Page 81 - Textos de Matemática Vol. 34

ON THE PROBLEM OF THE NEGATIVE SPECTRUM 71
λ1,...,λk, with k ≤ Mj −1, and let {e1,...,ek} denote an orthonormal system of eigenfunctions. Then, by Lemma 1.1 of [14, p. 489], we have
μMj (Gβj ) ≥ μk+1(Gβj )
≥inf(Gβjψ,ψ)W21(Rn) :ψ∈[e1,...,ek]⊥,∥ψ|W21(Rn)∥=1 = inf λ ∈ σ(Gβj )\{λ1,...,λk} ≥ 0.
But this contradicts (6.18). Therefore, Gβj has at least Mj eigenvalues less than zero, i.e.,
(6.19) Nβj ≥ Mj ≥ c4 2jd Ψ(2−j )−1 for βj = c3 2j(2−n+d) Ψ(2−j )−1. Denoting by Ψ♯d,n the Bruijn conjugate of Ψ(r1/(2−n+d))−1, we have
Ψ♯ (β−1) ≈ Ψ♯ 2−j(2−n+d) Ψ(2−j) ≈ Ψ(2−j), d,n j d,n
and then
so that (6.19) can be written as
βjΨ♯ (β−1)∼2j; d,n j
(6.20) Nβj ≥ c βd/(2−n+d) Ψ♯ (β−1)(n−2)/(2−n+d). j d,nj
It remains to prove that (6.20) holds not only for βj but for any β sufficiently large. Since the sequence (βj)j is strictly increasing after a finite number of terms, any sufficientlylargeβsatisfies1≤βj <β≤βj+1,forsomej≥j0. Forsuchaβ,we have
Qβ[ψj,l]≤1−βc12−j(2−n+d)Ψ(2−j)=1−c1c3ββ−1 <1−c1c3 <0, j
l = 1,...,Mj. Reasoning as above we may conclude also that
Nβ ≥ Mj ≥ c4 2jd Ψ(2−j)−1 ≥ c5 2(n−2)/(j+1) βj+1
≥ c6 βd/(d−n+2) Ψ♯ j+1 d,n
(6.21) ≥ c6 βd/(d−n+2) Ψ♯ d,n
(β−1 )(n−2)/(2−n+d) j+1
(β−1 )(n−2)/(2−n+d). j+1
Taking into account Proposition 2.1(ii),(iii), there exists a constant c > 0 such that
Ψ♯(β−1) d,n j+1 ≤ c max β−1 β,βj+1β−1 ≤ c max 1,βj+1β−1 ,
Ψ♯d,n (β −1 ) j +1
and, on the other hand, using again Proposition 2.1(ii),
βj+1β−1 = c3 2(j+1)(2−n+d) Ψ(2−(j+1))−1β−1
≤ c2j(2−n+d) Ψ(2−j)−1β−1 ≤ c′ βjβ−1 ≤ c′,