Page 77 - Textos de Matemática Vol. 34

ON THE PROBLEM OF THE NEGATIVE SPECTRUM 67
Compared with (6.9) we have only to justify the two first lines in (6.10). In what concerns trΓ it comes again from Definition 4.2, because
2−n−d=2−n+d+n−d and −1=−1+1. qtqt
As for id, it is a consequence of Theorem 5.1, because
2 − n + d − d 1 − 1 = 2 − n + d − d > 0.
Moreover, it holds
(6.11) ek(id) ∼ k Ψ♯d(k−1)−1−(2−n+d)/d Ψ♯d(k−1)−1, k ∈ N.
Using the last factorisation, the multiplication property of the entropy numbers
and (6.11), we obtain
e B:B(2−(n−d)/q,Ψ−1/q)(Rn)→B(2−(n−d)/q,Ψ−1/q)(Rn)
k t,∞ t,∞ ≤ c ∥b | Lr(Γ)∥ ek(id)
(6.12) ≤ c′ ∥b | Lr(Γ)∥ k Ψ♯d(k−1)−(2−n+d)/d Ψ♯d(k−1), k ∈ N,
where c′ is a positive constant independent of b and k.
To estimate from above Nβ, given by (6.3), we make use of the entropy version
of the Birman-Schwinger principle, cf. [15, 5.4.1]. We remark that this version of the Birman-Schwinger principle relies on Carl’s inequality and that B has the
same eigenvalues whether it is considered in W 1(Rn) or in B(2−(n−d)/q,Ψ−1/q )(Rn). 2 t,∞
Thus, we can use the estimate (6.12) in
Nβ =#σ(Gβ)∩(−∞,0]≤#k∈N:√2βek(B)≥1
to get (6.4), which is just what we wanted to prove. Remark 6.1. In the case Ψ ∼ 1, which corresponds to say that Γ is a d-set in Rn
if d ∈ (0, n), the estimate from above for Nβ in (6.4) is reduced to (6.13) Nβ ≤c∥b|Lr(Γ)∥βd/(2−n+d),
for all β > 0 and all b ∈ Lr(Γ). This was established by Triebel in [28, Theorem 31.3, p. 245].
Corollary 6.2. Under the same hypotheses of Theorem 6.1, but assuming further that Ψ♯d, the Bruijn conjugate of Ψ(r1/d)−1, satisfies
(6.14) Ψ♯d(ra) ∼ Ψ♯d(r)
tqr