# On a New Spacem2(M,A,ϕ,p) of Double Sequences

Authors: Cenap Duyar; Oğuz Oğur
Journal: Journal of Function Spaces and Applications
 (2013)

---

## Abstract

We introduce a new spacem2(M,A,ϕ,p) of double sequences related to p-absolute convergent double sequence space, combining an Orlicz function and an infinity double matrix. We study some properties of m2(M,A,ϕ,p) and obtain some inclusion relations involving m2(M,A,ϕ,p).


---

## Body

## 1. Introduction

Throughout this work,N and C denote the set of positive integers and complex numbers, respectively. A complex double sequence is a function x from N×N into C and briefly denoted by {xk,l}. Throughout this work, w and w2 denote the spaces of single complex sequences and double complex sequences, respectively. If, for all ε>0, there is nε∈N such that ∥xk,l-a∥X<ε where k>nε and l>nε, then a double sequence {xk,l} is said to be converging (in terms of Pringsheim) to a∈C. A real double sequence {xk,l} is nondecreasing, if xk,l≤xp,q for (k,l)<(p,q). A double series is infinity sum ∑k,l=1∞‍xk,l, and its convergence implies the convergence by |·| of partial sums sequence {Sn,m}, where Sn,m=∑k=1n‍∑l=1m‍xk,l (see [1–4]).A double sequence spaceE is said to be solid if {xk,lyk,l}∈E for all double sequences {yk,l} of scalars such that |yk,l|≤1 for all k,l∈N whenever {xk,l}∈E.Now let℘s be a family of subsets σ having most elements s in N. Also ℘s,t denotes the class of subsets σ=σ1×σ2 in N×N such that the element numbers of σ1 and σ2 are most s and t, respectively. Besides {ϕk,l} is taken as a nondecreasing double sequence of the positive real numbers such that
(1)kϕk+1,l≤(k+1)ϕk,l,lϕk,l+1≤(l+1)ϕk,l.
An Orlicz function is a function M:[0,∞)→[0,∞) which is continuous, nondecreasing, and convex with M(0)=0, M(x)>0 for x>0, and M(x)→∞ as x→∞.An Orlicz functionM can always be represented in the following integral form: M(x)=∫0xη(t)dt, where η is known as the kernel of M, is right differentiable for t≥0, η(0)=0, η(t)>0 for t>0, η is nondecreasing, and η(t)→∞ as t→∞.An Orlicz function is said to be satisfiedΔ2-condition for all values of u, if there exists a constant T>0 such that M(2u)≤TM(u) for all u≥0.Lindenstrauss and Tzafriri [5] used the idea of Orlicz functions to construct Orlicz sequence space
(2)ℓM={x={xk}∈w:∑k≥1‍M(|xk|ρ)<∞forsomeρ>0(|xk|ρ)}.
The sequence space ℓM is a Banach space according to the norm defined by
(3)∥x∥=inf{ρ>0:∑k≥1‍M(|xk|ρ)≤1}.
This space (ℓM,∥·∥) is called an Orlicz sequence space. The space ℓM is closely related to the space ℓp, which is an Orlicz sequence space with M(x)=xp for 1≤p<∞.The double sequence spaces in the various forms defined by Orlicz functions were introduced and studied by Khan and Tabassum in [6–12] and by Khan et al. in [13].The spacem(ϕ), introduced by Sargent in [14], is in the form
(4)m(ϕ)={∑k∈σ‍x={xk}∈w:∥x∥m(ϕ)=sups≥1,σ∈℘s1ϕs∑k∈σ‍|xk|<∞}.
Sargent studied some properties of this space and examined relationship between this space and lp-space. Similar sequence classes were studied by many mathematicians using Orlicz functions (see [15–17]).Later on, this space was investigated from sequence space point of view by Rath [18], Rath and Tripathy [19], Tripathy and Sen [20], Tripathy and Mahanta [17], and others. Recently Altun and Bilgin [15] introduced and studied the following sequence space m(M,A,ϕ,p).LetA=(aik) be an infinite matrix of complex numbers, M an Orlicz space, and p=(pi) a bounded sequence of positive real numbers such that 0<h=infpi≤pi≤H=suppi<∞. Then the space m(M,A,ϕ,p) is defined by
(5)m2(M,A,ϕ,p)={{xk}∈w:sups≥1,σ∈℘s1ϕs∑k∈σ‍M(|Ai(x)|ρ)pi<∞forsomeρ>0∑k∈σ‍M(|Ai(x)|ρ)pi},
where Ax=(Ai(x)) if Ai(x)=∑k=1∞‍aikxk converges for each i.Letx={xk,l} be a double sequence. A set S(x) is defined by
(6)S(x)={{xπ1(k),π2(k)}:π1andπ2arepermutationsofN}.IfS(x)⊆E for all x∈E, then E is said to be symmetric.In this work, we introduce the following sequence space.LetA=(aijkl) be an infinite double matrix of complex numbers, M an Orlicz function and p=(pij) bounded double sequence of positive real numbers such that 0<h=infpij≤pi≤H=suppij<∞. Then the space m2(M,A,ϕ,p) is defined by
(7)m2(M,A,ϕ,p)={x=(xkl)∈w2:sup{1ϕst∑i∈σ1‍∑j∈σ2‍M(|Aij(x)|ρ)pij:(s,t)≥(1,1),σ1×σ2∈℘st∑j∈σ2‍M(|Aij(x)|ρ)pij}<∞,forsomeρ>0∑j∈σ2‍M(|Aij(x)|ρ)pij},
where Ax=(Aij(x)) if Aij(x)=∑k,l=1∞‍aijklxkl converges for each (i,j)∈N×N.Also, we introduce and investigate the following space:(8)m2(M,ϕ,p)={∑l∈σ2‍M(|xk,l|ρ)pklx=(xkl)∈w2:sup{1ϕst∑k∈σ1‍∑l∈σ2‍M(|xk,l|ρ)pkl:(s,t)≥(1,1),σ1×σ2∈℘st∑l∈σ2‍M(|xk,l|ρ)pkl}<∞forsomeρ>0∑k∈σ1‍∑l∈σ2‍M(|xk,l|ρ)pkl}.In this work, we also use the following sequence spaces:(9)l(2)(M,A,ϕ,p)0={x={xk,l}∈w2:∑i,j=1∞‍M(|Aij(x)|ρ)pij<∞forsomeρ>0∑i,j=1∞‍M(|Aij(x)|ρ)pij},l(2)(M,A,ϕ,p)∞={x={xk,l}∈w2:sup(i,j)∈N×NM(|Aij(x)|ρ)pij<∞forsomeρ>0(|Aij(x)|ρ)pij}.The following inequality will be used throughout this paper:(10)|a+b|pqr≤max(1,2H-1)(|a|pqr+|b|pqr),
where a,b∈CandH=sup{pqr:(q,r)∈N×N}.


## 2. Main Results

Definition 1.
LetK(2) be a set of increasing positive integer binaries, namely, (k1,k2)<(l1,l2) if and only if k1<l1 and k2<l2, and E be a double sequence space. A K(2)-set space is a double sequence space, defined by
(11)λK(2)E={{xkm,lm}∈w2:{xk,l}∈E,(km,lm)∈K(2)}.
The canonical preimage of a double sequence {xkm,lm}∈λK(2)E is a double sequence {yk,l}∈w2 with
(12)yk,l={xk,l,(k,l)∈K(2)0,(k,l)∉K(2).
The canonical preimage of a set space λK(2)E is a set of canonical preimages of all elements in λK(2)E.Definition 2.
If a double sequence spaceE contains the canonical preimages of all set spaces, then E is said to be monotone.The following lemma is an easy result of the definitions.Lemma 3.
If a double sequence spaceE is solid, then E is monotone.Proposition 4.
The spacem2(M,A,ϕ,p) is a C-linear space.Proof.
Letx={xk,l}, y={yk,l} be in m2(M,A,ϕ,p) and α, β in C∖{0}. Then there exist positive numbers ρ1 and ρ2 such that
(13)sup{1ϕs,t∑i∈σ1‍∑j∈σ2‍M(|Aij(x)|ρ1)pij:(s,t)≥(1,1),σ1×σ2∈℘s,t(|Aij(x)|ρ1)pij}<∞,sup{1ϕs,t∑i∈σ1‍∑j∈σ2‍M(|Aij(y)|ρ2)pij:(s,t)≥(1,1),σ1×σ2∈℘s,t∑i∈σ1‍∑j∈σ2‍(|Aij(y)|ρ2)pij}<∞.
Let ρ3=max{2|α|ρ1,2|β|ρ2}. Using that M is nondecreasing convex function, we have
(14)∑i∈σ1‍∑j∈σ2‍M(|Aij(αx+βy)|ρ3)pij≤∑i∈σ1‍∑j∈σ2‍M(|∑k,l=1∞(αaijklxk,l+βaijklyk,l)|ρ3)pij≤∑i∈σ1‍∑j∈σ2‍M(|αAij(x)|2|α|ρ1+|βAij(y)|2|β|ρ2)pij=∑k∈σ1‍∑l∈σ2‍M(|Aij(x)|2ρ1+|Aij(y)|2ρ2)pij≤max(1,2H-1)(∑i∈σ1‍∑i∈σ2‍M(|Aij(x)|ρ1)pij+∑k∈σ1‍∑l∈σ2‍M(|Aij(y)|ρ2)pij).
Thus we can write
(15)sup{1ϕs,t∑i∈σ1‍∑j∈σ2‍M(|Aij(αx+βy)|ρ3)pij:(s,t)≥(1,1),σ1×σ2∈℘s,t∑i∈σ1‍∑j∈σ2‍M(|Aij(αx+βy)|ρ3)pij}≤max(1,2H-1)×{sup{1ϕs,t∑i∈σ1‍∑j∈σ2‍M(|Aij(x)|ρ1)pij:(s,t)≥(1,1),σ1×σ2∈℘s,t∑i∈σ1‍∑j∈σ2‍M(|Aij(x)|ρ1)pij}+sup{1ϕs,t∑k∈σ1‍∑l∈σ2‍M(|Aij(y)|ρ2)pij:(s,t)≥(1,1),σ1×σ2∈℘s,t∑i∈σ1‍∑j∈σ2‍M(|Aij(x)|ρ1)pij}}<∞.
This shows that {αxk,l+βyk,l}∈m2(M,A,ϕ,p). Hence m2(M,A,ϕ,p) is a linear space.Proposition 5.
The spacem2(M,A,ϕ,p) is a paranormed space with the paranorm
(16)g(x)=inf{ρpqr/H:[sup{1ϕs,t∑i∈σ1‍∑j∈σ2‍M(|Aij(x)|ρ)pij:(s,t)≥(1,1),σ1×σ2∈℘s,t∑j∈σ2‍M(|Aij(x)|ρ)pij}≤1]1/H,q∈N,r∈N∑j∈σ2‍M(|Aij(x)|ρ)pij}.Proof.
It is clear thatg(x)=g(-x) and g(x)=0 if x=0. If there are ρ1>0 and ρ2>0 such that
(17)sup{1ϕs,t∑i∈σ1‍∑j∈σ2‍M(|Aij(x)|ρ1)pij:(s,t)≥(1,1),σ1×σ2∈℘s,t∑i∈σ1‍∑j∈σ2‍M(|Aij(x)|ρ1)pij}≤1,sup{1ϕs,t∑i∈σ1‍∑j∈σ2‍M(|Aij(y)|ρ2)pij:(s,t)≥(1,1),σ1×σ2∈℘s,t∑i∈σ1‍∑j∈σ2‍M(|Aij(y)|ρ2)pij}≤1,
then
(18)sup{1ϕs,t∑i∈σ1‍∑j∈σ2‍M(|Aij(x+y)|ρ1+ρ2)pij:(s,t)≥(1,1),σ1×σ2∈℘s,t∑i∈σ1‍∑j∈σ2‍M(|Aij(x+y)|ρ1+ρ2)pij}≤sup{1ϕs,t∑i∈σ1‍∑j∈σ2‍M(|Aij(x)|ρ1+ρ2+|Aij(y)|ρ1+ρ2)pij:(s,t)≥(1,1),σ1×σ2∈℘s,t∑i∈σ1‍∑j∈σ2‍M(|Aij(x)|ρ1+ρ2+|Aij(y)|ρ1+ρ2)pij}≤(ρ1ρ1+ρ2)hsup{1ϕs,t∑i∈σ1‍∑j∈σ2‍M(|Aij(x)|ρ1)pij:(s,t)≥(1,1),σ1×σ2∈℘s,t∑i∈σ1‍∑j∈σ2‍M(|Aij(x)|ρ1)pij}+(ρ2ρ1+ρ2)hsup{1ϕs,t∑i∈σ1‍∑j∈σ2‍M(|Aij(x)|ρ1)pij:(s,t)≥(1,1),σ1×σ2∈℘s,t∑i∈σ1‍∑j∈σ2‍M(|Aij(x)|ρ1)pij},
where h=infpij. This shows that g(x+y)≤g(x)+g(y). Using this triangle inequality we can write
(19)g(λnxn-λx)≤g(λnxn-λnx)+g(λnx-λx).
Separately, we obtain
(20)g(λnxn-λnx)=inf{∑j∈σ2‍M(|Aij(λnxn-λnx)|ρn)pijρnpqr/H:[sup{1ϕs,t∑i∈σ1‍∑j∈σ2‍M(|Aij(λnxn-λnx)|ρn)pij:(s,t)≥(1,1),σ1×σ2∈℘s,t∑j∈σ2‍M(|Aij(λnxn-λnx)|ρn)pij}≤1∑i∈σ1‍∑j∈σ2‍M(|Aij(λnxn-λnx)|ρn)pij]1/H≤1,q∈N,r∈N∑j∈σ2‍M(|Aij(λnxn-λnx)|ρn)pij}=inf{ρnpqr/H:[sup{1ϕs,t∑i∈σ1‍∑j∈σ2‍M(|Aij(xn-x)|(ρn/|λn|))pij:(s,t)≥(1,1),σ1×σ2∈℘s,t∑i∈σ1‍∑j∈σ2‍M(|Aij(xn-x)|(ρn/|λn|))pij}≤1∑i∈σ1‍∑j∈σ2‍M(|Aij(xn-x)|(ρn/|λn|))pij]1/H≤1,q∈N,r∈N∑i∈σ1‍∑j∈σ2‍M(|Aij(xn-x)|(ρn/|λn|))pij}=inf{∑j∈σ2‍M(|Aij(xn-x)|ρn)pij(λnρn)pqr/H:[sup{1ϕs,t∑i∈σ1‍∑j∈σ2‍M(|Aij(xn-x)|ρn)pij:(s,t)≥(1,1),σ1×σ2∈℘s,t∑i∈σ1‍∑j∈σ2‍M(|Aij(xn-x)|ρn)pij}]1/H≤1,q∈N,r∈N∑i∈σ1‍∑j∈σ2‍M(|Aij(xn-x)|ρn)pij}≤max{|λn|h/H,|λn|}.inf{∑j∈σ2‍M(|Aij(xn-x)|ρn)pij(|λn|ρn)pqr/H:[sup{1ϕs,t∑i∈σ1‍∑j∈σ2‍M(|Aij(xn-x)|ρn)pij:(s,t)≥(1,1),σ1×σ2∈℘s,t∑i∈σ1‍∑j∈σ2‍M(|Aij(xn-x)|ρn)pij}]1/H≤1,q∈N,r∈N×∑i∈σ1‍∑j∈σ2‍M(|Aij(xn-x)|ρn)pij}=max{|λn|h/H,|λn|}·g(xn-x),g(λnx-λx)=inf{∑j∈σ2‍M(|Aij((λn-λ)x)|ρn)pijρnpqr/H:[sup{1ϕs,t∑i∈σ1‍∑j∈σ2‍M(|Aij((λn-λ)x)|ρn)pij:(s,t)≥(1,1),σ1×σ2∈℘s,t∑i∈σ1‍∑j∈σ2‍M(|Aij((λn-λ)x)|ρn)pij}]1/H≤1,q∈N,r∈N∑i∈σ1‍∑j∈σ2‍M(|Aij((λn-λ)x)|ρn)pij}=inf{∑j∈σ2‍M(|Aij(x)|(ρn/|λn-λ|))pijρnpqr/H:[sup{1ϕs,t∑i∈σ1‍∑j∈σ2‍M(|Aij(x)|(ρn/|λn-λ|))pij:(s,t)≥(1,1),σ1×σ2∈℘s,t∑i∈σ1‍∑j∈σ2‍M(|Aij(x)|(ρn/|λn-λ|))pij}]1/H≤1,q∈N,r∈N∑i∈σ1‍∑j∈σ2‍M(|Aij(x)|(ρn/|λn-λ|))pij}=inf{∑i∈σ1‍∑j∈σ2‍M(|Aij(x)|ρn)pij(|λn-λ|ρn)pqr/H:[sup{1ϕs,t∑i∈σ1‍∑j∈σ2‍M(|Aij(x)|ρn)pij:(s,t)≥(1,1),σ1×σ2∈℘s,t∑i∈σ1‍∑j∈σ2‍M(|Aij(x)|ρn)pij}]1/H≤1,q∈N,r∈N∑i∈σ1‍∑j∈σ2‍M(|Aij(x)|ρn)pij}≤max{|λn-λ|h/H,|λn-λ|}g(x).
Hence g(λnxn-λx)→0 where λn→λ and xn→x as n→∞. Consequently g is a paranom on m2(M,A,ϕ,p).Proposition 6.
The classm2(M,ϕ,p) of double sequences is solid.Proof.
Letα={αk,l} be a double sequence of scalars such that |αk,l|≤1 and y={yk,l}∈m2(M,ϕ,p). Then we can write
(21)sup{1ϕs,t∑k∈σ1‍∑l∈σ2‍M(|αk,lxk,l|ρ)pkl:(s,t)≥(1,1),σ1×σ2∈℘s,t∑k∈σ1‍∑l∈σ2‍M(|αk,lxk,l|ρ)pkl}≤sup{1ϕs,t∑k∈σ1‍∑l∈σ2‍M(|αk,l||yk,l|ρ)pkl:(s,t)≥(1,1),σ1×σ2∈℘s,t∑k∈σ1‍∑l∈σ2‍M(|αk,l||yk,l|ρ)pkl}≤sup{1ϕs,t∑k∈σ1‍∑l∈σ2‍M(|yk,l|ρ)pkl:(s,t)≥(1,1),σ1×σ2∈℘s,t∑k∈σ1‍∑l∈σ2‍M(|yk,l|ρ)pkl}.
This implies that{αk,lyk,l}∈m2(M,ϕ,p), and hence the class m2(M,ϕ,p) is solid.Corollary 7.
The spacem2(M,ϕ,p) is monotone.Theorem 8.
Letψ be another double sequence like ϕ. Then m2(M,A,ϕ,p)⊆m2(M,A,ψ,p) if and only if sup(s,t)≥(1,1)(ϕs,t/ψs,t)<∞.Proof.
LetK=sup(s,t)≥(1,1)(ϕs,t/ψs,t)<∞. Then ϕs,t≤K·ψs,t for all (s,t)≥(1,1). If x={xk,l}∈m2(M,A,ϕ,p), then
(22)sup{1ϕs,t∑i∈σ1‍∑j∈σ2‍M(|Aij(x)|ρ)pij:(s,t)≥(1,1),σ1×σ2∈℘s,t∑i∈σ1‍∑j∈σ2‍M(|Aij(x)|ρ)pij}<∞,
for some ρ>0. Thus
(23)sup{1Kψs,t∑k∈σ1‍∑l∈σ2‍M(|Aij(x)|ρ)pij:(s,t)≥(1,1),σ1×σ2∈℘s,t∑k∈σ1‍∑l∈σ2‍M(|Aij(x)|ρ)pij}<∞,
for some ρ>0, and hence x={xk,l}∈m2(M,A,ψ,p). This shows that m2(M,A,ϕ,p)⊆m2(M,A,ψ,p).
Conversely, letm2(M,A,ϕ,p)⊆m2(M,A,ψ,p). We say αs,t=ϕs,t/ψs,t for all (s,t)≥(1,1) and suppose sup(s,t)≥(1,1)αs,t=∞. Then there exists a subsequence {αsi,ti} of {αs,t} such that limi→∞αsi,ti=∞. If x={xk,l}∈m2(M,A,ϕ,p), then we have
(24)sup{1ψs,t∑i∈σ1‍∑j∈σ2‍M(|Aij(x)|ρ)pij:(s,t)≥(1,1),σ1×σ2∈℘s,t∑j∈σ2‍M(|Aij(x)|ρ)pij}=sup{αs,t1ϕs,t∑i∈σ1‍∑j∈σ2‍M(|Aij(x)|ρ)pij:(s,t)≥(1,1),σ1×σ2∈℘s,t∑j∈σ2‍M(|Aij(x)|ρ)pij}≥{supm≥1αsm,tm}·sup{1ϕsm,tm∑i∈σ1‍∑j∈σ2‍M(|Aij(x)|ρ)pij:σ1×σ2∈℘sm,tm∑j∈σ2‍M(|Aij(x)|ρ)pij}=∞.
This is a contradiction asx={xk,l}∉m2(M,A,ϕ,p). This completes the proof.Corollary 9.
m
2
(
M
,
A
,
ϕ
,
p
)
=
m
2
(
M
,
A
,
ψ
,
p
) if and only if sup(s,t)≥(1,1)αs,t<∞ and sup(s,t)≥(1,1)αs,t-1<∞.Theorem 10.
LetM, M1, and M2 be Orlicz functions satisfying Δ2-condition. Then (a)
m2(M1,A,ϕ,p)⊆m2(M∘M1,A,ϕ,p),(b)
m2(M1,A,ϕ,p)∩m2(M2,A,ϕ,p)⊆m2(M1+M2,A,ϕ,p).Proof.
(a) Letx={xk,l}∈m2(M1,A,ϕ,p). Then there exists ρ>0 such that
(25)sup{1ϕs,t∑i∈σ1‍∑j∈σ2‍M1(|Aij(x)|ρ)pij:(s,t)≥(1,1),σ1×σ2∈℘s,t∑j∈σ2‍M1(|Aij(x)|ρ)pij}<∞.By the continuity of M, we select a number δ with 0<δ<1 such that M(t)<ε, whenever 0≤t<δ, for arbitrary 0<ε<1. Now let yi,j=M1(|Aij(x)|/ρ). We can write
(26)∑i∈σ1‍∑j∈σ2‍M(yi,j)pij=&∑yi,j≤δ‍M(yi,j)pij+∑yi,j>δ‍M(yi,j)pij.
By the properties of M, we have
(27)∑yi,j≤δ‍M(yi,j)pij≤max{1,M(1)H}∑yi,j≤δ‍(yi,j)pij.
Again we can write(28)M(yi,j)<M(1+yi,jδ)<12M(2)+12M(2yi,jδ),
for yi,j>δ. If we use that M satisfies Δ2-condition, then we find
(29)M(yi,j)<12Tyi,jδM(2)+12Tyi,jδM(2)=Tyi,jδM(2),
and so
(30)∑yi,j>δ‍M(yi,j)pij≤max(1,(TδM(2))H)∑yi,j>δ‍yi,j.
Hence we get
(31)sup{1ϕs,t∑i∈σ1‍∑j∈σ2‍M(M1(|Aij(x)|ρ))pij:(s,t)≥(1,1),σ1×σ2∈℘s,t∑j∈σ2‍M(M1(|Aij(x)|ρ))pij}≤max{1,M(1)H}×sup{1ϕs,t∑i∈σ1‍∑j∈σ2‍(yi,j)pij:(s,t)≥(1,1),σ1×σ2∈℘s,t∑j∈σ2‍(yi,j)pij}+max(1,(TδM(2))H)×sup{1ϕs,t∑i∈σ1‍∑j∈σ2‍(yi,j)pij:(s,t)≥(1,1),σ1×σ2∈℘s,t∑j∈σ2‍(yi,j)pij}.
Finally, we have{xk,l}∈m2(M∘M1,ϕ), and hence m2(M1,ϕ)⊆m2(M∘M1,ϕ).(b)
Let{xk,l}∈m2(M1,A,ϕ,p)∩m2(M2,A,ϕ,p). Then there exists ρ>0 such that
(32)sup{1ϕs,t∑i∈σ1‍∑j∈σ2‍M1(|Aij(x)|ρ)pij:(s,t)≥(1,1),σ1×σ2∈℘s,t∑j∈σ2‍M1(|Aij(x)|ρ)pij}<∞,sup{1ϕs,t∑i∈σ1‍∑j∈σ2‍M2(|Aij(x)|ρ)pij:(s,t)≥(1,1),σ1×σ2∈℘s,t∑j∈σ2‍M1(|Aij(x)|ρ)pij}<∞.By the inequality
(33)∑i∈σ1‍∑j∈σ2‍(M1+M2)(|Aij(x)|ρ)pij≤max(1,2H-1)∑i∈σ1‍∑j∈σ2‍M1(|Aij(x)|ρ)pij+max(1,2H-1)∑i∈σ1‍∑j∈σ2‍M2(|Aij(x)|ρ)pij,
we have m2(M1,A,ϕ,p)∩m2(M2,A,ϕ,p)⊆m2(M1+M2,A,ϕ,p).Theorem 11.
(a)
l
(
2
)
(
M
,
A
,
ϕ
,
p
)
⊆
m
2
(
M
,
A
,
ϕ
,
p
)
⊆
l
(
2
)
(
M
,
A
,
ϕ
,
∞
).
(b)
m
2
(
M
,
A
,
ϕ
,
p
)
=
l
(
2
)
(
M
,
A
,
ϕ
,
p
) if and only if sup(s,t)≥(1,1)ϕs,t<∞.
(c)
m
2
(
M
,
A
,
ϕ
,
p
)
=
l
(
2
)
(
M
,
A
,
ϕ
,
∞
) if and only if sup(s,t)≥(1,1)(ϕs,t/s.t)>0.Proof.
(a) Letx={xk,l}∈l(2)(M,A,ϕ,p), and let a set B be defined as follows:
(34)B={1ϕs,t∑i∈σ1‍∑j∈σ2‍M(|Aij(x)|ρ)pij:(s,t)≥(1,1),σ1×σ2∈℘s,t∑j∈σ2‍M(|Aij(x)|ρ)pij}.
Since{ϕk,l} is a nondecreasing double sequence, {1/ϕk,l} is a nonincreasing double sequence. So we obtain
(35)1ϕ1,1∑i,j=1∞‍M(|Aij(x)|ρ)pij≥b,
for all b∈B and hence
(36)∞>M(|Aij(x)|ρ)pij≥ϕ1,1·supB.
Thus we have l(2)(M,A,ϕ,p)⊆m2(M,A,ϕ,p).
Conversely ifx={xk,l}∈m2(M,A,ϕ,p), then it is clear that
(37)supB≥1ϕ1,1M(|Aij(x)|ρ)pij,
for all (i,j)∈N×N, and hence
(38)supB≥1ϕ1,1sup(i,j)∈N×NM(|Aij(x)|ρ)pij.
This shows that if x={xk,l}∈m2(M,A,ϕ,p), then {xk,l}∈l(2)(M,A,ϕ,∞). Thus we have m2(M,A,ϕ,p)⊆l(2)(M,A,ϕ,∞).(b)
It is clear thatm2(M,A,ψ,p)=l(2)(M,A,ϕ,∞) where ψs,t=1 for all (s,t)∈N×N. Then we can write sup(s,t)≤(1,1)ϕs,t=sup(s,t)≤(1,1)(ϕs,t/ψs,t)<∞. By Theorem 8, we have m2(M,A,ϕ,p)⊆m2(M,A,ψ,p), and according to alternative (a)
(39)m2(M,A,ϕ,p)=l(2)(M,A,ϕ,p).(c)
Firstly we show thatm2(M,A,ψ,p)=l(2)(M,A,ϕ,∞) if ψs,t=s.t for all (s,t)∈N×N. Let x={xk,l}∈l(2)(M,A,ϕ,∞). Then for (s,t)≥(1,1) and σ1×σ2∈℘s,t, we can find some ρ>0 such that
(40)1s.t∑i∈σ1‍∑j∈σ2‍M(|Aij(x)|ρ)pij≤1s.ts.t.sup(i,j)∈N×NM(|Aij(x)|ρ)pij<∞.This gives the inclusion l(2)(M,A,ϕ,∞)⊆m2(M,A,ψ,p). Conversely let x={xk,l}∈m2(M,A,ψ,p). Then for (i,j)∈N×N, we can find some ρ>0 such that
(41)M(|Aij(x)|ρ)pij=1s.ts.t.M(|Aij(x)|ρ)pij≤sup{1ϕs,t∑i∈σ1‍∑j∈σ2‍M(|Aij(x)|ρ)pij:(s,t)≥(1,1),σ1×σ2∈℘s,t∑j∈σ2‍M(|Aij(x)|ρ)pij}<∞.
This shows that m2(M,A,ψ,p)⊆l(2)(M,A,ϕ,∞). By Theorem 8 and alternative (a), we can write m2(M,A,ϕ,p)=l(2)(M,A,ϕ,∞) if and only if sup(s,t)≥(1,1)(s.t/ϕs,t)<∞. This completes the proof.



---

*Source: 509613-2013-07-25.xml*