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Some topological and geometrical properties of new Banach sequence spaces

Abstract

In the present paper, we introduce a new band matrix F ˆ and define the sequence space

p ( F ˆ )= { x = ( x k ) ω : k | f k f k + 1 x k f k + 1 f k x k 1 | p < ; 1 p } ,

where f k is the k th Fibonacci number for every kN. We also establish some inclusion relations concerning this space and determine its α-, β-, γ-duals. Further, we characterize some matrix classes on the space p ( F ˆ ) and examine some geometric properties of this space.

MSC:11B39, 46A45, 46B45, 46B20.

1 Introduction

Let ω be the space of all real-valued sequences. Any vector subspace of ω is called a sequence space. By , c, c 0 and p (1p<), we denote the sets of all bounded, convergent, null sequences and p-absolutely convergent series, respectively. Also, we use the conventions that e=(1,1,) and e ( n ) is the sequence whose only non-zero term is 1 in the n th place for each nN, where N={0,1,2,}.

Let X and Y be two sequence spaces and A=( a n k ) be an infinite matrix of real numbers a n k , where n,kN. We write A=( a n k ) instead of A= ( a n k ) n , k = 0 . Then we say that A defines a matrix mapping from X into Y and we denote it by writing A:XY if for every sequence x= ( x k ) k = 0 X, the sequence Ax= { A n ( x ) } n = 0 , the A-transform of x, is in Y, where

A n (x)= k = 0 a n k x k (nN).
(1.1)

For simplicity in notation, here and in what follows, the summation without limits runs from 0 to ∞. Also, if xω, then we write x=( x k ) instead of x= ( x k ) k = 0 .

By (X,Y), we denote the class of all matrices A such that A:XY. Thus, A(X,Y) if and only if the series on the right-hand side of (1.1) converges for each nN and every xX and we have AxY for all xX.

The matrix domain X A of an infinite matrix A in a sequence space X is defined by

X A = { x = ( x k ) ω : A x X }
(1.2)

which is a sequence space.

The approach constructing a new sequence space by means of the matrix domain of a particular limitation method has recently been employed by several authors; see, for instance, [112].

Let Δ denote the matrix Δ=( Δ n k ) defined by

Δ n k ={ ( 1 ) n k ( n 1 k n ) , 0 ( 0 k < n 1  or  k > n )

or

Δ n k ={ ( 1 ) n k ( n k n + 1 ) , 0 ( 0 k < n  or  k > n + 1 ) .

In the literature, the matrix domain λ Δ is called the difference sequence space whenever λ is a normed or paranormed sequence space. The idea of difference sequence spaces was introduced by Kızmaz [13]. In 1981, Kızmaz [13] defined the sequence spaces

X(Δ)= { x = ( x k ) ω : ( x k x k + 1 ) X }

for X= , c and c 0 . The difference space b v p , consisting of all sequences ( x k ) such that ( x k x k 1 ) is in the sequence space p , was studied in the case 0<p<1 by Altay and Başar [14] and in the case 1p by Başar and Altay [4] and Çolak et al. [15]. The paranormed difference sequence space

Δλ(p)= { x = ( x k ) ω : ( x k x k + 1 ) λ ( p ) }

was examined by Ahmad and Mursaleen [16] and Malkowsky [17], where λ(p) is any of the paranormed spaces (p), c(p) and c 0 (p) defined by Simons [18] and Maddox [19].

Recently, Başar et al. [20] have defined the sequence spaces bv(u,p) and b v (u,p) by

bv(u,p)= { x = ( x k ) ω : k | u k ( x k x k 1 ) | p k < }

and

b v (u,p)= { x = ( x k ) ω : sup k N | u k ( x k x k 1 ) | p k < } ,

where u=( u k ) is an arbitrary fixed sequence and 0< p k H< for all kN. These spaces are generalization of the space b v p for 1p. Quite recently, Kirişçi and Başar [21] have introduced and studied the generalized difference sequence spaces

X ˆ = { x = ( x k ) ω : B ( r , s ) x X }

for X= , p , c and c 0 , where 1p< and B(r,s)x=(s x k 1 +r x k ) (r,s0). Following Kirişçi and Başar [21], Sönmez [22] has examined the sequence space X(B) as the set of all sequences whose B(r,s,t)-transforms are in the space X{ , p ,c, c 0 }, where B(r,s,t) denotes the triple band matrix B(r,s,t)={ b n k (r,s,t)} defined by

b n k (r,s,t)={ r ( n = k ) , s ( n = k + 1 ) , t ( n = k + 2 ) , 0 otherwise

for all n,kN and r,s,tR{0}. Also in [2334], the authors studied some difference sequence spaces.

In this paper, we define the Fibonacci difference matrix F ˆ by using the Fibonacci sequence { f n } n = 0 and introduce new sequence spaces p ( F ˆ ) and ( F ˆ ) related to the matrix domain of F ˆ in the sequence spaces p and , respectively, where 1p<. This study is organized as follows.

In Section 2, we give some notations and basic concepts including the Fibonacci sequence and a BK-space. In Section 3, we define a new band matrix with Fibonacci numbers and introduce the sequence spaces p ( F ˆ ) and ( F ˆ ). Also, we establish some inclusion relations concerning these spaces and construct the basis of the space p ( F ˆ ) for 1p<. In Section 4, we determine the α-, β-, γ-duals of the spaces p ( F ˆ ) and ( F ˆ ). In Section 5, we characterize the classes ( p ( F ˆ ),X) and ( ( F ˆ ),X), where 1p< and X is any of the spaces , 1 , c and c 0 . In the final section of the paper, we investigate some geometric properties of the space p ( F ˆ ) for 1<p<.

2 The Fibonacci difference sequence space p ( F ˆ )

Define the sequence { f n } n = 0 of Fibonacci numbers given by the linear recurrence relations

f 0 = f 1 =1and f n = f n 1 + f n 2 ,n2.

Fibonacci numbers have many interesting properties and applications in arts, sciences and architecture. For example, the ratio sequences of Fibonacci numbers converges to the golden ratio which is important in sciences and arts. Also, some basic properties of Fibonacci numbers [35] are given as follows:

Substituting for f n + 1 in Cassini’s formula yields f n 1 2 + f n f n 1 f n 2 = ( 1 ) n + 1 .

A sequence space X is called a FK-space if it is a complete linear metric space with continuous coordinates p n :XR (nN), where denotes the real field and p n (x)= x n for all x=( x k )X and every nN. A BK space is a normed FK space, that is, a BK-space is a Banach space with continuous coordinates. The space p (1p<) is a BK-space with x p = ( k = 0 | x k | p ) 1 / p and c 0 , c and are BK-spaces with x = sup k | x k |.

A sequence ( b n ) in a normed space X is called a Schauder basis for X if for every xX, there is a unique sequence ( α n ) of scalars such that x= n α n b n , i.e., lim m x n = 0 m α n b n =0.

The α-, β- and γ-duals of the sequence space X are respectively defined by

and

X γ = { a = ( a k ) ω : a x = ( a k x k ) b s  for all  x = ( x k ) X } ,

where cs and bs are the sequence spaces of all convergent and bounded series, respectively [36].

We assume throughout that p,q1 with p 1 + q 1 =1 and denote the collection of all finite subsets of by .

3 The Fibonacci difference sequence spaces p ( F ˆ ) and ( F ˆ )

In this section, we define the Fibonacci band matrix F ˆ =( f ˆ n k ) and introduce the sequence spaces p ( F ˆ ) and ( F ˆ ), where 1p<. Also, we present some inclusion theorems and construct the Schauder basis of the space p ( F ˆ ) for1p<.

Let f n be the n th Fibonacci number for every nN. Then we define the infinite matrix F ˆ =( f ˆ n k ) by

f ˆ n k ={ f n + 1 f n ( k = n 1 ) , f n f n + 1 ( k = n ) , 0 ( 0 k < n 1  or  k > n ) (n,kN).

Now, we introduce the Fibonacci difference sequence spaces p ( F ˆ ) and ( F ˆ ) as the set of all sequences such that their F ˆ -transforms are in the space p and , respectively, i.e.,

and

( F ˆ )= { x = ( x n ) ω : sup n N | f n f n + 1 x n f n + 1 f n x n 1 | < } .

With the notation of (1.2), the sequence spaces p ( F ˆ ) and ( F ˆ ) may be redefined by

p ( F ˆ )= ( p ) F ˆ (1p<)and ( F ˆ )= ( ) F ˆ .
(3.1)

Define the sequence y=( y n ), which will be frequently used, by the F ˆ -transform of a sequence x=( x n ), i.e.,

y n = F ˆ n (x)={ f 0 f 1 x 0 = x 0 ( n = 0 ) , f n f n + 1 x n f n + 1 f n x n 1 ( n 1 ) (nN).
(3.2)

Now, we may begin with the following theorem which is essential in the text.

Theorem 3.1 Let 1p. Then p ( F ˆ ) is a BK-space with the norm , that is,

x p ( F ˆ ) = ( n | F ˆ n ( x ) | p ) 1 / p (1p<)

and

x ( F ˆ ) = sup n N | F ˆ n (x)|.

Proof Since (3.1) holds, p and are BK-spaces with respect to their natural norms and the matrix F ˆ is a triangle; Theorem 4.3.12 of Wilansky [[37], p.63] gives the fact that the spaces p ( F ˆ ) and ( F ˆ ) are BK-spaces with the given norms, where 1p<. This completes the proof. □

Remark 3.2 One can easily check that the absolute property does not hold on the spaces p ( F ˆ ) and ( F ˆ ), that is, x p ( F ˆ ) | x | p ( F ˆ ) and x ( F ˆ ) | x | ( F ˆ ) for at least one sequence in the spaces p ( F ˆ ) and ( F ˆ ), and this shows that p ( F ˆ ) and ( F ˆ ) are the sequence spaces of non-absolute type, where |x|=(| x k |) and 1p<.

Theorem 3.3 The Fibonacci difference sequence space p ( F ˆ ) of non-absolute type is linearly isomorphic to the space p , that is, p ( F ˆ ) p for 1p.

Proof To prove this, we should show the existence of a linear bijection between the spaces p ( F ˆ ) and p for 1p. Consider the transformation T defined, with the notation of (3.2), from p ( F ˆ ) to p by xy=Tx. Then Tx=y= F ˆ x p for every x p ( F ˆ ). Also, the linearity of T is clear. Further, it is trivial that x=0 whenever Tx=0 and hence T is injective.

Furthermore, let y=( y k ) p for 1p and define the sequence x=( x k ) by

x k = j = 0 k f k + 1 2 f j f j + 1 y j (kN).
(3.3)

Then, in the cases 1p< and p=, we get

and

x ( F ˆ ) = sup k N | F ˆ k (x)|= y <,

respectively. Thus, we have x p ( F ˆ ) (1p). Hence, T is surjective and norm preserving. Consequently, T is a linear bijection which shows that the spaces p ( F ˆ ) and p are linearly isomorphic for 1p. This concludes the proof. □

Now, we give some inclusion relations concerning the space p ( F ˆ ).

Theorem 3.4 The inclusion p p ( F ˆ ) strictly holds for 1p.

Proof To prove the validity of the inclusion p p ( F ˆ ) for 1p, it suffices to show the existence of a number M>0 such that x p ( F ˆ ) M x p for every x p .

Let x p and 1<p. Since the inequalities f k f k + 1 1 and f k + 1 f k 2 hold for every kN, we obtain with the notation of (3.2),

k | F ˆ k (x) | p k 2 p 1 ( | x k | p + | 2 x k 1 | p ) 2 2 p 1 ( k | x k | p + k | x k 1 | p )

and

sup k N | F ˆ k (x)|3 sup k N | x k |,

which together yield, as expected,

(3.4)

for 1<p. Further, since the sequence x=( x k )=( f k + 1 2 )=(1, 2 2 , 3 2 , 5 2 ,) is in p ( F ˆ ) p , the inclusion p p ( F ˆ ) is strict for 1<p. Similarly, one can easily prove that inequality (3.4) also holds in the case p=1, and so we omit the details. This completes the proof. □

Theorem 3.5 Neither of the spaces b v p and p ( F ˆ ) includes the other one, where 1p<.

Proof Let e=(1,1,1,) and x=( x k )=( f k + 1 2 ). Then, since F ˆ x=(1,0,0,) p and Δx=(1, f 0 f 3 , f 1 f 4 ,, f k 1 f k + 2 ,) p , we conclude that x is in p ( F ˆ ) but not in b v p . Now, consider the equation

| f k f k + 1 f k + 1 f k |= | f k 2 f k + 1 2 | f k f k + 1 = | ( 1 ) k f k f k + 1 | f k f k + 1 (kN).

Then | ( 1 ) k f k f k + 1 |> f k f k + 1 whenever k is odd, which implies that the series k | f k f k + 1 f k + 1 f k | p is not convergent, where 1p<. Thus, F ˆ e=( f k f k + 1 f k + 1 f k ) is not in p for 1p<. Additionally, since Δe=(1,0,0,), the sequence e is in p . Hence, the sequence spaces p ( F ˆ ) and b v p overlap but neither contains the other, as asserted. □

Theorem 3.6 If 1p<s, then p ( F ˆ ) s ( F ˆ ).

Proof Let 1p<s and x p ( F ˆ ). Then we obtain from Theorem 3.1 that y p , where y is the sequence given by (3.2). Thus, the well-known inclusion p s yields y s . This means that x s ( F ˆ ) and hence, the inclusion p ( F ˆ ) s ( F ˆ ) holds. This completes the proof. □

Now, we give a sequence of the points of the space p ( F ˆ ) which forms a basis for the space p ( F ˆ ) (1p<).

Theorem 3.7 Let 1p< and define the sequence c ( k ) p ( F ˆ ) for every fixed kN by

( c ( k ) ) n ={ 0 ( n < k ) , f n + 1 2 f k f k + 1 ( n k ) (nN).
(3.5)

Then the sequence ( c ( k ) ) k = 0 is a basis for the space p ( F ˆ ), and every x p ( F ˆ ) has a unique representation of the form

x= k F ˆ k (x) c ( k ) .
(3.6)

Proof Let 1p<. Then it is obvious by (3.5) that F ˆ ( c ( k ) )= e ( k ) p (kN) and hence c ( k ) p ( F ˆ ) for all kN.

Further, let x p ( F ˆ ) be given. For every non-negative integer m, we put

x ( m ) = k = 0 m F ˆ k (x) c ( k ) .

Then we have that

F ˆ ( x ( m ) ) = k = 0 m F ˆ k (x) F ˆ ( c ( k ) ) = k = 0 m F ˆ k (x) e ( k )

and hence

F ˆ n ( x x ( m ) ) ={ 0 ( 0 n m ) , F ˆ n ( x ) ( n > m ) (n,mN).

Now, for any given ε>0, there is a non-negative integer m 0 such that

n = m 0 + 1 | F ˆ n (x) | p ( ε 2 ) p .

Therefore, we have for every m m 0 that

x x ( m ) p ( F ˆ ) = ( n = m + 1 | F ˆ n ( x ) | p ) 1 / p ( n = m 0 + 1 | F ˆ n ( x ) | p ) 1 / p ε 2 <ε,

which shows that lim m x x ( m ) p ( F ˆ ) =0 and hence x is represented as in (3.6).

Finally, let us show the uniqueness of the representation (3.6) of x p ( F ˆ ). For this, suppose that x= k μ k (x) c ( k ) . Since the linear transformation T defined from p ( F ˆ ) to p in the proof of Theorem 3.3 is continuous, we have

F ˆ n (x)= k μ k (x) F ˆ n ( c ( k ) ) = k μ k (x) δ n k = μ n (x)(nN).

Hence, the representation (3.6) of x p ( F ˆ ) is unique. This concludes the proof. □

4 The α-, β- and γ-duals of the space p ( F ˆ )

In this section, we determine the α-, β- and γ-duals of the sequence space p ( F ˆ ) of non-absolute type. Since the case p=1 can be proved by analogy, we omit the proof of that case and consider only the case 1<p in the proof of Theorems 4.5 and 4.6, respectively.

The following known results [38] are fundamental for our investigation.

Lemma 4.1 A=( a n k )( p , 1 ) if and only if

sup K F k | n K a n k |<,1<p.

Lemma 4.2 A=( a n k )( p ,c) if and only if

(4.1)
(4.2)

Lemma 4.3 A=( a n k )( ,c) if and only if (4.1) holds and

lim n k | a n k |= k | lim n a n k |.
(4.3)

Lemma 4.4 A=( a n k )( p , ) if and only if (4.2) holds with 1<p.

Theorem 4.5 The α-dual of the space p ( F ˆ ) is the set

d ˆ 1 = { a = ( a k ) ω : sup K F k | n K f n + 1 2 f k f k + 1 a n | q < } ,

where 1<p.

Proof Let 1<p. For any fixed sequence a=( a n )ω, we define the matrix B=( b n k ) by

b n k ={ f n + 1 2 f k f k + 1 a n ( 0 k n ) , 0 ( k > n )

for all n,kN. Also, for every x=( x n )ω, we put y= F ˆ x. Then it follows by (3.2) that

a n x n = k = 0 n f n + 1 2 f k f k + 1 a n y k = B n (y)(nN).
(4.4)

Thus, we observe by (4.4) that ax=( a n x n ) 1 whenever x p ( F ˆ ) if and only if By 1 whenever y p . Therefore, we derive by using Lemma 4.1 that

sup K F k | n K f n + 1 2 f k f k + 1 a n | q <,

which implies that ( p ( F ˆ ) ) α = d ˆ 1 . □

Theorem 4.6 Define the sets d ˆ 2 , d ˆ 3 and d ˆ 4 by

and

d ˆ 4 = { a = ( a k ) ω : lim n k = 0 n | j = k n f j + 1 2 f k f k + 1 a j | = k | j = k f j + 1 2 f k f k + 1 a j | < } .

Then ( p ( F ˆ ) ) β = d ˆ 2 d ˆ 3 and ( ( F ˆ ) ) β = d ˆ 2 d ˆ 4 , where 1<p<.

Proof Let a=( a k )ω and consider the equality

k = 0 n a k x k = k = 0 n a k ( j = 0 n f k + 1 2 f j f j + 1 y j ) = k = 0 n ( j = k n f j + 1 2 f k f k + 1 a j ) y k = D n (y),
(4.5)

where D=( d n k ) is defined by

d n k ={ n j = k f j + 1 2 f k f k + 1 a j ( 0 k n ) , 0 ( k > n ) , n,kN.

Then we deduce from Lemma 4.2 with (3.2) that ax=( a k x k )cs whenever x=( x k ) p ( F ˆ ) if and only if Dyc whenever y=( y k ) p . Thus, ( a k ) ( p ( F ˆ ) ) β if and only if ( a k ) d ˆ 2 and ( a k ) d ˆ 3 by (4.1) and (4.2), respectively. Hence, ( p ( F ˆ ) ) β = d ˆ 2 d ˆ 3 . It is clear that one can also prove the case p= by the technique used in the proof of the case 1<p< with Lemma 4.3 instead of Lemma 4.2. So, we leave the detailed proof to the reader. □

Theorem 4.7 ( p ( F ˆ ) ) γ = d ˆ 3 , where 1<p.

Proof This result can be obtained from Lemma 4.4 by using (4.5). □

5 Some matrix transformations related to the sequence space p ( F ˆ )

In this section, we characterize the classes ( p ( F ˆ ),X), where 1p and X is any of the spaces , 1 , c and c 0 .

For simplicity in notation, we write

a ˜ n k = j = k f j + 1 2 f k f k + 1 a n j

for all k,nN.

The following lemma is essential for our results.

Lemma 5.1 (see [[21], Theorem 4.1])

Let λ be an FK-space, U be a triangle, V be its inverse and μ be an arbitrary subset of ω. Then we have A=( a n k )( λ U ,μ) if and only if

C ( n ) = ( c m k ( n ) ) (λ,c) for all nN

and

C=( c n k )(λ,μ),

where

c m k ( n ) ={ j = k m a n j v j k ( 0 k m ) , 0 ( k > m )

and c n k = j = k a n j v j k for all k,m,nN.

Now, we list the following conditions:

(5.1)
(5.2)
(5.3)
(5.4)
(5.5)
(5.6)
(5.7)
(5.8)
(5.9)
(5.10)
(5.11)
(5.12)

Then, by combining Lemma 5.1 with the results in [38], we immediately derive the following results.

Theorem 5.2

  1. (a)

    A=( a n k )( 1 ( F ˆ ), ) if and only (5.2), (5.9) and (5.10) hold.

  2. (b)

    A=( a n k )( 1 ( F ˆ ),c) if and only if (5.2), (5.6), (5.9) and (5.10) hold.

  3. (c)

    A=( a n k )( 1 ( F ˆ ), c 0 ) if and only if (5.2), (5.6) with α ˜ k =0, (5.9) and (5.10) hold.

  4. (d)

    A=( a n k )( 1 ( F ˆ ), 1 ) if and only (5.2), (5.10) and (5.11) hold.

Theorem 5.3 Let 1<p<. Then we have

  1. (a)

    A=( a n k )( p ( F ˆ ), ) if and only if (5.1), (5.2) and (5.4) hold.

  2. (b)

    A=( a n k )( p ( F ˆ ),c) if and only if (5.1), (5.2), (5.4) and (5.6) hold.

  3. (c)

    A=( a n k )( p ( F ˆ ), c 0 ) if and only if (5.1), (5.2), (5.4) and (5.6) with α ˜ k =0 hold.

  4. (d)

    A=( a n k )( p ( F ˆ ), 1 ) if and only if (5.1), (5.2) and (5.5) hold.

Theorem 5.4

  1. (a)

    A=( a n k )( ( F ˆ ), ) if and only (5.2), (5.3) and (5.4) with q=1 hold.

  2. (b)

    A=( a n k )( ( F ˆ ),c) if and only (5.2), (5.3), (5.6) and (5.7) hold.

  3. (c)

    A=( a n k )( ( F ˆ ), c 0 ) if and only (5.2), (5.3) and (5.8) hold.

  4. (d)

    A=( a n k )( ( F ˆ ), 1 ) if and only (5.2), (5.3) and (5.12) hold.

6 Some geometric properties of the space p ( F ˆ ) (1<p<)

In this section, we study some geometric properties of the space p ( F ˆ ) for 1<p<.

For these properties, we refer to [3, 3947].

A Banach space X is said to have the Banach-Saks property if every bounded sequence ( x n ) in X admits a subsequence ( z n ) such that the sequence { t k (z)} is convergent in the norm in X [40], where

t k (z)= 1 k + 1 ( z 0 + z 1 ++ z k )(kN).
(6.1)

A Banach space X is said to have the weak Banach-Saks property whenever, given any weakly null sequence ( x n )X, there exists a subsequence ( z n ) of ( x n ) such that the sequence { t k (z)} is strongly convergent to zero.

In [43], García-Falset introduces the following coefficient:

R(X)=sup { lim inf n x n x : ( x n ) B ( X ) , x n w 0 , x B ( X ) } ,
(6.2)

where B(X) denotes the unit ball of X.

Remark 6.1 A Banach space X with R(X)<2 has the weak fixed point property [44].

Let 1<p<. A Banach space is said to have the Banach-Saks type p or the property ( B S ) p if every weakly null sequence ( x k ) has a subsequence ( x k l ) such that for some C>0,

l = 0 n x k l <C ( n + 1 ) 1 / p
(6.3)

for all nN ( see [45]).

Now, we may give the following results related to some geometric properties, mentioned above, of the space p ( F ˆ ), where 1<p<.

Theorem 6.2 Let 1<p<. Then the space p ( F ˆ ) has the Banach-Saks type p.

Proof Let ( ε n ) be a sequence of positive numbers for which ε n 1/2, and also let ( x n ) be a weakly null sequence in B( p ( F ˆ )). Set z 0 = x 0 =0 and z 1 = x n 1 = x 1 . Then there exists m 1 N such that

i = m 1 + 1 z 1 ( i ) e ( i ) p ( F ˆ ) < ε 1 .
(6.4)

Since ( x n ) being a weakly null sequence implies x n 0 coordinatewise, there is an n 2 N such that

i = 0 m 1 x n ( i ) e ( i ) p ( F ˆ ) < ε 1 ,

when n n 2 . Set z 2 = x n 2 . Then there exists an m 2 > m 1 such that

i = m 2 + 1 z 2 ( i ) e ( i ) p ( F ˆ ) < ε 2 .

Again using the fact that x n 0 coordinatewise, there exists an n 3 n 2 such that

i = 0 m 2 x n ( i ) e ( i ) p ( F ˆ ) < ε 2 ,

when n n 3 .

If we continue this process, we can find two increasing subsequences ( m i ) and ( n i ) such that

i = 0 m j x n ( i ) e ( i ) p ( F ˆ ) < ε j

for each n n j + 1 and

i = m j + 1 z j ( i ) e ( i ) p ( F ˆ ) < ε j ,

where b j = x n j . Hence,

j = 0 n z j p ( F ˆ ) = j = 0 n ( i = 0 m j 1 z j ( i ) e ( i ) + i = m j 1 + 1 m j z j ( i ) e ( i ) + i = m j + 1 z j ( i ) e ( i ) ) p ( F ˆ ) j = 0 n ( i = m j 1 + 1 m j z j ( i ) e ( i ) ) p ( F ˆ ) + 2 j = 0 n ε j .

On the other hand, it can be seen that x p ( F ˆ ) <1. Therefore, we have that

j = 0 n ( i = m j 1 + 1 m j z j ( i ) e ( i ) ) p ( F ˆ ) p = j = 0 n i = m j 1 + 1 m j | f i f i + 1 z j ( i ) f i + 1 f i z j ( i 1 ) | p j = 0 n i = 0 | f i f i + 1 z j ( i ) f i + 1 f i z j ( i 1 ) | p ( n + 1 ) .

Hence, we obtain

j = 0 n ( i = m j 1 + 1 m j z j ( i ) e ( i ) ) ( n + 1 ) 1 / p .

By using the fact that 1 ( n + 1 ) 1 / p for all nN and 1<p<, we have

j = 0 n z j p ( F ˆ ) ( n + 1 ) 1 / p +12 ( n + 1 ) 1 / p .

Hence, p ( F ˆ ) has the Banach-Saks type p. This concludes the proof. □

Remark 6.3 Note that R( p ( F ˆ ))=R( p )= 2 1 / p since p ( F ˆ ) is linearly isomorphic to p .

Hence, by Remarks 6.1 and 6.3, we have the following theorem.

Theorem 6.4 The space p ( F ˆ ) has the weak fixed point property, where 1<p<.

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Acknowledgements

The authors wish to thank the referee for his/her valuable suggestions, which improved the paper considerably.

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Correspondence to Emrah Evren Kara.

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Keywords

  • sequence spaces
  • Fibonacci numbers
  • difference matrix
  • α-, β-, γ-duals
  • matrix transformations
  • fixed point property
  • Banach-Saks type p