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Algebraic duals of some sets of difference sequences defined by Orlicz functions

Abstract

In this paper, we first give a description of some sets of sequences generated by difference operators and defined by Orlicz functions. Then their algebraic duals such as the α-, β-, γ- and null-duals are computed.

MSC:46A45, 47N40, 65J99, 46A20.

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 )?

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Correspondence to Iqbal H Jebril.

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Authors’ contributions

HD carried out the initial description of the spaces. IHJ proposed the structure of the paper. HD and IHJ both formulated and proved the results. All authors read and approved the final manuscript.

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Dutta, H., Jebril, I.H. Algebraic duals of some sets of difference sequences defined by Orlicz functions. J Inequal Appl 2014, 300 (2014). https://doi.org/10.1186/1029-242X-2014-300

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Keywords

  • difference sequences
  • Orlicz function
  • algebraic duals