In this work we introduce new spaces m2(F,ϕ,p) of double sequences defined by a double sequence of modulus functions, and we study some properties of this space.

1. Introduction

In this work, by w and w2, we denote the spaces of single complex sequences and double complex sequences, respectively. ℕ and ℂ denote the set of positive integers and complex numbers, respectively. If, for all ɛ>0, there is nɛ∈ℕ such that ∥xk,l-a∥X<ɛ where k>nɛ and l>nɛ, then a double sequence {xk,l} is said to be converge (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 [1–3]).

For 1≤p<∞, ℓp(2) denote the space of sequences x={xk,l} such that
(1)∑k,l=1∞|xk,l|p<∞.
(see [4]).

A double sequence x={xk,l} is said to be bounded if and only if supk,l|xk,l|<∞. The space of all bounded double sequences is denoted by ℓ∞(2). It is known that ℓ∞(2) is Banach space (see [5, 6]).

A double sequence space E is said to be normal if (ykl)∈E whenever |ykl|≤|xkl| for all k,l∈ℕ and (xkl)∈E.

The double sequence spaces in the various forms were introduced and studies by Khan and Tabassum in [7–14], by Khan in [15], and by Khan et al. in [16, 17].

Now let φs be a family of subsets σ having most elements s in ℕ. Also φs,t denote the class of subsets σ=σ1×σ2 in ℕ×ℕ such that the elements 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
(2)kϕk+1,l≤(k+1)ϕk,l,lϕk,l+1≤(l+1)ϕk,l.
(see [18]).

Let x={xk,l} be a double sequence. A set S(x) is defined by
(3)S(x)={{xπ1(k),π2(k)}:π1andπ2arepermutationsofℕ}.
A double sequence space E is said to be symmetric if u=(ukl)∈E and ∥u∥=∥x∥ whenever x=(xkl)∈E and u∈S(x).

A BK-space is a Banach sequence space E in which the coordinate maps are continuous.

A function f:[0,∞)→[0,∞) is said to be a modulus function if it satisfies the following:

f(x)=0 if and only if x=0;

f(x+y)≤f(x)+f(y) for all x,y∈[0,∞);

f is increasing;

f is continuous from right at 0.

It follows that f is continuous on [0,∞). The modulus function may be bounded or unbounded. For example, if we take f(x)=x/(x+1), then f(x) is bounded. But, for 0<p<1, f(x)=xp is not bounded.

The BK-spaces m(ϕ), introduced by Sargent in [19], is in the form
(4)m(ϕ)={x={xk}∈w:∥x∥m(ϕ)=sups≥1,σ∈φs1ϕs∑k∈σ|xk|<∞}.

Sargent studied some properties of this space and examined relationship between this space and lp-space.

The space m(ϕ) was extended to m(ϕ,p) by Tripathy and Sen [20] as follows:
(5)m(ϕ,p)={(∑k∈σ|xk|p)1/px={xk}∈w:∥x∥m(ϕ,p)w=sups≥1,σ∈φs1ϕs(∑k∈σ|xk|p)1/p<∞}.
Recently, Raj et al. [21] introduced and studied the following sequence space m(F,ϕ,p).

Let F=(fk) be a sequence of modulus functions. Then the space m(F,ϕ,p) is defined by
(6)m(F,ϕ,p)={x={xk}∈w:sups≥1,σ∈φs1ϕs(∑k∈σ[fk(|xk|ρ)]p)1/pwweww<∞,forsomeρ>0(∑k∈σ[fk(|xk|ρ)]p)1/p}.

In this work we introduce sequence spaces m2(F,ϕ,p) defined by
(7)m2(F,ϕ,p)={{∑i∈σ1∑j∈σ2[fi,j(|xi,j|ρ)]p}x=(xkl)∈w2:(ew)sup(s,t)≥(1,1)supσ1×σ2∈φst1ϕst{∑i∈σ1∑j∈σ2[fi,j(|xi,j|ρ)]p}1/p=(e)<∞forρ>0{∑i∈σ1∑j∈σ2[fi,j(|xi,j|ρ)]p}},
where F=(fi,j) is a double sequence of modulus functions.

2. Main Results

The result stated in the first theorem is not hard. So, we give it without proof.

Theorem 1.

The sequence space m2(F,ϕ,p) is a linear space.

Theorem 2.

The sequence spaces m2(F,ϕ,p) are complete.

Proof.

Let {x(i)} be a double Cauchy sequence in m2(F,ϕ,p) such that x(i)={xk,l(i)}k,l=1∞ for all i∈ℕ. Then
(8)sup(s,t)≥(1,1)supσ1×σ2∈φst1ϕst{∑k∈σ1∑l∈σ2[fk,l(|xk,l(i)|ρ)]p}1/p<∞
for some ρ>0 and for all i∈ℕ. For each ɛ>0, there exists a positive integer n0 such that
(9)∥x(i)-x(j)∥m2(F,ϕ,p)<ɛ
for all i,j≥n0. Hence
(10)sup(s,t)≥(1,1)supσ1×σ2∈φst1ϕst{∑k∈σ1∑l∈σ2[fk,l(|xk,l(i)-xk,l(j)|ρ)]p}1/p<ɛ
for some ρ>0 and for all i,j≥n0. This implies that
(11)|xk,l(i)-xk,l(j)|<εϕ1,1
for all i,j≥n0 and for each fixed (k,l)∈ℕ×ℕ. Hence {x(i)} is a Cauchy sequence in ℂ.

Then, there exists xk,l∈ℂ such that xk,l(i)→xk,l as i→∞ and let us define x=(xk,l). From (10), for each fixed (s,t),
(12){∑k∈σ1∑l∈σ2[fk,l(|xk,l(i)-xk,l(j)|ρ)]p}<εpϕstp
for some ρ>0, for all i,j≥n0 and σ1×σ2∈φst.

Letting j→∞, we get
(13){∑k∈σ1∑l∈σ2[fk,l(|xk,l(i)-xk,l|ρ)]p}<εpϕstp
for some ρ>0, for all i,j≥n0, and σ1×σ2∈φst. Thus we obtain
(14)sup(s,t)≥(1,1)supσ1×σ2∈φst1ϕst{∑k∈σ1∑l∈σ2[fk,l(|xk,l(i)-xk,l|ρ)]p}1/p<ɛ
for some ρ>0 and for all i,j≥n0. This implies that x(i)-x∈m2(F,ϕ,p) for all i,j≥n0.

Hence x=x(n0)+x-x(n0)∈m2(F,ϕ,p). By (14),
(15)∥x(i)-x∥m2(F,ϕ,p)<ɛ
for all i≥n0. This means that x(i)→x as i→∞. Hence m2(F,ϕ,p) is a Banach space.

Theorem 3.

The space m2(F,ϕ,p) is a BK-space.

Proof.

Suppose that {x(i)}∈m2(F,ϕ,p) with ∥x(i)-x∥m2(F,ϕ,p)→0 as i→∞. For each ɛ>0 there exists n0∈N such that
(16)∥x(i)-x∥m2(F,ϕ,p)<ɛ
for all i≥n0. Thus
(17)sup(s,t)≥(1,1)supσ1×σ2∈φst1ϕst{∑k∈σ1∑l∈σ2[fk,l(|xk,l(i)-xk,l|ρ)]p}1/p<ɛ
for some ρ>0 and for all i≥n0. Hence we obtain
(18)|xk,l(i)-xk,l|<εϕ1,1
for all i≥n0 and for all (k,l)∈ℕ×ℕ. This implies |xk,l(i)-xk,l|→0 as i→∞. This completes the proof.

Corollary 4.

The space m2(F,ϕ,p) is a symmetric space.

Proof.

Let x={xk,l}∈m2(F,ϕ,p) and let y={yk,l}∈S(x). Then we can write yk,l=xmk,ml. Thus we obtain
(19)∥x∥m2(F,ϕ,p)=∥y∥m2(F,ϕ,p).

Corollary 5.

The space m2(F,ϕ,p) is a normal space.

Proof.

It is obvious.

Theorem 6.

Consider
(20)m2(ϕ)⊆m2(F,ϕ,p).

Proof.

Suppose that x∈m2(ϕ). Then we have
(21)∥x∥m2(ϕ)=sup(s,t)≥(1,1)supσ1×σ2∈φst1ϕst{∑k∈σ1∑l∈σ2|xk,l|}∥x∥m2(ϕ)=K<∞.
Thus for each fixed (s,t) and for σ1×σ2∈φst,
(22)∑k∈σ1∑l∈σ2|xk,l|≤Kϕst
for some ρ>0. Hence
(23)sup(s,t)≥(1,1)supσ1×σ2∈φst1ϕst{∑k∈σ1∑l∈σ2[fk,l(|xk,l|ρ)]p}1/p≤K
for some ρ>0. This implies that x∈m2(F,ϕ,p). Hence m2(ϕ)⊆m2(F,ϕ,p).

Theorem 7.

m2(F,ϕ,p)⊆m2(F,ψ,p) if and only if sup(s,t)≥(1,1)(ϕst/ψst)<∞.

Proof.

Let K=sup(s,t)≥(1,1)(ϕst/ψst)<∞. Then ϕst≤Kψst for all (s,t)≥(1,1). If x∈m2(F,ϕ,p), then
(24)sup(s,t)≥(1,1)supσ1×σ2∈φst1ϕst{∑k∈σ1∑l∈σ2[fk,l(|xk,l|ρ)]p}1/p<∞
for some ρ>0. Thus
(25)sup(s,t)≥(1,1)supσ1×σ2∈φst1Kψst{∑k∈σ1∑l∈σ2[fk,l(|xk,l|ρ)]p}1/p<∞
for some ρ>0. Hence x∈m2(F,ψ,p). This shows that m2(F,ϕ,p)⊆m2(F,ψ,p). Conversely, let m2(F,ϕ,p)⊆m2(F,ψ,p). We define γs,t=ϕst/ψst. Let sup(s,t)≥(1,1)γs,t=∞. Then there exists a subsequence {γsi,ti} of {γs,t} such that γsi,ti→∞ as i→∞. Then for x∈m2(F,ϕ,p) we have
(26)sup(s,t)≥(1,1)supσ1×σ2∈φst1ψst{∑k∈σ1∑l∈σ2[fk,l(|xk,l|ρ)]p}1/p≥sup(s,t)≥(1,1)supσ1×σ2∈φstγs,tϕst{∑k∈σ1∑l∈σ2[fk,l(|xk,l|ρ)]p}1/p=∞
for some ρ>0. This is a contradiction as x∉m2(F,ψ,p) and this completes the proof.

Proposition 8.

Consider
(27)ℓp(2)⊆m2(F,ϕ,p)⊆ℓ∞(2).

Proof.

Clearly, ℓp(2)=m2(F,ψ,p), where ψst=1 for s,t=1,2,… when fk,l(x)=x and sup(s,t)≥(1,1)(ψst/ϕst)<∞ by nondecreasing (ϕst). Then, by Theorem 7, first inclusion is obtained. Suppose x∈m2(F,ϕ,p). Then
(28)sup(s,t)≥(1,1)supσ1×σ2∈φst1ϕst{∑k∈σ1∑l∈σ2[fk,l(|xk,l|ρ)]p}1/p=K<∞
for some ρ>0. Hence we obtain
(29)|xkl|≤Kϕ1,1
for all k,l∈ℕ. Thus x∈ℓ∞(2) and proof is completed.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

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