On fractional Hardy inequalities in convex sets

We prove a Hardy inequality on convex sets, for fractional Sobolev-Slobodecki\u{\i} spaces of order $(s,p)$. The proof is based on the fact that in a convex set the distance from the boundary is a superharmonic function, in a suitable sense. The result holds for every $1<p<\infty$ and $0<s<1$, with a constant which is stable as $s$ goes to $1$.


A quick overview on Hardy inequality.
Given an open set Ω ⊂ R N with Lipschitz boundary, we will use the notation A fundamental result in the theory of Sobolev spaces is the Hardy inequality dx ≤ˆΩ |∇u| 2 dx, for every u ∈ C ∞ 0 (Ω), see for example [20,Theorem 21.3]. It is well-known that for a convex set K ⊂ R N , such an inequality holds with the dimension-free universal constant see for example [8,Theorem 2]. Moreover, such a constant is sharp. In order to explain the aims and techniques of the present paper, it is useful to recall a proof of this fact. A well-known and very elegant way of proving (1.1) with sharp constant for convex sets, consists in mimicking Moser's logarithmic estimate for positive supersolutions of elliptic partial differential equations. The starting point is the observation that on a convex set K we have −∆d K ≥ 0, i.e. the distance function is superharmonic. More precisely, it holds (1.2)ˆK ∇d K , ∇ϕ dx ≥ 0, for every nonnegative ϕ ∈ C ∞ 0 (K).
By following Moser (see [19, page 586]), one can test the equation (1.2) with 1 where u ∈ C ∞ 0 (K). This gives We now use Young's inequality with the following choices where δ is an arbitrary positive real number. This leads tô which can be recast as 1 Of course, such a test function is not in C ∞ 0 (K). However, by a standard density argument, in (1.2) we can allow W 1,2 test functions with compact support. We avoid this unessential technicality for ease of readability.
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 sharp constant 1/4, once it is observed that |∇d K | = 1, a. e. in K. The latter implies that more generally for every 1 < p < ∞ we have −∆ p d K ≥ 0 in K, i.e. d K is p−superharmonic in the following sensê K |∇d K | p−2 ∇d K , ∇ϕ dx ≥ 0, for every nonnegative ϕ ∈ C ∞ 0 (K).
By testing this with ϕ = |u| p /d p−1 K and suitably adapting the proof above, one can prove the more general Hardy inequality for convex sets Once again, the constant appearing in (1.3) is sharp and independent of both K and the dimension N .

Main result.
The scope of the present paper is to prove a fractional version of Hardy inequality for convex sets, by adapting to the fractional setting the Moser-type proof presented above. An essential feature of our method is that the relevant constant appearing in the Hardy inequality is stable as the fractional order of differentiability s converges to 1, see Remark 1.2 below. More precisely, we prove the following Theorem 1.1 (Hardy inequality on convex sets). Let 1 < p < ∞ and 0 < s < 1. Let K ⊂ R N be an open convex set such that K = R N . Then for every u ∈ C ∞ 0 (K) we have for a computable constant C = C(N, p) > 0.
The constant obtained in (1.4) is very likely not sharp. However, a couple of comments are in order on this point. Remark 1.2 (Asymptotic behaviour in s of the constant). We recall that if u ∈ C ∞ 0 (K), then we have (see [ In this respect, we observe that the constant appearing in Theorem 1.1 has the correct asymptotic behaviour as s converges to 1: by passing to the limit in (1.4) as s goes to 1, we obtain the usual local Hardy inequality As for the limit s 0, we recall that (see [18,Theorem 3]) 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 asymptotic behaviour as s converges to 0.
. At a first glance, it may look strange that the constant C in Theorem 1.1 depends on N . Indeed, we have seen in (1.3) that in the local case such an inequality holds with the universal constant However, it is easily seen that C must depend on N in the fractional case. Indeed, we have already seen that passing to the limit in (1.4) as s goes to 1, we obtain the local Hardy inequality (1.5). As we already said, the sharp constant in the previous inequality is ((p − 1)/p) p , which means that we must have On the other hand, it is easily seen that α N,p converges to 0 as N goes to ∞. This shows that C in (1.4) must depend on N .
1.3. Method of proof. We now spend some words on the proof of Theorem 1.1. We first observe that there is an elementary proof of the inequality as pointed 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 point out that this method is purely nonlocal and does not have a local counterpart. Then it is not surprising that with this method we obtain a constant C N,p,s > 0 such that As explained in Remark 1.2, this means that the constant obtained in this way does not have the correct asymptotic dependence on s as this goes to 1. This suggests that this proof is not the correct one for s ∼ 1.
For this reason, in order to obtain a constant behaving as 1/(1 − s), we use a nonlocal variant of the Moser-type proof recalled at the beginning. This is based on the fact that for every 0 < s < 1 and 1 < p < ∞ we have in weak sense where (−∆ p ) s is the fractional p−Laplacian of order s. In other words, the function d s K is (s, p)−superharmonic in the following sense (see Proposition 3.2 below) for every nonnegative and smooth function ϕ, with compact support in K. Then we will test this inequality with ϕ = |u| p /d s (p−1) K . As in the local case, this trick is an essential feature in order to prove BMO regularity of the logarithm of positive supersolutions to the fractional p−Laplacian. This in turn is a crucial step in the proof of Hölder continuity of solutions to equations involving (−∆ p ) s . In this respect, this idea has already been exploited by Di Castro, Kuusi and Palatucci in [9, Lemma 1.3] (see also [16,Lemma 3.4] for the case p = 2). However, we observe that the computations 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 manipulations and a subtler pointwise inequality: these are contained in Lemma A.5, which is one of the main ingredients of the proof of Theorem 1.1. We refer to Remark A.7 below for a more detailed discussion on this point.
The quest for fractional Hardy inequalities is certainly not new. We list below some related contributions.
under suitable assumptions on the open Lipschitz set Ω ⊂ R N and some restrictions on the product s p, see also [11,Corollary 3].
Observe that in the right-hand side of (1.6), the fractional Sobolev seminorm is now computed on Ω × Ω, rather than on the whole R N × R N . However, as pointed out in [10], such a stronger inequality fails to hold for s p ≤ 1, whenever Ω is bounded.
On the other hand, when Ω is a half-space, inequality (1.6) holds for s p = 1. In this case, the sharp constant has been computed by Bogdan and Dyda in [1, Theorem 1] for p = 2 and by Frank and Seiringer in [12, Theorem 1.1] for a general 1 < p < ∞. We also mention that when s p > 1 and Ω ⊂ R N is an open convex set, inequality (1.6) with sharp constant (which is the same as in the half-space) has been proved by Loss and Sloane in [17,Theorem 1.2].
We point out that our proof is different from those of the aforementioned results and our Hardy inequality (1.4) holds without any restriction on the product s p.
1.4. Plan of the paper. We start with Section 2, containing the main notations, definitions and some technical results. In this part, the main point is Proposition 2.5. In Section 3 we show that, in a convex set K, the distance function d K raised to the power s is (s, p)−superharmonic, see Proposition 3.2. The proof of Theorem 1.1 is contained in Section 4. Finally, in Section 5 we highlight some applications of our main result. The paper is complemented with an Appendix, containing some pointwise inequalities which are crucially exploited in the proof of our main result.
For notational simplicity, for every 1 < p < ∞ we introduce the function J p : For α > 0, we also set If Ω ⊂ R N is an open set, for every 1 < p < ∞ and 0 < s < 1, we define The local version W s,p loc (Ω) is defined in the usual way. 2.2. Functional analytic facts. We start with the following for every nonnegative ϕ ∈ W s,p (Ω) with compact support in Ω; We observe that thanks to the assumptions on u, the double integral in (2.1) is finite for every admissible test function.
The following simple result is quite standard. We include the proof for completeness.
Proof. We observe that for every x ∈ Ω and y ∈ B r (x) we have

This gives immediatelŷ
Since Ω is bounded, we get the desired conclusion.
The following technical result will be used in the next section.
Proof. Let us call O the support of ϕ. We havë In order to treat the last integral, we observe that Thus we obtain We conclude by observing that In order to use a Moser-type argument for the proof of Theorem 1.1, we will need the following result to guarantee that a certain test function is admissible.
Proof. We start by observing that with simple manipulations we have In order to estimate the last integral, we set O = supp(u) and then take O such that O O Ω. We then obtain This gives the desired conclusion.

2.
3. An expedient estimate for convex sets. The following expedient result is a sort of fractional counterpart of the identity |∇d K | = 1 almost everywhere in K.
As explained in the Introduction, in the local case this is an essential ingredient in the proof of the Hardy inequality for convex sets. This will play an important role in our case as well.
Proposition 2.5. Let 1 < p < ∞ and 0 < s < 1. Let K ⊂ R N be an open bounded convex set. Then we havê Proof. We set for simplicity δ = d K (x), thus B δ (x) ⊂ K and we havê We now take x ∈ ∂K such that |x − x | = δ. For a given 0 < σ < 1, we consider the see Figure 1. By convexity of K, it is not difficult to see that for every 0 < σ < 1.
We can be more precise on this point. We denote by Π x the supporting hyperplane of K at the point x , orthogonal to x − x. Then for every y ∈ K, we denote by y the orthogonal projection of y on Π x . Thus by convexity we have We then observe that for every y ∈ Σ σ (x), it holds By using (2.2) and (2.3), we thus obtain By arbitrariness of 0 < σ < 1, we can take the supremum and get the conclusion.

Superharmonicity of the distance function
In this section we will prove that d s K in a convex set K is weakly (s, p)−superharmonic, see Definition 2.1. We start with the case of the half-space. The proof of the following fact can be found in [ |x − y| N +s p dy = 0, strongly in L 1 loc (H N + ). By appealing to the previous result and using the geometric properties of convex sets, we can prove the following Proposition 3.2. Let K ⊂ R N be an open bounded convex set. For 1 < p < ∞ and 0 < s < 1, we have that d s K is locally weakly (s, p)−superharmonic. Proof. We first observe that d s K is locally Lipschitz, bounded and vanishing outside K. Thus we have d s K ∈ W s,p loc (K) ∩ L p−1 s p (R N ). It is sufficient to prove that d s K verifies (2.1) for every ϕ ∈ C ∞ 0 (K) nonnegative. Thus, let ϕ ∈ C ∞ 0 (K) be nonnegative and call O K its support. We observe that the function is summable. Then by the Dominated Convergence Theorem, we havë 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 is summable on T ε . Thus we geẗ In the second equality, we used Fubini's Theorem and the fact that ϕ has support O. In order to conclude, we need to show that We now take 0 < ε 1 and x ∈ O, then we consider a point x ∈ ∂K such that d K (x) = |x − x |. We take a supporting hyperplane to K at x , up to a rigid motion we can suppose that this is given by {x ∈ R N : x N = 0} and that K ⊂ H N + . We observe that by convexity of K d K (y) ≤ d H N + (y), for y ∈ R N , see Figure 2. By exploiting these facts and the monotonicity of J p , we obtain for x ∈ Ô 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 multiplying the previous inequality by ϕ which is nonnegative, Figure 2. The distance of y from ∂K is smaller than its distance from the hyperplane.
integrating over O and using Lemma 3.1, we thus obtain This proves (3.1) and thus we get the desired conclusion.

Proof of Theorem 1.1
We divide the proof in two parts: we first prove

4.1.
Proof of inequality (4.1). In turn, we divide the proof in two cases: first we prove the result under the additional assumptions that K is bounded, then we extend it to general convex sets not coinciding with the whole space.
Case 1: bounded convex sets. By Proposition 3.2 we know that for every nonnegative ϕ ∈ W s,p (K) with compact support in K. Then we test with and ε > 0. By Lemma 2.4, we have that ϕ is admissible. Indeed, we already know that d s K ∈ W s,p loc (K) ∩ L ∞ (K). Moreover, for every ε > 0 the function Let us call O the support of u, then from (4.2) we have We first observe that We now need to estimate the double integral For this, we crucially exploit the fundamental inequality of Lemma A.5, with the choices

This entails
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 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 havë We now use the pointwise inequality (A.2), so to obtain By using this in (4.6) and then applying the expedient estimate of Proposition 2.5, we end up with where we have used the triangle inequality to replace the seminorm of |u| with that of u, i.e.
This concludes the proof of (4.1). We observe that where C 1 = C 1 (N, p) is the constant of Proposition 2.5, and C 2 , C 3 (which depend only on p) come from Lemma A.5.
Case 2: general convex sets. We now take K = R N an open unbounded convex set. For every R > 0 we set K R = K ∩ B R (0). Let us take u ∈ C ∞ 0 (K), then for every R large enough, we have u ∈ C ∞ 0 (K R ) as well. By using the previous case, we then get By observing that d K R ≤ d K , we then get the desired conclusion.

4.2.
Improved constant for s close to 0. For every x ∈ K, we take Then we can estimatë We observe that for every y ∈ R N \ B d K (x) (x 0 ), we have By convexity, we have that (see Figure 3) By joining the last two informations, we geẗ In conclusion, we have obtained for every u ∈ C ∞ 0 (K) By joining (4.1) and (4.9), we finally get By observing that the quantity A s p+1 + (1 − s) B is bounded from below by a positive constant independent of s, we finally get the desired inequality (1.4). and (4.9). It is easy to see that the constant C appearing in (1.4) is given by

Some consequences
For an open set Ω ⊂ R N , we define the homogeneous Sobolev-Slobodeckiȋ space D s,p 0 (Ω) as the completion of C ∞ 0 (Ω) with respect to the norm We also define the first eigenvalue of the fractional p−Laplacian of order s in Ω, i.e. This is the sharp constant in the fractional Poincaré inequality We observe that λ s 1,p (Ω) > 0 is equivalent to the continuity of the embedding D s,p 0 (Ω) → L p (Ω).
We highlight a couple of consequences of our main result, in terms of lower bounds on λ s 1,p . As usual, we pay particular attention to the factor s (1 − s).
Corollary 5.1. Let 1 < p < ∞ and 0 < s < 1. Let K ⊂ R N be an open convex set such that Then D s,p 0 (K) is a functional space, continuously embedded in L p (K). Moreover, it holds 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 K .
The quantity R K above is called inradius of K. Observe that this is the radius of the largest ball inscribed in K.
For a general open 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 1 , 2 ∈ R with 1 < 2 . For every open set such that we have where C is the same constant as in Theorem 1.1.

Appendix A. Some pointwise inequalities
We collect here some pointwise inequalities needed throughout the whole paper. The most important one is Lemma A.5. We recall the notation for t ∈ R.
Lemma A.1. Let 1 < p < ∞, for every a, b > 0 we have Equality holds if and only if a = b.
Proof. This is proved in [ Equality holds if and only if a = b.
Proof. We observe that if a = b there is nothing to prove. We then take a = b and without loss of generality we can suppose a > b. The seeked inequality is then equivalent to By setting t = b/a, this in turn is equivalent to prove that 1 − t 1 + t ≤ − log t, for 0 < t < 1.
By basic Calculus, it is easily seen that the function 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 for every a, b > 0 and 1 < p < ∞.
Lemma A.4. Let 0 < s < 1, then for every a, b > 0 we have We assume now that a > b and c > d, then by setting t = b/a ∈ (0, 1) and with t ∈ (0, 1) and A ∈ [0, 1). We study the function which is maximal for t = A. This in particular implies 2 We now distinguish two cases: A. Case 0 ≤ A ≤ 1/2. This is the simplest case. Indeed, we have Thus by using this and (A.6), we get which is (A.4) with C 2 = (C 3 − 1)/2 p+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. Case 1/2 < A < 1 and t ≥ A. This is easy, since we directly have and thus By using this and (A.6), we get 2 We observe that this is equivalent to which is a discrete version of Picone's inequality, see [2, Proposition 4.2].
By choosing ε = 1/2 and using that A − t ≤ 1 − t, we then obtain Once again, this is enough to get the desired conclusion, since Then we only need to choose C 2 = 1/2 p+2 in order to get (A.4). We thus concluded the proof.
Remark A.6 (The constants C 2 and C 3 ). An inspection of the proof reveals that in the case 1 < p ≤ 2, the constant C 3 can be chosen arbitrarily close to 1. Accordingly, we have and thus it degenerates to 0 as C 3 1. is replaced by |log a − log b| p min{c p , d p }.
The main difference is that in our inequality the terms c and d play a symmetric role. This means that the quantity (c p + d p ) is unchanged when we exchange the roles of c and d, while this is not the case for min{c p , d p }. This property is a crucial feature in order to prove Theorem 1.1: precisely, 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 property, i.e. one can not replace |log a − log b| p min{c p , d p }, by |log a − log b| p (c p + d p ).
Thus the inequality of [9] does not seem useful in order to prove Hardy inequality.
In the previous result, we needed the following inequality in order to deal with the case 1 < p ≤ 2.
Lemma A.8. Let 1 < p ≤ 2, for every s ∈ (0, 1) and A ∈ [0, 1] we have Proof. We rewrite We then observe that Indeed, the latter is equivalent to which is easily seen to be true. By using (A.11) in (A.10), we now obtain as desired.