1 Introduction

Let w denote the space of all scalar sequences, and any subspace of w is called a sequence space. Let , c and c 0 be the linear spaces of bounded, convergent and null sequences x=( x k ) with complex terms, respectively, normed by x = sup k | x k |, where kN={1,2,}, the set of positive integers.

Lindenstrauss and Tzafriri [1] used the Orlicz function and introduced the sequence space M as follows:

M = { ( x k ) w : k = 1 M ( | x k | ρ ) <  for some  ρ > 0 } .

They proved that M is a Banach space normed by

( x k ) =inf { ρ > 0 : k = 1 M ( | x k | ρ ) 1 } .

Throughout this section X will denote one of the sequence spaces , c and c 0 .

The notion of difference sequence spaces was introduced by Kizmaz [2]. For some other works on difference sequences, Orlicz functions and related literature, we refer to [38]. Let v=( v k ) be any fixed sequence of non-zero complex numbers. Et and Esi [9] generalized the above sequence spaces to the following sequence spaces:

X ( Δ v m ) = { x = ( x k ) : ( Δ v m x k ) X }

for X= ,c and  c 0 .

In this paper, for an Orlicz function M, we can have the following spaces in the line of the spaces studied by Mursaleen et al. [10]:

X ( M , Δ v m ) = { x = ( x k ) : ( Δ v m x k ) X ( M ) } .

In fact, we get the following spaces:

( M , Δ v m ) = { x = ( x k ) : sup k M ( | Δ v m x k | ρ ) <  for some  ρ > 0 } , c ( M , Δ v m ) = { x = ( x k ) : lim k M ( | Δ v m x k L | ρ ) = 0  for some  L  and  ρ > 0 } , c 0 ( M , Δ v m ) = { x = ( x k ) : lim k M ( | Δ v m x k | ρ ) = 0  for some  ρ > 0 } ,

where Δ v m x k = Δ v m 1 x k Δ v m 1 x k + 1 , Δ v m x k = i = 0 m ( 1 ) i ( m i ) v k + i x k + i for all kN.

Bektaş et al. [11] introduced the difference operator Δ v ( m ) and defined it as follows:

Δ v ( m ) x k = i = 0 m ( 1 ) i ( m i ) v k i x k i for all kN.

Using this difference operator, we can construct the following sequence space:

X ( M , Δ v ( m ) ) = { x = ( x k ) : ( Δ v ( m ) x k ) X ( M ) } .

The operator

Σ ( m ) :ww

is defined by

Σ ( 1 ) x k = j = 0 k x j (k=0,1,), Σ ( m ) = Σ ( 1 ) o Σ ( m 1 ) (m2)

and

Σ ( m ) o Δ ( m ) = Δ ( m ) o Σ ( m ) =id,the identity on w(see [9, 12]).

Now, for subsequent use, we slightly generalize the above definition as follows.

We define

Σ v ( m ) :ww

by

Σ v ( 1 ) x k = j = 0 k v j x j (k=0,1,), Σ v ( m ) = Σ v ( m ) o Σ v ( m 1 ) (m2)

and

v 1 Σ v ( m ) o Δ v ( m ) = v 1 Δ v ( m ) o Σ v ( m ) =id,the identity on w and  v 1 = ( v k 1 ) .

Now, for xX(M, Δ v m ), let us define

x Δ = i = 1 m | v i x i |+inf { ρ > 0 : sup k M ( | Δ v m x k | ρ ) 1 } .

It can be shown that (X(M, Δ v m ), Δ ) is a BK-space.

Again for xX(M, Δ v ( m ) ), let us define

x Δ =inf { ρ > 0 : sup k M ( | Δ v ( m ) x k | ρ ) 1 } .

It can be shown that (X(M, Δ v ( m ) ), Δ ) is a BK-space.

It is trivial that ( Δ v m x k )X(M) if and only if ( Δ v ( m ) x k )X(M). Also the norms Δ and Δ are equivalent.

Let us define the operator

D:X ( M , Δ v m ) X ( M , Δ v m )

by Dx=(0,0,, x m + 1 , x m + 2 ,), where x=( x 1 , x 2 ,, x m ,). It is trivial that D is a bounded linear operator on X(M, Δ v m ), X= , c and c 0 . Furthermore, the set

D [ X ( M , Δ v m ) ] =DX ( M , Δ v m ) = { x = ( x k ) : x X ( M , Δ v m ) , x 1 = x 2 = = x m = 0 }

is a subspace of X(M, Δ v m ) and normed by

x Δ =inf { ρ > 0 : sup k M ( | Δ v m x k | ρ ) 1 } in DX ( M , Δ v m ) .

DX(M, Δ v m ) and X(M) are equivalent as topological spaces since

Δ v m :DX ( M , Δ v m ) X(M),

defined by

Δ v m x=y= ( Δ v m x k ) ,
(1.1)

is a linear homeomorphism.

Moreover, obviously

Δ v ( m ) : X ( M , Δ v ( m ) ) X ( M ) , Δ v ( m ) x = y = ( Δ v ( m ) x k ) , Σ v ( m ) : X ( M ) X ( M , Δ v ( m ) ) , Σ v ( m ) x = y = ( Σ v ( m ) x k )

are isometric isomorphisms for X= ,c and  c 0 .

Hence (M, Δ v m ), c(M, Δ v m ) and c 0 (M, Δ v m ) are isometrically isomorphic to (M), c(M) and c 0 (M), respectively.

Moreover, X(M, Δ v i )X(M, Δ v m ) for i=0,1,,m1, which follows from the following inequality and convexity of M:

M ( | Δ v m x k | 2 ρ ) 1 2 M ( | Δ v m 1 x k | ρ ) + 1 2 M ( | Δ v m 1 x k + 1 | ρ ) .

Investigation of spaces is often combined with that of duals. The algebraic dual space is defined for all vector spaces. When defined for a topological vector space, there is a subspace of this dual space, corresponding to continuous linear functionals, which constitutes a continuous dual space. For any finite-dimensional normed vector space or topological vector space, such as Euclidean n-space, the continuous dual and the algebraic dual coincide. This is, however, false for any infinite-dimensional normed space. For some related literature on duality relevant to this paper, we refer to [9, 11, 13, 14]. Our results of this paper will generalize few existing results as well as generate some new results in the literature of algebraic duality within the field of functional analysis.

2 Computation of algebraic duals

In this section we compute the α-, β-, γ- and N-duals of the spaces (M, Δ v m ), c(M, Δ v m ) and c 0 (M, Δ v m ).

Definition 2.1 [13]

Let X be a sequence space and define

X α = { a = ( a k ) : k = 1 | a k x k | < , x X } , X β = { a = ( a k ) : k = 1 a k x k  is convergent , x X } , X γ = { a = ( a k ) : sup n | k = 1 n a k x k | < , x X } , X N = { a = ( a k ) : lim k a k x k = 0 , x X } ,

then X α , X β , X γ and X N are called the α-, β-, γ- and N-(or null) duals of X, respectively. It is known that if XY, then Y η X η for η=α,β,γ and N.

Lemma 2.1 [2]

Let m be a positive integer. Then there exist positive constants C 1 and C 2 such that

C 1 k m ( m + k k ) C 2 k m ,k=0,1,2,.

Lemma 2.2 x (M, Δ v m ) implies sup k M( | k 1 Δ v m 1 x k | ρ )< for some ρ>0.

Proof Let x (M, Δ v m ), then

sup k M ( | Δ v m 1 x k Δ v m 1 x k + 1 | ρ ) <for some ρ>0.

Then there exists U>0 such that

M ( | Δ v m 1 x k Δ v m 1 x k + 1 | ρ ) <Ufor all kN.

Taking η=kρ, for an arbitrary fixed positive integer k, by the subadditivity of modulus, the monotonicity and convexity of M:

M ( | Δ v m 1 x 1 Δ v m 1 x k + 1 | η ) 1 k l = 1 k M ( | Δ v m 1 x l Δ v m 1 x l + 1 | ρ ) <U.

Then the above inequality, the inequality

| Δ v m 1 x k + 1 | ( k + 1 ) ρ 1 k + 1 ( | Δ v m 1 x 1 | ρ + k | Δ v m 1 x 1 Δ v m 1 x k + 1 | k ρ )

and the convexity of M imply

M ( | Δ v m 1 x k + 1 | ( k + 1 ) ρ ) 1 k + 1 ( M ( | Δ v m 1 x 1 | ρ ) + k M ( | Δ v m 1 x 1 Δ v m 1 x k + 1 | k ρ ) ) max { M ( | Δ v m 1 x 1 | ρ ) , U } < .

Hence we have the desired result. □

Hence we have the following lemma.

Lemma 2.3

  1. (i)

    x (M, Δ v m ) implies sup k M( | k m v k x k | ρ )< for some ρ>0,

  2. (ii)

    x (M, Δ v m ) implies sup k k m | v k x k |<.

Theorem 2.4 Let M be an Orlicz function. Then

  1. (i)

    [ c 0 ( M , Δ v m ) ] α = [ c ( M , Δ v m ) ] α = [ ( M , Δ v m ) ] α = D 1 ,

  2. (ii)

    D 1 α = D 2 ,

where

D 1 = { a = ( a k ) : k = 1 k m | v k 1 a k | < } , D 2 = { b = ( b k ) : sup k k m | v k b k | < } .

Proof (i) Let a D 1 , then k = 1 k m | v k 1 a k |<. Now, for any x (M, Δ v m ), we have sup k k m | v k x k |<. Then we have

k = 1 | a k x k | sup k k m | v k x k | k = 1 k m | a k v k 1 |<.

Hence a [ ( M , Δ v m ) ] α .

Conversely, suppose that a [ X ( M , Δ v m ) ] α for X=c and  . Then k = 1 | a k x k |< for each xX(M, Δ v m ). So we can take

x k = k m v k 1 ,k1.

Then

k = 1 k m | v k 1 a k |= k = 1 | a k x k |<.

This implies that a D 1 .

Again suppose that a [ c 0 ( M , Δ v m ) ] α and a D 1 . Then there exists a strictly increasing sequence ( n i ) of positive integers n i with n 1 < n 2 < such that

k = n i + 1 n i + 1 k m | v k 1 a k |>i.

Define x c 0 (M, Δ v m ) by

x k = { 0 , 1 k n 1 , k m v k sgn a k / i , n i < k n i + 1 .

Then we have

k = 1 | a k x k | = k = n 1 + 1 n 2 | a k x k | + + k = n i + 1 n i + 1 | a k x k | + = k = n 1 + 1 n 2 k m | v k 1 a k | + + 1 i k = n i + 1 n i + 1 k m | v k 1 a k | + > 1 + 1 + = .

This contradicts a [ c 0 ( M , Δ v m ) ] α . Hence a D 1 . This completes the proof of (i).

  1. (ii)

    The proof is similar to that of part (i). □

If we take v k =1 for all kN in Theorem 2.4, then we obtain the following corollary.

Corollary 2.5 Let M be an Orlicz function. Then

  1. (i)

    [ c 0 ( M , Δ m ) ] α = [ c ( M , Δ m ) ] α = [ ( M , Δ m ) ] α = E 1 ,

  2. (ii)

    E 1 α = E 2 ,

where

E 1 = { a = ( a k ) : k = 1 k m | a k | < } , E 2 = { b = ( b k ) : sup k k m | b k | < } .

Theorem 2.6 Let M be an Orlicz function. Then [ c ( M , Δ v m ) ] N = [ ( M , Δ v m ) ] N = F 1 , where

F 1 = { a = ( a k ) : lim k k m v k 1 a k = 0 } .

Proof The proof is immediate using Lemma 2.3(ii). □

The following lemma will be used in the next theorem.

Lemma 2.7 [6]

Let ( p n ) be a sequence of positive numbers increasing monotonically to infinity.

  1. (i)

    If sup n | v = 1 n p v a v |<, then sup n | p n k = n + 1 a k |<.

  2. (ii)

    If k p k a k is convergent, then lim n p n k = n + 1 a k =0.

Theorem 2.8 Let M be an Orlicz function and c 0 + denote the set of all positive null sequences. Then

  1. (i)

    [ D ( M , Δ v m ) ] β = [ D c ( M , Δ v m ) ] β = G 1 ,

  2. (ii)

    [ D c 0 ( M , Δ v m ) ] β = G 2 ,

  3. (iii)

    [ D ( M , Δ v m ) ] γ = [ D c ( M , Δ v m ) ] γ = H 1 ,

  4. (iv)

    [ D c 0 ( M , Δ v m ) ] γ = H 2 ,

where

G 1 = { a = ( a k ) : k = 1 a k v k 1 j = 1 k m ( k j 1 m 1 )  is convergent , G 1 = k = 1 | j = k + 1 v j 1 a j | j = 1 k m + 1 ( k j 1 m 2 ) < } , G 2 = { a = ( a k ) : k = 1 a k v k 1 j = 1 k m ( k j 1 m 1 ) u j  converges and  G 2 = k = 1 | j = k + 1 v j 1 a j | j = 1 k m + 1 ( k j 1 m 2 ) u j < , u c 0 + } , H 1 = { a = ( a k ) : sup n | k = 1 n a k v k 1 j = 1 k m ( k j 1 m 1 ) | < , H 1 = k = 1 | j = k + 1 v j 1 a j | j = 1 k m + 1 ( k j 1 m 2 ) < } , H 2 = { a = ( a k ) : sup n | k = 1 n a k v k 1 j = 1 k m ( k j 1 m 1 ) u j | < , H 2 = k = 1 | j = k + 1 v j 1 a j | j = 1 k m + 1 ( k j 1 m 2 ) u j < , u c 0 + } .

Proof We give the proof for part (i) for D (M, Δ v m ), and the proof of other parts follows similarly using Lemma 2.7. For details, one may refer to [11].

For each xD (M, Δ v m ), there exists one and only one y=( y k ) (M) such that

x k = v k 1 j = 1 k m ( k j 1 m 1 ) y j , y 1 m = y 2 m == y 0 =0

for sufficiently large k by (1.1). Let a G 1 . Then, using the same technique as applied in [[11], p.429], we can show that a [ D ( M , Δ v m ) ] β .

Let a [ D ( M , Δ v m ) ] β . Again, using the same technique as applied in [[11], pp.429-430], we can show that a G 1 .

This completes the proof. □

3 Conclusion

Although we conclude this paper here, the following further suggestion remains open:

What is the N-dual of the space c 0 (M, Δ v m )?