34
L. H. F. de Andrade et al.
Suppose that ψ(αw) ↑ ∞, then for all K > 0, there exists δ > 0 such that
0 < |α − 1| < δ implies that ψ(αw) > K . Let λ
> λ be such that
T ϕ(c + w −
λ
u 0 )dμ < 1, taking K = λ
we have ϕ(c + αw − ψ(w)u 0 ) < ϕ(c + αw {w>0} −
λ
u 0 ) < ϕ(c + w {w>0} − λ
u 0 ), that is a μ-integrable function. Therefore by the
Dominated Convergence Theorem we have
lim
α↑1
T
ϕ(c + αw − λ
u 0 )dμ =
T
ϕ(c + w − λ
u 0 )dμ,
then
1 = lim
α↑1
T
ϕ(c + αw − ψ(αw)u 0 )dμ
≤ lim
α↑1
T
ϕ(c + αw − λ
u 0 )dμ =
T
ϕ(c + w − λ
u 0 )dμ < 1,
which is a contradiction.
If the Condition 2.2 is not satisfied, we elucidated the case of u ∈ ∂B
ϕ
c such that
T ϕ(c + u)dμ = ∞.
We can then make the following question: What about the case of u ∈ ∂B
ϕ
c such
that
T ϕ(c + u)dμ < ∞, how is the behavior of the normalizing function ψ?
This behavior is elucidated in the next proposition.
Proposition 2.7 Consider the deformed exponential function ϕ. Given u ∈ ∂B
ϕ
c such
that
T ϕ(c + u)dμ < ∞, we have that ψ(αu) → β, with β ∈ (0, ∞) as α ↑ 1.
Proof In fact, since
T ϕ(c + u)dμ < ∞, we have
T ϕ(c + u − λu 0 )dμ < ∞ for
all λ > 0. Suppose that ψ(αu) ↑ ∞, as α ↑ 1. Then, for all A > 0, there exists δ > 0,
such that 0 < 1 − α < δ ⇒ ψ(αu) > A.
Since
T ϕ(c + u − λu 0 )dμ < ∞ for all λ > 0, we have that there exists γ >
λ, such that
T ϕ(c + u − γ u 0 )dμ < 1. In particular, take A = γ . Then, from the
Dominated Convergence Theorem it follows that
1 = lim
α↑1
T
ϕ(c + αu − ψ(αu)u 0 )dμ
≤ lim
α↑1
T
ϕ(c + αu − γ u 0 )dμ
=
T
ϕ(c + u − γ u 0 )dμ
<1,
which is an absurd. Therefore, we obtain the desired result.