ON FRACTIONAL HARDY INEQUALITIES IN CONVEX SETS

Transcription

1 ON FRACTIONAL HARDY INEQUALITIES IN CONVEX SETS L. BRASCO AND E. CINTI Abstract. We rove a Hardy inequality on convex sets, for fractional Sobolev-Slobodeckiĭ saces of order (s,. The roof is based on the fact that in a convex set the distance from the boundary is a suerharmonic function, in a suitable sense. The result holds for every 1 < < and 0 < s < 1, with a constant which is stable as s goes to 1. Contents 1. Introduction A quick overview on Hardy inequality Main result Method of roof Plan of the aer 6 2. Preliminaries Notations Functional analytic facts An exedient estimate for convex sets 9 3. Suerharmonicity of the distance function Proof of Theorem Proof of inequality ( Imroved constant for s close to Some consequences 17 Aendix A. Some ointwise inequalities 18 References Introduction 1.1. A quick overview on Hardy inequality. Given an oen set Ω R N with Lischitz boundary, we will use the notation { inf x y, if x Ω, d Ω (x := y Ω 0, if x R N \ Ω. A fundamental result in the theory of Sobolev saces is the Hardy inequality u 2 (1.1 C Ω Ω d 2 dx u 2 dx, for every u C0 (Ω, Ω Ω Date: May 11, Mathematics Subject Classification. 39B72, 35R11, 46E35. ey words and hrases. Hardy inequality, nonlocal oerators, fractional Sobolev saces. 1

2 2 BRASCO AND CINTI see for examle [20, Theorem 21.3]. It is well-known that for a convex set R N, such an inequality holds with the dimension-free universal constant C = 1 4, see for examle [8, Theorem 2]. Moreover, such a constant is shar. In order to exlain the aims and techniques of the resent aer, it is useful to recall a roof of this fact. A well-known and very elegant way of roving (1.1 with shar constant for convex sets, consists in mimicking Moser s logarithmic estimate for ositive suersolutions of ellitic artial differential equations. The starting oint is the observation that on a convex set we have d 0, i.e. the distance function is suerharmonic. More recisely, it holds (1.2 d, ϕ dx 0, for every nonnegative ϕ C0 (. By following Moser (see [19, age 586], one can test the equation (1.2 with 1 where u C 0 that is (. This gives d 2 d d d We now use Young s inequality ϕ = u2 d,, u u dx 2 u 2 dx 2 d d d d 2 u 2 dx 0,, u u dx. with the following choices a, b a b 2 2, a, b RN, a = δ d d u and b = 1 δ u, where δ is an arbitrary ositive real number. This leads to d 2 d u 2 dx δ d 2 d u 2 dx + 1 δ which can be recast as δ (1 δ d d 2 u 2 dx u 2 dx. u 2 dx, 1 Of course, such a test function is not in C 0 (. However, by a standard density argument, in (1.2 we can allow W 1,2 test functions with comact suort. We avoid this unessential technicality for ease of readability.

3 FRACTIONAL HARDY INEQUALITY 3 It is now sufficient to observe that the term δ (1 δ in the left-hand side is maximal for δ = 1/2. This leads to the Hardy inequality with the claimed shar constant 1/4, once it is observed that d = 1, a. e. in. The latter imlies that more generally for every 1 < < we have d 0 in, i.e. d is suerharmonic in the following sense d 2 d, ϕ dx 0, for every nonnegative ϕ C0 (. By testing this with ϕ = u /d 1 and suitably adating the roof above, one can rove the more general Hardy inequality for convex sets ( 1 u (1.3 d dx u dx. Once again, the constant aearing in (1.3 is shar and indeendent of both and the dimension N Main result. The scoe of the resent aer is to rove a fractional version of Hardy inequality for convex sets, by adating to the fractional setting the Moser-tye roof resented above. An essential feature of our method is that the relevant constant aearing in the Hardy inequality is stable as the fractional order of differentiability s converges to 1, see Remark 1.2 below. More recisely, we rove the following Theorem 1.1 (Hardy inequality on convex sets. Let 1 < < and 0 < s < 1. Let R N be an oen convex set such that R N. Then for every u C0 ( we have C u RN u(x u(y (1.4 s (1 s d s dx dx dy, x y N+s RN for a comutable constant C = C(N, > 0. The constant obtained in (1.4 is very likely not shar. However, a coule of comments are in order on this oint. Remark 1.2 (Asymtotic behaviour in s of the constant. We recall that if u C0 (, then we have (see [21, Corollary 1.3] and [4, Proosition 2.8] RN u(x u(y lim(1 s s 1 x y N+s dx dy = α N, u dx, RN with α N, = 1 ω, e 1 dh N 1 (ω, e 1 = (1, 0,..., 0. S N 1 In this resect, we observe that the constant aearing in Theorem 1.1 has the correct asymtotic behaviour as s converges to 1: by assing to the limit in (1.4 as s goes to 1, we obtain the usual local Hardy inequality u (1.5 C d dx α N, u dx.

4 4 BRASCO AND CINTI As for the limit s 0, we recall that (see [18, Theorem 3] RN lim s u(x u(y s 0 x y N+s dx dy = β N, RN with u dx, β N, = 2 N ω N, and ω N is the volume of the N dimensional unit ball. Thus, in this case as well, our constant in (1.4 exhibits the correct asymtotic behaviour as s converges to 0. Remark 1.3 (Deendence on N of C. At a first glance, it may look strange that the constant C in Theorem 1.1 deends on N. Indeed, we have seen in (1.3 that in the local case such an inequality holds with the universal constant ( 1. However, it is easily seen that C must deend on N in the fractional case. Indeed, we have already seen that assing to the limit in (1.4 as s goes to 1, we obtain the local Hardy inequality (1.5. As we already said, the shar constant in the revious inequality is (( 1/, which means that we must have ( 1 C α N,, for every N 1. On the other hand, it is easily seen that α N, converges to 0 as N goes to. This shows that C in (1.4 must deend on N Method of roof. We now send some words on the roof of Theorem 1.1. We first observe that there is an elementary roof of the inequality u RN u(x u(y C N,,s d s dx dx dy, x y N+s RN as ointed out to us by Bart lomiej Dyda. This is based on geometric considerations and on the nonlocality of the double integral in the right-hand side. We detail this argument in Subsection 4.2 below. We oint out that this method is urely nonlocal and does not have a local counterart. Then it is not surrising that with this method we obtain a constant C N,,s > 0 such that C N,,s 1 as s 0, s while ( 1 C N,,s = o as s 1. 1 s As exlained in Remark 1.2, this means that the constant obtained in this way does not have the correct asymtotic deendence on s as this goes to 1. This suggests that this roof is not the correct one for s 1.

5 FRACTIONAL HARDY INEQUALITY 5 For this reason, in order to obtain a constant behaving as 1/(1 s, we use a nonlocal variant of the Moser-tye roof recalled at the beginning. This is based on the fact that for every 0 < s < 1 and 1 < < we have in weak sense ( s d s 0 in, where ( s is the fractional Lalacian of order s. In other words, the function d s is (s, suerharmonic in the following sense (see Proosition 3.2 below ( d (xs d (ys 2 (d (xs d (ys ϕ(x ϕ(y R N R N x y N+s dx dy 0, for every nonnegative and smooth function ϕ, with comact suort in. Then we will test this inequality with ϕ = u /d s ( 1. As in the local case, this trick is an essential feature in order to rove BMO regularity of the logarithm of ositive suersolutions to the fractional Lalacian. This in turn is a crucial ste in the roof of Hölder continuity of solutions to equations involving ( s. In this resect, this idea has already been exloited by Di Castro, uusi and Palatucci in [9, Lemma 1.3] (see also [16, Lemma 3.4] for the case = 2. However, we observe that the comutations in [9, Lemma 1.3] do not lead to the desired Hardy inequality, due to a lack of symmetry in x and y. For this, we need finer algebraic maniulations and a subtler ointwise inequality: these are contained in Lemma A.5, which is one of the main ingredients of the roof of Theorem 1.1. We refer to Remark A.7 below for a more detailed discussion on this oint. The quest for fractional Hardy inequalities is certainly not new. We list below some related contributions. Remark 1.4 (Comarison with known results. For the case of the whole sace, the following fractional Hardy inequality u RN C N,s, x s dx u(x u(y dx dy, s N, x y N+s RN R N has been roved by Maz ya and Shaoshnikova in [18, Theorem 2] and by Frank and Seiringer in [13, Theorem 1.1]. In [13], the shar value of the constant C N,s, is obtained. We also refer to [7, Theorem 1.4], [11, Theorem 1] and [14, Theorem 6.1] for some weighted versions of this inequality. As far as subsets Ω R N are concerned, we would like to mention that in [10, Theorem 1.1] Dyda roved (1.6 C Ω Ω u d s dx Ω Ω Ω u(x u(y x y N+s dx dy, for every u C 0 (Ω, under suitable assumtions on the oen Lischitz set Ω R N and some restrictions on the roduct s, see also [11, Corollary 3]. Observe that in the right-hand side of (1.6, the fractional Sobolev seminorm is now comuted on Ω Ω, rather than on the whole R N R N. However, as ointed out in [10], such a stronger inequality fails to hold for s 1, whenever Ω is bounded.

6 6 BRASCO AND CINTI On the other hand, when Ω is a half-sace, inequality (1.6 holds for s 1. In this case, the shar constant has been comuted by Bogdan and Dyda in [1, Theorem 1] for = 2 and by Frank and Seiringer in [12, Theorem 1.1] for a general 1 < <. We also mention that when s > 1 and Ω R N is an oen convex set, inequality (1.6 with shar constant (which is the same as in the half-sace has been roved by Loss and Sloane in [17, Theorem 1.2]. We oint out that our roof is different from those of the aforementioned results and our Hardy inequality (1.4 holds without any restriction on the roduct s Plan of the aer. We start with Section 2, containing the main notations, definitions and some technical results. In this art, the main oint is Proosition 2.5. In Section 3 we show that, in a convex set, the distance function d raised to the ower s is (s, suerharmonic, see Proosition 3.2. The roof of Theorem 1.1 is contained in Section 4. Finally, in Section 5 we highlight some alications of our main result. The aer is comlemented with an Aendix, containing some ointwise inequalities which are crucially exloited in the roof of our main result. Acknowledgments. We wish to thank Guido De Philiis for suggesting us the elegant geometric argument in the roof of Proosition 3.2. We also thank Tuomo uusi for a discussion which clarified some oints of his aer [9]. Xavier Cabré, Bart lomiej Dyda and Ruert L. Frank made some useful comments on a reliminary version of the aer, we warmly thank them. E. Cinti is suorted by the MINECO grant MTM C3-1-P, the ERC Advanced Grant 2013 n Comlex Patterns for Strongly Interacting Dynamical Systems - COMPAT and is art of the Catalan research grou 2014 SGR Both authors are members of the Gruo Nazionale er l Analisi Matematica, la Probabilità e le loro Alicazioni (GNAMPA of the Istituto Nazionale di Alta Matematica (INdAM. 2. Preliminaries 2.1. Notations. For x 0 R N and R > 0, we use the standard notation B R (x 0 = {x R N : x x 0 < R}. For notational simlicity, for every 1 < < we introduce the function J : R R defined by J (t = t 2 t, for t R. For α > 0, we also set L 1 α (R N = { u L 1 loc (RN : RN u(x 1 } dx < + (1 + x N+α If Ω R N is an oen set, for every 1 < < and 0 < s < 1, we define W s, (Ω = {u L (Ω : [u] W s, (Ω < + },.

7 FRACTIONAL HARDY INEQUALITY 7 where ( u(x u(y [u] W s, (Ω = dx dy Ω Ω x y N+s The local version W s, loc (Ω is defined in the usual way Functional analytic facts. We start with the following 1. Definition 2.1. Let 1 < < and 0 < s < 1. Let Ω R N be an oen set. We say that u W s, loc (Ω L 1 s (R N is: locally weakly (s, suerharmonic in Ω if ( J (u(x u(y ϕ(x ϕ(y (2.1 x y N+s dx dy 0, R N R N for every nonnegative ϕ W s, (Ω with comact suort in Ω; locally weakly (s, subharmonic in Ω if u is (s, suerharmonic in Ω; locally weakly (s, harmonic in Ω if it is both (s, suerharmonic and (s, subharmonic. We observe that thanks to the assumtions on u, the double integral in (2.1 is finite for every admissible test function. The following simle result is quite standard. We include the roof for comleteness. Lemma 2.2. Let Ω R N be a bounded measurable set. Then for every u Lα 1 (R N and r > 0 we have su x Ω R N \B r(x u(y 1 dy < +. x y N+α Proof. We observe that for every x Ω and y B r (x we have x y 1 + y x y 1 + x y + x = x x y r r x. This gives immediately u(y 1 ( r x N+s RN u(y 1 dy dy. x y N+s r (1 + y N+s R N \B r(x Since Ω is bounded, we get the desired conclusion. The following technical result will be used in the next section. Lemma 2.3. Let 1 < < and 0 < s < 1. Let Ω R N be an oen bounded set. Given u W s, loc (Ω L 1 s (R N, ϕ L (R N with comact suort in Ω and ε > 0, the function (x, y J (u(x u(y x y N+s ϕ(x, is summable on T ε := {(x, y R N R N : x y ε}.

8 8 BRASCO AND CINTI Proof. Let us call O the suort of ϕ. We have J (u(x u(y T ε x y N+s ϕ(x dx dy u(x u(y 1 = {(x,y O R N : x y ε} x y N+s ϕ(x dx dy ε N s u(x u(y 1 ϕ(x dx dy {(x,y O O : x y ε} x y N+s u(x 1 + u(y 1 + C x y N+s ϕ(x dx dy ε N s O 1 + C {(x,y O (R N \O : x y ε} ( O O {(x,y O (R N \O : x y ε} u(x u(y dx dy x y N+s In order to treat the last integral, we observe that Thus we obtain 1 ( O 1 ϕ dx u(x 1 + u(y 1 x y N+s ϕ(x dx dy. {(x, y O (R N \ O : x y ε} {(x, y O R N {(x,y O (R N \O : x y ε} We conclude by observing that su x O : x y ε}. u(x 1 + u(y 1 x y N+s ϕ(x dx dy ( u(x 1 ϕ(x O R N \B ε(x x y N+s dy dx ( u(y 1 ϕ(x + x y N+s dy dx O N ω N s ( + O R N \B ε(x R N \B ε(x ε s ( R N \B ε(x O u dx 1 ( O u(y 1 dy x y N+s u(y 1 dy < +, x y N+s 1 ϕ dx ϕ(x dx. thanks to the fact that u L 1 s (R N, see Lemma 2.2. In order to use a Moser tye argument for the roof of Theorem 1.1, we will need the following result to guarantee that a certain test function is admissible.

9 FRACTIONAL HARDY INEQUALITY 9 Lemma 2.4. Let 1 < < and 0 < s < 1. Let Ω R N be an oen bounded set. For every u W s, (Ω L (Ω with comact suort in Ω and v W s, loc (Ω L (Ω, we have u v W s, (Ω. Proof. We start by observing that with simle maniulations we have [u v] W s, (Ω u(x u(y v(x 2 1 Ω Ω x y N+s dx dy v(x v(y u(y Ω Ω x y N+s dx dy 2 1 v L (Ω [u] W s, (Ω + v(x v(y u(y 2 1 x y N+s dx dy. In order to estimate the last integral, we set O = su(u and then take O such that O O Ω. We then obtain v(x v(y u(y O O v(x v(y u(y Ω Ω x y N+s dx dy = x y N+s dx dy v(x v(y u(y + x y N+s dx dy (Ω\O O Ω Ω This gives the desired conclusion. u L (Ω [v] W s, (O 2 Ω O + dist(o, Ω \ O N+s u L (Ω v L (Ω An exedient estimate for convex sets. The following exedient result is a sort of fractional counterart of the identity d = 1 almost everywhere in. As exlained in the Introduction, in the local case this is an essential ingredient in the roof of the Hardy inequality for convex sets. This will lay an imortant role in our case as well. Proosition 2.5. Let 1 < < and 0 < s < 1. Let R N be an oen bounded convex set. Then we have d (x d (y x y N+s dy C 1 1 s d (x (1 s, for a. e. x, {y : d (y d (x} where C 1 = C 1 (N, > 0 is the constant C 1 = 1 su 0<σ<1 [ σ H N 1( {ω S N 1 : ω, e 1 > σ} ].

10 10 BRASCO AND CINTI Proof. We set for simlicity δ = d (x, thus B δ (x and we have d (x d (y d (x d (y x y N+s dy x y N+s dy. {y : d (y d (x} {y B δ (x : d (y d (x} We now take x such that x x = δ. For a given 0 < σ < 1, we consider the ortion Σ σ (x of B δ (x defined by { } y x Σ σ (x = y B δ (x : y x, x x x > σ, x see Figure 1. Figure 1. The set Σ σ (x and the suorting hyerlane Π x. By convexity of, it is not difficult to see that (2.2 Σ σ (x {y B δ (x : d (y d (x}, for every 0 < σ < 1. We can be more recise on this oint. We denote by Π x the suorting hyerlane of at the oint x, orthogonal to x x. Then for every y, we denote by y the orthogonal rojection of y on Π x. Thus by convexity we have d (y y y, for every y. We then observe that for every y Σ σ (x, it holds y x d (x = x x = y y + y x, x x x y x (2.3 x d (y + σ y x.

11 FRACTIONAL HARDY INEQUALITY 11 By using (2.2 and (2.3, we thus obtain d (x d (y {y : d (y d (x} x y N+s dy d (x d (y Σ σ(x x y N+s dy σ x y (1 s N dy Σ σ(x where f(σ > 0 is the quantity f(σ = δ = σ ({ } dh N 1 (ω ϱ 1+ (1 s dϱ ω S N 1 : ω, x x x >σ 0 x f(σ σ = (1 s δ (1 s, {ω S N 1 : ω,e 1 >σ} dh N 1 (ω. By arbitrariness of 0 < σ < 1, we can take the suremum and get the conclusion. 3. Suerharmonicity of the distance function In this section we will rove that d s in a convex set is weakly (s, suerharmonic, see Definition 2.1. We start with the case of the half-sace. The roof of the following fact can be found in [15, Lemma 3.2] (see also [5, Theorem 3.4.1] for the case = 2. Lemma 3.1. We set H N + = {x R N : x N > 0}. Let 1 < < and 0 < s < 1, then d s H N + is locally weakly (s, harmonic in H N +. Moreover, there holds J (d H N (x s d lim + H N (y s + ε 0 x y N+s dy = 0, strongly in L 1 loc (HN +. R N \B ε(x By aealing to the revious result and using the geometric roerties of convex sets, we can rove the following Proosition 3.2. Let R N be an oen bounded convex set. For 1 < < and 0 < s < 1, we have that d s is locally weakly (s, suerharmonic. Proof. We first observe that d s is locally Lischitz, bounded and vanishing outside. Thus we have d s W s, loc ( L 1 s (R N. It is sufficient to rove that d s verifies (2.1 for every ϕ C0 ( nonnegative. Thus, let ϕ C 0 ( be nonnegative and call O its suort. We observe that the function (x, y J (d (x s d (y s ( ϕ(x ϕ(y x y N+s,

12 12 BRASCO AND CINTI is summable. Then by the Dominated Convergence Theorem, we have ( J (d (xs d (ys ϕ(x ϕ(y R N R N x y N+s dx dy ( J (d = lim (xs d (ys ϕ(x ϕ(y ε 0 T ε x y N+s dx dy, where we set T ε = {(x, y R N R N : x y ε}. By Lemma 2.3 we have that for every fixed ε, the function (x, y J (d (x s d (y s x y N+s ϕ(x, is summable on T ε. Thus we get ( J (d (xs d (ys ϕ(x ϕ(y T ε x y N+s dx dy J (d (x s d (y s J (d (x s d (y s = T ε x y N+s ϕ(x dx dy T ε x y N+s ϕ(y dx dy ( J (d (x s d (y s = 2 x y N+s dy ϕ(x dx. O R N \B ε(x In the second equality, we used Fubini s Theorem and the fact that ϕ has suort O. In order to conclude, we need to show that ( J (d (x (3.1 lim ε 0 O s d (y s R N \B ε(x x y N+s dy ϕ(x dx 0. We now take 0 < ε 1 and x O, then we consider a oint x such that d (x = x x. We take a suorting hyerlane to at x, u to a rigid motion we can suose that this is given by {x R N : x N = 0} and that H N +. We observe that by convexity of and d (y d H N + (y, for y R N, d (x = x x = d H N + (x, see Figure 2. By exloiting these facts and the monotonicity of J, we obtain for x O R N \B ε(x J (d (x s d (y s J (d H N (x s d x y N+s dy + H N (y s + R N \B ε(x x y N+s dy. Observe that the last family of functions converges to 0 in L 1 (O as ε goes to 0, thanks to Lemma 3.1. Thus, by multilying the revious inequality by ϕ which is nonnegative,

13 FRACTIONAL HARDY INEQUALITY 13 Figure 2. The distance of y from is smaller than its distance from the hyerlane. integrating over O and using Lemma 3.1, we thus obtain ( J (d (x s d (y s lim ε 0 O R N \B ε(x x y N+s dy ϕ(x dx ( J (d H N (x lim ε 0 O s d + H N (y s + R N \B ε(x x y N+s dy ϕ(x dx = 0. This roves (3.1 and thus we get the desired conclusion. 4. Proof of Theorem 1.1 We divide the roof in two arts: we first rove s A u (4.1 1 s d s dx [u] W s, (R N, for every u C 0 (, with A = A(N, > 0. Then in the second art we show how to imrove the constant in the left-hand side for s close to 0 and obtain (1.4, by using elementary geometric considerations. The roof in this second art is essentially contained in [6, ages ], as ointed out to us by Bart lomiej Dyda Proof of inequality (4.1. In turn, we divide the roof in two cases: first we rove the result under the additional assumtions that is bounded, then we extend it to general convex sets not coinciding with the whole sace.

14 14 BRASCO AND CINTI Case 1: bounded convex sets. By Proosition 3.2 we know that J (d (x s d (y s (ϕ(x ϕ(y (4.2 x y N+s dx dy 0, R N R N for every nonnegative ϕ W s, ( with comact suort in. Then we test with ϕ = u (d s + ε 1, where u C0 ( and ε > 0. By Lemma 2.4, we have that ϕ is admissible. Indeed, we already know that d s W s, loc ( L (. Moreover, for every ε > 0 the function f(t = (t + ε 1 is Lischitz for t > 0, thus (d s + ε1 = f d s W s, loc ( L ( as well. Let us call O the suort of u, then from (4.2 we have J (d (x s d (y s ( u(x 0 x y N+s (d (x s + ε 1 u(y (d (y s + ε 1 dx dy (4.3 J (d (x s u(x + 2 x y N+s (d (x s dx dy. + ε 1 O (R N \ We first observe that J (d (x s u(x (4.4 x y N+s (d (x s dx dy + ε 1 O (R N \ O (R N \ u(x dx dy. x y N+s We now need to estimate the double integral J (d (x s d (y s ( u(x I := x y N+s (d (x s + ε 1 u(y (d (y s + ε 1 dx dy. For this, we crucially exloit the fundamental inequality of Lemma A.5, with the choices This entails (4.5 a = d (x s + ε, b = d (y s + ε, c = u(x, d = u(y. I C 2 + C 3 d (x s d (y s d (x s + d (y s + 2 ε u(x u(y x y N+s dx dy, ( u(x + u(y dx dy x y N+s where C 2 and C 3 are as in Lemma A.5. By using (4.5 in (4.3, together with (4.4, we obtain (4.6 C 2 d (x s d (y s d (x s + d (y s ( u(x + u(y dx dy x y N+s C [ ] 3 u W s, (R N.

15 FRACTIONAL HARDY INEQUALITY 15 To obtain (4.6, we also took the limit as ε goes to 0 and used Fatou s Lemma. We observe that by symmetry, we have d (x s d (y s d (x s + d (y s ( u(x + u(y dx dy x y N+s = 2 d (x s d (y s d (x s + d (y s u(x dx dy (4.7 x y N+s ( 2 d (x s d (y s dy d (x s + d (y s x y N+s u(x dx. {y : d (y d (x} We now use the ointwise inequality (A.2, so to obtain d (x s d (y s d (x s + d (y s ( u(x + u(y dx dy x y N+s ( s d (x d (y 2 1 x y N+s dy {y : d (y d (x} u(x d (x dx. By using this in (4.6 and then alying the exedient estimate of Proosition 2.5, we end u with s A u 1 s d s dx [u] W s, (R N, where we have used the triangle inequality to relace the seminorm of u with that of u, i.e. u(x u(y u(x u(y. This concludes the roof of (4.1. We observe that (4.8 A = C C 2 C 3, where C 1 = C 1 (N, is the constant of Proosition 2.5, and C 2, C 3 (which deend only on come from Lemma A.5. Case 2: general convex sets. We now take R N an oen unbounded convex set. For every R > 0 we set R = B R (0. Let us take u C0 (, then for every R large enough, we have u C0 ( R as well. By using the revious case, we then get s A u 1 s d s dx = s A u R 1 s R d s dx [u] W s, (R N. R By observing that d R d, we then get the desired conclusion.

16 16 BRASCO AND CINTI Figure 3. The set x in the second art of the roof of Theorem Imroved constant for s close to 0. For every x, we take x 0 R N \ such that x + x 0 x x 0 = 2 d (x and. 2 Then we can estimate RN u(x u(y ( dx dy u(x dy x y N+s x y N+s dx. RN We observe that for every y R N \ B d (x(x 0, we have (R N \\B d (x(x 0 x y x x 0 + x 0 y = 2 d (x + x 0 y 3 x 0 y. By convexity, we have that (see Figure 3 x := {y R N \ B d (x : y x 0, x x 0 < 0} (R N \ \ B d (x(x 0. By joining the last two informations, we get RN u(x u(y ( 1 N+s ( dx dy u(x x y N+s 3 RN ( 1 N+s ( = 3 ( + ϱ 1 s dϱ d (x = N ω N 2 1 s ( 1 3 x dy x 0 y N+s dx dh N 1 {ω S N 1 : ω,e 1 >0} N+s u(x dx u d s dx.

17 FRACTIONAL HARDY INEQUALITY 17 In conclusion, we have obtained for every u C0 ( B u (4.9 s d s dx [u] W s, (R N, with B = 1 N ω N 3 N. 2 By joining (4.1 and (4.9, we finally get A s +1 + (1 s B u 2 s (1 s d s dx [u] W s, (R N, for every u C 0 (. By observing that the quantity A s +1 + (1 s B is bounded from below by a ositive constant indeendent of s, we finally get the desired inequality (1.4. Remark 4.1. Let us set Φ(s = A s +1 + (1 s B, where A and B are defined in (4.8 and (4.9. It is easy to see that the constant C aearing in (1.4 is given by ( Φ(s C = min = 1 B 1 ( B 1/, if B < ( + 1 A, + 1 ( + 1A s (0,1 2 2 A, if B ( + 1 A. 5. Some consequences For an oen set Ω R N, we define the homogeneous Sobolev-Slobodeckiĭ sace D s, 0 (Ω as the comletion of C0 (Ω with resect to the norm inf u C0 (Ω u [u] W s, (R N. We also define the first eigenvalue of the fractional Lalacian of order s in Ω, i.e. } λ s 1,(Ω := {[u] W s,(rn : u dx = 1. Ω This is the shar constant in the fractional Poincaré inequality C u dx [u] W s, (R N, for every u C 0 (Ω. Ω We observe that λ s 1, (Ω > 0 is equivalent to the continuity of the embedding Ds, 0 (Ω L (Ω. We highlight a coule of consequences of our main result, in terms of lower bounds on. As usual, we ay articular attention to the factor s (1 s. λ s 1, Corollary 5.1. Let 1 < < and 0 < s < 1. Let R N be an oen convex set such that R := su d (x < +. x Then D s, 0 ( is a functional sace, continuously embedded in L (. Moreover, it holds C R s s (1 s λ s 1,(,

18 18 BRASCO AND CINTI where C is the same constant as in Theorem 1.1. Proof. This is a straightforward consequence of (1.4 and of the definition of R. The quantity R above is called inradius of. Observe that this is the radius of the largest ball inscribed in. For a general oen set, we have the following Corollary 5.2 (Poincaré inequality for sets bounded in one direction. Let ω 0 R N be such that ω 0 = 1 and let l 1, l 2 R with l 1 < l 2. For every oen set such that Ω {x R N : l 1 < x, ω 0 < l 2 }. we have ( 2 s C s (1 s λ s l 2 l 1,(Ω, 1 where C is the same constant as in Theorem 1.1. Proof. We set for simlicity S = {x R N : l 1 < x, ω 0 < l 2 }. Then by domain inclusion we directly obtain λ s 1, (Ω λs 1, (S. It is now sufficient to use Corollary 5.1 for the convex set S, for which R S = (l 2 l 1 /2. Aendix A. Some ointwise inequalities We collect here some ointwise inequalities needed throughout the whole aer. The most imortant one is Lemma A.5. We recall the notation J (t = t 2 t, for t R. Lemma A.1. Let 1 < <, for every a, b > 0 we have ( 1 J (a b b 1 1 a 1 ( 1 log b log a. Equality holds if and only if a = b. Proof. This is roved in [3, Lemma A.2 & Remark A.3]. Lemma A.2. For every a, b > 0 we have a b log a log b. a + b Equality holds if and only if a = b. Proof. We observe that if a = b there is nothing to rove. We then take a b and without loss of generality we can suose a > b. The seeked inequality is then equivalent to a b a + b log a, for 0 < b < a. b By setting t = b/a, this in turn is equivalent to rove that 1 t log t, for 0 < t < t

19 By basic Calculus, it is easily seen that the function FRACTIONAL HARDY INEQUALITY 19 ϕ(t = log t + 1 t 1 + t, is strictly increasing for t (0, 1 and ϕ(1 = 0. This gives the desired conclusion. Remark A.3. By combining Lemma A.1 and A.2, we also obtain ( 1 (A.1 J (a b b 1 1 a 1 ( 1 a b a + b, for every a, b > 0 and 1 < <. Lemma A.4. Let 0 < s < 1, then for every a, b > 0 we have (A.2 a s b s a s + b s s 2 a b max{a, b}. Proof. For a = b there is nothing to rove. Without loss of generality, we can assume a > b. By defining t = b/a (0, 1, inequality (A.2 is equivalent to rove 1 t s 1 + t s s (1 t. 2 We observe that by the below tangent roerty of concave functions, we have t s 1 + s (t 1 i. e. 1 t s s (1 t. By combining this with the trivial estimate 1 + t s 2, we get the conclusion. An essential ingredient in the roof of our main result has been the following ointwise inequality. Lemma A.5 (Fundamental inequality. Let 1 < < and let a, b, c, d R, with a, b > 0 and c, d 0. Then there exist two constants C 2 = C 2 ( > 0 and C 3 = C 3 ( > 1, such that ( c (A.3 J (a b a 1 d a b b 1 + C 2 a + b (c + d C 3 c d. Proof. We observe that for a = b there is nothing to rove, since the left-hand side vanishes. Without loss of generality, we can assume a > b. Also notice that if c d, then ( ( c (a b 1 a 1 d d b 1 (a b 1 a 1 d b ( 1 1 = d (a b 1 b 1 1 a 1 ( 1 d a b a + b ( 1 c + d a b 2 a + b, where in the second inequality we used (A.1. Thus inequality (A.3 holds with C 2 = ( 1/2 and C 3 > 0 arbitrary.

20 20 BRASCO AND CINTI We assume now that a > b and c > d, then by setting t = b/a (0, 1 and A = d/c [0, 1, inequality (A.3 is equivalent to ( ( (A.4 (1 t 1 1 A 1 t t 1 + C 2 (1 + A C 3 (1 A, 1 + t with t (0, 1 and A [0, 1. We study the function ( (A.5 Φ(t = (1 t 1 1 A t 1, t (0, 1, which is maximal for t = A. This in articular imlies 2 ( (A.6 (1 t 1 1 A t 1 (1 A. We now distinguish two cases: either 0 A 1 2 or 1 2 < A < 1. A. Case 0 A 1/2. This is the simlest case. Indeed, we have Thus by using this and (A.6, we get ( (1 t 1 1 A t 1 + C t 1 + t 1 and (1 A 1 2. ( 1 t (1 + A C 3 (1 A, 1 + t which is (A.4 with C 2 = (C 3 1/2 +1 and C 3 > 1 arbitrary. B. Case 1/2 < A < 1. Here in turn we consider two subcases: A t and 0 < t < A. B.1. and thus Case 1/2 < A < 1 and t A. This is easy, since we directly have By using this and (A.6, we get ( (1 t 1 1 A t 1 + C (1 t (1 A, ( 1 t (1 t (1 A. 1 + t 2 We observe that this is equivalent to ( c J (a b a d c d, 1 b 1 ( 1 t (1 + A C 3 (1 A, 1 + t which is a discrete version of Picone s inequality, see [2, Proosition 4.2].

21 FRACTIONAL HARDY INEQUALITY 21 which is (A.3 with C 2 = (C 3 1/2 and C 3 > 1 arbitrary. B.1. Case 1/2 < A < 1 and 0 < t < A. Here we need to study in more details the function Φ defined in (A.5. We have [ ] Φ (t = ( 1 (1 t A A = ( 1 (1 t 2 [ 2 1 t By an easy comutation, we can see that is monotone increasing, thus we get t (1 A t A Φ (t ( 1 (1 t 3 t t t +1 ] A t +1, t (0, 1. A t +1 ( 1 1, for 0 < t < A. A In articular, we get that Φ is concave on the interval (0, A. We use a second order Taylor exansion around the maximum oint t = A, i.e. (A.7 Φ(t = Φ(A + = (1 A + ( 1 A t A t Φ (s (s t ds (1 s 2 [ 2 1 s ] (1 A s A s +1 (s t ds, where we used that Φ (A = 0. In order to estimate the remainder term inside the integral, we distinguish once again two cases: if 1 < 2, then by using Lemma A.8 below and the fact that A 1/2, we get A [ ] 2 (1 s (1 2 A t 1 s s A s +1 (s t ds A = [( 2 s A ] 1 s A (1 s 2 s s (s t ds t ( 1 2 A ( 1 2 (1 t 2 t (1 s 2 (s t ds s A t (s t ds ( 1 = 2 +1 (1 t 2 (A t 2. By using the revious estimate in (A.7, we have (A.8 (1 t 1 ( 1 A t 1 (1 A ( (1 t 2 (A t 2.

22 22 BRASCO AND CINTI (A.9 It is now sufficient to observe that (1 t = (1 t 2 (1 t 2 (1 t 2 ( 2 (A t (1 A 2 2 (1 t 2 (A t (1 A and thus ( 1 t C 2 (1 + A 4 C 2 (1 t 2 (A t C 2 (1 A. 1 + t If we sum u (A.8 and (A.9 and choose C 3 > 1 arbitrary and C 2 = 1 { } 4 min ( 12 C 3 1, 2 +1, we get again the desired conclusion (A.4. if > 2, then by using that s A A t (1 s 2 [ 2 1 s (1 A s A A (1 s 2 (s t ds t s +1 A 2 (1 s 2 (s t ds. t ] A s +1 (s t ds The last integral can be exlicitly comuted, integrating by arts: we have A t (1 s 2 (s t ds = 1 1 (1 1 A 1 (A t + (1 t ( 1 1 ( 1 (1 A. We use Young s inequality to estimate the first term on the right-hand side 1 1 (1 A 1 (A t 1 1 ε 1 (1 A ε ( 1 (A t, with ε > 0. We use these estimates in (A.7. This in turn gives ( (1 t 1 1 A t 1 (1 A ε 1 1 (1 A + ε (A t (1 t (1 A.

23 FRACTIONAL HARDY INEQUALITY 23 By choosing ε = 1/2 and using that A t 1 t, we then obtain ( (1 t 1 1 A t 1 C 3 (1 A (1 t, with C 3 = Once again, this is enough to get the desired conclusion, since ( 1 t C 2 (1 + A 2 C 2 (1 t. 1 + t Then we only need to choose C 2 = 1/2 +2 in order to get (A.4. We thus concluded the roof. Remark A.6 (The constants C 2 and C 3. An insection of the roof reveals that in the case 1 < 2, the constant C 3 can be chosen arbitrarily close to 1. Accordingly, we have { C3 1 C 2 = min, 1 ( , C , 1 } 2 = 1 { } ( 1 2 min, C , 4 and thus it degenerates to 0 as C 3 1. Remark A.7. Inequality (A.3 looks similar to the ointwise inequality which can be found right before [9, equation (3.12, age 1289], where the term a b a + b (c + d, is relaced by log a log b min{c, d }. The main difference is that in our inequality the terms c and d lay a symmetric role. This means that the quantity (c + d is unchanged when we exchange the roles of c and d, while this is not the case for min{c, d }. This roerty is a crucial feature in order to rove Theorem 1.1: recisely, this is hidden in the estimate (4.7. On the other hand, it is easy to see that the inequality in [9] can not have this roerty, i.e. one can not relace by log a log b min{c, d }, log a log b (c + d. Thus the inequality of [9] does not seem useful in order to rove Hardy inequality. In the revious result, we needed the following inequality in order to deal with the case 1 < 2.

24 24 BRASCO AND CINTI Lemma A.8. Let 1 < 2, for every s (0, 1 and A [0, 1] we have Proof. We rewrite ( 2 s A 1 s A s ( 1 A. (A.10 ( 2 s A 1 s A s = (2 A s 1 s A s. We then observe that A s (A.11 A. 1 s Indeed, the latter is equivalent to A (1 + s s, which is easily seen to be true. By using (A.11 in (A.10, we now obtain as desired. ( 2 s A 1 s A s (2 A A s ( 1 A, References [1]. Bogdan, B. Dyda, The best constant in a fractional Hardy inequality, Math. Nachr., 284 (2011, [2] L. Brasco, G. Franzina, Convexity roerties of Dirichlet integrals and Picone-tye inequalities, odai Math. J., 37 (2014, [3] L. Brasco, E. Parini, The second eigenvalue of the fractional Lalacian, Adv. Calc. Var., 9 (2016, [4] L. Brasco, E. Parini, M. Squassina, Stability of variational eigenvalues for the fractional Lalacian, Discrete Contin. Dyn. Syst., 36 (2016, [5] C. Bucur, E. Valdinoci, Nonlocal Diffusion and Alications, Lecture Notes of the Unione Matematica Italiana, 20. Sringer, [Cham]; Unione Matematica Italiana, Bologna, [6] Z.-Q. Chen, R. Song, Hardy inequality for censored stable rocesses, Tohoku Math. J., 55 (2003, [7] E. Cinti, F. Ferrari, Geometric inequalities for fractional Lalace oerators and alications, NoDEA Nonlinear Differential Equations Al., 22 (2015, [8] E. B. Davies, A review of Hardy inequalities. The Maz ya anniversary collection, Vol. 2 (Rostock, 1998, 55 67, Oer. Theory Adv. Al., 110, Birkhäuser, Basel, [9] A. Di Castro, T. uusi, G. Palatucci, Local behavior of fractional minimizers, Ann. Inst. H. Poincaré Anal. Non Linéaire, 33 (2016, , 6, 23 [10] B. Dyda, A fractional order Hardy inequality, Illinois J. Math., 48 (2004, [11] B. Dyda, A. V. Vähäkangas, A framework for fractional Hardy inequalities, Ann. Acad. Sci. Fenn. Math., 39 (2014, [12] R. L. Frank, R. Seiringer, Shar fractional Hardy inequalities in half-saces. Around the research of Vladimir Maz ya. I, , Int. Math. Ser. (N. Y., 11, Sringer, New York, [13] R. L. Frank, R. Seiringer, Non-linear ground state reresentations and shar Hardy inequalities, J. Funct. Anal., 255 (2008,

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