Non-local sublinear problems: Existence, comparison, and radial symmetry

We establish a symmetry result for a non-autonomous overdetermined problem associated to a sublinear fractional equation. To this purpose we prove, in particular, that the solution of the corresponding Dirichlet problem is monotonically increasing with respect to the domain. We also obtain a strong minimum principle and a boundary-point lemma for linear fractional equations that may have an independent interest.

1. Introduction. We consider the boundary-value problem where Ω is a sufficiently smooth, bounded open set in R N , N ≥ 1, and the operator (−∆) s , s ∈ (0, 1), is the fractional Laplacian (see (2)). Due to the variational structure of the equation in (1), the existence of a bounded weak solution follows from the direct method of the calculus of variations under the assumptions mentioned in Proposition 1. The solution is sought in the function space X s 0 (Ω), which is the set of all functions u ∈ L 2 (R N ) vanishing identically in all of R N \ Ω and such that R 2N (u(x) − u(y)) 2 |x − y| N +2s dx dy < +∞.
The definition of such a function space (denoted there by X 0 ) is found, for instance, in [22, p. 70], and some of its properties are outlined in [22,Sections 2.2 and 2.3]. In the definition of X s 0 (Ω), the set Ω is allowed to be any open set (and may coincide with the whole space R N ). Furthermore, it is apparent that if Ω 1 ⊂ Ω 2 then X s 0 (Ω 1 ) ⊂ X s 0 (Ω 2 ). The space X s 0 (Ω) is a Hilbert space whose scalar product is given by |x − y| N +2s dx dy.
Note that the set Ω does not enter explicitly in the integral above, hence we write simply (u, v) instead of (u, v) X s 0 (Ω) . However, the norm of u in X s 0 (Ω) is denoted by u X s 0 (Ω) to keep it distinct from the norm in L 2 (Ω), which is also used in the sequel. We note in passing the useful equality (u(x) − u(y)) 2 |x − y| N +2s dx dy = Ω 2 (u(x) − u(y)) 2 |x − y| N +2s dx dy + 2 Ω×(R N \Ω) (u(x) − u(y)) 2 |x − y| N +2s dx dy.
A weak solution of problem (1) is a function u ∈ X s 0 (Ω) such that (the Lebesgue integral below is well defined and) for every v ∈ X s 0 (Ω). The constant C N,s is given by where Γ denotes Euler's gamma function (see the Appendix). The value of C N,s is chosen to make the fractional Laplacian the pseudodifferential operator whose symbol is |ξ| 2s : see [ [24]) implies, in particular, that for every u ∈ X s 0 (R N ) the function of x ∈ R given by u(x) − u(y) |x − y| N +s dy converges in L 2 (R N ) when ε → 0 + . The limiting function is denoted by (−∆) s/2 u, and its L 2 -norm is related to the norm u X s 0 (R N ) by the identity , which is found in [4,Proposition 3.6]. Replacing u with u ± v in the identity above, for u, v ∈ X s 0 (Ω), we obtain two equalities whose difference yields therefore equation (2) can be equivalently rewritten as A similar setting is adopted, for instance, in [21,Definition 2.1].
At the beginning of the present paper we establish some useful tools that will be used later and may have an independent interest. The first two tools are a strong minimum principle and a boundary-point lemma of Hopf's type for weak solutions w of the linear inequality in an open set Ω (possibly non-smooth, unbounded, and disconnected), where the coefficient b : Ω → R is a measurable function, bounded from below. The results extend [16, Theorem 2.5 and Lemma 2.7], respectively, where the case b ≡ 0 is considered (see also [17] for every non-negative η ∈ X s 0 (Ω). Theorem 1.1 (Strong minimum principle). Let Ω be an open set in R N , N ≥ 1, and let w ∈ X s 0 (R N ) be a weak solution of (4) in Ω satisfying w ≥ 0 a.e. in R N \ Ω. Suppose that the coefficient b is bounded from below. If either w ≥ 0 in Ω or b ≤ 0 in Ω, then either w vanishes a.e. in R N or for every compact subset K ⊂ Ω there exists a positive constant ε K > 0 such that w ≥ ε K a.e. in K.

Lemma 1.2 (Fractional Hopf's boundary-point lemma).
Let Ω be an open set in R N , N ≥ 1, and let z ∈ ∂Ω be a boundary point such that z ∈ ∂B R for some ball in Ω, then either w vanishes a.e. in R N or w can be modified on a negligible set to achieve lim inf Theorem 1.1 and Lemma 1.2 are proved in Section 2. The last tool is the following comparison principle, ensuring that the weak solution of the boundary-value problem (1) is monotone with respect to set inclusion. Let Ω be an open set in R N , N ≥ 1, and let f : Ω × (0, +∞) → (0, +∞) be a positive Carathéodory function (i.e., f (x, t) is measurable as a function of x for each given t, and continuous in t for almost all x) such that the ratio f (x, t) t is strictly decreasing in t ∈ (0, +∞) for almost every x. (6) Let u i ∈ X s 0 (R N ), i = 1, 2, be such that the integral below is finite and for all non-negative functions η ∈ X s 0 (Ω). Suppose that 1. both u 1 and u 2 are positive in Ω; The proof is found in Section 3. The result extends Theorem 20 of [18], where both u 1 and u 2 are supposed to be bounded and to vanish outside Ω. In Section 4 we then give a detailed existence and uniqueness proof of a weak solution of problem (1) assuming that the set Ω is bounded and sufficiently smooth, and the function f : R N × (0, +∞) → (0, +∞) is a positive Carathéodory function satisfying (6) and having the properties listed below. Now the variable x is let vary in the whole space R N in view of the subsequent application, where the domain Ω is an unknown of the problem. For every compact K ⊂ R N and every ε > 0 we require that there exists a constant A K,ε > 0 such that 0 < f (x, t) ≤ A K,ε + ε t for every (x, t) ∈ K × (0, +∞).
We also require that lim t→0 + f (x, t) t = +∞ locally uniformly with respect to x.
As a consequence of (8) we have lim t→+∞ f (x, t) t = 0 locally uniformly with respect to x.
Model cases are f (x, t) = t m , m ∈ (0, 1), as well as f (x, t) ≡ 1. Note that (8) prevents f (x, t) from becoming infinite as t → 0. When the function f is subject to different structural conditions, multiplicity results have been established: see, for instance, [20] and the references therein.
In Section 5 we turn to consider an overdetermined problem. Recall that if Ω is a sufficiently smooth, bounded domain, then the weak solution u of (1) is Hölder continuous in R N and satisfies lim x→z x∈Ω where ϕ ∈ C α (Ω) is a convenient Hölder continuous function (see Proposition 1). Denoting by ν the inner normal to ∂Ω at z, (10) implies where the fractional inner derivative (∂ ν ) s u is defined as If we do not know the shape of Ω, but we do know that the weak solution of (1) satisfies (11) for some radial function ϕ(z) = q(|z|), can we infer that Ω is a ball centered at the origin? Counterexamples show that the answer is negative, in general. Clearly, if ϕ ≡ constant and f does not depend on x, then the overdetermined problem (1)-(11) is autonomous. In such a case, which has been considered in [7], any solution u in a given domain Ω is readily transformed in the solution v(x) = u(x − x 0 ) of the corresponding problem in the domain Ω + x 0 , for every fixed x 0 ∈ R N . Hence we cannot say that the problem is solvable only if Ω is a ball centered at a prescribed point. Furthermore, a non-spherical domain Ω such that the solution of (1) satisfies (11) although f (·, t) and ϕ(·) are radial functions can be constructed as follows.
, and let f (x, t) =f (|x|, t) be any function satisfying (6), (8) and (9). For instance, we may simply take f ≡ 1. Then by Proposition 1 we have that problem (1) has a unique weak solution u. We now construct a convenient function ϕ(x) = q(|x|) such that the same solution u also satisfies (11) although u is clearly non-radial. Since f (x, t) is a radial function of the variable x = (x 1 , x 2 ), it is also symmetric with respect to each component x i , i = 1, 2. Hence the four functions obtained by arbitrarily selecting the two signs in u(±x 1 , ±x 2 ) are solutions of problem (1). By uniqueness, it follows that u is symmetric with respect to both x 1 and x 2 . But then it is legitimate to define q(r) = (∂ ν ) s u(z) for any z ∈ ∂Ω satisfying |z| = r because the derivative does not depend on the particular choice of z, by symmetry reasons. Thus, letting ϕ(z) = q(|z|), we may assert that u solves the overdetermined problem (1)- (11) with radial functions f (·, t) and ϕ(·) although the domain of the problem is not a disc.
Here we establish a sufficient condition to exclude counterexamples as above. Concerning regularity, we require that f (x, t) satisfies locally a Lipschitz condition from below in t in the sense that for every R 0 , t 0 > 0 there exists for almost every x ∈ B R0 (0) and for all t 1 , t 2 ∈ (0, t 0 ) satisfying t 1 < t 2 . We prove: Let Ω be a bounded open set of class C 1,1 in R N , N ≥ 2, containing the origin, and let f : R N ×(0, +∞) → (0, +∞) be a positive Carathéodory function having the form f (x, t) =f (|x|, t) and satisfying (6), (8), (9) and (12). Moreover, let q(r) be a positive function of the variable r > 0 such that for every ρ, λ > 0 the compound function is monotone non-decreasing in r > 0.
If the overdetermined problem has a weak solution, then Ω is a ball centered at the origin.
The proof is given in Section 5, and it is based on the comparison of u with suitable radial functions. Here we point out some cases in which assumption (13) is satisfied. The simplest case occurs when f (x, t) ≡ 1 and the ratio q(r)/r s is monotone non-decreasing in r > 0. This case is considered in [15], which deals with classical solutions. Theorem 1.5, instead, applies to weak solutions and does not require Ω to be connected, which is an assumption of [15, Theorem 1.3]. To give another example, assumption (13) is satisfied when f (x, t) = t m , m ∈ (0, 1), and the ratio (q(r)) 1−m r s(1+m) is monotone non-decreasing in r > 0.
This case is considered in [19] (where strict monotonicity is required). As a consequence of Theorem 1.5, we thus have: containing the origin, and let q(r) be a positive function of the variable r > 0 satisfying (15) for some m ∈ (0, 1). If the overdetermined problem has a weak solution, then Ω is a ball centered at the origin.

ANTONIO GRECO AND VINCENZINO MASCIA
3. the ratio q(r)/r s( 2 ε 0 −1) is monotone non-decreasing in r > 0, then assumption (13) is satisfied. This is easily checked, taking into account that the product r s q(r) is strictly increasing as a consequence of the third condition listed above. A similar setting is found in the paper [12], which deals with the subdiffusive p-Laplacian (a local, degenerate/singular operator). In particular, the third condition above may be considered as a fractional counterpart of condition (5) in [12].
is known explicitly (see [9,21]) and it is given by where the exponent + denotes the positive part, and γ N,s is the following constant: has a positive measure. If at least one of the following conditions holds true: Without loss of generality we may assume R = 1 and x 0 = 0. Let us write B 1 in place of B 1 (0) and let v n = 1 n v 1 , for shortness. Define η n = (v n − w) + and suppose, by contradiction, that for all n > 0 the set Furthermore, since w satisfies (4) we also have Since the function u n = v n − w coincides with η n in the set Ω n , by subtracting the last inequality from the preceding equality we obtain To reach a contradiction with (17), we will estimate from below the scalar product } has a positive measure comes into play. Indeed, the assumption implies the existence of a set S of positive measure such that w ≥ ε in S and S ⊂ B 1−ε (0) for some ε ∈ (0, 1). In particular, S ⊂ R N \Ω n for n large because v n tends uniformly to 0 in (19)) we may estimate (u n , η n ) as follows: where we have taken into account that |x − y| < 2 for (x, y) ∈ Ω n × S. Finally, since −u n (y) ≥ ε − v n (y) ≥ ε/2 in S for n large, we arrive at By comparing the last inequality with (17) we obtain which cannot hold as n → +∞. Hence the set Ω n must be negligible for n large, and the claim follows.
Proof of Theorem 1.1. In the case when b ≤ 0 in Ω, let us check that w ≥ 0 a.e. in Ω.
To this aim we suppose, by contradiction, that the negative part η(x) = w − (x) is positive in a set Ω − ⊂ Ω of positive measure. Then we have However, we also have Since the last integrand is positive in the domain of integration, and since Ω − has a positive measure by assumption, the inequality above implies that |R N \Ω − | = 0 and w(x) = w(y) a.e. in Ω 2 − . But then Ω − would have an infinite measure, and w would be a negative constant there, which is in contrast with w ∈ X s 0 (R N ) ⊂ L 2 (R N ). Hence we may assume w ≥ 0 a.e. in Ω in the sequel. To proceed in the proof of the theorem, we use the fact that Ω is open and write

ANTONIO GRECO AND VINCENZINO MASCIA
for some open balls B R k (x k ) ⊂ Ω. By Lemma 2.1, for each k the function w must either vanish a.e. or be positive a.e. in B R k (x k ). Hence we may write Ω = Ω 0 ∪ Ω + , where Ω 0 is the union of those B R k (x k ) where w vanishes almost identically, and Ω + the union of the remaining balls. Clearly, w = 0 a.e. in Ω 0 and w > 0 a.e. in Ω + , and therefore Ω 0 ∩ Ω + = ∅. Let us check that if Ω 0 = ∅ then Ω + = ∅. Assume that Ω 0 = ∅ and choose a non-negative η ∈ X s 0 (Ω 0 ) with a nonempty support. By (4) we have Furthermore, the scalar product (w, η) reduces to Since w(y) ≥ 0 in R N \ Ω 0 , we must have either R N \ Ω 0 = ∅ or w = 0 a.e. in R N \ Ω 0 . In both cases we have w = 0 a.e. in R N and consequently Ω + = ∅, as claimed. Hence, either w = 0 a.e. in R N or w > 0 a.e. in Ω. In the last case, every compact subset K ⊂ Ω admits a finite covering made up of a finite number n K of balls, still denoted by B R k (x k ), satisfying B 2R k (x k ) ⊂ Ω for k = 1, . . . , n K . By applying Lemma 2.1 to each ball B 2R k (x k ) we get a positive integer n k such that 3. Comparison principle. In order to establish Theorem 1.5 we need to compare a weak subsolution u 1 of problem (1) in an open set Ω 1 with a weak supersolution of the same problem in a possibly larger set Ω 2 . To this purpose we use Theorem 1.3, whose proof is given here. The argument is based on the following inequality.
Lemma 3.1. For every a 1 , b 1 ∈ R and every a 2 , b 2 > 0 we have Proof. Since the mapping t → t + is monotone non-decreasing, we have Letting p = (a 1 − a 2 )/a 2 and q = (b 1 − b 2 )/b 2 and multiplying the inequality by a 2 b 2 > 0 we obtain which is equivalent to (18).
Proof of Theorem 1.3. For ε ≥ 0 define G ε = { x ∈ R N | u 1 (x) > u 2 (x) + ε } ⊂ Ω and assume, contrary to the claim, that G 0 has a positive measure. In order to reach a contradiction we use the test functions v ε (x) = (u 1 − u 2 − ε) + (x) for ε ≥ 0, and Both v ε and w ε are essentially bounded because u 1 ∈ L ∞ (R N ) by assumption, and To this aim it suffices to observe that for all x, y ∈ R N , and therefore Since the right-hand side is finite, and since v ε = 0 in R N \ G ε , (20) follows. We also claim that w ε ∈ X s 0 (G ε ) for every ε > 0.

ANTONIO GRECO AND VINCENZINO MASCIA
By comparison with (24) we obtain Hence, by the monotone convergence theorem we get where the right-hand side is finite by assumption. However, by (6) we may write which contradicts (25). Hence the set G 0 must have measure zero, and the proof is complete.

Existence for the Dirichlet problem. The existence proof of a weak solution of problem (1) is based on the fact that equation (2) is the Euler equation in integral form of the functional
where F (x, u) is given by which is well defined for a.e. x ∈ Ω and for every u ∈ [0, +∞) by virtue of (8). In the negligible set of all x ∈ Ω such that f (x, ·) is not a continuous function, we let F (x, u) = 0 for all u ∈ [0, +∞). The domain of the functional J is the set (X s 0 (Ω)) + = { u ∈ X s 0 (Ω) | u ≥ 0 a.e. in Ω }. Proposition 1. Let Ω be a bounded open set of class C 1,1 in R N , N ≥ 2, and let f : R N × (0, +∞) → (0, +∞) be a Carathéodory function satisfying (6), (8) and (9). Then (i) Problem (1) has a unique weak solution u ∈ X s 0 (Ω) ∩ L ∞ (Ω). (ii) The solution u belongs to the Hölder class C s (R N ). (iii) The ratio u(x)/(dist(x, ∂Ω)) s is in the Hölder class C α (Ω) for some α ∈ (0, 1).

Proof.
Step 1. Two useful estimates. Observe that for every function u ∈ (X s 0 (Ω)) + ⊂ L 2 (Ω) the compound function F (x, u(x)) is measurable: see [3, Section 8.3 C, pp. 284-285]. As a consequence of (8) and the boundedness of Ω, for every ε > 0 there exists a constant A ε such that and therefore Letting t = u(x) ≥ 0 we get the pointwise estimate By integration over Ω, and since u L 1 (Ω) ≤ |Ω| 1/2 u L 2 (Ω) we obtain The estimate above implies, in particular, that the functional J[u] in (26) has a finite value for each u ∈ (X s 0 (Ω)) + .
Step 3. Existence of a minimizer in (X s 0 (Ω)) + . Since J is coercive, any minimizing sequence (made up of non-negative functions) for the functional J must be bounded in X s 0 (Ω) and therefore it must also have a converging subsequence in L 2 (Ω). By the weak compactness theorem in the Hilbert space X s 0 (Ω) we may also extract a sub-subsequence (u n ) n≥1 converging to some u 0 ∈ X s 0 (Ω) in the weak topology of X s 0 (Ω). Since the dual space (L 2 (Ω)) * is included in (X s 0 (Ω)) * , we also have u n u 0 in L 2 (Ω), hence u 0 must coincide with the strong limit of u n in L 2 (Ω) whose existence was asserted above. Passing to a further subsequence (u n k ) k≥1 we may assume that u n k (x) → u 0 (x) a.e. in Ω and there exists a function h ∈ L 2 (Ω) such that u n k (x) ≤ h(x) a.e. in Ω for all k ≥ 1, and consequently there exists a summable function ψ(x) such that in Ω for all k ≥ 1 (cf. (28)). Hence, Lebesgue's dominated convergence theorem applies and we have Recalling that the norm in a Hilbert space is weakly lower semicontinuous, we also have u 0 X s 0 (Ω) ≤ lim inf k→+∞ u n k X s 0 (Ω) , and consequently Thus, u 0 is a minimizer of J.
Step 4. The minimum of J over (X s 0 (Ω)) + is negative, and therefore the null function is not a minimizer. To see this, choose a bounded function u 1 , not vanishing identically, in the space (X s 0 (Ω)) + . Let u t (x) = tu 1 (x) for t > 0. We aim to verify that J[tu 1 ] < 0 when t is small. By (9), for every M > 0 there exists t M > 0 such that f (x, t) > M t for almost all x ∈ Ω and for all t ∈ (0, t M ). But then F (x, tu 1 (x)) ≥ M t 2 u 1 (x)/2 whenever tu 1 (x) ∈ [0, t M ) (apart from a negligible subset of Ω) and therefore .
We may choose M so large that the difference in parentheses is negative, thus getting J[u t ] < 0 for t in a small interval (0, t 0 ), hence min u∈(X s 0 (Ω)) + J[u] < 0.
Step 5. Every minimizer u ∈ (X s 0 (Ω)) + satisfies the variational inequality (−∆) s u ≥ 0 in the weak sense in Ω. To check this, take an arbitrary function η ∈ (X s 0 (Ω)) + and a real number t > 0. We have Since the function F (x, u) is strictly increasing in the variable u for almost every x, the integral above is non-negative. Dividing the inequality by t we arrive at Letting t → 0 + we obtain (u, η) ≥ 0, which is what we had to prove.
Step 6. Every minimizer u ∈ (X s 0 (Ω)) + keeps far from zero on each compact subset K ⊂ Ω. This follows by letting w = u and b = 0 in the strong minimum principle (Theorem 1.1), taking into account that u does not vanish a.e. in Ω.
Step 7. Every minimizer u of the functional J in X s 0 (Ω) satisfies (2). To prove this, we first take η in the space C ∞ c (Ω) of all functions η ∈ C ∞ (R N ) whose support is a compact subset of Ω. Since the minimizer u keeps far from zero in the support of η, the function u t = u + tη keeps positive as long as |t| is sufficiently small, hence the function ϕ(t) = J[u t ] has a minimum at t = 0. We have By (27), and since u, η ∈ L 2 (Ω), we may compute ϕ (0) by differentiating under the sign of integral, which yields But since ϕ (0) = 0, we obtain (2) restricted to the case when v = η ∈ C ∞ c (Ω). Finally, since the space C ∞ c (Ω) is a dense subset of X s 0 (Ω) (see [8,Theorem 6]), equality (2) follows in its full generality.
Step 8. Conclusion. The preceding steps show the existence of a weak solution u of problem (1) in the space X s 0 (Ω). Moreover, by (8) the assumption in [16, Lemma 3.2] is satisfied with q = 2, and therefore we have u ∈ L ∞ (Ω). Uniqueness follows now from the comparison principle (Theorem 1.3), and the proof of Claim (i) is complete. Finally, since the compound function f (x, u(x)) is also essentially bounded in Ω, the results in [21, Proposition 1.1 and Theorem 1.2] apply, whence Claim (ii) and Claim (iii) follow. To this purpose recall that by Remark 1 a solution of (2) is also a weak solution of (−∆) s u = f (x, u(x)) in the sense of [21, Definition 2.1].
Remark 2. Problem (1) may be seen as a non-local counterpart of the problems investigated in the classical papers [1,5]. However, assumptions (5) and (6) of [1] as well as (1) and (2) in [5] involve the first eigenvalue λ 1 of the domain Ω and therefore they are domain-dependent. Proposition 1, instead, asserts the existence of a unique solution in every bounded open set Ω. This is particularly useful in the proof of Theorem 1.5, which makes use of radial solutions in balls whose radii are arbitrary.
5. Constrained symmetry. The proof of Theorem 1.5 is based on the comparison with suitable radial functions. The argument and some of its variants have been used in a series of papers on local problems (see, for instance, [11,12,13,14] and the references listed there) and in [15] for a non-local problem. Here we present a further improvement, which is effective under more general conditions. See [10] for a characterization of the ellipsoid through a (local) overdetermined problem.
Proof of Theorem 1.5. Assume that the overdetermined problem (14) is solvable in some bounded open set Ω 0, and define Suppose, contrary to the claim, that R 1 < R 2 . In order to reach a contradiction, denote by u 1 ∈ X s 0 (B 1 ) ∩ L ∞ (Ω) ⊂ X s 0 (Ω) the weak solution of (−∆) s u 1 = f (x, u 1 ), u 1 > 0 in B 1 , where B 1 is the ball of radius R 1 centered at 0. Existence and uniqueness of such a solution follow from Proposition 1. Since f (x, t) is a radial function of x by assumption, problem (29) is invariant under rotations: this and uniqueness imply that u 1 is a radial function. Using the comparison principle (Theorem 1.3) in Ω = B 1 we get u 1 ≤ u in R N . Now let us consider a point z 1 ∈ ∂Ω ∩ ∂B 1 . The last condition in (14) implies that the solution u satisfies (∂ ν ) s u(z 1 ) = q(R 1 ). Note that the inner normal ν 1 = −|z 1 | −1 z 1 to ∂B 1 at z 1 coincides with the inner normal ν to ∂Ω. Since the fractional derivative (∂ ν1 ) s u 1 exists by Proposition 1, and both u and u 1 are continuous at z 1 , we arrive at To proceed further, let B 2 be the ball of radius R 2 centered at 0. Instead of considering the radial solution u 2 ∈ X s 0 (B 2 ) which solves (29) where the subscript 1 is replaced with 2 , define the rescaled radial function v(x) = a u 1 (R 1 x/R 2 ) in R N , where the coefficient a is given by .

ANTONIO GRECO AND VINCENZINO MASCIA
The use of v in place of u 2 is a refinement of the argument in the literature (cf. [12,13,14,15]) and allows for less demanding assumptions. Since the inner fractional derivative of v on ∂B 2 is given by where ν 2 = −|z| −1 z denotes the inner normal to ∂B 2 at z, from (30) we get Furthermore, since u 1 is the weak solution of (29), by computation we find that v is the weak solution of the problem To complete the proof, we have to compare v and u. Choose x ∈ B 2 and define ρ = R 1 |x| and λ = R s 1 q(R 1 ) v(x). By assumption (13) we get This and (32) show that v satisfies (−∆) s v ≥ f (x, v) in the weak sense in B 2 . But then by Theorem 1.3 we must have u ≤ v in R N , and therefore at every point z 2 ∈ ∂Ω∩∂B 2 . Of course, the inner normal ν to ∂Ω coincides with the inner normal ν 2 to ∂B 2 at z 2 . Now the third condition in (14), together with (31), shows that (∂ ν ) s u(z 2 ) = (∂ ν ) s v(z 2 ): hence we are in a position to apply Hopf's lemma (Lemma 1.2). The function w = v − u ≥ 0 in R N satisfies (∂ ν ) s w(z 2 ) = 0 as well as (−∆) s w ≥ b(x) w(x) in the weak sense in Ω, where b : Ω → R is given by Since f satisfies (12), and v is bounded, the function b(x) is bounded from below, and by Lemma 1.2 we deduce that u = v in R N . But since the solution u vanishes in R N \ Ω, and v is positive in B 2 ⊃ Ω, we deduce Ω = B 2 and the proof is complete.
Appendix. Two equivalent expressions of C N,s . The constant C N,s is written in different forms by different authors: compare, for instance, [2,Remark 3.11] and [4, (3.2)]. Here we prove that the two expressions in (3) coincide. The proof is based on some identities involving the Bessel functions J ν and the gamma function, which are found, for instance, in the classical reference [6].
Proposition 2. For all N ≥ 1 and all s ∈ (0, 1) the following equality holds: Proof. Let us start with special case N = 1. In this case we have x 1 = x and, after an integration by parts, the integral in (33) reduces to The last integral can be expressed in terms of the gamma function. To this aim, recall that the Bessel function J 1 2 (x) is related to sin x as follows: (see formula (14) in [6, Section 7.11, p. 79]). Hence we can apply the identity which is valid when −ν < α < 3 2 and α + ν > 0 (see formula (19) in [6, Section 7.7.3, p. 49]). Letting α = 3 2 − 2s and ν = 1 2 we obtain Plugging this into (34) we obtain (33) in dimension N = 1. To manage the case N ≥ 2 we need to integrate over spherical surfaces of different dimensions. To this purpose, let us denote by B N R the N -dimensional ball in R N centered at 0 and of radius R. By the change of variables ρ = |x| and u = |x| −1 x, x = 0, we get where u 1 denotes the first component of the unit vector u ∈ ∂B N 1 , and dω is the surface element. For fixed u 1 > 0, integration by parts followed by the change of variable t = ρ u 1 yields To manage the last integral, observe that for every φ ∈ (0, π 2 ), the set of all u ∈ ∂B N 518 ANTONIO GRECO AND VINCENZINO MASCIA the spherical surface is given by By applying formula (19) in [6, Vol. 1, Section 1.5.1, p. 10] to evaluate the last integral, we obtain π 2 0 (sin φ) 2s (cos φ) N −2 dφ = Γ( 1 2 + s) Γ( N −1 2 ) 2 Γ( N 2 + s) and therefore .
Inserting this expression into (37), the claim follows.