Introduction

Throughout this paper, w, χ, and Λ denote the classes of all, gai, and analytic scalar valued single sequences, respectively. We write w2 for the set of all complex sequences (x mn ), wherem,nN, the set of positive integers. Then, w2 is a linear space under the coordinatewise addition and scalar multiplication.

Some initial works on double sequence spaces is found in Bromwich[1]. Later on, they were investigated by Hardy[2], Moricz[3], Moricz and Rhoades[4], Basarir and Solankan[5], Tripathy[6], Turkmenoglu[7], and many others. We procure the following sets of double sequences:

M u ( t ) : = x mn w 2 : sup m , n N x mn t mn < , C p ( t ) : = x mn w 2 : p lim m , n x mn t mn = 1 for some C , C 0 p ( t ) : = x mn w 2 : p lim m , n x mn t mn = 1 , L u ( t ) : = x mn w 2 : m = 1 n = 1 x mn t mn < , C bp ( t ) : = C p ( t ) M u ( t ) and C 0 bp ( t ) = C 0 p ( t ) M u ( t ) ,

where t = (t mn ) is the sequence of strictly positive reals t mn for allm,nN and p − limm,n denotes the limit in the Pringsheim’s sense. In the case where t mn  = 1 for allm,nN, M u (t), C p (t), C 0 p (t), L u (t), C bp (t), and C 0 bp (t) reduce to the sets M u , C p , C 0 p , L u , C bp , and C 0 bp , respectively. Now, we may summarize the knowledge given in some document related to the double sequence spaces. Gö khan and Colak[8, 9] have proved that M u (t) and C p (t), C bp (t) are complete paranormed spaces of double sequences and gave the α−, β−, γ− duals of the spaces M u (t) and C bp (t). Quite recently, in her PhD thesis, Zeltser[10] has essentially studied both the theory of topological double sequence spaces and the theory of summability of double sequences. Mursaleen and Edely[11], and Tripathy[6] have independently introduced the statistical convergence and Cauchy for double sequences and given the relation between statistical convergent and strongly Cesà ro summable double sequences. Altay and BaŞar[12] have defined the spacesBS,BS(t), CS p , CS bp , CS r , andBV of double sequences consisting of all double series whose sequence of partial sums is in the spaces M u , M u (t), C p , C bp , C r , and L u , respectively, and also examined some properties of those sequence spaces and determined the α− duals of the spacesBS,BV, CS bp , and the β(ϑ)− duals of the spaces CS bp and CS r of double series. Basar and Sever[13] have introduced the Banach space L q of double sequences corresponding to the well-known space q of single sequences and examined some properties of the space L q . Quite recently, Subramanian and Misra[14] have studied the space χ M 2 p , q , u of double sequences and gave some inclusion relations.

The class of sequences which is strongly Cesà ro summable with respect to a modulus was introduced by Maddox[15] as an extension of the definition of strongly Cesà ro summable sequences. Connor[16] further extended this definition to a definition of strong A- summability with respect to a modulus, where A = (an,k) is a non-negative regular matrix, and established some connections between strong A- summability, strong A- summability with respect to a modulus, and A- statistical convergence. In[17], the notion of convergence of double sequences was presented by Pringsheim. Also, in[18, 19], and[20], the four-dimensional matrix transformation ( Ax ) k , = m = 1 n = 1 a k mn x mn was studied extensively by Hamilton.

We need the following inequality in the sequel of the paper. For a, b, ≥ 0 and 0 < p < 1, we have

( a + b ) p a p + b p .
(1)

The double series m , n = 1 x mn is called convergent if and only if the double sequence (s mn ) is convergent, where s mn = i , j = 1 m , n x ij (m,nN). A sequence x = (x mn ) is said to be double analytic if sup mn |x mn |1/m+n < . The vector space of all double analytic sequences will be denoted by Λ2. A sequence x = (x mn ) is called double gai sequence if ((m + n)!|x mn |)1/m+n → 0 as m,n. The double gai sequences will be denoted by χ2. Let ϕ = {all finite sequences}.

Consider a double sequence x = (x ij ). The (m,n)th section x[m,n] of the sequence is defined by x [ m , n ] = i , j = 0 m , n x ij ij for allm,nN, where ij denotes the double sequence whose only non-zero term is a 1 i + j ! in the (i,j)th place for eachi,jN.

A Fréchet coordinate space (FK-space or a metric space) X is said to have an AK property if ( mn ) is a Schauder basis for X, or equivalently x[m,n] → x. An FDK-space is a double sequence space endowed with a complete metrizable space, locally convex topology under which the coordinate mappingsx=( x k )( x mn )(m,nN) are also continuous.

Let M and Φ be mutually complementary modulus functions. Then, we have

  1. (1)

    For all u,y ≥ 0,

    uyM(u)+Φ(y),(Young’s inequality; see [21]).
    (2)
  2. (2)

    For all u ≥ 0,

    uη(u)=M(u)+Φ η ( u ) .
    (3)
  3. (3)

    For all u ≥ 0 and 0 < λ < 1,

    M λ u λM(u).
    (4)

Lindenstrauss and Tzafriri[22] used the idea of Orlicz function to construct Orlicz sequence space

M = x w : k = 1 M x k ρ < , for some ρ > 0 .

The space M with the norm

x = inf ρ > 0 : k = 1 M x k ρ 1

becomes a Banach space which is called an Orlicz sequence space. For M(t) = tp(1 ≤ p < ), the spaces M coincide with the classical sequence space p .

A sequence f = (f mn ) of modulus function is called a Musielak-modulus function. A sequence g = (g mn ) defined by

g mn ( v ) = sup v u f mn ( u ) : u 0 , m , n = 1 , 2 ,

is called the complementary function of a Musielak-modulus function f. For a given Musielak modulus function f, the Musielak-modulus sequence space t f and its subspace h f are defined, respectively, as follows:

t f = x w 2 : I f x mn 1 / m + n 0 as m , n

and

h f = x w 2 : I f x mn 1 / m + n 0 as m , n ,

where I f is a convex modular defined by

I f ( x ) = m = 1 n = 1 f mn x mn 1 / m + n , x = x mn t f .

We consider that t f is equipped with the Luxemburg metric

d x , y = sup mn inf m = 1 n = 1 f mn × x mn 1 / m + n mn 1 .

If X is a sequence space, we give the following definitions:

  1. (1)

    X = the continuous dual of X;

  2. (2)
    X α = a = ( a mn ) : m , n = 1 a mn x mn < , for each x X

    ;

  3. (3)
    X β = a = ( a mn ) : m , n = 1 a mn x mn is convergent, for each x X

    ;

  4. (4)
    X γ = a = ( a mn ) : sup mn 1 m , n = 1 M , N a mn x mn < , for each x X

    ;

  5. (5)

    let X be an FK-space  ⊃ ϕ, then X f = {f( mn ) : f ∈ X };

  6. (6)

    X δ = {a = (a mn ) : sup mn  |a mn x mn |1/m+n < , for each x ∈ X},

where Xα, Xβ, and Xγ are called α− (or Kö the-Toeplitz) dual of X, β− (or generalized Kö the-Toeplitz) dual of X, γ− dual of X, and δ− dual of X, respectively. Xα is defined by Kantham and Gupta[21]. It is clear that Xα ⊂ Xβ and Xα ⊂ Xγ, but Xβ ⊂ Xγ does not hold since the sequence of partial sums of a double convergent series needs not to be bounded.

The notion of difference sequence spaces (for single sequences) was introduced by Kizmaz[23] as follows:

Z Δ = x = x k w : Δ x k Z

for Z = c,c0 and , where Δx k  = x k xk+1 for allkN.

Here, c, c0, and denote the classes of convergent, null, and bounded scalar valued single sequences, respectively. The difference sequence space bv p of the classical space p is introduced and studied in the case 1 ≤ p ≤  and in the case 0 < p < 1 by Altay and BaŞar in[12]. The spaces c(Δ), c0(Δ), (Δ), and bv p are Banach spaces normed by

x = x 1 + sup k 1 Δ x k and x bv p = k = 1 x k p 1 / p , 1 p < .

Later on, the notion was further investigated by many others. We now introduce the following difference double sequence spaces defined by

Z Δ = x = x mn w 2 : Δ x mn Z ,

where Z = Λ2,χ2 and Δx mn  = (x mn  − xmn+1) − (xm+1n − xm+1n+1) = x mn  − xmn+1 − xm+1n + xm+1n+1 for allm,nN.

Definition and preliminaries

LetnN and X be a real vector space of dimension w, where n ≤ w. A real valued function d p ( x 1 ,, x n )=( d 1 ( x 1 ),, d n ( x n )) p on X satisfying the following four conditions:

  1. (1)
    ( d 1 ( x 1 ),, d n ( x n )) p =0

    if and and only if d 1 ( x 1 ),, d n ( x n ) are linearly dependent,

  2. (2)
    ( d 1 ( x 1 ),, d n ( x n )) p

    is invariant under permutation,

  3. (3)
    (α d 1 ( x 1 ),, d n ( x n )) p = α ( d 1 ( x 1 ),, d n ( x n )) p ,α
  4. (4)

    d p ((x 1,y 1),(x 2,y 2)⋯(x n ,y n )) = (d X (x 1,x 2,⋯x n )p + d Y (y 1,y 2,⋯y n )p)1/p for 1 ≤ p < ; (or)

  5. (5)

    d((x 1,y 1),(x 2,y 2),⋯(x n ,y n )) := sup{d X (x 1,x 2,⋯x n ),d Y (y 1,y 2,⋯y n )}, for x 1,x 2,⋯x n  ∈ X,y 1,y 2,⋯y n  ∈ Y which is called the p product metric of the Cartesian product of n metric spaces is the p norm of the n-vector of the norms of the n subspaces.

A trivial example of the p product metric of the n metric space is the p norm space which isX=R equipped with the following Euclidean metric in the product space:

( d 1 ( x 1 ) , , d n ( x n ) ) E = sup | det ( d mn x mn ) | = sup d 11 x 11 d 12 x 12 d 1 n x 1 n d 21 x 21 d 22 x 22 d 2 n x 1 n d n 1 x n 1 d n 2 x n 2 d nn x nn ,

where x i = x i 1 , x in R n for each i = 1,2,⋯n.

If every Cauchy sequence in X converges to some L ∈ X, then X is said to be complete with respect to the p- metric. Any complete p- metric space is said to be p- Banach metric space.

Let X be a linear metric space. A functionw:XR is called paranorm if

  1. (1)

    w(x) ≥ 0 for all x ∈ X;

  2. (2)

    w(−x) = w(x) for all x ∈ X,

  3. (3)

    w(x + y) ≤ w(x) + w(y) for all x,y ∈ X;

  4. (4)

    If (σ mn ) is a sequence of scalars with σ mn  → σ as m,n → , and (x mn ) is a sequence of vectors with w(x mn  − x) → 0 as m,n → , then w(σ mn x mn  − σ x) → 0 as m,n → .

A paranorm w for which w(x) = 0 implies x = 0 is called a total paranorm, and the pair (X,w) is called a total paranormed space. It is well known that the metric of any linear metric space is given by some total paranorm (see[24], Theorem 10.4.2, p.183).

The notion of λ− double gai and double analytic sequences is as follows: Letλ= λ mn m , n = 0 be a strictly increasing sequence of positive real numbers tending to infinity, that is,

0 < λ 0 < λ 1 < and λ mn as m , n

and that a sequence x = (x mn ) ∈ w2 is λ− convergent to 0, called a the λ− limit of x, if μ mn (x) → 0 as m,n → , where

μ mn ( x ) = 1 φ rs m σ , σ P rs n σ , σ P rs × λ m , n λ m , n + 1 λ m + 1 , n + λ m + 1 , n + 1 x mn 1 / m + n .

The sequence x = (x mn ) ∈ w2 is λ− double analytic if sup uv  |μ mn (x)| < . If lim mn x mn  = 0 in the ordinary sense of convergence, then

lim mn 1 φ rs m σ , σ P rs n σ , σ P rs λ m , n λ m , n + 1 λ m + 1 , n + λ m + 1 , n + 1 m + n ! x mn 0 1 / m + n = 0 .

This implies that it yields lim uv μ mn (x) = 0, and hence, x = (x mn ) ∈ w2 is λ− convergent to 0. Let f = (f mn ) be a Musielak-modulus function, (X,∥(d(x1),d(x2),⋯ ,d(xn−1))∥ p ) be a p-metric space, and q = (q mn ) be double analytic sequence of strictly positive real numbers. By w2(p − X), we denote the space of all sequences as (X,∥(d(x1),d(x2),⋯ ,d(xn−1))∥ p ). The following inequality will be used throughout the paper. If 0 ≤ q mn  ≤ supq mn  = H,K = m a x(1,2H−1), then

a mn + b mn q mn K a mn q mn + b mn q mn
(5)

for all m,n and a mn , b mn C. Also, a q mn max 1 , a H for allaC.

In the present paper, we define the following sequence spaces:

χ f μ 2 q , d x 1 , d x 2 , , d x n 1 p φ = lim mn f mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p q mn = 0 , Λ f μ 2 q , d x 1 , d x 2 , , d x n 1 p φ = sup mn f mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p q mn < .

If we take f mn (x) = x, we get

χ μ 2 q , d x 1 , d x 2 , , d x n 1 p φ = lim mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p q mn = 0 , Λ μ 2 q , d x 1 , d x 2 , , d x n 1 p φ = sup mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p q mn < .

If we take q = (q mn ) = 1

χ f μ 2 , d x 1 , d x 2 , , d x n 1 p φ = lim mn f mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p = 0 , Λ f μ 2 , d x 1 , d x 2 , , d x n 1 p φ = sup mn f mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p < .

In the present paper, we plan to study some topological properties and inclusion relation between the above defined sequence spaces, χ f μ 2 q , d x 1 , d x 2 , , d x n 1 p φ and Λ f μ 2 q , d x 1 , d x 2 , , d x n 1 p φ , which we shall discuss in this paper.

Main results

Theorem 1

Let f = ( f mn ) be a Musielak-modulus function and q = (q mn ) be a double analytic sequence of strictly positive real numbers; the sequence spaces χ f μ 2 q , d x 1 , d x 2 , , d x n 1 p φ and Λ f μ 2 q , d x 1 , d x 2 , , d x n 1 p φ are linear spaces.

Proof

It is routine verification. Therefore, the proof is omitted. □

Theorem 2

Let f = ( f mn ) be a Musielak-modulus function and q = (q mn ) be a double analytic sequence of strictly positive real numbers; the sequence space χ f μ 2 q , d x 1 , d x 2 , , d x n 1 p φ is a paranormed space with respect to the paranorm defined by

g ( x ) = inf f mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ q mn 1 / H 1 ,

where H = max(1,sup mn q mn  < ).

Proof

Clearly, g(x) ≥ 0 forx= x mn χ f μ 2 q , d x 1 , d x 2 , , d x n 1 p V 2 . Since f mn (0) = 0, we get g(0) = 0.

Conversely, suppose that g(x) = 0, then

inf f mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ q mn 1 / H 1 = 0 .

Suppose that μ mn (x) ≠ 0 for eachm,nN. Then, μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ . It follows that f mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p V 2 q mn 1 / H which is a contradiction. Therefore, μ mn (x) = 0. Let

f mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ q mn 1 / H 1

and

f mn μ mn ( y ) , d x 1 , d x 2 , , d x n 1 p φ q mn 1 / H 1 .

Then, by using Minkowski’s inequality, we have

f mn μ mn x + y , d x 1 , d x 2 , , d x n 1 p φ q mn 1 / H f mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ q mn 1 / H + f mn μ mn ( y ) , d x 1 , d x 2 , , d x n 1 p φ q mn 1 / H .

So, we have

g x + y = inf f mn μ mn x + y , d x 1 , d x 2 , , d x n 1 p φ q mn 1 / H 1 inf f mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ q mn 1 / H 1 + inf f mn μ mn ( y ) , d x 1 , d x 2 , , d x n 1 p φ q mn 1 / H 1 .

Therefore,

g x + y g ( x ) + g ( y ) .

Finally, to prove that the scalar multiplication is continuous, let λ be any complex number. By definition,

g λ x = inf f mn μ mn λ x , d x 1 , d x 2 , , d x n 1 p φ q mn 1 / H 1 .

Then,

g λ x = inf ( ( λ t ) q mn / H : f mn μ mn x , d x 1 , d x 2 , , d x n 1 p φ q mn 1 / H 1 ,

wheret= 1 λ . Since λ q mn max 1 , λ sup p mn , we have

g λ x max 1 , λ sup p mn × inf t q mn / H : f mn μ mn λ x , d x 1 , d x 2 , , d x n 1 p φ q mn 1 / H 1 .

Theorem 3

The βdual space of χ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ β = Λ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ .

Proof

First, we observe that

χ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ β Γ f μ 2 q , μ mn x , d x 1 , d x 2 , , d x n 1 p φ .

Therefore,

Γ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ β χ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ β .

But

Γ f μ 2 q β Λ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ .

Hence,

Λ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ χ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ β .
(6)

Next, we show that

χ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ β Λ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ .

Lety= y mn χ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ β . Considerf(x)= m = 1 n = 1 x mn y mn with

x = x mn χ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ x = λ mn λ mn + 1 λ m + 1 n λ m + 1 n + 1 = 0 0 0 0 0 0 0 0 0 0 0 0 φ rs Δ λ mn m + n ! φ rs Δ λ mn m + n ! 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 φ rs Δ λ mn m + n ! φ rs Δ λ mn m + n ! 0 0 0 0 0 0
f mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ = 0 0 0 0 0 0 0 0 0 , 0 0 0 f mn φ rs Δ λ mn m + n ! f mn φ rs Δ λ mn m + n ! 0 0 0 f mn φ rs Δ λ mn m + n ! f mn φ rs Δ λ mn m + n ! 0 0 0 0 0 , 0 .

Hence, it converges to zero.

Therefore,

λ mn λ mn + 1 λ m + 1 n λ m + 1 n + 1 χ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ .

Hence, d((λ mn  − λmn+1) − (λm+1n − λm+1n+1),0) = 1. But

y mn f d λ mn λ mn + 1 λ m + 1 n λ m + 1 n + 1 , 0 f · 1 <

for each m,n. Thus, (y mn ) is a p- metric paranormed space of double analytic sequence and, hence, an p- metric double analytic sequence.

In other words.y Λ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ . But y = (y mn ) is arbitrary in χ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ β .Therefore,

χ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ β Λ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ .
(7)

From (6) and (7), we get

χ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ β = Λ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ .

Theorem 4

The dual space of χ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ is Λ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ . In other words. χ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ = Λ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ .

Proof

We recall that

λ mn = 0 0 0 0 0 0 0 0 0 0 φ rs Δ λ mn m + n ! 0 0 0 0 0

with φ rs Δ λ mn m + n ! in the (m,n)th position and zeros elsewhere,

χ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ = 0 . . . 0 . . . 0 f φ rs Δ λ mn m + n ! 1 / m + n . 0 ( m , n ) th 0 . . 0

which is a p- metric of double gai sequence. Hence,

x χ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ .f ( x ) = m , n = 1 x mn y mn

withx χ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ andf χ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ , where χ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ is the dual space of χ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ .

Takex=( x mn ) χ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ . Then,

y mn f d( φ rs ,0)<m,n.
(8)

Thus, (y mn ) is a p- metric of the double analytic sequence and an p- metric of double analytic sequence. In other words,y Λ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ . Therefore,

χ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ = Λ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ .

This completes the proof. □

Theorem 5

(1) If the sequence (f mn ) satisfies uniform Δ2condition, then

χ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ α = χ g 2 q μ , μ uv ( x ) , d x 1 , d x 2 , , d x n 1 p φ .
  1. (2)

    If the sequence (g mn ) satisfies uniform Δ2− condition, then

    χ g 2 q μ , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ α = χ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ .

Proof

Let the sequence ( f mn ) satisfies uniform Δ2− condition; we get

χ g 2 q μ , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ χ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ α .
(9)

To prove the inclusion

χ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ α χ g 2 q μ , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ ,

leta χ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ α . Then, for all {x mn } with x mn χ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ , we have

m = 1 n = 1 x mn a mn <.
(10)

Since the sequence ( f mn ) satisfies the uniform Δ2− condition and then

y mn χ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ ,

we get m = 1 n = 1 φ rs y mn a mn Δ λ mn m + n ! <. by (10). Thus, φ rs a mn χ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ = χ g 2 q μ , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ , and hence, a mn χ g 2 q μ , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ . This gives that

χ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ α χ g 2 q μ , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ .
(11)

We are granted with (9) and (11) that

χ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ α = χ g 2 q μ , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ .
  1. (3)

    Similarly, one can prove that

    χ g 2 q μ , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ α χ f μ 2 q , μ mn x , d x 1 , d x 2 , , d x n 1 p φ

if the sequence (g mn ) satisfies the uniform Δ2− condition. □

Proposition 1

If 0 < q mn  < p mn  < ∞ for each m and m, then

Λ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ Λ f μ 2 p , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ .

Proof

Letx= x mn Λ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ . We have

sup mn Λ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ < .

This implies that

Λ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ < 1

for sufficiently large value of m and n. Since f mn s are non-decreasing, we get

sup mn Λ f μ 2 p , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ sup mn Λ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ .

Thus,x= x mn Λ f μ 2 p , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ . □

Proposition 2

(1) If 0 < infq mn  ≤ q mn  < 1, then

Λ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ Λ f μ 2 , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ .

(2) If 1 ≤ q mn  ≤ supq mn  < , then Λ f μ 2 , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ Λ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ .

Proof

Letx= x mn Λ f μ 2 , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ . Since 0 < infq mn  ≤ 1, we have

sup uv Λ f μ 2 , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ Λ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ ,

and hence

x = x mn Λ f μ 2 , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ .
  1. (3)

    Let q mn for each (m,n) and sup mn q mn  < .

Letx= x mn Λ f μ 2 , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ . Then, for each 0 < ε < 1, there exists a positive integerN such that

sup uv Λ f μ 2 , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ ε < 1 ,

for all m,n ≥ N. This implies that

sup mn Λ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ sup mn Λ f μ 2 , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ .

Thus,x= x mn Λ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ . □

Proposition 3

Let f = f mn and f = f mn be sequences of Musielak functions; we have

Λ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ × Λ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ × Λ f + f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ .

Proof

The proof is easy, so we omit it. □

Proposition 4

For any sequence of Musielak functions f = (f mn ) and q = (q mn ) be double analytic sequence of strictly positive real numbers. Then,

χ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ Λ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ .

Proof

The proof is easy, so we omit it. □

Proposition 5

The sequence space Λ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ is solid.

Proof

Letx= x mn Λ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ , i.e.,

sup mn Λ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ < .

Let (α mn ) be double sequence of scalars such that |α mn | ≤ 1 for all m,n ∈ N × N. Then, we get

sup mn Λ f μ 2 q , μ mn α x , d x 1 , d x 2 , , d x n 1 p φ sup mn Λ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ .

Proposition 6

The sequence space Λ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ is monotone.

Proof

The proof follows from Proposition 5. □

Proposition 7

If f = (f mn ) is any Musielak function, then

Λ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ Λ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ

if and only if sup r , s 1 φ rs φ rs <.

Proof

Letx Λ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ andN= sup r , s 1 φ rs φ rs <. Then, we get

Λ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ rs = N Λ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ rs = 0 .

Thus,x Λ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ . Conversely, suppose that

Λ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ Λ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ

andx Λ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ .

Then, Λ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ <εfor every ε > 0. Suppose that sup r , s 1 φ rs φ rs =, then there exists a sequence of members (rs jk ) such that lim j , k φ jk φ jk =. Hence, we have Λ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ rs =. Therefore,x Λ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ , which is a contradiction. □

Proposition 8

If f = (f mn ) is any Musielak function, then

Λ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ = Λ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ

if and only if sup r , s 1 φ rs φ rs <, sup r , s 1 φ rs φ rs >.

Proof

It is easy to prove, so we omit it. □

Proposition 9

The sequence space χ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ is not solid.

Proof

The result follows from the following example. Consider

x = x mn = 1 1 1 1 1 1 1 1 1 χ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ .

Let

α mn = 1 m + n 1 m + n 1 m + n 1 m + n 1 m + n 1 m + n 1 m + n 1 m + n 1 m + n ,

for allm,nN. Then, α mn x mn χ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ . Hence, χ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ is not solid. □

Proposition 10

The sequence space χ f μ 2 q , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ is not monotone.

Proof

The proof follows from Proposition 9. □

Generalized four-dimensional infinite matrix sequence spaces

LetA= a k mn be a four-dimensional infinite matrix of complex numbers. Then, we haveA(x)= ( Ax ) k = m = 1 n = 1 a k mn x mn which converges for each k,.

In this section, we introduce the following sequence spaces:

χ f μ 2 qA , d x 1 , d x 2 , , d x n 1 p φ = lim mn f mn A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p q mn = 0 , Λ f μ 2 qA , d x 1 , d x 2 , , d x n 1 p φ = sup mn f mn A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p q mn < .

If we take f mn (x) = x, we get

χ μ 2 qA , d x 1 , d x 2 , , d x n 1 p φ = lim mn A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p q mn = 0 , Λ μ 2 qA , d x 1 , d x 2 , , d x n 1 p φ = sup mn A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p q mn < .

If we take q = (q mn ) = 1,

χ f μ 2 A , d x 1 , d x 2 , , d x n 1 p φ = lim mn f mn A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p = 0 , Λ f μ 2 A , d x 1 , d x 2 , , d x n 1 p φ = sup mn f mn A mn μ mn x , d x 1 , d x 2 , , d x n 1 p < .

Theorem 6

For a Musielak-modulus function, f = (f mn ). Then, the sequence spaces χ f μ 2 qA , d x 1 , d x 2 , , d x n 1 p φ and Λ f μ 2 qA , d x 1 , d x 2 , , d x n 1 p φ are linear spaces over the set of complex numbersC.

Proof

It is routine verification. Therefore, the proof is omitted. □

Theorem 7

For any Musielak-modulus function f = (f mn ) and a double analytic sequence q = (q mn ) of strictly positive real numbers, the space χ f μ 2 qA , d x 1 , d x 2 , , d x n 1 p φ is a topological linear space paranormed by

g ( x ) = inf f mn A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ q mn 1 / H 1 ,

where H = max(1,sup mn q mn  < ).

Proof

Clearly, g(x) ≥ 0 forx= x mn χ f μ 2 qA , d x 1 , d x 2 , , d x n 1 p V 2 . Since f mn (0) = 0, we get g(0) = 0. Conversely, suppose that g(x) = 0, theninf f mn A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ q mn 1 / H 1=0. Suppose that A mn μ mn (x) ≠ 0 for eachm,nN, then

A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ .
(12)

It follows that f mn A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p V 2 q mn 1 / H which is a contradiction. Therefore, A mn μ mn (x) = 0. Let f mn A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ q mn 1 / H 1 and f mn A mn μ mn ( y ) , d x 1 , d x 2 , , d x n 1 p φ q mn 1 / H 1.

Then, by using Minkowski’s inequality, we have

f mn A mn μ mn x + y , d x 1 , d x 2 , , d x n 1 p φ q mn 1 / H f mn A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ q mn 1 / H + f mn A mn μ mn ( y ) , d x 1 , d x 2 , , d x n 1 p φ q mn 1 / H .

So, we have

g x + y = inf f mn A mn μ mn x + y , d x 1 , d x 2 , , d x n 1 p φ q mn 1 / H 1 inf f mn A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ q mn 1 / H 1 + inf f mn A mn μ mn ( y ) , d x 1 , d x 2 , , d x n 1 p φ q mn 1 / H 1 .

Therefore,

g x + y g ( x ) + g ( y ) .

Finally, to prove that the scalar multiplication is continuous, let λ be any complex number. By definition,

g λ x = inf f mn A mn μ mn λ x , d x 1 , d x 2 , , d x n 1 p φ q mn 1 / H 1 .

Then,

g λ x = inf ( ( λ t ) q mn / H : f mn A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ q mn 1 / H 1 ,

wheret= 1 λ . Since λ q mn max 1 , λ supp mn , we have

g λ x max 1 , λ supp mn inf t q mn / H : f mn A mn μ mn λ x , d x 1 , d x 2 , , d x n 1 p φ q mn 1 / H 1 .

Theorem 8

The βdual space of χ f μ 2 qA , A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ β = Λ f μ 2 qA , A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ .

Proof

First, we observe that

χ f μ 2 qA , A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ β Γ f μ 2 qA , A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ .

Therefore,

Γ f μ 2 qA , A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ β χ f μ 2 qA , A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ β .

But

Γ f μ 2 qA β Λ f μ 2 qA , A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ .

Hence,

Λ f μ 2 qA , A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ χ f μ 2 qA , A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ β .

Next, we show that

χ f μ 2 qA , A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ β Λ f μ 2 qA , A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ .

Lety= y mn χ f μ 2 qA , A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ β . Considerf(x)= m = 1 n = 1 x mn y mn with

x = x mn χ f μ 2 qA , A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ x = λ mn λ mn + 1 λ m + 1 n λ m + 1 n + 1 = 0 0 0 0 0 0 0 0 0 0 0 0 φ rs a k mn Δ λ mn m + n ! φ rs a k mn Δ λ mn m + n ! ... 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 φ rs a k mn Δ λ mn m + n ! φ rs a k mn Δ λ mn m + n ! ... 0 0 0 0 0 0 f mn A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ = 0 0 ... 0 0 ... 0 0 0 ... 0 0 , ... 0 . . . 0 0 ... f mn φ rs a k mn Δ λ mn m + n ! f mn φ rs a k mn Δ λ mn m + n ! ... 0 0 0 ... f mn φ rs a k mn Δ λ mn m + n ! f mn φ rs a k mn Δ λ mn m + n ! ... 0 0 0 ... 0 0 , ... 0 .

Hence, converges to zero.

Therefore,

λ mn λ mn + 1 λ m + 1 n λ m + 1 n + 1 × χ f μ 2 qA , A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ .

Hence,d a k mn λ mn λ mn + 1 λ m + 1 n λ m + 1 n + 1 , 0 =1. However, y mn fd a k mn λ mn λ mn + 1 λ m + 1 n λ m + 1 n + 1 , 0 f·1< for each m,n. Thus, (y mn ) is a p- metric paranormed space of double analytic sequence and, hence, an p- metric double analytic sequence.

In other words,y Λ f μ 2 qA , μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ . However, y = (y mn ) is arbitrary in χ f μ 2 q , A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ β . Therefore,

χ f μ 2 qA , A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ β Λ f μ 2 qA , A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ .
(13)

From (12) and (13), we get

χ f μ 2 qA , A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ β = Λ f μ 2 qA , A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ .

Theorem 9

The dual space of χ f μ 2 qA , A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ is Λ f μ 2 qA , A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ . In other words,

χ f μ 2 qA , A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ = Λ f μ 2 qA , A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ .

Proof

We recall that

λ mn = 0 0 0 0 0 0 0 0 0 0 φ rs a k mn Δ λ mn ( m + n ) ! 0 0 0 0 0

with φ rs a k mn Δ λ mn ( m + n ) ! in the (m,n)th position and zero elsewhere,

χ f μ 2 qA , A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ = 0 . . . 0 0 f φ rs a k mn Δ λ mn ( m + n ) ! 1 / m + n . 0 ( m , n ) th 0 . . 0

which is a p- metric of double gai sequence. Hence,

x χ f μ 2 qA , A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ .f ( x ) = m , n = 1 x mn y mn

with

x χ f μ 2 qA , A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ

and

f χ f μ 2 qA , A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ ,

where

χ f μ 2 qA , A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ

is the dual space of χ f μ 2 qA , A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ .

Takex=( x mn ) χ f μ 2 qA , A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ . Then,

y mn f d( φ rs ,0)<m,n.
(14)

Thus, (y mn ) is a p- metric of double analytic sequence and, hence, an p- metric of double analytic sequence. In other words,y Λ f μ 2 qA , A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ . Therefore,

χ f μ 2 qA , A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ = Λ f μ 2 q , A mn μ mn ( x ) , d x 1 , d x 2 , , d x n 1 p φ .