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 , for all k ∈ ℕ, we have
□
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 and ϱ(ax) ≤ aϱ(x);
-
(ii)
if a > 1, then ϱ(x) ≤ ;
-
(iii)
if a ≥ 1, then ϱ(x) ≤ aϱ(x) ≤ ϱ(ax).
Proof. (i) Let 0 < a < 1. Then we have
By convexity of modular ϱ, we have ϱ(ax) ≤ aϱ(x), so (i) is obtained.
(ii) Let a > 1. Then
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 . By (i) and (iii) of Proposition 2.2, we have
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) < aM < 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). □
Proposition 2.4. For any x ∈ ces(p)(q), we have
-
(i)
if 0 < a < 1 and ||x|| > a, then ϱ(x) > aM;
-
(ii)
if a ≥ 1 and ||x|| < a, then ϱ(x) < aM .
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 = aM . If a > 1, then by Proposition 2.2(ii), we obtain that . □
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 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 → ∞.
Proof. Let ε > 0 be given. Since , there exists k0 ∈ ℕ such that
(2.1)
Since and x
n
(i) → x(i) as n → ∞ for all i ∈ ℕ there exists n0 ∈ ℕ such that
(2.2)
for all n ≥ n0 and
(2.3)
for all n ≥ n0. It follow from (2.1), (2.2), and (2.3), for all n ≥ n0 we have
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.
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., , hence by Lemma 1.2 there exists δ ∈ (0, 1) independent of x such that ϱ(x) > δ. Also, by and Lemma 1.1 asserts that there exists δ1 ∈ (0, δ) such that
(2.4)
whenever, ϱ(y) ≤ 1 and ϱ(z) ≤ δ1. Choose k0 ∈ ℕ such that
(2.5)
So, we have
(2.6)
which implies that
(2.7)
Since , then there exists n0 ∈ ℕ such that
(2.8)
for all n > n0, since weak convergence implies coordinatewise convergence. Again, by , then there exists n1 ∈ ℕ such that
(2.9)
for all n > n1 where p
k
≤ M for all k ∈ ℕ. Hence, by the triangle inequality of the norm, we get
(2.10)
It follows by the definition of || · ||, we have
(2.11)
implies that
(2.12)
for all n > n1. By inequality (2.4), (2.5), (2.8), and (2.12), yields for any n > n1 that
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.