“…(♯ = N) As in the previous case, if Σ • ⊂ Σ then g Σ• λ = SL λ 1 Σ• with 1 Σ• ∈ H 1/2−ǫ Σ (Γ), ǫ > 0. Since γ 1 DL λ ∈ B(H 1/2 (Γ), H −1/2 (Γ)) is coercive (see, e.g., [17, Lemma 3.2]), R Σ γ 1 DL λ R * Σ ∈ B(H 1/(Γ), H −1/2 (Σ)) is coercive as well (see [16, Remark 5.2]) and hence (R Σ γ 1 DL λ R * Σ ) −1 ∈ B(H −1/2 (Σ), H 1/(Γ)) (see[16, Remark 4.6]); then, by the mapping properties of γ 1 DL λ (see[7, Theorem 3]), one gets (R Σ γ 1 DL λ R * Σ ) −1 ∈ B(H s−1/2 (Σ), H s+1/2 Σ (Γ)), s ∈ [0, 1/2]. Since both g Σ• λ and DL λ φ solve (−∆ + λ)u = 0 in R n \Σ • with boundary condition R Σ γ 1 u = R Σ γ 1 g Σ• λ whenever φ = (R Σ γ 1 DL λ R * Σ ) −1 R Σ γ 1 g Σ• λand such a problem has a unique radiating solution (see[1, Theorem 3.3]), one gets g Σ• λ = DL λ φ = G N λ φ, φ ∈ H s+1/2 Σ (Γ).…”