ) is unique. Needless to say, this solution does not have to belong to any of the classes U g and L g : it may ignore pointwise described boundary values. An obstacle problem makes it possible to control the boundary values in the procedure.
Let the function g act as an obstacle and consider the problem of minimizing the integral (1.2) among all functions u ≥ g in R n belonging to W s,p (R n ) with boundary values u−g ∈ W s,p 0 ( ). Let v denote the unique minimizer. It is decisive that v − g ∈ W s,p 0 ( ) and that v ≥ g. In addition, v is a local weak supersolution in and therefore v ∈ U g .
Take an exhaustion of with regular domains D j : To us it is enough that each D j satisfies the exterior sphere condition so that the Poisson Hence the pointwise limit W = lim V j exists and, in addition, the V j s converges locally weakly in W s,p (R n ) and locally in L p . Repeating the arguments in the proof of Proposition 21, one can prove that, being locally the limit of solutions, W itself is a local weak solution in . We is a local weak solution in and satisfies u − g ∈ W s,p 0 ( ). There is only one such function W (Theorem 8). Notice carefully that W may fail to belong to the upper class U g . But what counts now is that each V j ∈ U g , and so, at every point Repeating the procedure using subsolutions from below we arrive at with the same unique solution W to the Dirichlet problem with boundary values in Sobolev's sense. We conclude that h g = h g .

Remark 27
The proof above reveals that the exterior limit holds for every ξ ∈ ∂ . Thus the correct boundary value g(ξ ) may fail only for the interior limit.

Changing the Right-Hand Side
In this section we show how one can transform a supersolution of L s p u ≤ 1 to a supersolution of L s p u ≤ 0 by pulling it down far away.
is a supersolution of L s p u ≤ 0 in B 1 , provided that the constant M is large enough. Here η is a smooth function satisfying and 0 ≤ η ≤ 1.
Proof Let φ be a non-negative test function in C ∞ 0 (B 1 ). We claim that from which the desired result can be read off. To this end, we now compare the corresponding integrals. First of all, we can split the integrals as since φ has compact support in B 1 . The integrals over B 2 × B 2 remain the same. Notice that (φ(x) − φ(y)) dxdy = φ(x) dxdy when y ∈ B 2 . We note that The inequality above was valid pointwise, sinceũ(x) −ũ(y) ≥ u(x) − u(y) when x ∈ B 1 , y ∈ B 4 \ B 2 . The last integral can be estimated as where we used Lemma 28 and the fact that u(x) − u(y) ≥ −2, taking M ≥ max(3, 2 u L ∞ (R n ) ). The claim follows.

Regular Points
As a corollary of the resolutivity theorem, we obtain Theorem 2, according to which the right-hand side f of the equation L s p u = f has no influence of the regularity of a boundary point. This is an immediate consequence of the proposition below. We keep s, p, and fixed, but change the right-hand side. Proof Suppose first that ξ 0 is regular with respect to −1. Denote by u ± the (unique Perron) solutions of L s p u ± = − ± 1 in , u ± = g in R n \ , and by u the solution of Then u − ≤ u ≤ u + . Since ξ 0 is regular with respect to −1 it is also regular with respect to +1. Hence, so that ξ 0 is regular with respect to −f . Now suppose ξ 0 is regular with respect to −f . Let u be the solution of Then v ≥ u so that lim inf since ξ 0 is regular with respect to −f . Now considerṽ = v + Mη, where η ≥ 0 is a smooth function supported at a positive distance . Then, as in Lemma 29, for M large enough L s pṽ ≥ 1 andṽ =g = g + Mη in R n \ . Let w be the solution of Then w ≥ṽ so that Therefore, ξ 0 is regular with respect to −1.

Explicit "Barriers"
In the non-linear case there are few explicit examples of sp-harmonic functions in the literature. In Lemma 3.7 in [8] we find that in R n . (The computations diverge without the restriction on s.) See also [14] for explicit computations when p = 2.

Concave Functions
since the integrand is an odd function with respect to the variable x − y, given that the integral at infinity converges, 6 of course. Hence the concave function is a supersolution. 6 A careful arrangement of the calculation shows that for every non-negative φ ∈ C ∞ 0 (R n ).

Infimal Convolution
In passing, we mention that the so-called infimal convolution preserves the inequality L s p v ≤ −1. This holds under fairly general assumptions.

The Positive Part
The function is a solution to the equation L s p u = 0 in the half-space n · x = n 1 x 1 + · · · + n n x n > 0.
See Lemma 3.1 in [18]. We shall also need a strict supersolution of this type. We begin with the one-dimensional case.

Lemma 31 In one dimension
where the constant C(β, s, p) > 0 for 0 < β < s. In the case β = s we have Proof Let u(y) = (y + ) β . Keep 0 < x < ∞. We split the integral in the formula into three parts: The part y ≤ 0: where we have substituted y = tx, dy = xdt.
In the last integral we substitute Let us first consider the case 0 < β < s. In that case there is no problem about convergence, and letting ε → 0 some terms cancel so that only the following identity is left: where x > 0. It is plain that the integral has the same sign as β − s, since the factor [t p(s−β) − 1] determines this. This settles the case β < s. If s = β most terms cancel immediately, but in this case the integral becomes It approaches zero as ε → 0, although the complete integral diverges at the endpoint t = 1 if s > 1 − 1 p . This proves that L s p (x + ) s = 0 when x > 0.
In order to use x s + as a minorant, we modify it near infinity so that it becomes the bounded function:

Proposition 32
There is a number δ > 0 such that Proof The only change in the previous calculations is that the right-hand member 0 of the equation L s p = 0 should be replaced by twice the negative quantity Upon a partial integration this correction becomes twice the integral When x L is small enough, the integral is of the magnitude s −1 ( x L ) s and the correction is of the magnitude 2 p−1 sp L −s . The desired estimate follows.
Proof The infimum over all the half-plane supersolutions n · (n − x) β + , where n is the exterior unit normal, is a supersolution in their common domain. (If β = s, one even gets a strict supersolution.)

A Radial "Barrier"
The proof of Theorem 15 about the fundamental barrier function was based on the following result about the radial function Lemma 35 Let 0 < β < s. There is a δ = δ(r 0 , s, β, p, n) > 0 such that L s p ω(r) ≤ −1 when r 0 < r < r 0 + δ.
Writing L s p ω(r) in spherical coordinates and integrating away the azimuth φ we obtain the expression 2 · 2π In toto, we obtain by adding the integrals left in the three cases It is of utmost importance that this integral is strictly negative when 0 < β < s. It will dominate over the other terms provided that r − r 0 is small. The next one is II = Notice that this integral is uniformly small when 0 < r − r 0 < δ = a small number. The third term is The factor (r − r 0 ) in front of the term III makes it small compared to the main term. This term is problematic, because the integral would not converge at t = 0 without the expression in braces. We can write 7 where the Mean Value Theorem was used for the function (1 + y) 1−sp and −t < ξ < t + 2r 0 t r−r 0 . Thus we have the estimate where we have gained one power of t. Split the integral in IV as where the above estimate shows that we now can make the quantity (r − r 0 ) σ 0 · · · dt as small as we wish, by adjusting σ . The remaining integral over [σ, 1] multiplied by r − r 0 approaches zero, as r → r 0 .
In conclusion, the main term dominates in some small interval r 0 < r < r 0 + δ. This was the case sp = 1.
The case sp = 1 requires minor modifications. Doing the same calculations again, the final formula (8.1) has again four integrals. The main term I is the same. So is II. But in III and IV one has to change the expressions of the type 1 1 − sp . . .
using the limit procedure After these replacements, the proof goes as above.
One dimension The function ω 1 (x) = (|x| − 1) β + of one variable satisfies the equation below, when |x| > 1. We omit the calculations, which are of the same kind as in the three dimensional situation, although shorter. We remark that the function (|x| 2 − 1) β + seems to 7 The case sp ≤ 1 is urgent, but the estimation also works for sp > 1. be nicer, but it offers us no advantages in the calculations, on the contrary it produces longer expressions.
The first integral is strictly negative when 0 < β < s and the two other integrals approach zero as x → 1 + 0 or x → −1 − 0. Again we get a strict supersolution in a thin "shell" 1 < |x| < 1 + δ.