Open Access

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

Journal of Inequalities and Applications20142014:300

https://doi.org/10.1186/1029-242X-2014-300

Received: 29 May 2014

Accepted: 8 July 2014

Published: 19 August 2014

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.

Keywords

difference sequencesOrlicz functionalgebraic duals

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 k N = { 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 k N .

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  k N .
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 ) : w w
is defined by
Σ ( 1 ) x k = j = 0 k x j ( k = 0 , 1 , ) , Σ ( m ) = Σ ( 1 ) o Σ ( m 1 ) ( m 2 )
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 ) : w w
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 ) ( m 2 )
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 x X ( 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 x X ( 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 D x = ( 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 ) ] = D X ( 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  D X ( M , Δ v m ) .
D X ( M , Δ v m ) and X ( M ) are equivalent as topological spaces since
Δ v m : D X ( 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 , , m 1 , 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 X Y , 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 ) < U for all  k N .
Taking η = k ρ , 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 x X ( M , Δ v m ) . So we can take
x k = k m v k 1 , k 1 .
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 k N 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 x D ( 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 ) ?

Declarations

Authors’ Affiliations

(1)
Department of Mathematics, Gauhati University
(2)
Department of Mathematics, Taibah University

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Copyright

© Dutta and Jebril; licensee Springer. 2014

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