32
L. H. F. de Andrade et al.
Clearly, we have that c (t, v m (t)) < ∞ and 2
−m
c (t, v m (t)) ≥ c−λ m u 0
(t, v m (t)). By Lemma 2.3, there exist an increasing sequence {m n } of indices and
a sequence {A n } of pairwise disjoint, measurable sets such that I c (v m n χ A n ) = 1.
Taking λ n = λ m , u n = v m n and A n , we obtain (2.11).
Next proposition ensures that, assuming Condition 2.2 is not satisfied, we have
that the boundary of B
ϕ
c is not empty.
Proposition 2.5 If the deformed exponential function ϕ does not satisfy Condition 2.2, then c /
∈ 2 .
Proof Suppose that Condition 2.2 does not hold. Take λ > 0. Then, there exists a
n 0 ∈ N, such that λ > λ n , for all n ≥ n 0 .
By Proposition 2.4 and Lemma 2.2, we can take u =
∞
n=n 0
u n χ An . Since u ∈ L
ϕ
c
and I c (u) = ∞, it follows the result.
Next section analyzes the behavior of the normalizing function near the boundary
of the parametrization domain in case the considered points are inside and outside
the Musielak–Orlicz class.
2.3.3 Condition 2.2 is Not Satisfied: Behavior
of the Normalizing Function
In this section we prove that if Condition 2.2 is not satisfied it is still possible to find
u ∈ ∂B
ϕ
c such that
T ϕ(c + u)dμ = ∞ and the normalizing function ψ converges
to some finite value β. Moreover, we verify that for points u belonging to Musielak–
Orlicz class L
c , regardless of occurrence of Condition 2.2, the normalizing function
converges.
In the next proposition we study the behavior of the ψ in the case the boundary
points are not in the Musielak–Orlicz class.
Proposition 2.6 Assuming that Condition 2.2 is not satisfied in the definition of ϕfunction, then there exists u ∈ ∂B
ϕ
c such that
T ϕ(c + u)dμ = ∞ but ψ(αu) → β,
with β ∈ (0, ∞), as α ↑ 1.
Proof Let {λ n }, {u n } and {A n } as in Lemma 2.4. Given any λ > 0, take n 0 ≥ 1
such that λ ≥ λ n for all n ≥ n 0 . Denote B = T \
∞
n=n 0
A n , then we define u =
∞
n=n 0
u n χ A n . From Eq. (2.11), it follows that
L. H. F. de Andrade et al.
Clearly, we have that c (t, v m (t)) < ∞ and 2
−m
c (t, v m (t)) ≥ c−λ m u 0
(t, v m (t)). By Lemma 2.3, there exist an increasing sequence {m n } of indices and
a sequence {A n } of pairwise disjoint, measurable sets such that I c (v m n χ A n ) = 1.
Taking λ n = λ m , u n = v m n and A n , we obtain (2.11).
Next proposition ensures that, assuming Condition 2.2 is not satisfied, we have
that the boundary of B
ϕ
c is not empty.
Proposition 2.5 If the deformed exponential function ϕ does not satisfy Condition 2.2, then c /
∈ 2 .
Proof Suppose that Condition 2.2 does not hold. Take λ > 0. Then, there exists a
n 0 ∈ N, such that λ > λ n , for all n ≥ n 0 .
By Proposition 2.4 and Lemma 2.2, we can take u =
∞
n=n 0
u n χ An . Since u ∈ L
ϕ
c
and I c (u) = ∞, it follows the result.
Next section analyzes the behavior of the normalizing function near the boundary
of the parametrization domain in case the considered points are inside and outside
the Musielak–Orlicz class.
2.3.3 Condition 2.2 is Not Satisfied: Behavior
of the Normalizing Function
In this section we prove that if Condition 2.2 is not satisfied it is still possible to find
u ∈ ∂B
ϕ
c such that
T ϕ(c + u)dμ = ∞ and the normalizing function ψ converges
to some finite value β. Moreover, we verify that for points u belonging to Musielak–
Orlicz class L
c , regardless of occurrence of Condition 2.2, the normalizing function
converges.
In the next proposition we study the behavior of the ψ in the case the boundary
points are not in the Musielak–Orlicz class.
Proposition 2.6 Assuming that Condition 2.2 is not satisfied in the definition of ϕfunction, then there exists u ∈ ∂B
ϕ
c such that
T ϕ(c + u)dμ = ∞ but ψ(αu) → β,
with β ∈ (0, ∞), as α ↑ 1.
Proof Let {λ n }, {u n } and {A n } as in Lemma 2.4. Given any λ > 0, take n 0 ≥ 1
such that λ ≥ λ n for all n ≥ n 0 . Denote B = T \
∞
n=n 0
A n , then we define u =
∞
n=n 0
u n χ A n . From Eq. (2.11), it follows that