where c is a positive constant depending only on Ω, p, q, η and Σ k . Hence by (3.7)we concludeµ λ (Ω, Σ k ) ≤ 1 + r1 − cr1 + cε1 − cε( (N − k)24)+ τ + cr1 − r .Taking the limit in ε, then in r and then in τ, the claim follows.Step 2: We claim that there exists ˜λ ∈ R such that µ˜λ(Ω, Σ k ) ≥ (N−k)24.Thanks to Lemma 3.1, the proof uses a standard argument of cut-off function andintegration by parts (see [2]) and we can obtain∫∫∫δ −2 u 2 q dx ≤ |∇u| 2 p dx + C δ −2 u 2 η dx ∀u ∈ Cc ∞ (Ω),ΩΩfor some constant C > 0. We skip the details. The claim now follows by choosing˜λ = −CΩFinally, noticing that µ λ (Ω, Σ k ) is decreasing in λ, we can set}(3.9) λ ∗ (N − k)2:= sup{λ ∈ R : µ λ (Ω, Σ k ) =4so that µ λ (Ω, Σ k ) < (N−k)24for all λ > λ ∗ .4 Non-existence resultLemma 4.1 Let Ω be a smooth bounded domain of R N , N ≥ 3, and let Σ k be asmooth closed submanifold of ∂Ω of dimension k **with** 1 ≤ k ≤ N − 2. Then, thereexist bounded smooth domains Ω ± such that Ω + ⊂ Ω ⊂ Ω − and∂Ω + ∩ Ω = ∂Ω − ∩ Ω = Σ k .Proof. Consider the mapsx ↦→ g ± (x) := d ∂Ω (x) ± 1 2 δ2 (x),where d ∂Ω is the distance function to ∂Ω. For some β 1 > 0 small, g ± are smooth inΩ β1and since |∇g ± | ≥ C > 0 on Σ k , by the implicit function theorem, the sets{x ∈ Ω β : g ± = 0}20

are smooth (N − 1)-**dimensional** submanifolds of R N , for some β > 0 small. Inaddition, by construction, they can be taken to be part of the boundaries of smoothbounded domains Ω ± **with** Ω + ⊂ Ω ⊂ Ω − and such that∂Ω + ∩ Ω = ∂Ω − ∩ Ω = Σ k .Now, we prove the following non-existence result.Theorem 4.2 Let Ω be a smooth bounded domain of R N and let Σ k be a smoothclosed submanifold of ∂Ω of dimension k **with** 1 ≤ k ≤ N − 2 and let λ ≥ 0.Assume that p, q and η satisfy (1.2) and (1.3). Suppose that u ∈ H0 1 (Ω) ∩ C(Ω) is anon-negative function satisfying(4.1) −div(p∇u) −If ∫ √ 1Σ kdσ = +∞ then u ≡ 0.1−p(σ)/q(σ)(N − k)2qδ −2 u ≥ −ληδ −2 u in Ω.4Proof. We first assume that p ≡ 1. Let Ω + be the set given by Lemma 4.1. Wewill use the notations in Section 2 **with** U = Ω + and M = ∂Ω + . For β > 0 small wedefineΩ + β := {x ∈ Ω+ :δ(x) < β}.We suppose by contradiction that u does not vanish identically near Σ k and satisfies(4.1) so that u > 0 in Ω β by the maximum principle, for some β > 0 small.Consider the subsolution V ε defined in Lemma 2.3 which satisfies(4.2) L λ V ε ≤ 0 in Ω + β, ∀ε ∈ (0, 1).Notice that ∂Ω + β ∩ Ω+ ⊂ Ω thus, for β > 0 small, we can choose R > 0 (independenton ε) so thatR V ε ≤ R V 0 ≤ u on ∂Ω + β ∩ Ω+ ∀ε ∈ (0, 1).Again by Lemma 2.3, setting v ε = R V ε − u, it turns out that v ε+ = max(v ε , 0) ∈H0 1(Ω+ β ) because V ε = 0 on ∂Ω + β \ ∂Ω+ β ∩ Ω+ . Moreover by (4.1) and (4.2),L λ v ε ≤ 0 in Ω + β, ∀ε ∈ (0, 1).21