Hitchhiker s guide to the fractional Sobolev spaces

Transcription

1 Hitchhiker s guide to the fractional Sobolev spaces Eleonora Di Nezza a, Giampiero Palatucci a,b,,, Enrico Valdinoci a,c,2 a Dipartimento di Matematica, Università di Roma Tor Vergata - Via della Ricerca Scientifica,, 0033 Roma, Italy b Dipartimento di Matematica, Università degli Studi di Parma, Campus - Viale delle Scienze, 53/A, 4324 Parma, Italy c Dipartimento di Matematica, Università degli Studi di Milano - Via Saldini, 50, 2033 Milano, Italy Abstract This paper deals with the fractional Sobolev spaces W s,p. We analyze the relations among some of their possible definitions and their role in the trace theory. We prove continuous and compact embeddings, investigating the problem of the extension domains and other regularity results. Most of the results we present here are probably well known to the experts, but we believe that our proofs are original and we do not make use of any interpolation techniques nor pass through the theory of Besov spaces. We also present some counterexamples in non-lipschitz domains. Keywords: Fractional Sobolev spaces, Gagliardo norm, fractional Laplacian, nonlocal energy, Sobolev embeddings, Riesz potential 200 MSC: Primary 46E35, Secondary 35S30, 35S05 Corresponding author addresses: dinezza@mat.uniroma2.it (Eleonora Di Nezza), giampiero.palatucci@unimes.fr (Giampiero Palatucci), enrico@math.utexas.edu (Enrico Valdinoci) Giampiero Palatucci has been supported by Istituto Nazionale di Alta Matematica F. Severi (Indam) and by ERC grant Vectorial problems. 2 Enrico Valdinoci has been supported by the ERC grant ɛ Elliptic Pde s and Symmetry of Interfaces and Layers for Odd Nonlinearities and the FIRB project A&B Analysis and Beyond. 3 These notes grew out of a few lectures given in an undergraduate class held at the Università di Roma Tor Vergata. It is a pleasure to thank the students for their warm interest, their sharp observations and their precious feedback. Preprint submitted to Elsevier June 9, 20

2 Contents Introduction 2 2 The fractional Sobolev space W s,p 4 3 The space H s and the fractional Laplacian operator 4 Asymptotics of the constant C(n, s) 2 5 Extending a W s,p () function to the whole of R n 32 6 Fractional Sobolev inequalities 38 7 Compact embeddings 48 8 Hölder regularity 53 9 Some counterexamples in non-lipschitz domains 58 References 64. Introduction These pages are for students and young researchers of all ages who may like to hitchhike their way from to s (0, ). To wit, for anybody who, only endowed with some basic undergraduate analysis course (and knowing where his towel is), would like to pick up some quick, crash and essentially self-contained information on the fractional Sobolev spaces W s,p. The reasons for such a hitchhiker to start this adventurous trip might be of different kind: (s)he could be driven by mathematical curiosity, or could be tempted by the many applications that fractional calculus seems to have recently experienced. In a sense, fractional Sobolev spaces have been a classical topic in functional and harmonic analysis all along, and some important books, such as [59, 90] treat the topic in detail. On the other hand, fractional spaces, and the corresponding nonlocal 2

3 equations, are now experiencing impressive applications in different subjects, such as, among others, the thin obstacle problem [87, 69], optimization [37], finance [26], phase transitions [2, 4, 88, 40, 45], stratified materials [83, 23, 24], anomalous diffusion [68, 98, 65], crystal dislocation [92, 47, 8], soft thin films [57], semipermeable membranes and flame propagation [5], conservation laws [9], ultra-relativistic limits of quantum mechanics [4], quasi-geostrophic flows [64, 27, 2], multiple scattering [36, 25, 50], minimal surfaces [6, 20], materials science [4], water waves [8, 00, 99, 32, 29, 74, 33, 34, 3, 30, 42, 5, 75, 35], elliptic problems with measure data [7, 54], non-uniformly elliptic problems [39], gradient potential theory [72] and singular set of minima of variational functionals [70, 56]. Don t panic, instead, see also [86, 87] for further motivation. For these reasons, we thought that it could be of some interest to write down these notes or, more frankly, we wrote them just because if you really want to understand something, the best way is to try and explain it to someone else. Some words may be needed to clarify the style of these pages have been gathered. We made the effort of making a rigorous exposition, starting from scratch, trying to use the least amount of technology and with the simplest, low-profile language we could use since capital letters were always the best way of dealing with things you didn t have a good answer to. Differently from many other references, we make no use of Besov spaces 4 or interpolation techniques, in order to make the arguments as elementary as possible and the exposition suitable for everybody, since when you are a student or whatever, and you can t afford a car, or a plane fare, or even a train fare, all you can do is hope that someone will stop and pick you up, and it s nice to think that one could, even here and now, be whisked away just by hitchhiking. Of course, by dropping fine technologies and powerful tools, we will miss several very important features, and we apologize for this. So, we highly 4 About this, we would like to quote [53], according to which The paradox of Besov spaces is that the very thing that makes them so successful also makes them very difficult to present and to learn. 3

4 recommend all the excellent, classical books on the topic, such as [59, 90,, 93, 94, 0, 80, 9, 67, 60], and the many references given therein. Without them, our reader would remain just a hitchhiker, losing the opportunity of performing the next crucial step towards a full mastering of the subject and becoming the captain of a spaceship. In fact, compared to other Guides, this one is not definitive, and it is a very evenly edited book and contains many passages that simply seemed to its editors a good idea at the time. In any case, of course, we know that we cannot solve any major problems just with potatoes it s fun to try and see how far one can get though. In this sense, while most of the results we present here are probably well known to the experts, we believe that the exposition is somewhat original. These are the topics we cover. In Section 2, we define the fractional Sobolev spaces W s,p via the Gagliardo approach and we investigate some of their basic properties. In Section 3 we focus on the Hilbert case p = 2, dealing with its relation with the fractional Laplacian, and letting the principal value integral definition interplay with the definition in the Fourier space. Then, in Section 4 we analyze the asymptotic behavior of the constant factor that appears in the definition of the fractional Laplacian. Section 5 is devoted to the extension problem of a function in W s,p () to W s,p (R n ): technically, this is slightly more complicated than the classical analogue for integer Sobolev spaces, since the extension interacts with the values taken by the function in via the Gagliardo norm and the computations have to take care of it. Sobolev inequalities and continuous embeddings are dealt with in Section 6, while Section 7 is devoted to compact embeddings. Then, in Section 8, we point out that functions in W s,p are continuous when sp is large enough. In Section 9, we present some counterexamples in non-lipschitz domains. After that, we hope that our hitchhiker reader has enjoyed his trip from the integer Sobolev spaces to the fractional ones, with the advantages of being able to get more quickly from one place to another - particularly when the place you arrived at had probably become, as a result of this, very similar to the place you had left. The above sentences written in old-fashioned fonts are Douglas 4

5 Adams s of course, and we took the latitude of adapting their meanings to our purposes. The rest of these pages are written in a more conventional, may be boring, but hopefully rigorous, style. 2. The fractional Sobolev space W s,p This section is devoted to the definition of the fractional Sobolev spaces. No prerequisite is needed. We just recall the definition of the Fourier transform of a distribution. First, consider the Schwartz space S of rapidly decaying C functions in R n. The topology of this space is generated by the seminorms p N (ϕ) = sup x R n ( + x ) N α N D α ϕ(x), N = 0,, 2,..., where ϕ S (R n ). Let S (R n ) be the set of all tempered distributions, that is the topological dual of S (R n ). As usual, for any ϕ S (R n ), we denote by F ϕ(ξ) = e iξ x ϕ(x) dx (2π) n/2 R n the Fourier transform of ϕ and we recall that one can extend F from S (R n ) to S (R n ). Let be a general, possibly non smooth, open set in R n. For any real s > 0 and for any p [, ), we want to define the fractional Sobolev spaces W s,p (). In the literature, fractional Sobolev-type spaces are also called Aronszajn, Gagliardo or Slobodeckij spaces, by the name of the ones who introduced them, almost simultaneously (see [3, 44, 89]). We start by fixing the fractional exponent s in (0, ). For any p [, + ), we define W s,p () as follows { } W s,p () := u L p () : u(x) u(y) L p ( ) x y n p +s ; (2.) i.e, an intermediary Banach space between L p () and W,p (), endowed with the natural norm ( ) u W s,p () := u p u(x) u(y) p p dx + dx dy, (2.2) x y n+sp 5

6 where the term ( [u] W s,p () := is the so-called Gagliardo (semi )norm of u. ) u(x) u(y) p p dx dy x y n+sp It is worth noticing that, as in the classical case with s being an integer, the space W s,p is continuously embedded in W s,p when s s, as next result points out. Proposition 2.. Let p [, + ) and 0 < s s <. Let be an open set in R n and u : R be a measurable function. Then u W s,p () C u W s,p () for some suitable positive constant C = C(n, s, p). In particular, W s,p () W s,p (). Proof. First, { x y } u(x) p dx dy x y n+sp ( z ) dz u(x) p dx z n+sp C(n, s, p) u p L p (), where we used the fact that the kernel / z n+sp is integrable since n+sp > n. Taking into account the above estimate, it follows u(x) u(y) p dx dy x y n+sp { x y } 2 p On the other hand, u(x) u(y) p dx dy x y n+sp { x y <} { x y } u(x) p + u(y) p dx dy x y n+sp 2 p C(n, s, p) u p L p (). (2.3) 6 { x y <} u(x) u(y) p x y n+s p dx dy. (2.4)

7 Thus, combining (2.3) with (2.4), we get u(x) u(y) p dx dy x y n+sp 2 p C(n, s, p) u p L p () + u(x) u(y) p x y n+s p dx dy and so u p W s,p () ( 2 p C(n, s, p) + ) u p L p () + u(x) u(y) p x y n+s p dx dy C(n, s, p) u p W s,p (), which gives the desired estimate, up to relabeling the constant C(n, p, s). We will show in the forthcoming Proposition 2.2 that the result in Proposition 2. holds also in the limit case, namely when s =, but for this we have to take into account the regularity of (see Example 9.). As usual, for any k N and α (0, ], we say that is of class C k,α if there exists M > 0 such that for any x there exists a ball B = B r (x), r > 0, and an isomorphism T : Q B such that T C k,α (Q), T C k,α (B), T (Q + ) = B, T (Q 0 ) = B where and T C k,α (Q) + T C k,α (B) M, Q := { x = (x, x n ) R n R : x < and x n < }, Q + := { x = (x, x n ) R n R : x < and 0 < x n < } and Q 0 := {x Q : x n = 0}. We have the following result. 7

8 Proposition 2.2. Let p [, + ) and s (0, ). Let be an open set in R n of class C 0, with bounded boundary and u : R be a measurable function. Then u W s,p () C u W,p () (2.5) for some suitable positive constant C = C(n, s, p). In particular, W,p () W s,p (). Proof. Let u W,p (). Thanks to the regularity assumptions on the domain, we can extend u to a function ũ : R n R such that ũ W,p (R n ) and ũ W,p (R n ) C u W,p () for a suitable constant C (see, e.g., [49, Theorem 7.25]). Now, using the change of variable z = y x and the Hölder inequality, we have u(x) u(y) p dx dy { x y <} x y n+sp u(x) u(z + x) B p dz dx = z n+sp u(x) u(z + x) B p dz dx z p z n+(s )p ( ) p u(x + tz) dt dz dx B R n B B z n p +s ũ(x + tz) p z n+p(s ) ũ p L p (R n ) z n+p(s ) dt dz dt dz dx C (n, s, p) ũ p L p (R n ) C 2 (n, s, p) u p W,p (). (2.6) Also, by (2.3), { x y } u(x) u(y) p x y n+sp dx dy C(n, s, p) u p L p (). (2.7) 8

9 Therefore, from (2.6) and (2.7) we get estimate (2.5). We remark that the Lipschitz assumption in Proposition 2.2 cannot be completely dropped (see Example 9. in Section 9); we also refer to the forthcoming Section 5, in which we discuss the extension problem in W s,p. Let us come back to the definition of the space W s,p (). Before going ahead, it is worth explaining why the definition in (2.) cannot be plainly extended to the case s. Suppose that is a connected open set in R n, then any measurable function u : R such that u(x) u(u) p x y n+sp dx dy < + is actually constant (see [0, Proposition 2]). This fact is a matter of scaling and it is strictly related to the following result that holds for any u in W,p (): u(x) u(y) p lim s ( s) dx dy = C x y n+sp u p dx (2.8) for a suitable positive constant C depending only on n and p (see []). In the same spirit, in [66], Maz ja and Shaposhnikova proved that, for a function u 0<s< W s,p (R n ), it yields Rn lim s u(x) u(y) p dx dy = C s 0 + R x y n n+sp 2 u p dx, (2.9) R n for a suitable positive constant C 2 depending 5 only on n and p. When s > and it is not an integer we write s = m + σ, where m is an integer and σ (0, ). In this case the space W s,p () consists of those equivalence classes of functions u W m,p () whose distributional derivatives D α u, with α = m, belong to W σ,p (), namely { } W s,p () := u W m,p () : D α u W σ,p () for any α s.t. α = m (2.0) 5 For the sake of simplicity, in the definition of the fractional Sobolev spaces and those of the corresponding norms in (2.) and (2.2) we avoided any normalization constant. In view of (2.8) and (2.9), it is worthing notice that, in order to recover the classical W,p and L p spaces, one may consider to add a factor C(n, p, s) s( s) in front of the double integral in (2.2). 9

10 and this is a Banach space with respect to the norm u W s,p () := u p W m,p () + p D α u p W σ,p (). (2.) α =m Clearly, if s = m is an integer, the space W s,p () coincides with the Sobolev space W m,p (). Corollary 2.3. Let p [, + ) and s, s >. Let be an open set in R n of class C 0,. Then, if s s, we have W s,p () W s,p (). Proof. We write s = k + σ and s = k + σ, with k, k integers and σ, σ (0, ). In the case k = k, we can use Proposition 2. in order to conclude that W s,p () is continuously embedded in W s,p (). On the other hand, if k k +, using Proposition 2. and Proposition 2.2 we have the following chain W k +σ,p () W k,p () W k+,p () W k+σ,p (). The proof is complete. As in the classic case with s being an integer, any function in the fractional Sobolev space W s,p (R n ) can be approximated by a sequence of smooth functions with compact support. Theorem 2.4. For any s > 0, the space C 0 (R n ) of smooth functions with compact support is dense in W s,p (R n ). A proof can be found in [, Theorem 7.38]. Let W s,p 0 () denote the closure of C 0 () in the norm W s,p () defined in (2.). Note that, in view of Theorem 2.4, we have W s,p 0 (R n ) = W s,p (R n ), (2.2) but in general, for R n, W s,p () W s,p 0 (), i.e. C0 () is not dense in W s,p (). Furthermore, it is clear that the same inclusions stated in Proposition 2., Proposition 2.2 and Corollary 2.3 hold for the spaces W s,p 0 (). 0

11 Remark 2.5. For s < 0 and p (, ), we can define W s,p () as the dual space of W s,q 0 () where /p + /q =. Notice that, in this case, the space W s,p () is actually a space of distributions on, since it is the dual of a space having C0 () as density subset. Finally, it is worth noticing that the fractional Sobolev spaces play an important role in the trace theory. Precisely, for any p (, + ), assume that the open set R n is sufficiently smooth, then the space of traces T u on of u in W,p () is characterized by T u < + (see [43]). W p,p ( ) Moreover, the trace operator T is surjective from W,p () onto W p,p ( ). In the quadratic case p = 2, the situation simplifies considerably, as we will see in the next section and a proof of the above trace embedding can be find in the forthcoming Proposition The space H s and the fractional Laplacian operator In this section, we focus on the case p = 2. This is quite an important case since the fractional Sobolev spaces W s,2 (R n ) and W s,2 0 (R n ) turn out to be Hilbert spaces. They are usually denoted by H s (R n ) and H0(R s n ), respectively. Moreover, they are strictly related to the fractional Laplacian operator ( ) s (see Proposition 3.6), where, for any u S and s (0, ), ( ) s is defined as ( ) s u(x) u(y) u(x) = C(n, s) P.V. dy (3.) R x y n n+2s u(x) u(y) = C(n, s) lim dy. ε 0 + x y n+2s C B ε(x) Here P.V. is a commonly used abbreviation for in the principal value sense (as defined by the latter equation) and C(n, s) is a dimensional constant that depends on n and s, precisely given by ( cos(ζ ) C(n, s) = R ζ n n+2s The choice of this constant is motived by Proposition 3.3. ) dζ. (3.2)

12 Remark 3.. Due to the singularity of the kernel, the right hand-side of (3.) is not well defined in general. In the case s (0, /2) the integral in (3.) is not really singular near x. Indeed, for any u S, we have u(x) u(y) dy R x y n n+2s x y C B R x y dy + u n+2s L (R n ) dy C B R x y n+2s ( ) = C dy + dy x y n+2s x y n+2s ( R = C B R 0 dρ + ρ 2s + R C B R dρ ρ 2s+ ) < + where C is a positive constant depending only on the dimension and on the L norm of u. Now, we show that one may write the singular integral in (3.) as a weighted second order differential quotient. Lemma 3.2. Let s (0, ) and let ( ) s be the fractional Laplacian operator defined by (3.). Then, for any u S, ( ) s u(x) = 2 C(n, s) u(x + y) + u(x y) 2u(x) dy, x R n. R y n n+2s (3.3) Proof. The equivalence of the definitions in (3.) and (3.3) immediately follows by the standard changing variable formula. Indeed, by choosing z = y x, we have ( ) s u(y) u(x) u(x) = C(n, s) P.V. dy R x y n n+2s u(x + z) u(x) = C(n, s) P.V. dz. (3.4) R z n n+2s Moreover, by substituting z = z in last term of the above equality, we have u(x + z) u(x) u(x z) u(x) P.V. dz = P.V. d z. (3.5) R z n n+2s R z n n+2s 2

13 and so after relabeling z as z u(x + z) u(x) 2P.V. dz R z n n+2s u(x + z) u(x) = P.V. R z n n+2s = P.V. u(x z) u(x) dz + P.V. dz R z n n+2s R n u(x + z) + u(x z) 2u(x) z n+2s dz. (3.6) Therefore, if we rename z as y in (3.4) and (3.6), we can write the fractional Laplacian operator in (3.) as ( ) s u(x) = 2 C(n, s) P.V. u(x + y) + u(x y) 2u(x) dy. R y n n+2s The above representation is useful to remove the singularity of the integral at the origin. Indeed, for any smooth function u, a second order Taylor expansion yields u(x + y) + u(x y) 2u(x) y n+2s D2 u L y n+2s 2, which is integrable near 0 (for any fixed s (0, )). Therefore, since u S, one can get rid of the P.V. and write (3.3). 3.. An approach via the Fourier transform Now, we take into account an alternative definition of the space H s (R n ) = W s,2 (R n ) via the Fourier transform. Precisely, we may define { } Ĥ s (R n ) = u L 2 (R n ) : ( + ξ 2s ) F u(ξ) 2 dξ < + (3.7) R n and we observe that the above definition, unlike the ones via the Gagliardo norm in (2.2), is valid also for any real s. We may also use an analogous definition for the case s < 0 by setting { } Ĥ s (R n ) = u S (R n ) : ( + ξ 2 ) s F u(ξ) 2 dξ < +, R n 3

14 although in this case the space Ĥs (R n ) is not a subset of L 2 (R n ) and, in order to use the Fourier transform, one has to start from an element of S (R n ), (see also Remark 2.5). The equivalence of the space Ĥs (R n ) defined in (3.7) with the one defined in the previous section via the Gagliardo norm (see (2.)) is stated and proven in the forthcoming Proposition 3.4. First, we will prove that the fractional Laplacian ( ) s can be viewed as a pseudo-differential operator of symbol ξ 2s. The proof is standard and it can be found in many papers (see, for instance, [9, Chapter 6]). We will follow the one in [97] (see Section 3), in which is shown how singular integrals naturally arise as a continuous limit of discrete long jump random walks. Proposition 3.3. Let s (0, ) and let ( ) s : S L 2 (R n ) be the fractional Laplacian operator defined by (3.). Then, for any u S, ( ) s u = F ( ξ 2s (F u)) ξ R n. (3.8) Proof. In view of Lemma 3.2, we may use the definition via the weighted second order differential quotient in (3.3). We denote by L u the integral in (3.3), that is L u(x) = 2 C(n, s) u(x + y) + u(x y) 2u(x) dy, R y n n+2s with C(n, s) as in (3.2). L is a linear operator and we are looking for its symbol (or multiplier ), that is a function S : R n R such that We want to prove that L u = F (S(F u)). (3.9) S(ξ) = ξ 2s, (3.0) where we denoted by ξ the frequency variable. To this scope, we point out that u(x + y) + u(x y) 2u(x) y n+2s ( ) 4 χ B (y) y 2 n 2s sup D 2 u + χ R n \B (y) y n 2s sup u B (x) ) C (χ B (y) y 2 n 2s ( + x n+ ) + χ R n \B (y) y n 2s L (R 2n ). 4 R n

15 Consequently, by the Fubini-Tonelli s Theorem, we can exchange the integral in y with the Fourier transform in x. Thus, we apply the Fourier transform in the variable x in (3.9) and we obtain S(ξ)(F u)(ξ) = F (L u) = 2 C(n, s) F (u(x + y) + u(x y) 2u(x)) R y n n+2s = 2 C(n, s) e iξ y + e iξ y 2 dy(f u)(ξ) R y n n+2s cos(ξ y) = C(n, s) dy(f u)(ξ). (3.) R y n n+2s Hence, in order to obtain (3.0), it suffices to show that R n cos(ξ y) y n+2s dy = C(n, s) ξ 2s. (3.2) To check this, first we observe that, if ζ = (ζ,..., ζ n ) R n, we have near ζ = 0. Thus, cos ζ ζ n+2s ζ 2 ζ n+2s ζ n 2+2s cos ζ dζ is finite and positive. (3.3) R ζ n n+2s Now, we consider the function I : R n R defined as follows cos (ξ y) I(ξ) = dy. R y n n+2s We have that I is rotationally invariant, that is dy I(ξ) = I( ξ e ), (3.4) where e denotes the first direction vector in R n. Indeed, when n =, then we can deduce (3.4) by the fact that I( ξ) = I(ξ). When n 2, 5

16 we consider a rotation R for which R( ξ e ) = ξ and we denote by R T transpose. Then, by substituting ỹ = R T y, we obtain ( cos (R( ξ e )) y ) I(ξ) = dy R y n n+2s ( cos ( ξ e ) (R T y) ) = dy R y n n+2s ( cos ( ξ e ) ỹ ) = dỹ = I( ξ e R ỹ n n+2s ), its which proves (3.4). As a consequence of (3.3) and (3.4), the substitution ζ = ξ y gives that I(ξ) = I( ξ e ) cos ( ξ y ) = dy R y n n+2s = cos ζ ξ n R n ζ/ ξ n+2s dζ = C(n, s) ξ 2s. where we recall that C(n, s) cos(ζ ) is equal to dζ by (3.2). R ζ n n+2s Hence, we deduce (3.2) and then the proof is complete. Proposition 3.4. Let s (0, ). Then the fractional Sobolev space H s (R n ) defined in Section 2 coincides with Ĥs (R n ) defined in (3.7). In particular, for any u H s (R n ) [u] 2 H s (R n ) = 2C(n, s) R n ξ 2s F u(ξ) 2 dξ. where C(n, s) is defined by (3.2). Proof. For every fixed y R n, by changing of variable choosing z = x y, 6

17 we get (Rn ) u(x) u(y) 2 dx dy = x y n+2s R n = = = R n R n R n R n where Plancherel Formula has been used. Rn u(z + y) u(y) 2 dz dy z ( n+2s ) 2 u(z + y) u(y) R z n n/2+s dy dz 2 u(z + ) u( ) z n/2+s dz L 2 (R n ) ( ) u(z + ) u( ) 2 F dz, z n/2+s L 2 (R n ) Now, using (3.2) we obtain ( ) u(z + ) u( ) 2 F dz R z n n/2+s L 2 (R n ) Rn e iξ z 2 = F u(ξ) 2 dξ dz This completes the proof. R n = 2 R n z n+2s ( cos ξ z) R z n n+2s F u(ξ) 2 dz dξ = 2C(n, s) R n ξ 2s F u(ξ) 2 dξ. Remark 3.5. The equivalence of the spaces H s and Ĥs stated in Proposition 3.4 relies on Plancherel Formula. As well known, unless p = q = 2, one cannot go forward and backward between an L p and an L q via Fourier transform (see, for instance, the sharp inequality in [5] for the case < p < 2 and q equal to the conjugate exponent p/(p ) ). That is why the general fractional space defined via Fourier transform for < p < and s > 0, say H s,p (R n ), does not coincide with the fractional Sobolev spaces W s,p (R n ) and will be not discussed here (see, e.g., [0]). Finally, we are able to prove the relation between the fractional Laplacian operator ( ) s and the fractional Sobolev space H s. 7

18 Proposition 3.6. Let s (0, ) and let u H s (R n ). Then, where C(n, s) is defined by (3.2). [u] 2 H s (R n ) = 2C(n, s) ( ) s 2 u 2 L 2 (R n ). (3.5) Proof. The equality in (3.5) plainly follows from Proposition 3.3 and Proposition 3.4. Indeed, ( ) s 2 u 2 L 2 (R n ) = F ( ) s 2 u 2 L 2 (R n ) = ξ s F u 2 L 2 (R n ) = 2 C(n, s)[u]2 H s (R n ). Remark 3.7. In the same way as the fractional Laplacian ( ) s is related to the space W ( s,2 as its Euler-Lagrange equation or from the formula u 2 W = u( ) s u dx ), a more general integral operator can be defined s,2 that is related to the space W s,p for any p (see the recent paper [52]). Armed with the definition of H s (R n ) via the Fourier transform, we can easily analyze the traces of the Sobolev functions (see the forthcoming Proposition 3.8). We will follow Sections 3, 5 and 6 in [9]. Let R n be an open set with continuous boundary. Denote by T the trace operator, namely the linear operator defined by the uniformly continuous extension of the operator of restriction to for functions in D(), that is the space of functions C 0 (R n ) restricted 6 to. Now, for any x = (x, x n ) R n and for any u S (R n ), we denote by v S (R n ) the restriction of u on the hyperplane x n = 0, that is v(x ) = u(x, 0) x R n. (3.6) Then, we have F v(ξ ) = R F u(ξ, ξ n ) dξ n ξ R n, (3.7) 6 Notice that we cannot simply take T as the restriction operator to the boundary, since the restriction to a set of measure 0 (like the set ) is not defined for functions which are not smooth enough. 8

19 where, for the sake of simplicity, we keep the same symbol F for both the Fourier transform in n and in n variables. To check (3.7), we write F v(ξ ) = = (2π) n 2 (2π) n 2 R n e iξ x R n e iξ x v(x ) dx u(x, 0) dx. (3.8) On the other hand, we have F u(ξ, ξ n ) dξ n R = = = R (2π) n 2 (2π) n 2 (2π) n 2 e i (ξ,ξ n) (x,x n) u(x, x n ) dx dx n dξ n R n [ ] e iξn xn u(x, x (2π) n ) dx n dξ n dx 2 R R R n e iξ x R n e iξ x [ u(x, 0) ] dx, where the last equality follows by transforming and anti-transforming u in the last variable, and this coincides with (3.8). Now, we are in position to characterize the traces of the function in H s (R n ), as stated in the following proposition. Proposition 3.8. ([9, Lemma 6.]). Let s > /2, then any function u H s (R n ) has a trace v on the hyperplane { x n = 0 }, such that v H s 2 (R n ). Also, the trace operator T is surjective from H s (R n ) onto H s 2 (R n ). Proof. In order to prove the first claim, it suffices to show that there exists an universal constant C such that, for any u S (R n ) and any v defined as in (3.6), v H s 2 (R n ) C u H s (R n ). (3.9) 9

20 By taking into account (3.7), the Cauchy-Schwarz inequality yields ( ) ( ) F v(ξ ) 2 ( + ξ 2 ) s F u(ξ, ξ n ) 2 dξ n dξ n. (3.20) R R ( + ξ 2 ) s Using the changing of variable formula by setting ξ n = t + ξ 2, we have R dξ n ( + ξ 2 ) s = R ( + ξ 2) /2 ( ( + ξ 2 )( + t 2 ) ) s dt = ( + ξ ) 2 s R ( + t 2 ) s dt = C(s) ( + ξ 2) 2 s, (3.2) dt where C(s) := < + since s > /2. R ( + t 2 ) s Combining (3.20) with (3.2) and integrating in ξ R n, we obtain ( + ξ 2) s 2 F v(ξ ) 2 dξ R n ( C(s) ) + ξ 2 s F u(ξ, ξ n ) 2 dξ n dξ, R n R that is (3.9). Now, we will prove the surjectivity of the trace operator T. For this, we show that for any v H s 2 (R n ) the function u defined by ( ) F u(ξ, ξ n ) = F v(ξ ξ n ) ϕ + ξ 2 + ξ, (3.22) 2 with ϕ C 0 (R) and R R ϕ(t) dt =, is such that u H s (R n ) and T u = v. Indeed, we integrate (3.22) with respect to ξ n R, we substitute ξ n = t + ξ 2 and we obtain ( ) F u(ξ, ξ n ) dξ n = F v(ξ ξ n ) ϕ + ξ 2 + ξ dξ 2 n = R R F v(ξ ) ϕ(t) dt = F v(ξ ) (3.23) 20

21 and this implies v = T u because of (3.7). The proof of the H s -boundedness of u is straightforward. In fact, from (3.22), for any ξ R n, we have ( ) + ξ 2 s F u(ξ, ξ n ) 2 dξ n R = R ( + ξ 2 ) s F v(ξ ) 2 ϕ ( ) ξ 2 n + ξ 2 + ξ dξ 2 n = C ( + ξ 2 ) s 2 F v(ξ ) 2, (3.24) where we used again the changing of variable formula with ξ n = t + ξ 2 ( and the constant C is given by ) + t 2 s ϕ(t) 2 dt. Finally, we obtain that u H s (R n ) by integrating (3.24) in ξ R n. R Remark 3.9. We conclude this section by recalling that the fractional Laplacian ( ) s, which is a nonlocal operator on functions defined in R n, may be reduced to a local, possibly singular or degenerate, operator on functions sitting in the higher dimensional half-space R n+ We have ( ( ) s u(x) = C lim t t 0 2s U + = R n (0, + ). ) (x, t), t where the function U : R n+ + R solves div(t 2s U) = 0 in R n+ + and U(x, 0) = u(x) in R n. This approach was pointed out by Caffarelli and Silvestre in [9]; see, in particular, Section 3.2 there, where was also given an equivalent definition of the H s (R n )-norm: R n ξ 2s F u 2 dξ = C R n+ + U 2 t 2s dx dt. The cited results turn out to be very fruitful in order to recover an elliptic PDE approach in a nonlocal framework, and they have recently been used very often (see, e.g., [8, 88, 3, 7, 78], etc.). 2

22 4. Asymptotics of the constant C(n, s) In this section, we go into detail on the constant factor C(n, s) that appears in the definition of the fractional Laplacian (see (3.)), by analyzing its asymptotic behavior as s and s 0 +. This is relevant if one wants to recover the Sobolev norms of the spaces H (R n ) and L 2 (R n ) by starting from the one of H s (R n ). We recall that in Section 3, the constant C(n, s) has been defined by ( cos(ζ ) C(n, s) = R ζ n n+2s ) dζ. Precisely, we are interested in analyzing the asymptotic behavior as s 0 + and s of a scaling of the quantity in the right hand-side of the above formula. By changing variable η = ζ / ζ, we have cos(ζ ) cos(ζ ) dζ = dζ dζ R ζ n n+2s R R ζ n n+2s ( + ζ 2 / ζ 2 ) n+2s 2 cos(ζ ) = dη dζ R R ζ n +2s ( + η 2 ) n+2s 2 A(n, s) B(s) = s( s) where and 7 B(s) = s( s) A(n, s) = dη (4.) R n ( + η 2 ) n+2s 2 R cos t t +2s dt. (4.2) Proposition 4.. For any n >, let A and B be defined by (4.) and (4.2) respectively. The following statements hold: 7 Of course, when n = (4.) reduces to A(n, s) =, so we will just consider the case n >. 22

23 + (i) lim A(n, s) = ω ρ n 2 n 2 dρ < + ; s 0 ( + ρ 2 ) n (ii) lim A(n, s) = ω ρ n 2 n 2 dρ < + ; s 0 + ( + ρ 2 ) n 2 (iii) (iv) lim s B(s) = 2 ; lim B(s) =, s where ω n 2 denotes (n 2)-dimensional measure of the unit sphere S n 2. As a consequence, and ( C(n, s) + lim s s( s) = ωn ρ n 2 ( + ρ 2 ) n 2 C(n, s) + lim (ω s 0 + s( s) = ρ n 2 n 2 ( + ρ 2 ) n 2 Proof. First, by polar coordinates, for any s (0, ), we get + dη ρ n 2 = ω R n ( + η 2 ) n+2s n 2 dρ. 2 0 ( + ρ 2 ) n+2s 2 Now, observe that for any s (0, ) and any ρ 0, we have 0 ) dρ (4.3) + dρ). (4.4) ρ n 2 ( + ρ 2 ) n+2s 2 ρ n 2 ( + ρ 2 ) n 2 and the function in the right hand-side of the above inequality belongs to L ((0, + )) for any n >. Then, the Dominated Convergence Theorem yields and lim s A(n, s) = ω n ρ n 2 ( + ρ 2 ) n 2 + lim A(n, s) = ω ρ n 2 n 2 s 0 + ( + ρ 2 ) n dρ dρ. 23

24 This proves (i) and (ii). Now, we want to prove (iii). First, we split the integral in (4.2) as follows cos t cos t cos t dt = dt + dt. t +2s t +2s t +2s R Also, we have that 0 t t < t cos t + dt 4 t +2s t dt = 2 +2s s and t < cos t t +2s dt t < t 2 t 3 dt C dt = 2C 2 t +2s t < t +2s 3 2s, for some suitable positive constant C. From the above estimates it follows that cos t lim s( s) dt = 0 s t +2s t and cos t lim s( s) s t < t +2s dt = lim s( s) s t < t 2 dt. 2 t +2s Hence, we get lim s s B(s) = lim ( s( s) 0 ) t 2s s( s) dt = lim s 2( s) = 2. Similarly, we can prove (iv). For this we notice that 0 t < cos t t +2s dt C 0 t 2s dt which yields cos t lim s( s) dt = 0. s 0 + t < t +2s 24

25 Now, we observe that for any k N, k, we have 2(k+)π cos t dt 2kπ t+2s = 2kπ+π cos t 2kπ+π cos(τ + π) dt + 2kπ t+2s 2kπ (τ + π) 2kπ+π ( ) = cos t 2kπ t dt +2s (t + π) +2s 2kπ+π t +2s (t + π) +2s dt = = 2kπ 2kπ+π 2kπ 2kπ+π 2kπ 2kπ+π 2kπ 2kπ+π 2kπ 2kπ+π 2kπ As a consequence, + cos t t +2s dt (t + π) +2s t +2s dt t +2s (t + π) +2s ( π t +2s (t + π) +2s 3π(t + π) 2s dt t +2s (t + π) +2s 3π t(t + π) dt 3π t 2 dt C k 2. 2π t dt + log(2π) + up to relabeling the constant C > 0. It follows that cos t dt t +2s t t t +2s dt k= +2s dτ ) ( + 2s)(t + ϑ) 2s dϑ dt + 2(k+)π k= = 2kπ C k 2 C, = 2 cos t t cos t t +2s dt dt +2s t + cos t dt t+2s C

26 and then lim s( s) s 0 + t cos t t +2s dt = lim s( s) dt. s 0 + t t +2s Hence, we can conclude that lim B(s) = s 0 + lim s( s) s 0 + = lim 2s( s) s 0 + 2s( s) = lim s 0 + 2s t + =. t +2s dt t 2s dt Finally, (4.3) and (4.4) easily follow combining the previous estimates and recalling that s( s) C(n, s) = A(n, s)b(s). The proof is complete. Corollary 4.2. For any n >, let C(n, s) be defined by (3.2). following statements hold: (i) C(n, s) lim s s( s) = 4n ; ω n The (ii) C(n, s) lim s 0 + s( s) = 2. ω n where ω n denotes the (n )-dimensional measure of the unit sphere S n. Proof. For any θ R such that θ > n, let us define E n (θ) := + 0 ρ n 2 ( + ρ 2 ) θ 2 dρ. 26

27 Observe that the assumption on the parameter θ ensures the convergence of the integral. Furthermore, integrating by parts we get Then, we set E n (θ) = = = + n θ n (ρ n ) ( + ρ 2 ) θ 2 ρ n ( + ρ 2 ) θ+2 2 dρ dρ θ n E n+2(θ + 2). (4.5) and + I n () := E n (n + 2) = 0 + I n (0) := E n (n) = 0 In view of (4.5), it follows that I n () and I n (0) way, since ρ n 2 ( + ρ 2 ) n 2 ρ n 2 ( + ρ 2 ) n 2 + dρ dρ. can be obtained in a recursive I () n+2 = E n+2 (n + 4) = n n + 2 E n(n + 2) = n n + 2 I() n (4.6) and I (0) n+2 = E n+2 (n + 2) = n n E n(n) = n n I(0) n. (4.7) and Now we claim that I () n = ω n 2nω n 2 (4.8) I (0) n = ω n 2ω n 2. (4.9) We will prove the previous identities by induction. We start by noticing that the inductive basis are satisfied, since + I () 2 = 0 ( + ρ 2 ) dρ = π + 2 4, ρ I() 3 = 0 ( + ρ 2 ) 5 2 dρ = 3 27

28 and + I (0) 2 = 0 ( + ρ 2 ) dρ = π + 2, ρ I(0) 3 = 0 ( + ρ 2 ) 3 2 dρ =. Now, using (4.6) and (4.7), respectively, it is clear that in order to check the inductive steps, it suffices to verify that = n ω n. (4.0) ω n n ω n 2 ω n+ We claim that the above formula plainly follows from a classical recursive formula on ω n, that is ω n = 2π n ω n 2. (4.) To prove this, let us denote by ϖ n the Lebesgue measure of the n- dimensional unit ball and let us fix the notation x = ( x, x ) R n 2 R 2. By integrating on R n 2 and then using polar coordinates in R 2, we see that ( ) ϖ n = dx = d x dx x 2 = ϖ n 2 x x x 2 x 2 ( x 2) (n 2) 2 dx = 2πϖ n 2 ρ ( ρ 2) (n 2) 2 dρ = 2πϖ n 2. (4.2) n Moreover, by polar coordinates in R n, ϖ n = x 0 dx = ω n Thus, we use (4.3) and (4.2) and we obtain ω n = nϖ n = 2πϖ n 2 = 2πω n 3 n 2, 0 ρ n dρ = ω n n. (4.3) which is (4.), up to replacing n with n. In turn, (4.) implies (4.0) and so (4.8) and (4.9). 28

29 and Finally, using (4.8), (4.9) and Proposition 4. we can conclude that as desired 8. C(n, s) lim s s( s) = 2 ω n 2 I n () C(n, s) lim s 0 + s( s) = ω n 2 I n (0) = 4n ω n = 2 ω n, Remark 4.3. It is worth noticing that when p = 2 we recover the constants C and C 2 in (2.8) and (2.9), respectively. In fact, in this case it is known that C = 2 S n ξ 2 dσ(ξ) = 2n n i= ξ i 2 dσ(ξ) = ω n S 2n n and C 2 = ω n (see [] and [66]). Then, by Proposition 3.5 and Corollary 4.2 it follows that Rn u(x) u(y) 2 lim s ( s) dx dy x y n+2s R n = lim s 2( s)c(n, s) ξ s F u 2 L 2 (R n ) = ω n 2n u 2 L 2 (R n ) and lim s s 0 + R n Rn u(x) u(y) 2 = C u 2 H (R n ) x y n+2s dx dy = lim s 0 + 2sC(n, s) ξ s F u 2 L 2 (R n ) = ω n u 2 L 2 (R n ) = C 2 u 2 L 2 (R n ). 8 Another (less elementary) way to obtain this result is to notice that E n (θ) = 2 B ((n )/2, (θ n )/2), where B is the Beta function. 29

30 We will conclude this section with the following proposition that one could plainly deduce from Proposition 3.3. We prefer to provide a direct proof, based on Lemma 3.2, in order to show the consistency in the definition of the constant C(n, s). Proposition 4.4. Let n >. For any u C 0 (R n ) the following statements hold: (i) lim s 0 +( ) s u = u; (ii) lim s ( ) s u = u. Proof. Fix x R n, R 0 > 0 such that supp u B R0 and set R = R 0 + x +. First, u(x + y) + u(x y) 2u(x) dy y B R y n+2s u C 2 (R n ) BR 2 dy y n+2s ω n u C 2 (R n ) R 0 ρ 2s dρ = ω n u C 2 (R n )R 2 2s. (4.4) 2( s) Furthermore, observe that y R yields x±y y x R x > R 0 and consequently u(x ± y) = 0. Therefore, u(x + y) + u(x y) 2u(x) dy = u(x) dy 2 R n \B R y n+2s R n \B R y n+2s = ω n u(x) + R ρ 2s+ dρ = ω n R 2s u(x). (4.5) 2s Now, by (4.4) and Corollary 4.2, we have s) u(x + y) + u(x y) 2u(x) lim C(n, dy = 0 s B R y n+2s 30

31 and so we get, recalling Lemma 3.2, s) u(x + y) + u(x y) 2u(x) lim u = lim C(n, s 0 +( )s s R n \B R y n+2s C(n, s)ω n R 2s = lim u(x) = u(x), s 0 + 2s dy where the last identities follow from (4.5) and again Corollary 4.2. This proves (i). Similarly, we can prove (ii). In this case, when s goes to, we have no contribution outside the unit ball, as the following estimate shows u(x + y) + u(x y) 2u(x) dy R n \B y n+2s 4 u L (R n ) dy y n+2s 4ω n u L (R n ) = 2ω n u L s (R n ). R n \B + ρ 2s+ dρ As a consequence (recalling Corollary 4.2), we get s) u(x + y) + u(x y) 2u(x) lim C(n, dy = 0. (4.6) s 2 R n \B y n+2s On the other hand, we have u(x + y) + u(x y) 2u(x) D 2 u(x)y y dy B y n+2s 3 u C 3 (R n ) ω n u C 3 (R n ) = ω n u C 3 (R n ) 3 2s B y 3 y 0 n+2s dy ρ 2s 2 dρ

32 and this implies that s) lim C(n, s 2 u(x + y) + u(x y) 2u(x) dy B y n+2s s) = lim C(n, s 2 Now, notice that if i j then iju(x)y 2 i y j dy = iju(x)ỹ 2 i ỹ j dỹ, B B where ỹ k = y k for any k j and ỹ j = y j, and thus B D 2 u(x)y y y n+2s dy. (4.7) B 2 iju(x)y i y j dy = 0. (4.8) Also, up to permutations, for any fixed i, we get iiu(x)y 2 2 i dy = 2 y 2 i B y iiu(x) n+2s B y dy = y n+2s 2 2 iiu(x) dy B y n+2s n = 2 iiu(x) yj n B 2 dy = 2 iiu(x) y y n+2s n B 2 dy y n+2s j= = 2 iiu(x) ω n 2n( s). (4.9) Finally, combining (4.6), (4.7), (4.8), (4.9), Lemma 3.2 and Corollary 4.2, we can conclude s) u(x + y) + u(x y) 2u(x) lim u = lim C(n, dy s ( )s s 2 B y n+2s s) D 2 u(x)y y = lim C(n, dy s 2 B y n+2s s) n = lim C(n, iiu(x)y 2 i 2 dy s 2 i= B y n+2s = lim C(n, s)ω n n 2 s 4n( s) iiu(x) = u(x). i= 32

33 5. Extending a W s,p () function to the whole of R n As well known when s is an integer, under certain regularity assumptions on the domain, any function in W s,p () may be extended to a function in W s,p (R n ). Extension results are quite important in applications and are necessary in order to improve some embeddings theorems, in the classic case as well as in the fractional case (see Section 6 and Section 7 in the following). For any s (0, ) and any p [, ), we say that an open set R n is an extension domain for W s,p if there exists a positive constant C = C(n, p, s, ) such that: for every function u W s,p () there exists ũ W s,p (R n ) with ũ(x) = u(x) for all x and ũ W s,p (R n ) C u W s,p (). In general, an arbitrary open set is not an extension domain for W s,p. To the authors knowledge, the problem of characterizing the class of sets that are extension domains for W s,p is open 9. When s is an integer, we cite [58] for a complete characterization in the special case s =, p = 2 and n = 2, and we refer the interested reader to the recent book by Leoni [60], in which this problem is very well discussed (see, in particular, Chapter and Chapter 2 there). In this section, we will show that any open set of class C 0, with bounded boundary is an extension domain for W s,p. We start with some preliminary lemmas, in which we will construct the extension to the whole of R n of a function u defined on in two separated cases: when the function u is identically zero in a neighborhood of the boundary and when coincides with the half-space R n +. Lemma 5.. Let be an open set in R n and u a function in W s,p () with s (0, ) and p [, + ). If there exists a compact subset K such that u 0 in \ K, then the extension function ũ defined as { u(x) x, ũ(x) = (5.) 0 x R n \ 9 While revising this paper, we were informed that an answer to this question has been given by Zhou, by analyzing the link between extension domains in W s,p and the measure density condition (see [02]). 33

34 belongs to W s,p (R n ) and ũ W s,p (R n ) C u W s,p (), where C is a suitable positive constant depending on n, p, s, K and. Proof. Clearly ũ L p (R n ). Hence, it remains to verify that the Gagliardo norm of ũ in R n is bounded by the one of u in. Using the symmetry of the integral in the Gagliardo norm with respect to x and y and the fact that ũ 0 in R n \, we can split as follows R n Rn ũ(x) ũ(y) p x y n+sp dx dy = +2 u(x) u(y) p dx dy (5.2) x y ( n+sp ) u(x) p dy dx, x y n+sp R n \ where the first term in the right hand-side of (5.2) is finite since u W s,p (). Furthermore, for any y R n \ K, and so ( u(x) p x y = χ K(x) u(x) p χ n+sp x y n+sp K (x) u(x) p sup x K x y n+sp R n \ ) u(x) p dy dx x y n+sp R n \ dist(y, K) n+sp dy u p L p (). (5.3) Note that the integral in (5.3) is finite since dist(, K) α > 0 and n + sp > n. Combining (5.2) with (5.3), we get where C = C(n, s, p, K). ũ W s,p (R n ) C u W s,p () Lemma 5.2. Let be an open set in R n, symmetric with respect to the coordinate x n, and consider the sets + = {x : x n > 0} and = {x : x n 0}. Let u be a function in W s,p ( + ), with s (0, ) and p [, + ). Define ū(x) = { u(x, x n ) x n 0, u(x, x n ) x n < 0. (5.4) Then ū belongs to W s,p () and ū W s,p () 4 u W s,p ( + ). 34

35 Proof. By splitting the integrals and changing variable ˆx = (x, x n ), we get ū p L p () = u(x) p dx + u(ˆx, ˆx n ) p d x = 2 u p L p ( + ). (5.5) + + Also, if x R n + and y C R n + then (x n y n ) 2 (x n + y n ) 2 and therefore ū(x) ū(y) p u(x) u(y) dx dy = x y n+sp + + p dx dy x y n+sp u(x) u(y +2 + C, y n ) p dx dy + x y n+sp u(x + C + C, x n ) u(y, y n ) p dx dy + x y n+sp This concludes the proof. 4 u p W s,p ( + ). Now, a truncation lemma near. Lemma 5.3. Let be an open set in R n, s (0, ) and p [, + ). Let us consider u W s,p () and ψ C 0, (), 0 ψ. Then ψu W s,p () and ψ u W s,p () C u W s,p (), (5.6) where C = C(n, p, s, ). Proof. It is clear that ψ u L p () u L p () since ψ. Furthermore, adding and subtracting the factor ψ(x)u(y), we get ψ(x) u(x) ψ(y) u(y) p dx dy x y n+sp ( 2 p ψ(x) u(x) ψ(x) u(y) p dx dy x y n+sp ) ψ(x) u(y) ψ(y) u(y) p + dx dy x y n+sp ( 2 p u(x) u(y) p dx dy (5.7) x y n+sp ) u(x) p ψ(x) ψ(y) p + dx dy. x y n+sp 35

36 Since ψ belongs to C 0, (), we have u(x) p ψ(x) ψ(y) p dx dy Λ p x y n+sp + x y x y u(x) p x y p x y n+sp dx dy u(x) p dx dy x y n+sp C u p L p (), (5.8) where Λ denotes the Lipschitz constant of ψ and C is a positive constant depending on n, p and s. Note that the last inequality follows from the fact that the kernel x y n+( s)p is summable with respect to y if x y since n + (s )p < n and, on the other hand, the kernel x y n sp is summable when x y since n + sp > n. Finally, combining (5.7) with (5.8), we obtain estimate (5.6). Now, we are ready to prove the main theorem of this section, that states that every open Lipschitz set with bounded boundary is an extension domain for W s,p. Theorem 5.4. Let p [, + ), s (0, ) and R n be an open set of class C 0, with bounded boundary 0. Then W s,p () is continuously embedded in W s,p (R n ), namely for any u W s,p () there exists ũ W s,p (R n ) such that ũ = u and ũ W s,p (R n ) C u W s,p () where C = C(n, p, s, ). Proof. Since is compact, we can find a finite number of balls B j such k k that B j and so we can write R n = B j (R n \ ). j= If we consider this covering, there exists a partition of unity related to it, i.e. there exist k + smooth functions ψ 0, ψ,..., ψ k such that j= 0 Motivated by an interesting remark of the anonymous Referee, we point out that it should be expected that the Lipschitz assumption on the boundary of may be weakened when s (0, ), since in the case s = 0 clearly no regularity at all is needed for the extension problem. 36

37 spt ψ 0 R n \, spt ψ j B j for any j {,..., k}, 0 ψ j for any k j {0,..., k} and ψ j =. Clearly, j=0 u = k ψ j u. j=0 By Lemma 5.3, we know that ψ 0 u belongs to W s,p (). Furthermore, since ψ 0 u 0 in a neighborhood of, we can extend it to the whole of R n, by setting { ψ 0 u(x) x, ψ 0 u(x) = 0 x R n \ and ψ 0 u W s,p (R n ). Precisely ψ 0 u W s,p (R n ) C ψ 0 u W s,p () C u W s,p (), (5.9) where C = C(n, s, p, ) (possibly different step by step, see Lemma 5. and Lemma 5.3). For any j {,..., k}, let us consider u Bj and set v j (y) := u (T j (y)) for any y Q +, where T j : Q B j is the isomorphism of class C 0, defined in Section 2. Note that such a T j exists by the regularity assumption on the domain. Now, we state that v j W s,p (Q + ). Indeed, using the standard changing variable formula by setting x = T j (ˆx) we have v(ˆx) v(ŷ) Q + Q+ p dˆx dŷ ˆx ŷ n+sp u(t j (ˆx)) u(t j (ŷ)) = Q + Q+ p dˆx dŷ ˆx ŷ n+sp u(x) u(y) p = B j C B j B j B j 37 T j (x) T (y) j det(t n+sp j )dx dy u(x) u(y) p x y n+sp dx dy, (5.0)

38 where (5.0) follows from the fact that T j is bi-lipschitz. Moreover, using Lemma 5.2 we can extend v j to all Q so that the extension v j belongs to W s,p (Q) and v j W s,p (Q) 4 v j W s,p (Q + ). We set w j (x) := v j ( T j (x) ) for any x B j. Since T j is bi-lipschitz, by arguing as above it follows that w j W s,p (B j ). Note that w j u (and consequently ψ j w j ψ j u) on B j. By definition ψ j w j has compact support in B j and therefore, as done for ψ 0 u, we can consider the extension ψ j w j to all R n in such a way that ψ j w j W s,p (R n ). Also, using Lemma 5., Lemma 5.2, Lemma 5.3 and estimate (5.0) we get ψ j w j W s,p (R n ) C ψ j w j W s,p (B j ) C w j W s,p (B j ) C v j W s,p (Q) C v j W s,p (Q + ) C u W s,p ( B j ), (5.) where C = C(n, p, s, ) and it is possibly different step by step. Finally, let ũ = ψ 0 u + k ψ j w j be the extension of u defined on all R n. By construction, it is clear that ũ = u and, combining (5.9) with (5.), we get with C = C(n, p, s, ). j= ũ W s,p (R n ) C u W s,p () Corollary 5.5. Let p [, + ), s (0, ) and be an open set in R n of class C 0, with bounded boundary. Then for any u W s,p (), there exists a sequence {u n } C 0 (R n ) such that u n u as n + in W s,p (), i.e., lim u n u W n + s,p () = 0. 38

39 Proof. The proof follows directly by Theorem 2.4 and Theorem Fractional Sobolev inequalities In this section, we provide an elementary proof of a Sobolev-type inequality involving the fractional norm W s,p (see Theorem 6.5 below). The original proof is contained in the Appendix of [84] and it deals with the case p = 2 (see, in particular, Theorem 7 there). We note that when p = 2 and s [/2, ) some of the statements may be strengthened (see [0]). We also note that more general embeddings for the spaces W s,p can be obtained by interpolation techniques and by passing through Besov spaces; see, for instance, [6, 7, 95, 96, 63]. For a more comprehensive treatment of fractional Sobolev-type inequalities we refer to [6, 62, 2,, 9] and the references therein. We remark that the proof here is self-contained. Moreover, we will not make use of Besov or fancy interpolation spaces. In order to prove the Sobolev-type inequality in forthcoming Theorem 6.5, we need some preliminary results. The first of them is an elementary estimate involving the measure of finite measurable sets E in R n as stated in the following lemma (see [85, Lemma A.] and also [20, Corollary 24 and 25]). Lemma 6.. Fix x R n. Let p [, + ), s (0, ) and E R n measurable set with finite measure. Then, dy x y C n+sp E sp/n, C E be a for a suitable constant C = C(n, p, s) > 0. Proof. We set and then it follows ( E ρ := ω n ) n (C E) B ρ (x) = B ρ (x) E B ρ (x) = E E B ρ (x) = E C B ρ (x). 39

40 Therefore, dy = x y n+sp C E (C E) B ρ(x) (C E) B ρ(x) dy x y + dy n+sp (C E) C B ρ(x) x y n+sp dy ρ n+sp + = (C E) B ρ(x) + ρ n+sp = E C B ρ(x) + ρ n+sp = E C B ρ(x) C B ρ(x) dy x y n+sp + dy x y n+sp. (C E) C B ρ(x) (C E) C B ρ(x) (C E) C B ρ(x) dy x y n+sp dy x y n+sp dy x y n+sp (C E) C B ρ(x) dy x y n+sp The desired result easily follows by using polar coordinates centered at x. Now, we recall a general statement about a useful summability property (see [84, Lemma 5]. For related results, see also [38, Lemma 4]). Lemma 6.2. Let s (0, ) and p [, + ) such that sp < n. Fix T > ; let N Z and Then, a k be a bounded, nonnegative, decreasing sequence with a k = 0 for any k N. k Z a (n sp)/n k T k C a k+ a sp/n k T k, k Z a k 0 for a suitable constant C = C(n, p, s, T ) > 0, independent of N. Proof. By (6.), (6.) both k Z a (n sp)/n k T k and a k+ a sp/n k T k are convergent series. (6.2) k Z a k 0 40

41 Moreover, since a k is nonnegative and decreasing, we have that if a k = 0, then a k+ = 0. Accordingly, a (n sp)/n k+ T k = a (n sp)/n k+ T k. k Z k Z a k 0 Therefore, we may use the Hölder inequality with exponents α := n/sp and β := n/(n sp) by arguing as follows. T k Z a (n sp)/n k T k = a (n sp)/n k+ T k k Z = k Z a k 0 = k Z a k 0 ( k Z ( k Z a (n sp)/n k+ T k ( )( ) a sp/(nβ) k T k/α a /β k+ a sp/(nβ) k T k/β ) ( ) /α α a sp/(nβ) k T k/α a (n sp)/n k ) sp/n T k k Z a k 0 k Z a k 0 ( a /β a k+ a sp/n k So, recalling (6.2), we obtain the desired result. k+ a sp/(nβ) k T k ) β T k/β (n sp)/n We use the above tools to deal with the measure theoretic properties of the level sets of the functions (see [84, Lemma 6]). Lemma 6.3. Let s (0, ) and p [, + ) such that sp < n. Let For any k Z let Then, R n f L (R n ) be compactly supported. (6.3) Rn f(x) f(y) p. /β a k := { f > 2 k }. (6.4) x y n+sp dx dy C k Z a k 0 4 a k+ a sp/n k 2 pk,

42 for a suitable constant C = C(n, p, s) > 0. Proof. Notice that f(x) f(y) f(x) f(y), and so, by possibly replacing f with f, we may consider the case in which f 0. We define A k := { f > 2 k }. (6.5) We remark that A k+ A k, hence We define Notice that a k+ a k. (6.6) D k := A k \ A k+ = {2 k < f 2 k+ } and d k := D k. d k and a k are bounded and they become zero when k is large enough, (6.7) thanks to (6.3). Also, we observe that the D k s are disjoint, that D l = C A k+ (6.8) l Z l k and that D l = A k. (6.9) l Z l k As a consequence of (6.9), we have that a k = l Z l k d l (6.0) and so d l. (6.) d k = a k l Z l k+ 42