In this paper, we introduce an order φ-function on a Riesz space. Further, we construct a Riesz valued sequence spaces by using the φ-function and obtain the condition to characterize these spaces.

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Numerical Functional Analysis and Optimization
ISSN: 0163-0563 (Print) 1532-2467 (Online) Journal homepage: http://www.tandfonline.com/loi/lnfa20
Some Characterizations of Riesz Valued Sequence
Spaces generated by an Order ϕ-Function
E. Herawati, Supama, M. Mursaleen & M. Nasution
To cite this article: E. Herawati, Supama, M. Mursaleen & M. Nasution (2017): Some
Characterizations of Riesz Valued Sequence Spaces generated by an Order ϕ-Function, Numerical
Functional Analysis and Optimization, DOI: 10.1080/01630563.2017.1351987
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Some Characterizations of Riesz Valued Sequence Spaces generated by an Order
φ-Function
E. Herawati1, Supama2, M. Mursaleen3, and M. Nasution4
1Department of Mathematics, University of Sumatera, Utara, Indonesia
2Department of Mathematics, Gadjah Mada University, Yogyakarta, Indonesia
3Department of Mathematics, Aligarh Muslim University, India
4Department of Mathematics, University of North Sumatera, Indonesia
Abstract
In this paper, we introduce an order φ-function on a Riesz space. Further, we construct a Riesz
valued sequence spaces by using the φ-function and obtain the condition to characterize these
spaces.
KEYWORDS: Order φ-function; σ-order closed; modular; order convergence
2010 Mathematics Subject Classification: Primary 39B82; Secondary 44B20; 46C05
Received 12 August 2016; Accepted 4 July 2017
M. Mursaleen mursaleenm@gmail.com Department of Mathematics, Aligarh Muslim University,
India.
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1. INTRODUCTION AND PRELIMINARIES
A concept of a vector valued sequence space (VVSS) was introduced by some authors based on
the work of Pietch [13]. By using the Orlicz function, Ghosh and Srivastava [4] and Kolk [8]
constructed some VVSS (See, e.g. [1–3, 9, 10, 12]). They observed some topological properties of
the spaces, as well. Walsh [14] introduced a certain type of VVSS and applied the space to study of
K-absolutely summing and majorizing operators for Banach lattices. Later on, Gupta and Kaushal
[5] and Herawati et al. [6, 7] discussed about Riesz valued sequence spaces.
A Riesz space was introduced in [11] and some basic notions related to Riesz space refer to [11, 15].
In this paper, we introduce an order φ-function from a Riesz space into its positive cone and discuss
a Riesz valued sequence space generated by the order φ-function. Notice that the space can be
equipped with a norm and an order modular defined by the order φ -function. Henceforth, we
observe some notions, such as σ-order closed of subspace, σ-Fatou property, diagonal property,
and σ-order continuous norm. We also consider some relationships between the norm and the order
modular.
In this section, we mention the silent features of Riesz spaces and Riesz valued sequence spaces
required for our present work. However, for unexplained terms in these theories, the reader is urged
to look into [11, 15].
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Let E be a Riesz space. A sequence (fk) ⊂ E is said to be decreasing (resp. increasing), denoted
by fk ↓ resp.fk ↑ if fk ≥ fk+1 (resp. fk ≤ fk+1) for every k ∈ N. We denote fk ↓ f0 resp.fk ↑ f0 ,
whenever fk ↓ and f0 = infk fk resp.fk ↑ and f0 = supk fk . The ideal A in E is said to be a σ-
ideal whenever (fk) ⊂ A and that 0 ≤ fk ↑ f0 imply f0 ∈ A. A sequence (fk) ⊂ E is said to be
order convergent to some f0 ∈ E, denoted by fk
o
−→k f0, if there exists a decreasing sequence (pk)
satisying pk ↓ 0 in E such that |fk − f0|≤ pk for all k ∈ N. In this case, f0 is called an order limit of
the sequence (fk).
A sequence (fk) ⊂ E is said to be relatively uniformly convergent to some f0 ∈ E, denoted by
fk
o
−→k f0(ru), if there exists u ∈ E+ and a sequence of numbers εk ↓ 0 such that |fk −f0|≤ εku for
every k. A sequence (fk) in E is said to be an order Cauchy sequence, if there exists a decreasing
sequence (pk) satisying pk ↓ 0 in E such that |fj − fk|≤ pk for all j ≥ k ≥ 1. The Riesz space E is
said to be order complete, if every order Cauchy sequence in E has an order limit.
Any sequence fk
o
−→k 0 in a Riesz space E is said to be stable, if there exists a sequence of a real
numbers (λk) such that 0 ≤ λk ↑ ∞ and λkfk
o
−→k 0. Relatively uniform convergence in Riesz
space is stable, since for any sequence fk
o
−→k 0(ru), there exists a sequence positive real numbers
(εk) and u ∈ E+ such that εk ↓ 0 and |fk|≤ εku for all k and so for λk = 1√
εk
satisfies the condition
of the stable property. A Riesz space E is said to have diagonal property (for order convergence),
if given that
fn,k
o
−→k fn
o
−→n f0
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hold in E, there exists a “diagonal sequence”
fn,k(n) : n = 1, 2, . . .
converging to f0, i.e., fn,k(n)
o
−→n f0.
In any Archimedean Riesz space E, there is a positive correlation between a stable sequence and
the diagonal property. The correlation is stated in the following theorem, where its proof can be
seen, for e.g. in [11].
Theorem 1.1. Let E be an Archimedean, then the following statements are equivalent.
(i) E has the diagonal property.
(ii) Any sequence (fk) in E which is order convergent to 0 is stable and for any sequence (fk) ⊂ E+,
there exists a sequence (λk) of positive numbers such that (λkfk) is bounded above.
A Riesz space E equipped with a norm · is called a normed Riesz space if |f| ≤ |g| in E implies
f ≤ g . Further, a normed Riesz space E is said to satisfy the σ-Fatou property, if 0 ≤ fk ↑ f0
in E implies fk ↑ f0 . The norm · in a normed Riesz space E is said to be σ-order continuous,
if fk ↓ f0 or fk ↑ f0 in E implies fk − f0 → 0.
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Let E be a Riesz space. The set of all E-valued sequences will be denoted by (E). A sequence
fk ∈ (E) is said to be finite if there exists an N ∈ N such that fk = 0 for every k > N. The set
of all finite E-valued sequences will be denoted by Ø(E). Any subspace X(E) in (E) containing
Ø(E) is called a Riesz valued sequence space or an RVSS in short. It is easy to check that any
RVSS X(E) in (E) is a Riesz space with respect to the coordinate-wise ordering. The followings
are examples of RVSS (See, for e.g. [5, 6]).
ℓ(E) = f = (fk) ∈ (E) : (∃f0 ∈ E),
n
k=1
|fk|
o
−→n f0 .
c(E) = f = (fk) ∈ (E) : (∃f0 ∈ E), fk
o
−→k f0 .
ℓ∞(E) = f = (fk) ∈ (E) : (∃f0 ∈ E), sup
k
{fk} = f0 .
c0(E) = f = (fk) ∈ (E) : fk
o
−→k 0 .
We observed the following theorems.
Theorem 1.2. If E is an order complete Riesz space, then (E) and Ø(E) are also order complete
(E).
Theorem 1.3. A Riesz subspace X(E) of (E) is order complete if and only if it is an ideal of
(E).
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It is easy to check that ℓ(E), c0(E) and ℓ∞(E) are ideals of (E), while c(E) is a Riesz subspace
of (E) which is not an ideal.
2. RIESZ VALUED SEQUENCE SPACES DEFINED BY ORDER φ-FUNCTION
In this section, we introduce the complete normed Riesz spaces Xα(E, ϕ) and X∃(E, ϕ). Let E be a
Riesz space and E+ be a positive cone of E, i.e. E+ = {x ∈ E : 0 ≤ x}. A function ϕ : E → E+
is called an order φ-function if
(i) ϕ(f) = 0 if and only if f = 0,
(ii) ϕ is increasing on E+,
(iii) ϕ is order continuous on E, and
(iv) ϕ(−f) = ϕ(f) for every f ∈ E.
The order φ-function ϕ is said to satisfy the 2 -condition, denoted ϕ ∈ 2, if there exists a
constant M > 0 such that ϕ 2f ≤ Mϕ(f) for every f ∈ E+.
Let ϕ be an order φ-function on E and let α be a positive real number. For a RVSS X(E) ⊂ (E)
we define the following sets
(1) Xα(E, ϕ) = f = (fk) ∈ (E) : f
α ∈ X(E) ,
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(2) X∃(E, ϕ) = f = (fk) ∈ (E) : (∃α > 0) f
α ∈ X(E) ,
where (f) = ϕ(fk) . It is easy to prove that Xα(E, ϕ) =
α>0
X∃(E, ϕ). The set X∃(E, ϕ) may not
be a sequence space. However, a routine verification shows that if X(E) is an ideal space and ϕ is
a convex order φ-function, then X∃(E, ϕ) is an ideal space of (E). Moreover, X∃(E, ϕ) is a Riesz
space with respect to the coordinate-wise ordering and Xα(E, ϕ) is an ideal of X∃(E, ϕ).
Let E be a Riesz space. A function ρ : E → [0, ∞) is called a modular if for every v, w ∈ E we
have
(i) ρ(v) = 0 if and only if v = 0
(ii) ρ(v) = ρ(|v|)
(iii) ρ(v ∨ w) ≤ ρ(v) + ρ(w)
(iv) If 0 ≤ v ≤ w then ρ(v) ≤ ρ(w).
The Riesz space E equipped with a modular ρ is called a modular Riesz space. For any convex
order φ-function ϕ and any ideal normed space X(E), the space X∃(E, ϕ) is a moduar Riesz space
with respect to the modular
ρϕ(f) =
(f) if (f) ∈ X(E)
∞ otherwise
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Furthermore, the space is a normed Riesz space with respect to the Luxemburg norm
f ϕ = inf ε > 0 : ρϕ
f
ε
≤ 1 .
Lemma 2.1. Let ϕ be a convex order φ-function. If X(E) is a normed Riesz space satisfied the
σ-Fatou property, then X∃(E, ϕ) equipped with the norm · ϕ satisfies the σ-Fatou property as
well.
Proof. Suppose it is not true. Then there exists a sequence (f(n)) ⊂ X∃(E, ϕ) and f ∈ (E) with
0 ≤ f(n) ↑n f, but f(n)
ϕ n f ϕ. Since f(n)
ϕ ↑n< ∞, then there exists r ∈ R+ such that
r = sup
n≥1
f(n)
ϕ . Therefore, for every n ∈ N, we have
ρϕ
f(n)
r+1 ≤ ρϕ
f(n)
f(n)
φ
≤ 1.
Since f(n)
r+1 ↑ f
r and X(E) satisfies the σ-Fatou property, then
ρϕ
f
r+1 = f
r+1 = lim
n→∞
f(n)
r+1 ≤ 1.
So, f φ ≤ r + 1. Since 0 ≤ f(n) ↑n f and f(n)
φ n f φ, then there exists ε0 > 0 such that
for every k ∈ N there exixts nk > k such that f(nk)
φ < f φ − ε0. Therefore,
ρφ
f(nk)
f φ − ε0
< ρφ
f(nk)
f(nk)
φ
≤ 1 (2.1)
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Since f φ − ε0
−1
f(nk) ≤ f φ − ε0
−1
f for every k ∈ N and X(E) satisfies the σ-Fatou
property, then (2.1) implies
f
f φ−ε0
= lim
k→∞
f(nk)
f(nk)
φ
≤ 1.
So, we have f φ < f φ − ε0, that is a contradiction. Therefore f(n)
φ→ f φ.
It can be proved that any normed Riesz space which satisfy the σ-Fatou property is complete.
Hence, X∃(E, ϕ) is a Banach lattice. In the following theorem we show that under a certain
assumption, Xα(E, ϕ) is an Archimedean σ-ideal of X∃(E, ϕ).
Theorem 2.2. Let ϕ be a convex order φ-function on a σ-Dedekind complete Riesz space E and
X(E) an ideal of (E). Then for α > 0, Xα(E, ϕ) is an Archimedean σ-ideal of X∃(E, ϕ).
Proof. It has been mentioned before that Xα(E, ϕ) is an ideal of X∃(E, ϕ). It is guaranteed by the
convexity of ϕ. For proving that Xα(E, ϕ) is a σ-ideal of X∃(E, ϕ), we take any sequence (f(n))
in Xα(E, ϕ) that satisfies 0 ≤ f(n) ↑ f and prove that f ∈ Xα(E, φ). Since is increasing,
then (f(n)) ↑≤ (f) in X(E). Therefore, by following the assumption that X(E) is σ-Dedekind
complete, there exists u ∈ X(E) such that (f(n)) ↑ u and u ≤ (f). Since (f(n)) ≤ u for
every n and is an increasing function on X(E)+, then f(n) ≤ −1(u) for every n. These imply
f ≤ −1(u) or (f) ≤ u. So, we have (f(n)) ↑ (f) in X(E). And hence, f ∈ Xα(E, φ).
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Now, we are going to prove that Xα(E, φ) is an Archimedean. To do that take any sequence f, g ∈
Xα(E, ϕ) such that 0 ≤ f ≤ 1
ng for every n ∈ N. Since is increasing, then 0 ≤ (f) ≤ 1
n (g)
in X(E). Since X(E) is σ-Dedekind complete, then 1
n (g) ↓ (f). Since (f) ≤ 1
n (g) ↓ 0, then
(f) = 0. So we get f = 0.
3. SOME PROPERTIES OF Xα(E, ϕ)
Theorem 3.1. Let ϕ be a convex order φ-function. If X(E) is a normed Riesz space which satisfies
the σ-Fatou property, then any sequence in Xα(E, ϕ) which is order convergent to 0 is stable.
Proof. Take any sequence (f(n)) ⊂ Xα(E, ϕ) such that f(n) o
−→n 0. Since f(n) o
−→n 0 in Xα(E, φ),
then there exists p(n) ↓ 0 in Xα(E, ϕ) such that |f(n)|≤ p(n) for every n ∈ N. Since p(n) ↓ 0, then
p(n) ↓ 0. Since Xα(E, ϕ) is a Banach lattice, then there exists a subse quence (qk) = (p(nk)) of
(p(n)) such that qk = p(nk) < k−1.2−k. Now, for any n ∈ N we define a positive integer λn by
λn =
1 if n < n1
k for nk ≤ n ≤ nk+1 (k = 1, 2, . . .
then 0 < λn ↑ ∞. Since, λnp(n) ≤ kp(nk) = kqk for any for nk ≤ n ≤ nk+1, then λnp(n) → 0 in
Xα(E, ϕ). So, we get λnp(n) o
−→n 0 and the assertion follows.
Theorem 3.2. Let ϕ be a convex order φ-function. If X(E) is a normed Riesz space satisfies the
σ-Fatou property, then Xα(E, φ) has the diagonal property.
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Proof. Take any sequence (f(n)) ⊂ Xα(E, ϕ)+. It is sufficient to prove the existence of positive
numbers λn (n = 1, 2, . . . ) such that the sequence (λnf(n)) is bounded above in Xα(E, ϕ+). Since
1
α > 0, then there exists a sequence of positive numbers (λn) = 1/2nα such that (λnf(n)) ≤
fn
α ∈ X(E). Since X(E) is an ideal, then (λnf(n)) ∈ X(E). Since E has an order unit u, then there
exists γ > 0 such that
(λnf(n)) = 1
2nα λnf(n) ≤ 1
2n
f(n)
α ≤ γ
2n .
This implies 0 ≤ λnf(n) ≤ 1
2n
−1(λu), for some real number λ. Therefore,
∞
n=1
λnf(n) < ∞. Since
Xα(E, ϕ)+ is a Banach lattice, then there exists w ∈ Xα(E, ϕ)+ such that
m
n=1
λnf(n) ↑ w. Hence,
(λnf(n)) is bounded above.
Theorem 3.3. Let X(E) ⊂ (E) be an ideal. If ϕ is an order φ-function satisfies 2-condition,
then
Xα(E, ϕ) = X∃(E, ϕ).
Proof. It is sufficient to show that the following inclusion holds
X∃(E, ϕ) ⊆ Xα(E, ϕ).
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Take any sequence f = (fk) in X∃(E, ϕ), then there exists α1 > 0 such that f
α1
∈ X(E). If
α ≥ α1, then f
α ≤ f
α1
. Since X(E) is an ideal, then f ∈ Xα(E, ϕ).
On the other hand, if α < α1, then by Archimedean there exists n0 ∈ N such that 1
α ≤ 2n0 1
α0
. So,
|f|
α ≤ 2n0 |f|
α1
≤ K f
δ ,
for some K > 0. This follows from the assumption that φ satisfies the 2-condition. Since X(E) is
an ideal in (E), f ∈ Xα(E, ϕ). These completes the proof.
Corollary 3.4. If ϕ ∈ 2, then
(i) The space Xα(E, ϕ) is a Banach lattice with respect to the Luxemburg norm . ϕ.
(ii) The space X∃(E, ϕ) is an Archimedean regular Riesz space.
Lemma 3.5. Let ϕ ∈ △2 and (f(n)) a sequence in Xα(E, ϕ). Then ρϕ(f(n)) →n 0 holds if and only
if f(n)
ϕ →n 0.
Proof. Let ε ∈ (0, 1], then there exists n0 ∈ N such that 1/ε < 2n0. Since ϕ ∈ △2, then there exists
M > such that
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0 ≤ ρφ
f(n)
ε ≤ ρφ(2n0f(n)) ≤ Mn0ρφ(f(n))
If ρφ(f(n)) →n 0, then there exists n1 ∈ N such that ρφ(f(n)) < ǫ for every n ≥ n1. So, ρφ
f(n)
ε < ǫ
for every n ≥ n1. This implies f(n)
ϕ →n 0.
Now, assume f(n)
ϕ →n 0, then for every ε ∈ (0, 1) there exists n2 ∈ N such that f(n)
ϕ≤ ε for
every n ≥ n2. So, we have ρϕ(f(n)) < ε for every n ≥ n2, or ρϕ(f(n)) →n 0.
Lemma 3.6. If ϕ ∈ △2 and (f(n)) is a sequence in Xα(E, ϕ), then f(n)
ϕ →n 1 if and only if
ρϕ(f(n)) →n 1.
Proof. Suppose that ρϕ (f(n)) →n 1. If ρϕ (f(n)) ≤ 1, then ρϕ(f) ≤ f(n)
ϕ and if ρϕ(f(n)) > 1
then f(n)
ϕ ≤ 1. Therefore f ϕ−1 < |ρϕ(f(n)) − 1| and the assertion follows. Now, assume
that f(n)
φ →n 1. We consider two cases.
Case 1. Take f(nk)
φ ↑k 1, for any subsequence (f(nk)) of (f(n)). We can assume f(nk)
φ ≥ 1/2.
Since φ ∈ 2, then the sequence ρφ(2f(n)) is bounded, that is, there exists M > 0 such that
ρφ(2f(n)) ≤ M for every n ∈ N. Suppose ρφ(f(n)) n 1, then there exists δ1 > 0 such that for
every k ∈ N there exists nk > k such that ρφ(f(nk)) ≤ 1−δ1. Take ank = 1
f(nk)
φ
−1 then ank →k 0
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and ank ≤ 1 for every k ∈ N. So we have
1 = ρφ
f(nk)
f(nk)
= ρφ (ank + 1)f(nk)
= ρφ 2ank f(nk)
+ (ank − 1)f(nk)
≤ ank ρφ 2f(nk)
+ (an − 1)ρφ f(nk)
.
Therefore
1 ≤ ank M + (1 − an) ρφ(f(nk)) ≤ ank M + (1 − ank ) (1 − δ1) −→ 1 − δ1
This leads a contradiction.
Case 2. Take f(nk)
φ ↓k for any subsequence (f(nk)) of (f(n)). Assume that f(nk)) φ ≤ 2 for
every k ∈ N. then ρφ
f(nk)
2 ≤ 1 for every k ∈ N. Assume for contrary that ρφ(f(n)) n 1, then
there exists δ2 > 0 such that for every k ∈ N there exists nk > k such that ρφ(f(nk)) > 1+δ2. Since
1 − 1
f(nk)
φ
≥ 0 dan 1
f(nk)
φ
≥ 0, then
1 + δ2 ≤ ρφ(f(nk)
) ≤ ρφ 1 −
1
f(nk)
φ
f(nk)
2
+
1
f(nk)
φ
f(nk)
2
≤ 1 −
1
f(n)
φ
ρφ(
f(nk)
2
) +
1
f(nk)
φ
ρφ
f(nk)
2
= 1
This is impossible. So, the proof is completed.
Theorem 3.7. If ϕ ∈ 2 and X(E) has a σ-order continuous norm, then Xα(E, φ) possesses the
same property.
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Proof. Since ϕ ∈ 2, by Corollary 3.4 we get that the space Xα(E, φ) is an Archimedean Banach
lattice. Let (fn) be a sequence such that fn ↑≤ g in Xα(E, φ). We will show that the sequence (fn)
converges to g in the norm · φ.
Let us define the set D = g ∈ Xα(E, φ) : fn ≤ g, for all n ∈ N . Let k ∈ N and 0 ≤ ku ≤ g − fn
for all n and all g ∈ D. Then fn ≤ g − ku for all n, so g − ku ∈ D. Since for any (fixed) n ∈ N,
0 ≤ ku ≤ g − fn and Xα(E, φ) is an Archimedean, then u = 0. So, g − fn ↓ 0. Since ϕ is an
increasing function, then (g − fn) ↓ 0 in X(E). And finally, since X(E) has a σ-order continuous
norm, then (g − fn) → 0 and by using the condition- 2 of ϕ, we get g − fn
φ→ 0.
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