By Artstein-Avidan S.

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**Example text**

9. Proof. 4, we set N1,ε = sup k ∈ N : ε2 ωk < α0 − ε 1−γ 4 . By the asymptotics of ωk we have N1,ε ∼ (129) 2√ α0 ε as ε → 0. By the definition of α0 , N1,ε and the smoothness of α → µα there holds µε2 ωk < −2ε 1−γ 2 for k ≤ N1,ε . 10 implies µk,ε < −ε 1−γ 2 for k ≤ N1,ε and ε → 0. Consider the following subspace M1 ⊆ HΣε M1 = span {φk (εy1 )uk,ε (y ), k = 0, . . , N1,ε } . N1,ε 0 Clearly dim(M1 ) = N1,ε + 1, and for any u = u 2 Sε = 1 ε N1,ε βk2 , (TSε u, u)HSε = 0 1 ε βk φk uk,ε there holds N1,ε µk,ε βk2 which implies 0 (TSε u, u)HSε 1−γ ≤ −ε 2 .

Multipeak solutions for a semilinear Neumann problem, Duke Math. J. , 84 (1996), pp. 739-769. [17] Gui, C. , On multiple mixed interior and boundary peak solutions for some singularly perturbed Neumann problems, Canad. J. Math. 52 (2000), no. 3, 522–538. , Multiple boundary peak solutions for some singularly perturbed Neumann problems, Ann. Inst. H. Poincar Anal. Non Linaire 17 (2000), no. 1, 47–82. , Perturbation theory for linear operators. Second edition. Grundlehren der Mathematischen Wissenschaften, Band 132.

Note that, by the normalization of vj,ε (116) v 2 HSε 1 = ε 1 −ξ 2ε αj2 . ,∞ as −1 1 2 Cε fv (x1 ) = ∞ αj φj (x1 ) = 0 βk ψk (x1 ). 9 and (109)) ε−ξ Cε−1 v˜(y) = C 0 βk Ψk (εy1 , y ) + 0 βk ψk (εy1 )vk (y ). 5. Since u3 is orthogonal to H2 , we get (u3 , v)HΣε = (u3 , A1 )HΣε + (u3 , A2 )HΣε + (u3 , A3 )HΣε + (u3 , A4 )HΣε + (u3 , A5 )HΣε . We prove now that Ai holds HSε A1 2HSε is small for every i = 1, . . , 5. 4 there 1 ≤ C ε −1 1 2 Cε αj2 (1 + ε2 λj ) vj − v0 2 j,ε 2 2 ≤ CC (1 + C ) v 2 HSε .