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1 Introduction 2

2 The fractional Sobolev W s,p 5

3 The space Hs and the fractional Laplacian operator 11

4 Asymptotics of the constant C(n,s) 22

5 Extending a W s,p(Ω) to the whole of Rn 33

6 Fractional Sobolev inequalities 39

7 Compact embeddings 49

8 H¨older regularity 54

9 Some counterexamples in non-Lipschitz domains 59

References 65

1. Introduction These pages are for students and young researchers of all ages who may like to hitchhike their way from 1 to s ∈ (0, 1). 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 equations, are now experiencing impressive applications in different sub- jects, such as, among others, the thin [87, 69], opti- mization [37], finance [26], phase transitions [2, 14, 88, 40, 45], stratified materials [83, 23, 24], anomalous diffusion [68, 98, 65], crystal disloca- tion [92, 47, 8], soft thin films [57], semipermeable membranes and flame propagation [15], conservation laws [9], ultra-relativistic limits of quan- tum mechanics [41], quasi-geostrophic flows [64, 27, 21], multiple scatter- ing [36, 25, 50], minimal surfaces [16, 20], materials science [4], water waves [81, 100, 99, 32, 29, 74, 33, 34, 31, 30, 42, 51, 75, 35], elliptic problems with measure data [71, 54], non-uniformly elliptic problems [39], gradient [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 spaces4 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

4About 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 recommend all the excellent, classical books on the topic, such as [59, 90, 1, 93, 94, 101, 80, 91, 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(Ω) to W s,p(Rn): technically, this is slightly more complicated than the classi- cal analogue for integer Sobolev spaces, since the extension interacts with the values taken by the function in Ω via the Gagliardo norm and the computations have to take care of it. Sobolev inequalities and continuous embeddings are dealt with in Sec- tion 6, while Section 7 is devoted to compact embeddings. Then, in Sec- tion 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 do- mains. 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 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 S of rapidly decaying C∞ functions in Rn. The topology of this space is generated by the

N α pN (ϕ) = sup (1 + |x|) |D ϕ(x)| , N =0, 1, 2, ... , x∈Rn |αX|≤N where ϕ ∈ S (Rn). Let S ′(Rn) be the set of all tempered distributions, that is the topological dual of S (Rn). As usual, for any ϕ ∈ S (Rn), we denote by F 1 −iξ·x ϕ(ξ) = n/2 e ϕ(x) dx (2π) Rn Z the Fourier transform of ϕ and we recall that one can extend F from S (Rn) to S ′(Rn). Let Ω be a general, possibly non smooth, open set in Rn. For any real s > 0 and for any p ∈ [1, ∞), we want to define the fractional Sobolev spaces W s,p(Ω). 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, 1). For any p ∈ [1, +∞), we define W s,p(Ω) as follows

s,p p |u(x) − u(y)| p W (Ω) := u ∈ L (Ω) : n +s ∈ L (Ω × Ω) ; (2.1) ( |x − y| p ) i.e, an intermediary between Lp(Ω) and W 1,p(Ω), endowed with the natural norm

1 p p p |u(x) − u(y)| kuk s,p := |u| dx + dxdy , (2.2) W (Ω) |x − y|n+sp ZΩ ZΩ ZΩ  5 where the term 1 |u(x) − u(y)|p p [u] s,p := dxdy W (Ω) |x − y|n+sp ZΩ ZΩ  is the so-called Gagliardo (semi )norm of u. 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 2.1. Let p ∈ [1, +∞) and 0 < s ≤ s′ < 1. Let Ω be an open set in Rn and u : Ω → R be a measurable function. Then

kukW s,p(Ω) ≤ CkukW s′,p(Ω) for some suitable positive constant C = C(n,s,p) ≥ 1. In particular,

′ W s ,p(Ω) ⊆ W s,p(Ω) . Proof. First, |u(x)|p 1 dxdy ≤ dz |u(x)|p dx |x − y|n+sp |z|n+sp ZΩ ZΩ ∩ {|x−y|≥1} ZΩ Z|z|≥1  p ≤ C(n,s,p)kukLp(Ω) , where we used the fact that the kernel 1/|z|n+sp is integrable since n+sp > n. Taking into account the above estimate, it follows |u(x) − u(y)|p dxdy |x − y|n+sp ZΩ ZΩ ∩ {|x−y|≥1} |u(x)|p + |u(y)|p ≤ 2p−1 dxdy |x − y|n+sp ZΩ ZΩ ∩{|x−y|≥1} p p ≤ 2 C(n,s,p)kukLp(Ω) . (2.3) On the other hand, |u(x) − u(y)|p |u(x) − u(y)|p dxdy ≤ dx dy . |x − y|n+sp |x − y|n+s′p ZΩ ZΩ ∩ {|x−y|<1} ZΩ ZΩ ∩ {|x−y|<1} (2.4)

6 Thus, combining (2.3) with (2.4), we get

|u(x) − u(y)|p dxdy |x − y|n+sp ZΩ ZΩ |u(x) − u(y)|p ≤ 2pC(n,s,p)kukp + dxdy Lp(Ω) |x − y|n+s′p ZΩ ZΩ and so

p p p p |u(x) − u(y)| kuk s,p ≤ 2 C(n,s,p)+1 kuk p + dxdy W (Ω) L (Ω) |x − y|n+s′p ZΩ ZΩ  p ≤ C(n,s,p)kuk ′ , W s ,p(Ω) 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 2.1 holds also in the limit case, namely when s′ = 1, but for this we have to take into account the regularity of ∂Ω (see Example 9.1). As usual, for any k ∈ N and α ∈ (0, 1], we say that Ω is of class Ck,α if there exists M > 0 such that for any x ∈ ∂Ω there exists a ball B = Br(x), r> 0, and an isomorphism T : Q → B such that

k,α −1 k,α T ∈ C (Q), T ∈ C (B), T (Q+)= B ∩ Ω, T (Q0)= B ∩ ∂Ω

−1 and kT kCk,α(Q) + kT kCk,α(B) ≤ M, where

′ n−1 ′ Q := x =(x , xn) ∈ R × R : |x | < 1 and |xn| < 1 ,

 ′ n−1 ′ Q+ := x =(x , xn) ∈ R × R : |x | < 1 and 0 < xn < 1

 and Q0 := {x ∈ Q : xn =0} .

We have the following result.

7 Proposition 2.2. Let p ∈ [1, +∞) and s ∈ (0, 1). Let Ω be an open set in Rn of class C0,1 with bounded boundary and u : Ω → R be a measurable function. Then kukW s,p(Ω) ≤ CkukW 1,p(Ω) (2.5) for some suitable positive constant C = C(n,s,p) ≥ 1. In particular, W 1,p(Ω) ⊆ W s,p(Ω) . Proof. Let u ∈ W 1,p(Ω). Thanks to the regularity assumptions on the domain Ω, we can extend u to a functionu ˜ : Rn → R such thatu ˜ ∈ 1,p n W (R ) and ku˜kW 1,p(Rn) ≤ CkukW 1,p(Ω) for a suitable constant C (see, e.g., [49, Theorem 7.25]). Now, using the change of variable z = y − x and the H¨older inequality, we have |u(x) − u(y)|p dxdy |x − y|n+sp ZΩ ZΩ ∩ {|x−y|<1} |u(x) − u(z + x)|p ≤ dz dx |z|n+sp ZΩ ZB1 |u(x) − u(z + x)|p 1 = dz dx |z|p |z|n+(s−1)p ZΩ ZB1 p 1 |∇u(x + tz)| ≤ n +s−1 dt dz dx Ω B 0 |z| p Z Z 1 Z ! 1 |∇u˜(x + tz)|p ≤ n+p(s−1) dtdz dx Rn |z| Z ZB1 Z0 1 p k∇u˜k p Rn ≤ L ( ) dtdz |z|n+p(s−1) ZB1 Z0 p ≤ C1(n,s,p)k∇u˜kLp(Rn)

p ≤ C2(n,s,p)kukW 1,p(Ω) . (2.6)

Also, by (2.3), |u(x) − u(y)|p dxdy ≤ C(n,s,p)kukp . (2.7) |x − y|n+sp Lp(Ω) ZΩ ZΩ ∩ {|x−y|≥1} 8 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 9.1 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(Ω). Before going ahead, it is worth explaining why the definition in (2.1) cannot be plainly extended to the case s ≥ 1. Suppose that Ω is a connected open set in Rn, then any measurable function u : Ω → R such that |u(x) − u(u)|p dxdy < +∞ |x − y|n+sp ZΩ ZΩ is actually constant (see [10, 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 1,p(Ω): p |u(x) − u(y)| p lim (1 − s) dxdy = C1 |∇u| dx (2.8) s→1− |x − y|n+sp ZΩ ZΩ ZΩ for a suitable positive constant C1 depending only on n and p (see [11]). In the same spirit, in [66], Maz’ja and Shaposhnikova proved that, for s,p Rn a function u ∈ 0 1 and it is not an integer we write s = m + σ, where m is an integer and σ ∈ (0, 1). In this case the space W s,p(Ω) consists of those equivalence classes of functions u ∈ W m,p(Ω) whose distributional Dαu, with |α| = m, belong to W σ,p(Ω), namely W s,p(Ω) := u ∈ W m,p(Ω) : Dαu ∈ W σ,p(Ω) for any α s.t. |α| = m n (2.10)o

5For the sake of simplicity, in the definition of the fractional Sobolev spaces and those of the corresponding norms in (2.1) 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 1,p and Lp spaces, one may consider to add a factor C(n,p,s) ≈ s(1 − s) in front of the double integral in (2.2).

9 and this is a Banach space with respect to the norm

1 p p α p kuk s,p := kuk + kD uk . (2.11) W (Ω)  W m,p(Ω) W σ,p(Ω) |αX|=m   Clearly, if s = m is an integer, the space W s,p(Ω) coincides with the Sobolev space W m,p(Ω).

Corollary 2.3. Let p ∈ [1, +∞) and s,s′ > 1. Let Ω be an open set in Rn of class C0,1. Then, if s′ ≥ s, we have

′ W s ,p(Ω) ⊆ W s,p(Ω).

Proof. We write s = k + σ and s′ = k′ + σ′, with k,k′ integers and σ, σ′ ∈ (0, 1). In the case k′ = k, we can use Proposition 2.1 in order to conclude that W s′,p(Ω) is continuously embedded in W s,p(Ω). On the other hand, if k′ ≥ k + 1, using Proposition 2.1 and Proposition 2.2 we have the following chain ′ ′ ′ W k +σ ,p(Ω) ⊆ W k ,p(Ω) ⊆ W k+1,p(Ω) ⊆ W k+σ,p(Ω) . The proof is complete.

As in the classic case with s being an integer, any function in the fractional Sobolev space W s,p(Rn) can be approximated by a sequence of smooth functions with compact .

∞ Rn Theorem 2.4. For any s> 0, the space C0 ( ) of smooth functions with compact support is dense in W s,p(Rn).

A proof can be found in [1, Theorem 7.38]. s,p ∞ Let W0 (Ω) denote the closure of C0 (Ω) in the norm k·kW s,p(Ω) defined in (2.11). Note that, in view of Theorem 2.4, we have

s,p Rn s,p Rn W0 ( )= W ( ) , (2.12)

Rn s,p s,p ∞ but in general, for Ω ⊂ , W (Ω) =6 W0 (Ω), i.e. C0 (Ω) is not dense in W s,p(Ω). Furthermore, it is clear that the same inclusions stated in Propo- s,p sition 2.1, Proposition 2.2 and Corollary 2.3 hold for the spaces W0 (Ω).

10 Remark 2.5. For s< 0 and p ∈ (1, ∞), we can define W s,p(Ω) as the dual −s,q space of W0 (Ω) where 1/p +1/q = 1. Notice that, in this case, the space W s,p(Ω) is actually a space of distributions on Ω, since it is the dual of a ∞ space having C0 (Ω) 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 ∈ (1, +∞), assume that the open set Ω ⊆ Rn is sufficiently smooth, then the space of traces Tu 1,p on ∂Ω of u in W (Ω) is characterized by kTuk 1− 1 ,p < +∞ (see [43]). W p (∂Ω) 1− 1 ,p Moreover, the T is surjective from W 1,p(Ω) onto W p (∂Ω). 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 can be find in the forthcoming Proposition 3.8.

3. The space Hs and the fractional Laplacian operator In this section, we focus on the case p = 2. This is quite an important s,2 Rn s,2 Rn case since the fractional Sobolev spaces W ( ) and W0 ( ) turn out s Rn s Rn to be Hilbert spaces. They are usually denoted by H ( ) and H0 ( ), respectively. Moreover, they are strictly related to the fractional Laplacian operator (−∆)s (see Proposition 3.6), where, for any u ∈ S and s ∈ (0, 1), (−∆)s is defined as

s u(x) − u(y) (−∆) u(x) = C(n, s) P.V. n+2s dy (3.1) Rn |x − y| Z u(x) − u(y) = C(n, s) lim n+2s dy. ε→0+ C |x − y| Z Bε(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

−1 1 − cos(ζ1) C(n, s)= n+2s dζ . (3.2) Rn |ζ| Z  The choice of this constant is motived by Proposition 3.3.

11 Remark 3.1. Due to the singularity of the kernel, the right hand-side of (3.1) is not well defined in general. In the case s ∈ (0, 1/2) the integral in (3.1) is not really singular near x. Indeed, for any u ∈ S , we have |u(x) − u(y)| n+2s dy Rn |x − y| Z |x − y| 1 ∞ Rn ≤ C n+2s dy + kukL ( ) n+2s dy |x − y| C |x − y| ZBR Z BR 1 1 = C n+2s−1 dy + n+2s dy |x − y| C |x − y| ZBR Z BR  R 1 +∞ 1 = C dρ + dρ < +∞ |ρ|2s |ρ|2s+1 Z0 ZR  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 (3.1) as a weighted second order differential quotient. Lemma 3.2. Let s ∈ (0, 1) and let (−∆)s be the fractional Laplacian operator defined by (3.1). Then, for any u ∈ S ,

s 1 u(x + y)+ u(x − y) − 2u(x) Rn (−∆) u(x)= − C(n, s) n+2s dy, ∀x ∈ . 2 Rn |y| Z (3.3) Proof. The equivalence of the definitions in (3.1) 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. n+2s dy Rn |x − y| Z u(x + z) − u(x) = − C(n, s) P.V. n+2s dz. (3.4) Rn |z| Z Moreover, by substitutingz ˜ = −z in last term of the above equality, we have u(x + z) − u(x) u(x − z˜) − u(x) P.V. n+2s dz = P.V. n+2s dz.˜ (3.5) Rn |z| Rn |z˜| Z Z 12 and so after relabelingz ˜ as z

u(x + z) − u(x) 2P.V. n+2s dz Rn |z| Z u(x + z) − u(x) u(x − z) − u(x) = P.V. n+2s dz + P.V. n+2s dz Rn |z| Rn |z| Z Z u(x + z)+ u(x − z) − 2u(x) = P.V. n+2s dz. (3.6) Rn |z| Z Therefore, if we rename z as y in (3.4) and (3.6), we can write the fractional Laplacian operator in (3.1) as

s 1 u(x + y)+ u(x − y) − 2u(x) (−∆) u(x) = − C(n, s) P.V. n+2s dy. 2 Rn |y| Z 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

2 u(x + y)+ u(x − y) − 2u(x) kD uk ∞ ≤ L , |y|n+2s |y|n+2s−2 which is integrable near 0 (for any fixed s ∈ (0, 1)). Therefore, since u ∈ S , one can get rid of the P.V. and write (3.3).

3.1. An approach via the Fourier transform Now, we take into account an alternative definition of the space Hs(Rn)= W s,2(Rn) via the Fourier transform. Precisely, we may define

Hˆ s(Rn)= u ∈ L2(Rn): (1 + |ξ|2s)|F u(ξ)|2 dξ < +∞ (3.7) Rn  Z  and we observe that the above definition, unlike the ones via the Gagliardo norm in (2.2), is valid also for any real s ≥ 1. We may also use an analogous definition for the case s< 0 by setting

Hˆ s(Rn)= u ∈ S ′(Rn): (1 + |ξ|2)s|F u(ξ)|2 dξ < +∞ , Rn  Z  13 although in this case the space Hˆ s(Rn) is not a subset of L2(Rn) and, in order to use the Fourier transform, one has to start from an element of S ′(Rn), (see also Remark 2.5). The equivalence of the space Hˆ s(Rn) defined in (3.7) with the one defined in the previous section via the Gagliardo norm (see (2.1)) 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, [91, Chapter 16]). 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, 1) and let (−∆)s : S → L2(Rn) be the fractional Laplacian operator defined by (3.1). Then, for any u ∈ S , (−∆)su = F −1(|ξ|2s(F u)) ∀ξ ∈ Rn. (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 1 u(x + y)+ u(x − y) − 2u(x) u(x)= − C(n, s) n+2s dy, 2 Rn |y| Z 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 : Rn → R such that L u = F −1(S(F u)). (3.9) We want to prove that S(ξ) = |ξ|2s, (3.10) 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 2−n−2s 2 −n−2s n ≤ 4 χB1 (y)|y| sup |D u| + χR \B1 (y)|y| sup |u| Rn B1(x)   2−n−2s n+1 −1 −n−2s 1 2n n R ≤ C χB1 (y)|y| (1 + |x| ) + χR \B1 (y)|y| ∈ L ( ).   14 Consequently, by the Fubini-Tonelli’s Theorem, we can exchange the in- tegral 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) 1 F (u(x + y)+ u(x − y) − 2u(x)) = − C(n, s) n+2s dy 2 Rn |y| Z iξ·y −iξ·y 1 e + e − 2 F = − C(n, s) n+2s dy( u)(ξ) 2 Rn |y| Z 1 − cos(ξ · y) F = C(n, s) n+2s dy( u)(ξ). (3.11) Rn |y| Z Hence, in order to obtain (3.10), it suffices to show that

1 − cos(ξ · y) −1 2s n+2s dy = C(n, s) |ξ| . (3.12) Rn |y| Z n To check this, first we observe that, if ζ =(ζ1, ..., ζn) ∈ R , we have

1 − cos ζ |ζ |2 1 1 ≤ 1 ≤ |ζ|n+2s |ζ|n+2s |ζ|n−2+2s near ζ = 0. Thus,

1 − cos ζ1 n+2s dζ is finite and positive. (3.13) Rn |ζ| Z Now, we consider the function I : Rn → R defined as follows

1 − cos(ξ · y) I(ξ)= n+2s dy. Rn |y| Z We have that I is rotationally invariant, that is

I(ξ)= I(|ξ|e1), (3.14)

n where e1 denotes the first direction vector in R . Indeed, when n = 1, then we can deduce (3.14) by the fact that I(−ξ) = I(ξ). When n ≥ 2,

15 T we consider a rotation R for which R(|ξ|e1) = ξ and we denote by R its transpose. Then, by substitutingy ˜ = RT y, we obtain

1 − cos (R(|ξ|e1)) · y I(ξ) = n+2s dy Rn |y|  Z T 1 − cos (|ξ|e1) · (R y) = n+2s dy Rn |y| Z  1 − cos (|ξ|e1) · y˜ = n+2s dy˜ = I(|ξ|e1), Rn |y˜| Z  which proves (3.14). As a consequence of (3.13) and (3.14), the substitution ζ = |ξ|y gives that

I(ξ) = I(|ξ|e1)

1 − cos(|ξ|y1) = n+2s dy Rn |y| Z 1 1 − cos ζ1 −1 2s = n n+2s dζ = C(n, s) |ξ| . |ξ| Rn Z ζ/|ξ|

− 1 1 − cos(ζ1) where we recall that C(n, s) is equal to n+2s dζ by (3.2). Rn |ζ| Hence, we deduce (3.12) and then the proof isZ complete.

Proposition 3.4. Let s ∈ (0, 1). Then the fractional Sobolev space Hs(Rn) defined in Section 2 coincides with Hˆ s(Rn) defined in (3.7). In particular, for any u ∈ Hs(Rn)

2 −1 2s F 2 [u]Hs(Rn) = 2C(n, s) |ξ| | u(ξ)| dξ. Rn Z where C(n, s) is defined by (3.2).

Proof. For every fixed y ∈ Rn, by changing of variable choosing z = x − y,

16 we get |u(x) − u(y)|2 |u(z + y) − u(y)|2 n+2s dx dy = n+2s dz dy Rn Rn |x − y| Rn Rn |z| Z Z  Z Z u(z + y) − u(y) 2 = n/2+s dy dz Rn Rn |z| Z Z !

u(z + ·) − u(·) 2

= n/2+s dz Rn |z| 2 n Z L (R ) 2 u(z + ·) − u(·) F = n/2+s dz, Rn |z| 2 n Z   L (R )

where Plancherel Formula has been used. Now, using (3.12) we obtain

2 F u(z + ·) − u(·) n/2+s dz Rn |z| 2 n Z   L (R )

iξ·z 2 |e − 1| F 2 = n+2s | u(ξ)| dξ dz Rn Rn |z| Z Z (1 − cos ξ · z) F 2 = 2 n+2s | u(ξ)| dz dξ Rn Rn |z| Z Z = 2C(n, s)−1 |ξ|2s |F u(ξ)|2 dξ. Rn Z This completes the proof.

Remark 3.5. The equivalence of the spaces Hs and Hˆ s stated in Proposi- tion 3.4 relies on Plancherel Formula. As well known, unless p = q = 2, one cannot go forward and backward between an Lp and an Lq via Fourier transform (see, for instance, the sharp inequality in [5] for the case 1 < p< 2 and q equal to the conjugate exponent p/(p − 1) ). That is why the general fractional space defined via Fourier transform for 1 0, say Hs,p(Rn), does not coincide with the fractional Sobolev spaces W s,p(Rn) and will be not discussed here (see, e.g., [101]). Finally, we are able to prove the relation between the fractional Lapla- cian operator (−∆)s and the fractional Sobolev space Hs.

17 Proposition 3.6. Let s ∈ (0, 1) and let u ∈ Hs(Rn). Then,

s 2 −1 2 2 [u]Hs(Rn) = 2C(n, s) k(−∆) ukL2(Rn). (3.15) where C(n, s) is defined by (3.2). Proof. The equality in (3.15) plainly follows from Proposition 3.3 and Proposition 3.4. Indeed,

s s 2 2 F 2 2 sF 2 k(−∆) ukL2(Rn) = k (−∆) ukL2(Rn) = k|ξ| ukL2(Rn)

1 2 = C(n, s)[u] s Rn . 2 H ( )

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 2 s kuk s,2 = u(−∆) u dx , a more general integral operator can be defined W that is related to the space W s,p for any p (see the recent paper [52]). R  Armed with the definition of Hs(Rn) via the Fourier transform, we can easily analyze the traces of the Sobolev functions (see the forthcom- ing Proposition 3.8). We will follow Sections 13, 15 and 16 in [91]. Let Ω ⊆ Rn be an open set with continuous boundary ∂Ω. Denote by T the trace operator, namely the linear operator defined by the uniformly continuous extension of the operator of restriction to ∂Ω for functions ∞ Rn 6 in D(Ω), that is the space of functions C0 ( ) restricted to Ω. ′ n n Now, for any x = (x , xn) ∈ R and for any u ∈ S (R ), we denote by n−1 v ∈ S (R ) the restriction of u on the hyperplane xn = 0, that is

v(x′)= u(x′, 0) ∀x′ ∈ Rn−1. (3.16)

Then, we have

′ ′ ′ n−1 F v(ξ ) = F u(ξ ,ξn) dξn ∀ξ ∈ R , (3.17) R Z

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 ∂Ω) is not defined for functions which are not smooth enough.

18 where, for the sake of simplicity, we keep the same symbol F for both the Fourier transform in n − 1 and in n variables. To check (3.17), we write

F ′ 1 −iξ′·x′ ′ ′ v(ξ ) = n−1 e v(x ) dx 2 Rn−1 (2π) Z 1 −iξ′·x′ ′ ′ = n−1 e u(x , 0) dx . (3.18) 2 Rn−1 (2π) Z

On the other hand, we have

′ F u(ξ ,ξn) dξn R Z

′ ′ 1 −i (ξ ,ξn)·(x ,xn) ′ ′ = n e u(x , xn) dx dxn dξn R (2π) 2 Rn Z Z

′ ′ 1 −iξ ·x 1 −iξn·xn ′ ′ = n−1 e 1 e u(x , xn) dxndξn dx 2 Rn−1 π 2 R R (2π) Z "(2 ) Z Z #

1 −iξ′·x′ ′ ′ = n−1 e u(x , 0) dx , 2 Rn−1 (2π) Z   where the last equality follows by transforming and anti-transforming u in the last variable, and this coincides with (3.18). Now, we are in position to characterize the traces of the function in Hs(Rn), as stated in the following proposition.

Proposition 3.8. ([91, Lemma 16.1]). Let s > 1/2, then any function s n u ∈ H (R ) has a trace v on the hyperplane xn = 0 , such that v ∈ s− 1 n−1 s n H 2 R T H R ( ). Also, the trace operator is surjective from ( ) onto s− 1 n−1 H 2 (R ).

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 (Rn) and any v defined as in (3.16),

kvk s− 1 ≤ CkukHs(Rn). (3.19) H 2 (Rn−1)

19 By taking into account (3.17), the Cauchy-Schwarz inequality yields

F ′ 2 2 s F ′ 2 dξn | v(ξ )| ≤ (1 + |ξ| ) | u(ξ ,ξn)| dξn 2 s . (3.20) R R (1 + |ξ| ) Z  Z  ′ 2 Using the changing of variable formula by setting ξn = t 1+ |ξ | , we have

p 1 ′ 2 1/2 ′ 2 −s dξn 1+ |ξ | 1+ |ξ | 2 s = s dt = 2 s dt R (1 + |ξ| ) R (1 + |ξ′|2)(1 + t2) R (1 + t ) Z Z  Z  1 −s = C(s) 1+ |ξ′|2 2 ,  (3.21)

dt  where C(s) := 2 s < +∞ since s> 1/2. R (1 + t ) Z Combining (3.20) with (3.21) and integrating in ξ′ ∈ Rn−1, we obtain

s− 1 1+ |ξ′|2 2 |F v(ξ′)|2 dξ′ Rn−1 Z  2 s ′ 2 ′ ≤ C(s) 1+ |ξ| |F u(ξ ,ξn)| dξn dξ , Rn−1 R Z Z  that is (3.19). Now, we will prove the surjectivity of the trace operator T . For this, s− 1 n−1 we show that for any v ∈ H 2 (R ) the function u defined by

′ ′ ξn 1 F u(ξ ,ξn) = F v(ξ ) ϕ , (3.22) 1+ |ξ′|2 ! 1+ |ξ′|2 p p ∞ R s Rn with ϕ ∈ C0 ( ) and ϕ(t) dt = 1, is such that u ∈ H ( ) and Tu = v. R Z Indeed, we integrate (3.22) with respect to ξn ∈ R, we substitute ξn = t 1+ |ξ′|2 and we obtain p ′ ′ ξn 1 F u(ξ ,ξn) dξn = F v(ξ ) ϕ dξn R R 1+ |ξ′|2 1+ |ξ′|2 Z Z ! p p = F v(ξ′) ϕ(t) dt = F v(ξ′) (3.23) R Z 20 and this implies v = Tu because of (3.17). The proof of the Hs-boundedness of u is straightforward. In fact, from (3.22), for any ξ′ ∈ Rn−1, we have

2 s ′ 2 1+ |ξ| |F u(ξ ,ξn)| dξn R Z  2 2 s F ′ 2 ξn 1 = 1+ |ξ| | v(ξ )| ϕ ′ 2 dξn R 1+ |ξ′|2 1+ |ξ | Z !  1 ′ 2 s− 2 F ′ 2 p = C 1+ |ξ | ) | v(ξ )| , (3.24)

′ 2 where we used again the changing of variable formula with ξn = t 1+ |ξ | s and the constant C is given by 1+ t2 |ϕ(t)|2 dt. Finally,p we obtain R s n Z ′ n−1 that u ∈ H (R ) by integrating (3.24) in ξ ∈ R .

Remark 3.9. We conclude this section by recalling that the fractional Lapla- cian (−∆)s, which is a nonlocal operator on functions defined in Rn, may be reduced to a local, possibly singular or degenerate, operator on func- Rn+1 Rn tions sitting in the higher dimensional half-space + = × (0, +∞). We have ∂U (−∆)su(x) = −C lim t1−2s (x, t) , t→0 ∂t   Rn+1 R 1−2s Rn+1 where the function U : + → solves div(t ∇U) = 0 in + and U(x, 0) = u(x) in Rn. This approach was pointed out by Caffarelli and Silvestre in [19]; see, in particular, Section 3.2 there, where was also given an equivalent definition of the Hs(Rn)-norm:

|ξ|2s|F u|2 dξ = C |∇U|2t1−2s dx dt. Rn Rn+1 Z Z + 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., [18, 88, 13, 17, 78], etc.).

21 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 (3.1)), by analyzing its asymptotic behavior as s → 1− and s → 0+. This is relevant if one wants to recover the Sobolev norms of the spaces H1(Rn) and L2(Rn) by starting from the one of Hs(Rn). We recall that in Section 3, the constant C(n, s) has been defined by

−1 1 − cos(ζ1) C(n, s)= n+2s dζ . Rn |ζ| Z  Precisely, we are interested in analyzing the asymptotic behavior as s → 0+ and s → 1− of a scaling of the quantity in the right hand-side of the above formula. ′ ′ By changing variable η = ζ /|ζ1|, we have

1 − cos(ζ1) 1 − cos(ζ1) 1 ′ n+2s dζ = n+2s n+2s dζ dζ1 Rn |ζ| R Rn−1 |ζ | ′ 2 2 2 Z Z Z 1 (1 + |ζ | /|ζ1| ) 1 − cos(ζ1) 1 ′ = 1+2s n+2s dη dζ1 R Rn−1 |ζ | ′ 2 2 Z Z 1 (1 + |η | ) A(n, s) B(s) = s(1 − s) where 1 ′ A(n, s)= n+2s dη (4.1) Rn−1 ′ 2 2 Z (1 + |η | ) and7 1 − cos t B(s)= s(1 − s) 1+2s dt. (4.2) R |t| Z

Proposition 4.1. For any n> 1, let A and B be defined by (4.1) and (4.2) respectively. The following statements hold:

7Of course, when n = 1 (4.1) reduces to A(n,s) = 1, so we will just consider the case n> 1.

22 +∞ ρn−2 (i) lim A(n, s)= ωn−2 n +1 dρ < +∞; s→1− (1 + ρ2) 2 Z0 +∞ ρn−2 (ii) lim A(n, s)= ωn−2 n dρ < +∞; s→0+ (1 + ρ2) 2 Z0 1 (iii) lim B(s)= ; s→1− 2 (iv) lim B(s)=1, s→0+

n−2 where ωn−2 denotes (n − 2)-dimensional measure of the S . As a consequence,

+∞ n−2 −1 C(n, s) ωn−2 ρ lim = n +1 dρ (4.3) s→1− s(1 − s) 2 (1 + ρ2) 2  Z0  and C(n, s) +∞ ρn−2 −1 lim = ωn−2 n dρ . (4.4) s→0+ s(1 − s) (1 + ρ2) 2  Z0  Proof. First, by polar coordinates, for any s ∈ (0, 1), we get

+∞ n−2 1 ′ ρ n+2s dη = ωn−2 n+2s dρ. Rn−1 ′ 2 2 2 2 Z (1 + |η | ) Z0 (1 + ρ ) Now, observe that for any s ∈ (0, 1) and any ρ ≥ 0, we have ρn−2 ρn−2 ≤ n n+2s 2 (1 + ρ2) 2 (1 + ρ ) 2 and the function in the right hand-side of the above inequality belongs to L1((0, +∞)) for any n> 1. Then, the Dominated Convergence Theorem yields +∞ ρn−2 lim A(n, s) = ωn−2 n +1 dρ s→1− (1 + ρ2) 2 Z0 and +∞ ρn−2 lim A(n, s) = ωn−2 n dρ. s→0+ (1 + ρ2) 2 Z0 23 This proves (i) and (ii). Now, we want to prove (iii). First, we split the integral in (4.2) as follows 1 − cos t 1 − cos t 1 − cos t 1+2s dt = 1+2s dt + 1+2s dt. R |t| |t| |t| Z Z|t|<1 Z|t|≥1 Also, we have that

1 − cos t +∞ 1 2 0 ≤ dt ≤ 4 dt = |t|1+2s t1+2s s Z|t|≥1 Z1 and 1 − cos t t2 |t|3 2C dt − dt ≤ C dt = , |t|1+2s 2|t|1+2s |t|1+2s 3 − 2s Z|t|<1 Z|t|<1 Z|t|<1 for some suitable positive constant C. From the above estimates it follows that 1 − cos t lim s(1 − s) dt =0 s→1− |t|1+2s Z|t|≥1 and 1 − cos t t2 lim s(1 − s) dt = lim s(1 − s) dt. s→1− |t|1+2s s→1− 2|t|1+2s Z|t|<1 Z|t|<1 Hence, we get

1 s(1 − s) 1 lim B(s) = lim s(1 − s) t1−2s dt = lim = . s→1− s→1− s→1− 2(1 − s) 2 Z0 

Similarly, we can prove (iv). For this we notice that

1 − cos t 1 0 ≤ dt ≤ C t1−2s dt |t|1+2s Z|t|<1 Z0 which yields 1 − cos t lim s(1 − s) dt = 0. s→0+ |t|1+2s Z|t|<1 24 Now, we observe that for any k ∈ N, k ≥ 1, we have

2(k+1)π cos t 2kπ+π cos t 2kπ+π cos(τ + π) dt = dt + dτ t1+2s t1+2s (τ + π)1+2s Z2kπ Z2kπ Z2kπ

2kπ+π 1 1 = cos t − dt t1+2s (t + π)1+2s Z2kπ   2kπ+π 1 1 ≤ − dt t1+2s (t + π)1+2s Z2kπ 2kπ+π (t + π)1+2s − t1+2s = dt t1+2s(t + π)1+2s Z2kπ 2kπ+π 1 π = (1+2s)(t + ϑ)2s dϑ dt t1+2s(t + π)1+2s Z2kπ Z0  2kπ+π 3π(t + π)2s ≤ dt t1+2s(t + π)1+2s Z2kπ 2kπ+π 3π ≤ dt t(t + π) Z2kπ 2kπ+π 3π C ≤ dt ≤ . t2 k2 Z2kπ As a consequence,

+∞ cos t 2π 1 +∞ 2(k+1)π cos t dt ≤ dt + dt t1+2s t t1+2s Z1 Z1 k=1 Z2kπ X + ∞ C ≤ log(2π)+ ≤ C, k2 Xk=1 up to relabeling the constant C > 0. It follows that 1 − cos t 1 cos t dt − dt = dt |t|1+2s |t|1+2s |t|1+2s Z|t|≥1 Z|t|≥1 Z|t|≥1

+∞ cos t = 2 dt ≤ C t1+2s Z1

25 and then 1 − cos t 1 lim s(1 − s) dt = lim s(1 − s) dt. s→0+ |t|1+2s s→0+ |t|1+2s Z|t|≥1 Z|t|≥1

Hence, we can conclude that 1 lim B(s) = lim s(1 − s) dt s→0+ s→0+ |t|1+2s Z|t|≥1 +∞ = lim 2s(1 − s) t−1−2s dt s→0+ Z1 2s(1 − s) = lim =1. s→0+ 2s

Finally, (4.3) and (4.4) easily follow combining the previous estimates and recalling that s(1 − s) C(n, s)= . A(n, s)B(s) The proof is complete.

Corollary 4.2. For any n > 1, let C(n, s) be defined by (3.2). The following statements hold: C(n, s) 4n (i) lim = ; s→1− s(1 − s) ωn−1 C(n, s) 2 (ii) lim = . s→0+ s(1 − s) ωn−1

n−1 where ωn−1 denotes the (n−1)-dimensional measure of the unit sphere S .

Proof. For any θ ∈ R such that θ > n − 1, let us define

+∞ ρn−2 En(θ) := dρ. 2 θ Z0 (1 + ρ ) 2

26 Observe that the assumption on the parameter θ ensures the convergence of the integral. Furthermore, integrating by parts we get

1 +∞ (ρn−1)′ En(θ) = θ dρ n − 1 2 2 Z0 (1 + ρ ) θ +∞ ρn = θ+2 dρ n − 1 2 2 Z0 (1 + ρ ) θ = E (θ + 2). (4.5) n − 1 n+2 Then, we set

+∞ n−2 (1) ρ In := En(n +2) = n +1 dρ (1 + ρ2) 2 Z0 and +∞ n−2 (0) ρ In := En(n)= n dρ. (1 + ρ2) 2 Z0 (1) (0) In view of (4.5), it follows that In and In can be obtained in a recursive way, since n − 1 n − 1 I(1) = E (n +4)= E (n +2) = I(1) (4.6) n+2 n+2 n +2 n n +2 n and n − 1 n − 1 I(0) = E (n +2)= E (n)= I(0). (4.7) n+2 n+2 n n n n

Now we claim that (1) ωn−1 In = (4.8) 2nωn−2 and (0) ωn−1 In = . (4.9) 2ωn−2 We will prove the previous identities by induction. We start by noticing that the inductive basis are satisfied, since

+∞ +∞ (1) 1 π (1) ρ 1 I2 = 2 2 dρ = , I3 = 5 dρ = (1 + ρ ) 4 2 2 3 Z0 Z0 (1 + ρ ) 27 and

+∞ +∞ (0) 1 π (0) ρ I2 = 2 dρ = , I3 = 3 dρ =1. 0 (1 + ρ ) 2 0 ρ2 2 Z Z (1 + ) Now, using (4.6) and (4.7), respectively, it is clear that in order to check the inductive steps, it suffices to verify that ω n − 1 ω n+1 = n−1 . (4.10) ωn n ωn−2 We claim that the above formula plainly follows from a classical recursive formula on ωn, that is 2π ω = ω . (4.11) n n − 1 n−2

To prove this, let us denote by ̟n the Lebesgue measure of the n- dimensional unit ball and let us fix the notation x = (˜x, x′) ∈ Rn−2 × R2. By integrating on Rn−2 and then using polar coordinates in R2, we see that

′ ̟n = dx = dx˜ dx 2 ′ 2 ′ 2 Z|x| ≤1 Z|x |≤1 Z|x˜| ≤1−|x |  (n−2) ′ 2 2 ′ = ̟n−2 1 −|x | dx ′ Z|x |≤1  1 (n−2) 2π̟ = 2π̟ ρ 1 − ρ2 2 dρ = n−2 . (4.12) n−2 n Z0  Moreover, by polar coordinates in Rn,

1 ω ̟ = dx = ω ρn−1 dρ = n−1 . (4.13) n n−1 n Z|x|≤1 Z0 Thus, we use (4.13) and (4.12) and we obtain 2πω ω = n̟ =2π̟ = n−3 , n−1 n n−2 n − 2 which is (4.11), up to replacing n with n − 1. In turn, (4.11) implies (4.10) and so (4.8) and (4.9).

28 Finally, using (4.8), (4.9) and Proposition 4.1 we can conclude that C(n, s) 2 4n lim = = s→1− (1) s(1 − s) ωn−2In ωn−1 and C(n, s) 1 2 lim = = , s→0+ (0) s(1 − s) ωn−2In ωn−1 as desired8.

Remark 4.3. It is worth noticing that when p = 2 we recover the constants C1 and C2 in (2.8) and (2.9), respectively. In fact, in this case it is known that n 1 2 1 2 ωn−1 C1 = |ξ1| dσ(ξ)= |ξi| dσ(ξ)= 2 n−1 2n n−1 2n S i=1 S Z X Z and C2 = ωn−1 (see [11] and [66]). Then, by Proposition 3.15 and Corol- lary 4.2 it follows that |u(x) − u(y)|2 lim (1 − s) n+2s dxdy s→1− Rn Rn |x − y| Z Z −1 sF 2 = lim 2(1 − s)C(n, s) k|ξ| ukL2(Rn) s→1−

ωn−1 2 = k∇uk 2 n 2n L (R )

2 = C1kukH1(Rn) and 2 |u(x) − u(y)| −1 sF 2 lim s n+2s dxdy = lim 2sC(n, s) k|ξ| ukL2(Rn) s→0+ Rn Rn |x − y| s→0+ Z Z 2 = ωn−1kukL2(Rn)

2 = C2kukL2(Rn).

8 Another (less elementary) way to obtain this result is to notice that En(θ) = 2 B ((n − 1)/2, (θ − n − 1)/2), where B is the Beta function.

29 We will conclude this section with the following proposition that one could plainly deduce from Proposition 3.3. We prefer to provide a di- rect proof, based on Lemma 3.2, in order to show the consistency in the definition of the constant C(n, s).

∞ Rn Proposition 4.4. Let n> 1. For any u ∈ C0 ( ) the following statements hold:

s (i) lims→0+ (−∆) u = u;

s (ii) lims→1− (−∆) u = −∆u. Rn Proof. Fix x ∈ , R0 > 0 such that supp u ⊆ BR0 and set R = R0 +|x|+1. First,

u(x + y)+ u(x − y) − 2u(x) |y|2 dy ≤ kuk 2 Rn dy |y|n+2s C ( ) |y|n+2s ZBR ZBR

R 1 ≤ ω kuk 2 Rn dρ n−1 C ( ) ρ2s−1 Z0 2−2s ω kuk 2 Rn R = n−1 C ( ) . (4.14) 2(1 − s)

Furthermore, observe that |y| ≥ R yields |x±y|≥|y|−|x| ≥ R−|x| > R0 and consequently u(x ± y) = 0. Therefore,

1 u(x + y)+ u(x − y) − 2u(x) 1 − n+2s dy = u(x) n+2s dy 2 Rn |y| Rn |y| Z \BR Z \BR +∞ 1 = ω u(x) dρ n−1 ρ2s+1 ZR ω R−2s = n−1 u(x). (4.15) 2s Now, by (4.14) and Corollary 4.2, we have

C(n, s) u(x + y)+ u(x − y) − 2u(x) lim − dy =0 s→0+ 2 |y|n+2s ZBR

30 and so we get, recalling Lemma 3.2,

s C(n, s) u(x + y)+ u(x − y) − 2u(x) lim (−∆) u = lim − n+2s dy s→0+ s→0+ 2 Rn |y| Z \BR C(n, s)ω R−2s = lim n−1 u(x) = u(x), s→0+ 2s where the last identities follow from (4.15) and again Corollary 4.2. This proves (i). Similarly, we can prove (ii). In this case, when s goes to 1, we have no contribution outside the unit ball, as the following estimate shows

u(x + y)+ u(x − y) − 2u(x) n+2s dy Rn |y| Z \B1 1 ∞ Rn ≤ 4kukL ( ) n+2s dy Rn |y| Z \B1 +∞ 1 ≤ 4ω kuk ∞ Rn dρ n−1 L ( ) ρ2s+1 Z1 2ωn−1 = kuk ∞ Rn . s L ( ) As a consequence (recalling Corollary 4.2), we get

C(n, s) u(x + y)+ u(x − y) − 2u(x) lim − n+2s dy =0. (4.16) s→1− 2 Rn |y| Z \B1

On the other hand, we have

u(x + y)+ u(x − y) − 2u(x) − D2u(x)y · y dy |y|n+2s ZB1 3 |y| ≤kuk 3 Rn dy C ( ) |y|n+2s ZB1 1 1 ≤ ω kuk 3 Rn dρ n−1 C ( ) ρ2s−2 Z0 ω kuk 3 Rn = n−1 C ( ) 3 − 2s

31 and this implies that C(n, s) u(x + y)+ u(x − y) − 2u(x) lim − dy s→1− 2 |y|n+2s ZB1 C(n, s) D2u(x)y · y = lim − dy. (4.17) s→1− 2 |y|n+2s ZB1

Now, notice that if i =6 j then

2 2 ∂iju(x)yi · yj dy = − ∂iju(x)˜yi · y˜j dy,˜ ZB1 ZB1 wherey ˜k = yk for any k =6 j andy ˜j = −yj, and thus

2 ∂iju(x)yi · yj dy =0. (4.18) ZB1 Also, up to permutations, for any fixed i, we get ∂2 u(x)y2 y2 y2 ii i dy = ∂2 u(x) i dy = ∂2 u(x) 1 dy |y|n+2s ii |y|n+2s ii |y|n+2s ZB1 ZB1 ZB1 ∂2 u(x) n y2 ∂2 u(x) |y|2 = ii j dy = ii dy n |y|n+2s n |y|n+2s j=1 B1 B1 X Z Z ∂2 u(x) ω = ii n−1 . (4.19) 2n(1 − s) Finally, combining (4.16), (4.17), (4.18), (4.19), Lemma 3.2 and Corol- lary 4.2, we can conclude C(n, s) u(x + y)+ u(x − y) − 2u(x) lim (−∆)su = lim − dy s→1− s→1− 2 |y|n+2s ZB1 C(n, s) D2u(x)y · y = lim − dy s→1− 2 |y|n+2s ZB1 C(n, s) n ∂2 u(x)y2 = lim − ii i dy s→1− 2 |y|n+2s i=1 ZB1 X n C(n, s)ωn−1 2 = lim − ∂iiu(x) = −∆u(x). s→1− 4n(1 − s) i=1 X

32 5. Extending a W s,p(Ω) function to the whole of Rn As well known when s is an integer, under certain regularity assump- tions on the domain Ω, any function in W s,p(Ω) may be extended to a function in W s,p(Rn). Extension results are quite important in applica- tions 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, 1) and any p ∈ [1, ∞), we say that an open set Ω ⊆ Rn is an extension domain for W s,p if there exists a positive constant C = C(n, p, s, Ω) such that: for every function u ∈ W s,p(Ω) there exists s,p n u˜ ∈ W (R ) withu ˜(x)= u(x) for all x ∈ Ω and ku˜kW s,p(Rn) ≤ CkukW s,p(Ω).

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 open9. When s is an integer, we cite [58] for a complete characterization in the special case s = 1, 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 11 and Chapter 12 there). In this section, we will show that any open set Ω of class C0,1 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 Rn of a function u defined on Ω in two separated cases: when the function u is identically zero in a neighborhood of the Rn boundary ∂Ω and when Ω coincides with the half-space +. Lemma 5.1. Let Ω be an open set in Rn and u a function in W s,p(Ω) with s ∈ (0, 1) and p ∈ [1, +∞). If there exists a compact subset K ⊂ Ω such that u ≡ 0 in Ω \ K, then the extension function u˜ defined as

u(x) x ∈ Ω , u˜(x)= n (5.1) ( 0 x ∈ R \ Ω

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 [102]).

33 belongs to W s,p(Rn) and

ku˜kW s,p(Rn) ≤ CkukW s,p(Ω), where C is a suitable positive constant depending on n, p, s, K and Ω. Proof. Clearlyu ˜ ∈ Lp(Rn). Hence, it remains to verify that the Gagliardo norm ofu ˜ in Rn is bounded by the one of u in Ω. Using the symmetry of the integral in the Gagliardo norm with respect to x and y and the fact thatu ˜ ≡ 0 in Rn \ Ω, we can split as follows |u˜(x) − u˜(y)|p |u(x) − u(y)|p n+sp dxdy = n+sp dxdy (5.2) Rn Rn |x − y| |x − y| Z Z ZΩ ZΩ |u(x)|p +2 n+sp dy dx, Rn |x − y| ZΩ Z \Ω  where the first term in the right hand-side of (5.2) is finite since u ∈ W s,p(Ω). Furthermore, for any y ∈ Rn \ K, p p |u(x)| χK (x)|u(x)| p 1 n+sp = n+sp ≤ χK(x)|u(x)| sup n+sp |x − y| |x − y| x∈K |x − y| and so p |u(x)| 1 p n+sp dy dx ≤ n+sp dy kukLp(Ω) . (5.3) Rn |x − y| Rn dist(y,∂K) ZΩ Z \Ω  Z \Ω Note that the integral in (5.3) is finite since dist(∂Ω,∂K) ≥ α > 0 and n + sp>n. Combining (5.2) with (5.3), we get

ku˜kW s,p(Rn) ≤ CkukW s,p(Ω) where C = C(n,s,p,K).

Lemma 5.2. Let Ω be an open set in Rn, symmetric with respect to the coordinate xn, and consider the sets Ω+ = {x ∈ Ω: xn > 0} and s,p Ω− = {x ∈ Ω: xn ≤ 0}. Let u be a function in W (Ω+), with s ∈ (0, 1) and p ∈ [1, +∞). Define

′ u(x , xn) xn ≥ 0 , u¯(x)= ′ (5.4) (u(x , −xn) xn < 0 . Then u¯ belongs to W s,p(Ω) and

s,p s,p ku¯kW (Ω) ≤ 4kukW (Ω+) .

34 ′ Proof. By splitting the integrals and changing variablex ˆ = (x , −xn), we get

p p ′ p p ku¯k p = |u(x)| dx + |u(ˆx , xˆ )| dx˜ = 2kuk p . (5.5) L (Ω) n L (Ω+) ZΩ+ ZΩ+ Rn C Rn 2 2 Also, if x ∈ + and y ∈ + then (xn − yn) ≥ (xn + yn) and therefore |u¯(x) − u¯(y)|p |u(x) − u(y)|p dxdy = dxdy |x − y|n+sp |x − y|n+sp ZΩ ZΩ ZΩ+ ZΩ+ ′ p |u(x) − u(y , −yn)| +2 n+sp dxdy C |x − y| ZΩ+ Z Ω+ ′ ′ p |u(x , −xn) − u(y , −yn)| + n+sp dxdy C C |x − y| Z Ω+ Z Ω+ p ≤ 4kuk s,p . W (Ω+) This concludes the proof.

Now, a truncation lemma near ∂Ω. Lemma 5.3. Let Ω be an open set in Rn, s ∈ (0, 1) and p ∈ [1, +∞). Let us consider u ∈ W s,p(Ω) and ψ ∈ C0,1(Ω), 0 ≤ ψ ≤ 1. Then ψu ∈ W s,p(Ω) and kψukW s,p(Ω) ≤ CkukW s,p(Ω), (5.6) where C = C(n, p, s, Ω).

Proof. It is clear that kψukLp(Ω) ≤ kukLp(Ω) since |ψ| ≤ 1. Furthermore, adding and subtracting the factor ψ(x)u(y), we get |ψ(x) u(x) − ψ(y) u(y)|p dxdy |x − y|n+sp ZΩ ZΩ |ψ(x) u(x) − ψ(x) u(y)|p ≤ 2p−1 dxdy |x − y|n+sp  ZΩ ZΩ |ψ(x) u(y) − ψ(y) u(y)|p + dxdy |x − y|n+sp ZΩ ZΩ  |u(x) − u(y)|p ≤ 2p−1 dxdy (5.7) |x − y|n+sp  ZΩ ZΩ |u(x)|p |ψ(x) − ψ(y)|p + dxdy . |x − y|n+sp ZΩ ZΩ  35 Since ψ belongs to C0,1(Ω), we have

|u(x)|p |ψ(x) − ψ(y)|p |u(x)|p |x − y|p dxdy ≤ Λp dxdy |x − y|n+sp |x − y|n+sp ZΩ ZΩ ZΩ ZΩ∩|x−y|≤1 |u(x)|p + dxdy |x − y|n+sp ZΩ ZΩ∩|x−y|≥1 ˜ p ≤ CkukLp(Ω), (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+(1−s)p is summable with respect to y if |x − y| ≤ 1 since n +(s − 1)p < n and, on the other hand, the kernel |x − y|−n−sp is summable when |x − y| ≥ 1 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 Ω with bounded boundary is an extension domain for W s,p.

Theorem 5.4. Let p ∈ [1, +∞), s ∈ (0, 1) and Ω ⊆ Rn be an open set of class C0,1 with bounded boundary10. Then W s,p(Ω) is continuously embedded in W s,p(Rn), namely for any u ∈ W s,p(Ω) there exists u˜ ∈ W s,p(Rn) such that u˜|Ω = u and ku˜kW s,p(Rn) ≤ CkukW s,p(Ω) where C = C(n, p, s, Ω).

Proof. Since ∂Ω is compact, we can find a finite number of balls Bj such k k n n that ∂Ω ⊂ Bj and so we can write R = Bj ∪ (R \ ∂Ω). j=1 j=1 [ [ If we consider this covering, there exists a partition of unity related to it, i.e. there exist k + 1 smooth functions ψ0, ψ1,..., ψk such that

10 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 Ω may be weakened when s ∈ (0, 1), since in the case s = 0 clearly no regularity at all is needed for the extension problem.

36 n spt ψ0 ⊂ R \ ∂Ω, spt ψj ⊂ Bj for any j ∈ {1, ..., k}, 0 ≤ ψj ≤ 1 for any k j ∈{0, ..., k} and ψj = 1. Clearly, j=0 X k

u = ψju . j=0 X s,p By Lemma 5.3, we know that ψ0 u belongs to W (Ω). Furthermore, since n ψ0 u ≡ 0 in a neighborhood of ∂Ω, we can extend it to the whole of R , by setting

ψ0 u(x) x ∈ Ω, ψ0 u(x)= n ( 0 x ∈ R \ Ω

s,p n g and ψ0 u ∈ W (R ). Precisely

g kψ0 ukW s,p(Rn) ≤ C kψ0 ukW s,p(Ω) ≤ C kukW s,p(Ω), (5.9) where C = C(n,g s, p, Ω) (possibly different step by step, see Lemma 5.1 and Lemma 5.3).

For any j ∈{1, ..., k}, let us consider u|Bj ∩Ω and set

vj(y) := u (Tj(y)) forany y ∈ Q+,

0,1 where Tj : Q → Bj is the isomorphism of class C defined in Section 2. Note that such a Tj exists by the regularity assumption on the domain Ω. s,p Now, we state that vj ∈ W (Q+). Indeed, using the standard changing variable formula by setting x = Tj(ˆx) we have |v(ˆx) − v(ˆy)|p dxdˆ yˆ |xˆ − yˆ|n+sp ZQ+ ZQ+ |u(T (ˆx)) − u(T (ˆy))|p = j j dxdˆ yˆ |xˆ − yˆ|n+sp ZQ+ ZQ+ |u(x) − u(y)|p = det(T −1)dxdy |T −1(x) − T −1(y)|n+sp j ZBj ∩Ω ZBj ∩Ω j j |u(x) − u(y)|p ≤ C dx dy, (5.10) |x − y|n+sp ZBj ∩Ω ZBj ∩Ω 37 where (5.10) follows from the fact that Tj is bi-Lipschitz. Moreover, using Lemma 5.2 we can extend vj to all Q so that the extensionv ¯j belongs to W s,p(Q) and

s,p s,p kv¯jkW (Q) ≤ 4kvjkW (Q+).

We set −1 wj(x) :=v ¯j Tj (x) for any x ∈ Bj.  Since Tj is bi-Lipschitz, by arguing as above it follows that wj ∈ s,p W (Bj). Note that wj ≡ u (and consequently ψj wj ≡ ψj u) on Bj ∩ Ω. By definition ψj wj has compact support in Bj and therefore, as done for ] n ψ0 u, we can consider the extension ψj wj to all R in such a way that ] s,p n ψj wj ∈ W (R ). Also, using Lemma 5.1, Lemma 5.2, Lemma 5.3 and estimate (5.10) we get ] s,p n s,p s,p kψj wjkW (R ) ≤ Ckψj wjkW (Bj ) ≤ C kwjkW (Bj )

s,p s,p ≤ Ckv¯jkW (Q) ≤ CkvjkW (Q+)

s,p ≤ CkukW (Ω∩Bj ), (5.11) where C = C(n, p, s, Ω) and it is possibly different step by step. Finally, let k ] u˜ = ψ0 u + ψj wj j=1 X be the extension of u defined ong all Rn. By construction, it is clear that u˜|Ω = u and, combining (5.9) with (5.11), we get

ku˜kW s,p(Rn) ≤ CkukW s,p(Ω) with C = C(n, p, s, Ω).

Corollary 5.5. Let p ∈ [1, +∞), s ∈ (0, 1) and Ω be an open set in Rn of class C0,1 with bounded boundary. Then for any u ∈ W s,p(Ω), there exists ∞ Rn s,p a sequence {un} ∈ C0 ( ) such that un → u as n → +∞ in W (Ω), i.e.,

lim kun − ukW s,p(Ω) =0 . n→+∞

38 Proof. The proof follows directly by Theorem 2.4 and Theorem 5.4.

6. Fractional Sobolev inequalities In this section, we provide an elementary proof of a Sobolev-type in- equality involving the fractional norm k·kW 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 ∈ [1/2, 1) some of the statements may be strengthened (see [10]). 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 [61, 62, 12, 1, 91] 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 Theo- rem 6.5, we need some preliminary results. The first of them is an el- ementary estimate involving the measure of finite measurable sets E in Rn as stated in the following lemma (see [85, Lemma A.1] and also [20, Corollary 24 and 25]).

Lemma 6.1. Fix x ∈ Rn. Let p ∈ [1, +∞), s ∈ (0, 1) and E ⊂ Rn be a measurable set with finite measure. Then,

dy −sp/n n+sp ≥ C |E| , C |x − y| Z E for a suitable constant C = C(n,p,s) > 0.

Proof. We set 1 |E| n ρ := ω  n  and then it follows

|(C E) ∩ Bρ(x)| = |Bρ(x)|−|E ∩ Bρ(x)| = |E|−|E ∩ Bρ(x)|

= |E ∩ C Bρ(x)|.

39 Therefore, dy dy dy n+sp = n+sp + n+sp C |x − y| C |x − y| C C |x − y| Z E Z( E)∩Bρ(x) Z( E)∩ Bρ(x) dy dy ≥ n+sp + n+sp C ρ C C |x − y| Z( E)∩Bρ(x) Z( E)∩ Bρ(x) |(C E) ∩ Bρ(x)| dy = n+sp + n+sp ρ C C |x − y| Z( E)∩ Bρ(x) |E ∩ C Bρ(x)| dy = n+sp + n+sp ρ C C |x − y| Z( E)∩ Bρ(x) dy dy ≥ n+sp + n+sp C |x − y| C C |x − y| ZE∩ Bρ(x) Z( E)∩ Bρ(x) dy = n+sp . C |x − y| Z Bρ(x) 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, 1) and p ∈ [1, +∞) such that sp< n. Fix T > 1; let N ∈ Z and

ak be a bounded, nonnegative, decreasing sequence (6.1)

with ak =0 for any k ≥ N. Then, (n−sp)/n k −sp/n k ak T ≤ C ak+1ak T , k∈Z k∈Z X aXk=06 for a suitable constant C = C(n,p,s,T ) > 0, independent of N. Proof. By (6.1),

(n−sp)/n k −sp/n k both ak T and ak+1ak T are convergent series. (6.2) k∈Z k∈Z X aXk=06

40 Moreover, since ak is nonnegative and decreasing, we have that if ak = 0, then ak+1 = 0. Accordingly,

(n−sp)/n k (n−sp)/n k ak+1 T = ak+1 T . k∈Z k∈Z X aXk=06 Therefore, we may use the H¨older inequality with exponents α := n/sp and β := n/(n − sp) by arguing as follows.

1 (n−sp)/n k (n−sp)/n k ak T = ak+1 T T Z Z Xk∈ Xk∈ (n−sp)/n k = ak+1 T k∈Z aXk=06 sp/(nβ) k/α 1/β −sp/(nβ) k/β = ak T ak+1ak T k∈Z aXk=06    1/β 1/α α β sp/(nβ) k/α 1/β −sp/(nβ) k/β ≤ ak T  ak+1ak T  k∈Z ! k∈Z   a =06   X  Xk   (n−sp)/n  sp/n (n−sp)/n k −sp/n k ≤ ak T  ak+1ak T  . k∈Z ! k∈Z X  aXk=06    So, recalling (6.2), we obtain the desired result. 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, 1) and p ∈ [1, +∞) such that sp 2 } . (6.4) Then, p |f(x) − f(y)| −sp/n pk n+sp dxdy ≥ C ak+1ak 2 , Rn Rn |x − y| Z Z k∈Z aXk=06

41 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 k Ak := {|f| > 2 }. (6.5)

We remark that Ak+1 ⊆ Ak, hence

ak+1 ≤ ak. (6.6)

We define

k k+1 Dk := Ak \ Ak+1 = {2 < f ≤ 2 } and dk := |Dk|.

Notice that

dk and ak are bounded and they become zero when k is large enough, (6.7) thanks to (6.3). Also, we observe that the Dk’s are disjoint, that

Dℓ = C Ak+1 (6.8) ℓ∈Z [ℓ≤k and that Dℓ = Ak. (6.9) ℓ∈Z [ℓ≥k As a consequence of (6.9), we have that

ak = dℓ (6.10) ℓ∈Z Xℓ≥k and so dk = ak − dℓ. (6.11) ℓ∈Z ℓX≥k+1

42 We stress that the series in (6.10) is convergent, due to (6.7), thus so is the series in (6.11). Similarly, we can define the convergent series

pℓ −sp/n S := 2 aℓ−1 dℓ. (6.12) ℓ∈Z aℓX−1=06 −sp/n −sp/n We notice that Dk ⊆ Ak ⊆ Ak−1, hence ai−1 dℓ ≤ ai−1 aℓ−1. Therefore Z −sp/n (i, ℓ) ∈ s.t. ai−1 =6 0 and ai−1 dℓ =06 (6.13) n o ⊆ (i, ℓ) ∈ Z s.t. aℓ−1 =06 . We use (6.13) and (6.6) in the followingn computation: o

−sp/n pi −sp/n 2 ai−1 dℓ = 2 ai−1 dℓ i∈Z ℓ∈Z i∈Z ℓ∈Z ai−1=06 ℓ≥i+1 ai−1=06 ℓ≥i+1 X X X sp/nX a d =06 i−1 ℓ pi −sp/n ≤ 2 ai−1 dℓ i∈Z ℓ∈Z X ℓX≥i+1 aℓ−1=06 pi −sp/n = 2 ai−1 dℓ ℓ∈Z i∈Z aℓX−1=06 iX≤ℓ−1 pi −sp/n ≤ 2 aℓ−1 dℓ ℓ∈Z i∈Z aℓX−1=06 iX≤ℓ−1 +∞ p(ℓ−1) −pk −sp/n = 2 2 aℓ−1 dℓ ≤ S. (6.14) ℓ∈Z k=0 aℓX−1=06 X

Now, we fix i ∈ Z and x ∈ Di: then, for any j ∈ Z with j ≤ i − 2 and any y ∈ Dj we have that |f(x) − f(y)| ≥ 2i − 2j+1 ≥ 2i − 2i−1 = 2i−1 and therefore, recalling (6.8), |f(x) − f(y)|p dy dy ≥ 2p(i−1) |x − y|n+sp |x − y|n+sp j∈Z ZDj j∈Z ZDj jX≤i−2 jX≤i−2 p(i−1) dy = 2 n+sp . C |x − y| Z Ai−1 43 This and Lemma 6.1 imply that, for any i ∈ Z and any x ∈ Di, we have that |f(x) − f(y)|p dy ≥ c 2pia−sp/n, |x − y|n+sp o i−1 j∈Z ZDj jX≤i−2 for a suitable co > 0. As a consequence, for any i ∈ Z, |f(x) − f(y)|p dxdy ≥ c 2pia−sp/nd . (6.15) |x − y|n+sp o i−1 i j∈Z ZDi×Dj jX≤i−2 Therefore, by (6.11), we conclude that, for any i ∈ Z,

|f(x) − f(y)|p dxdy ≥ c 2pia−sp/na − 2pia−sp/nd . |x − y|n+sp o  i−1 i i−1 ℓ j∈Z ZDi×Dj ℓ∈Z jX≤i−2  ℓX≥i+1   (6.16)

By (6.12) and (6.15), we have that |f(x) − f(y)|p dxdy ≥ c S. (6.17) |x − y|n+sp o i∈Z j∈Z ZDi×Dj aiX−1=06 jX≤i−2 Then, using (6.16), (6.14) and (6.17), |f(x) − f(y)|p dxdy |x − y|n+sp i∈Z j∈Z ZDi×Dj aiX−1=06 jX≤i−2

pi −sp/n pi −sp/n ≥ co 2 ai−1 ai − 2 ai−1 dℓ i∈Z i∈Z ℓ∈Z  aiX−1=06 aiX−1=06 ℓX≥i+1   

pi −sp/n ≥ co 2 ai−1 ai − S i∈Z  aiX−1=06    |f(x) − f(y)|p ≥ c 2pia−sp/na − dx dy. o i−1 i |x − y|n+sp i∈Z i∈Z j∈Z ZDi×Dj aiX−1=06 aiX−1=06 jX≤i−2

44 That is, by taking the last term to the left hand side, |f(x) − f(y)|p dxdy ≥ c 2pia−sp/na , (6.18) |x − y|n+sp o i−1 i i∈Z j∈Z ZDi×Dj i∈Z aiX−1=06 jX≤i−2 aiX−1=06 up to relabeling the constant c0. On the other hand, by symmetry, |f(x) − f(y)|p n+sp dxdy Rn Rn |x − y| Z × |f(x) − f(y)|p = dxdy |x − y|n+sp i,j∈Z Di×Dj X Z |f(x) − f(y)|p ≥ 2 dxdy |x − y|n+sp i,j∈Z ZDi×Dj Xj

Lemma 6.4. Let q ∈ [1, ∞). Let f : Rn → R be a measurable function. For any N ∈ N, let

n fN (x) := max min{f(x), N}, −N ∀x ∈ R . (6.20)

Then  lim kfN kLq(Rn) = kfkLq(Rn). N→+∞

Proof. We denote by |f|N the function obtained by cutting |f| at level N. We have that |f|N = |fN | and so, by Fatou Lemma, we obtain that

1 1 q q q q lim inf kfN kLq(Rn) = lim inf |f|N ≥ |f| = kfkLq(Rn). N→+∞ N→+∞ Rn n Z  ZR 

The reverse inequality easily follows by the fact that |f|N (x) ≤ |f(x)| for any x ∈ Rn.

45 Taking into account the previous lemmas, we are able to give an elemen- tary proof of the Sobolev-type inequality stated in the following theorem.

Theorem 6.5. Let s ∈ (0, 1) and p ∈ [1, +∞) such that sp < n. Then there exists a positive constant C = C(n,p,s) such that, for any measurable and compactly supported function f : Rn → R, we have

p p |f(x) − f(y)| kfkLp⋆ (Rn) ≤ C n+sp dx dy. (6.21) Rn Rn |x − y| Z Z where p⋆ = p⋆(n, s) is the so-called “fractional critical exponent” and it is equal to np/(n − sp). Consequently, the space W s,p(Rn) is continuously embedded in Lq(Rn) for any q ∈ [p,p⋆]. Proof. First, we note that if the right hand side of (6.21) is unbounded then the claim in the theorem plainly follows. Thus, we may suppose that f is such that |f(x) − f(y)|p n+sp dxdy < +∞. (6.22) Rn Rn |x − y| Z Z Moreover, we can suppose, without loss of generality, that

f ∈ L∞(Rn). (6.23)

Indeed, if (6.22) holds for bounded functions, then it holds also for the function fN , obtained by any (possibly unbounded) f by cutting at levels −N and +N (see (6.20)). Therefore, by Lemma 6.4 and the fact that (6.22) together with the Dominated Convergence Theorem imply

p p |fN (x) − fN (y)| |f(x) − f(y)| lim n+sp dxdy = n+sp dx dy, N→+∞ Rn Rn |x − y| Rn Rn |x − y| Z Z Z Z we obtain estimate (6.21) for the function f.

Now, take ak and Ak defined by (6.4) and (6.5), respectively. We have

p⋆ p⋆ k+1 p⋆ kfkLp⋆ (Rn) = |f(x)| dx ≤ (2 ) dx Z Ak\Ak+1 Z Ak\Ak+1 Xk∈ Z Xk∈ Z (k+1)p⋆ ≤ 2 ak. Z Xk∈ 46 That is, p/p⋆ p p kp⋆ kfkLp⋆ (Rn) ≤ 2 2 ak . Z ! Xk∈ Thus, since p/p⋆ = (n − sp)/n = 1 − sp/n < 1,

p p kp (n−sp)/n kfkLp⋆ (Rn) ≤ 2 2 ak (6.24) Z Xk∈ and, then, by choosing T =2p, Lemma 6.2 yields

sp p kp − n kfkLp⋆ (Rn) ≤ C 2 ak+1ak , (6.25) k∈Z aXk=06 for a suitable constant C depending on n, p and s. Finally, it suffices to apply Lemma 6.3 and we obtain the desired result, up to relabeling the constant C in (6.25). Furthermore, the embedding for q ∈ (p,p⋆) follows from standard applica- tion of H¨older inequality. Remark 6.6. From Lemma 6.1, it follows that

dxdy (n−sp)/n n+sp ≥ c(n, s) |E| (6.26) C |x − y| ZE Z E for all measurable sets E with finite measure. On the other hand, we see that (6.21) reduces to (6.26) when f = χE, so (6.26) (and thus Lemma 6.1) may be seen as a Sobolev-type inequality for sets.

The above embedding does not generally hold for the space W s,p(Ω) since it not always possible to extend a function f ∈ W s,p(Ω) to a function f˜ ∈ W s,p(Rn). In order to be allowed to do that, we should require further regularity assumptions on Ω (see Section 5).

Theorem 6.7. Let s ∈ (0, 1) and p ∈ [1, +∞) such that sp < n. Let Ω ⊆ Rn be an extension domain for W s,p. Then there exists a positive constant C = C(n, p, s, Ω) such that, for any f ∈ W s,p(Ω), we have

kfkLq(Ω) ≤ CkfkW s,p(Ω), (6.27)

47 for any q ∈ [p,p⋆]; i.e., the space W s,p(Ω) is continuously embedded in Lq(Ω) for any q ∈ [p,p⋆]. If, in addition, Ω is bounded, then the space W s,p(Ω) is continuously embedded in Lq(Ω) for any q ∈ [1,p⋆]. Proof. Let f ∈ W s,p(Ω). Since Ω ⊆ Rn is an extension domain for W s,p, then there exists a constant C1 = C1(n, p, s, Ω) > 0 such that ˜ kfkW s,p(Rn) ≤ C1kfkW s,p(Ω), (6.28) with f˜ such that f˜(x)= f(x) for x a.e. in Ω. On the other hand, by Theorem 6.5, the space W s,p(Rn) is continuously q n ⋆ embedded in L (R ) for any q ∈ [p,p ]; i.e., there exists a constant C2 = C2(n,p,s) > 0 such that ˜ ˜ kfkLq (Rn) ≤ C2kfkW s,p(Rn). (6.29) Combining (6.28) with (6.29), we get ˜ ˜ ˜ kfkLq(Ω) = kfkLq(Ω) ≤ kfkLq(Rn) ≤ C2kfkW s,p(Rn)

≤ C2C1kfkW s,p(Ω), that gives the inequality in (6.27), by choosing C = C2C1. In the case of Ω being bounded, the embedding for q ∈ [1,p) plainly follows from (6.27), by using the H¨older inequality.

Remark 6.8. In the critical case q = p⋆ the constant C in Theorem 6.7 does not depend on Ω: this is a consequence of (6.21) and of the extension property of Ω.

6.1. The case sp = n We note that when sp → n the critical exponent p⋆ goes to ∞ and so it is not surprising that, in this case, if f is in W s,p then f belongs to Lq for any q, as stated in the following two theorems. Theorem 6.9. Let s ∈ (0, 1) and p ∈ [1, +∞) such that sp = n. Then there exists a positive constant C = C(n,p,s) such that, for any measurable and compactly supported function f : Rn → R, we have

kfkLq(Rn) ≤ CkfkW s,p(Rn), (6.30)

48 for any q ∈ [p, ∞); i.e., the space W s,p(Rn) is continuously embedded in Lq(Rn) for any q ∈ [p, ∞).

Theorem 6.10. Let s ∈ (0, 1) and p ∈ [1, +∞) such that sp = n. Let Ω ⊆ Rn be an extension domain for W s,p. Then there exists a positive constant C = C(n, p, s, Ω) such that, for any f ∈ W s,p(Ω), we have

kfkLq(Ω) ≤ CkfkW s,p(Ω), (6.31) for any q ∈ [p, ∞); i.e., the space W s,p(Ω) is continuously embedded in Lq(Ω) for any q ∈ [p, ∞). If, in addition, Ω is bounded, then the space W s,p(Ω) is continuously embedded in Lq(Ω) for any q ∈ [1, ∞). The proofs can be obtained by simply combining Proposition 2.1 with Theorem 6.5 and Theorem 6.7, respectively.

7. Compact embeddings In this section, we state and prove some compactness results involving the fractional spaces W s,p(Ω) in bounded domains. The main proof is a modification of the one of the classical Riesz-Frechet-Kolmogorov Theorem (see [55, 79]) and, again, it is self-contained and it does not require to use Besov or other interpolation spaces, nor Fourier transform and flows (see [28, Theorem 1.5]). We refer to [77, Lemma 6.11] for the case p = q = 2.

Theorem 7.1. Let s ∈ (0, 1), p ∈ [1, +∞), q ∈ [1,p], Ω ⊂ Rn be a bounded extension domain for W s,p and T be a bounded subset of Lp(Ω). Suppose that |f(x) − f(y)|p sup dxdy < +∞. T |x − y|n+sp f∈ ZΩ ZΩ Then T is pre-compact in Lq(Ω). Proof. We want to show that T is totally bounded in Lq(Ω), i.e., for any q ε ∈ (0, 1) there exist β1,...,βM ∈ L (Ω) such that for any f ∈ T there exists j ∈{1,...,M} such that

kf − βjkLq(Ω) ≤ ε. (7.1)

49 Since Ω is an extension domain, there exists a function f˜ in W s,p(Rn) ˜ such that kfkW s,p(Rn) ≤ CkfkW s,p(Ω). Thus, for any cube Q containing Ω, we have ˜ ˜ kfkW s,p(Q) ≤ kfkW s,p(Rn) ≤ CkfkW s,p(Ω). Observe that, since Q is a bounded open set, f˜ belongs also to Lq(Q) for any q ∈ [1,p]. Now, for any ε ∈ (0, 1), we let

|f˜(x) − f˜(y)|p ˜ q Co := 1 + sup kfkL (Q) + sup n+sp dx dy, f∈T f∈T Q Q |x − y| 1 Z Z s n ε ε ρ q ρ = ρ := and η = η := , ε 1 n+sp ε q 2 2Co n 2p !

11 and we take a collection of disjoints cubes Q1,...,QN of side ρ such that

N

Ω ⊆ Q = Qj. j=1 [ For any x ∈ Ω, we define

j(x) as the unique integer in {1,...,N} for which x ∈ Qj(x). (7.2)

Also, for any f ∈ T , let 1 P (f)(x) := f˜(y) dy. |Qj(x)| ZQj(x) Notice that

P (f + g)= P (f)+ P (g) for any f,g ∈ T and that P (f) is constant, say equal to qj(f), in any Qj, for j ∈{1,...,N}. Therefore, we can define

n/q N R(f) := ρ q1(f),...,qN (f) ∈ R  11To be precise, for this one needs to take ε ∈ (0, 1) arbitrarily small and such that the ration between the side of Q and ρε is integer.

50 and consider the spatial q-norm in RN as

1 N q q N kvkq := |vj| , for any v ∈ R . j=1 ! X We observe that R(f + g)= R(f)+ R(g). Moreover,

N q q kP (f)kLq(Ω) = |P (f)(x)| dx j=1 Qj ∩Ω X Z N kR(f)kq ≤ ρn |q (f)|q = kR(f)kq ≤ q . (7.3) j q ρn j=1 X Also, by H¨older inequality,

N N q q n q 1 ˜ kR(f)kq = ρ |qj(f)| = n(q−1) f(y) dy ρ Qj j=1 j=1 Z X X N ˜ q ˜ q ˜ q ≤ |f(y)| dy = |f(y)| dy = kfkLq(Q). j=1 Qj Q X Z Z In particular, q sup kR(f)kq ≤ Co, f∈T that is, the set R(T ) is bounded in RN (with respect to the q-norm of RN as well as to any equivalent norm of RN ) and so, since it is finite N dimensional, it is totally bounded. Therefore, there exist b1,...,bM ∈ R such that M

R(T ) ⊆ Bη(bi), (7.4) i=1 [ N where the balls Bη are taken in the q-norm of R .

For any i ∈{1,...,M}, we write the coordinates of bi as

N bi =(bi,1,...,bi,N ) ∈ R .

For any x ∈ Ω, we set −n/q βi(x) := ρ bi,j(x),

51 where j(x) is as in (7.2). Notice that βi is constant on Qj, i.e. if x ∈ Qj then − n P (βi)(x)= ρ q bi,j = βi(x) (7.5) − n and so qj(βi)= ρ q bi,j; thus

R(βi)= bi. (7.6) Furthermore, for any f ∈ T N q q kf − P (f)kLq(Ω) = |f(x) − P (f)(x)| dx j=1 Qj ∩Ω X Z q N 1 = f(x) − f˜(y) dy dx Qj ∩Ω |Qj| Qj j=1 Z Z X N q 1 ˜ = q f(x) − f(y) dy dx Qj ∩Ω |Qj| Qj j=1 Z Z X q 1 N ≤ |f(x) − f˜(y)| dy dx. (7.7) ρnq j=1 Qj ∩Ω " Qj # X Z Z Now for any fixed j ∈ 1, ··· , N, by H¨older inequality with p and p/(p − 1) we get q 1 |f(x) − f˜(y)| dy ρnq "ZQj # q p 1 q(p−1) p ≤ |Q | p f(x) − f˜(y) dy ρnq j "ZQj #

q 1 p = f(x) − f˜(y) p dy ρnq/p "ZQj #

q ˜ p p 1 n+sp q q (n+sp) |f(x) − f(y)| ≤ n( 2 ) p ρ p dy ρnq/p |x − y|n+sp "ZQj # q ˜ p p n+sp q sq |f(x) − f(y)| ≤ n( 2 ) p ρ dy . (7.8) |x − y|n+sp "ZQ # 52 Hence, combining (7.7) with (7.8), we obtain that

q ˜ ˜ p p q n+sp q sq |f(x) − f(y)| kf − P (f)k ≤ n( 2 ) p ρ dy dx Lq(Ω) |x − y|n+sp ZQ "ZQ # q ˜ ˜ p p n+sp q sq |f(x) − f(y)| ≤ n( 2 ) p ρ dy dx (7.9) |x − y|n+sp "ZQ ZQ # q n+sp q sq ε ≤ C n( 2 ) p ρ = . o 2q where (7.9) follows from Jensen inequality since t 7→ |t|q/p is a concave function for any fixed p and q such that q/p ≤ 1. Consequently, for any j ∈{1, ..., M}, recalling (7.3) and (7.5)

kf − βjkLq(Ω) ≤ kf − P (f)kLq(Ω) + kP (βj) − βjkLq(Ω) + kP (f − βj)kLq (Ω) ε kR(f) − R(β )k ≤ + j q . (7.10) 2 ρn/q

Now, given any f ∈ T , we recall (7.4) and (7.6) and we take j ∈ {1,...,M} such that R(f) ∈ Bη(bj). Then, (7.5) and (7.10) give that

ε kR(f) − bjkq ε η kf − β k q ≤ + ≤ + = ε. (7.11) j L (Ω) 2 ρn/q 2 ρn/q

This proves (7.1), as desired.

Corollary 7.2. Let s ∈ (0, 1) and p ∈ [1, +∞) such that sp < n. Let q ∈ [1,p⋆), Ω ⊆ Rn be a bounded extension domain for W s,p and T be a bounded subset of Lp(Ω). Suppose that

|f(x) − f(y)|p sup dxdy < +∞. T |x − y|n+sp f∈ ZΩ ZΩ Then T is pre-compact in Lq(Ω).

Proof. First, note that for 1 ≤ q ≤ p the compactness follows from Theo- rem 7.1.

53 For any q ∈ (p,p⋆), we may take θ = θ(p,p⋆, q) ∈ (0, 1) such that ⋆ 1/q = θ/p +1 − θ/p , thus for any f ∈ T and βj with j ∈{1, ..., N} as in the theorem above, using H¨older inequality with p/(θq) and p⋆/((1 − θ)q), we get

1/q qθ q(1−θ) kf − βjkLq(Ω) = |f − βj| |f − βj| dx ZΩ  θ/p (1−θ)/p⋆ p p⋆ ≤ |f − βj| dx |f − βj| dx ZΩ  ZΩ  1−θ θ = kf − βjkLp⋆(Ω) kf − βjkLp(Ω)

1−θ θ ˜ θ ≤ Ckf − βjkW s,p(Ω) kf − βjkLp(Ω) ≤ Cε , where the last inequalities comes directly from (7.11) and the continuos embedding (see Theorem 6.7). Remark 7.3. As well known in the classical case s = 1 (and, more generally, when s is an integer), also in the fractional case the lack of compactness for the critical embedding (q = p⋆) is not surprising, because of translation and dilation invariance (see [76] for various results in this direction, for any 0

Notice that the regularity assumption on Ω in Theorem 7.1 and Corol- lary 7.2 cannot be dropped (see Example 9.2 in Section 9).

8. H¨older regularity In this section we will show certain regularity properties for functions in W s,p(Ω) when sp > n and Ω is an extension domain for W s,p with no external cusps. For instance, one may take Ω any Lipschitz domain (recall Theorem 5.4). The main result is stated in the forthcoming Theorem 8.2. First, we need a simple technical lemma, whose proof can be found in [46] (for instance).

54 Lemma 8.1. ([46, Lemma 2.2]). Let p ∈ [1, +∞) and sp ∈ (n, n + p]. Let Ω ⊂ Rn be a domain with no external cusps and f be a function in s,p ′ ′ W (Ω). Then, for any x0 ∈ Ω and R, R , with 0 < R

(sp−n)/np |hfiBR(x0)∩Ω −hfiBR′ (x0)∩Ω| ≤ c [f]p,sp |BR(x0) ∩ Ω| (8.1) where 1 p −sp p [f]p,sp := sup ρ |f(x) −hfiBρ(x0)∩Ω| dx x0∈Ω ρ>0 ZBρ(x0)∩Ω ! and 1 hfi := f(x)dx. Bρ(x0)∩Ω |B (x ) ∩ Ω| ρ 0 ZBρ(x0)∩Ω

Theorem 8.2. Let Ω ⊆ Rn be an extension domain for W s,p with no external cusps and let p ∈ [1, +∞), s ∈ (0, 1) such that sp>n. Then, there exists C > 0, depending on n, s, p and Ω, such that

1 p p p |f(x) − f(y)| kfk 0,α ≤ C kfk + dxdy , (8.2) C (Ω) Lp(Ω) |x − y|n+sp  ZΩ ZΩ  for any f ∈ Lp(Ω), with α := (sp − n)/p. Proof. In the following, we will denote by C suitable positive quantities, possibly different from line to line, and possibly depending on p and s. First, we notice that if the right hand side of (8.2) is not finite, then we are done. Thus, we may suppose that |f(x) − f(y)|p dxdy ≤ C, |x − y|n+sp ZΩ ZΩ for some C > 0. Second, since Ω is an extension domain for W s,p, we can extend any f ˜ ˜ to a function f such that kfkW s,p(Rn) ≤ CkfkW s,p(Ω). Now, for any bounded measurable set U ⊂ Rn, we consider the average value of the function f˜ in U, given by 1 hf˜i := f˜(x) dx. U |U| ZU 55 For any ξ ∈ Rn, the H¨older inequality yields

p p 1 1 ξ −hf˜i = ξ − f˜(y) dy ≤ |ξ − f˜(y)|p dy. U |U|p |U| ZU ZU

˜ Accordingly, by taking xo ∈ Ω and U := Br(xo), ξ := f(x) and integrating over Br(xo), we obtain that

˜ ˜ p |f(x) −hfiBr(xo)| dx ZBr(xo) 1 ≤ |f˜(x) − f˜(y)|p dx dy. |B (x )| r o ZBr(xo) ZBr(xo)

Hence, since |x − y| ≤ 2r for any x, y ∈ Br(xo), we deduce that

˜ ˜ p |f(x) −hfiBr(xo)| dx ZBr(xo) (2r)n+sp |f˜(x) − f˜(y)|p ≤ dxdy |B (x )| |x − y|n+sp r o ZBr(xo) ZBr(xo) n+sp sp p 2 r Ckfk s,p ≤ W (Ω) , (8.3) |B1| that implies p p [f]p,sp ≤ CkfkW s,p(Ω), (8.4) for a suitable constant C. Now, we will show that f is a continuos function. Taking into account

(8.1), it follows that the sequence of functions x → hfiBR(x)∩Ω converges uniformly in x ∈ Ω when R → 0. In particular the limit function g will be continuos and the same holds for f, since by Lebesgue theorem we have that 1 lim f(y) dy = f(x) for almost every x ∈ Ω. R→0 |B (x) ∩ Ω| R ZBR(x)∩Ω

Now, take any x, y ∈ Ω and set R = |x − y|. We have ˜ ˜ ˜ ˜ |f(x) − f(y)|≤|f(x) −hfiB2R(x)| + |hfiB2R(x) −hfiB2R(y)| + |hfiB2R(y) − f(y)|

56 We can estimate the first and the third term of right hand-side of the above inequality using Lemma 8.1. Indeed, getting the limit in (8.1) as R′ → 0 and writing 2R instead of R, for any x ∈ Ω we get

˜ (sp−n)/np (sp−n)/p |hfiB2R(x) − f(x)| ≤ c [f]p,sp|B2R(x)| ≤ C[f]p,sp R (8.5)

(sp−n)/p where the constant C is given by c 2 /|B1|. On the other hand, ˜ ˜ ˜ ˜ ˜ |hfiB2R(x) −hfiB2R(y)|≤ |f(z) −hfiB2R(x)| + |f(z) −hfiB2R(y)| and so, integrating on z ∈ B2R(x) ∩ B2R(y), we have ˜ ˜ |B2R(x) ∩ B2R(y)||hfiB2R(x) −hfiB2R(y)|

˜ ˜ ≤ |f(z) −hfiB2R(x)| dz ZB2R(x)∩B2R(y) ˜ ˜ + |f(z) −hfiB2R(y)| dz ZB2R(x)∩B2R(y) ˜ ˜ ˜ ˜ ≤ |f(z) −hfiB2R(x)| dz + |f(z) −hfiB2R(y)| dz. ZB2R(x) ZB2R(y)

Furthermore, since BR(x) ∪ BR(y) ⊂ B2R(x) ∩ B2R(y) , we have  |BR(x)|≤|B2R(x) ∩ B2R(y)| and |BR(y)|≤|B2R(x) ∩ B2R(y)| and so 1 |hf˜i −hf˜i | ≤ |f˜(z) −hf˜i | dz B2R(x) B2R(y) |B (x)| B2R(x) R ZB2R(x) 1 + |f˜(z) −hf˜i | dz. |B (y)| B2R(y) R ZB2R(y)

57 An application of the H¨older inequality gives 1 |f˜(z) −hf˜i | dz |B (x)| B2R(x) R ZB2R(x) |B (x)|(p−1)/p 1/p ≤ 2R |f˜(z) −hf˜i |p dz |B (x)| B2R(x) R ZB2R(x)  (p−1)/p |B2R(x)| s ≤ (2R) [f]p,sp |BR(x)|

(sp−n)/p ≤ C [f]p,sp R . (8.6) Analogously, we obtain 1 |f˜(z) −hf˜i | dz ≤ C [f] R(sp−n)/p . (8.7) |B (y)| B2R(y) p,sp R ZB2R(y)

Combining (8.5), (8.6) with (8.7) it follows (sp−n)/p |f(x) − f(y)| ≤ C [f]p,sp |x − y| , (8.8) up to relabeling the constant C. Therefore, by taking into account (8.4), we can conclude that f ∈ C0,α(Ω), with α =(sp − n)/p. Finally, taking R0 < diam(Ω) (note that the latter can be possibly infinity), using estimate in (8.5) and the H¨older inequality we have, for any x ∈ Ω, |f(x)| ≤ |hf˜i | + |f(x) −hf˜i | BR0 (x) BR0 (x) C α p ≤ 1/p kfkL (Ω) + c [f]p,sp |BR0 (x)| . (8.9) |BR0 (x)| Hence, by (8.4), (8.8) and (8.9), we get |f(x) − f(y)| 0,α ∞ kfkC (Ω) = kfkL (Ω) + sup α x,y∈Ω |x − y| x6=y

≤ C kfkLp(Ω) + [f]p,sp  ≤ CkfkW s,p(Ω) . for a suitable positive constant C.

58 Remark 8.3. The estimate in (8.3) says that f belongs to the Campanato space L p,λ, with λ := sp, (see [22] and, e.g., [46, Definition 2.4]). Then, the conclusion in the proof of Theorem 8.2 is actually an application of the Campanato Isomorphism (see, for instance, [46, Theorem 2.9]).

Just for a matter of curiosity, we observe that, according to the def- inition (2.1), the fractional Sobolev space W s,∞(Ω) could be view as the space of functions

|u(x) − u(y)| u ∈ L∞(Ω) : ∈ L∞(Ω × Ω) , |x − y|s   but this space just boils down to C0,s(Ω), that is consistent with the H¨older embedding proved in this section; i.e., taking formally p = ∞ in Theorem 8.2, the function u belongs to C0,s(Ω).

9. Some counterexamples in non-Lipschitz domains When the domain Ω is not Lipschitz, some interesting things happen, as next examples show.

Example 9.1. Let s ∈ (0, 1). We will construct a function u in W 1,p(Ω) that does not belong to W s,p(Ω), providing a counterexample to Proposition 2.2 when the domain is not Lipschitz. Take any p ∈ (1/s, +∞). (9.1) Due to (9.1), we can fix p +1 κ> . (9.2) sp − 1 We remark that κ> 1. Let us consider the cusp in the plane

κ C := (x1, x2) with x1 ≤ 0 and |x2|≤|x1| and take polar coordinates on R2 \ C, say ρ = ρ(x) ∈ (0 , +∞) and θ = 2 θ(x) ∈ (−π, π), with x =(x1, x2) ∈ R \ C.

59 We define the function u(x) := ρ(x)θ(x) and the heart-shaped do- 2 main Ω := (R \ C) ∩ B1, with B1 being the unit ball centered in the origin. Then, u ∈ W 1,p(Ω) \ W s,p(Ω). To check this, we observe that x ∂ ρ = (2ρ)−1∂ ρ2 = (2ρ)−1∂ (x2 + x2) = 1 x1 x1 x1 1 2 ρ and, in the same way, x ∂ ρ = 2 . x2 ρ Accordingly, x2 1 = ∂ x = ∂ (ρ cos θ) = ∂ ρ cos θ − ρ sin θ∂ θ = 1 − x ∂ θ x1 1 x1 x1 x1 ρ2 2 x1 x2 = 1 − 2 − x ∂ θ. ρ2 2 x1 That is x ∂ θ = − 2 . x1 ρ2

By exchanging the roles of x1 and x2 (with some care on the sign of the derivatives of the trigonometric functions), one also obtains x ∂ θ = 1 . x2 ρ2 Therefore,

−1 −1 ∂x1 u = ρ (x1θ − x2) and ∂x2 u = ρ (x2θ + x1) and so |∇u|2 = θ2 +1 ≤ π2 +1. This shows that u ∈ W 1,p(Ω). On the other hand, let us fix r ∈ (0, 1), to be taken arbitrarily small N κ at the end, and let us define r0 := r and, for any j ∈ , rj+1 := rj − rj . By induction, one sees that rj is strictly decreasing, that rj > 0 and so rj ∈ (0,r) ⊂ (0, 1). Accordingly, we can define

ℓ := lim rj ∈ [0, 1]. j→+∞

60 By construction

κ κ ℓ = lim rj+1 = lim rj − rj = ℓ − ℓ , j→+∞ j→+∞ hence ℓ = 0. As a consequence,

+∞ N N κ κ rj = lim rj = lim rj − rj+1 N→+∞ N→+∞ j=0 j=0 j=0 X X X = lim r0 − rN+1 = r. (9.3) N→+∞ We define

2 2 Dj := (x, y) ∈ R × R s.t. x1,y1 ∈ (−rj, −rj+1), κ κ κ κ  x2 ∈ (|x1| , 2|x1| ) and − y2 ∈ (|y1| , 2|y1| ) . We observe that

2 2 Ω × Ω ⊇ (x, y) ∈ R × R s.t. x1,y1 ∈ (−r, 0), κ κ κ κ  x2 ∈ (|x1| , 2|x1| ) and − y2 ∈ (|y1| , 2|y1| ) +∞

⊇ Dj, j=0 [ and the union is disjoint. Also, r r = r (1 − rκ−1) ≥ r (1 − rκ−1) ≥ j , j+1 j j j 2 for small r. Hence, if (x, y) ∈Dj,

|x1| ≤ rj ≤ 2rj+1 ≤ 2|y1| and, analogously, |y1| ≤ 2|x1|.

Moreover, if (x, y) ∈Dj,

κ κ κ κ κ |x1 − y1| ≤ rj − rj+1 = rj ≤ 2 rj+1 ≤ 2 |x1| and κ κ κ+2 κ |x2 − y2|≤|x2| + |y2| ≤ 2|x1| +2|y1| ≤ 2 |x1| .

61 As a consequence, if (x, y) ∈Dj,

κ+3 κ |x − y| ≤ 2 |x1| .

Notice also that, when (x, y) ∈Dj, we have θ(x) ≥ π/2 and θ(y) ≤ −π/2, so π ρ(x) π |x | u(x) − u(y) ≥ u(x) ≥ ≥ 1 . 2 2 As a consequence, for any (x, y) ∈Dj, |u(x) − u(y)|p ≥ c|x |p−κ(2+sp), |x − y|2+sp 1 for some c> 0. Therefore, p |u(x) − u(y)| p−κ(2+sp) 2+sp dxdy ≥ c|x1| dxdy Dj |x − y| Dj ZZ ZZ κ κ −rj+1 −rj+1 2|x1| −|y1| p−κ(2+sp) = c dx1 dy1 dx2 dy2|x1| κ κ Z−rj Z−rj Z|x1| Z−2|y1| −rj+1 −rj+1 p−κ(2+sp) κ κ = c dx1 dy1|x1| |x1| |y1| Z−rj Z−rj −rj+1 −rj+1 −κ p−κsp ≥ c 2 dx1 dy1|x1| Z−rj Z−rj −rj+1 −rj+1 −κ p−κsp ≥ c 2 dx1 dy1rj Z−rj Z−rj −κ p−κsp+2κ −κ κ−α = c 2 rj = c 2 rj , with α := κ(sp − 1) − p> 1, (9.4) thanks to (9.2). In particular, |u(x) − u(y)|p dxdy ≥ c 2−κr−αrκ |x − y|2+sp j ZZDj and so, by summing up and exploiting (9.3),

|u(x) − u(y)|p +∞ |u(x) − u(y)|p dxdy ≥ dxdy ≥ c 2−κr1−α. |x − y|2+sp |x − y|2+sp Ω Ω j=0 Dj Z Z X ZZ 62 By taking r as small as we wish and recalling (9.4), we obtain that

|u(x) − u(y)|p dxdy =+∞, |x − y|2+sp ZΩ ZΩ so u 6∈ W s,p(Ω). ✷

Example 9.2. Let s ∈ (0, 1). We will construct a sequence of functions s,p {fn} bounded in W (Ω) that does not admit any convergent subsequence in Lq(Ω), providing a counterexample to Theorem 7.1 when the domain is not Lipschitz. We follow an observation by [82]. For the sake of simplicity, fix n = k p = q = 2. We take ak := 1/C for a constant C > 10 and we consider the ∞ N 2 set Ω = k=1 Bk where, for any k ∈ , Bk denotes the ball of radius ak centered in ak. Notice that S 2 2 ak → 0 as k →∞ and ak − ak > ak+1 + ak+1. Thus, Ω is the union of disjoint balls, it is bounded and it is not a Lipschitz domain. For any n ∈ N, we define the function fn : Ω → R as follows

− 1 −2 π 2 an x ∈ Bn, fn(x)= ( 0 x ∈ Ω \ Bn .

We observe that we cannot extract any subsequence convergent in L2(Ω) from the sequence of functions {fn}, because fn(x) → 0 as n → +∞, for any fixed x ∈ Ω but

2 2 −1 −4 kfnkL2(Ω) = |fn(x)| dx = π an dx = 1. ZΩ ZBn

s Now, we compute the H norm of fn in Ω. We have |f (x) − f (y)|2 π−1a−4 n n dxdy = 2 n dxdy |x − y|2+2s |x − y|2+2s ZΩ ZΩ ZΩ\Bn ZBn −4 −1 an = 2π 2+2s dx dy. (9.5) Bk Bn |x − y| Xk6=n Z Z 63 Thanks to the choice of {ak} we have that |a − a | |a2 + a2| = a2 + a2 ≤ n k . n k n k 2

Thus, since x ∈ Bn, y ∈ Bk, it follows 2 2 2 2 |x − y| ≥ |an − an − (ak + ak)| = |an − ak − (an + ak)| |a − a | ≥ |a − a |−|a2 + a2|≥|a − a | − n k n k n k n k 2 |a − a | = n k . 2 Therefore, a−4 a−4 n dxdy ≤ 22+2s n dxdy |x − y|2+2s |a − a |2+2s ZBk ZBn ZBk ZBn n k 4 2+2s 2 ak = 2 π 2+2s . (9.6) |an − ak| Also, if m ≥ j + 1 we have 1 1 1 1 a a − a ≥ a − a = − = 1 − ≥ j . (9.7) j m j j+1 Cj Cj+1 Cj C 2   Therefore, combining (9.7) with (9.5) and (9.6), we get 2 |fn(x) − fn(y)| 2+2s dxdy Ω Ω |x − y| Z Z 4 3+2s ak ≤ 2 π 2+2s |an − ak| Xk6=n 4 4 3+2s ak ak = 2 π 2+2s + 2+2s (ak − an) (an − ak) ! Xkn 4 4 5+4s ak ak ≤ 2 π 2+2s + 2+2s ak an ! Xkn 1 k ≤ 26+4sπ a2−2s = 26+4sπ < +∞. k C2−2s Xk6=n Xk6=n   s This shows that {fn} is bounded in H (Ω). ✷

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74