Page 78 - Textos de Matemática Vol. 34

68 SUSANA D. MOURA
for any a > 0, then we have the same assertions of that theorem and, moreover,
 d −1  ∥b|Lr(Γ)∥β d+1 Ψ (∥b|Lr(Γ)∥β)−1 d+1 ,

n = 1,
n = 2, 2−n+d, n>2,
Nβ ≤ c
for any large enough β and for all b ∈ Lr(Γ).
∥b|Lr(Γ)∥β,
 d

 ∥b|L (Γ)∥β 2−n+d (1/Ψ♯)♯ (∥b|L (Γ)∥β)− d
n−2
r drn−2 Proof. The case n = 2 plainly follows from (6.4).
Now let n = 1 and consider the function f defined by
f(x) := x(1+d)/d Ψ♯d(x−1)1/d, x ∈ [1,∞).
From (6.14),
and, by Proposition 2.1(iv)(v), for its asymptotic inverse g holds
f(x) ∼ x(1+d)/d Ψ♯d(x−(1+d))1/d
g(x) ≈ xd/(1+d) (Ψ♯d)♯(x−d)1/(1+d) ≈ x(1+d)/d Ψ(x−1)−1/(1+d)
(x → ∞). Note that by Proposition 2.1(iii), g is asymptotically equivalent to a non-decreasing
function. Thus, if then
f(k) ≤ c1 ∥b|Lr(Γ)∥β, k ≤ c2 gc1 ∥b|Lr(Γ)∥β
d −1 ≤ c3 ∥b | Lr(Γ)∥ β d+1 Ψ (c1∥b | Lr(Γ)∥ β)−1 d+1
d −1 ≤ c4 ∥b|Lr(Γ)∥β d+1 Ψ (∥b|Lr(Γ)∥β)−1 d+1
for any k and β sufficiently large, which, in view of (6.4), gives the desired estimate. Finally, suppose n > 2. Now let
f(x) := x(2−n+d)/d Ψ♯d(x−1)−(n−2)/d, x ∈ [1,∞). Using the same arguments as above,
f(x) ∼ x(2−n+d)/d Ψ♯d(x−(2−n+d)/(n−2))−(n−2)/d,
whose asymptotic inverse g is asymptotically equivalent to a non-decreasing func- tion and
g(x) ≈ xd/(2−n+d) (1/Ψ♯d)♯(x−d/(n−2))(n−2)/((2−n+d)) ≈ x(1+d)/d Ψ(x−1)−1/(1+d) (x → ∞).