# Mappings of Type Special Space of Sequences

Authors: Awad A. Bakery
Journal: Journal of Function Spaces
 (2016)

---

## Abstract

We give sufficient conditions on a special space of sequences defined by Mohamed and Bakery (2013) such that the finite rank operators are dense in the complete space of operators whose approximation numbers belong to this sequence space. Hence, under a few conditions, every compact operator would be approximated by finite rank operators. We apply it on the sequence space defined by Tripathy and Mahanta (2003). Our results match those known forp-absolutely summable sequences of reals.


---

## Body

## 1. Introduction and Basic Definitions

Byω and L
(
V
,
W
), we will denote the spaces of all real sequences and all bounded linear operators between two Banach spaces V into W, respectively. In [1], Pietsch, by using the approximation numbers andp-absolutely summable sequences of real numbers, formed the operator ideals. In [2], Mohamed and Bakery have considered the space l
M, when M
(
t
)
=
t
p
(
0
<
p
<
∞
), which matches especially l
p. A subclass U of L
=
{
L
(
V
,
W
)
} is an operator ideal if its components verify the following conditions:(i)
The spaceF
(
V
,
W
) of all finite rank operators is a subset of U
(
V
,
W
).
(ii)
The spaceU
(
V
,
W
) is linear.
(iii)
For two Banach spacesV
0 and W
0, if T
∈
L
(
V
0
,
V
), S
∈
U
(
V
,
W
), and R
∈
L
(
W
,
W
0
), then R
S
T
∈
U
(
V
0
,
W
0
). See [3, 4].An Orlicz function is a functionM
:
[
0
,
∞
)
→
[
0
,
∞
) which is convex, positive, nondecreasing, and continuous, where M
(
0
)
=
0 and l
i
m
x
→
∞
M
(
x
)
=
∞. An Orlicz function M is said to satisfy Δ
2-condition for all values of x
≥
0 if there exists a constant k
>
0, such that M
(
2
x
)
≤
k
M
(
x
). Lindenstrauss and Tzafriri [5] used the idea of an Orlicz function to define Orlicz sequence spaces as follows: (1)
l
M
=
x
∈
ω
:
∃
λ
>
0
w
i
t
h
ρ
λ
x
=
∑
k
=
1
∞
M
λ
x
k
<
∞
.
(
l
M
,
x
) is a Banach space, where x
=
i
n
f
{
λ
>
0
:
ρ
(
x
/
λ
)
≤
1
}
. The space l
p is an Orlicz sequence space with M
(
x
)
=
x
p for 1
≤
p
<
∞.Remark 1.
For any Orlicz functionM, we have M
(
λ
x
)
≤
λ
M
(
x
), for all λ with 0
<
λ
<
1.LetP
s be the class of all subsets of N
=
{
0,1
,
2
,
…
} that do not contain more than s number of elements and let {
ϕ
n
} be a nondecreasing sequence of positive reals such that n
ϕ
n
+
1
≤
(
n
+
1
)
ϕ
n, for all n
∈
N. Tripathy and Mahanta [6] defined and studied the following sequence space: (2)
m
ϕ
,
M
=
x
=
x
k
∈
ω
:
∃
ζ
>
0
w
i
t
h
s
u
p
s
≥
1
,
σ
∈
P
s
1
ϕ
s
∑
k
∈
σ
M
x
k
ζ
<
∞
,with the norm (3)
ρ
x
=
i
n
f
ζ
>
0
:
s
u
p
s
≥
1
,
σ
∈
P
s
1
ϕ
s
∑
k
∈
σ
M
x
k
ζ
≤
1
.Lemma 2.
(i)l
M
⊆
m
(
ϕ
,
M
).
(ii)l
M
=
m
(
ϕ
,
M
) if and only if s
u
p
s
ϕ
s
<
∞.As of late, different classes of sequences have been presented using Orlicz functions by Braha [7], Raj and Sharma Sunil [8], Raj et al. [9], and many others ([10–13]).Definition 3 (see [14, 15]).
A special space of sequences (sss) is a linear spaceE with the following: (1)
e
n
∈
E for all n
∈
N, where e
n
=
{
0,0
,
…
,
1,0
,
0
,
…
} with 1 appearing at nth place for all n
∈
N.
(2)
E is s
o
l
i
d
′
′.
(3)
(
x
0
,
x
0
,
x
1
,
x
1
,
…
)
∈
E, if (
x
n
)
n
∈
N
∈
E.
A premodular (sss)E
ρ is a (sss) and there is a function ρ
:
E
→
[
0
,
∞
[ with the following: (i)
ρ
(
x
)
≥
0, for each x
∈
E and ρ
(
x
)
=
0
⇔
x
=
θ, where θ is the zero element of E.
(ii)
ρ satisfies Δ
2-condition.
(iii)
For eachx
,
y
∈
E, ρ
(
x
+
y
)
≤
k
(
ρ
(
x
)
+
ρ
(
y
)
) holds for some k
≥
1.
(iv)
The spaceE is ρ-solid; that is, ρ
(
(
x
n
)
)
≤
ρ
(
(
y
n
)
), whenever |
x
n
|
≤
|
y
n
|, for all n
∈
N.
(v)
For some numbersk
0
≥
1, the inequality ρ
(
(
x
n
)
n
∈
N
)
≤
ρ
(
(
x
0
,
x
0
,
x
1
,
x
1
,
…
)
)
≤
k
0
ρ
(
(
x
n
)
n
∈
N
) holds.
(vi)
F
¯
=
E
ρ; that is, the set of all finite sequences F is ρ-dense in E.
(vii)
For eachλ
>
0, there is a constant ξ
>
0 such that ρ
(
λ
,
0,0
,
0
,
…
)
≥
ξ
λ
ρ
(
1,0
,
0,0
,
…
).Condition (ii) says thatρ is continuous at θ. The function ρ defines a metrizable topology in E and the linear space E enriched with this topology is denoted by E
ρ.Definition 4 (see [16]).
Consider the following:(4)
U
E
a
p
p
≔
U
E
a
p
p
V
,
W
,where (5)
U
E
a
p
p
V
,
W
≔
T
∈
L
V
,
W
:
α
n
T
n
∈
N
∈
E
.Theorem 5 (see [2]).
IfE is a (sss), then U
E
a
p
p is an operator ideal.We explain some results related to the operator spaces.


## 2. Main Results

In this part, we give sufficient conditions onE such that the finite rank operators are dense in the complete space of operators U
E
a
p
p
(
V
,
W
).Lemma 6.
IfE
ρ is a premodular (sss) and (
x
n
)
∈
E
ρ is a decreasing sequence of positive reals, then (6)
ρ
0,0
,
0
,
…
,
0
︷
2
n
,
x
n
,
x
n
+
1
,
x
n
+
2
,
…
≤
k
0
ρ
0,0
,
0
,
…
,
0
︷
n
,
x
n
,
x
n
+
1
,
x
n
+
2
,
…
.Proof.
By using Definition3, conditions (iv) and (v), and since the elements of E are decreasing, we get (7)
ρ
0,0
,
0
,
…
,
0
︷
2
n
,
x
n
,
x
n
+
1
,
x
n
+
2
,
…
≤
ρ
0,0
,
0
,
…
,
0
︷
2
n
,
x
n
,
x
n
,
x
n
+
1
,
x
n
+
1
,
…
≤
k
0
ρ
0,0
,
0
,
…
,
0
︷
n
,
x
n
,
x
n
+
1
,
x
n
+
2
,
…
.Theorem 7.
LetE
ρ be a premodular (sss); then F
(
V
,
W
)
g
¯
=
U
E
ρ
a
p
p
(
V
,
W
), where g
(
T
)
=
ρ
(
α
n
(
T
)
n
∈
N
).Proof.
To prove thatF
(
V
,
W
)
⊆
U
E
a
p
p
(
V
,
W
), since e
m
∈
E for each m
∈
N, from the linearity of E and T
∈
F
(
V
,
W
), then finitely many elements of (
α
n
(
T
)
)
n
∈
N are different from zero. Hence, T
∈
U
E
a
p
p
(
V
,
W
). For the other inclusion U
E
a
p
p
(
V
,
W
)
⊆
F
(
V
,
W
)
¯, let T
∈
U
E
a
p
p
(
V
,
W
) and, from the definition of approximation numbers, there is N
∈
N, N
>
0, A
N with r
a
n
k
(
A
N
)
≤
N and also (8)
T
-
A
N
≤
2
α
N
T
.Since α
N
(
T
)
→
0 as N
→
∞, then T
-
A
N
→
0 as N
→
∞; we have to prove that ρ
(
(
α
n
(
T
-
A
N
)
)
n
∈
N
)
→
0 as N
→
∞, by taking N
=
8
η, where η is a natural number. From Definition 3, condition (iii), we have (9)
d
T
,
A
N
=
ρ
α
n
T
-
A
N
n
∈
N
=
ρ
α
0
T
-
A
N
,
α
1
T
-
A
N
,
…
,
α
8
η
-
1
T
-
A
N
,
0,0
,
0
,
…
+
0,0
,
0
,
…
,
0
︷
8
η
,
α
8
η
T
-
A
N
,
α
8
η
+
1
T
-
A
N
,
…
,
α
12
η
-
1
T
-
A
N
,
0,0
,
0
,
…
+
0,0
,
0
,
…
,
0
︷
12
η
,
α
12
η
T
-
A
N
,
α
12
η
+
1
T
-
A
N
,
…
≤
k
2
ρ
α
0
T
-
A
N
,
α
1
T
-
A
N
,
…
,
α
8
η
-
1
T
-
A
N
,
0,0
,
0
,
…
+
ρ
0,0
,
0
,
…
,
0
︷
8
η
,
α
8
η
T
-
A
N
,
α
8
η
+
1
T
-
A
N
,
…
,
α
12
η
-
1
T
-
A
N
,
0,0
,
0
,
…
+
ρ
0,0
,
0
,
…
,
0
︷
12
η
,
α
12
η
T
-
A
N
,
α
12
η
+
1
T
-
A
N
,
…
=
k
2
I
1
N
+
I
2
η
+
I
3
η
.Since α
n
(
A
N
)
=
0 for n
≥
N, then (10)
α
n
T
-
A
N
≤
α
n
-
N
T
.By using Lemma 6, inequality (10), and Definition 3, condition (v), we get (11)
I
3
η
=
ρ
0,0
,
0
,
…
,
0
︷
12
η
,
α
12
η
T
-
A
N
,
α
12
η
+
1
T
-
A
N
,
…
≤
ρ
0,0
,
0
,
…
,
0
︷
12
η
,
α
4
η
T
,
α
4
η
+
1
T
,
…
≤
k
0
ρ
0,0
,
0
,
…
,
0
︷
6
η
,
α
4
η
T
,
α
4
η
+
1
T
,
…
≤
k
0
2
ρ
0,0
,
0
,
…
,
0
︷
3
η
,
α
4
η
T
,
α
4
η
+
1
T
,
…
≤
k
0
2
ρ
0,0
,
0
,
…
,
0
︷
3
η
,
α
3
η
T
,
α
3
η
+
1
T
,
…
=
k
0
2
ρ
α
n
T
n
=
3
η
∞
⟶
0
a
s
η
⟶
∞
.Now, using Lemma 6 and Definition 3, condition (v), we have (12)
I
2
η
=
ρ
0,0
,
0
,
…
,
0
︷
8
η
,
α
8
η
T
-
A
N
,
α
8
η
+
1
T
-
A
N
,
…
,
α
12
η
-
1
T
-
A
N
,
0,0
,
0
,
…
≤
k
0
ρ
0,0
,
0
,
…
,
0
︷
4
η
,
α
8
η
T
-
A
N
,
α
8
η
+
1
T
-
A
N
,
…
,
α
12
η
-
1
T
-
A
N
,
0,0
,
0
,
…
≤
k
0
ρ
α
0
T
-
A
N
,
α
1
T
-
A
N
,
…
,
α
8
η
-
1
T
-
A
N
,
0,0
,
0
,
…
=
k
0
I
1
N
.Finally, we have to show that I
1
(
N
)
→
0 as N
→
∞. Since T
∈
U
E
a
p
p
(
V
,
W
) and ρ is continuous at θ, we have ρ
(
α
k
(
T
)
)
k
=
n
∞
→
0 as n
→
∞. Then, for each ε
>
0, there exists N
0
(
ε
) such that for all n
≥
N
0
(
ε
) we have (13)
ρ
α
k
T
k
=
n
∞
<
ε
.By taking ε
1
=
ε
/
3
l
k for each n
≥
N
0
(
ε
1
) and using inequality (13) and Definition 3, conditions (ii) and (iii), then we have (14)
I
1
N
=
ρ
α
0
T
-
A
N
,
α
1
T
-
A
N
,
…
,
α
N
-
1
T
-
A
N
,
0,0
,
0
,
…
≤
k
ρ
α
0
T
-
A
N
,
α
1
T
-
A
N
,
…
,
α
N
0
-
1
T
-
A
N
,
0,0
,
0
,
…
+
ρ
0,0
,
0
,
…
,
0
︷
N
0
,
α
N
0
T
-
A
N
,
α
N
0
+
1
T
-
A
N
,
…
,
α
N
-
1
T
-
A
N
,
0,0
,
0
,
…
≤
k
ρ
T
-
A
N
,
T
-
A
N
,
…
,
T
-
A
N
,
0,0
,
0
,
…
+
ρ
0,0
,
0
,
…
,
0
︷
N
0
,
2
α
N
T
,
2
α
N
T
,
…
,
2
α
N
T
,
0,0
,
0
,
…
≤
k
T
-
A
N
l
ρ
1,1
,
1
,
…
,
1
︷
N
0
,
0,0
,
0
,
…
+
2
l
ρ
0,0
,
0
,
…
,
0
︷
N
0
,
α
N
0
T
,
α
N
0
+
1
T
,
…
,
α
N
T
,
0,0
,
0
,
…
≤
k
T
-
A
N
l
k
1
ε
+
2
l
ε
1
,where k
1
(
ε
)
=
ρ
(
1,1
,
1
,
…
,
1
︷
N
0
,
0,0
,
0
,
…
), and since T
-
A
N
→
0
a
s
N
→
∞, then for each ε
>
0 there exists N such that T
-
A
N
k
1
(
ε
)
≤
ε
1, for that we have I
1
(
N
)
≤
k
[
l
ε
1
+
2
l
ε
1
]
=
3
k
l
ε
1
=
ε
. This completes the proof.We give here the sufficient conditions on the sequence spacesm
(
ϕ
,
M
) such that the class of all bounded linear operators between any arbitrary Banach spaces with (α
n
(
T
)
)
n
∈
N in these sequence spaces form an ideal operator; the ideal of the finite rank operators in the class of Banach spaces is dense in U
m
(
ϕ
,
M
)
a
p
p
(
V
,
W
).Theorem 8.
LetM be an Orlicz function satisfying Δ
2-condition. Then,(a)
U
m
(
ϕ
,
M
)
a
p
p is an operator ideal,
(b)
F
(
V
,
W
)
¯
=
U
m
(
ϕ
,
M
)
a
p
p
(
V
,
W
).Proof.
We first prove that the spacem
(
ϕ
,
M
) is a (sss).
(1) Letλ
1
,
λ
2
∈
R and x
,
y
∈
m
(
ϕ
,
M
); then there exist ζ
1
>
0 and ζ
2
>
0 such that (15)
s
u
p
s
≥
1
,
σ
∈
P
s
1
ϕ
s
∑
k
∈
σ
M
x
k
ζ
1
<
∞
,
s
u
p
s
≥
1
,
σ
∈
P
s
1
ϕ
s
∑
k
∈
σ
M
x
k
ζ
2
<
∞
.Let ζ
3
=
m
a
x
(
2
|
λ
1
|
ζ
1
,
2
|
λ
2
|
ζ
2
). Since M is nondecreasing convex function with Δ
2-condition, we have (16)
∑
k
∈
σ
M
λ
1
x
k
+
λ
2
y
k
ζ
3
≤
1
2
∑
k
∈
σ
M
x
k
ζ
1
+
∑
k
∈
σ
M
y
k
ζ
2
.So, we get (17)
s
u
p
s
≥
1
,
σ
∈
P
s
1
ϕ
s
∑
k
∈
σ
M
λ
1
x
k
+
λ
2
y
k
ζ
3
≤
1
2
s
u
p
s
≥
1
,
σ
∈
P
s
1
ϕ
s
∑
k
∈
σ
M
x
k
ζ
1
+
s
u
p
s
≥
1
,
σ
∈
P
s
1
ϕ
s
∑
k
∈
σ
M
y
k
ζ
2
.Thus, λ
1
x
+
λ
2
y
∈
m
(
ϕ
,
M
). Hence, m
(
ϕ
,
M
) is a linear space over the field of real numbers. Also, since e
n
∈
l
M and l
M
⊆
m
(
ϕ
,
M
), we have e
n
∈
m
(
ϕ
,
M
) for all n
∈
N.
(2) Letx
∈
ω and y
=
(
y
k
)
k
=
0
∞
∈
m
(
ϕ
,
M
) with |
x
k
|
≤
|
y
k
| for each k
∈
N; since M is nondecreasing, then we get (18)
s
u
p
s
≥
1
,
σ
∈
P
s
1
ϕ
s
∑
k
∈
σ
M
x
k
ζ
≤
s
u
p
s
≥
1
,
σ
∈
P
s
1
ϕ
s
∑
k
∈
σ
M
y
k
ζ
<
∞
;then x
=
(
x
k
)
k
=
0
∞
∈
m
(
ϕ
,
M
).
(3) Letx
=
(
x
k
)
k
=
0
∞
∈
m
(
ϕ
,
M
); then we have (19)
s
u
p
s
≥
1
,
σ
∈
P
s
1
ϕ
s
∑
k
∈
σ
M
x
k
/
2
ζ
≤
2
s
u
p
s
≥
1
,
σ
∈
P
s
1
ϕ
s
∑
k
∈
σ
M
x
k
ζ
<
∞
;then x
=
(
x
[
k
/
2
]
)
k
=
0
∞
∈
m
(
ϕ
,
M
).
Finally, we have proved that the spacem
(
ϕ
,
M
) with ρ
(
x
) is a premodular (sss).
(i) Clearly,ρ
(
x
)
≥
0 for all x
∈
m
(
ϕ
,
M
) and ρ
(
x
)
=
0
⇔
x
=
θ.
(ii) Letλ
∈
R and x
∈
m
(
ϕ
,
M
); then for λ
≠
0 we have (20)
ρ
λ
x
=
i
n
f
ζ
>
0
:
s
u
p
s
≥
1
,
σ
∈
P
s
1
ϕ
s
∑
k
∈
σ
M
λ
x
k
ζ
≤
1
=
i
n
f
λ
μ
>
0
:
s
u
p
s
≥
1
,
σ
∈
P
s
1
ϕ
s
∑
k
∈
σ
M
x
k
μ
≤
1
,where μ
=
ζ
/
|
λ
|. Thus, ρ
(
λ
x
)
=
|
λ
|
i
n
f
{
μ
>
0
:
s
u
p
s
≥
1
,
σ
∈
P
s
1
/
ϕ
s
∑
k
∈
σ
‍
M
(
|
x
k
|
/
μ
)
≤
1
}
=
|
λ
|
ρ
(
x
)
.
Also, forλ
=
0, we have ρ
(
λ
x
)
=
λ
ρ
(
x
)
=
0
.
(iii) Letx
,
y
∈
m
(
ϕ
,
M
); then there exist ζ
1
>
0 and ζ
2
>
0 such that (21)
s
u
p
s
≥
1
,
σ
∈
P
s
1
ϕ
s
∑
k
∈
σ
M
x
k
ζ
1
≤
1
,
s
u
p
s
≥
1
,
σ
∈
P
s
1
ϕ
s
∑
k
∈
σ
M
x
k
ζ
2
≤
1
.Let ζ
=
ζ
1
+
ζ
2, and since M is nondecreasing and convex, then we have (22)
sup
s
≥
1
,
σ
∈
P
s
⁡
1
ϕ
s
∑
k
∈
σ
M
x
k
+
y
k
ζ
≤
sup
s
≥
1
,
σ
∈
P
s
⁡
1
ϕ
s
∑
k
∈
σ
M
x
k
+
y
k
ζ
1
+
ζ
2
≤
sup
s
≥
1
,
σ
∈
P
s
⁡
1
ϕ
s
∑
k
∈
σ
ζ
1
ζ
1
+
ζ
2
M
x
k
ζ
1
+
ζ
2
ζ
1
+
ζ
2
M
y
k
ζ
2
≤
ζ
1
ζ
1
+
ζ
2
sup
s
≥
1
,
σ
∈
P
s
⁡
1
ϕ
s
∑
k
∈
σ
M
x
k
ζ
1
+
ζ
2
ζ
1
+
ζ
2
∑
k
∈
σ
M
y
k
ζ
2
≤
1
.Since ζ’s are nonnegative, we have (23)
ρ
x
+
y
=
i
n
f
ζ
>
0
:
s
u
p
s
≥
1
,
σ
∈
P
s
1
ϕ
s
∑
k
∈
σ
M
x
k
+
y
k
ζ
≤
1
≤
i
n
f
ζ
1
>
0
:
s
u
p
s
≥
1
,
σ
∈
P
s
1
ϕ
s
∑
k
∈
σ
M
x
k
ζ
1
≤
1
+
i
n
f
ζ
2
>
0
:
s
u
p
s
≥
1
,
σ
∈
P
s
1
ϕ
s
∑
k
∈
σ
M
y
k
ζ
2
≤
1
=
ρ
x
+
ρ
y
.
(iv) Let|
x
k
|
≤
|
y
k
| for each k
∈
N, and since M is nondecreasing, then we get (24)
s
u
p
s
≥
1
,
σ
∈
P
s
1
ϕ
s
∑
k
∈
σ
M
x
k
ζ
≤
s
u
p
s
≥
1
,
σ
∈
P
s
1
ϕ
s
∑
k
∈
σ
M
y
k
ζ
;thus, (25)
i
n
f
ζ
>
0
:
s
u
p
s
≥
1
,
σ
∈
P
s
1
ϕ
s
∑
k
∈
σ
M
x
k
ζ
≤
i
n
f
ζ
>
0
:
s
u
p
s
≥
1
,
σ
∈
P
s
1
ϕ
s
∑
k
∈
σ
M
y
k
ζ
.So, ρ
(
x
)
≤
ρ
(
y
).
(v) Since(26)
s
u
p
s
≥
1
,
σ
∈
P
s
1
ϕ
s
∑
k
∈
σ
M
x
k
/
2
ζ
≤
2
s
u
p
s
≥
1
,
σ
∈
P
s
1
ϕ
s
∑
k
∈
σ
M
x
k
ζ
,we have (27)
i
n
f
ζ
>
0
:
s
u
p
s
≥
1
,
σ
∈
P
s
1
ϕ
s
∑
k
∈
σ
M
x
k
/
2
ζ
≤
2
i
n
f
ζ
>
0
:
s
u
p
s
≥
1
,
σ
∈
P
s
1
ϕ
s
∑
k
∈
σ
M
x
k
ζ
.So, ρ
(
(
x
k
)
)
≤
ρ
(
(
x
[
k
/
2
]
)
)
≤
2
ρ
(
(
x
k
)
).
(vi) For eachx
=
(
x
k
)
k
=
0
∞
∈
m
(
ϕ
,
M
), then (28)
ρ
x
k
k
=
0
∞
=
i
n
f
ζ
>
0
:
s
u
p
s
≥
1
,
σ
∈
P
s
1
ϕ
s
∑
k
∈
σ
M
x
k
/
2
ζ
<
∞
;we can find t
∈
N such that ρ
(
(
x
k
)
k
=
t
∞
)
<
∞. This means the set of all finite sequences is ρ-dense in m
(
ϕ
,
M
).
(vii) For anyλ
>
0, there exists a constant ζ
∈
]
0,1
] such that (29)
ρ
λ
,
0,0
,
0
,
…
≥
ζ
λ
ρ
1,0
,
0,0
,
…
.By using Theorems 5 and 7, the proof follows.As special cases of the above theorem, we obtain the following corollaries.Corollary 9.
Ifsup
⁡
s
ϕ
s
<
∞, one gets that (a)
U
l
M
a
p
p is an operator ideal,
(b)
F
(
V
,
W
)
¯
=
U
l
M
a
p
p
(
V
,
W
).Corollary 10.
Ifsup
⁡
s
ϕ
s
<
∞ and M
(
t
)
=
t
p with 0
<
p
<
∞, one gets that (a)
U
l
p
a
p
p is an operator ideal,
(b)
F
(
V
,
W
)
¯
=
U
l
p
a
p
p
(
V
,
W
). See [1].Theorem 11.
IfE
ρ is a premodular (sss), then U
E
ρ
a
p
p
(
V
,
W
) is complete.Proof.
Let(
T
m
) be a Cauchy sequence in U
E
ρ
a
p
p
(
V
,
W
); then, by using Definition 3, condition (vii), and since U
E
ρ
a
p
p
(
V
,
W
)
⊆
L
(
V
,
W
), we get (30)
ρ
α
n
T
i
-
T
j
n
∈
N
≥
ρ
α
0
T
i
-
T
j
,
0,0
,
0
,
…
=
ρ
T
i
-
T
j
,
0,0
,
0
,
…
≥
ξ
T
i
-
T
j
ρ
1,0
,
0,0
,
…
;then (
T
m
) is also a Cauchy sequence in L
(
V
,
W
). Since the space L
(
V
,
W
) is a Banach space, then there exists T
∈
L
(
V
,
W
) such that T
m
-
T
→
0, as m
→
∞, and since (
α
n
(
T
m
)
)
n
∈
N
∈
E, for each m
∈
N, then from Definition 3, conditions (iii) and (v), and since ρ is continuous at θ, we have (31)
ρ
α
n
T
n
∈
N
=
ρ
α
n
T
-
T
m
+
T
m
n
∈
N
≤
k
ρ
α
n
/
2
T
-
T
m
n
∈
N
+
k
ρ
α
n
/
2
T
m
n
∈
N
≤
k
ρ
T
m
-
T
n
∈
N
+
k
ρ
α
n
T
m
n
∈
N
<
ε
;we get (
α
n
(
T
)
)
n
∈
N
∈
E, and then T
∈
U
E
ρ
a
p
p
(
V
,
W
). This finishes the proof.By applying Theorem11 on m
(
ϕ
,
M
), we can easily conclude the next corollaries.Corollary 12.
Pick up an Orlicz functionM which satisfies Δ
2-condition. Then, M is continuous from the right at 0 and U
m
(
ϕ
,
M
)
a
p
p
(
V
,
W
) is complete.Corollary 13.
Pick up an Orlicz functionM which satisfies Δ
2-condition with sup
⁡
s
ϕ
s
<
∞. Then, M is continuous from the right at 0 and U
l
M
a
p
p
(
X
,
Y
) is complete.Corollary 14.
U
l
p
a
p
p
(
V
,
W
) is complete if M
(
t
)
=
t
p and p
∈
(
0
,
∞
).



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*Source: 4372471-2016-08-28.xml*