Georgian Mathematical Journal Volume 10 (2003), Number 1, 193–200

ON SOME CONVEXITY PROPERTIES OF GENERALIZED CESÁRO SEQUENCE SPACES

SUTHEP SUANTAI

Abstract. We define a generalized Cesáro sequence space and consider it equipped with the Luxemburg norm under which it is a Banach space, and we show that it is locally uniformly rotund.

2000 Mathematics Subject Classification: 46E30, 46E40, 46B20.

Key words and phrases: Generalized Cesáro sequence spaces, Luxemburg norm, extreme point, locally uniformly rotund point, property (H), convex modular.

1. Preliminaries

For a Banach space X, we denote by S(X) and B(X) the unit sphere and unit ball of X, respectively. A point x_{0} ∈ S(X) is called

a) an extreme point if for every x, y ∈ S(X) the equality 2x_{0} = x + y implies x = y;

b) a locally uniformly rotund point (LUR-point for short) if for any sequence (x_{n}) in B(X) such that ∥x_{n} + x∥ → 2 as n → ∞ there holds ∥x_{n} − x∥ → 0 as n → ∞;

c) an H-point if for any sequence (x_{n}) in X such that ∥x_{n}∥ → 1 as n → ∞, the weak convergence of (x_{n}) to x_{0} (write x_{n} →^{w} x_{0}) implies that ∥x_{n} − x∥ → 0 as n → ∞.

A Banach space X is said to be rotund (R) if every point of S(X) is an extreme point.

If every x ∈ S(X) is a LUR-point, then X is said to be locally uniformly rotund (LUR).

X is said to possess property (H) provided every point of S(X) is an H-point. For these geometric notions and their role in Mathematics we refer to the

monographs [1], [6], [12] and [13]. Some of them were studied for Orlicz spaces in [1], [7], [8], [12] and [14].

Let X be a real vector space. A functional ϱ : X → [0, ∞] is called a modular if it satisfies the 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.

ISSN 1072-947X / $8.00 / \begin{matrix}c \\ ◯ \end{matrix} Heldermann Verlag www.heldermann.de

If ϱ is a modular in X, we define

X_{ϱ} = {x ∈ X : lim_{λ→0^{+}} ϱ(λx) = 0},

and X_{ϱ}^{∗} = {x ∈ X : ϱ(λx) < ∞ for some λ > 0}.

It is clear that X_{ϱ} ⊆ X_{ϱ}^{∗}. If ϱ is a convex modular, for x ∈ X_{ϱ} we define

∥x∥ = inf \{λ > 0 : ϱ(\frac{x}{λ}) ≤ 1\}.  (1.1)

Orlicz [13] proved that if ϱ is a convex modular in X, then X_{ϱ} = X_{ϱ}^{∗} and ∥.|| is a norm on X_{ϱ} for which it is a Banach space. The norm ∥.∥ defined as in (1.1) is called the Luxemburg norm.

A modular ϱ on X is called

(a) right-continuous if lim_{λ→1^{+}} ϱ(λx) = ϱ(x) for all x ∈ X_{ϱ};

(b) left-continuous if lim_{λ→1^{−}} ϱ(λx) = ϱ(x) for all x ∈ X_{ϱ};

(c) continuous if it is both left-continuous and right-continuous.

The following known results gave some relationships between the modular ϱ and the Luxemburg norm ∥.∥ on X_{ϱ}.

Theorem 1.1. Let ϱ be a convex modular on X and let x ∈ X_{ϱ} and (x_{n}) a sequence in X_{ϱ}. Then ∥x_{n} − x∥ → 0 as n → ∞ if and only if ϱ(λ(x_{n} − x)) → 0 as n → ∞ for every λ > 0.

Proof. See [11, Theorem 1.3]. □

Theorem 1.2. Let ϱ be a convex modular on X and x ∈ X_{ϱ}.

(i) If ϱ is right-continuous, then ∥x∥ < 1 if and only if ϱ(x) < 1.

(ii) If ϱ is left-continuous, then ∥x∥ ≤ 1 if and only if ϱ(x) ≤ 1.

(iii) If ϱ is continuous, then ∥x∥ = 1 if and only if ϱ(x) = 1.

Proof. See [11, Theorem 1.4]. □

Let us denote by l^{0} the space of all real sequences. For 1 ≤ p < ∞, the Cesáro sequence space (ces_{p}, for short) is defined by

ces_{p} = \{x ∈ l^{0} : ∑_{n=1}^{∞} (\frac{1}{n} ∑_{i=1}^{n} |x(i)|)^{p} < ∞\}

equipped with the norm

∥x∥ = ( ∑_{n=1}^{∞} (\frac{1}{n} ∑_{i=1}^{n} |x(i)|)^{p})^{\frac{1}{p}} .

This space was introduced by J.S. Shue [16]. It is useful in the theory of matrix operators and others (see [9] and [10]). Some geometric properties of the Cesáro sequence space ces_{p} were studied by many mathematicians. It is known that ces_{p} is LUR and possesses property (H) (see [10] ). Y. A. Cui and H. Hudzik [2] proved that ces_{p} has the Banach-Saks property, and it was shown in [5] that ces_{p} has property (β).

Now let p = (p_{k}) be a sequence of positive real numbers with p_{k} ≥ 1 for all k ∈ ℕ. The Nakano sequence space l(p) is defined by

l(p) = {x ∈ l^{0} : σ(λx) < ∞ for some λ > 0},

where σ(x) = ∑_{i=1}^{∞} |x(i)|^{p_{i}}. We consider the space l(p) equipped with the norm

∥x∥ = inf \{λ > 0 : σ(\frac{x}{λ}) ≤ 1\}

under which it is a Banach space. If p = (p_{k}) is bounded, we have

l(p) = {x ∈ l^{0} : ∑_{i=1}^{∞} |x(i)|^{p_{i}} < ∞}.

Several geometric properties of l(p) were studied in [1] and [4].

The generalized Cesáro sequence space ces(p) is defined by

ces(p) = {x ∈ l^{0} : ϱ(λx) < ∞ for some λ > 0},

where ϱ(x) = ∑_{n=1}^{∞}(\frac{1}{n} ∑_{i=1}^{n} |x(i)|)^{p_{n}}. We consider this space equipped with the so-called Luxemburg norm

∥x∥ = inf \{λ > 0 : ϱ(\frac{x}{λ}) ≤ 1\}

under which it is a Banach space. If p = (p_{k}) is bounded, we have

ces(p) = \{x = x(i) : ∑_{n=1}^{∞} (\frac{1}{n} ∑_{i=1}^{n} |x(i)|)^{p_{n}} < ∞\}.

W. Sanhan [15] proved that ces(p) is nonsquare when p_{k} > 1 for all k ∈ ℕ. In this paper, we show that the Cesáro sequence space ces(p) equipped with the Luxemburg norm is LUR and has property (H) when p = (p_{k}) is bounded with p_{k} > 1 for all k ∈ ℕ.

Throughout this paper we assume that p = (p_{k}) is bounded with p_{k} > 1 for all k ∈ ℕ, and M = sup_{k} p_{k}.

2. Main Results

We begin by giving some basic properties of the modular ϱ on the space ces(p). By the convexity of the function t → |t|^{p_{k}}, for every k ∈ ℕ we have that ϱ is a convex modular. So we have the following proposition.

Proposition 2.1. The functional ϱ on the Cesáro sequence space ces(p) is a convex modular.

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

(i) if 0 < a < 1, then a^{M}ϱ(\frac{x}{a}) ≤ 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. All assertions are clearly obtained by the definition of ϱ. □

Proposition 2.3. The modular ϱ on ces(p) is continuous.

Proof. For λ > 1, by Proposition 2.2 (ii) and (iii), we have

ϱ(x) ≤ λϱ(x) ≤ ϱ(λx) ≤ λ^{M}ϱ(x).  (2.1)

By taking λ → 1^{+} in (2.1), we have lim_{λ→1^{+}} ϱ(λx) = ϱ(x). Thus ϱ is rightcontinuous. If 0 < λ < 1, by Proposition 2.2 (i), we have

λ^{M}ϱ(x) ≤ ϱ(λx) ≤ λϱ(x)  (2.2)

By taking λ → 1^{−} in (2.2), we have that lim_{λ→1^{−}} ϱ(λx) = ϱ(x), hence ϱ is left-continuous. Thus ϱ is continuous. □

Next, we give some relationships between the modular ϱ and the Luxemburg norm on ces(p).

Proposition 2.4. For any x ∈ ces(p), 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, (vi) if 0 < a < 1 and ∥x|| > a, then ϱ(x) > a^{M}, and (vii) if a ≥ 1 and ∥x∥ < a, then ϱ(x) < a^{M}.

Proof. If ∥x∥ ≤ 1 , it follows by the convexity and continuity of ϱ that ϱ(x) = ϱ(∥x∥\frac{x}{∥x∥}) ≤ ∥x∥ϱ(\frac{x}{∥x∥}) ≤ ∥x∥. So (i) is obtained. If ∥x∥ > 1, then there

is ε_{0} > 0 such that ∥x∥ − ε > 1 for all ε ∈ (0, ε_{0}). Consequently, ϱ(x) =

ϱ((∥x∥ − ε)\frac{x}{∥x∥ − ε}) ≥ (∥x∥ − ε)ϱ(\frac{x}{∥x∥ − ε}) > ∥x∥ − ε, so (ii) is satisfied. It

is clear that (iii), (iv) and (v) follow by Theorem 1.2, and properties (vi) and (vii) follow by Proposition 2.2. □

Proposition 2.5. Let (x_{n}) be a sequence in ces(p).

(i) If ∥x_{n}∥ → 1 as n → ∞, then ϱ(x_{n}) → 1 as n → ∞.

(ii) ∥x_{n}∥ → 0 as n → ∞ if and only if ϱ(x_{n}) → 0 as n → ∞.

Proof. (i) Suppose ∥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 (vi) and (vii), 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) It follows from Theorem 1.1 that if ∥x_{n}∥ → 0 as n → ∞, then ϱ(x_{n}) → 0 as n → ∞. Conversely, suppose ∥x_{n}∥ ↛ 0 as n → ∞. Then there is ϵ ∈ (0, 1) and a subsequence (x_{n_{k}}) of (x_{n}) such that ∥x_{n_{k}}∥ > ϵ for all k ∈ ℕ. By Proprosition 2.4 (vi), we have ϱ(x_{n_{k}}) > ϵ^{M} for all k ∈ ℕ. This implies ϱ(x_{n}) ↛ 0 as n → ∞. □

Proposition 2.6. Let (x_{n}) ⊆ B(l(p)) and (y_{n}) ⊆ B(l(p)). If σ(\frac{x_{n} + y_{n}}{2} ) → 1, then x_{n}(i) − y_{n}(i) → 0 as n → ∞ for all i ∈ ℕ.

Proof. We first note that if x ∈ B(ℓ(p), then σ(x) ≤ 1. Supose that x_{n}(i) − y(i) ↛ 0 as n → ∞ for some i ∈ ℕ. Without loss of generality we may assume that i = 1, and then assume without loss of generality (passing to a subsequence if necessary) that, for some ϵ > 0,

|x_{n}(1) − y_{n}(1)|^{p_{1}} ≥ ϵ ∀ n ∈ ℕ.

Thus

2^{p_{1}}(|x_{n}(1)|^{p_{1}} + |y_{n}(1)|^{p_{1}}) ≥ ϵ ∀ n ∈ ℕ.  (2.3)

Since the function t → |t|^{p_{1}} is uniformly convex, there exists δ > 0 such that

|\frac{x_{n}(1) + y_{n}(1)}{2}|^{p_{1}} ≤ (1 − δ)(\frac{|x_{n}(1)|^{p_{1}} + |y_{n}(1)|^{p_{1}}}{2} ) ∀ n ∈ ℕ.  (2.4)

It follows from (2.3) and (2.4) that for each n ∈ ℕ,

σ(\frac{x_{n} + y_{n}}{2} ) = ∑_{i=1}^{∞} |\frac{x_{n}(i) + y_{n}(i)}{2}|^{p_{i}}

= |\frac{x_{n}(1) + y_{n}(1)}{2}|^{p_{1}} + ∑_{i=2}^{∞} |\frac{x_{n}(i) + y_{n}(i)}{2}|^{p_{i}}

≤ (1 − δ)(\frac{|x_{n}(1)|^{p_{1}} + |y_{n}(1)|^{p_{1}}}{2} ) + \frac{1}{2} ∑_{i=2}^{∞} |x_{n}(i)|^{p_{i}} + \frac{1}{2} ∑_{i=2}^{∞} |y_{n}(i)|^{p_{i}}

= \frac{1}{2} σ(x_{n}) + \frac{1}{2} σ(y_{n}) − δ(\frac{|x_{n}(1)|^{p_{1}} + |y_{n}(1)|^{p_{1}}}{2} )

≤ \frac{1}{2} + \frac{1}{2} − δ \frac{ϵ}{2^{p_{1}+1}} = 1 − δ \frac{ϵ}{2^{p_{1}+1}} .

This implies that σ(\frac{x_{n} + y_{n}}{2} ) ↛ 1 as n → ∞, a contradiction, which finishes the proof. □

Proposition 2.7. Let (x_{n}) ⊆ B(ces(p)) and x ∈ S(ces(p)). If ϱ(\frac{x_{n} + x}{2}) → 1 as n → ∞, then x_{n}(i) → x(i) as n → ∞ for all i ∈ ℕ.

Proof. For each n ∈ ℕ and i ∈ ℕ, let

s_{n}(i) = \begin{cases}sgn(x_{n}(i) + x(i)) & if x_{n}(i) + x(i) ≠ 0, \\ 1 & if x_{n}(i) + x(i) = 0. \end{cases}

Hence we have

1 ← ϱ(\frac{x_{n} + x}{2}) = ∑_{k=1}^{∞} (\frac{1}{k} ∑_{i=1}^{k} |\frac{x_{n}(i) + x(i)}{2}|)^{p_{k}}

= ∑_{k=1}^{∞} (\frac{1}{k} ∑_{i=1}^{k} s_{n}(i)\frac{x_{n}(i)}{2} + \frac{1}{k} ∑_{i=1}^{k} s_{n}(i)\frac{x(i)}{2})^{p_{k}}.  (2.5)

Let a_{n}(k) = \frac{1}{k} ∑_{i=1}^{k} s_{n}(i)x_{n}(i) and b_{n}(k) = \frac{1}{k} ∑_{i=1}^{k} s_{n}(i)x(i) for all n, k ∈ ℕ. Then (a_{n}) ∈ l(p) and (b_{n}) ∈ l(p), and from (2.5) we have

σ(\frac{a_{n} + b_{n}}{2} ) → 1 as n → ∞.

Form Proposition 2.6 we have

a_{n}(i) − b_{n}(i) → 0 as n → ∞  (2.6)

for all i ∈ ℕ. Now we shall show that x_{n}(k) → x(k) as n → ∞ for all k ∈ ℕ. From (2.6) we have

s_{n}(1)x_{n}(1) − s_{n}(1)x(1) → 0 as n → ∞.

This implies x_{n}(1) → x(1) as n → ∞. Assume that x_{n}(i) → x(i) as n → ∞ for all i ≤ k − 1. Then we have

s_{n}(i)(x_{n}(i) − x(i)) → 0 as n → ∞  (2.7)

for all i ≤ k−1. Since s_{n}(k)(x_{n}(k)−x(k)) = k(a_{n}(k)−b_{n}(k))−∑_{i=1}^{k−1} s_{n}(i)(x_{n}(i)− x(i)), it follows from (2.6) and (2.7) that s_{n}(k)(x_{n}(k) − x(k)) → 0 as n → ∞. This implies x_{n}(k) → x(k) as n → ∞. So we have by induction that x_{n}(k) → x(k) as n → ∞ for all k ∈ ℕ. □

Theorem 2.8. The space ces(p) is LUR.

Proof. Let (x_{n}) ⊆ B(ces(p)) and x ∈ S(ces(p)) be such that ∥x_{n} + x∥ → 2

as n → ∞. Then ∥\frac{x_{n} + x}{2}∥ → 1 as n → ∞. By Proposition 2.5 (i) we have

ϱ(\frac{x_{n} + x}{2}) → 1 as n → ∞. By Proposition 2.7 we have x_{n}(i) → x(i) as n → ∞ for all i ∈ ℕ.

Now let ϵ > 0 be given. Then there exist k_{0} ∈ ℕ and n_{0} ∈ ℕ such that

∑_{k=k_{0}+1}^{∞} (\frac{1}{k} ∑_{i=1}^{k} |x(i)|)^{p_{k}} < \frac{ϵ}{3} \frac{1}{2^{M+1}} ,  (2.8)

∑_{k=1}^{k_{0}} (\frac{1}{k} ∑_{i=1}^{k} |x_{n}(i) − x(i)|)^{p_{k}} < \frac{ϵ}{3} for all n ≥ n_{0},  (2.9)

∑_{k=1}^{k_{0}} (\frac{1}{k} ∑_{i=1}^{k} |x_{n}(i)|)^{p_{k}} > ∑_{k=1}^{k_{0}} (\frac{1}{k} ∑_{i=1}^{k} |x(i)|)^{p_{k}} − \frac{ϵ}{3} \frac{1}{2^{M}} for all n ≥ n_{0}. (2.10)

By Proposition 2.4 (i) and (iii) we have ϱ(x_{n}) ≤ 1 for all n ∈ ℕ and ϱ(x) = 1. From these together with (2.8), (2.9), (2.10) and the fact that (a + b)^{p_{k}} ≤ 2^{p_{k}}(a^{p_{k}} + b^{p_{k}}) for a, b ≥ 0 we have that for all n ≥ n_{0},

ϱ(x_{n} − x) = ∑_{k=1}^{∞} (\frac{1}{k} ∑_{i=1}^{k} |x_{n}(i) − x(i)|)^{p_{k}}

= ∑_{k=1}^{k_{0}} (\frac{1}{k} ∑_{i=1}^{k} |x_{n}(i) − x(i)|)^{p_{k}} + ∑_{k=k_{0}+1}^{∞} (\frac{1}{k} ∑_{i=1}^{k} |x_{n}(i) − x(i)|)^{p_{k}}

< \frac{ϵ}{3} + 2^{M}( ∑_{k=k_{0}+1}^{∞} (\frac{1}{k} ∑_{i=1}^{k} |x_{n}(i)|)^{p_{k}} + ∑_{k=k_{0}+1}^{∞} (\frac{1}{k} ∑_{i=1}^{k} |x(i)|)^{p_{k}})

= \frac{ϵ}{3} + 2^{M}(ϱ(x_{n}) − ∑_{k=1}^{k_{0}} (\frac{1}{k} ∑_{i=1}^{k} |x_{n}(i)|)^{p_{k}} + ∑_{k=k_{0}+1}^{∞} (\frac{1}{k} ∑_{i=1}^{k} |x(i)|)^{p_{k}})

≤ \frac{ϵ}{3} + 2^{M}(1 − ∑_{k=1}^{k_{0}} (\frac{1}{k} ∑_{i=1}^{k} |x_{n}(i)|)^{p_{k}} + ∑_{k=k_{0}+1}^{∞} (\frac{1}{k} ∑_{i=1}^{k} |x(i)|)^{p_{k}})

< \frac{ϵ}{3} + 2^{M}(1 − ∑_{k=1}^{k_{0}} (\frac{1}{k} ∑_{i=1}^{k} |x(i)|)^{p_{k}} + \frac{ϵ}{3} \frac{1}{2^{M}} + ∑_{k=k_{0}+1}^{∞} (\frac{1}{k} ∑_{i=1}^{k} |x(i)|)^{p_{k}})

= \frac{ϵ}{3} +2^{M}(ϱ(x)−∑_{k=1}^{k_{0}} (\frac{1}{k} ∑_{i=1}^{k} |x(i)|)^{p_{k}} + \frac{ϵ}{3} \frac{1}{2^{M}} + ∑_{k=k_{0}+1}^{∞} (\frac{1}{k} ∑_{i=1}^{k} |x(i)|)^{p_{k}})

= \frac{ϵ}{3} + 2^{M}( ∑_{k=k_{0}+1}^{∞} (\frac{1}{k} ∑_{i=1}^{k} |x(i)|)^{p_{k}} + \frac{ϵ}{3} \frac{1}{2^{M}} + ∑_{k=k_{0}+1}^{∞} (\frac{1}{k} ∑_{i=1}^{k} |x(i)|)^{p_{k}})

= \frac{ϵ}{3} + 2^{M}(2 ∑_{k=k_{0}+1}^{∞} (\frac{1}{k} ∑_{i=1}^{k} |x(i)|)^{p_{k}} + \frac{ϵ}{3} \frac{1}{2^{M}} )

= \frac{ϵ}{3} + 2^{M+1} ∑_{k=k_{0}+1}^{∞} (\frac{1}{k} ∑_{i=1}^{k} |x(i)|)^{p_{k}} + \frac{ϵ}{3} < \frac{ϵ}{3} + \frac{ϵ}{3} + \frac{ϵ}{3} = ϵ.

This shows that ϱ(x_{n} − x) → 0 as n → ∞. By Proposition 2.5(ii) we have ∥x_{n} − x∥ → 0 as n → ∞. This completes the proof of the theorem. □

It is known in general that a locally uniformly rotund space has property (H). So we have the following result.

Corollary 2.9. The space ces(p) possesses property (H).

Acknowledgements

The author would like to thank the Thailand Research Fund for the financial support and is very indebted to the referee for his valuable comments and suggestions.

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(Received 20.11.2001; revised 27.05.2002)

Author’s address:

Department of Mathematics Faculty of Science Chiang Mai University, Chiang Mai, 50200 Thailand E-mail: suantai@yahoo.com