Ahmed YOUSSFI

National School of Applied Sciences Sidi Mohamed Ben Abdellah University-Fez Laboratory of Mathematical Analysis and Applications My Abdellah Avenue, Road Imouzer, P.O. Box 72 F`es-Principale, 30 000, Fez, Morocco

E-mail: ahmed.youssfi@gmail.com; ahmed.youssfi@usmba.ac.ma

Ghoulam OULD MOHAMED MAHMOUD

National School of Applied Sciences Sidi Mohamed Ben Abdellah University-Fez My Abdellah Avenue, Road Imouzer, P.O. Box 72 F`es-Principale, 30 000, Fez, Morocco

E-mail: ghoulam.ouldmohamedmahmoud@usmba.ac.ma

Abstract We study the existence and the regularity of solutions for a class of nonlocal equations involving the fractional Laplacian operator with singular nonlinearity and Radon measure data.

Key words fractional Laplacian; singular elliptic equations; measure data

1 Introduction

Lately, problems involving nonlocal operators and singular terms have recently received considerable attention in the literature. A good amount of investigations have focused on the existence and/or regularity of solutions to such problems governed by the fractional Laplacian with a singularity due to a negative power of the unknown or described by a potential, see for instance, [1, 2, 4, 7, 8, 12]and related papers.

A prototype of nonlocal operators is the fractional Laplacian operator of the form (??)s,0 < s < 1, which is actually the infinitesimal generator of the radially symmetric and sstable Lévy processes [6]. Fractional Laplacian operators naturally arise from a wide range of applications. They appear, for instance, in thin obstacle problems [14], crystal dislocation[18],phase transition [30]and others.

In this paper, we are interested in the existence and regularity of solutions to the following Dirichlet problem

where ? is an open bounded subset in RN, N > 2s, of class C0,1, s ∈ (0,1), γ > 0, f is a non-negative function on ?, μ is a non-negative bounded Radon measure on ? and (??)sis the fractional Laplacian operator of order 2s defined by

where “P.V.” stands for the integral in the principal value sense and α(N,s) is a positive renormalizing constant, depending only on N and s, given by

so that the identity (??)su = F?1(|ξ|2sFu), ξ ∈ RN,s ∈ (0,1) and u ∈ S(RN) holds, where Fu stands for the Fourier transform of u belonging to the Schwartz class S(RN) (cf. [23]).More details on the operator (??)sand the asymptotic behaviour of α(N,s) can be found in[17].

The case s = 1 corresponds to the classical Laplacian operator. If further μ = 0, an important result is due to Lazer-McKenna[24]. Under regularity assumptions on ? and f, the authors present an obstruction to the existence of an energy solution. In fact, such a solution lying inshould exists if and only if γ < 3 while it is not inif γ > 1. As far as problem with L1-data are concerned, the threshold 3 essentially due to the boundedness of the datum was sharpened in [32]while in [11]the existence of a distributional solution u is proved.In fact, it is proved in [11]that if γ < 1 and f ∈ Lm(?), 1wherewhileif f ∈ Lm(?) withIn the case where f ∈ L1(?), if γ = 1 then u ∈whileif γ > 1. We note that in the latter case, the boundary datum is only assumed in a weaker sense than the usual one of traces, that isLet us point out here that solutions with infinite energy may exist if γ > 1 even for smooth data ([24]).

The nonhomogeneous case (i.e.,has been considered. In [26]the authors studied the existence of weak solutions for the problem

where f ∈ L1(?) and μ is a bounded Radon measure. They prove the existence of a weak solution u of the problem (1.2) such thatfor everywhen γ ≤ 1 while if γ > 1, u ∈for everywith the regularityTkbeing the truncation function at levels±k. Other related singular equations can be found for instance in[13, 15, 21, 27, 31].

Regarding nonlocal problems,the study of(1.2)withμ=0 was extended in[7,12]where the Laplacian is substituted by the fractional Laplacian01. The authors obtain some existence and regularity results for the solutions depending on the summability of the datum f and γ (splitting in the cases γ <1, γ =1, γ >1). Some fractional equations with measure data are studied in [5, 20, 28].

It is our purpose in this paper, to consider the problem (1.1) in the nonlocal framework and prove existence results of solutions to problem(1.1)with μ a bounded Radon measure and data f ∈ L1(?). We use an approximation method that consists in analyzing the sequence of approximated problems truncating the datum f and the singular termand approximatingμ by smooth functions, obtaining non singular problems with L∞-data whose approximated solutions uncan be obtained by a direct application of the Schauder fixed point theorem. We faced many difficulties in dealing with the nonlocal problem (1.1), but the main one is how to get estimations in appropriate fractional Sobolev spaces.

Observe that in the local setting, if the approximated solutions are such that the sequence{?un}nis uniformly bounded in the Marcinkiewicz spacethen we conclude that the sequence {un}nis uniformly bounded in the Sobolev spacesfor every(see [9]).

However, we underline here that given the fractional structure of the operator of the principal part, we can not retrieve the gradient of the approximate solutions and so appears the problem of getting a priori estimates in some fractional Sobolev spaces. To overcome this difficulty, we first prove the key result Lemma 4.1 (Section 4) and use suitable test functions and algebraic inequalities that enable us to get appropriate a priori estimates in both cases γ ≤ 1 and γ >1.

The paper is organized as follows: in Section 2 we give some basic notations and necessary results that we will use in the accomplishment of the paper. We also give the main results. In Section 3, we construct a series of approximate problems to which we show the existence and uniqueness of the approximate solution. In Section 4, we prove some a priori estimates of the approximate solutions in fractional Sobolev spaces. Section 5, is devoted to the proof of the main results (Theorems 2.7 and 2.8). While in Section 6 we give a regularity result. Finally, in Appendix we expose and prove the technical and functional results that we used in the previous sections.

2 Some Useful Notations and Main Results

In this section we provide some basic facts about fractional Sobolev spaces. We refer to[10, 16, 17, 29]for more details. Let ? be an open subset in RNand let C? :=RN?. For any 0

Ws,q(?), also known as Aronszajn, Gagliardo or Slobodeckij spaces, is a Banach space when equipped with the natural norm

It can be regarded as an intermediate space between Lq(?)and W1,q(?). Recall that the space Ws,q(?) is reflexive for all q >1 (see [22, Theorem 6.8.4]). We point out that if 0

Here and in the sequel Supp f stands for the support of the function f.is a Banach space under the norm

If ? is bounded and is of class C0,1,we can give a fractional version of the Poincaré inequality in1 ≤ q < +∞, whose proof in the case where q = 2 can be found in [3]. For the convenience of the reader, we are giving the proof here.

Lemma 2.1(fractional Poincaré-type inequality) Let ? be a bounded open subset of RNof class C0,1, 1 ≤ q <+∞ and let 0

ProofLetObserve first that the above inequality holds if ? = 0. Assume thata nd set

We shall prove that λ(?) > 0. To do so, we argue by contradiction assuming that λ(?) = 0.Thus, there exists a sequence {?n} ofsuch that

It follows that

By virtue of [17, Corollary 7.2], there exists a function f and a subsequence of {?n}, still indexed by n, such that

Therefore,

Applying Fatou’s lemma, we get

Thus, we have f ∈ Ws,q(?). On the other hand, in view of (2.2) we can write

Hence, ?n→ f in Ws,q(?) and so f ∈By (2.2), the function f has a constant value on ?. The only possible value is f ≡ 0 which yields a contradiction with the fact that

So, we get

Applying the inequality (2.3) for ?nand passing to the limit, we conclude the result.

Under the same assumptions of Lemma 2.1, the Banach spacecan be endowed with the norm

In the case where q =2, we noteEndowed with the inner product

It is worth recalling that for any u and ? belonging to Hs(RN), we have the following duality product

Thus, it can be seen that

is a continuous and symmetric operator defined on Hs(RN).

In the particular case, if u and ? belong to Hs(RN) with u= ? =0, on C?, we have

where Q := R2N(C? × C?). For N > 2s we define the fractional Sobolev critical exponentThe following result is a fractional version of the Sobolev inequality which provides a continuous embedding ofin the critical Lebesgue spaceThe proof can be found in [17, 29].

Theorem 2.2(Fractional Sobolev embedding) Let 02s. Then,there exists a constant S(N,s) depending only on N and s, such that for all

Remark 2.3In particular,if ? is an open bounded subset in RNof class C0,1with N >2s and 0

Indeed, by [17, Theorem 5.4]we can write

The result follows then by Theorem 2.2 and Lemma 2.1.

We will prove some estimates in the usual Marcinkiewicz space Mq(?), 0 0 satisfying

for every t>0. Here and in what follows, meas(E)denotes the Lebesgue measure of a measurable subset E of ?. It is worth recalling the following connection between Marcinkiewicz and Lebesgue spaces

for every 1 < q < ∞ and 0 < ε ≤ q ? 1 (see for instance [22]). We will also use the following truncation functions Tkand Gk, k >0, defined for every s ∈R by

We denote by Mb(?) the space of all bounded Radon measures on ?. The norm of a measureμ ∈ Mb(?) is given by

Definition 2.4We say that the sequence of measurable functions {μn} is converging weakly to μ in the sense of the measures if

In what follows we make use of the following technical algebraic inequalities.

Lemma 2.5i) Let α >0. For every x, y ≥ 0 one has

ii) Let 0< α <1. For every x, y ≥ 0 withone has

iii) Let α ≥ 1. Then

where cαis a constant depending only on α.

Taking into account that less regular data are involved,the classical notion of finite energy solution cannot be used. Instead, we shall consider the notion of weak solution whose meaning is defined as follows.

Definition 2.6Let f ∈ L1(?) and let μ be a non-negative bounded Radon measure. By a weak solution of problem (1.1), we mean a measurable function u satisfying

Theorem 2.7Let ? be an open bounded subset in RNof class C0,1with N > 2s and 0 < s < 1. Let 0 < γ ≤ 1 and let f ∈ L1(?). Then the problem (1.1) admits a weak solutionfor everyand for every s1

Theorem 2.8Let ? be an open bounded subset in RNof class C0,1with N > 2s and 0 < s < 1. Let γ > 1 and let f ∈ L1(?). Then the problem (1.1) admits a weak solutionfor everyfor all s1< s. Furthermore,for every k >0.

We point out that the inclusionholds for any s2< s1(see [17]).Therefore, the range of s1in both Theorem 2.7 and Theorem 2.8 can be that of the set of the exponents s1close to s. Indeed, we can consider s1to be such thatSo that when s tends to 1 one has also s1tends to 1?. In addition, letting s tends to 1?the operator(??)sis nothing but the standard Laplacian. So that the equation in (1.1) becomes

and then the results in both Theorem 2.7 and Theorem 2.8 covers those obtained in [26].

3 Approximated Problems: Existence Result and Comparison Principle

Consider the sequence of approximate problems

where fn=Tn(f)is the truncation at level n of f andμnis a sequence of bounded non-negative smooth functions in L1(?) converging weakly to μ in the sense of the measures.

We shall prove that for every fixed integer n ∈N, the problem (3.1) admits a unique weak solution unin the following sense :

Lemma 3.1For each integer n ∈N, the problem (3.1) admits a non-negative weak solution

ProofLet n ∈ N be fixed and let v ∈ L2(?). We define the map

where w =S(v) is the weak solution to the following problem

The existence of w can be derived by classical minimization argument. Indeed,sinceμn∈ L∞(?), we already know (see [12, Lemma 2.1]) that problem (3.2) has a unique weak solutionwhere

in the following sense

Hence, by density arguments it follows thatThus,As regards the uniqueness of w inwe suppose there exist two solutionsSumming up the both equations satisfied by w1and w2respectively,we get(??)s(w1?w2)=0.Thus, taking (w1?w2) as a test function in this last equation and then integrating over RN,we obtain

So we get w1(x) = w2(x), for almost every x ∈ ?. Since w1= w2= 0 on RN?, we get w1(x)=w2(x) for almost every x ∈RN. Furthermore, by the comparison principle [8, Lemma 2.1]we get w ≥0. Now, inserting w as a test function in (3.2) we obtain

with C′and C(n,s,N,?) are independent of v, so that the ball of radius C′(nγ+1+C(n)) is invariant under S in

Now,using the Schauder’s fixed point theorem over S to prove the existence and uniqueness of solution of (3.1), we need to verify the continuity and compactness of S as an operator from

First, we go to prove the continuity of S as an operator from L2(?) to L2(?). Let us consider a sequence vkthat converges to v in L2(?), then up to a subsequence, we have

Denoting wk=S(vk) and w =S(v), we have

Taking wk(x)?w(x)as a test function in(3.5)and(3.6)respectively,then subtracting term at term the both resulting equations and using Hlder’s inequality we arrive at

Applying the fractional Sobolev embedding and Hlder’s inequality with the exponentswe get

then by the dominated convergence theorem we conclude that

So S is continuous from L2(?) to L2(?) and it follows that S is continuous fromto

Now,we prove that S is compact fromlet us consider a sequence{vk}k∈Nsuch thatthen by the compact embeddingin Lr(?)for every(see [17, Corollary 7.2]), we have

Denoting wk=S(vk) and w =S(v), by (3.3) we have

where C is a constant not depending on k, then by the previous compact embedding and by the continuity of S on L2(?) we get

So, by the uniqueness of the limit we have=w. In view of the previous equations (3.5) and(3.6) we have

Taking wk? w as a test function in the previous equation, using Hlder’s inequality and (3.7)we obtain

It follows that

Hence, S is a compact operator fromand therefore by Schauder’s fixed point theorem there existssuch that un=S(un). This means that unis a weak solution to the problem

In addition, since the right hand side of belongs to L∞(?) by [25]we obtain un∈ L∞(?).

Lemma 3.2(comparison principle) The sequence {un}n∈Nis such that for every subset ω ?? ? there exists a positive constant cω, independent on n, such that

ProofConsider the following problem

In [7], the authors proved the existence of a weak solution vnof (3.9) such that

for every x ∈ ω and for every n ∈ N. Here the constant cωis independent on n. On the other hand, we have

we obtain the following inequality

Now, taking (vn?un)+as a test function in (3.10) and then integrating over RN, we get

Observe that for any function g : RN→R the following inequality

holds true for every x, y ∈RN, where g+=max(g,0). Therefore, we obtain

which implies that un≥ vnin ? and so

for every x ∈ ω and for every n ∈ N.

Remark 3.3Lemma 3.2 shows that the problem (3.1) has a unique solution. Indeed,if unand wnare two solutions of problem (3.1), then as above taking (un?wn)+as a test function in the problem satisfied by(un? wn),we conclude that un≤ wnin ? and again taking(wn? un)+as a test function we get wn≤ unin ?. Hence, follows un=wnin ?.

4 A Priori Estimates in Fractional Sobolev Spaces

In order to prove the existence of solutions for problem (1.1), we first need some a priori estimates on un. We start by proving the following lemma that we will use in both cases γ ≤ 1 and γ >1.

Lemma 4.1Let vn∈be a sequence that satisfies the following assumptions

1) The sequence {vn}nis uniformly bounded in Lr(?), for all

2) For any sufficient small θ ∈ (0,1)

where C is a constant not depending on n and wn= vn+1. Then the sequence {vn}nis uniformly bounded in the fractional Sobolev spacefor everyand for all s1

ProofWe shall prove that the sequence {vn} is uniformly bounded in the fractional Sobolev spacefor everyand for all s1< s. That is there is a constant C not depending on n such that

To this aim, let q <2 which will be chosen in a few lines. We can write

Pointing out that the quantity in the middle of the product inside the integral can be written as follows

Applying the Young inequality with the exponentsand θ +1, we have

Here, |SN?1| stands for the Lebesgue measure of the unit sphere in RN. By x/y symmetry,there exists a constant C, not depending on n, such that

Now we choose θ > 0 in order to getThat isTo ensure the existence of θ we must have 2N ? 2q(N ? s)>0 which yieldsWe then conclude that(4.1) is fulfilled and the sequence {vn} is uniformly bounded infor everyand for all s1

4.1 The case γ ≤ 1

Lemma 4.2Letbe the solution of the problem(3.1). If 0< γ ≤ 1, then the sequence {un} is uniformly bounded infor everyand for all s1

ProofLet k ≥1 be fixed. By Lemma 6.4 (in Appendix) the function Tk(un) is an admissible test function in (3.1). Thus, inserting it in (3.1) we obtain

By using Proposition 6.2 (in Appendix), we get

For the left hand side, observing that on the set {un≥k}, we have Tk(un)=k, we get

which yields

Thus, the sequence {un} is uniformly bounded inand then so it is in Lr(?), for allLet s1∈ (0,s) be fixed. For every x ≥ 0 we define the function

Observe that the function φ satisfies

The function φ(un)is an admissible test function in(3.1). So that inserting it as a test function in (3.1) we obtain

Being φ non-decreasing and ? × ? ? Q, the integral in the left-hand side can be treated as follows

So that we obtain

where we have set wn= un+1. Therefore, by Lemma 4.1 with 0 < θ ≤ γ the sequence {un}is uniformly bounded infor everyand for all s1

4.2 The case γ >1

Lemma 4.3Let f ∈ L1(?) and let unbe the solution of (3.1). For k >0 and γ >1 the sequenceis uniformly bounded in

ProofLet us fix k >0. Insertingas a test function in (3.1), we get

where C1=is a constant not depending on n. By applying Proposition 6.2 (in Appendix) and Lemma 2.5, we have

Therefore, we obtain

The proof is then achieved.

Lemma 4.4Let unbe the solution of the problem (3.1). If γ > 1, then the sequence{un} is uniformly bounded infor everyand for all s1

ProofFor every ω ?? ?, for alland for all s1< s, we shall prove that there exists a constant C =C(q,s1,w), not depending on n, such that

We begin by proving the left estimate in (4.5). Let k0≥1 be fixed. Let q < 2 and s1< s.Using the fact that un=Tk0(un)+Gk0(un), we can write

which implies

So,it is sufficient to prove that{Gk0(un)}nand{Tk0(un)}nare uniformly bounded inrespectively. We begin by proving that Gk0(un)is uniformly bounded infor alland for all s1k0we take Tk(Gk0(un)) as a test function in (3.1) and use the fact that Gk0(un)=0 on {un≤k0}, we obtain

where C1=is a constant not depending on n. Using the decomposition of unas un=Tk0(un)+Gk0(un), we can write

Let us observe that since Tk0and Tk(Gk0) are non-decreasing functions, we get

Hence, it follows

In the right-hand side of the above inequality, we decompose Gk0(un) as follows Gk0(un(x))=Gk(Gk0(un(x))) +Tk(Gk0(un(x))) and we apply Proposition 6.2 (in Appendix) with α = 1 obtaining

Hence,using the fractional Sobolev inequality,we get again the inequality(4.3)for the function Gk0(un) that is

which implies that {Gk0(un)}nis uniformly bounded in Lr(?) for every

Let φ be the function defined in (4.4). Observe that for every 0 < θ < 1 the function φ enjoys the following properties

Inserting φ(Gk0(un)) as a test function in (3.1) we get

Then, writing the decomposition un= Tk0(un)+ Gk0(un) and using the fact that Tk0and φ(Gk0) are non-decreasing functions, we obtain

where we have set wn= Gk0(un)+1. Thus, Lemma 4.1 ensures that the sequence {Gk0(un)}is uniformly bounded infor alland for all s1

Now, we shall prove that {Tk0(un)}nis uniformly bounded inTo do so, we insertas a test function in (3.1) obtaining

By Lemma 2.5 (item iii)) there exists a constant cγ>0, depending only on γ such that

Let now ω be a compact subset in ?. By Proposition 6.2 (in Appendix) we can write

Pointing out that by Lemma 3.2 we have Tk0(un(x))≥ min(k0,cω) for every x ∈ ω, we obtain

which proves that {Tk0(un)}nis uniformly bounded in

We now prove the second estimate in (4.5). Forand s1

we conclude the result. In fact, for every γ > 0 the sequence {un} is uniformly bounded in Lq(?) for all 1 ≤

5 Proof of the Main Results

In this section, we show that in both cases γ ≤ 1 and γ >1, the problem (1.1) has a weak solution obtained as the limit of approximate solutions {un}nof the problem (3.1).

5.1 The case γ ≤ 1

Proofof Theorem 2.7. By virtue of Lemma 4.2 and the compact embedding ofin L1(?) (see [17, Corollary 7.2]), there exist a subsequence of {un}nstill indexed by n and a measurable functionsuch that

Let u the function such that u=v in ? and u=0 in RN?. Thus, un→ u a.e. in RNwhich implieswe have

We need that the term |x ? y|ρN+(1+ρ)(2s?s1)vanishes from within the integral. To get this, it is sufficient to have (1+ ρ)(1+s1?2s)? ρN ≥ 0. To this aim, we consider s1to be very close

Let ρ > 0 be a small enough real number that we will choose later. For any of s. Precisely, we impose on s1the condition

We point out that with this range of values of s1and with the assumption N > 2s, we easily get

Thus, the fact that (1+ ρ)(1+s1? 2s)? ρN ≥ 0 is equivalent to 0< ρ ≤Hence,we get

where diam(?) stands for the diameter of ?. Now we have to make a choice of ρ which enables us to use the uniform boundedness of {un}ninfor everyThis is the case if 1+Finally, we choose ρ to be such that

Therefore, there is a constant C >0 not depending on n such that

Consequently by De La Valle Poussin and Dunford-Pettis theorems the sequence

is equi-integrable in L1(? × ?). Now, takingas a test function in (3.1) we get

We split the integral in left-hand side into three integrals as follows

By Vitali’s lemma we have

For the second integral I2in (5.2), we start noticing that since un(y) = ?(y) = 0 for every y ∈ C? we can write

Since Supp ? is a compact subset in ?, we have

Therefore, an easy computation leads to

As a consequence of the convergence in norm of the sequence {un} in L1(?) there exist a subsequence of {un} still indexed by n and a positive function g in L1(?) such that

which enables us to get

We observe that by (5.3) the function (x,y)→belongs to L1(?×C?)

Thus, by the dominated convergence theorem, we have

For the third integral I3in (5.2), we can follow exactly the same lines as above using the x/y symmetry. We then conclude that

obtaining by the dominated convergence theorem

and in the last term in (5.1), by the convergence of μnto μ we have

Finally, passing to the limit as n → +∞, we obtain

5.2 The case γ >1

Proofof Theorem 2.8. By virtue of Lemma 4.4, there exist a subsequence of {un}nstill indexed by n and a measurable functionsuch that

So that defining the function u by u=v in ? and u=0 in RN?, one has

Let K be a compact subset of ? such that Supp ? ? K and dist(Supp ?,?K)>0. The integral in the left-hand side of the previous equality can be splitted as

As in the proof of the Theorem 2.7, the same ideas allow to obtain

6 Regularity of Solutions

Now, we prove some regularities of the solution u of the problem (1.1).

Proposition 6.1Assume that μ is a Radon measure, f ∈ L1(?) and 0 < γ ≤ 1. Then the solution u of the problem (1.1) obtained by approximation is such that

ProofWe follow closely the lines in [25]. By (4.3) and Theorem 2.7, we can apply Fatou’s Lemma, we conclude that u ∈ Lr(?), for everyNow, we will prove thatis bounded in the Marcinkiewicz spaceWe fix β >0 and for any positive k ≥1, we have

By using [25, Corollary 1]and Lemma 4.2, we get

By using (4.3), we have

Appendix

In this Appendix we give the functional and technical results we have used in the previous sections. We start with the following inequality whose proof in the cases where α = 1 can be found [25]. Here we give a simple proof based on the monotony of the truncation functions.

Proposition 6.2Let α ≥ 1 and let v :RN→ R be a positive measurable function. Then for every k >0 and for every (x,y)∈ RN× RN

ProofLet x, y ∈RNbe arbitrary. Without loss of generality we can assume that v(x) ≥ v(y). Since the functionsare non-decreasing on R, we have

The next result, well known in classical Sobolev spaces, provides a necessary condition for a function to belong to the fractional Sobolev space

Lemma 6.3Let ? be an open set in RNof class C0,1with bounded boundary,1 ≤ p<+∞and let 0

ProofLet u ∈ Ws,p(?) be a function with Supp u be a compact subset included in ?.Then there exists an open set ω such that

Then by [17, Corollary 5.5], there exists a sequence{un}nof functionssuch that

Using the fact that ?u=u on ?, we obtain

For the second part of the normwe can write it as follows.

where we have set

Thus,in order to prove that ?unconverges to u in Ws,p(?),it is sufficient to prove that up to a subsequence,{Fn(x,y)}converges to F(x,y)in norm in Lp(?×?). Since, up to a subsequence still indexed by n, unconverges almost everywhere to u, we obtain

The norm convergence of unto u in Ws,p(?), yields

According to (6.1) and the norm convergence of {un} in Lp(?), there exist a subsequence of{un} still indexed by n and two positive functions g in L1(? × ?) and h in L1(?) such that

So that writing

We need to prove that the function in the second term in the right-hand side in (6.2) belongs to L1(? × ?). To do so we can write

where Clipstands for the Lipschitz constant of ? and |SN?1| stands for the Lebesgue measure of the surface area of the unit N-sphere SN?1of RN. Applying the dominated convergence theorem, we conclude our claim and thus follows

Lemma 6.4Let ? be an open set in RNof class C0,1with bounded boundary,1 ≤ p<+∞and let 0 < s < 1. Let φ : R → R be a uniformly Lipschitz function, with φ(0) = 0. Then for everyone has

ProofLet us denote by K the Lipschitz constant of φ and letThere exists a sequence {un} offunctions which converges to u in norm in Ws,p(?). That is there exists n0∈ N such that for all n ∈ N with n ≥ n0one has

Defining vn= φ(un), Gn(x,y) = un(x)? un(y) and G(x,y) = u(x)? u(y), we can write for every n ≥n0

C0and C1are two constants not depending on n. Thus,{vn}is uniformly bounded in Ws,p(?).Since by φ(0)=0 the function vnis compactly supported in ?,so that by Lemma 6.3 we obtainNow, we prove that

Since the sequence {un} converges to u in norm in Ws,p(?), then for a subsequence of {un},still indexed by n, we have

Then, it follows

Furthermore,

On the other hand we can write

where we noted

In order to show that vnconverges to φ(u) in Ws,p(?), it sufficient to prove that for a subsequence of {Fn(x,y)}n≥1, still denoted by {Fn(x,y)}n≥1,By the almost everywhere convergence of vnto φ(u), we have

Observe that the norm convergence of unto u in Ws,p(?) implies

So that since

the sequence{|Fn(x,y)|p}nis then equi-integrable. Applying Vitali’s theorem we getwhich in turn impliesSince the sequence{vn} belongs to the closed spaceforces the limit φ(u) to belong to