28
L. H. F. de Andrade et al.
We have that not all functions u ∈ ∂B
ϕ
c satisfy
T ϕ(c + u)dμ < ∞ or
T ϕ(c +
u)dμ = ∞ as we can see in the next proposition, which is a consequence of the
Lemma 2.1.
Proposition 2.2 [26, Proposition 5] The boundary of B
ϕ
c is non-empty if and only if
the Musielak–Orlicz function c = ϕ(t, c(t) + u) − ϕ(t, c(t)) does not satisfy the
2 -condition. Moreover, in any of these cases, there exist functions w ∗ and w
∗ in
∂B
ϕ
c such that
T ϕ(c + w ∗ )dμ < ∞ and
T ϕ(c + w
∗
)dμ = ∞.
Proof Given non-negative functions u ∗ and u
∗ in L
ϕ
c satisfying (2.7) and (2.8) in
2.1, we consider the functions
w ∗ = u ∗ −
T
u ∗ ϕ
+ (c)dμ
T
u 0 ϕ
+ (c)dμ
u 0 ,
and
w
∗
= u
∗
−
T
u
∗
ϕ
+ (c)dμ
T
u 0 ϕ
+ (c)dμ
u 0 ,
which are in B
ϕ
c . Next, we show that w ∗ is in ∂B
ϕ
c and satisfies
T ϕ(c + w ∗ )dμ < ∞.
For any 0 ≤ λ ≤ 1, it is clear that
T
ϕ(c + λw ∗ )dμ ≤
T
ϕ(c + λu ∗ )dμ < ∞.
Now suppose that
T ϕ(c + λ 0 w ∗ )dμ < ∞ for some λ 0 > 1. In the view of
1 ≤
T ϕ(c + λ 0 w ∗ ) dμ < ∞, we can find α 0 ≥ 0 such that
T ϕ(c + λ 0 w ∗ − α 0 u 0 )
dμ = 1. By the definition of u 0 , fixed any measurable function c such that
T ϕ( c)dμ = 1, we have that
T ϕ( c + αu 0 )dμ < ∞ for all α ∈ R. Hence, considering c = c + λ 0 w ∗ − α 0 u 0 and
α = λ 0
T
u ∗ ϕ
+ (c)dμ
T
u 0 ϕ
+ (c)dμ
+ α 0 .
Hence, we obtain that
T ϕ(c + λ 0 u ∗ )dμ =
T ϕ( c + αu 0 )dμ < ∞, which is a
contradiction. Consequently,
T ϕ(c + λw ∗ )dμ = ∞ for all λ > 1, and w ∗ belongs
to ∂B
ϕ
c and satisfies
T ϕ(c + w ∗ )dμ < ∞.
Proceeding as above, we show that
T ϕ(c + λw
∗
)dμ < ∞ for all 0 ≤ λ < 1,
and
T ϕ(c + λw
∗
)dμ = ∞ for all λ ≥ 1. This result implies that w
∗ belongs to
∂B
ϕ
c and is such that
T ϕ(c + w
∗
)dμ = ∞.
The behavior of ψ depends on whether
T ϕ(c + u) dμ < ∞ holds or not. We
analyze such behavior in the next proposition.
Proposition 2.3 Let u be a function in the boundary of B
ϕ
c . For α ∈ [0, 1), denote
ψ u (α) := ψ(αu). If
T ϕ(c + u) dμ < ∞ then ψ u (α) = ψ(αu) converges to some
β ∈ (0, ∞) as λ ↑ 1. On other side, if u is such that
T ϕ(c + u)dμ = ∞, then
ψ(αu) → ∞ as α ↑ 1.
L. H. F. de Andrade et al.
We have that not all functions u ∈ ∂B
ϕ
c satisfy
T ϕ(c + u)dμ < ∞ or
T ϕ(c +
u)dμ = ∞ as we can see in the next proposition, which is a consequence of the
Lemma 2.1.
Proposition 2.2 [26, Proposition 5] The boundary of B
ϕ
c is non-empty if and only if
the Musielak–Orlicz function c = ϕ(t, c(t) + u) − ϕ(t, c(t)) does not satisfy the
2 -condition. Moreover, in any of these cases, there exist functions w ∗ and w
∗ in
∂B
ϕ
c such that
T ϕ(c + w ∗ )dμ < ∞ and
T ϕ(c + w
∗
)dμ = ∞.
Proof Given non-negative functions u ∗ and u
∗ in L
ϕ
c satisfying (2.7) and (2.8) in
2.1, we consider the functions
w ∗ = u ∗ −
T
u ∗ ϕ
+ (c)dμ
T
u 0 ϕ
+ (c)dμ
u 0 ,
and
w
∗
= u
∗
−
T
u
∗
ϕ
+ (c)dμ
T
u 0 ϕ
+ (c)dμ
u 0 ,
which are in B
ϕ
c . Next, we show that w ∗ is in ∂B
ϕ
c and satisfies
T ϕ(c + w ∗ )dμ < ∞.
For any 0 ≤ λ ≤ 1, it is clear that
T
ϕ(c + λw ∗ )dμ ≤
T
ϕ(c + λu ∗ )dμ < ∞.
Now suppose that
T ϕ(c + λ 0 w ∗ )dμ < ∞ for some λ 0 > 1. In the view of
1 ≤
T ϕ(c + λ 0 w ∗ ) dμ < ∞, we can find α 0 ≥ 0 such that
T ϕ(c + λ 0 w ∗ − α 0 u 0 )
dμ = 1. By the definition of u 0 , fixed any measurable function c such that
T ϕ( c)dμ = 1, we have that
T ϕ( c + αu 0 )dμ < ∞ for all α ∈ R. Hence, considering c = c + λ 0 w ∗ − α 0 u 0 and
α = λ 0
T
u ∗ ϕ
+ (c)dμ
T
u 0 ϕ
+ (c)dμ
+ α 0 .
Hence, we obtain that
T ϕ(c + λ 0 u ∗ )dμ =
T ϕ( c + αu 0 )dμ < ∞, which is a
contradiction. Consequently,
T ϕ(c + λw ∗ )dμ = ∞ for all λ > 1, and w ∗ belongs
to ∂B
ϕ
c and satisfies
T ϕ(c + w ∗ )dμ < ∞.
Proceeding as above, we show that
T ϕ(c + λw
∗
)dμ < ∞ for all 0 ≤ λ < 1,
and
T ϕ(c + λw
∗
)dμ = ∞ for all λ ≥ 1. This result implies that w
∗ belongs to
∂B
ϕ
c and is such that
T ϕ(c + w
∗
)dμ = ∞.
The behavior of ψ depends on whether
T ϕ(c + u) dμ < ∞ holds or not. We
analyze such behavior in the next proposition.
Proposition 2.3 Let u be a function in the boundary of B
ϕ
c . For α ∈ [0, 1), denote
ψ u (α) := ψ(αu). If
T ϕ(c + u) dμ < ∞ then ψ u (α) = ψ(αu) converges to some
β ∈ (0, ∞) as λ ↑ 1. On other side, if u is such that
T ϕ(c + u)dμ = ∞, then
ψ(αu) → ∞ as α ↑ 1.