Generalized maximum principle and evaluation of the first eigenvalue for Heisenberg type operators with discontinuous coefficients.

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Generalized maximum principle and evaluation of the first eigenvalue for Heisenberg type operators with discontinuous coefficients. M. Chicco Università di Genova Dipartimento di Metodi e Modelli Matematici Piazzale Kennedy pad.d, 16129 Genova M.R. Lancia Università di Roma La Sapienza Dipartimento di Metodi e Modelli Matematici per le Scienze Applicate Via A. Scarpa 10, 00161 Roma Sommario. Precedenti risultati riguardanti il principio di massimo generalizzato e la valutazione del primo autovalore per operatori uniformemente ellittici di tipo variazionale vengono estesi agli operatori subellittici di tipo Heisenberg non simmetrici e a coefficienti discontinui. Introduction In the mathematical literature the strong maximum principle for degenerate operators with nonnegative characteristic form and regular coefficients has been studied by Bony ([4]); for a complete discussion see also ([20]). In his results the usual assumption on the sign of the coefficient c in the operator L is adopted. Following this philosophy, maximum principles can be also achieved for Heisenberg-type operators ([17]). In this paper we prove a generalized maximum principle and we compute the first eigenvalue for linear second order subelliptic operators whose principal part is in divergence form with respect to the Heisenberg vector fields with discontinuous coefficients. 1

The generalized maximum principle differs from the ordinary one as follows. In the ordinary maximum principle usually one states that if a subsolution (of the elliptic equation) is m on Ω, then it is also max(0, m) in the interior of Ω. This property is valid provided one assumes the coefficient c of L to be non negative a. e. in Ω. In the present work we want to prove a generalized maximum principle: if a subsolution u is non-positive on Ω, then it is non-positive also in the interior of Ω. This property, as proved in Theorem 3.1, is valid iff there exists (at least) a subsolution w such that w 0 in Ω and a(w, v) < 0 for some test function v 0 in Ω (for the definition of the bilinear form a(, ), associated with the subelliptic operator L, see (2.1)). These results were proved in [6] for uniformly elliptic divergence form equations, and they are extended in the present note to the case of Heisenberg type subelliptic operators. We consider a local solution of the equation Lu = 0 in Ω where Ω is an open bounded connected set of R 2n+1, with smooth boundary. The operator L is given by Lu = j=1 X j [ i=1 a ij (x)x i u + d j (x)u] + i=1 b i (x)x i u + c(x)u where X j are the Heisenberg vector fields in R 2n+1, X j is the L2 adjoint of X j. The operator L is assumed to be uniformly subelliptic. More precisely we shall assume the following: (A) a ij are measurable functions on R 2n+1, and there exist positive constants µ, M such that µ M and µ ξ 2 i,j=1 for all x R 2n+1 and ξ R 2n (B) there exists q > 2n + 2 such that for i = 1,..., 2n. a ij (x)ξ i ξ j M ξ 2 b i L q (Ω), d i L q (Ω), c L q/2 (Ω) 2

Under these assumptions we can prove the following Theorem 3.1 Let B R Ω with 192R < R. Every non positive subsolution u on B R satisfies a generalized maximum principle iff there exists a negative subsolution in B R. Here by B R we denote the intrinsic balls defined in (1.7) and R is a suitable value of the radius (see proposition 2.1). The lower bound on R is motivated by the use of Harnack inequality (proved in ([18]) in lemma 2.5 and corollary 2.6. The previous results enable us to give in theorem 4.2 a characterization of the first eigenvalue having largest real part: Theorem 4.2 Let λ 1 denote the eigenvalue of Problem (4.1) having the largest real part. Then λ 1 is real and it turns out a(w, v) λ 1 = sup inf : w S 2 (B R ), w < 0 in B R (w, v) L 2 (B R ) v S 2 (B R ),v>0 where S 2 (B R ) = { u L 2 (B R ) : X i u L 2 (B R ), i = 1,..., 2n } (see Definition (1.2) below). In this paper, we extend to the non euclidean context of the Heisenberg group the results proved in [6, 7] for the case of linear second order elliptic partial differential equations in divergence form with discontinuous coefficients. Following the proofs in [6] we use a local Harnack inequality proved in [18], for more general Hörmander vector fields (see also [2]), and we had to adapt some proofs in [6] to our case. Actually, we use some results about connections between Hausdorff measure and the analogue of p capacity with respect to the Heisenberg vector fields (here we will not give the proofs which can be found in [16]). In section 1 we describe the main tools we shall make use of. We introduce the Heisenberg group and the associated Sobolev spaces. We recall that R 2n+1, equipped with the distance intrinsically associated with the Heisenberg vector fields, is a homogeneous space in the sense of abstract harmonic analysis [8] with homogeneous dimension N = 2n + 2. Then, we state the results about the connections between Hausdorff measure and 2 capacity, where the 2 capacity is defined with respect to the Heisenberg vector fields. In section 2, we recall some preliminary results proved in [18] and we deduce some straightforward consequences which, together with the relation 3

between Hausdorff measure and 2 capacity, will turn out essential to the proof of corollary 2.6. In section 3 we prove the generalized maximum principle, where the technique follows the lines of [6] in the euclidean context. Finally, in section 4 we can give a characterization of the first eigenvalue having largest real part. 1 Notations Consider the euclidean space R 2n+1, whose elements we denote by x = (x 1,..., x n, y 1,..., y n, t), equipped with the multiplication law x x = (x 1 + x 1,..., x n + x n, y 1 + y 1,..., y n + y n, t + t + 2 (x iy i x i y i)). (1.1) It is a group whose identity is the origin and where the inverse is given by x 1 = ( x 1,..., x n, y 1,..., y n, t) The space R 2n+1 with the structure (1.1) is the Heisenberg group denoted by H n [22]. The non isotropic dilations δ x = (δx 1,... δx n, δy 1,... δy n, δ 2 t) (δ R, x H n ) (1.2) are automorphisms of H n. The nonnegative function ( n ) 1/4 ρ(x) = (x 2 i + yi 2 ) 2 + t 2 (1.3) i=1 defines a norm for the Heisenberg group, in particular it is homogeneous of degree 1 with respect to the dilations (1.2), i.e. ρ(δ x) = δ ρ(x) (1.4) for every x H n and δ R. Moreover, there exist positive constants c 1 and c 2 such that c 1 x < ρ(x) < c 2 x 1/2 (1.5) for every x in a bounded set of H n, where x denotes the Euclidean norm in R 2n+1. By (1.4) and (1.5) it follows that the function d defined by d(x, x ) = ρ(x 1 x) (1.6) 4

is a distance in H n, topologically equivalent to the Euclidean one and left invariant with respect to the law (1.1). By using the distance d we define the intrinsic balls B(x, R) = B R (x) = {x H n : d(x, x ) < R} (1.7) The Lebesgue measure dx = dx 1... dx n dy 1... dy n dt is invariant with respect to the translations (1.1) so that for every x H n and R > 0 we have B R (x) = B R (0) (1.8) Since the jacobian of the dilations (1.2) is given by J δ = δ 2n+2 (1.9) from (1.8) and (1.9) it follows that for every x H n and R > 0 B R (x) = R 2n+2 B 1 (1.10) In our context we need a vector field basis which is invariant with respect to the translations (1.1). Such a basis is given by X i = x i + 2y i t 2x i n y n i t for i = 1,..., n for i = n + 1,..., 2n (1.11) T = t For the vector fields (1.11) we have the commutative law [X i, X i+n ] = 4T for every i = 1,..., n (1.12) while the other commutators vanish. We recall that a commutator of two vector fields V 1 and V 2 is the new vector field given by [V 1, V 2 ] = V 1 V 2 V 2 V 1 (1.13) therefore, X 1,..., X 2n are a basis for the Lie algebra of the vector fields invariant with respect to (1.1). Moreover they are homogeneous of degree 1 5

with respect to the dilations (1.2), whereas, by (1.12), T is homogeneous of degree 2 i.e. and X i (u(δ x)) = δ((x i u)(δ x)) (i = 1,..., 2n) (1.14) T (u(δ x)) = δ 2 ((T u)(δ x)). (1.15) The family of the intrinsic balls B R (x) reflects the nonisotropic nature of the Heisenberg vector fields. The space R 2n+1, equipped with the distance d, acquires the structure of a space of homogeneous type of dimension N = 2n + 2 in the sense of Coifman and Weiss [8] Let Ω be a bounded open subset of R 2n+1. The following covering lemma holds (see [8] ch.3 lemma 1.1) Lemma 1.1 For every ε (0, 1), the ball B(x, r), x Ω, 0 < r < r 0, can be covered by the union of balls B(y i, εr) with y i B(x, r), for i = 1, 2,..., l such that l < cε α for suitable constants c > 0 and α > 0. The constants c and α are independent on x, r and ε and depend only on c 0. This dimension will play a crucial role in view of the close connection between capacity and Hausdorff dimension which we shall describe later. Following [10, 22] we give the Definition 1.2 We denote by S 2 (Ω) the Sobolev type space of the functions u L 2 (Ω), such that the distribution derivatives X i u belong to L 2 (Ω) for i = 1,..., 2n. The norm in S 2 (Ω) is given by u 2 = Ω ( u 2 + i=1 X i u 2 )dx (1.16) The closure of C 0 (Ω) in the above norm is denoted by S 2 (Ω). By S 2 loc (Ω) we mean the set of functions u which belong to S 2 (Ω ) for every Ω Ω. In the following, we set for brevity Xu 2 L 2 (Ω) := Ω i=1 X i u 2. We recall the following properties of the above Sobolev spaces: 6

Poincaré inequality (see [12]). For every B R (x) R 2n+1 and u S 2 (B R (x)) we have u 2 dx cr 2 X j u 2 dx (1.17) B R (x) B R (x) j=1 Sobolev inequality (see [13]). There exists a constant S > 0 such that for every u S 2 (R 2n+1 ) we have where ( ) 2/2 u 2 dx S 2n+1 R R 2n+1 j=1 2 = 2N ( = 2n + 2 ) N 2 n X j u 2 dx, (1.18) Compact embedding (see [10]). For every bounded domain Ω the Sobolev space S 2 (Ω) is compactly embedded into L p (Ω), for every p < 2. Let E R 2n+1 a compact set. We define the capacity of E as cap 2 E = (1.19) = inf X j f 2 dx : f C0 (R 2n+1 ); f 1 on E R 2n+1 j=1 For the reader s convenience we now recall some results whose proofs can be found in [16]. In the sequel, we denote by H s ( ) the s-dimensional Hausdorff measure in the metric space (R 2n+1, d). Proposition 1.3 Let A denote a compact set in H n and B(x, r) Ω then the following properties hold: i) cap 2 (B(x, r)) = r N 2 B 1 ; ii) cap 2 (A) ch N 2 (A); where c is a costant depending only on N. Theorem 1.4 Let A denote a compact set in H n, if H N 2 (A) < then cap 2 (A) = 0 7

Proof. See [16]. Theorem 1.5 Let A H n be a compact subset. If cap 2 (A) = 0 then H s (A) = 0 for every s > N 2. Proof. See [16]. Definition 1.6 Let Ω denote a bounded connected set and E Ω. Let u S 2 (Ω). The function u is nonnegative in the sense of S 2 (Ω), or briefly u 0 on E in S 2 (Ω) if there exists a sequence {u n } C (Ω) such that { un (x) 0, x E u n (x) u in S 2 (Ω) Remark 1 If u S 2 (Ω) and v S 2 (Ω) we say that u v on E in S 2 (Ω) if u v 0 on E in S 2 (Ω). The following proposition can be easily proved ([11]): Proposition 1.7 If u S 2 (Ω) and u k a.e. in Ω then u k on K in the sense of S 2 (Ω) for every compact set K Ω. Remark 2 It is easy to see, as in the euclidean context, that if u S 2 (Ω) and E Ω is a compact subset and u = 0 in the sense of S 2 (Ω) then u = 0 in the sense of the capacity defined in (1.19). 2 Preliminary results In this section we recall, in the case of interest for us, some results proved in [18] and we deduce some consequence which will turn out essential to our purposes. 2.1 Variational formulation of the problem Let Ω be a bounded domain in R 2n+1. differential operator Let us consider the following Lu = X j [ a ij (x)x i u + d j (x)u] + j=1 i=1 i=1 8 b i (x)x i u + c(x)u

where X 1, X 2,..., X 2n are the Heisenberg vector fields and X j is the L2 adjoint of X j. We make the following assumptions: A) a ij (i, j = 1,..., 2n) are 2n 2n measurable functions on Ω that satisfy the following conditions: i) there exists M > 0 such that a ij (x) M for a. e. x Ω; ii) there exists µ > 0 such that µ ξ 2 i,j=1 a ij ξ i ξ j for almost every x Ω and every ξ R m ; B) b i L N (Ω), i = 1,..., 2n; C) c L N/2 (Ω) and d j L N (Ω), j = 1,..., 2n where N = 2n + 2. Let a be the bilinear form defined by: a(u, v) = (a ij X i ux j v + b i (X i u)v + d j ux j v + cuv dx (2.1) Ω i,j u S 2 loc (Ω), v S 2 (Ω). In [18] it was proved the following: Proposition 2.1 Let B R = B R (x) Ω, R R 0 /2. Then: i) a is continuous in S 2 (Ω) S 2 (Ω) ii) there exists R with 0 R R 0 /2 and ν > 0 depending on the structural costants and on S, M, µ, b i, d i, c but not depending on x Ω, such that a is coercive on S 2 (B R ) for every R R: a(u, u) ν u S 2 (BR ) for every u S 2 (B R ). (2.2) 9

(Here R 0 is a suitable constant defined in [18]). As pointed out in [18] (see remark 2.2), Proposition 2.1 still holds when the hypothesis C) is replaced by C ) d j L q (Ω) (j = 1, 2,..., 2n), c L q/2 (Ω), q > N which cannot be removed in order to prove Harnack inequality and Theorem 2.3. Therefore from now on we will assume C ) to hold. Definition 2.2 We say that u Sloc 2 (Ω) is a local solution of the equation Lu = 0 (2.3) if for every B R Ω and for every v S 2 (B R ) it turns out a(u, v) = B R (a ij X i ux j v + b i (X i u)v + d j ux j v + cuv dx (2.4) i,j If in addition u S 2 (Ω) we say that u is a solution. We say that u is a local subsolution if a(u, ϕ) 0 ϕ S 2 (B R ), ϕ 0. (2.5) In [18] a Harnack inequality is proved. Theorem 2.3 Let u Sloc 2 (Ω) be a positive local solution of (2.3). There exists C > 0, such that for every B R with B 192R Ω and 192R < R we have sup u C inf u. (2.6) B R B R From Harnack inequality and theorem (3.2) in [18] the following result can be deduced: Corollary 2.4 Let u be a local positive solution in Ω of Lu = 2n i=1 X i f i with f i L p (Ω), p > N then sup u K( inf u + f i L B(x,r) B(x,r) p (B R ) R 1 N/p ), where K does not depend on B R. 10

From the above corollary it follows Lemma 2.5 Let u S 2 (Ω), u 0 in Ω, u non identically zero in Ω, a(u, v) = 0 for every v S 2 (Ω). Then for every B R Ω with 192R < R Proof: see the proof of corollary (8.1) in [21]. sup B R u < 0 (2.7) Corollary 2.6 Let w S 2 (Ω), w 0 in Ω, w not identically zero in Ω, a(w, v) 0 for every v S 2 (Ω), v 0. Then, for every B R Ω, with 192R < R we have Proof. Let us consider the following subsets of Ω: sup B R w < 0 (2.8) B 1 := {x Ω : there exists a neighborhood U of x s.t. w = 0 a.e. in U} B 2 := {x Ω : there exists a neighborhood U of x s.t. ess sup U w < 0} B 3 := Ω\(B 1 B 2 ) = {x Ω : for every neighborhood of x there exist a subset A 1 s.t. A 1 > 0 where w = 0 a.e. and a subset A 2 with A 2 > 0 where w < 0 a.e. } It turns out that our hypothesis implies that B 1 cannot coincide with Ω. If B 2 Ω, by compactness arguments we have that for every compact set D Ω it follows ess sup D w < 0, whence the thesis. Therefore, if both B 1 and B 2 are not empty, then B 3 also is not empty, due to the fact that Ω is a connected set. Finally we can restrict ourselves to consider only the last case. Let x 0 B 3. Then there exists a ball B(x 0, r) with radius r small enough such that w / S 2 (B(x 0, r)). In fact, if this were not the case, by contradiction there would exist a r 0 > 0 s.t. w S 2 (B(x 0, r)) for every r (0, r 0 ). Thus, by virtue of remark 2, w = 0 in the capacity sense on the boundary B(x 0, r) of every ball B(x 0, r) for r (0, r 0 ). Theorem 1.5 yields that if cap 2 ( B(x 0, r)) = 0 then H N 1 ( B(x 0, r)) = 0 for every r (r, r 0 ). By means of the intrinsic coarea formula (see (17) in [9], or [5]) we have that w(x) Xρ dx = 0 B(x 0,r 0 ) 11

where Xρ = ( 2n i=1 (X i ρ) 2 ) 1/2. Taking into account that Xρ > 0 a.e. we deduce that w(x) = 0 a.e. in B(x 0, r 0 ), a contradiction. Therefore we can choose r s.t. w / S 2 (B(x 0, r)) and s.t. the form a(, ) is coercive in S 2 (B(x 0, r)). (As pointed out in [18] this happens whenever the measure of B(x 0, r) is suitably small). Then if we consider the solution u of the Dirichlet problem a(u, v) = 0 v S 2 (B(x 0, r)) u w S 2 (B(x 0, r)) (the existence easily follows as in thm.(3.3) of [21]), we can prove that w u a.e., on B(x 0, r) (2.9) In fact, if we set g = max(w u, 0), the function g turns out to be in S 2 (B(x 0, r)), g 0 a.e. and a(g, g) 0 whence g 0 a.e. in B(x 0, r). In the same way we can prove that u 0 on B(x 0, r) and that u is not identically zero on B(x 0, r) because u w S 2 (B(x 0, r)) and w / S 2 (B(x 0, r)). Therefore by lemma 2.5 we have sup u < 0 for every B(x, r) B(x 0, r) B(x, r) If by chance x 0 B(x, r) then it would follow that sup B(x, r) w = 0, thus contradicting (2.9). 3 The generalized maximum principle In this section we prove that every non positive subsolution u on B R satisfies a generalized maximum principle iff there exists a negative subsolution in B R. Theorem 3.1 The two following statements are equivalent: a) for every u S 2 loc (Ω) such that u 0 on B R, B R Ω, 192R < R and a(u, v) 0 v S 2 (B R ), v 0, then u 0 in B R. b) There exists a function w S 2 loc (Ω) such that w 0 in B R Ω, a(w, v) 0 v S 2 (B R ), v 0 in B R, and a(w, v) < 0 at least for one function v S 2 (B R ), v 0 a.e. in B R. 12

Proof. We prove that a) b). For the sake of simplicity, let f C 0 (Ω) f < 0. Let us consider the boundary value problem a(w, v) = Ω fvdx v S 2 (B R ) w S 2 (B R ) there exists a unique solution w S 2 (B R ) for every B R Ω with R < R. Let w denote the solution, then it satisfies b). In fact, condition a) and the hypothesis on f yield w 0 in B R, a(u, w) < 0 if v S 2 (B R ), v > 0 in B R Let us now prove that b) a). Let u S 2 loc (Ω), u 0 on B R, a(u, v) 0 for every v S 2 (B R ), v 0; we want to show that u 0 in B R. Let us consider the function w k (x) = max(u + kw, 0), where k R and w is the function which satisfies condition b). Let A R (k) = {x B R : w k (x) > 0} k 0 = inf{k : w k (x) = 0 a.e. in B R }; (3.1) if we show that k 0 0 then we get our result. We proceed by contradiction. Let us assume that k 0 > 0. Let B R (x), 192R < R, so that by corollary (2.6) we claim that sup BR w < 0; moreover it can be shown (as in [21] sect.5) that sup BR u < +. Therefore there exists a h R such that whence obviously We now prove that w k = 0 in B R for k h lim A R(k) = 0. (3.2) k + lim A R (k) = 0 (3.3) k k 0 even though k 0 is finite. If (3.3) did not hold, there would exist a set H B R, H > 0 such that u+k 0 w = 0 in H and u+k 0 w 0 in B R. Then let B r B R such that B r H > 0, Corollary 2.6 applied to the function u + k 0 w yields u+k 0 w = 0 in B r, so that by covering arguments (see [8]), u+k 0 w = 0 in B R, 13

i.e. u = k 0 w. Hence a(w, v) = 0 v S 2 (B R ) which contradicts condition b) and proves (3.3). On the other hand, as the function R k A R (k) is decreasing, from (3.3) we get A R (k 0 ) = 0 whence w k0 = 0 a.e. in B R. Therefore arguing as before we have a contradiction and then k 0 0. Corollary 3.2 Let us assume condition a) of Theorem 3.1 to hold. Let µ > 0 and let z be in S 2 (Ω), z 0 on B R, for every B R Ω such that Then z is 0 in B R. a(z, v) + µ(z, v) L 2 (B R ) 0 v S 2 (B R ), v 0. Proof. By virtue of theorem 3.1 there exists a function w S 2 (B R ) s.t. w 0 in B R Ω, a(w, v) < 0 v S 2 (B R ), v > 0 in B R. Then a(w, v) + µ(w, v) L 2 (B R ) < 0, v S 2 (B R ), v > 0, so that Theorem 3.1 applied to the form a(w, v) + µ(w, v) gives the result. 4 Characterization of the eigenvalues In this section, we give a characterization of the eigenvalues of the problem a(w, v) + λ(w, v) L 2 (Ω) = 0, v S 2 (Ω) u S (4.1) 2 (Ω) Before stating our result, we recall a theorem which will be useful later (see [14]), for the sake of clarity. Theorem 4.1 (Kreǐn and Rutman) Let X be a Banach space and T denote a completely continuous operator from X into X such that T maps a cone K into K and T has at least a non zero eigenvalue. Then T has a positive eigenvalue λ 1 such that λ 1 λ for every λ eigenvalue of T. Theorem 4.2 Let λ 1 denote the eigenvalue of problem (4.1) having the largest real part. Then λ 1 is real and it turns out λ 1 = sup inf v S 2 (B R ),v>0 a(w, v) (w, v) L 2 (B R ) : w S 2 (B R ), w < 0 in B R (4.2) 14

Proof. The proof will take four steps. Step 1). We prove that λ 1 R. Suppose that µ does not belong to the spectrum of (4.1) and consider the operator defined as: G µ : L 2 (B R ) S 2 (B R ) a(g µ u, v) + µ(g µ u, v) L 2 (B R ) = (u, v) L 2 (B R ) v S 2 (B R ). (4.3) Taking into account that S 2 (B R ) is compactly embedded in L 2 (B R ) (see [3, 12]) for every B R Ω such that R < R, the operator G µ is compact in L 2 (B R ). Let µ be large enough such that there exists a positive constant µ 0 with a(z, z) + µ z 2 L 2 (B R ) µ 0 z 2 S 2 (BR ) z S 2 (B R ) (4.4) (see [18] and [21]). If (4.4) holds, µ does not belong to the spectrum; moreover if u L 2 (B R ), u 0 in B R then G µ u 0 in B R. In fact, set ũ = max(g µ u, 0), v = ũ in (4.3); taking (4.4) into account we get µ 0 ũ 2 S 2 (BR ) a(ũ, ũ) + µ ũ L 2 (B R ) = (u, ũ) L 2 (B R ) 0. Then ũ = 0 in B R and G µ u 0 in B R. invariant the cone K defined by Thus the operator G µ leaves K = {u L 2 (B R ) : u 0 in B R } The operator G µ, under assumption (4.4), satisfies the hypothesis of Kreǐn-Rutman Theorem, therefore there exists a real eigenvalue t 1 of G µ such that t t 1 for every eigenvalue t of G µ (4.5) Moreover, it is well known that if λ is an eigenvalue of the problem (4.1) the number t = (µ λ) is an eigenvalue of the operator G µ and viceversa. If we set t 1 = (µ λ 1 ) 1, (4.5) yields µ λ µ λ 1 (4.6) for every λ eigenvalue of (4.1). Then taking into account that (4.6) holds under the hypothesis that µ is large enough, we let µ diverge at +, thus obtaining: Re(λ) λ 1 (4.7) 15

for every λ eigenvalue of (4.1). This guarantees the existence of an eigenvalue λ 1 having maximal real part. We are now ready to make the second step. Step 2). We set λ = sup inf v S 2 (B R ),v>0 a(w, v) (w, v) L 2 (B R ) : w S 2 (B R ), w < 0 in B R and we remark that λ >. By contradiction, if λ = take any real p; then there exists a function w S 2 (B R ), w < 0 in B R, such that inf v S 2 (B R ),v>0 a(w, v) (w, v) L 2 (B R ) > p whence a(w, v) + p(w, v) L 2 (B R ) < 0, v S 2 (B R ), v > 0. From Theorem 3.1 it follows that neither p is an eigenvalue of problem (4.1) nor any number t > p. But problem (4.1), like the compact operator G µ defined in (4.3), has at least an eigenvalue: this is a contradiction, so λ is real. We prove now that λ is an eigenvalue of problem (4.1). By contradiction, if λ were not an eigenvalue, there would exist the operator G λ, which satisfies the property G λ u 0 in B R for every u L 2 (B R ), u 0 in B R. In fact, by the definition of λ and Theorem (3.1), it can be shown that lim G µ u G λ u L µ λ 2 (B R ) = 0 + and G µ u 0 in B R if µ > λ. If λ were not an eigenvalue and if 0 < λ µ < G λ 1 then µ would not be an eigenvalue so that we would have G µ = G λ [I (λ µ)g λ ] 1 = (λ µ) j G j+1 λ j=0 Hence, with this choice of µ it turns out that G µ u 0 in B R whenever u 0 in B R. If w = G µ u with u L 2 (B R ), u < 0 it turns out that a(w, v) + µ(w, v) L 2 (B R ) < 0, v S 2 (B R ), v > 0 in B R so that w < 0 in B R, by virtue of Corollary 2.6. It follows that inf v S 2 (B R ),v>0 a(w, v) (w, v) L 2 (B R ) 16 µ

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