Large $s$-harmonic functions and boundary blow-up solutions for the fractional Laplacian

We present a notion of weak solution for the Dirichlet problem driven by the fractional Laplacian, following the Stampacchia theory. Then, we study semilinear problems on bounded domains $\Omega$ with two different boundary conditions at the same time: the shape of the solution outside $\Omega$ and a weighted limit to the boundary. We allow the nonlinearity to be positive or negative and we look for solutions blowing up at the boundary. Our starting observation is the existence of s-harmonic functions which explode at the boundary: these will be used both as supersolutions in the case of negative nonlinearity and as subsolutions in the positive case.


Introduction
In this paper we study a suitable notion of weak solution to semilinear problems driven by the fractional Laplacian (−∆) s , i.e. the integral operator defined as (see e.g. Di Nezza, Palatucci and Valdinoci [14] for an introduction) where A(n, s) is a normalizing constant 3 . In order to do this, we will need to develop a theory for the Dirichlet problem for the fractional Laplacian with measure data (see Karlsen, Petitta and Ulusoy [18] and Chen and Véron [8], for earlier results in this direction). We pay particular attention to those solutions having an explosive behaviour at the boundary of the prescribed domain, known in the literature as large solutions or also boundary blow-up solutions.
Let us recall that in the classical setting (see Axler, Bourdon and Ramey [1, Theorem 6.9]), to any nonnegative Borel measure µ ∈ M(∂B) on ∂B it is possible to associate, via the representation through the Poisson kernel, a harmonic function in B with µ as its trace on the boundary. Conversely, any positive harmonic function on the ball B has a trace on ∂B that is a nonnegative Borel measure (see [1,Theorem 6.15]).
When studying the semilinear problem for the Laplacian, solutions can achieve the boundary datum +∞ on the whole boundary. More precisely, take Ω a bounded smooth domain and f nondecreasing such that f (0) = 0. According to the works of Keller [19] and Osserman [26], the equation has a solution if and only if f satisfies the so called Keller-Osserman condition, that is see also Dumont, Dupaigne, Goubet and Rădulescu [15] for the case of oscillating nonlinearity. The case of nonsmooth domains is delicate, see in particular the work of Dhersin and Le Gall [13] for the case f (u) = u 2 . For the same nonlinearity, and for Ω = B, Mselati [24] completely classified positive solutions in terms of their boundary trace, which can be +∞ on one part of the boundary and a measure that doesn't charge sets of zero boundary capacity on the remaining part. See the upcoming book by Marcus and Véron for further developments in this direction.
In the fractional context, our starting point is that large solutions arise even in linear problems. In particular, it is possible to provide large s-harmonic functions, i.e. functions satisfying See Lemma 3.2.4 below. Moreover, letting σ → 1 − s we recover the following example found in Bogdan, Byczkowski, Kulczycki, Ryznar, Song, and Vondraček [5], qualitatively different from the previous one: The function u 1−s so defined satisfies and shows how problems where only outer values are prescribed are ill-posed in the classical sense. Different kinds of boundary conditions have to be taken into account: indeed, in the first case we have an s-harmonic function associated to the prescribed data of u σ outside B; in the second case all the mass of the boundary datum concentrates on ∂B. This means that we need a notion of weak solution that can deal at the same time with these two different boundary data, one on the complement of the domain and the other one on its boundary. Recently, Felmer and Quaas [17] and Chen, Felmer and Quaas [7] have shown the existence of large solutions to problems of the form both under assumptions of the explosion of the datum g at ∂Ω and when g = 0. In all cases they need to assume p > 2s + 1. Our approach allows to deal with general equations of the form with no necessary assumptions on the sign and the growth of the nonlinearity f and to provide large solutions in both cases where u is prescribed outside of Ω or only at ∂Ω as a measure.
The definition of s-harmonicity is given via a mean value property, namely Definition 1.1.1. We say that a measurable nonnegative function u : R n → [0, +∞] is s-harmonic on an open set Ω ⊆ R n if u ∈ C(Ω) and for any x ∈ Ω and 0 < r < dist(x, ∂Ω) c(n, s) r 2s |y − x| n (|y − x| 2 − r 2 ) s u(y) dy = (η r * u) (x).

An integration by parts formula
For any two functions u, v in the Schwartz class S, the self-adjointness of the operator (−∆) s entails this follows from the representation of the fractional Laplacian via the Fourier transform, see [14, Paragraph 3.1]. By splitting R n into two domains of integration Proposition 1.2.1.
Let Ω ⊆ R n open and bounded. Let C 2s+ε (Ω) = {v : R n → R such that v ∈ C(Ω) and for any K compactly supported in Ω, there exists α = α(K, v) such that v ∈ C 2s+α (K)}. If u ∈ C 2s+ε (Ω) ∩ L ∞ (R n ) and v = 0 in R n \ Ω, v ∈ C 2s+ε (Ω) ∩ C s (R n ) and (−∆) The proof can be found in the Appendix. From now on the set Ω ⊆ R n will be an open bounded domain with C 1,1 boundary. More generally, we will prove in Section 3 the following Proposition 1.2.2. Let δ(x) = dist(x, ∂Ω) for any x ∈ R n , and u ∈ C 2s+ε loc (Ω) such that Let -G Ω (x, y), x, y ∈ Ω, x = y, be the Green function of the fractional Laplacian on Ω, that is s G Ω (x, y), x ∈ Ω, y ∈ R n \ Ω, be the corresponding Poisson kernel, -for x ∈ Ω, θ ∈ ∂Ω, 5 M Ω (x, θ) = lim y → θ y∈Ω G(x, y) δ(y) s , -for θ ∈ ∂Ω, where the limit is well-defined in view of Lemma 3.1.5 below. 4 the construction of H can be found in the proof of Theorem 2.0.17 below 5 this is a readaptation of the Martin kernel of Ω Then the integration by parts formula and v ≡ 0 in R n \ Ω. Such a limit exists and is continuous in θ in view of Lemma 3.3.1. In addition, we have the representation formula The paper is organized as follows. Section 2 relates Definition 1.1.1 with the fractional Laplacian (−∆) s . Section 3 recalls some facts on Green functions and Poisson kernels and it studies the linear Dirichlet problem both in the pointwise and in a weak sense. Section 4 deals with the nonlinear problem. Now, let us outline the main results in Section 3 and Section 4.

The Dirichlet problem
For a fixed x ∈ Ω the Poisson kernel satisfies for some constant c > 0 independent of y, as it will be shown later on. In particular any Dirichlet condition u = g in R n \ Ω satisfying (11) below is admissible in the representation formula (10). We prove the following and h ∈ C(∂Ω). Then, the function defined by setting x ∈ R n \ Ω (12) belongs to C 2s+ε loc (Ω), fulfills (8), and u is the only pointwise solution of Here Eu has been defined in Proposition 1. (12), with the latter notation we would like to stress on the fact that the boundary ∂Ω plays an important role in this setting. Remark 1.2.5. Since ∂Ω ∈ C 1,1 , we will exploit the behaviour of the Green function and the Poisson kernel described by [9, Theorem 2.10 and equation (2.13) resp.]: there exist c 1 = c 1 (Ω, s) > 0, c 2 = c 2 (s, Ω) such that 6 compare also with equation (32) below and 1 Remark 1.2.6 (Construction of large s-harmonic functions). The case when f = 0 in Theorem 1.2.3 corresponds to s-harmonic functions: when h ≡ 0 then u automatically explodes somewhere on the boundary (by definition of E), while if h ≡ 0 then large s-harmonic functions can be built as follows. Take any positive g satisfying (11) with lim δ(x)↓0 g(x) = +∞ and let g N = min{g, N }, N ∈ N. By Theorem 1.2.3, the corresponding solutions u N ∈ C(Ω). In particular, u N = N on ∂Ω and by the Maximum Principle {u N } N is increasing. Hypotheses (11) guarantees a uniform bound on {u N } N . Then increases to a s-harmonic function u such that Next, in view of Theorem 1.2.3, we introduce the test function space and starting from the integration by parts formula (9), we introduce the following notion of weak solution Definition 1.2.7. Given three Radon measures λ ∈ M(Ω), µ ∈ M(R n \ Ω) and ν ∈ M(∂Ω), such that The integrals in the definition are finite for any φ ∈ T (Ω) in view of Lemma 3.3.1.
In addition, we have the representation formula for some constant C = C(n, s, Ω) > 0.
We will conclude Section 3 by showing that for s < β < 1 + s.
Moreover, there exist a constant c = c(n, s, Ω) > 0 and c = c(n, s, Ω) such that the solution of for any g satisfying (11) such that g, g are decreasing functions in 0 + and lim t↓0 g(t) = +∞.

The nonlinear problem
We consider nonlinearities f : Ω × R → R satisfying hypotheses and all positive boundary data g that satisfy (11). After having constructed large s-harmonic functions, we first prove the following preliminary Theorem 1.2.10. Let f : Ω × R → R be a function satisfying f.1). Let g : R n \ Ω → R be a measurable bounded function. Assume the nonlinear problem on ∂Ω admits a subsolution u ∈ L 1 (Ω) and a supersolution u ∈ L 1 (Ω) in the weak sense Assume also u ≤ u in Ω, and u, u ∈ L ∞ (Ω) ∩ C(Ω). Then the above nonlinear problem has a weak solution u in the sense of Definition 1.2.7 satisfying In addition, for all x ∈ Ω, then there is a unique solution, • if not, there is a unique minimal solution u 1 , that is a solution u 1 such that u ≤ u 1 ≤ v for any other supersolution v ≥ u.
In case our boundary datum g is a nonnegative bounded function, then Theorem 1.2.10 provides a unique solution, since we may consider u = sup g and u = 0. Then we attack directly the problem with unbounded boundary values, and we are especially interested in those data exploding on ∂Ω. The existence of large s-harmonic functions turns out to be the key ingredient to prove all the following theorems, that is, Depending on the nature of the nonlinearity f one can be more precise about the Dirichlet values of u. Namely, Theorem 1.2.12 (Damping term). Let f : Ω × R → R be a function satisfying f.1) and f.2), and g : R n \ Ω → [0, +∞] a measurable function satisfying (11); let also be h ∈ C(∂Ω), h ≥ 0. The semilinear problem • the problem doesn't admit any weak solution if there exist b 1 , T > 0 such that If, in addition, f is increasing in the second variable then the problem admits only one positive solution.
Theorem 1.2.14 (Superlinear source). Let f : Ω × R → R be a function satisfying f.1) and f.2). For 0 < β < 1 − s, consider problems Existence. If there exist a 1 , a 2 , T > 0 and p ≥ 1, such that x ∈ Ω, t > T. We finally note that, with the definition of weak solution we are dealing with, the nonexistence of a weak solution implies complete blow-up, meaning that: then we say there is complete blow-up.

Notations
In the following we will always use the following notations: δ(x) for dist(x, ∂Ω) once Ω ⊆ R n has been fixed, M(Ω), when Ω ⊆ R n , for the space of measures on Ω, H, for the n − 1 dimensional Hausdorff measure, dropping the "n − 1" subscript whenever there is no ambiguity, f ∧ g, when f, g are two functions, for the function min{f, g}, C 2s+ε (Ω) = {v : R n → R and for any K compactly supported in Ω, there exists α = α(K, v) such that v ∈ C 2s+α (K)}.

A mean value formula
Definition 1.1.1 of s-harmonicity turns out to be equivalent to have a null fractional Laplacian. Since we couldn't find a precise reference for this, we provide a proof. Indeed, on the one hand we have that any function u which is s-harmonic in an open set Ω solves indeed condition (5) can be rewritten, using (4), and therefore A(n, s) .
Indeed, by dominated convergence, far from x it is Now, any function u s-harmonic in Ω is smooth in Ω: this follows from the representation through the Poisson kernel on balls, given in Theorem 1.2.3, and the smoothness of the Poisson kernel, see formula (35) below. Since u is a smooth function, a Taylor expansion when |y − x| < 1 Theorem 2.0.17. Let u : R n → R a measurable function such that for some open set Ω ⊆ R n is u ∈ C 2s+ε (Ω). Also, suppose that R n |u(y)| 1 + |y| n+2s dy < +∞.
Then for any x ∈ Ω and r > 0 such that B r (x) ⊆ Ω, one has u(x) = CBr η r (y) u(x − y) dy + γ(n, s, r)(−∆) s u(z), γ(n, s, r) = Γ(n/2) 2 2s Γ n+2s 2 Γ(1 + s) r 2s (18) for some z = z(x, s) ∈ B r (x). and H solves in the pointwise sense We claim that H satisfies equality Finally, as in 1., note that Since where we have used the integration by parts formula (1.2.1) and the definition of Γ s . On the one hand while on the other hand so that we can let j → +∞ in equality (19). Collecting the information so far, we have for some |z| ≤ r, by continuity of (−∆) s u in B r and since v > 0 in B r . The constant γ(n, s, r) appearing in the statement equals to Br v.
Let us compute γ(n, s, r) = Br v. If we consider the solution ϕ δ to and we apply formula (20) to ϕ δ in place of u to entail The solution ϕ δ is explicitly known (see [28, equation (1.4)] and references therein) and given by Hence, by letting δ ↓ 0, Remark 2.0.18. The asymptotics as s ↑ 1 of (18) are studied in Appendix C.

Linear theory for the fractional Dirichlet problem
Assume Ω ⊆ R n is open and bounded, with C 1,1 boundary.

Preliminaries on fractional Green functions, Poisson kernels and Martin kernels
Consider the function G Ω : Ω × R n → R built as the family of solutions to the problems This function can be written as the sum where Γ s is the fundamental solution to the fractional Laplacian, and [29,Proposition 2.7]): from this we deduce that h(x, ·) ∈ C 2s+ε (Ω) ∩ C s (R n ) and then also H(x, ·) ∈ C 2s+ε (Ω) ∩ C s (R n ). i) G Ω is continuous in Ω × Ω except on the diagonal, where its singularity is inherited by the singularity iv) (−∆) s G Ω (x, y), x ∈ Ω, y ∈ CΩ is given by the formula Proof. We prove all conclusions step by step. Proof of ii). First of all, we use the estimate and γ s is a regularization of Γ s as in Lemma 3.1.1; for the inequality we refer to [28, Proposition 1.1]. We deduce that, for y sufficiently close to ∂Ω, it is Then, for y ∈ CΩ, loc (Ω) ∩ C(Ω) and so we can apply Proposition 1.2.1. Hence, This formula is a Green's representation formula: G Ω is the Green function while its fractional Laplacian is the Poisson kernel.
For this reason we borrow the usual notation of the Martin kernel.
Then for any θ * ∈ ∂Ω Eu(θ * ) := lim Proof. Denote by i.e. L is a bounded quantity. Indeed, referring to estimates on the Green function in [9, equation (2.13)], we have inequalities Thus, Describe ∂Ω as a graph in a neighborhood of 0, i.e.
Let us now write, Suppose without loss of generality that θ * = 0 and denote by ω the modulus of continuity of h: 3.2 Linear theory for smooth data: proof of Theorem 1.2.3 We start by stating Then u ≤ 0 in Ω.

Proof.
Call Ω + = {x ∈ R n : u > 0}: Ω + is an open set contained in Ω. Assume by contradiction that Ω + = ∅. By continuity of u in Ω + , there exists x 0 ∈ Ω + such that u(x 0 ) = max{u(x) : x ∈ Ω + }, but this point x 0 will be also a global maximum for u since outside Ω + the function u is nonpositive. Thus contradicting our hypotheses. Therefore Ω + is empty.
By splitting g into its positive and negative part, it suffices to prove Theorem 1.2.3 in the case where g ≥ 0. So, from now on we will deal with nonnegative boundary data g : CΩ → [0, +∞) which are measurable functions with Note that, in view of equation (25), Proof of Theorem 1.2.3. We split the proof by building the solution associated to each datum f , g and h separately.
First case: f, h ≡ 0. We present here a readaptation of [21, Lemma 1.13]. The function u defined by equation (12) is continuous in Ω as an application of the Dominated Convergence Theorem and inequality (33). The continuity up to the boundary is postponed to Paragraph 3.5.2.
For the sake of clarity we divide the proof in four steps: for special forms of g, for g regular enough, for g bounded and finally for any other g.
Step 2. If g ∈ C ∞ (CΩ) and supp g is bounded, then g admits an extensiong ∈ C ∞ c (R n ) and (see [21, Lemma 1.1]) there exists an absolutely continuous measure ν, with density Ψ, such that Denote by ν Ω the measure obtained by restricting ν to Ω, i.e. ν Ω (A) = ν(A ∩ Ω) for any measurable A and ν Ω has density Ψ Ω = Ψχ Ω , and for any fixed y ∈ CΩ as a consequence of G Ω (·, y) ∈ C(Ω) and equation (22). Define γ = ν − ν Ω + ν Ω which is a measure supported in CΩ. Then, when x ∈ CΩ, where we have used (23). Therefore so that we can apply the previous step of the proof.
Step 3. For g ∈ L ∞ (CΩ), consider a sequence {g N } n∈N ⊆ C ∞ (CΩ) uniformly bounded and converging pointwisely to g. The corresponding sequence of s-harmonic functions u N converges to u, since by Dominated Convergence. Then, again by the Dominated Convergence theorem we have i.e. u is s-harmonic in Ω.
Step 4. For a general measurable nonnegative g it suffices now to consider an increasing sequence g N converging to g, e.g. g N = min{g, N }. Then the corresponding sequence of s-harmonic functions u N converges to u. Moreover, the sequence {u N } N is increasing: Then, thanks to the Monotone Convergence theorem we have Uniqueness. Finally, if g ∈ C(CΩ), the solution we have built is the only solution in C(Ω) as an application of Lemma 3.2.1. It is not possible in this case to have a pointwise solution of Indeed, if we set g N = min{g, N }, N ∈ N, then g N converges monotonically to g, and Second case: g, h ≡ 0. Use the construction of G Ω to write for x ∈ Ω the first addend is a function u 1 (x) which solves (−∆) s u 1 = f χ Ω in R n , let us turn to the second one: According to the Step 4 above, u 2 solves Finally, u ∈ C 2s+α (Ω) thanks to [ Remark 3.2.3. Note that these computations give an alternative integral representation to the one provided in equation (21) for u, meaning that we have both so we must conclude that G Ω (x, y) = G Ω (y, x) for x, y ∈ Ω, x = y.
Third case: f, g ≡ 0. The function u(x) = ∂Ω M (x, θ) h(θ) dH(θ) is s-harmonic: to show this we use both the construction of M (x, θ) and the mean value formula (18). Using the notations of (28), for any where the equality (−∆) s u ε = f ε holds throughout Ω in view of (34). Letting ε ↓ 0 we have both This implies that we have equality i.e. u is s-harmonic.
Lemma 3.2.4 (An explicit example on the ball). The functions where c(n, s) is given by (4).
Proof. According to [21, equation (1.6.11')] and in view to the computations due to Riesz [27], the Poisson kernel for the ball B of radius 1 and centered at 0 has the explicit expression We construct here the s-harmonic function induced by data Indeed, it can be explicitly computed We are interested in letting σ → 1 − s. Obviously, in CB everywhere in R n \ ∂B. s-harmonicity is preserved, since for x ∈ B and any r ∈ (0, 1 − |x|), Then on the one hand On the other hand . Splitting x ∈ CB in spherical coordinates, i.e. x = ρθ, ρ = |x| ∈ (1, +∞) and |θ| = 1, and denoting by φ ∈ C(Ω) the function satisfying (−∆) and has a decay at infinity which is comparable to that of ρ −n . Therefore, as σ → 1 − s, Indeed, by the definition of c(n, s) in (3), it is −c(n, s + σ) and, since in (36) the product ψ 1 (ρ) ρ n−2 ∈ C([1, +∞)), and so u solves in CB Eu = c(n, 1/2) on ∂B.

The linear Dirichlet problem: an L 1 theory
We define L 1 solutions for the Dirichlet problem, in the spirit of Stampacchia [30]. A proper functional space in which to consider test functions is the following.
Proof of Theorem 1.2.8. In case ν = 0, we claim that the solution is given by formula Take φ ∈ T (Ω): again using (37). Then, we claim that the function This, along with the first part of the proof, proves our thesis. Take φ ∈ T (Ω) and call ψ = (−∆) The uniqueness is due to Lemma 3.3.4 below. Theorem 3.4.1 below proves the estimate on the L 1 norm of the solution.

Lemma 3.3.4 (Maximum Principle). Let u ∈ L 1 (Ω) be a solution of
Then u ≤ 0 a.e. in R n .

Regularity theory
on ∂Ω which we know to satisfy 0 ≤ ζ(x) ≤ C δ(x) s in Ω, see [28]. Note also that, by approximating ζ with functions in T (Ω) and by 37, for x ∈ CΩ and therefore, when x ∈ CΩ and δ(x) < 1, Furthermore, D s ζ is a well-defined function on ∂Ω, again thanks to the results in [28, Proposition 1.1]. Indeed, consider an increasing sequence {ψ k } k∈N ⊆ C ∞ c (Ω), such that 0 ≤ ψ k ≤ 1, ψ k ↑ 1 in Ω. Call φ k the function in T (Ω) associated with ψ k , i.e. (−∆) s φ k | Ω = ψ k . In this setting Finally, we underline how D s ζ ∈ L ∞ (Ω), thanks to (31). We split the rest of the proof by using the integral representation of u: the mass induced by the right-hand side is the one induced by the external datum finally the mass due to the boundary behavior Note that the smoothness of the domain is needed only to make the last point go through, and we can repeat the proof in case ν = 0 without requiring ∂Ω ∈ C 1,1 .
To gain higher integrability on a solution, the first step we take is the following Proof. Call as usual ψ = (−∆) s φ| Ω ∈ C ∞ c (Ω). Let us work formally where 1 p + 1 q = 1. Thus we need only to understand for what values of p we are not writing a trivial inequality. The main tool here is inequality which holds in C 1,1 domains, see [9, equation 2.14]. We have then which is uniformly bounded in x for p < n n−s . This condition on p becomes q > n s . In view of this last lemma, we are able to provide the following theorem, which is the fractional counterpart of a classical result (see e.g. [16,Proposition A.9.1]).
has a finite L p -norm controlled by Proof. For any ψ ∈ C ∞ c (Ω), let φ ∈ T (Ω) be chosen in such a way that (−∆) s φ| Ω = ψ. We have where q is the conjugate exponent of p, according to Lemma 3.4.2. By density of C ∞ c (Ω) in L q (Ω) and the isometry between L p (Ω) and the dual space of L q (Ω), we obtain our thesis.
3.5 Asymptotic behaviour at the boundary 3.5.1 Right-hand side blowing up at the boundary: proof of (16) In this paragraph we study the boundary behaviour of the solution u to the problem and by using (14), up to multiplicative constants, Set ε := δ(x) and x = (1 − ε)e 1 , θ := y |y| , r := δ(y) and y = (1 − r)θ, and rewrite Split the angular variable in θ = (θ 1 , θ ) where θ 1 = θ, e 1 and θ 2 1 + |θ | 2 = 1: then, for a general F , and we write From now on we will drop all multiplicative constants and all inequalities will have to be interpreted to hold up to constants. Let us apply a first change of variables where σ = 1 − θ 1 . We compute now the integral in the variable σ. Let σ be defined by equality and let σ * = max{σ , 0}.
The quantity σ * equals 0 if and only if and it is easy to verify that this happens whenever Remark 3.5.1. t 2 (ε) < 1 1−ε , since for small ε > 0 which is true for any positive ε < 1. Hence We split now integral (40) into four pieces, as following and we treat each of them separately: • for the first one we have and therefore the first integral is less than note now that t 2 (ε) − t 1 (ε) ∼ ε and 1 1−ε − 1 ∼ ε and thus (42) is of magnitude ε −β+2s ; • for the second integral we have and therefore the second integral is less than and (43) is of magnitude ε −β+2s ; • for the third integral we have ε −β+2s for s < β < 1 + s; • for the fourth integral we have so that the fourth integral is less than and (45) is of magnitude Resuming the information collected so far, what we have gained is that, up to constants, This establish an upper bound for the solutions. Note now that the integral (40) works also as a lower bound, of course up to constants. Using the split expression (41) we entail where we have used only the expression with the third integral in (41). We claim that where the inequality is intended to hold up to constants. In case (47) holds and β = s we have The integral on the second line is a bounded quantity as ε ↓ 0 since β < 1 + s. Now, we are left with We still have to prove (47): note that an integration by parts yields, for n ≥ 4, so that we can show (47) only in dimensions n = 2, 3 and deduce the same conclusions for any other value of n by integrating by parts a suitable number of times. For n = 2 which completely proves our claim (47).
Then, using the Fubini's Theorem, we write Split the integration on ∂Ω into the integration on Γ := {θ ∈ ∂Ω : |θ − θ * | < δ 1 } and ∂Ω \ Γ, and choose δ 1 > 0 small enough to have a C 1,1 diffeomorphism We build now where we suppose e 1 ∈ Γ and φ(e 1 ) = θ * . With this change of variables (48) becomes which, since |Dϕ(ω)| is a bounded continuous quantity far from 0, is bounded below and above in terms of i.e. we are brought back to the spherical case.

Boundary continuity of s-harmonic functions
Consider g : CΩ → R and and think of letting x → θ ∈ ∂Ω. Suppose that for any small ε > 0 there exists δ > 0 such that |g(y) − g(θ)| < ε for any y ∈ CΩ ∩ B δ (θ). Then The first addend satisfies For the second one we exploit (13): and by arbitrarily choosing ε, we conclude lim x→θ u(x) = g(θ). (17) We study here the rate of divergence of

Explosion rate of large s-harmonic functions: proof of
on ∂Ω in case g explodes at ∂Ω.
Remark 3.5.2. The asymptotic behaviour of u depends only on the values of g near the boundary, since we can split g = gχ {d<η} + gχ {d≥η} and the second addend has a null contribution on the boundary, in view of Paragraph 3.5.2. Therefore in our computations we will suppose that g(y) = 0 for δ(y) > η.
In the further assumption that g explodes like a power, i.e. there exist η, k, K > 0 for which (the choice σ < 1−s is in order to have (32), see (13) above) our proof doesn't require heavy computations and it is as follows.
Dropping multiplicative constants in inequalities and for Ω η = {y ∈ CΩ : δ(y) < η}: Similarly one can treat also the lower bound: The limit we have computed above is the continuity up to the boundary of u solution of Denote now by ← − g (r) = sup δ(x)=r g(x). Splitting the integral in the θ variable into two integrals in the variables (θ 1 , θ ) where θ 2 1 + |θ | 2 = |θ| 2 = 1, up to constants we obtain Define M := 1+r 1−ε > 1 and look at the inner integral: The integral from −1 to 0 contributes by a bounded quantity so that we are left with Our claim now is that this last expression is controlled by g(ε) as ε ↓ 0. Since g is exploding in 0, for small ε it is g(τ ε) ≤ g(ε) for τ > 1 and For the other integral To compute the limit as ε ↓ 0 we use a Taylor expansion: where we have denoted by G(ε) = ε −s g(s). Thus .

Indeed
which is guaranteed by the fact that G is integrable in a neighbourhood of 0. These computations show that, in the case of the ball, the explosion rate of the s-harmonic function induced by a large boundary datum is the almost the same as the rate of the datum itself.
Note now that up to (49) the same computations provide a lower estimate for u if we substitute g with g(r) = inf δ(x)=r g(x). Then where the last inequality is (47). Finally we need only to repeat the above computations replacing g with g and other minor modifications.
In the case of a general smooth domain, we can reduce to the spherical case as we did to conclude Paragraph 3.5.1.

.10
The proof is a simple readaptation of the result by Clément and Sweers [10].
Existence. We can reduce the problem to homogeneous boundary condition, indeed by considering the solution of we can think of solving the problem therefore from now on we will suppose g ≡ 0. Note also that since v is continuous and bounded then ( for every x ∈ Ω, u ∈ R : the function F (x, u) is continuous and bounded on Ω × R, by hypothesis f.1). We can write a solution of as a fixed-point of the map obtained as the composition The first map sends L ∞ (Ω) in a bounded subset of L ∞ (Ω), by continuity of F and boundedness of u, u.
The second map is compact since w ∈ C s (R n ), thanks to the results in [28,Proposition 1.1]. Then the composition admits a fixed point in view of the Shauder Fixed Point Theorem.
Note that a solution to the original problem lying between u and u in Ω, is also a solution of ( * ). Moreover, any solution of ( * ) is between u and u. Indeed consider A := {x ∈ Ω : u(x) > u(x)}, which is open by the continuity of u and u. For any ψ ∈ C ∞ c (A), ψ ≥ 0, with the corresponding φ ∈ T (A): which implies u ≤ u in A, by positivity of ψ, proving A = ∅.
Uniqueness. If we have two continuous solutions u and w Defining Ω 1 = {x ∈ Ω : w(x) < u(x)}, thanks to the monotony of f ,

Proof of Theorem 1.2.11
In the case of negative right-hand side, Theorem 1.2.11 follows from Theorem 1.2.12. So, assume the right-hand side is positive and consider We look for a suitable shape g of v outside Ω and exploding at ∂Ω: the large s-harmonic function v 0 induced by g in Ω will be a subsolution of our equation, and in particular will imply that the blow-up condition at ∂Ω is fulfilled. Then, in order to prove the existence part, we need a supersolution.
Consider F : R → R continuous, increasing and such that F (t) ≥ f (x, t) for any t ≥ 0: for example, where c = c(n, s, Ω) is the constant of equation (17) giving the upper control of large s-harmonic functions in terms of the boundary datum (see Paragraph 3.5.3) and and Such g satisfies hypothesis (32), since when δ(x) is small Denote by w := v − v 0 : our claim is that problem admits a solution w. Indeed, we have a subsolution which is the function constant to 0 in Ω and χ Ω turns out to be a supersolution. To show this we consider the problem

Damping term: proof of Theorem 1.2.12
For any N ∈ N, denote by g N = min{g, N }. Also, with the notation of equation (28), for a small parameter r > 0 denote by f r (y) = f r (ρ, θ) = h(θ) ϕ(ρ/r) K r , K r = 1 1 + s Ωr ϕ(δ(y)/r) δ(y) s dy, and recall that this is an approximation of the h boundary datum. Finally call u N,r the minimal solution of provided by Theorem 1.2.10. Note that for any r > 0, the sequence {u N,r } N ∈N we obtain is increasing in N : indeed, u N +1,r is a supersolution for the problem defining u N,r , since it has larger boundary values and the minimality property on u N,r gives u N,r ≤ u N +1,r . Moreover, {u N,r } N ∈N is bounded by the function u 0 r associated with the linear problem with data g and f r , i.e.
Therefore u N,r admits a pointwise limit in R n . Call u r this limit: obviously u r = g in CΩ. Take any then where we have used the Fatou lemma and the continuity of the map t → f (x, t). This means that u r is a subsolution. We are left to prove that u r is also a supersolution. Call Ω = suppψ ⊂⊂ Ω and build a sequence {Ω k } k∈N such that Ω ⊆ Ω k ⊆ Ω and Ω k Ω. Since ψ ∈ C ∞ c (Ω k ) for any k, then the we can build the sequence of functions φ k ∈ T (Ω k ) induced by ψ: this sequence is increasing and converges pointwisely to φ. Moreover, for any k, since (−∆) letting both sides of the inequality pass to the limit as k ↑ +∞ we obtain . This means that u r is both a sub-and a supersolution and it solves since, in this case, it is obviously f r ≡ 0.
We want now to push r ↓ 0, under the additional assumption We claim that the family {u r } r is uniformly bounded and equicontinuous on every compact subset of Ω: there exist then a u ∈ L 1 (Ω) (since |u r (x)| ≤ Cδ(x) s−1 for a C independent on r) and a sequence {r k } k∈N such that r k → 0 as k → +∞ and u r k → u a.e. in Ω and uniformly on compact subsets. Then for any φ ∈ T (Ω) and p(s − 1) + s > −1 by hypothesis. We still have to prove that our claim is true. The uniform boundedness on compact subsets of {u r } r is a consequence of inequalities and the convergence of v r to u 0 is uniform in compact subsets of Ω. Then implies the equicontinuity.
Note that this proof exploits the negativity of the right-hand side only in considering the s-harmonic function induced by g and h as a supersolution of problem With minor modifications to the proof we can state Lemma 4.3.2. Let f : Ω × R → [0, +∞) be a function satisfying f.1) and f.2). Let g : CΩ → R + be a measurable function satisfying (11) and h ∈ C(∂Ω), h ≥ 0. Assume the nonlinear problem admits a subsolution u ∈ L 1 (Ω) and a supersolution u ∈ L 1 (Ω). Assume also u ≤ u in Ω. Then the above nonlinear problem has a weak solution u ∈ L 1 (Ω) satisfying Proof. Replace, in the above proof, the function u 0 with the supersolution u.

Sublinear nonnegative nonlinearity: proof of Theorem 1.2.13
We first prove a Lemma which will make the proof easily go through.
is solvable.
Proof. We can equivalently solve the integral equation where u 0 is the s-harmonic function induced by g and h in Ω. Define the map K : The condition g ≥ m in CΩ entails u 0 ≥ m in Ω; also, for any w ∈ L 1 (Ω), w ≥ 0 implies Kw ≥ u 0 ≥ m, therefore K sends the subset D m := {w ∈ L 1 (Ω) : w ≥ m} of L 1 (Ω) into itself. Moreover, for u, v ∈ D m i.e. K is a contraction on D m , and K has a fixed point in D m .
In general, for the problem we have a subsolution which is the s-harmonic function satisfying the boundary conditions. But we are now able to provide a supersolution: this can be done by setting g m = max{g, m} and by solving, for some large value of m, It is sufficient to apply the classical iteration scheme starting from the s-harmonic function u 0 and with iteration step In such a way we build an increasing sequence {u k } k∈N ⊆ L 1 (Ω) which is uniformly bounded from above by u. Indeed, on the one hand we have that

Superlinear nonnegative nonlinearity: proof of Theorem 1.2.14
We give the proof for problem while for the other one it is sufficient to replace β with 1 − s and repeat the same computations.
To treat the case of a general nonlinearity we use again the equivalent integral equation u(x) = u 0 (x) + λ Ω G Ω (x, y) f (y, u(y)) dy, where u 0 is the s-harmonic function induced in Ω by the boundary data. In this case the computations in Section 3.5 on the rate of explosion at the boundary turn out to be very useful. Indeed on the one hand we have that u 0 inherits its explosion from the boundary data g and h: briefly, in our case Since u 0 is a subsolution, our first goal is to build a supersolution and we build it of the form in Ω, γ > 0 ζ = 0 in CΩ, Eζ = 0 on ∂Ω.
This is an admissible choice for γ provided γ < 1 + s, i.e. only if p > 1 + s 1 − s ; in case p doesn't satisfy this lower bound then and we are in the previous case. Finally, if qβ > 1 + s then a solution u should satisfy, whenever δ(x) < 1, f (x, u(x)) ≥ bu(x) q ≥ bu 0 (x) q ≥ cδ(x) −qβ which would imply Ω G Ω (x, y) f (y, u(y)) dy = +∞, x ∈ Ω, which means that the problem is not solvable. 8 for p < 1 we are actually in the case of the previous paragraph A Proof of Proposition 1.
Since (−∆) s v ∈ C(R n \ Ω), see [29, Proposition 2.4], we infer that We give now a pointwise estimate on (−∆) s α k (x), x ∈ B 1/k . Since α k ∈ C ∞ c (R n ), we can write (see Using (53), the L 1 -norm of (−∆) s v k can be estimated by Cδ(y) s k n+2s ∆α L ∞ (R n ) dy dx + so that, by the Fatou's Lemma we have Note now that − − → 0, and this concludes the proof.

B The Liouville theorem for s-harmonic functions
Following the proof of the classic Liouville Theorem due to Nelson [25], it is possible to prove the analogous result for the fractional Laplacian.
Theorem B.0.1. Let u : R n → R be a function which is s-harmonic throughout R n . Then, if u is bounded in R n , it is constant.