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

Chirasak Mongkolkeha; Poom Kumam

doi:10.1186/1029-242X-2012-76

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

Chirasak Mongkolkeha¹ and
Poom Kumam¹Email author

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 \to \infty} inf | | x_{n} | | \leq lim_{n \to \infty} 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 \leq lim_{n \to \infty} inf | | x_{n} + x | | .

For example, the space in [3–5] 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 \in l^{0} : \sum_{n = 1}^{\infty} {(\frac{1}{n} \sum_{i = 0}^{n} | x (i) |)}^{p} < \infty\}

equipped with the norm

| | x | | = {({\sum_{n = 1}^{\infty} (\frac{1}{n} \sum_{i = 1}^{n} | x (i) |)}^{p})}^{\frac{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 ∈ ℕ by

ce s_{(p)} = {x \in l^{0} : ϱ (λ x) < \infty for some λ > 0},

where

ϱ (x) = \sum_{n = 1}^{\infty} {(\frac{1}{n} \sum_{i = 1}^{k} | x (i) |)}^{p_{n}}

equipped with the Luxemburg norm

| | x | | = inf \{ε > 0 : ϱ (\frac{x}{ε}) \leq 1\} .

In the case when p_k = p, 1 ≤ p < ∞ for all k ∈ ℕ, 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 \in l^{0} : {(\sum_{k = 1}^{\infty} {(\frac{1}{Q_{k}} \sum_{i = 1}^{k} | q_{i} x (i) |)}^{p})}^{1 / p} < \infty\},

where $Q_{k} = \sum_{k = 1}^{n} q_{k},$ n ∈ ℕ. If q_k = 1 for all k ∈ ℕ, 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 ∈ ℕ by

ce s_{(p)} (q) = {x \in l^{0} : ϱ (λ x) < \infty for some λ > 0},

where

ϱ (x) = \sum_{k = 1}^{\infty} {(\frac{1}{Q_{k}} \sum_{i = 1}^{k} | q_{i} x (i) |)}^{p_{k}}

with

Q_{k} = \sum_{k = 1}^{n} q_{k}

and consider ces_(p)(q) equipped with the Luxemburg norm

| | x | | = inf \{ε > 0 : ϱ (\frac{x}{ε}) \leq 1\} .

Thus, we see that p_k = p, 1 ≤ p < ∞ for all k ∈ ℕ, then ces_(p)(q) reduces to ces_p(q) and if q_k = 1 for all k ∈ ℕ, then ces_(p)(q) reduces to ces_(p). Throughout this article, for x ∈ l⁰, i ∈ ℕ, we denote

\begin{aligned} e_{i} = (\overset{i - 1}{\overset{⏞}{0, 0, \dots, 0,}} 1, 0, 0, 0, \dots), \\ x |_{i} = (x (1), x (2), x (3), \dots, x (i), 0, 0, 0, \dots), \\ x |_{ℕ - i} = (0, 0, 0, \dots, x (i + 1), x (i + 2), \dots), \end{aligned}

and M = sup _k p_k with p_k > 1 for all k ∈ ℕ. 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;

(i)

ρ(x) = 0 if and only if x = 0;
(ii)

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

(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 \in X : ρ (λ x) \to 0 as λ \to 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) \leq 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 $(ρ \in Δ_{2}^{s})$ .

Lemma 1.1. [[12], Lemma 2.1] If

ρ \in 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 $ρ \in Δ_{2}^{s}$ , 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 \mapsto | t |^{p_{k}}

, for all k ∈ ℕ, we have

\begin{aligned} ϱ (α x + β y) & = \sum_{k = 1}^{\infty} {(\frac{1}{Q_{k}} \sum_{i = 1}^{k} | α q_{i} x (i) + β q_{i} y (i) |)}^{p_{k}} \\ \leq \sum_{k = 1}^{\infty} {(α \frac{1}{Q_{k}} \sum_{i = 1}^{k} | q_{i} x (i) | + β \frac{1}{Q_{k}} \sum_{i = 1}^{k} | q_{i} y (i) |)}^{p_{k}} \\ \leq α \sum_{k = 1}^{\infty} {(\frac{1}{Q_{k}} \sum_{i = 1}^{k} | q_{i} x (i) |)}^{p_{k}} + β \sum_{k = 1}^{\infty} {(\frac{1}{Q_{k}} \sum_{i = 1}^{k} | q_{i} y (i) |)}^{p_{k}} \\ = α ϱ (x) + β ϱ (y) . \end{aligned}

□

Proposition 2.2. For x ∈ ces_(p)(q), the modular ϱ on ces_(p)(q) satisfies the following properties:

(i)

if 0 < a < 1, then $a^{M} ϱ (\frac{x}{a}) \leq ϱ (x)$ and ϱ(ax) ≤ aϱ(x);
(ii)

if a > 1, then ϱ(x) ≤ $a^{M} ϱ (\frac{x}{a})$ ;
(iii)

if a ≥ 1, then ϱ(x) ≤ aϱ(x) ≤ ϱ(ax).

Proof. (i) Let 0 < a < 1. Then we have

\begin{aligned} ϱ (x) & = \sum_{k = 1}^{\infty} {(\frac{1}{Q_{k}} \sum_{i = 1}^{k} | q_{i} x (i) |)}^{p_{k}} \\ = \sum_{k = 1}^{\infty} {(\frac{a}{Q_{k}} \sum_{i = 1}^{k} |\frac{q_{i} x (i)}{a}|)}^{p_{k}} \\ = \sum_{k = 1}^{\infty} {a^{p_{k}} (\frac{1}{Q_{k}} \sum_{i = 1}^{k} |\frac{q_{i} x (i)}{a}|)}^{p_{k}} \\ \geq \sum_{k = 1}^{\infty} {a^{M} (\frac{1}{Q_{k}} \sum_{i = 1}^{k} |\frac{q_{i} x (i)}{a}|)}^{p_{k}} \\ = a^{M} \sum_{k = 1}^{\infty} {(\frac{1}{Q_{k}} \sum_{i = 1}^{k} |\frac{q_{i} x (i)}{a}|)}^{p_{k}} \\ = a^{M} ϱ (\frac{x}{a}) . \end{aligned}

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

(ii) Let a > 1. Then

\begin{aligned} ϱ (x) & = \sum_{k = 1}^{\infty} {(\frac{1}{Q_{k}} \sum_{i = 1}^{k} | q_{i} x (i) |)}^{p_{k}} \\ = \sum_{k = 1}^{\infty} {a^{p_{k}} (\frac{1}{Q_{k}} \sum_{i = 1}^{k} |\frac{q_{i} x (i)}{a}|)}^{p_{k}} \\ \leq a^{M} \sum_{k = 1}^{\infty} {(\frac{1}{Q_{k}} \sum_{i = 1}^{k} |\frac{q_{i} x (i)}{a}|)}^{p_{k}} \\ = a^{M} ϱ (\frac{x}{a}) . \end{aligned}

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

Proposition 2.3. For any x ∈ ces_(p)(q), we have

(i)

if ||x|| < 1, then ϱ(x) ≤ ||x||;
(ii)

if ||x|| > 1, then ϱ(x) ≥ ||x||;
(iii)

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

||x|| < 1 if and only if ϱ(x) < 1;
(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 (\frac{x}{λ}) \leq 1

. By (i) and (iii) of Proposition 2.2, we have

\begin{aligned} ϱ (x) & \leq ϱ (\frac{(| | x | | + ε)}{λ} x) \\ = ϱ ((| | x | | + ε) \frac{x}{λ}) \\ \leq (| | x | | + ε) ϱ (\frac{x}{λ}) \\ \leq | | x | | + ε, \end{aligned}

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

(ii) Let ε > 0 such that $0 < ε < \frac{| | x | | - 1}{| | x | |}$ , then 0 < (1 - ε)||x|| ≤ ||x||. By definition of ||.|| and Proposition 2.2(i), we have $1 < ϱ (\frac{x}{(1 - ε) | | x | |}) < \frac{x}{(1 - ε) | | x | |} ϱ (x)$ , so (1 - ε)||x|| < ϱ(x) for all $ε \in (0, \frac{| | x | | - 1}{| | x | |})$ 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 $ϱ (\frac{x}{λ}) \leq 1$ . By Proposition 2.2(ii), we have $ϱ (x) \leq λ^{M} ϱ (\frac{x}{λ}) \leq λ^{M} < {(1 + ε)}^{M}$ , so ${(ϱ (x))}^{\frac{1}{M}} < 1 + ε$ 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 $ϱ (\frac{x}{a}) \leq \frac{1}{a^{M}} ϱ (x) < 1$ . 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

(i)

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

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

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

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

If a = 1, it is obvious that ϱ(x) < 1 = a^M . If a > 1, then by Proposition 2.2(ii), we obtain that $ϱ (x) \leq a^{M} ϱ (\frac{x}{a}) < a^{M}$ . □

Proposition 2.5. Let (x_n ) be a sequence in ces_(p)(q).

(i)

If ||x_n || → 1 as n → ∞, then ϱ(x_n ) → 1 as n → ∞.
(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 ∈ 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 ∈ ℕ. By Proposition 2.4(i) we obtain $ϱ (x_{n_{k}}) > {(ε)}^{M}$ for all k ∈ ℕ. 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 ∈ ℕ, then x_n → x as n → ∞.

Proof. Let ε > 0 be given. Since

ϱ (x) = \sum_{k = 1}^{\infty} {(\frac{1}{Q_{k}} \sum_{i = 1}^{k} | q_{i} x_{n} (i) |)}^{p_{k}} < \infty

, there exists k₀ ∈ ℕ such that

\sum_{k = k_{0} + 1}^{\infty} {(\frac{1}{Q_{k}} \sum_{i = 1}^{k} | q_{i} x_{n} (i) |)}^{p_{k}} < \frac{ε}{3 \cdot 2^{M + 1}} .

(2.1)

Since

ϱ (x_{n}) - \sum_{k = 1}^{k_{0}} {(\frac{1}{Q_{k}} \sum_{i = 1}^{k} | q_{i} x_{n} (i) |)}^{p_{k}} \to ϱ (x) - \sum_{k = 1}^{k_{0}} {(\frac{1}{Q_{k}} \sum_{i = 1}^{k} | q_{i} x (i) |)}^{p_{k}}

and x_n (i) → x(i) as n → ∞ for all i ∈ ℕ there exists n₀ ∈ ℕ such that

ϱ (x_{n}) - \sum_{k = 1}^{k_{0}} {(\frac{1}{Q_{k}} \sum_{i = 1}^{k} | q_{i} x_{n} (i) |)}^{p_{k}} < ϱ (x) - \sum_{k = 1}^{k_{0}} {(\frac{1}{Q_{k}} \sum_{i = 1}^{k} | q_{i} x (i) |)}^{p_{k}} + \frac{ε}{3 \cdot 2^{M}}

(2.2)

for all n ≥ n₀ and

\sum_{k = 1}^{k_{0}} {(\frac{1}{Q_{k}} \sum_{i = 1}^{k} | q_{i} x_{n} (i) - q_{i} x (i) |)}^{p_{k}} < \frac{ε}{3},

(2.3)

for all n ≥ n₀. It follow from (2.1), (2.2), and (2.3), for all n ≥ n₀ we have

\begin{aligned} ϱ (x_{n} - x) & = \sum_{k = 1}^{\infty} {(\frac{1}{Q_{k}} \sum_{i = 1}^{k} | q_{i} x_{n} (i) - q_{i} x (i) |)}^{p_{k}} \\ = \sum_{k = 1}^{k_{0}} {(\frac{1}{Q_{k}} \sum_{i = 1}^{k} | q_{i} x_{n} (i) - q_{i} x (i) |)}^{p_{k}} + \sum_{k = k_{0} + 1}^{\infty} {(\frac{1}{Q_{k}} \sum_{i = 1}^{k} | q_{i} x_{n} (i) - q_{i} x (i) |)}^{p_{k}} \\ < \frac{ε}{3} + 2^{M} ({\sum_{k = k_{0} + 1}^{\infty} (\frac{1}{Q_{k}} \sum_{i = 1}^{k} | q_{i} x_{n} (i)) |)}^{p_{k}} + {\sum_{k = k_{0} + 1}^{\infty} (\frac{1}{Q_{k}} \sum_{i = 1}^{k} | q_{i} x (i)) |)}^{p_{k}}) \\ = \frac{ε}{3} + 2^{M} (ϱ (x_{n}) - {\sum_{k = 1}^{k_{0}} (\frac{1}{Q_{k}} \sum_{i = 1}^{k} | q_{i} x_{n} (i) |)}^{p_{k}} + {\sum_{k = k_{0} + 1}^{\infty} (\frac{1}{Q_{k}} \sum_{i = 1}^{k} | q_{i} x (i)) |)}^{p_{k}}) \\ < \frac{ε}{3} + 2^{M} (ϱ (x) - {\sum_{k = 1}^{k_{0}} (\frac{1}{Q_{k}} \sum_{i = 1}^{k} | q_{i} x (i) |)}^{p_{k}} + \frac{ε}{3 \cdot 2^{M}} + {\sum_{k = k_{0} + 1}^{\infty} (\frac{1}{Q_{k}} \sum_{i = 1}^{k} | q_{i} x (i)) |)}^{p_{k}}) \\ = \frac{ε}{3} + 2^{M} ({\sum_{k = k_{0} + 1}^{\infty} (\frac{1}{Q_{k}} \sum_{i = 1}^{k} | q_{i} x (i) |)}^{p_{k}} + \frac{ε}{3 \cdot 2^{M}} + {\sum_{k = k_{0} + 1}^{\infty} (\frac{1}{Q_{k}} \sum_{i = 1}^{k} | q_{i} x (i)) |)}^{p_{k}}) \\ = \frac{ε}{3} + 2^{M} (2 {\sum_{k = k_{0} + 1}^{\infty} (\frac{1}{Q_{k}} \sum_{i = 1}^{k} | q_{i} x (i) |)}^{p_{k}} + \frac{ε}{3 \cdot 2^{M}}) \\ < \frac{ε}{3} + \frac{ε}{3} + \frac{ε}{3} \\ = ε . \end{aligned}

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 $x_{n} \overset{w}{\to} x$ 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) → ℝ 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 ∈ ℕ. 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 \in Δ_{2}^{s}

, hence by Lemma 1.2 there exists δ ∈ (0, 1) independent of x such that ϱ(x) > δ. Also, by

ϱ \in Δ_{2}^{s}

and Lemma 1.1 asserts that there exists δ₁ ∈ (0, δ) such that

| ϱ (y + z) - ϱ (y) | < \frac{δ}{4}

(2.4)

whenever, ϱ(y) ≤ 1 and ϱ(z) ≤ δ₁. Choose k₀ ∈ ℕ such that

\sum_{k = k_{0} + 1}^{\infty} {(\frac{1}{Q_{k}} \sum_{i = k_{0} + 1}^{k} | q_{i} x (i) |)}^{p_{k}} < \sum_{k = k_{0} + 1}^{\infty} {(\frac{1}{Q_{k}} \sum_{i = 1}^{k} | q_{i} x (i) |)}^{p_{k}} < \frac{δ_{1}}{4} .

(2.5)

So, we have

\begin{aligned} δ < \sum_{k = 1}^{k_{0}} {(\frac{1}{Q_{k}} \sum_{i = 1}^{k} | q_{i} x (i) |)}^{p_{k}} + \sum_{k = k_{0} + 1}^{\infty} {(\frac{1}{Q_{k}} \sum_{i = 1}^{k} | q_{i} x (i) |)}^{p_{k}} \\ \leq \sum_{k = 1}^{k_{0}} {(\frac{1}{Q_{k}} \sum_{i = 1}^{k} | q_{i} x (i) |)}^{p_{k}} + \frac{δ_{1}}{4}, \end{aligned}

(2.6)

which implies that

\begin{aligned} \sum_{k = 1}^{k_{0}} {(\frac{1}{Q_{k}} \sum_{i = 1}^{k} | q_{i} x (i) |)}^{p_{k}} & > δ - \frac{δ_{1}}{4} \\ > δ - \frac{δ}{4} \\ = \frac{3 δ}{4} . \end{aligned}

(2.7)

Since

x_{n} \overset{w}{\to} 0

, then there exists n₀ ∈ ℕ such that

\frac{3 δ}{4} \leq \sum_{k = 1}^{k_{0}} {(\frac{1}{Q_{k}} \sum_{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} \overset{w}{\to} 0

, then there exists n₁ ∈ ℕ such that

| | x_{n |_{k_{o}}} | | < 1 - {(1 - \frac{δ}{4})}^{\frac{1}{M}}

(2.9)

for all n > n₁ where p_k ≤ M for all k ∈ ℕ. Hence, by the triangle inequality of the norm, we get

| | x_{n |_{ℕ - k_{o}}} | | > {(1 - \frac{δ}{4})}^{\frac{1}{M}} .

(2.10)

It follows by the definition of || · ||, we have

\begin{aligned} 1 & \leq ϱ (\frac{x_{n |_{ℕ - k_{o}}}}{{(1 - \frac{δ}{4})}^{\frac{1}{M}}}) \\ = \sum_{k = k_{0} + 1}^{\infty} {(\frac{\frac{1}{Q_{k}} \sum_{i = k_{0} + 1}^{k} | q_{i} x_{n} (i) |}{{(1 - \frac{δ}{4})}^{\frac{1}{M}}})}^{p_{k}} \\ \leq {(\frac{1}{{(1 - \frac{δ}{4})}^{\frac{1}{M}}})}^{M} \sum_{k = k_{0} + 1}^{\infty} {(\frac{1}{Q_{k}} \sum_{i = k_{0} + 1}^{k} | q_{i} x_{n} (i) |)}^{p_{k}} \end{aligned}

(2.11)

implies that

\sum_{k = k_{0} + 1}^{\infty} {(\frac{1}{Q_{k}} \sum_{i = k_{0} + 1}^{\infty} | q_{i} x_{n} (i) |)}^{p_{k}} \geq 1 - \frac{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

\begin{aligned} ϱ (x_{n} + x) & = \sum_{k = 1}^{k_{0}} {(\frac{1}{Q_{k}} \sum_{i = 1}^{k} | q_{i} x_{n} (i) + q_{i} x (i) |)}^{p_{k}} + \sum_{k = k_{0} + 1}^{\infty} {(\frac{1}{Q_{k}} \sum_{i = 1}^{k} | q_{i} x_{n} (i) + q_{i} x (i) |)}^{p_{k}} \\ > \sum_{k = 1}^{k_{0}} {(\frac{1}{Q_{k}} \sum_{i = 1}^{k} | q_{i} x_{n} (i) + q_{i} x (i) |)}^{p_{k}} + \sum_{k = k_{0} + 1}^{\infty} {(\frac{1}{Q_{k}} \sum_{i = k_{0} + 1}^{k} | q_{i} x_{n} (i) + q_{i} x (i) |)}^{p_{k}} \\ \geq \frac{3 δ}{4} + \sum_{k = k_{0} + 1}^{\infty} {(\frac{1}{Q_{k}} \sum_{i = k_{0} + 1}^{k} | q_{i} x_{n} (i) |)}^{p_{k}} - \frac{δ}{4} \\ \geq \frac{3 δ}{4} + (1 - \frac{δ}{4}) - \frac{δ}{4} \\ \geq 1 + \frac{δ}{4} . \end{aligned}

Since $ϱ \in Δ_{2}^{s}$ and by Lemma 1.3 there exists τ depending on δ only such that || x_n +x || ≥ 1+τ, which implies that $lim_{n \to \infty} inf | | x_{n} + x | | \geq 1 + τ$ , 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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