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On H-property and uniform Opial property of generalized cesàro sequence spaces

Journal of Inequalities and Applications20122012:76

https://doi.org/10.1186/1029-242X-2012-76

Received: 29 October 2011

Accepted: 30 March 2012

Published: 30 March 2012

Abstract

In this article, we define the generalized cesàro sequence spaces ces(p)(q) and consider it equipped with the Luxemburg norm. We show that the spaces ces(p)(q) has the H-property and Uniform Opial property. The results of this article, we improve and extend some results of Petrot and Suantai.

Keywords

generalized Cesàro sequence spaces H-property uniform Opial property

1. Introduction

Let (X, || · ||) be a real Banach space and let B(X) (resp., S(X)) be a closed unit ball (resp., the unit sphere) of X. A point x S(X) is an H-point of B(X) if for any sequence (x n ) in X such that ||x n || 1 as n → ∞, the week convergence of (x n ) to x implies that ||x n - x|| 0 as n → ∞. If every point in S(X) is an H-point of B(X), then X is said to have the property (H). A Banach space X is said to have the Opial property (see [1]), if every weakly null sequence (x n ) in X satisfies
lim n inf | | x n | | lim n inf | | x n - x | | ,
for every x X \{0}. Opial proved in [1] that the sequence space l p (1 < p < ∞) have this property but L p [0, π](p ≠ 2, 1 < p < ∞) do not have it. A Banach space X is said to have the uniform Opial property (see [2]), if for each ε > 0 there exists τ > 0 such that for any weakly null sequence (x n ) in S(X) and x X with || x || > ε there holds
1 + τ lim n inf | | x n + x | | .

For example, the space in [35] have the uniform Opial property.

Let l0 be the space of all real sequences. For 1 ≤ p < ∞, the Cesàro sequence space (ces p , for short) is defined by
ce s p = x l 0 : n = 1 1 n i = 0 n | x ( i ) | p <
equipped with the norm
| | x | | = n = 1 1 n i = 1 n | x ( i ) | p 1 p
(1.1)
This space was first introduced by Shiue [6]. It is useful in the theory of matrix operators and others (see [7, 8]). Suantai [9, 10] defined the generalized Cesàro sequence space ces(p)when p = (p k ) is a bounded sequence of positive real numbers with p k 1 for all k by
ce s ( p ) = { x l 0 : ϱ ( λ x ) < for some  λ > 0 } ,
where
ϱ ( x ) = n = 1 1 n i = 1 k | x ( i ) | p n
equipped with the Luxemburg norm
| | x | | = inf ε > 0 : ϱ x ε 1 .
In the case when p k = p, 1 ≤ p < ∞ for all k , the generalized Cesàro sequence space ces(p)is the Cesàro sequence space ces p and the Luxemburg norm is expressed by the formula (1.1). Khan [11] defined the generalized Cesàro sequence space for 1 ≤ p < ∞ with q = q k is a bounded sequence of positive real numbers by
ce s p ( q ) = x l 0 : k = 1 1 Q k i = 1 k | q i x ( i ) | p 1 / p < ,

where Q k = k = 1 n q k , n . If q k = 1 for all k , then ces p (q) reduces to ces p .

In this article, we define the generalized Cesàro sequence space for a bounded sequence p = (p k ) and q = q k of positive real numbers with p k ≥ 1 and q k ≥ 1 for all k by
ce s ( p ) ( q ) = { x l 0 : ϱ ( λ x ) < for some  λ > 0 } ,
where
ϱ ( x ) = k = 1 1 Q k i = 1 k | q i x ( i ) | p k
with Q k = k = 1 n q k and consider ces(p)(q) equipped with the Luxemburg norm
| | x | | = inf ε > 0 : ϱ x ε 1 .
Thus, we see that p k = p, 1 ≤ p < ∞ for all k , then ces(p)(q) reduces to ces p (q) and if q k = 1 for all k , then ces(p)(q) reduces to ces(p). Throughout this article, for x l0, i , we denote
e i = ( 0 , 0 , , 0 , i - 1 1 , 0 , 0 , 0 , ) , x | i = ( x ( 1 ) , x ( 2 ) , x ( 3 ) , , x ( i ) , 0 , 0 , 0 , ) , x | - i = ( 0 , 0 , 0 , , x ( i + 1 ) , x ( i + 2 ) , ) ,
and M = sup k p k with p k > 1 for all k . First, we start with a brief recollection of basic concepts and facts in modular space. For a real vector space X, a function ρ: X → [0, ] is called a modular if it satisfies the following conditions;
  1. (i)

    ρ(x) = 0 if and only if x = 0;

     
  2. (ii)

    ρ(αx) = ρ(x) for all scalar α with |α| = 1;

     
  3. (iii)

    ρ(αx + βy) ≤ ρ(x) + ρ(y), for all x, y X and all α, β ≥ 0 with α + β = 1.

     
The modular ρ is called convex if
  1. (iv)

    ρ(αx + βy) ≤ αρ(x) + βρ(y), for all x, y X and all α, β ≥ 0 with α + β = 1.

     
For modular ρ on X, the space
X ρ = { x X : ρ ( λ x ) 0 as λ 0 + }

is called the modular space.

A sequence (x n ) in X ρ is called modular convergent to x X ρ if there exists a λ > 0 such that ρ(λ(x n - x)) 0 as n → ∞.

A modular ρ is said to satisfy the Δ2-condition (ρ Δ2) if for any ε > 0 there exist a constants K ≥ 2 and a > 0 such that
ρ ( 2 u ) K ρ ( u ) + ε

for all u X ρ with ρ(u) ≤ a.

If ρ satisfies the Δ2-condition for any a > 0 with K ≥ 2 dependent on a, we say that ρ the strong Δ2-condition ( ρ Δ 2 s ) .

Lemma 1.1. [[12], Lemma 2.1] If ρ Δ 2 s , then for any L > 0 and ε > 0, there exists δ = δ(L, ε) > 0 such that
| ρ ( u + v ) - ρ ( u ) | < ε ,

whenever u, v X ρ with ρ(u) ≤ L, and ρ(v) ≤ δ.

Lemma 1.2. [[12], Lemma 2.3] Convergences in norm and in modular are equivalent in X ρ if ρ Δ2.

Lemma 1.3. [[12], Lemma 2.4] If ρ Δ 2 s , then for any ε > 0 there exists δ = δ(ε) > 0 such that || x || 1 + δ, whenever ρ(x) 1 + ε.

2. Main results

In this section, we prove the property H and uniform Opial property in generalized Cesàro sequence space ces(p)(q). First, we give some results which are very important for our con-sideration.

Proposition 2.1. The functional ϱ is a convex modular on ces(p)(q).

Proof. Let x, y ces(p)(q). It is obvious that ϱ(x) = 0 if and only if x = 0 and ϱ(αx) = ϱ(x) for scalar α with |α| = 1. Let α ≥ 0, β ≥ 0 with α + β = 1. By the convexity of the function t | t | p k , for all k , we have
ϱ ( α x + β y ) = k = 1 1 Q k i = 1 k | α q i x ( i ) + β q i y ( i ) | p k k = 1 α 1 Q k i = 1 k | q i x ( i ) | + β 1 Q k i = 1 k | q i y ( i ) | p k α k = 1 1 Q k i = 1 k | q i x ( i ) | p k + β k = 1 1 Q k i = 1 k | q i y ( i ) | p k = α ϱ ( x ) + β ϱ ( y ) .
Proposition 2.2. For x ces(p)(q), the modular ϱ on ces(p)(q) satisfies the following properties:
  1. (i)

    if 0 < a < 1, then a M ϱ ( x a ) ϱ ( x ) and ϱ(ax) ≤ aϱ(x);

     
  2. (ii)

    if a > 1, then ϱ(x) a M ϱ ( x a ) ;

     
  3. (iii)

    if a ≥ 1, then ϱ(x) ≤ aϱ(x) ≤ ϱ(ax).

     
Proof. (i) Let 0 < a < 1. Then we have
ϱ ( x ) = k = 1 1 Q k i = 1 k | q i x ( i ) | p k = k = 1 a Q k i = 1 k q i x ( i ) a p k = k = 1 a p k 1 Q k i = 1 k q i x ( i ) a p k k = 1 a M 1 Q k i = 1 k q i x ( i ) a p k = a M k = 1 1 Q k i = 1 k q i x ( i ) a p k = a M ϱ x a .

By convexity of modular ϱ, we have ϱ(ax) ≤ aϱ(x), so (i) is obtained.

(ii) Let a > 1. Then
ϱ ( x ) = k = 1 1 Q k i = 1 k | q i x ( i ) | p k = k = 1 a p k 1 Q k i = 1 k q i x ( i ) a p k a M k = 1 1 Q k i = 1 k q i x ( i ) a p k = a M ϱ x a .

Hence (ii) is satisfies. (iii) follows from the convexity of ϱ. □

Proposition 2.3. For any x ces(p)(q), we have
  1. (i)

    if ||x|| < 1, then ϱ(x) ||x||;

     
  2. (ii)

    if ||x|| > 1, then ϱ(x) ||x||;

     
  3. (iii)

    ||x|| = 1 if and only if ϱ(x) = 1;

     
  4. (iv)

    ||x|| < 1 if and only if ϱ(x) < 1;

     
  5. (v)

    ||x|| > 1 if and only if ϱ(x) > 1.

     
Proof. (i) Let ε > 0 be such that 0 < ε < 1 - ||x||, so ||x|| + ε < 1. By the definition of ||·||, then there exits λ > 0 such that ||x|| + ε > λ and ϱ ( x λ ) 1 . By (i) and (iii) of Proposition 2.2, we have
ϱ ( x ) ϱ ( | | x | | + ε ) λ x = ϱ ( | | x | | + ε ) x λ ( | | x | | + ε ) ϱ x λ | | x | | + ε ,

which implies that ϱ(x) ||x||. Hence (i) is satisfies.

(ii) Let ε > 0 such that 0 < ε < | | x | | - 1 | | x | | , then 0 < (1 - ε)||x|| ||x||. By definition of ||.|| and Proposition 2.2(i), we have 1 < ϱ ( x ( 1 - ε ) | | x | | ) < x ( 1 - ε ) | | x | | ϱ ( x ) , so (1 - ε)||x|| < ϱ(x) for all ε ( 0 , | | x | | - 1 | | x | | ) which implies that ||x|| ≤ ϱ(x).

(iii) Assume that ||x|| = 1. Let ε > 0 then there exits λ > 0 such that 1 + ε > λ > ||x|| and ϱ ( x λ ) 1 . By Proposition 2.2(ii), we have ϱ ( x ) λ M ϱ ( x λ ) λ M < ( 1 + ε ) M , so ( ϱ ( x ) ) 1 M < 1 + ε for all ε > 0 which implies that ϱ(x) 1. If ϱ(x) < 1, let a (0, 1) such that ϱ(x) < a M < 1. From Proposition 2.2(i), we have ϱ ( x a ) 1 a M ϱ ( x ) < 1 . Hence ||x|| ≤ a < 1, which is contradiction. Thus, we have ϱ(x) = 1.

Conversely, assume that ϱ(x) = 1. By definition of ||·||, we conclude that ||x|| 1. If ||x|| < 1, then we have by (i) that ϱ(x) ||x|| < 1, which is contradiction, so we obtain that ||x|| = 1. (iv) follows from (i) and (iii), (v) follows from (iii) and (iv). □

Proposition 2.4. For any x ces(p)(q), we have
  1. (i)

    if 0 < a < 1 and ||x|| > a, then ϱ(x) > a M ;

     
  2. (ii)

    if a ≥ 1 and ||x|| < a, then ϱ(x) < a M .

     

Proof. (i) Let 0 < a < 1 and ||x|| > a. Then | | x a | | > 1 , by Proposition 2.3(v), we have ϱ ( x a ) > 1 . Hence by Proposition 2.2(i), we have ϱ ( x ) a M ϱ ( x a ) > a M , so we obtain (i).

(ii) Suppose a ≥ 1 and ||x|| < a. Then | | x a | | < 1 , by Proposition 2.3(iv), we have ϱ ( x a ) < 1 .

If a = 1, it is obvious that ϱ(x) < 1 = a M . If a > 1, then by Proposition 2.2(ii), we obtain that ϱ ( x ) a M ϱ ( x a ) < a M . □

Proposition 2.5. Let (x n ) be a sequence in ces(p)(q).
  1. (i)

    If ||x n || 1 as n∞, then ϱ(x n ) 1 as n → ∞.

     
  2. (ii)

    If ϱ(x n ) 0 as n∞, then ||x n || 0 as n → ∞.

     

Proof. (i) Assume that ||x n || → 1 as n → ∞. Let ε (0, 1). Then there exists N such that 1 - ε < ||x n || < 1 + ε for all n ≥ N. By Proposition 2.4, we have (1 - ε) M < ϱ(x n ) < (1 + ε) M for all n ≥ N, which implies that ϱ(x n ) 1 as n → ∞.

(ii) Suppose that ||x n || 0 as n → ∞. Then there exists ε (0, 1) and a subsequence ( x n k ) of (x n ) such that | | x n k | | > ε for all k . By Proposition 2.4(i) we obtain ϱ ( x n k ) > ( ε ) M for all k . This implies that ϱ(x n ) 0 as n → ∞. □

Lemma 2.6. Let x ces(p)(q) and (x n ) ces(p)(q). If ϱ(x n ) → ϱ(x) as n → ∞ and x n (i) → x(i) as n → ∞ for all i , then x n → x as n → ∞.

Proof. Let ε > 0 be given. Since ϱ ( x ) = k = 1 1 Q k i = 1 k | q i x n ( i ) | p k < , there exists k0 such that
k = k 0 + 1 1 Q k i = 1 k | q i x n ( i ) | p k < ε 3 2 M + 1 .
(2.1)
Since ϱ ( x n ) - k = 1 k 0 1 Q k i = 1 k | q i x n ( i ) | p k ϱ ( x ) - k = 1 k 0 1 Q k i = 1 k | q i x ( i ) | p k and x n (i) → x(i) as n → ∞ for all i there exists n0 such that
ϱ ( x n ) - k = 1 k 0 1 Q k i = 1 k | q i x n ( i ) | p k < ϱ ( x ) - k = 1 k 0 1 Q k i = 1 k | q i x ( i ) | p k + ε 3 2 M
(2.2)
for all n ≥ n0 and
k = 1 k 0 1 Q k i = 1 k | q i x n ( i ) - q i x ( i ) | p k < ε 3 ,
(2.3)
for all n ≥ n0. It follow from (2.1), (2.2), and (2.3), for all n ≥ n0 we have
ϱ ( x n - x ) = k = 1 1 Q k i = 1 k | q i x n ( i ) - q i x ( i ) | p k = k = 1 k 0 1 Q k i = 1 k | q i x n ( i ) - q i x ( i ) | p k + k = k 0 + 1 1 Q k i = 1 k | q i x n ( i ) - q i x ( i ) | p k < ε 3 + 2 M k = k 0 + 1 1 Q k i = 1 k | q i x n ( i ) ) | p k + k = k 0 + 1 1 Q k i = 1 k | q i x ( i ) ) | p k = ε 3 + 2 M ϱ ( x n ) - k = 1 k 0 1 Q k i = 1 k | q i x n ( i ) | p k + k = k 0 + 1 1 Q k i = 1 k | q i x ( i ) ) | p k < ε 3 + 2 M ϱ ( x ) - k = 1 k 0 1 Q k i = 1 k | q i x ( i ) | p k + ε 3 2 M + k = k 0 + 1 1 Q k i = 1 k | q i x ( i ) ) | p k = ε 3 + 2 M k = k 0 + 1 1 Q k i = 1 k | q i x ( i ) | p k + ε 3 2 M + k = k 0 + 1 1 Q k i = 1 k | q i x ( i ) ) | p k = ε 3 + 2 M 2 k = k 0 + 1 1 Q k i = 1 k | q i x ( i ) | p k + ε 3 2 M < ε 3 + ε 3 + ε 3 = ε .

This show that ϱ(x n -x) 0 as n → ∞. Hence, by Proposition 2.5(ii), we have ||x n -x|| 0 as → ∞. □

Theorem 2.7. The space ces(p)(q) has the property (H).

Proof. Let x S(ces(p)(q)) and (x n ) ces(p)(q) such that ||x n || 1 and x n w x as n → ∞. By Proposition 2.3(iii), we have ϱ(x) = 1, so it follow form Proposition 2.5(i), we get ϱ(x n ) → ϱ(x) as n → ∞. Since the mapping π i : ces(p)(q) defined by π i (y) = y(i), is a continuous linear functional on ces(p)(q), it follow that x n (i) → x(i) as n → ∞ for all i . Thus by Lemma 2.6, we obtain x n → x as n → ∞, and hence the space ces(p)(q) has the property (H). □

Corollary 2.8. For any 1 < p < ∞, the space ces p (q) has the property (H).

Corollary 2.9. [9, Theorem 2.6] The space ces(p)has the property (H).

Corollary 2.10. For any 1 < p < ∞, the space ces p has the property (H).

Theorem 2.11. The space ces(p)(q) has uniform Opial property.

Proof. Take any ε > 0 and x ces(p)(q) with ||x|| ≥ ε. Let (x n ) be weakly null sequence in S(ces(p)(q)). By sup k p k < ∞, i.e., ϱ Δ 2 s , hence by Lemma 1.2 there exists δ (0, 1) independent of x such that ϱ(x) > δ. Also, by ϱ Δ 2 s and Lemma 1.1 asserts that there exists δ1 (0, δ) such that
| ϱ ( y + z ) - ϱ ( y ) | < δ 4
(2.4)
whenever, ϱ(y) 1 and ϱ(z) ≤ δ1. Choose k0 such that
k = k 0 + 1 1 Q k i = k 0 + 1 k | q i x ( i ) | p k < k = k 0 + 1 1 Q k i = 1 k | q i x ( i ) | p k < δ 1 4 .
(2.5)
So, we have
δ < k = 1 k 0 1 Q k i = 1 k | q i x ( i ) | p k + k = k 0 + 1 1 Q k i = 1 k | q i x ( i ) | p k k = 1 k 0 1 Q k i = 1 k | q i x ( i ) | p k + δ 1 4 ,
(2.6)
which implies that
k = 1 k 0 1 Q k i = 1 k | q i x ( i ) | p k > δ - δ 1 4 > δ - δ 4 = 3 δ 4 .
(2.7)
Since x n w 0 , then there exists n0 such that
3 δ 4 k = 1 k 0 1 Q k i = 1 k | q i x n ( i ) + q i x ( i ) | p k
(2.8)
for all n > n0, since weak convergence implies coordinatewise convergence. Again, by x n w 0 , then there exists n1 such that
| | x n | k o | | < 1 - 1 - δ 4 1 M
(2.9)
for all n > n1 where p k ≤ M for all k . Hence, by the triangle inequality of the norm, we get
| | x n | - k o | | > 1 - δ 4 1 M .
(2.10)
It follows by the definition of || · ||, we have
1 ϱ x n | - k o 1 - δ 4 1 M = k = k 0 + 1 1 Q k i = k 0 + 1 k | q i x n ( i ) | 1 - δ 4 1 M p k 1 1 - δ 4 1 M M k = k 0 + 1 1 Q k i = k 0 + 1 k | q i x n ( i ) | p k
(2.11)
implies that
k = k 0 + 1 1 Q k i = k 0 + 1 | q i x n ( i ) | p k 1 - δ 4
(2.12)
for all n > n1. By inequality (2.4), (2.5), (2.8), and (2.12), yields for any n > n1 that
ϱ ( x n + x ) = k = 1 k 0 1 Q k i = 1 k | q i x n ( i ) + q i x ( i ) | p k + k = k 0 + 1 1 Q k i = 1 k | q i x n ( i ) + q i x ( i ) | p k > k = 1 k 0 1 Q k i = 1 k | q i x n ( i ) + q i x ( i ) | p k + k = k 0 + 1 1 Q k i = k 0 + 1 k | q i x n ( i ) + q i x ( i ) | p k 3 δ 4 + k = k 0 + 1 1 Q k i = k 0 + 1 k | q i x n ( i ) | p k - δ 4 3 δ 4 + 1 - δ 4 - δ 4 1 + δ 4 .

Since ϱ Δ 2 s and by Lemma 1.3 there exists τ depending on δ only such that || x n +x || 1+τ, which implies that lim n inf | | x n + x | | 1 + τ , hence the proof is complete. □

Corollary 2.12. For any 1 < p < ∞, the space ces p (q) has the uniform Opial property.

Corollary 2.13. [5, Theorem 2.6] The space ces(p)has the uniform Opial property.

Corollary 2.14. [4, Theorem 2] For any 1 < p < ∞, the space ces p has the uniform Opial property.

Declarations

Acknowledgements

Mr. Chirasak Mongkolkeha was supported from the Thailand Research Fund through the Royal Golden Jubilee Program under Grant PHD/0029/2553 for the Ph.D program at KMUTT, Thailand. Moreover, the authors would like to thank the Higher Education Research Promotion and National Research University Project of Thailand, Office of the Higher Education Commission (NRU-CSEC No.54000267) for financial support.

Authors’ Affiliations

(1)
Department of Mathematics, Faculty of Science, King Mongkut's University of Technology Thonburi (KMUTT) Bangmod

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Copyright

© Mongkolkeha and Kumam; licensee Springer. 2012

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