A fractional eigenvalue problem in $\mathbb{R}^N$

We prove that a linear fractional operator with an asymptotically constant lower order term in the whole space admits eigenvalues.


Introduction
It is well known that the spectrum of the Laplace operator in R N is purely continuous, that is σ(−∆) = [0, ∞). On the other hand, considering operators of the form where g : R N → R satisfies suitable growth conditions, one may hope to apply the standard approach in Hilbert spaces and prove the existence of a principal eigenvalue (see [4] and [6], [9], [12], [13] for examples in a Banach setting) or of a diverging sequence of eigenvalues (see [1], or [17] when g(x) → 0 at infinity).
In the recent paper [10], the authors consider operators of the form −∆u + βg(x)u, where g ≈ 1 at infinity, and study the associated spectrum. Of course, in this case the situation is different, since no compactness argument can be invoked, but they prove that eigenvalues do exist. Inspired by their result, we consider a related situation for the following eigenvalue fractional Laplacian problem in R N : Here N > 2s, s ∈ (0, 1), β > 0 is a parameter and (−∆) s denotes the fractional Laplacian defined through the Fourier transform in the following way: for any f ∈ S (R N ) with Fourier transform F f =f , we define, modulo a positive multiplicative constant depending on N and s, see [19]. When u is sufficiently regular, a pointwise expression of the fractional Laplacian is also available, namely for some C N,s > 0. Here, we will not go into further details about the functional setting of the problem, postponing these aspects to Section 2. We only remark that we look for couples (λ, u) with λ ∈ R and u = 0 which satisfy (1). As in [10], we assume to deal with a function g which is not constant and tends to 1 at infinity. More precisely, denoting by m(S) the measure of a set S ⊂ R N , we make the following assumptions on g: Although the assumptions on g are rather weak, we show that also in this case the operator admits eigenvalues, which are all below β, see Theorem 2.2 below. We emphasize the fact that, in spite of this difference with the case of −∆ in bounded domains, we prove the existence of a first eigenfunction which preserves the same nice properties of the first eigenfunction of −∆ in bounded domains. More precisely, we prove that: • the first eigenvalue λ 1 is simple, • the associated eigenfunction is strictly positive in R N , • any eigenfunction associated to any other eigenvalue is nodal, i.e. signchanging, • there exists a different minimax characterization of the second eigenvalue. Of course, as usual when eigenvalues are found via the Krasnoselskii genus, as we do, we don't know if other eigenvalues can show up.
Completing the previous description, in analogy with the case of −∆ in R N , we observe the following property, which is well-known, and which we prove using an elegant Pohozaev identity.
Remark 1. Of course, we reduce to problem (3) when g = 1 in R N , and this shows that the condition "m x ∈ R N : g(x) < 1 > 0" in (g 1 ) is necessary and sufficient to get eigenfunctions for L β .
The spectral properties that we prove here are the starting point of further investigations about existence results for nonlinear fractional problems in the whole of R N , see [3].

Functional setting and eigenvalues
As usual, we are interested in functions satisfying (1) in a weak sense. For this, let us recall some definitions: for s ∈ (0, 1) we denote by H s (R N ) the Sobolev space of fractional order s defined as denotes the Gagliardo seminorm of u. Moreover, we set · 2 = · L 2 (R N ) . Now, by the Plancharel Theorem, u 2 = û 2 , and by (2) for every u ∈ H s (R N ). As a consequence, β being positive, H s (R N ) can be endowed with the norm for all u ∈ H s (R N ), see [5].
On the other hand, by [8, Lemma 3.1], for every u ∈ H s (R N ) and some positive constant c N,s . Then, by (4) and (5), after setting c N,s = 1, we immediately get In light of the previous considerations, we are now ready to give the following Of course, in view of (6), the previous identity can be written as In order to state our main result, let us introduce some notions. First, we set Then, for every k ∈ N, we introduce the families of sets where γ(A) denotes the Krasnoselskii genus of a symmetric set A, defined as see [16]. Associatd to Σ k we set Finally, we define Now, we are ready to state our main result: Then (1) {λ k } k is a nondecreasing sequence and λ k ≤ β for all k ∈ N.
(2) If λ k < β, then λ k is an eigenvalue of the operator L β , namely there exist an eigenfunction u k ∈ Σ k such that for any ϕ ∈ H s (R N ) (3) If λ 1 is an eigenvalue, then it is simple and the associated eigenfunction e 1 is strictly positive. Moreover, eigenfunctions associated to eigenvalues different from λ 1 are nodal, i.e. they change sign. (4) If Γ k < β, then λ k < β, and hence λ k is an eigenvalue.
. ., and from the very definition of λ k we have that {λ k } k is a nondecreasing sequence. Now, let us take A ∈ Σ k , and Since, from Lebesgue's dominated convergence theorem, Let us notice that such a limit is uniform in u ∈ A, since A ∈ Σ k is compact. Moreover, also A t ∈ Σ k for every t > 0, so that λ k ≤ sup v∈At Φ(v), and, as a consequence, and the claim holds.
(2) Let us suppose that λ k < β. If Φ| Σ satisfies the (P S) condition 1 , then λ k is a critical value of Φ| Σ (see [2, Theorem 10.9]), and there exists u ∈ Σ k such that for all ϕ ∈ H s (R N ). Now, suppose that (P S) does not hold. By the Ekeland Principle (see [20,Theorem 8.5]) we can find a (P S) sequence for Φ| Σ k at level λ k : this means that there exist a sequence {u n } n ⊂ Σ k and a sequence {µ n } n ⊂ R such that By (10), it is clear that {u n } n is bounded in H s (R N ). Thus, taking ϕ = u n in (11), we obtain This limit, together with (10) and the fact that {u n } n ⊂ Σ k , implies that lim n→∞ µ n = λ k .
Since H s (R N ) is a Hilbert space, we can assume that N −2s is the critical fractional Sobolev exponent (see [15]). Now, fixed ϕ ∈ C ∞ 0 (R N ). From (12) we have

Thus (11) implies
for all ϕ ∈ C ∞ 0 (R N ), and, by density, for all ϕ ∈ H s (R N ). This shows that u solves (1) with λ = λ k . Now, we show that u ≡ 0. By the Concentration-Compactness Principle, either there exist R > 0 and ν > 0 such that (up to subsequences)  (14) holds. By (g 1 ) we have that for all ǫ > 0 there exist R ǫ > 0 such that 0 ≤ 1 − g(x) < ǫ if |x| > R ǫ , and from (14) 0 From (15) we immediately get Choosing ϕ = u n as test function in (11) and passing to the limit, we obtain β = λ k − c for some constant c ≥ 0. Hence, β ≤ λ k , which contradicts our assumption. Summing up, we have shown that λ k is an eigenvalue with associated eigenfunction u ≡ 0.
(3) First, let us notice that we can always find a positive eigenfunction associated to λ 1 . Indeed, for all u ∈ H s (R N ), by the triangle inequality we have Thus, if e 1 is an eigenfunction associated to λ 1 , and without loss of generality we can assume |e 1 | p = 1, we have that so that also |e 1 | ≥ 0 is an eigenfunction, as well. Thus, |e 1 | is a solution of (7), and by the regularity result of [7, Theorem 3.4], we get that |e 1 | ∈ C 0,µ (R N ) for some µ ∈ (0, 1). Moreover, thanks to the Harnack's inequality in [18,Theorem 3.1], we conclude that |e 1 | is strictly positive in R N . This implies that all signed solutions of (7) with k = 1 are strictly positive (or negative) in the whole of R N . Now let u be another eigenfunction associated to λ 1 . Then, for all R > 0 there exists Since e 1 − χ R u can be assumed to be a signed eigenfunction associated to λ 1 , then the previous identity implies that e 1 −χ R u = 0. Now, if R 1 < R 2 , we find e 1 = χ R1 u in B R1 and e 1 = χ R2 u in B R2 . Hence, χ R2 = χ R1 = χ. This implies that e 1 = χu in the whole space R N , i.e. the first eigenvalue is simple.
Finally, let us suppose that φ is an eigenfunction associated to an eigenvalue λ > λ 1 , and let us suppose that φ is signed, for example φ is nonnegative, that is φ − := max{0, −φ} ≡ 0. We know that if v, w are eigenfunctions associated to eigenvalues µ = ν, then so that, using the fact that < L β v, w >=< L β w, v >, we get In particular, we have which is absurd, because both e 1 and φ are nonnegative and non zero. Then φ − ≡ 0. A similar argument can be used to prove that φ + ≡ 0. In conclusion, φ is sign changing.
(4) Let us suppose that Γ k < β. By definition of Γ k , there exist a compact and symmetric set A ∈ M k such that

Let us define
Of course, A * ⊂ Σ and A * ∈ Σ k , in fact A * is compact and γ(A * ) ≥ k. Now, for any u * ∈ A * , we have and thus λ k ≤ sup We conclude this section with the Proof of Proposition 1. Suppose u ∈ H s (R N ) is a solution of (3), then u satisfies the Pohozaev identity (see [14,Proposition 4.1]). By (5), we immediately get If u ≡ 0 and µ = 0, then that is 2s = 0, which is absurd. If µ = 0, we get [u] H s (R N ) = 0, so thatû ≡ 0, and so u ≡ 0.
2.1. Another characterization for λ 2 . For further investigations where the spectrum of L β plays a rôle, we think it is useful to provide another characterization of the second eigenvalue λ 2 in terms of minimax values: Proposition 2. Let us set .
Proof. Take h ∈ Σ * and define the maph : Set A =h(S 1 ); then A ∈ Σ 2 . In fact, A is compact, becauseh is continuous; moreover, A is symmetric, forh is odd; finally, γ(A) ≥ 2: if we suppose by contradiction that γ(A) = 1, there should exist an odd function H ∈ C 0 (A, R \ {0}), but A is a connected set, being the image of a connected set through a continuous function, while H (A) is disconnected, and a contradiction arises. So, by definition of λ 2 , we have that and hence By Theorem 2.2.(1), we know that λ 2 ≤ β, and using arguments similar to those used therein, we can prove that λ * ≤ β. Thus, if λ 2 = β, then λ 2 = λ * = β.