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Some geometric properties of N(λ,p)-spaces

Abstract

In this paper, we introduce the sequence spaces N(λ,p) and we show some geometric properties of that spaces. The main purpose of this paper is to show that N(λ,p) is a Banach space and has the rotund property, the Kadec-Klee property, the uniform Opial property, the (β)-property, the k-NUC property and the Banach-Saks property of type p.

MSC:46A45, 40C05, 46B45, 46A35.

1 Introduction

By ω, we denote the space of all real valued sequences. Any vector subspace of ω is called a sequence space. We write l , c, and c 0 for the spaces of all bounded, convergent and null sequences, respectively. Also by bs, cs, l 1 , and l , we denote the spaces of all bounded, convergent, absolutely convergent and p-absolutely convergent series, respectively; where 1<p<. Assume here and after that ( p k ) be a bounded sequence of strictly positive real numbers with sup{ p k }=H and M=max{1,H}. Then, the linear space l(p) was defined by Maddox [1] (see also Simons [2] and Nakano [3]) as follows:

l(p)= { x = ( x k ) ω : k | x k | p k < } (0< p k H<)

which is complete paranormed space paranormed by

g(x)= ( k | x k | p k ) 1 M .

For simplicity in notation, here and in what follows, the summation without limits runs from 1 to ∞.

In [4] was introduced the following numerical sequence λ= ( λ k ) k = 0 , which is a strictly increasing sequence of positive real numbers tending to infinity, as k, that is

0< λ 0 < λ 1 <and λ k as k.

We will introduce the following sequence space:

N(λ,p)= { x = ( x n ) ω : k = 1 ( 1 λ k i = 1 k | ( λ i λ i 1 ) x i | ) p k < } .

For λ k =k, we obtain the Cesaro sequence space ces(p) (see [5]). If λ k =k and p k =p, then N(λ,p)= ces p (see [6]). In case where p k =p for all kN, then we will denote N(λ,p)= N p . Some results related to the geometric properties of sequence spaces are given in [79].

2 Topological properties

Theorem 2.1 The paranorm on N(λ,p) is given by the relation

h(x)= ( k = 1 ( 1 λ k i = 1 k | ( λ i λ i 1 ) x i | ) p k ) 1 M ,

where M=max{1,H} and H=sup p k .

3 Geometrical properties

In this section we will show some geometric properties of the N(λ,p)-spaces, such as the (β)-property, the k-NUC property, the Banach-Saks property of type p, and the (H)-property. It is well known that these properties are most important in Banach spaces (see [10, 11] and [1]).

Definition 3.1 A Banach space X is said to be k-nearly uniformly convex (k-NUC) if for any ϵ>0, there exists a δ>0 such that for any sequence ( x n )B(X) with sep( x n )ϵ, there are n 1 , n 2 ,, n k N such that

x n 1 + x n 2 + + x n k k <1δ,

where sep( x n )=inf{ x n x m :nm}>ϵ.

Definition 3.2 A Banach space X has property (β) if and only if for each r>0 and ϵ>0, there exists a δ>0 such that for each element xB(X) and each sequence x n B(X) with sep( x n )ϵ, there is an index k for each

x + x k 2 δ.

Definition 3.3 A Banach space X is said to have the Banach-Saks property type 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

for all nN.

Definition 3.4 Let X be a real vector space. A functional σ:X[0,) is called a modular if

  1. (1)

    σ(x)=0 if and only if x=θ,

  2. (2)

    σ(αx)=σ(x) for all scalars α with |α|=1,

  3. (3)

    σ(αx+βy)σ(x)+σ(y) for all x,yX and α,β>0 with α+β=1,

  4. (4)

    the modular σ is called convex if σ(αx+βy)ασ(x)+βσ(y) for all x,yX and α,β>0, with α+β=1.

A modular σ is called:

  1. (5)

    right continuous if lim α 1 + σ(αx)=σ(x) for all x X σ ,

  2. (6)

    left continuous if lim α 1 σ(αx)=σ(x) for all x X σ ,

  3. (7)

    continuous if it is both right and left continuous,

where X σ ={xX: lim α 0 + σ(αx)=0}. We define σ p on N(λ,p) as follows:

σ p (x)= ( k = 1 ( 1 λ k i = 0 k | ( λ i λ i 1 ) x i | ) p k ) ,

where λ 1 =0.

If p k 1, for all kN, by the convexity of the function t | t | p k , for all kN, σ p defined above is a modular convex in the N(λ,p).

Definition 3.5 A modular σ p is said to satisfy the δ 2 -conditions if for every ϵ>0, there exist constant M>0 and m>0 such that

σ p (2t)M σ p (t)+ϵ
(3.1)

for all t X σ p with σ p (t)m.

Lemma 3.6 ([12])

If σ p satisfies the δ 2 -conditions, then for any A>0 and ϵ>0, there exists δ>0 such that

| σ p (t+w) σ p (t)|<ϵ
(3.2)

whenever t,w X σ p with σ p (t)A and σ p (w)δ.

Theorem 3.7 ([12])

  1. (1)

    If σ p satisfies the δ 2 -conditions, then for any x X σ p , x=1 if and only if σ p (x)=1.

  2. (2)

    If σ p satisfies the δ 2 -conditions, then for any sequence ( x n ) X σ p , x n 0 if and only if σ p ( x n )0.

Theorem 3.8 If σ p satisfies the δ 2 -conditions, then for any ϵ(0,1), there exists δ(0,1) such that σ p (x)1ϵ implies x1δ.

Proof The proof of the theorem follows directly from the above two facts. □

Theorem 3.9 For any xN(λ,p) and ϵ(0,1), there exists δ(0,1), such that σ p (x)1ϵ implies x1δ.

Proof The proof of the theorem follows directly from Theorem 3.8. □

Proposition 3.10 If p k 1, for all kN, then the modular function σ p , on N(λ,p), satisfies the following conditions:

  1. (1)

    If 0<α1, then α M σ p ( x α ) σ p (x) and σ p (αx)α σ p (x).

  2. (2)

    If α1, then σ p (x) α M σ p ( x α ).

  3. (3)

    If α1, then σ p (x)α σ p ( x α ).

  4. (4)

    The modular function σ p (x) is continuous on N(λ,p).

Proof The proof of the proposition is similar to Proposition 2.1 in [13]. □

Now we will define the following two norms (the first is known as the Luxemburg norm and the second as the Amemiya norm) in N(λ,p):

x L =inf { α > 0 : σ p ( x α ) 1 }
(3.3)

and

x A = inf α > 0 1 α { 1 + σ p ( α x ) } .
(3.4)

Proposition 3.11 Let xN(λ,p). Then the following relations are satisfied:

  1. (1)

    If x L <1, then σ p (x) x L .

  2. (2)

    If x L >1, then σ p (x) x L .

  3. (3)

    x L =1 if and only if σ p (x)=1.

  4. (4)

    x L <1 if and only if σ p (x)<1.

  5. (5)

    x L >1 if and only if σ p (x)>1.

Proof (1) Let xN(λ,p) and x L <1. Let also ϵ>0 such that 0<ϵ<1 x L . On the other hand from the definition of the norm by relation (3.3) we find that there exists a α>0 such that x L +ϵ>α and σ p ( x α )1. From the above relations and property (1) of Proposition 3.10, we obtain

x L + ϵ α >1

and

σ p (x) x L + ϵ α σ p (x)= x L + ϵ α σ p ( α x α ) x L + ϵ α α σ p ( x α ) x L +ϵ.

The previous statement is valid for every ϵ>0, from which it follows that σ p (x) x L .

  1. (2)

    In this case we will choose ϵ>0 such that 0<ϵ<1 1 x L , and we obtain 1<(1ϵ) x L < x L . Now using into consideration definition of the norm (3.3) and relation (1) of Proposition 3.10, we get

    1< σ p ( x ( 1 ϵ ) x L ) 1 ( 1 ϵ ) x L σ p (x)(1ϵ) x L σ p (x)

for every ϵ(0,1 1 x L ). Finally we have proved that x L σ p (x).

  1. (3)

    Since σ p (x) is continuous function (see [14]), this property follows immediately.

  2. (4)

    Follows from properties (1) and (3).

  3. (5)

    Follows from properties (2) and (3). □

Theorem 3.12 N(λ,p) is a Banach space under the Luxemburg and Amemiya norms.

Proof We will prove that N(λ,p) is a Banach space under the Luxemburg norm. In a similar way we can prove that N(λ,p) is a Banach space under the Amemiya norm. In what follows we need to show that every Cauchy sequence in N(λ,p) is convergent according to the Luxemburg norm. Let { x k n } be any Cauchy sequence in N(λ,p) and ϵ(0,1). Thus there exists a natural number n 0 , such that for any n,m n 0 we get x ( n ) x ( m ) L <ϵ. From Proposition 3.11 we get

σ p ( x ( n ) x ( m ) ) x ( n ) x ( m ) L <ϵ
(3.5)

for all n,m n 0 . This implies that

k = 1 ( 1 λ k i = 1 k | ( λ i λ i 1 ) ( x i ( n ) x i ( m ) ) | ) p k <ϵ.
(3.6)

For each fixed k and for all n,m n 0 ,

1 λ k i = 1 k |( λ i λ i 1 ) ( x i ( n ) x i ( m ) ) |<ϵ.

Hence ( y k ( n ) ) k = ( 1 λ k i = 1 k | ( λ i λ i 1 ) x i ( n ) | ) k is a Cauchy sequence in . Since is a complete normed space, there exists a ( y k ) k = ( 1 λ k i = 1 k | ( λ i λ i 1 ) x i | ) k R such that ( y k ( n ) ) y k as n. Therefore, as n, by relation (3.6) we have

k = 1 ( 1 λ k i = 1 k | ( λ i λ i 1 ) ( x i x i ( m ) ) | ) p k <ϵ

for all m n 0 . In the sequel we will show that ( y k ) is a sequence form N(λ,p). From Proposition 3.10 and relation (3.5) we have

lim n σ p ( x ( n ) x ( m ) ) = σ p ( x x ( m ) ) x x ( m ) L <ϵ

for all m n 0 . This implies that ( x ( n ) )x as m. So we have x= x ( n ) ( x ( n ) x)N(λ,p). This proves that N(λ,p) is a complete normed space under the Luxemburg norm. □

In what follows we will show results related to the Luxemburg norm, and for this reason we will denote it just .

Theorem 3.13 The space N(λ,p) is rotund if and only if p k >1 for all kN.

Proof Let N(λ,p) be rotund and choose kN such that p k =1. Consider the two sequences given by

x= ( 0 , 0 , , 0 , λ k 2 k | λ k λ k 1 | , 0 , 0 , )

and

y= ( 0 , 0 , , 0 , 2 λ k 3 k | λ k λ k 1 | , 0 , 0 , ) .

Then obviously xy and

σ p (x)= σ p (y)= σ p ( x + y 2 ) =1.

Then from Proposition 3.11, property (3), it follows that x,y, x + y 2 S[N(λ,p)], which leads to the contradiction that the sequence space N(λ,p) is not rotund. Hence p k >1, for all kN.

Conversely, let xS[N(λ,p)] and y,zS[N(λ,p)] such that x= y + z 2 . By the convexity of σ p and property (3) from Proposition 3.11, we have

1= σ p (x) σ p ( y ) + σ p ( z ) 2 1 2 + 1 2 =1,

which gives σ p (y)= σ p (z)=1 and

σ p (x)= σ p ( y ) + σ p ( z ) 2 .
(3.7)

From the previous relation we obtain

k = 1 ( 1 λ k i = 1 k | ( λ i λ i 1 ) x i | ) p k = 1 2 ( k = 1 ( 1 λ k i = 1 k | ( λ i λ i 1 ) y i | ) p k + k = 1 ( 1 λ k i = 1 k | ( λ i λ i 1 ) z i | ) p k ) .

Since x= y + z 2 , we get

k = 1 ( 1 λ k i = 1 k | ( λ i λ i 1 ) ( y i + z i ) | ) p k = 1 2 ( k = 1 ( 1 λ k i = 1 k | ( λ i λ i 1 ) y i | ) p k + k = 1 ( 1 λ k i = 1 k | ( λ i λ i 1 ) z i | ) p k ) .

This implies that

( 1 λ k i = 1 k | ( λ i λ i 1 ) ( y i + z i ) | ) p k = 1 2 ( ( 1 λ k i = 1 k | ( λ i λ i 1 ) y i | ) p k + ( 1 λ k i = 1 k | ( λ i λ i 1 ) z i | ) p k ) .
(3.8)

From the previous relation we get y i = z i for all iN, respectively, z=y. That is, the sequence space N(λ,p) is rotund. □

In what follows we will give two facts without proof because their proofs follow directly from Proposition 3.10 and Proposition 3.11.

Theorem 3.14 Let xN(λ,p). Then the following statements hold:

  1. (i)

    For 0<α<1 and x>α we get σ p (x)> α M .

  2. (ii)

    If α1 and x<α, then we have σ p (x)< α M .

Theorem 3.15 Let ( x n ) be a sequence in N(λ,p). Then the following statements hold:

  1. (i)

    lim n x n =1 implies lim n σ p ( x n )=1.

  2. (ii)

    lim n σ p ( x n )=0 implies lim n x n =0.

Theorem 3.16 Let xN(λ,p) and ( x ( n ) )N(λ,p). If σ p ( x ( n ) ) σ p (x) as n and x k ( n ) x k as n for all kN, then x ( n ) x as n.

Proof The proof of the theorem is similar to Theorem 2.9 in [13]. □

Theorem 3.17 The sequence space N(λ,p) has the Kadec-Klee property.

Proof It is enough to prove that every weakly convergent sequence on the unit sphere is convergent in norm. Let xN(λ,p) and ( x ( n ) )N(λ,p) such that x ( n ) 1 and x ( n ) w x be given. From the properties of Theorem 3.15 it follows that σ p ( x ( n ) )1 as n. On the other hand, from Proposition 3.11, we get σ p (x)=1. Therefore we have σ p ( x ( n ) ) σ p (x), as n. Since x ( n ) w x and p k (x)= x k is a continuous functional, x k ( n ) x k as n and for kN. Now the proof of the theorem follows from Theorem 3.16. □

Theorem 3.18 For any 1<p<, the space N p has the uniform Opial property.

We omit this proof.

To prove the following theorem we will use the same technique given in [15] and will consider that lim n inf p n >1.

Theorem 3.19 The Banach space N(λ,p) has the k-NUC property for every k2.

Proof Let ϵ>0 and ( x n )B(N(λ,p)) with sep( x n )ϵ. For each mN, let

x n m = ( 0 , 0 , , 0 m 1 , x n ( m ) , x ( m + 1 ) , ) .
(3.9)

Since for each iN, ( x n ( i ) ) n = 1 is bounded, by the diagonal method (see [16]), we find that for each mN, we can find a subsequence ( x n j ) of ( x n ) such that ( x n j (i)) converges for each iN, 1im. Therefore, there exists an increasing sequence of positive integers ( t m ) such that sep( ( x n j m ) j > t m )ϵ. Hence, there is a sequence of positive integers ( r m ) m N with r 1 < r 2 < r 3 < such that x r m m ϵ 2 for all mN. Then by Theorem 3.15, we may assume that there exists η>0 such that

σ p ( x r m m ) ηfor all mN.
(3.10)

Let α>0 be such that 1<α< lim inf n p n . For fixed integer k2, let ϵ 1 =( ( k α 1 1 ) ( k 1 ) k α ) η 2 . Then by Lemma 3.6, there is a δ>0 such that

| σ p (u+v) σ p (u)| ϵ 1 ,
(3.11)

whenever σ p (u)1 and σ p (v)δ. Since by Proposition 3.11, property (1), we get σ p ( x n )1, nN. Then there exist positive integers m i (i=1,2,,k1) with m 1 < m 2 << m k 1 such that σ p ( x p i m i )δ and α p j for all j m k 1 . Define m k = m k 1 +1. By (3.10), we have σ p ( x r m k m k )η. Let s i =i for 1ik1 and s k = r m k . From relations (3.10), (3.11), and the convexity of the function f i (u)= | u | p i (iN), we have

σ p ( x s 1 + x s 2 + + x s k k ) = n = 1 ( 1 λ n i = 1 n | ( λ i λ i 1 ) x s 1 ( i ) + x s 2 ( i ) + + x s k ( i ) k | ) p n = n = 1 m 1 ( 1 λ n i = 1 n | ( λ i λ i 1 ) x s 1 ( i ) + x s 2 ( i ) + + x s k ( i ) k | ) p n + n = m 1 + 1 ( 1 λ n i = 1 n | ( λ i λ i 1 ) x s 1 ( i ) + x s 2 ( i ) + + x s k ( i ) k | ) p n ;

from (3.11) we get

n = 1 m 1 ( 1 λ n i = 1 n | ( λ i λ i 1 ) x s 1 ( i ) + x s 2 ( i ) + + x s k ( i ) k | ) p n + n = m 1 + 1 ( 1 λ n i = 1 n | ( λ i λ i 1 ) x s 2 ( i ) + + x s k ( i ) k | ) p n + ϵ 1

from the convexity of f i (u)= | u | p i (iN), it follows that

n = 1 m 1 1 k j = 1 k ( 1 λ k i = 1 n | ( λ i λ i 1 ) x s j ( i ) | ) p n + n = m 1 + 1 m 2 ( 1 λ n i = 1 n | ( λ i λ i 1 ) x s 2 ( i ) + x s 3 ( i ) + + x s k ( i ) k | ) p n + n = m 2 + 1 ( 1 λ n i = 1 n | ( λ i λ i 1 ) x s 2 ( i ) + x s 3 ( i ) + + x s k ( i ) k | ) p n + ϵ 1 n = 1 m 1 1 k j = 1 k ( 1 λ k i = 1 n | ( λ i λ i 1 ) x s j ( i ) | ) p n + n = m 1 + 1 m 2 ( 1 λ n i = 1 n | ( λ i λ i 1 ) x s 2 ( i ) + x s 3 ( i ) + + x s k ( i ) k | ) p n + n = m 2 + 1 ( 1 λ n i = 1 n | ( λ i λ i 1 ) x s 3 ( i ) + x s 4 ( i ) + + x s k ( i ) k | ) p n + 2 ϵ 1 n = 1 m 1 1 k j = 1 k ( 1 λ k i = 1 n | ( λ i λ i 1 ) x s j ( i ) | ) p n + n = m 1 + 1 m 2 1 k j = 2 k ( 1 λ k i = 1 n | ( λ i λ i 1 ) x s j ( i ) | ) p n + n = m 2 + 1 m 3 1 k j = 3 k ( 1 λ k i = 1 n | ( λ i λ i 1 ) x s j ( i ) | ) p n + n = m 3 + 1 m 4 1 k j = 4 k ( 1 λ k i = 1 n | ( λ i λ i 1 ) x s j ( i ) | ) p n + + n = m k 1 + 1 m k 1 k j = k 1 k ( 1 λ k i = 1 n | ( λ i λ i 1 ) x s j ( i ) | ) p n + n = m k + 1 ( 1 λ k i = 1 n | ( λ i λ i 1 ) x s k ( i ) k | ) p n + ( k 1 ) ϵ 1 σ p ( x s 1 ) + σ p ( x s 2 ) + + σ p ( x s k 1 ) k + 1 k n = 1 m k ( 1 λ k i = 1 n | ( λ i λ i 1 ) x s k ( i ) | ) p n + n = m k + 1 ( 1 λ k i = 1 n | ( λ i λ i 1 ) x s k ( i ) k | ) p n + ( k 1 ) ϵ 1 k 1 k + 1 k n = 1 m k ( 1 λ k i = 1 n | ( λ i λ i 1 ) x s k ( i ) | ) p n + 1 k α n = m k + 1 ( 1 λ k i = 1 n | ( λ i λ i 1 ) x s k ( i ) | ) p n + ( k 1 ) ϵ 1 1 1 k + 1 k [ 1 n = m k + 1 ( 1 λ k i = 1 n | ( λ i λ i 1 ) x s k ( i ) k | ) p n ] + 1 k α n = m k + 1 ( 1 λ k i = 1 n | ( λ i λ i 1 ) x s k ( i ) k | ) p n + ( k 1 ) ϵ 1 1 + ( k 1 ) ϵ 1 ( k α 1 1 k α ) n = m k + 1 ( 1 λ k i = 1 n | ( λ i λ i 1 ) x s k ( i ) | ) p n 1 + ( k 1 ) ϵ 1 ( k α 1 1 k α ) η = 1 ( k α 1 1 k α ) η 2 .

Now from Theorem 3.9, there exists a ϑ>0 such that

x s 1 + x s 2 + + x s k k <1ϑ.

 □

The proof of the following results we omit.

Theorem 3.20 The Banach space N(λ,p) has the (β)-property.

Theorem 3.21 The Banach space N(λ,p) has the Banach-Saks property of type p.

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Braha, N.L. Some geometric properties of N(λ,p)-spaces. J Inequal Appl 2014, 112 (2014). https://doi.org/10.1186/1029-242X-2014-112

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Keywords

  • normed sequence spaces
  • rotund property
  • Kadec-Klee property
  • uniform Opial property
  • (β)-property
  • k-NUC property
  • Banach-Saks property of type p