A POSITIVE BOUND STATE FOR AN ASYMPTOTICALLY LINEAR OR SUPERLINEAR SCHR¨ODINGER EQUATION IN EXTERIOR DOMAINS

. We establish the existence of a positive solution for semilinear elliptic equation in exterior domains where N ≥ 2, Ω is an open subset of R N and R N \ Ω is bounded and not empty but there is no restriction on its size, nor any symmetry assumption. The nonlinear term f is a non homogeneous, asymptotically linear or superlinear function at inﬁnity. Moreover, the potential V is a positive function, not necessarily symmetric. The existence of a solution is established in situations where this problem does not have a ground state.


1.
Introduction. Our goal in this paper is to show the existence of a positive bound state solution for problem (P V ) when a ground state can not be obtained. Using a new approach recently developed byÉvéquoz and Weth [20], Clapp and Maia [14] and Maia and Pellacci [25] a positive solution is found, extending the existence results obtained in the celebrated papers of Benci and Cerami [5] and Bahri and Lions [3], for general non homogeneous nonlinearities, either superlinear or asymptotically linear at infinity in an exterior domain.
The study of solitary waves of nonlinear Schrodinger equations or of nonlinear Klein-Gordon equations is modeled by (P V ) with Ω = R N . Likewise, exterior boundary-value problems may be associated with models of steady-state flows in fluid dynamics (see [21]) and electrostatic problem of capacitors ( see [16],Volume 1, Chapter II), for instance.
The primary works applying variational methods to find solutions of problems like (P V ) report to the 80's and 90's with the articles of Benci and Cerami [5] and Bahri and Lions [3]. The method applied in both works was finding critical points of a functional constrained on a manifold and absorbing a Lagrange multiplier by the homogeneity of the nonlinear term f (u) = |u| p−2 u where p ∈ (2, 2 * ) and 2 * = 2N N −2 if N ≥ 3, 2 * = ∞ if N = 2 in order to obtain a positive solution of the Euler equation in (P V ).
One of the main challenges of trying to apply the usual variational method when Ω is an unbounded domain is the lack of compactness of the Sobolev embeddings. In order to circumvent this difficulty, a deeper study of the obstruction for compactness was performed by Benci and Cerami in [5] and a clever description was obtained of what happens when a Palais-Smale sequence does not converge to its weak limit (for details see [11] and references therein). Problem (P V ) with f (u) = |u| p−2 u, p ∈ (2, 2 * ) was solved in the case that the ground state does not exist first in [5] in the autonomous case V (x) = λ a positive constant, proving the existence of a positive solution with some restriction on the size of the hole R N \ Ω, and posteriorly that condition was eliminated in [3] and existence was proved for potentials V which decay to a constant potential V ∞ at infinity. In the same spirit, this problem has been extensively studied if Ω is an exterior domain for power non-linearity f (u) = |u| p−2 u in recent years (see [3]). If the non-linear term f is not a pure power with respect to u, there are few contributions in the literature. In particular, the existence of solution is proved in [13], using topological methods, in the case that f is super-linear and depends on the spatial variable but the asymptotic nonlinearity f ∞ , of the autonomous problem, must satisfy a convexity assumption.
In the case Ω is spherically symmetric about some point, benefiting from the strength of the symmetry property, this problem can be solved on H 1 rad (Ω) (subspace of radial functions in R N ) which embeds compactly in L p (Ω), if p ∈ (2, 2 * ). This idea was exploited by Berestycki and Lions in [6], Coffman and Marcus in [15] and Esteban and Lions in [19] when Ω is the complement of a ball. However, symmetry of Ω does not help if we don't have radial symmetry in V (x). This is the case in our problem (P V ) where we do not assume any symmetry, neither in Ω nor in V (x).
In the past five decades a different approach has been successfully applied in order to obtain solutions for this class of problems with no symmetry assumption. The so-called Nehari method, [27] and [28], which consists of finding solutions of (P V ) which are critical points of a functional associated with the equation in (P V ), restricted to the Nehari manifold. This method has been extensively used in the last years in order to find ground state solutions as well as sign changing solutions of nonlinear elliptic problems in R N and exterior domains (see [17,24,29] and references therein). When finding a solution which is a minimum of the functional restricted to the Nehari manifold, the Lagrange multiplier is proved to be zero, yielding that the constrained critical point is in fact a free critical point of the functional, and that the manifold is a natural constraint. This allows to solve the problem for non-homogeneous nonlinearties because the multiplier does not have to be absorbed in the construction of a solution for the equation. Most importantly, this approach enables to avoid the use of a technical algebraic inequality (a + b) p ≥ a p +b p +(p−1)(a p−1 b+ab p−1 ) largely applied in the case f (u) = |u| p−2 u ( [2,3,12]). We follow these ideas, closely related to the arguments found in [14] and [25], for general nonlinearities f which satisfy the assumption that f (s)/s is increasing. In this setting, not all functions u = 0 are projectable on the Nehari manifold , however the class of functions which are good for projections in this environment is enough to pursue the argument.
Our main contribution in this paper is extending the result of Bahri and Lions [3] for non-homogeneous f , with no symmetry assumption on V or Ω. Note that here Ω is an general open set in R N , with a bounded complement set, which might included several holes in a ball or just a point set {0}. Moreover, we allow the nonlinear f to be a less smooth function just in C 1 , improving the hypotheses in [14] and [25] where it was considered in C 3 for technical reasons (see Lemma 3.3 in [14]). Likewise, the work of [20] provided some useful tools and insight for estimates, even though their problem is for super-linear f in the whole R N and uses the generalized Nehari manifold.
To our knowledge the results we present here are new and extend the previous works in the literature for a class of problems in exterior domains. We consider the elliptic problem , Ω is open and R N \ Ω ⊆ B K (0) the ball of radius K and center at the origin in R N , in fact R N \ Ω is bounded, u ∈ H 1 0 (Ω) and V is a potential satisfying the conditions: The conditions that we consider on the nonlinearity f are the following: is unique, up to translations.
s s > 0 for s > 0. It is straightforward to verify that the superlinear model nonlinearity f (s) = s p , s > 0 with p ∈ (1, 2 * − 1), and the asymptotically linear model nonlinearity

Remark 3.
The assumption (U') ψ(s) := −V∞s+f (s) sf (s)−f (s) is non decreasing in s ∈ (σ, +∞) where σ is the unique positive number such that f (σ) σ = V ∞ , guarantees that the positive solution to the problem (P ∞ ) is unique (see [26], theorem 1 or 2792 ALIREZA KHATIB AND LILIANE A. MAIA [30], theorem 1). It may be replaced by any other assumption which guarantees the uniqueness of positive ground state solution.
The main result of this work is the following The paper is organized as follows. In section 2, we formulate the variational setting and present some preliminary results. Section 3 is dedicated to compactness condition. In section 4, applying a topological argument, which involves the barycenter map, we show that I V has a positive critical value.

2.
Variational setting and exponential decay estimate. Note that by Remark 2, f (s) > 0 for s > 0, and we shall consider the extended f (s) := −f (−s) for s < 0, so without loss of generality we may suppose that f is odd and establish the existence of positive solution for it, which in particular will be a positive solution of the problem with the original f . We will use the following notation: Our assumptions on V imply that . Ω is a norm in H 1 0 (Ω) which is equivalent to the standard one.We write and since V ∞ > 0, then . is a norm in H 1 (R N ) which is equivalent to the standard one. If u ∈ H 1 0 (Ω) we may define u = 0 in R N \ Ω and in fact H 1 0 (Ω) ⊂ H 1 (R N ) (see [9], Proposition 9.18).
The solutions of problem (P V ) are critical points of the functional Also we denote in the same way I ∞ , J ∞ , N ∞ and c ∞ for the definition with the constant V ∞ in place of V (x). Let w be the unique positive radial solution of (P ∞ ), see [6,7,30]. It is well known, see [22] that there are positive constants C 1 and C 2 , Hereafter C will denote a positive constant, not necessarily the same one. The following lemma gives informations about the Nehari manifold N V which are, by now, standard (see [14] Lemma 2.1). We include them here for the sake of completeness.
Lemma 2.1. (a)There exists > 0 such that u Ω ≥ for every u ∈ N V . (b)N V is a closed C 1 -submanifold of H 1 0 (Ω) and natural constraint for I V . (c)If u ∈ N V , the function t → I V (tu) is strictly increasing in (0, 1] and strictly decreasing in (1, ∞). In particular, Proof. (a) Property (f 2 ) and the Sobolev embedding theorem imply that If u ∈ N V then J V (u) = 0 and so C u p 2 +1 u 2 > 1 and as for every u ∈ N V . This implies that 0 is a regular value of J V : . It also implies that u is not on the tangent space of N V at u and, therefore, that N V is a natural constraint for I V . ( By property (f 5 ) we have that f (u) u is strictly increasing for u ∈ (0, ∞) and strictly decreasing for u ∈ (−∞, 0). Therefore d dt I V (tu) > 0 if t ∈ (0, 1) and d dt I V (tu) > 0 if t ∈ (1, ∞). This proves (c).
The next lemma is the key that allows us to take the non-linear f a less smooth function in C 1 , which improved the hypotheses in [14] and [25] where it was considered in C 3 .
Proof. The inequality is obviously satisfied if u = 0 or v = 0. By (f 5 ), f is increasing, which yields Moreover by (f 2 ) for every 0 < ν < p 1 − 1 we have and thenC ρ := sup 0<u≤ρ By the symmetry in u and v, the same estimate holds for 0 < u ≤ v, and the proof is complete.
Now we present a sequence of lemmas which will be used to show Theorem 2.10, that states that N V is not empty. In fact, it shows that some interesting convex combination of two copies of translated solution of the limit problem are projected into N V . As before, C will always denote a positive constant, not necessarily the same one.
The second inequality is obtained in a similar way.
Proof. The function φ is continuous and As w is a solution of problem (P ∞ ) we have that There are two cases to study: and if λ = 0, lim r→∞ φ(1, r) = −∞ and this case is settled.
Now, let y 0 ∈ R N with |y 0 | = 1 and B 2 (y 0 ) := {x ∈ R N : |x − y 0 | ≤ 2}. We write for each y ∈ ∂B 2 (y 0 ) and R > 0, Proof. In order to prove the first estimate let 2K < 1 2 R, so that Now by (2) and (3) we have The proofs of the other estimates are similar.
Now by (2), (4) and Lemma 2.3 we have Similarly we can prove the second estimate. and Proof. In order to prove the first inequality, it follows from (V 2 ) and estimative (2) that > 0, so we can write (8) as As On the other hand, |x + Ry 0 | ≥ R|y 0 | − σR = (1 − σ)R for x in B σR (0) and by making a change of variables, we have The proof of the first inequality is complete. Similarly we can prove the second estimate.
Finally, in order to prove (7) we may repeat the above argument for B σR (Ry) ∪ B σR (Ry 0 ) rather than B σR (Ry 0 ) . Note that |x − Ry 0 |, |x − Ry| > σR in Γ := R N \{B σR (Ry) ∪ B σR (Ry 0 )}; performing a change of variables and applying Lemma 2.3 In what follows we exploit the ideas of Bahri and Li in [2] by working with a convex combination of two translated copies of w, the ground state solution of (P ∞ ) (see also [14], [20] and [25]). Define where ψ ∈ C ∞ (R N ) is continuous radially symmetric and increasing cut-off function Note that here as before K is the radius of the sphere B K (0) which contains R N \Ω. We can consider U R λ,y ∈ H 1 (R N ) by extending U R λ,y ≡ 0 outside Ω. Proof. First of all, for R sufficiently large we claim that therefore and by the claim we have and this shows the lemma. Now, in order to complete the proof we have to show the claim. To obtain the first estimate (12) we use Lemma 2.5 with q = 2, To prove the second estimate (13) we have ψ ∈ C ∞ , then there exists positive constants C 3 and C 4 such that and so by Lemma 2.5 with q = 2, √ V∞R as claimed. The proof of (14) and (15) are similar.
By Lemma 2.8 the first part of (17) is equal to o R (1), then it is enough to show that .
By (f 2 ), Lemma 2.5 and the inequality ( Our assumptions do not guarantee that every u ∈ H 1 0 (Ω)\{0} admits a projection onto N V . However, the following lemma says that U R λ,y does admit a projection onto N V if R is sufficiently large.
Theorem 2.10. There exist R 0 > 0, T 0 > 2 such that for each R ≥ R 0 , y ∈ ∂B 2 (y 0 ) and λ ∈ [0, 1], there exists a unique T R λ,y such that T R λ,y U R λ,y ∈ N V , T R λ,y ∈ [0, T 0 ] and T R λ,y is a continuous function of the variables λ, y and R. In particular for λ = 1/2 we have T R 1 2 ,y → 2 as R → ∞ uniformly in y ∈ ∂B 2 (y 0 ). Proof. First note that, for each u ∈ H 1 0 (Ω), u > 0, property (f 5 ) implies that is strictly decreasing in r ∈ (0, ∞). Therefore, if there exists r u ∈ (0, ∞) such that J V (r u u) = 0, this number will be unique. Observe also that J V (ru) > 0 for r small enough. Next, we will show that, for R large enough and some T 0 > 0, This implies that there exists T R λ,y ∈ [0, T 0 ) such that J V (T R λ,y U R λ,y ) = 0, i.e. T R λ,y U R λ,y ∈ N V . Let us prove (18). For u, v ∈ H 1 (R N ), u, v > 0, and r ∈ (0, ∞), by using (f 5 ) we have

ALIREZA KHATIB AND LILIANE A. MAIA
Setting u := λw R 0 and v := (1−λ)w R y , performing a change of variable and applying Lemma 2.4 we conclude that where o R (1) → 0 as R → ∞, uniformly in y ∈ ∂B 2 (y 0 ) and λ ∈ [0, 1]. Also S 0 < 0 as in Lemma 2.4. Now since and by Lemma 2.7 and Lemma 2.9 Hence, there exists R 0 > 0 such that This proves (18), and so we have shown that also by the symmetry of ϕ(u, v) in u and v, we get similarly and from the two previous estimates we have Now, by Lemma 2.6 and (19) we can write 1) since w is a solution of (P ∞ ). So, by Lemma 2.9 we have Therefore, by (20), Lemma 2.7 and and the lemma is proved.
3. Compactness results. Our goal in this section is to investigate the lack of compactness due to the fact that the problem is in an unbounded domain Ω. In order to do so, we present a sequence of lemmas that will culminate in Corollary 1, which presents a range of levels where I V satisfies the Palais-Smale condition. Also note that the uniqueness of the positive solution of the limit problem is going to play a role in the arguments.
First of all note that d ≥ 0, since Now fix D > d. Assume, by contradiction, that u k → ∞ and set v k := t k u k with t k = 2 √ D u k . By Lemma 2.1 (c), for k large enough we have that By using hypothesis (f 2 ), we get that Hence, v = 0 and there exists a subset Λ of positive measure in B 1 (0) such that v(x) = 0 for every x ∈ Λ. It follows that |ũ k (x)| → ∞ for every x ∈ Λ. Property So, from property (f 3 ) and Fatou's lemma, we conclude that This is a contradiction.
Proof. Let u k ∈ N V be such that I V (u k ) → c V . By Lemma 3.1, after passing to a subsequence, we have that (u k ) is bounded in H 1 0 (Ω). From Lemma 2.1(a) and by property (f 2 ) we obtain This inequality, together with Lions lemma (considering the extention of u k to H 1 (R N )), implies that there exist δ > 0 and a sequence (y k ) in R N such that Setũ k (x) := u k (x + y k ) After passing to a subsequenceũ k u weakly in H 1 (R N ), u k → u in L 2 loc (R N ) andũ k (x) → u(x) a.e. in R N . Therefore, Hence, u = 0 and there exists a subset Λ of positive measure in B 1 (0) such that u(x) = 0 for every x ∈ Λ. Property (f 5 ) implies that 1 2 So, from the Fatou's lemma, we conclude that as claimed. By repeating this argument we obtain c ∞ > 0.
Note that if c V is attained, then problem (P V ) has a non-zero ground state solution from this previous lemma. Proof. If u is a solution of P V then where u + := max{u, 0} and u − := min{u, 0} and so u ± ∈ N V . Now if u + = 0 and u − = 0 then This proves the lemma.
The functional I V satisfies the Palais-Smale condition on N V at the level d if every (P S) d -sequence for I V on N V contains a convergent subsequence.

Remark 4.
We can write ∇I V (u) for the gradient of I V at u, as

Now by the Hahn-Banach Theorem, there is a continuous linear functional
The next Lemma shows that N V is a natural constraint. Lemma 3.4. Every (P S) d -sequence (u k ) for I V restricted to N V contains a subsequence which is a (P S) d -sequence for I V in H 1 0 (Ω). Proof. Let (u k ) be (P S) d -sequence for I V on N V . By Lemma (3.1), after passing to a subsequence, we have that (u k ) is bounded in By assumption (f 4 ), the Sobolev embedding and Hölder's inequality, for any v in , This proves that (∇J V (u k )) is bounded.

ALIREZA KHATIB AND LILIANE A. MAIA
As |∇J V (u k )u k | ≤ ∇J V (u k ) u k < C, after passing to a subsequence, we have that |J V (u k )u k | → ρ ≥ 0. We show that ρ > 0.
From Lemma (2.1)(a) and by property (f 2 ) we obtain This inequality, together with Lions' lemma, implies that there exist δ > 0 and a sequence (y k ) in R N such that Setũ k (x) := u k (x + y k ) After passing to a subsequenceũ k u weakly in a.e. in R N . Therefore, Hence, u = 0 and there exists a subset Λ of positive measure in B 1 (0) such that u(x) = 0 for every x ∈ Λ. Property ( Taking the inner product of (21) with u k we obtain and so t k → 0 and from (21) we deduce ∇I V (u k ) → 0 as ∇ N V I V (u k ) → 0 and this proves the lemma.
Replacing u k by a subsequence if necessary, there exist a solution u 0 de (P V ), a number m ∈ N, m functions w 1 , . . . , w m in H 1 (R N ) and m sequences of points (y j k ) ∈ R N , 1 ≤ j ≤ m, satisfying: a) u k → u 0 in H 1 0 (Ω) or b) w j are nontrivial solutions of the limit problem (P ∞ ); c) |y j k | → +∞ and |y j k − y i k | → +∞ i = j; Proof. The proof is almost the same as that in [25] Lemma 4.5, just observing that )(u 1 k ) 2 and using (V 2 ). Lemma 3.6. Problem (P ∞ ) does not have a solution u such that I ∞ (u) ∈ (c ∞ , 2c ∞ ) Proof. Under our assumptions on f including that f is odd, the limit problem (P ∞ ) has a positive solution w with I ∞ (w) = c ∞ [6]. If u is a solution of P ∞ such that I ∞ (u) ∈ [c ∞ , 2c ∞ ) then, by Lemma (3.3), u does not change sign and, by [7], it is radially symmetric. By assumption (U ) problem (P ∞ ) has a unique positive solution and therefore u = ±w, up to a translation. Hence, I ∞ (u) = c ∞ .
Corollary 1 (Compactness). If c V is not attained, then c V ≥ c ∞ and I V satisfies the Palais-Smale condition on N V at every level d ∈ (c ∞ , 2c ∞ ).
Proof. Let (u k ) be a (P S) d -sequence for I V on N V . By Lemmas 3.1 and Lemmas 3.4, after passing to a subsequence, we have that (u k ) is a bounded (P S) d -sequence for I V . By the definition c V := inf u∈N V I V (u), there exists (u j ) ∈ N V such that I V (u j ) → c V . Now by the Ekeland variational principle there exists (ũ j ) ∈ N V such that I V (ũ j ) → c V and I V (ũ j ) → 0 (Theorem 8.5 [32]). Now by the Splitting lemma If d ∈ (c ∞ , 2c ∞ ) and (u k ) does not have a convergent subsequence then, by the Splitting lemma, then in both cases, m < 2 and so m = 1. The hypothesis 2c ∞ > d ≥ (m + 1)c ∞ implies that it is not possible to occur m = 1 and u 0 = 0, there for u 0 = 0, which implies I V (u n ) → I ∞ (w 1 ) = d. By (U ) the solution is unique and so w 1 = w that yields there exists a solution w of P ∞ with d = I ∞ (w), which contradicts Lemma 3.6. Hence, I V satisfies the Palais-Smale condition on N V at every d ∈ (c ∞ , 2c ∞ ).

4.
Existence of a positive bound state solution. Let us define, for R > 0, |y 0 | = 1 and y ∈ ∂B 2 (y 0 ), Note that in principle ε R = ε R (y) is dependent on y, but we are going to show that the estimates on ε R are independent of y.
Hereafter, we present three lemmas that prepare for the essential proposition 1, which gives an upper bounded for the functional I V on a convenient subset of N V . Lemma 4.1. There exists C > 0 such that for all y ∈ ∂B 2 (y 0 ).
Proof. ¿From property (f 2 ), performing a change of variable, we have that

ALIREZA KHATIB AND LILIANE A. MAIA
As p 2 ≥ p 1 > 1, using estimates (2), Lemma 2.3 and Lemma 2.6 with p = p 1 and q = 1 we obtain that so the lemma is proved.

Lemma 4.2.
There exists C > 0 such that for all s, t ≥ 1 2 , y ∈ ∂B 2 (y 0 ) and R large enough, Proof. For |x| < 1 and R large enough we have Now, by (f 5 ), (24) and the decay estimates (2) there exists C > 0 such that V∞R .
If we set s, t = 1 in above lemma we have for all s ∈ [0, b], y ∈ ∂B 2 (y 0 ) and R large enough.
We have that

ALIREZA KHATIB AND LILIANE A. MAIA
The sum in line (26) is equal to I ∞ (sw R 0 ) + o(ε R ). Indeed, But by Lemma 2.5 with q = 2 and (16) we have On the other hand, the mean value theorem, (f 2 ) and Lemma 2.5 yield 0 is the least energy solution of the limit problem (P ∞ ) and by Lemma 2.1 (c) we have I ∞ (sw R 0 ) ≤ c ∞ and similarly for the sum in line (27), so (26) + (27) ≤ 2c ∞ + o(ε R ).
Remark 6. Note that assumption γ > 2 √ V ∞ in (V 2 ) is sharp if one looks for an upper bound in Lemma 2.7 with exponential decay of order e −2 √ V∞R . In [14] there is a constant 2 missing in the exponential term which bounds ε R . Lemma 4.4. For any δ > 0, there exists R 2 > 0 such that I V (T R λ,y U R λ,y ) < c ∞ + δ, for λ = 0 and every y ∈ ∂B 2 (y 0 ) and R ≥ R 2 . In particular, c V ≤ c ∞ .
Proof. By Lemma 2.10, T R λ,y is bounded uniformly in λ, y and R. As w R y is a ground state solution of problem (P ∞ ), using Lemma 2.1(c) and Lemma 2.7, we obtain where o R (1) → 0 as R → ∞, uniformly in y ∈ ∂B 2 (y 0 ).
Note that β(u) = 0 if u is radial. Barycenter maps have been constructed in [4]. where I c V = {u ∈ H 1 0 (Ω), I V (u) ≤ c}. Proof. If c V is not attained, Corollary 1 and Lemma 4.4 imply that c V = c ∞ . Arguing by contradiction, assume that for each k ∈ N there exists v k ∈ N V such that I V (v k ) < c V + 1 k and β(v k ) = 0. By Ekeland variational principle [18], there exists a (P S) d -sequence (u k ) for I V on N V at the level d = c V such that u k −v k → 0 [[32], Theorem 8.5]. By Lemmas 3.4 and 3.1, after passing to a subsequence, we have that (u k ) is a bounded (P S) d -sequence for I V . As c V is not attained, Lemma 3.5 (splitting) implies that there exists a sequence (y k ) in R N such that |y k | → ∞ and u k − w(· − y k ) → 0, where w is the (positive or negative) radial ground state of (P ∞ ). Settingṽ k (x) := v k (x + y k ), and using properties (35) and the continuity of the barycenter, we conclude that −y k = β(v k ) − y k = β(ṽ k ) → β(w) = 0 yielding a contradiction.
We have constructed all the tools in order to apply a topological argument analogous to that found in [14] to prove our main result. For the sake of completeness we recall the argument and prove theorem 1.1.
Proof of Theorem 1.1. If c V is attained by I V at some u ∈ N V then, by Lemma 3.2, u is a nontrivial solution of problem (P V ). So assume that c V is not attained. Then, by Lemma 4.5, c V = c ∞ . We will show that I V has a critical value in (c ∞ , 2c ∞ ). By Lemma 4.5, we may fix δ ∈ (0, c∞ 4 ) such that β(u) = 0, ∀u ∈ N V ∩ I c∞+δ V . Proposition 1 and Lemma 4.4 allow us to choose R > 0 sufficiently large and its corresponding η R = η ∈ (0, c 4 ) such that I V (T R λ,y U R λ,y ) ≤ 2c ∞ − η for all λ ∈ [0, 1] and all y ∈ ∂B 2 (y 0 ) c ∞ + δ for λ = 0 and all y ∈ ∂B 2 (y 0 ).
Arguing by contradiction, assume that I V does not have a critical value in (c ∞ , 2c ∞ ). As, by Corollary 1, I V satisfies the Palais-Smale condition on N V at every level in (c ∞ , 2c ∞ ), there exists ε > 0 such that . At this point we may use a Deformation Lemma for C 1 manifold found in [8], which yields a continuous function such that ρ(u) = u for all u ∈ N V ∩ I c∞+δ V (see [8]). Now we define Γ(x) := (β • ρ • α)(x). By Lemma 4.5 Γ(x) = 0 and so the function h : B 2 (y 0 ) → ∂B 2 (0) given by h(x) := 2 Γ(x) |Γ(x)| is well defined and continuous. Furthermore, if y ∈ ∂B 2 (y 0 ), then it holds α(y) = T R 0,y U R 0,y = T R 0,y w R y ∈ N Ω ∩ I c∞+δ Ω and hence (β • ρ • α)(y) = β(T R 0,y w R y ) = Ry. Therefore, if we consider the homeomorphismh : ∂B 2 (y 0 ) → ∂B 2 (0) defined bỹ h(y) := 2y |y| , then (h −1 • h)(y) = y for every y ∈ ∂B 2 (y 0 ) and by Brouwer Fixed Point Theorem such a retraction does not exist, thus I V must have a critical point u ∈ N V with I V (u) ∈ (c ∞ , 2c ∞ ). By Lemma 3.3, u does not change sign and, since f is odd, −u is also a solution of (P V ). This proves that problem (P V ) has a positive solution.