Weighted Hardy inequality with higher dimensional ... - CAPDE

Multiplying the above **inequality** by v ε+ and integrating by parts yields∫∫∫|∇v +Ω + ε | 2 (N − k)2dx − δ −2 q|v +4βΩ + ε | 2 dx + λ ηδ −2 |vβΩ + ε + | 2 dx ≤ 0.βBut then Lemma 3.1 implies that v ε + = 0 in Ω + βprovided β small enough because|η| ≤ Cδ near Σ k . Therefore u ≥ R V ε for every ε ∈ (0, 1). In particular u ≥ R V 0 .Hence we obtain from Lemma 2.3 that∫u 2 ∫∞ >δ 2 ≥ R2Ω + βΩ + βV 20δ 2≥ ∫√1 − q(σ)dσΣ k1which leads to a contradiction. We deduce that u ≡ 0 in Ω + β. Thus by the maximumprinciple u ≡ 0 in Ω.For the general case p ≠ 1, we argue as in [3] by setting(4.3) ũ = √ pu.This function satisfies−∆ũ −(N − k)24qp δ−2 ũ ≥ −λ η (p δ−2 ũ + − ∆p )2p + |∇p|24p 2 ũ in Ω.Hence since p ∈ C 2 (Ω) and p > 0 in Ω, we get the same conclusions as in the casep ≡ 1 and q replaced by q/p.5 Existence of minimizers for µ λ (Ω, Σ k )Theorem 5.1 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. Assume that p, q andη satisfy (1.2) and (1.3). Then µ λ (Ω, Σ k ) is achieved for every λ < λ ∗ .Proof. The proof follows the same argument of [2] by taking into account the factthat η = 0 on Σ k so we skip it.22