By Heinonen J., Kilpelftine T.

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**Extra info for A -superharmonic functions and supersolutions of degenerate elliptic equations**

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2,ε ) term by term. 1 in [10]) it follows that ei ε2 f2 (εs) ε ˜ 1,ε ) − Sε (ψ˜1,ε ) Sε (Ψ = g˜11 ε4 (f2 )2 ψ˜1,ε + iε3 f2 ψ˜1,ε + 2iε2 f2 ∂s ψ˜1,ε + 2i l i g˜1l ε2 f2 ∂l ψ˜1,ε + √ ∂A g A1 det g˜ 34 det g˜ ε2 f2 ψ˜1,ε . 1 of [10], multiplying the last equation by ei ei f˜(εs) ε = ei A1 f˜0 (εs) ε ei f˜0 (εs) ε ε2 f2 (εs) ε one obtains ˜ 1,ε ) − Sε (ψ˜1,ε ) Sε (Ψ = ei f˜(εs) ε ˜ 1,ε ) − ei Sε (Ψ f˜0 (εs) ε Sε (ψ˜1,ε ) ˜ 1 := A1,0 + A1,r,e + A1,r,o + A1,i,e + A1,i,o + A1,1 , = A1,0 + A (136) where A1,0 = 2ε2 f f2 hU ; A1,r,e = ε3 f2 [2f1 hU + 4 H, Φ f hU + 2f wr,e ] , A1,r,o (137) = ε3 f2 [4 H, z f hU + 2f wr,o ] ; A1,i,e = iε3 f2 hU + 2iε3 f2 [f wi,e + h U + hk ∇U · z] ; A1,1 = A1,i,o = 2iε3 f f2 wi,o (f2 )2 R4 (Φ, Φ ) + f2 R5 (Φ, Φ ) + f2 Φ R4 (Φ, Φ ) + f2 R4 (Φ, Φ ).

L ε ε ε ξ1 ξ2 0 Since the numerator is symmetric in ξ1 , ξ2 , the infimum of the above ratio is realized by some ξ0 , so by (184) and the latter formula we find ∂ν ν 1 ≥ + ∂ε ε ε (185) L (2α2 k 2 0 + 4f αkQ3,α )ξ02 L 2 ξ 0 0 + O(δ 2 ) 1 1 ≥ ν + inf 2α2 k 2 + 4f αkQ3,α − Cδ 2 . ε ε [0,L] Notice that for ν and δ sufficiently small, the coefficient of 1ε in the above formula is positive and uniformly bounded away from zero. From (61) and the asymptotics in (59) (which follows from the Weyl’s formula), 2 one can show that ΠY3,δ Λ0 has a number of negative eigenvalues of order δε .

On Φ and f2 we assume the following conditions for some constants c1 , c2 to be determined later (121) Φ H2 ≤ c1 ε; f2 H2 ≤ c2 . 3 and ϕj , ωj as in (55), we also assume that (122) Φ ∈ span h p+1 4 ϕj : j = 0, . . , δ ε 1 f2 ∈ span h 2 ωj : j = 0, . . , ; δ ε . To deal with the resonance phenomenon mentioned in the introduction, related to the components in K3,δ of the approximate kernel, we add to the approximate solutions a function vδ like (123) vδ = β(εs)Zα(εs) + iξ(εs)Wα(εs) (see (54) and the lines after), with β, ξ given by δ2 ε (124) β(εs) = δ2 ε bj βj (εs); ξ= 2 j=− δε bj ξj (εs), 2 j=− δε where, we recall, ξj solves (59) and is related to βj by (60).