On H-property and uniform Opial property of generalized cesàro sequence spaces
© Mongkolkeha and Kumam; licensee Springer. 2012
Received: 29 October 2011
Accepted: 30 March 2012
Published: 30 March 2012
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.
Keywordsgeneralized Cesàro sequence spaces H-property uniform Opial property
where n ∈ ℕ. If q k = 1 for all k ∈ ℕ, then ces p (q) reduces to ces p .
ρ(x) = 0 if and only if x = 0;
ρ(αx) = ρ(x) for all scalar α with |α| = 1;
ρ(αx + βy) ≤ ρ(x) + ρ(y), for all x, y ∈ X and all α, β ≥ 0 with α + β = 1.
ρ(αx + βy) ≤ αρ(x) + βρ(y), for all x, y ∈ X and all α, β ≥ 0 with α + β = 1.
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 → ∞.
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.
whenever u, v ∈ X ρ with ρ(u) ≤ L, and ρ(v) ≤ δ.
Lemma 1.2. [, Lemma 2.3] Convergences in norm and in modular are equivalent in X ρ if ρ ∈ Δ2.
Lemma 1.3. [, Lemma 2.4] If, 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).
if 0 < a < 1, then and ϱ(ax) ≤ aϱ(x);
if a > 1, then ϱ(x) ≤ ;
if a ≥ 1, then ϱ(x) ≤ aϱ(x) ≤ ϱ(ax).
By convexity of modular ϱ, we have ϱ(ax) ≤ aϱ(x), so (i) is obtained.
Hence (ii) is satisfies. (iii) follows from the convexity of ϱ. □
if ||x|| < 1, then ϱ(x) ≤ ||x||;
if ||x|| > 1, then ϱ(x) ≥ ||x||;
||x|| = 1 if and only if ϱ(x) = 1;
||x|| < 1 if and only if ϱ(x) < 1;
||x|| > 1 if and only if ϱ(x) > 1.
which implies that ϱ(x) ≤ ||x||. Hence (i) is satisfies.
(ii) Let ε > 0 such that , then 0 < (1 - ε)||x|| ≤ ||x||. By definition of ||.|| and Proposition 2.2(i), we have , so (1 - ε)||x|| < ϱ(x) for all 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 . By Proposition 2.2(ii), we have , so 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 . 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). □
if 0 < a < 1 and ||x|| > a, then ϱ(x) > a M ;
if a ≥ 1 and ||x|| < a, then ϱ(x) < a M .
Proof. (i) Let 0 < a < 1 and ||x|| > a. Then , by Proposition 2.3(v), we have . Hence by Proposition 2.2(i), we have , so we obtain (i).
(ii) Suppose a ≥ 1 and ||x|| < a. Then , by Proposition 2.3(iv), we have .
If a = 1, it is obvious that ϱ(x) < 1 = a M . If a > 1, then by Proposition 2.2(ii), we obtain that . □
If ||x n || → 1 as n → ∞, then ϱ(x n ) → 1 as n → ∞.
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 of (x n ) such that for all k ∈ ℕ. By Proposition 2.4(i) we obtain 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 → ∞.
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 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.
Since and by Lemma 1.3 there exists τ depending on δ only such that || x n +x || ≥ 1+τ, which implies that , 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.
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.
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