cal-pisane 8pisane podwojne podwojne

by

S. Chen, Y. Cui, H. Hudzik and B. Sims

Henryk Hudzik

Abstract

.

Keywords and Phrases: Cesaro sequence space, packing constant. AMS Classification: 46B20, 46E30.

1. INTRODUCTION

The aim of this chapter is to present criteria for the most important geometric properties related to the metric fixed point theory in some classes of Banach function lattices, mainly in Orlicz spaces and Cesaro sequence spaces. We also give some informations about respective results for Musielak-Orlicz spaces, Orlicz Lorentz spaces and Calderon-Lozanovsky spaces.

Orlicz Spaces

0. Some general facts

We denote by ℝ, ℝ_{+} and N the sets of real numbers, nonnegative real numbers and natural numbers respectively. A mapping Φ: ℝ → ℝ_{+} is said to be an Orlicz function if:

Φ is even, continuous, convex and vanisking only at zero,

lim_{u→0}(Φ(u)/u) = 0 and lim_{u→∞}(Φ(u)/u) = ∞. It is well known (see [60], [61],

[63] and [69]) that Φ is an Orlicz function if and only if Φ(u) = ∫_{0}^{|u|} p(t)dt, where p is the right derivative of Φ satisfying the conditions: p is right-continuous and nondecreasing on ℝ_{+}, p(t) > 0 whenever t > 0, p(0) = 0 and lim_{t→∞} p(t) = ∞. Hence it follows immediately that

(0.1)  \frac{1}{2}p(\frac{u}{2}) ≤ \frac{Φ(u)}{u} ≤ p(u)  (u > 0).

By the convexity of Φ and Φ(0) = 0, we get

(0.2)  Φ(αu) < αΦ(u) (0 < α < 1, u > 0),

which yields

\frac{Φ(u)}{u} < \frac{Φ(v)}{v} (0 < u < v).

For the function p satisfying conditions (iii),(iv) and (v), we define

q(s) = sup{t > 0: p(t) ≤ s} = inf{t > 0: p(t) > s},

calling the right-inverse function of p. It is easy to show that q also satisfies conditions (iii), (iv) and (v). If Φ is an Orlicz function with the right derivative p and q is the right-inverse function of p, then the function

Ψ(v) = ∫_{0}^{|v|} q(s)ds

is called the complementary function of Φ (or the Young conjugate of Φ). It is well known (see [10], [60], [61], [63] and [69]) that we have the Young inequality

uv ≤ Φ(u) + Ψ(v) (u, v ≥ 0)

and that the equality

(0.3)  uv = Φ(u) + Ψ(v)

holds for u ≥ 0 if and only if v ∈ [p_{−}(u), p(u)], where p_{−} is the lefth derivative of Φ. Sometimes Orlicz functions are defined only by condition (i). It is easy to see that (Φ(u)/u) → 0 as u → 0 is equivalent to the fact that Ψ vanishes only at zero and (Φ(u)/u) → ∞ as u → ∞ is equivalent to the fact that Ψ has only finite values.

Example 0.1. Let Φ be an Orlicz function. If Ψ_{1} is the Young conjugate of the function Φ_{1} defined on ℝ by Φ_{1}(u) = aΦ(bu), where a, b are fixed positive numbers and p is the right derivative of Φ on ℝ_{+}, then the right derivative of Φ_{1}, is p_{1}(t) = abp(bt) and so its right-inverse function is

q_{1}(s) = \frac{1}{b}q(\frac{1}{ab}s),

where q is the right derivative of Ψ, and Ψ is the Young conjugate of Φ. Hence

Ψ_{1}(v) = ∫_{0}^{|v|} q_{1}(s)ds = a ∫_{0}^{|v|/(ab)} q(s)ds = aΨ(\frac{|v|}{ab}).

Example 0.2. Let Ψ_{1}, Ψ_{2} be the Young conjugates of Orlicz functions Φ_{1} and Φ_{2}, respectively. Suppose that

Φ_{1}(u) ≤ Φ_{2}(u) (u ≥ u_{0} ≥ 0).

Consider the relationship between Ψ_{1} and Ψ_{2}. By the Young inequality and equality (0.3), we have

Ψ_{2}(q_{2}(v)) + Ψ_{2}(v) = q_{2}(v)v ≤ Ψ_{1}(q_{2}(v)) + Ψ_{1}(v) (v ≥ 0).

Hence by

Φ_{2}(q_{2}(v)) ≥ Ψ_{1}(q_{2}(v)) (q_{2}(v) ≥ u_{0}),

we obtain

Ψ_{2}(v) ≤ Ψ_{1}(v) (q_{2}(v) ≥ u_{0}).

Let (T, Σ, μ) denote a nonatomics complete and finite measure space and L^{0} = L^{0}(T, Σ, μ) denote the space of all (equivalence classes of) Σ-measurable real functions defined on T. Given Orlicz function Φ, we define on L^{0} a convex modular I_{Φ} by

I_{Φ}(x) = ∫_{T} Φ(x(t))dμ.

The Orlicz space L^{Φ} generated by Φ is the set of those x ∈ L^{0} that I_{Φ}(λx) < ∞ for some λ > 0. If l^{0} is the space of all real sequences x = (x(i))_{i=1}^{∞}, then the modular I_{Φ} is defined on l^{0} by

I_{Φ}(x) = ∑_{i=1}^{∞} Φ(x(i))

and the corresponding space

l^{Φ} = {x ∈ l^{0}: I_{Φ}(λx) < ∞ for some λ > 0} is called the Orlicz sequence space. We also define

\begin{aligned}E^{Φ} & = \{x ∈ L^{0}: I_{Φ}(λx) < ∞ λ > 0\}, \\ h^{Φ} & = \{x ∈ l^{0}: I_{Φ}(λx) < ∞ λ > 0\}. \end{aligned}

Definition 0.1. An Orlicz function Φ is said to satisfy condition Δ_{2}(Φ ∈ Δ_{2}) if there exist u_{0} > 0 and K > 2 such that Φ(2u) ≤ KΦ(u)(u ≥ u_{0}). Lemma 0.1. Let Φ be an Orlicz function and Ψ be its Young conjugate. Then the following are equivalent:

0Φ ∈ Δ_{2}.

0There exist l > 1, u_{0} > 0 and K > 1 such that

(0.4)  Φ(lu) ≤ KΦ(u) (u > u_{0})

0For any l_{1} > 1 and u_{1} > 0 there exists K′ > 0 such that holds for l = l_{1}, u_{0} = u_{1} and K = K′. 0For any l_{2} > 1 and u_{2} > 0 there exists ε in the interval (0, 1) such that

(0.5)  Φ((1 + ε)u) ≤ l_{2}Φ(u) (u > u_{2}).

0For any l_{3} > 1 there exist v_{0} > 0 and δ > 0 such that

(0.6)  Ψ(l_{3}v) ≥ (l_{3} + δ)Ψ(v) (v ≥ v_{0}).

0There exist l_{3} > 1, v_{0} > 0 and δ > 0 such that holds.

P r o o f. The implication 1^{∘} ⇒ 2^{∘} is obvious. 2^{∘} ⇒ 3^{∘}. Given l_{1} > 1, choose an integer α such that l^{α} > l_{1}. Then by (0.4),

Φ(l_{1}u) ≤ Φ(l^{α}u) ≤ K^{α}Φ(u) (u ≥ u_{0}).

Hence, if u_{1} ≥ u_{0}, then K^{α} is a candidate for K′. If u_{1} < u_{0}, then, we choose K′ = max(K^{α}, K_{0}), where

K_{0} = max{Φ(l_{1}u)/Φ(u): u ∈ [u_{1}, u_{0}]}.

3^{∘} ⇒ 4^{∘}. For l_{2} > 1 and u_{2} > 0, by 3^{∘}, there exists K′ > l_{2} such that

Φ(2u) ≤ K′Φ(u) (u ≥ u_{2}).

Take ε = (l_{2} − 1)/(K′ − 1). Then 0 < ε < 1 and by the convexity of Φ,

\begin{aligned}Φ((1 + ε)u) & = Φ((1 − ε)u + 2εu) ≤ (1 − ε)Φ(u) + εΦ(2u), \\ & ≤ (1 − ε)Φ(u) + εK′Φ(u) = l_{2}Φ(u) (u ≥ u_{2}). \end{aligned}

4^{∘} ⇒ 5^{∘}. For any l_{3} > 1 and v_{0} > 0, choose u_{2} ∈ (0, q(v_{0})]. Then inequality (0.5) and Examples 1 and 2 imply

Ψ(v) ≤ \frac{1}{l_{3}} Φ(\frac{l_{3}}{1 + ε}v) (v ≥ v_{0}).

Hence it follows that

l_{3}Ψ(v) ≤ Ψ(\frac{l_{3}}{1 + ε}v) ≤ \frac{1}{1 + ε}Ψ(l_{3}(v)) (v ≥ v_{0}).

By setting δ = l_{3}ε, we get (0.6). The implication 5^{∘} ⇒ 6^{∘} is trivial.

6^{∘} ⇒ 1^{∘}. Let β = (l_{3} + δ)/l_{3}. Then (0.6) can be written in the form

\frac{1}{βl_{3}} Ψ(l_{3}v) ≥ Ψ(v) (v ≥ v_{0}).

Choose n ∈ N such that β^{n} ≥ 2 and set K = β^{n}l_{3}^{n}. Then by Examples 0.1 and 0.2, we get

Φ(2u) ≤ Φ(β^{n}u) ≤ β^{n}l_{3}^{n}Φ(u) = KΦ(u) (u ≥ u_{0}).

Corollary 0.1. Set v = 0 in the inequality which defines uniform convexity of an Orlicz function Φ. Then inequality is satisfied for Φ in place of Ψ. Therefore, by Lemma 0.2, Ψ ∈ Δ_{2} (respectively at ∞ or on ℝ_{+}) whenever it is uniformly convex respectively at ∞ or on ℝ_{+}.

1. Normal structure, weakly normal structure, weak sum property, sum property and uniformly normal structure

Let X be a Banach space and {x_{n}} a sequence in X such that x_{i} ≠ x_{j} whenever i ≠ j. If for any x ∈ co {x_{n}}, the convex hull of {x_{n}}, the limit △(x) = lim_{n} ∥x − x_{n}∥ > 0 exists and △(x) is affine on co {x_{n}}, then {x_{n}} is called a limit affine sequence. If, in particular, △(x) is equal to a constant on co {x_{n}}, then {x_{n}} is called a limit constant sequence. If X contains no (weakly convergent) limit affine sequence {x_{n}} satisfying △(x_{n}) ↑, then it is said to have the (weak) sum-property. X is said to have (weakly) normal structure (NS (WNS)) if it contains no (weakly convergent) limit constant sequence. The original definition of the (weakly) normal structure is given in the following equivalent way: X has (weakly) normal structure if for any nonsingleton (weakly compact) nonempty bounded closed convex subset C of X, there exists x ∈ C such that

r_{c}(x) := sup{∥x − y∥: y ∈ C} < diam (C) := sup{∥u − v∥: u, v ∈ C}.

Normal structure will be denoted by NS and weakly normal structure by w-NS. If moreover, there exists h ∈ (0, 1) such that for each nonsingleton nonempty bounded closed convex subset C, there exists x ∈ C such that r_{c}(x) ≤ (1−h) diam (C), then X is said to have uniformly normal structure (UNS). The above concepts are introduced as a powerful tool in fixed point theory. For instance, if X has weakly normal structure, then it has weakly fixed point property (w-FPP), i.e., any nonexpansive self-mapping defined on a weakly compact convex nonempty subset of X has a fixed point (see [54], [27] and [33]).

In this section we will consider Orlicz spaces over a finite nonatomic measure space only.

Theorem 1.1. Let X be equal to one of the space L_{Φ}, L_{Φ}^{0}, l_{Φ} or l_{Φ}^{0}. Then X has UNS iff it is reflexive.

To prove this theorem, we need the following lemmas.

Lemma 1.2. Suppose Φ ∈ △_{2}. Then for any β > 1 and ε > 0, there exists K ≥ 2 such that for all x ∈ L_{Φ},

I_{Φ}(βx) ≤ KI_{Φ}(x) + ε

P r o o f. Let α > 0 satisfy Ψ(βα)μT < ε. Then since Φ ∈ △_{2}, there exists K ≥ 2 such that Φ(βu) ≤ KΦ(u) for all u ≥ α. For given x ∈ L_{Φ},set F = {t ∈ T : |x(t)| ≥ α}. Then

I_{Φ}(βx) = I_{Φ}(βx|_{F}) + I_{Φ}(βx|_{T \F}) ≤ KI_{Φ}(x|_{F}) + Φ(βα)μ(T \ F) ≤ KI_{Φ}(x) + ε,

where x|_{F} = xχ_{F}.

Lemma 1.3. Assume Φ ∈ △_{2} ∩ ∇_{2}. Then for any α > 0, there exist c > 1 and δ > 0 such that

Φ(\frac{u + v}{2} ) ≤ \frac{1 − δ}{2} [Φ(u) + Φ(v)]

whenever |u| ≥ α and |u| ≥ c|v|,or uv ≤ 0.

P r o o f. By Lemma 0.1, there exist γ > 0,and ε ∈ (0, 1/2) such that

Φ(\frac{w}{2} ) ≤ \frac{1 − γ}{2} Φ(w) (|w| ≥ α)

and

Φ((1 + ε)w) ≤ \frac{2}{2 − γ} Φ(w) (|w| ≥ α).

Set

c = \frac{1}{ε}

,

δ = 1 − \frac{2 − 2γ}{2 − γ}

. Then |u| ≥ α, |u| ≥ c|u| or uv ≤ 0 imply

\begin{aligned}Φ(\frac{u + v}{2} ) & ≤ Φ(\frac{1 + c^{−1}}{2} u) ≤ \frac{1 − γ}{2} Φ((1 + \frac{1}{c})u) \\ & ≤ \frac{1 − γ}{2} \frac{2}{2 − γ} Φ(u) ≤ \frac{1 − δ}{2} [Φ(u) + Φ(v)]. \end{aligned}

Lemma 1.4. If a Banach space X does not have UNS, then for each n ∈ N and ε > 0, there exists a family {x_{i} : 1 ≤ i ≤ n + 1} in X such that

∥x_{j}∥ ≤ 1, ∥x_{i} − x_{j}∥ ≤ 1 (1 ≤ i ≤ j ≤ n + 1)

and

∥x_{m+1} − \frac{1}{m} ∑_{i=1}^{m} x_{i}∥ > 1 − ε (m = 1, ⋯, n)

.

P r o o f. By the assumption, there exists a bounded nonempty closed convex subset C of X such that for each z ∈ C, there exists x ∈ C satisfying ∥z − x∥ > (1−ε)diam C. Without loss of generality ,we may assume that 0 ∈ C and diamC = 1, i.e., ∥x∥ ≤ 1 and ∥x − y∥ ≤ 1 for all x, y ∈ C. Pick x_{1} ∈ C arbitrarily. Then, by the hypothesis, there exists x_{2} ∈ C such that ∥x_{2} − x_{1}∥ > 1 − ε. Since C is convex, \frac{x_{1}+x_{2}}{2} ∈ C. Therefore, there exists x_{3} ∈ C such that ∥x_{3} − \frac{x_{1}+x_{2}}{2} ∥ > 1 − ε, and so on, by induction, we can choose the desired system of elements.

P r o o f of Theorem 1.1. We only prove the theorem for X = L_{Φ}. The proofs for other spaces are analogous. Since all Banach spaces with UNS are reflexive,we only need to show the sufficiency. By Lemmas 1.2 and 1.3, there exist K ≥ 2, b > 0, c > 1 and δ > 0 such that

(1.1)  I_{Φ}(2x) ≤ KI_{Φ}(x) + \frac{1}{8} (x ∈ L_{Φ}),
(1.2)  Φ(b)μT ≤ \frac{1}{8K}

and

(1.3.)  Φ(\frac{u + v}{2} ) ≤ \frac{1 − δ}{2} [Φ(u) + Φ(v)] (|u| ≥ b, |u| ≥ c|v|, uv ≤ 0)

Select an integer p > 16c^{2}K^{2} and set n = 8p. If L_{Φ} fails to have UNS, then Lemma 1.4 and Φ ∈ △_{2} yield the existence of {x_{i}}, 1 ≤ i ≤ n + 1, such that

(1.4)  I_{Φ}(x_{i}) ≤ 1, I_{Φ}(x_{i} − x_{j}) ≤ 1 (1 ≤ i ≤ j ≤ n + 1)
(1.5)  1 ≥ I_{Φ}(x_{m+1} − \frac{1}{m} ∑_{i=1}^{m} x_{i}) > 1 − \frac{δ}{4n^{2}K} .

Now, we first introduce some notatiions. Set u_{i}(t) = x_{n+1}(t) − x_{i}(t)(i ≤ n) and for each t ∈ T, rearrange {u_{i}(t)}_{i≤n} into {y_{s}(t) = u_{i_{s(t)}}(t)} such that y_{1}(t) ≤ ⋯ ≤ y_{n}(t). Then it is not difficult to check that each y_{s}(t) is Σ-measurable. Moreover, define

\begin{aligned}x(t) & = \frac{[y_{4p}(t) + y_{4p+1}(t)]}{2}, x_{0}(t) = \frac{2}{n} ∑_{i=1}^{n} |u_{i}(t)|, \\ I(t) & = \{i ≤ n : |u_{i}(t)| > c|x(t)| c|u_{i}(t)| < x(t) u_{i}(t)x(t) ≤ 0\}, \\ A & = \{t ∈ T : I(t) 4p \}, B = T \ A. \end{aligned}

Then

(16)  |x(t)| ≤ max{|y_{s}(t)|, |y_{4p+s}(t)|} ≤ x_{0}(t).

Moreover, (1.1), (1.4) and the convexity of Φ imply

(1.7)  I_{Φ}(x_{0}) ≤ KI_{Φ}(\frac{1}{n} ∑_{i=1}^{n} |u_{i}|) + \frac{1}{8} ≤ \frac{K}{n} ∑_{i=1}^{n} I_{Φ}(u_{i}) + \frac{1}{8} < K + \frac{1}{8}.

In the first step,we show that

(1.8)  ∫_{B} Φ(\frac{x_{1}(t) − x_{2}(t)}{2})dt > \frac{1}{2K} .

Since (1.4) and (1.1) yield

\frac{7}{8} < 1 − \frac{δ}{4n^{2}K} < I_{Φ}(x_{1} − x_{2}) ≤ KI_{Φ}(\frac{x_{1} − x_{2}}{2} ) < K + \frac{1}{8}

i.e., I_{Φ}(\frac{x_{1}−x_{2}}{2} ) > \frac{3}{4K}, to verify (1.8) it suffices to show that

∫_{A} Φ(\frac{x_{1}(t) − x_{2}(t)}{2})dt < \frac{1}{4K} .

For this purpose, we first check that t ∈ A implies

|y_{s}(t)| > c|y_{4p+s}(t)| c|y_{s}(t)| < |y_{4p+s}(t)| y_{s}(t)y_{4p+s}(t) ≤ 0

for each s ≤ 4p. In fact, if there exist some j ≤ 4p and t ∈ A such that none of the above three inequalities holds, then we get

c^{−1}y_{4p+j}(t) ≤ y_{j}(t) ≤ cy_{4p+j} c^{−1}y_{4p+j}(t) ≥ y_{j}(t) ≥ cy_{4p+j}(t)
Since x(t) is between y_{j}(t) and y_{4p+j}(t), we derive that
c^{−1}x(t) ≤ y_{s}(t) ≤ cx(t) c^{−1}x(t) ≥ y_{s}(t) ≥ cx(t)

for all s = j, j +1, ⋯, 4p+j, which contradicts the definition of A. Hence if we define for each s ≤ 4p,

A(s) = {t ∈ A : max{|y_{s}(t)|, |y_{4p+s}(t)|} > b},

then (1.3) and the convexity of Φ imply

1 − \frac{δ}{4n^{2}K} < I_{Φ}(x_{n+1} − \frac{1}{n} ∑_{i=1}^{n} x_{i})

= I_{Φ}(\frac{2}{n} ∑_{s=1}^{4p} \frac{y_{s} + Y_{4p+s}}{2} ) ≤ \frac{2}{n} ∑_{s=1}^{4p} I_{Φ}(\frac{y_{s} + Y_{4p+s}}{2} )

≤ \frac{2}{n} ∑_{s=1}^{4p} \frac{1}{2} ∫_{G/A(s)} [Φ(y_{s}(t)) + Φ(y_{4p+s}(t))]dt

+ \frac{2}{n} ∑_{s=1}^{4p} \frac{1}{2}(1 − δ) ∫_{A(s)} [Φ(y_{s}(t)) + Φ(y_{4p+s}(t))]dt

= \frac{1}{n} ∑_{i=1}^{n} I_{Φ}(u_{i}) − \frac{δ}{n} ∑_{s=1}^{4p} ∫_{A(s)} [Φ(y_{s}(t)) + Φ(y_{4p+s}(t))]dt.

It follows from (1.4) that

(1.9)  ∑_{s=1}^{4p} ∫_{A(s)} [Φ(y_{s}(t)) + Φ(y_{4p+s}(t))]dt < \frac{1}{4nK} < \frac{1}{8K} .

Now, we define

\begin{aligned}D_{i} & = \{t ∈ A : |u_{i}(t) > b\} (i = 1, 2), \\ B_{i}(s) & = \{t ∈ A : u_{i}(t) = y_{s}(t) y_{4p+s}(t)\} (i = 1, 2). \end{aligned}

Then from (1.2), (1.9) and the fact that ∪_{s=1}^{4p}B_{i}(s) = A, D_{i} ∩ B_{i}(s) ⊂ A(s)(i = 1, 2), we derive

∫_{A} Φ(\frac{x_{1}(t) − x_{2}(t)}{2})dt = ∫_{A} Φ(\frac{u_{1}(t) − u_{2}(t)}{2})dt ≤ \frac{1}{2} ∑_{i=1}^{2} ∫_{A} Φ(u_{i}(t))dt

≤ \frac{1}{2} ∑_{i=1}^{2} ∫_{D_{i}} Φ(u_{i}(t))dt + Φ(b)μA

= \frac{1}{2} ∑_{i=1}^{2} ∑_{s=1}^{4p} ∫_{D_{i}∩B_{i}(s)} Φ(u_{i}(t))dt + Φ(b)μA

≤ \frac{1}{2} ∑_{i=1}^{2} ∑_{s=1}^{4p} ∫_{D_{i}∩B_{i}(s)} [Φ(y_{s}(t) + Φ(y_{4p+s}(t))]dt + Φ(b)μA

< \frac{1}{8K} + \frac{1}{8K} = \frac{1}{4K} .

This ends the proof of inequality (1.8).

In the second step, we set for each i = 3, ⋯, n − 1,

G(i) = {t ∈ B : |x_{s}(t) − x_{i}(t)| ≤ c|x(t)|/p s > i s ≤ n}.

Then

(1.10)  ⋃_{i=3}^{n−1} G(i) = B.

In fact, for any t ∈ B = T \ A, by the definition of A, there exist at least five u_{i}(t) such that their distance from each other is no more than c|x(t)|/p, and thus, there exist i, j; 3 ≤ i < j ≤ n, such that

|u_{i}(t) − u_{j}(t)| ≤ c|x(t)| ≤ c|x(t)|/p,

i.e. t ∈ G(i).This proves (1.10). Now, we define

D(3) = G(3), D(i) = G(i) / ⋃_{k=3}^{i−1} G(k) (i = 4, ⋯, n − 1).

Then {D(i)}_{i=3}^{n−1} are pairwise disjoint and ∪_{i=3}^{n−1}D(i) = B. Let, for each i = 3, ⋯, n− 1 and each t ∈ D(i),

i′(t) = i, i′′(t) = max{k ≤ n : |x_{k}(t) − x_{i}(t)| ≤ c|x_{(}t)|/p}.

Then i′(t), i′′(t) are well defined by the definition of G(i) and i′(t) < i′′(t). Next, we construct two Σ-measurable functions as follows:

x′(t) = ∑_{i=3}^{n−1} x_{i′(t)}(t)_{χ_{D(i)}}(t), x′′(t) = ∑_{i=3}^{n−1} x_{i′′(t)}(t)_{χ_{D(i)}}(t).

Then by (1.6) and the definition of i′(t), i′′(t),

(1.11)  |x′(t) − x′′(t)| ≤ c|x(t)|/p ≤ cx_{0}(t)/p.

Since (3.8) and the convexity of M imply

1_{\overline{2 ∫_{B} [Φ(x′′(t)−x_{1}(t))+Φ(x′′(t)−x_{2}(t))]dt≥∫_{B} Φ( \frac{x_{1}(t)−x_{2}(t)}{2} dt> \frac{1}{2K} ,}}

without loss of generality, we assume that

(1.12)  ∫_{B} Φ(x′′(t) − x_{1}(t))dt > \frac{1}{2K} .

Finally, let

E = {t ∈ B : |x′′(t) − x_{1}(t)| ≥ max{b, c^{2}x_{0}(t)/p}}.

Then by (1.11), t ∈ E implies |x′′(t) − x_{1}(t)| ≥ c^{2}x_{0}(t)/p ≥ |x′(t) − x′′(t)|. It follows from (1.3) that if t ∈ E, then

(1.13) Φ(\frac{x′′(t) − x′(t) + x′′(t) − x_{1}(t)}{2}) ≤ \frac{1 − δ}{2} [Φ(x′′(t)−x′(t))+Φ(x′′(t)−x_{1}(t))].

Moreover, (1.12), (1.3), (1.1), (1.7) and the inequality c^{2}/p < (4K)^{−2} < 1 imply that

(1.14) ∫_{E} Φ(x′′(t) − x_{1}(t))dt = ∫_{B} Φ(x′′(t) − x_{1}(t))dt − ∫_{D/E} Φ(x′′(t) − x_{1}(t))dt

> \frac{1}{2K} − [ ∫_{B/E} Φ(\frac{c^{2}x_{0}(t)}{p})dt + Φ(b)μ(B/E)]

≥ \frac{1}{2K} − [\frac{c^{2}}{p} ∫_{G} Φ(x_{0}(t))dt + \frac{1}{8K}]

≥ \frac{1}{2K} − [\frac{c^{2}}{p} (K + \frac{1}{8}) + \frac{1}{8K}]

> \frac{1}{2K} − \frac{1}{8K} − \frac{1}{8K} = \frac{1}{4K} .

In light of (1.13) and the convexity of Φ, for all t ∈ E, we have

∑_{m=2}^{n} Φ(\frac{1}{m − 1} ∑_{k=1}^{m−1}(x_{m}(t) − x_{k}(t)))

= ∑_{{2≤m≤n \atop m≠i”(t)}} Φ(\frac{1}{m − 1} ∑_{k=1}^{m−1}(x_{m}(t) − x_{k}(t)))

+ Φ(\frac{1}{i”(t) − 1}[ ∑_{{2≤k≤i”(t)−1 \atop k≠i′(t)}}(x′′(t) − x_{k}(t)) + 2\frac{x′′(t) − x′(t) + x′′(t) − x_{1}(t)}{2}])

≤ ∑_{{2≤m≤n \atop m≠i”(t)}} Φ(\frac{1}{m − 1} ∑_{k=1}^{m−1}(x_{m}(t) − x_{k}(t)))

+ \frac{1}{i”(t) − 1}[ ∑_{{2≤k≤i”(t)−1 \atop k≠i′(t)}} Φ(x′′(t) − x_{k}(t)) + 2Φ(\frac{x′′(t) − x′(t) + x′′(t) − x_{1}(t)}{2})]

≤ ∑_{m=2}^{n} \frac{1}{m − 1} ∑_{k=1}^{m−1} Φ(x′′(t) − x_{k}(t))

− \frac{δ}{i”(t) − 1}[Φ(x′′(t) − x′(t)) + Φ(x′′(t) − x_{1}(t))].

Combining this with (1.5), (1.14) and x′′(t) − 1 ≤ n − 1, we deduce

1 − \frac{δ}{4n^{2}K} < \frac{1}{n − 1} ∑_{m=2}^{n} I_{Φ}(x_{m} − \frac{1}{m − 1} ∑_{k=1}^{m−1} x_{k})

≤ \frac{1}{n − 1} ∑_{m=2}^{n} \frac{1}{m − 1} ∑_{k=1}^{m−1} I_{Φ}(x_{m} − x_{k})

− \frac{δ}{n(n − 1)} ∫_{E} [Φ(x′′(t) − x′(t)) + Φ(x′′(t) − x_{1}(t))]dt < 1 − \frac{δ}{4n^{2}K} .

This contradiction completes the proof. Lemma 1.5. Suppose x_{n} ∈ B(L_{Φ}) for n ∈ ℕ and x_{n} → x in measure. Then

x ∈ B(L_{Φ}).

P r o o f. Since I_{Φ}(x_{n}) ≤ 1 for all n ∈ ℕ, by the Fatou Lemma, we have I_{Φ}(x) ≤ lim inf_{n} I_{Φ}(x_{n}) ≤ 1.

Theorem 1.2. The spaces L_{Φ}^{0} and l_{Φ}^{0} have the weak sum-property, so they have WNS.

We will proove this theorem together with the next theorem.

Lemma 1.6. If {x_{n}} is a bounded sequence in L_{Φ}^{0}, k_{n} ∈ K(x_{n}) for n = 1, 2, …, and k_{n} → ∞, then x_{n} → 0 in measure.

P r o o f. For each σ > 0, define G_{n} = {t ∈ T: |x_{n}(t)| ≥ σ}. Then

∥x_{n}∥^{0} = \frac{1}{k_{n}} [1 + I_{Φ}(k_{n}x_{n})] ≥ \frac{1}{k_{n}} Φ(k_{n}σ)μ(G_{n}).

Applying the fact that Φ(u)/u → ∞ as u → ∞, we get μ(G_{n}) → 0.

Lemma 1.7. If {x_{n}} is a bounded sequence in L_{Φ} and it converges to zero in measure, then x_{n} → 0 E_{Ψ}-weakly, where Ψ is the Young conjugate of Φ. If {x_{n}} is a bounded sequence in l_{Φ} and it converges to zero coordinatewise, then x_{n} → 0 h_{Ψ}-weakly.

P r o o f. We only prove (i) because the proof of (ii) is analogous. Suppose that ∥x_{n}∥ ≤ K for any n ∈ ℕ. For any v ∈ E_{Ψ} and ε > 0, choose δ > 0 such that E ∈ Σ and μ(E) < δ imply ∥xχ_{E}∥_{Ψ}^{0} < ε. Since x_{n} → 0 in measure, we cand find G_{n} ∈ Σ with μ(G_{n}) < δ such that |x_{n}(t)| < ε on T \ G_{n} for all large n. Hence, for such n,

\begin{aligned}|⟨v, x_{n}⟩| & ≤ ∫_{T \G_{n}} |x_{n}(t)v(t)|dμ \\ & ≤ ε∥χ_{T} ∥ ∥v∥_{Ψ}^{0} + ∥x_{n}∥ ∥vχ_{G_{n}}∥_{Ψ}^{0} ≤ ε∥χ_{T} ∥ ∥v∥_{Ψ}^{0} + εK. \end{aligned}

Since ε > 0 is arbitrary, this shows that ⟨v, x_{n}⟩ → 0. In the following by SAI of Φ we denote an interval [a, b] such that Φ is affine on [a, b] but it is not affine neither on [a − ε, a] nor on [b, b + ε], where ε > 0.

Theorem 1.3. Let X be one of the spaces L_{Φ}^{0}, E_{Φ}^{0}, l_{Φ}^{0} or h_{Φ}^{0}. Then the following are equivalent: X has the sum-property. X has NS. There exist a > 0 and c > 1 such that for any SAI [u, v] of Φ, u > a ⇒ v ≤ cu (when X = L_{Φ}^{0} or E_{Φ}^{0}), 0 < u ≤ a ⇒ v ≤ cu (when X = l_{Φ}^{0} or h_{Φ}^{0}).

P r o o f of Theorems 1.2 and 1.3. We only prove the theorems for X = L_{Φ}^{0} and X = E_{Φ}^{0}. Let {x_{n}} be a limit affine sequence in L_{Φ}^{0} and k_{ij} ∈ K(x_{i} −x_{j})(i ≠ j). First we show that there exists a subsequence N_{1} of N such that for any j ∈ N_{1}, {k_{ij} : i ∈ N_{1}} is bounded. Indeed, if {k_{ij} : i ∈ N} is bounded for all j ∈ N, then we set N_{1} = N. Otherwise, there exist some m ∈ N and a subsequence I of N such that k_{im} → ∞ as i(∈ I) → ∞. This shows that{x_{i} : i ∈ I} converges to x_{m} in measure according to Lemma 1.6. Since x_{i} ≠ x_{j} for all i ≠ j, by the same reason, we get that N_{1} = I \ {m} satisfies our requirement. By the diagonal method, we can pick a subsequence N_{2} of N_{1} such that k_{ij} → k < ∞ as i(∈ N_{2}) → ∞ for each j ∈ N_{1}. We claim that k_{j} → ∞ as i(∈ N_{2}) → ∞. In fact, if this is not true, then N_{2} contains a subsequence N_{3} such that k_{j} → k < ∞ as j(∈ N_{3}) → ∞. Therefore, for all n, i, j ∈ N_{3} with n ≠ i, j,

∥x_{n} − x_{i}∥^{0} + ∥x_{n} − x_{j}∥^{0} − ∥2x_{n} − x_{i} − x_{j}∥^{0} ≥ \frac{1}{k_{ni}} [1 + I_{Φ}(k_{ni}(x_{n} − x_{i}))] + \frac{1}{k_{nj}} [1 + I_{Φ}(k_{nj}(x_{n} − x_{j}))]

− \frac{k_{ni} + k_{nj}}{k_{ni}k_{nj}} [1 + I_{Φ}(\frac{k_{ni}k_{nj}}{k_{ni} + k_{nj}} (2x_{n} − x_{i} − x_{j}))]

= ∫_{G} \{\frac{1}{k_{ni}} Φ(k_{ni}(x_{(}t) − x_{i}(t))) + \frac{1}{k_{nj}} Φ(k_{nj}(x_{n}(t) − x_{j}(t)))

− \frac{k_{ni} + k_{nj}}{k_{ni}k_{nj}} Φ(\frac{k_{ni}k_{nj}}{k_{ni} + k_{nj}} (2x_{n}(t) − x_{i}(t) − x_{j}(t)))\}dt.

(1.15).

Denote the last integrand in (1.15) by f_{n}^{ij}(t). Then f_{n}^{ij} ≥ 0 for all t ∈ T since Φ is convex. Recall that △(x) is affine on co{x_{k}}. By letting n → ∞ we get ∫_{G} f_{n}^{ij}(t)dt → 0, and thus f_{n}^{ij}(t) → 0 in measure. Hence, the diagonal method allows

us to find a subsequence N_{4} of N_{3} such that f_{n}^{ij}(t) → 0 μ -a.e. on T as n(∈ N_{4}) → ∞ for all i, j ∈ N_{3}. Now, for each t ∈ T, we pick a subsequence {n_{γ} = n_{γ}(t)} of N_{4} such that

(1.16)  |v(t)| = lim inf_{n∈N_{4}} |x_{n}(t)|, lim_{γ} x_{n_{γ}} (t) = v(t).

Then, by the Fatou Lemma, |v(t)| < ∞ μ-a.e. on T, analogously to the proof of Lemma (1.12). Let γ → ∞. Then the convexity of Φ gives

0 = lim_{γ} f_{n_{γ}}^{ij} (t) = \frac{1}{k_{i}} Φ(k_{i}(v(t) − x_{i}(t))) + \frac{1}{k_{j}} Φ(k_{j}(v(t) − x_{j}(t))) −\frac{k_{i}+k_{j}}{k_{i}k_{j}} Φ(\frac{k_{i}k_{j}}{k_{i}+k−_{j}} (2v(t) − x_{i}(t) − x_{j}(t)))

(1.17)

μ-a.e. on T. Since for μ-a.e. t ∈ T, (1.17) holds for all i, j ∈ N_{3}, by replacing j by n_{γ} in (1.17) and letting γ → ∞, we have for μ-a.e. t ∈ T,

(1.18)  \frac{1}{k_{i}} Φ(k_{i}v(t) − x_{i}(t))) = \frac{k_{i} + k_{j}}{k_{i}k_{j}} Φ(\frac{k_{i}k_{j}}{k_{i} + k_{j}} (v(t) − x_{i}(t))).

Since 0 < \frac{k}{K_{i}+k_{j}} < 1, by (1.7), condition (1.18) holds only for v(t) = x_{i}(t). This means v = x_{i} for all i ∈ N_{3}, contradicting the assumption that x_{i} ≠ x_{j} whenever i ≠ j. This contradiction proves that k_{j} → ∞ as j(∈ N_{3}) → ∞. Now, we prove (iii) ⇒ (i) in Theorem 1.3. If L_{Φ}^{0} does not have the sum-property, then there exists a limit affine sequence {y_{n}} in L_{Φ}^{0} such that △(y_{n}) ↑. By the above discussion, {y_{n}} contains a subsequence {x_{n}} satisfying k_{ij} → k_{j} < ∞ as i → ∞, and k_{j} → ∞ as j → ∞, where k_{ij} ∈ K(x_{i} − x_{j}), i ≠ j. Since Φ(u)/u → ∞ as u → ∞, for the constant a > 0 in (iii), we can find b > a such that

Φ(\frac{a + b}{2}) < \frac{Φ(a) + Φ(b)}{2}.

Since Φ is convex, by (iii),

(1.19)  Φ(αu + (1 − α)v) < αΦ(u) + (1 − α)Φ(v)

for all 0 < α < 1 and all u ≤ a, v ≥ b or u ≥ a, v ≥ cu. If we define v(t) as in (1.16), then by (1.16) and (1.19), for μ-a.e. t ∈ T, if k_{i}|v(t) − x_{i}(t)| ≤ α, then k_{j}|v(t) − x_{j}(t)| ≤ b; if k_{i}|v(t) − x_{i}(t)| ≥ α, then k_{j}|v(t) − x_{j}(t)| ≤ ck_{i}|v(t) − x_{i}(t)|. Therefore, for μ-a.e. t ∈ T,

(1.20)  k_{j}|v(t) − x_{j}(t)| ≤ max{b, ck_{j}|v(t) − x_{i}(t)|} =: u_{i}(t).

By the Fatou Lemma,

(1.21)  △(x_{j}) ≥ k_{j}^{−1}[1 + I_{Φ}(k_{j}(v − x_{j}))] ≥ ∥c − x_{j}∥^{0}.

Thus v − x_{j} ∈ L_{Φ}, whence u_{i} ∈ L_{Φ}. Since △ > 0, lim inf_{j} ∥v − x_{j}∥^{0} =: γ > 0. It follows from (1.20) that k_{j} = ∥k_{j}(v − x_{j}∥^{0}/∥v − x_{j}∥^{0} ≤ ∥u_{i}∥^{0}/∥v − x_{j}∥^{0}. Letting j → ∞, we get a contradiction:∞ = ∥u_{i}∥^{0}/γ < ∞.

Now, we turn to Theorem 1.2. If L_{Φ}^{0} does not have the weak sum-property, then by (3.15), there exists a weakly convergent (to zero) limit affine sequence (x_{n}) with ∥x_{n}∥^{0} → 1 and △(x_{n}) → 1. By the first part of the proof, passing to a subsequence if necessary, we may assume that k_{ij} → k_{j} < ∞ as i → ∞ and k_{j} → ∞ as j → ∞, where k_{ij} ∈ K(x_{i} − x_{j}). It follows from (1.21) that x_{j} → v in measure (verified as in the first part of the proof). Therefore, by Lemma 1.7, v = 0. We may also assume that x_{j} → 0 μ-a.e. on T. We prove the theorem by showing that lim △(x_{j}) ≥ \frac{4}{3}, which contradicts the assumption △(x_{j}) → 1. For each j ∈ N, we choose a set G_{j} ∈ ∑ such that x_{j} is bounded on G_{j} and

k_{j}^{−1}[1 + I_{Φ}(k_{j}x_{j}χ_{G_{j}})] > k_{j}^{−1}[1 + I_{Φ}(k_{j}x_{j})] − 1/k_{j}.

Then by (1.21),

\begin{aligned}△(x_{j}) & ≥ k_{j}^{−1}[1 + I_{Φ}(k_{j}x_{j})] = k_{j}^{−1}[1 + I_{Φ}(k_{j}x_{j}χ_{G_{j}})] + k_{j}^{−1}[1 + I_{Φ}(k_{j}x_{j}χ_{T \G_{j}}) − \frac{1}{k_{j}} \\ & > k_{j}^{−1}[1 + I_{Φ}(k_{j}x_{j}] − \frac{1}{k_{j}} + ∥x_{j}χ_{T \G_{j}}∥^{0} − \frac{1}{k_{j}} ≥ ∥x_{j}∥^{0} + ∥x_{j}χ_{T \G_{j}}∥^{)} − \frac{2}{k_{j}} , \end{aligned}

i.e.

(1.22)  ∥x_{j}χ_{T \G_{j}}∥ < △(x_{j}) − ∥x_{j}∥^{0} − \frac{2}{k_{j}} .

It follows that

(1.23)  ∥x_{j}χ_{G_{j}}∥^{0} ≥ ∥x_{j}∥^{0} − ∥x_{j}χ_{T \G_{j}}∥^{0} > 2∥x_{j}∥^{0} − △(x_{j}) − \frac{2}{k_{j}} .

Since x_{j} is bounded on G_{j}, there exists δ = δ(j) > 0 such that

(1.24)  ∥x_{j}χ_{E}∥^{0} < \frac{1}{k_{j}}  E ⊂ G_{j} μE < δ.

Since x_{i} → 0 μ-a.e. on T, there exists F ∈ ∑ with μF < δ such that x_{i} → 0 uniformly on T \ F. Hence, there exists I = I(j) ∈ N such that for all i > I,

(1.25)  ∥x_{i}χ_{T \F} ∥^{0} < \frac{1}{k_{j}} .

It follows that

(1.26)  ∥x_{j}χ_{F} ∥^{0} ≥ ∥x_{i}∥^{0} − ∥x_{i}χ_{T \F} ∥^{0} > ∥x_{i}∥^{0} − \frac{1}{k_{j}} .

Hence, by (1.22)–(1.26),

∥x_{i} − x_{j}∥^{0} = k_{ij}^{−1}[1 + I_{Φ}(k_{ij}(x_{i} − x_{j})χ_{T \(T \F)})]

+ k_{ij}^{−1}[1 + I_{Φ}(k_{ij}(x_{i} − x_{j})χ_{T/F})] − \frac{1}{k_{ij}}

≥ ∥(x_{i} − x_{j})χ_{T \(T \F)}∥^{0} + ∥(x_{i} − x_{j})χ_{T \F} ∥ − \frac{1}{k_{ij}}

≥ ∥x_{i}χ_{T \(T \F)}∥^{0} − ∥x_{j}χ_{T \(T \F)}∥^{0}

+ ∥x_{j}χ_{G_{j}/F} ∥^{0} − ∥x_{i}χ_{G_{j}/F} ∥^{0} − \frac{1}{k_{ij}}

= ∥x_{i}χ_{T \(T \F)}∥^{0} − ∥x_{j}χ_{T \G_{j}} + x_{j}χ_{G_{j}∩F} ∥^{0}

+ ∥x_{j}χ_{G_{j}} − x_{j}χ_{G_{j}∩F} ∥^{0} − ∥x_{i}χ_{G_{j}\F} ∥^{0} − \frac{1}{k_{ij}}

> (∥x_{i}∥^{0} − \frac{1}{k_{j}}) − (△(x_{j}) − ∥x_{j}∥^{0} + \frac{2}{k_{j}} + \frac{1}{k_{j}})

+ (2∥x_{j}∥^{0} − △(x_{j}) − \frac{2}{k_{j}} − \frac{1}{k_{j}}) − \frac{1}{k_{j}} − \frac{1}{k_{ij}}

= ∥x_{i}∥^{0} + 3∥x_{j}∥^{0} − 2△(x_{j}) − \frac{8}{k_{j}} − \frac{1}{k_{ij}} .

Letting i → ∞, we have △(x_{j}) ≥ 1+3∥x_{j}∥^{0} −2△(x_{j})− \frac{9}{k_{j}}. Hence, lim_{j} △(x_{j}) ≥ \frac{4}{3}. Finally, we prove the implication (ii)⇒(iii) of Theorem 1.3. If (iii) does not hold, then there exist sequences {u_{j}}, {v_{j}} such that Φ(u_{1})μT > 1, u_{j+1} > 2^{j}u_{j}, v_{j} > 2^{j}u_{j} and p(u) is a constant on [u_{j}, v_{j}], j ∈ N. By the first two assumptions, we can choose disjoint sets G_{j} ∈ Σ such that μ(T \ ⋃_{j∈N} G_{j}) > 0 and

(1.27)  2^{−j} = u_{j}p(u_{j})μG_{j} = [Φ(u_{j}) + Ψ(p(u_{j}))]μG_{j}.

Hence, we can find u_{0} large enough so that there is G_{0} ⊂ T \ ⋃_{j} G_{j} satisfying

(1.28)  ∑_{j∈N} Ψ(p(u_{j}))μG_{j} + Ψ(p(u_{0}))μG_{0} = 1.

Define

v = ∑_{j≥0} p(u_{j})χ_{G_{j}}, x_{n} = u_{0}χ_{G_{0}} + ∑_{j≤n} v_{j}χ_{G_{j}} + ∑_{j>n} u_{j}χ_{G_{j}}.

Then by (1.28), I_{Φ}(v) = 1, whence v ∈ L_{Φ}^{∗} and ∥v∥_{Ψ} = 1. First we show that

x_{n} ∈ E_{Φ} for any n ∈ N. Given arbitrary K > 1, choose J > n such that 2^{J} > K. Then for all j > J, we have v_{j} > 2^{j}u_{j} > Ku_{j} > u_{j}. Therefore,

\begin{aligned}∑_{j>J} Φ(Ku_{j})μG_{j} & = ∑_{j>J}[Ku_{j}p(Ku_{j}) − Ψ(p(Ku_{j}))]μG_{J} < ∑_{j>J} Ku_{j}p(Ku_{j})μG_{j} \\ & = ∑_{j>J} Ku_{j}p(u_{j})μG_{j} = K ∑_{j>J} 2^{−1} < ∞. \end{aligned}

This implies that I_{Φ}(Kx_{n}) < ∞. Since K > 1 is arbitrary, we have x_{n} ∈ E_{Φ}. Let k_{n} = ∥x_{n}∥^{0} and y_{n} = \frac{x_{n}}{k_{n}}. Then y_{n} ∈ E_{Φ} and ∥y_{n}∥^{0} = 1. By (1.28),

\begin{aligned}∥y_{n}∥^{0} & ≥ ⟨v, y_{n}⟩ = k_{n}^{−1}[u_{0}p(u_{0})μG_{0} + ∑_{j≤n} v_{j}p(u_{j})μG_{j} + ∑_{j>n} u_{j}p(u_{j})μG_{j}] \\ & = k_{n}^{−1}[ϱ_{Ψ}(v) + I_{Φ}(k_{n}y_{n})] ≥ ∥y_{n}∥^{0} = 1. \end{aligned}

Moreover, since

k_{n} = ∥x_{n}∥^{0} ≥ ⟨v, x_{n}⟩ > ∑_{j≤n} v_{j}p(u_{j})μG_{j} ≥ ∑_{j≤n} 2^{j}u_{j}p(u_{j})μG_{j} = n,

we have k_{n} → ∞ as n → ∞. We complete the proof by showing that △ = 2 on co(y_{n}). Indeed, for any y ∈ co(y_{n}), there exist λ_{i} ≥ 0 with ∑_{i≤m} λ_{i} = 1 such that y = ∑_{i≤m} λ_{i}y_{i}. Since ⟨v, y_{n}⟩ = 1, we have ⟨v, y⟩ = ∑_{i≤m} λ_{i}⟨v, y_{n}⟩ = 1. For any ε > 0, since y ∈ E_{Φ}, there exists I > m such that ∥yχ_{F} ∥^{0} < ε, where F = ∪_{i≤I}G_{i}. In view of x_{n}(t) ≤ max{v_{I}, u_{0}} on G\F and k_{n} → ∞ as n → ∞, we can find n_{0} ∈ G_{i} such that ∥y_{n}χ_{T \F} ∥^{0} < ε for all n > n_{0}. Define v_{0} = vχ_{T \F} − vχ_{F}. Then ∥v_{0}∥_{Ψ} and for n > n_{0},

2 ≥ ∥y∥^{0} + ∥y_{n}∥^{0} ≥ ∥y − y_{n}∥^{0} ≥ ⟨v_{0}, y − y_{n}⟩
= ⟨v_{0}, yχ_{T \F} ⟩ + ⟨v_{0}, yχ_{F} ⟩ − ⟨v_{0}, y_{n}χ_{T \F} ⟩ − ⟨v, y_{n}χ_{F} ⟩
= ⟨v, yχ_{T \F} ⟩ − ⟨v, yχ_{F} ⟩ − ⟨v, y_{n}χ_{T \F} ⟩ + ⟨v, y_{n}χ_{F} ⟩
⟨v, y⟩ − 2⟨v, yχ_{F} ⟩ − 2⟨v, y_{n}χ_{T \F} ⟩ + ⟨v, y_{n}⟩
> 1 − 2∥yχ_{F} ∥^{0} − 2∥y_{n}χ_{T \F} ∥ − 2∥y_{n}χ_{T \F} ∥^{0} + 1 > 2 − 4ε,

which shows that △(y) = 2.

Theorem 1.4. Let X be one of the spaces L_{Φ}, E_{Φ}, l_{Φ} or h_{Φ}. Then the following are equivalent: X has the sum-property. X has WNS. M ∈ △_{2}.

P r o o f. This time, we prove the theorem for X = l_{Φ} and X = h_{Φ}. (i) ⇒ (ii).The implication is trivial. (ii)⇒ (iii). If Φ∉△_{2}, then there exist a_{k} ↓ 0 such that Φ(α_{1}) < ε and Φ((1+ \frac{1}{k})α_{k}) > 2^{k}Φ(α_{k})(k ∈ N), where 0 < ε < 1 is a given constant. For each k ∈ N, choose an integer m_{k} such that

m_{k}Φ(α_{k}) ≤ \frac{ε}{2^{k}} , (m_{k} + 1)Φ(α_{k}) > \frac{ε}{2^{k}}

and define

x_{n}(i) = α_{n} ∑_{i−1}^{m_{n}} e_{i+s_{n}} (n ∈ N),

where {e_{i}} is the natural basis of c_{0} and s_{n} = ∑_{i=1}^{n−1} m_{i}. Obviously, {x_{n}} have mutually disjoint supports, and so,

I_{Φ}(x_{i} − x_{j}) ≤ \frac{ε}{2^{i}} + \frac{ε}{2^{j}} < 1(i ≠ j). Moreover, for any v > 1, it is easy to check that I_{Φ}(vx_{n}) → ∞ as n → ∞. Therefore, for any n ∈ N, △(x_{n}) = 1 and △(x) = 1 for all x ∈ co{x_{n}}. Clearly, x_{n} → 0 l_{Φ} -weakly, i.e., x_{n} → 0 weakly in h_{Φ}. This means that {x_{n}} is a weakly convergent limit constant sequence, thus, h_{Φ} does not have the WNS. (iii) ⇒ (i). Assume that l_{Φ} has a limit affine sequence {x_{n}} with △(x_{n}) ↑ △′. By the diagonal method, we can find a subsequence of {x_{n}}, again denoted by {x_{n}}, such that x_{n} → x coordinatewise. By Lemma 1.5, x ∈ l_{Φ}. Hence, we may assume that x_{n} → 0 coordinatewise and that △′ = lim △(x_{n}) > 0. For any i, j ∈ N, since △ is affine on co{x_{n}},

lim_{n} ∥2x_{n} − x_{i} − x_{j}∥ = △(x_{i}) + △(x_{j}).

Hence, as Φ ∈ △_{2},

(1.29)  lim_{n} I_{Φ}(\frac{x_{n} − x_{i}}{△(x_{i})} ) = lim_{n} I_{Φ}(\frac{x_{n} − x_{j}}{△(x_{j})} ) = lim_{n} I_{Φ}(\frac{2x_{n} − x_{i} − x_{j}}{△(x_{i}) + △(x_{j})}) = 1.

Let λ_{ij} = \frac{△(x_{i})}{△(x_{i})+△(x_{j})}. Then by the convexity of Φ,

λ_{ij}I_{Φ}(\frac{x_{n} − x_{i}}{△(x_{i})} ) + (1 − λ_{ij})I_{Φ}(\frac{x_{n} − x_{j}}{△(x_{j})} ) − I_{Φ}(\frac{2x_{n} − x_{i} − x_{j}}{△(x_{i}) + △(x_{j})}) ≥ λ_{ij}Φ(\frac{x_{n}(k)−x_{i}(k)}{△(x_{i})}) + (1 − λ_{ij})Φ(\frac{x_{n}(k)−x_{j}(k)}{△(x_{j})})

− Φ(\frac{2x_{n}(k) − x_{i}(k) − x_{j}(k)}{△(x_{i}) + △(x_{j})}) ≥ 0 (k ∈ N).(1.30)

Recall that x_{n} → 0 coodinatewise. By letting n → ∞, we conclude from (1.19) and (1.30) that

λ_{ij}Φ(\frac{x_{i}(k)}{△(x_{i})}) + (1 − λ_{ij})Φ(\frac{x_{j}}{△(x_{j})}) = Φ(\frac{x_{i}(k) + x_{j}(k)}{△(x_{i}) + △(x_{j})}) (k ∈ N).

Letting j → ∞, this equality becomes the equality

\frac{△(x_{i})}{△(x_{i}) + △′} Φ(\frac{x_{i}(k)}{△(x_{i})}) = Φ(\frac{x_{i}(k)}{△(x_{i}) + △′}) (k ∈ N).

But 0 < △(x_{i})/(△(x_{i}) + △′) < 1 by (1.7), so the above equality holds only for x_{i}(k) = 0. This means that x_{i} = 0 (i ∈ N), contradicting the assumption that {x_{n}} is a limit affine sequence. To end this section, we present a different sufficient condition for h_{Φ} to have the weakly fixed point property.

Lemma 1.8. The space h_{Φ} has the weak orthogonality property, i.e. for any sequence {x_{n}} in h_{Φ} such that x_{n} → 0 weakly, there holds

lim inf_{n} lim inf_{m} ∥|x_{n}| ∧ |x_{Φ}|∥ = 0,

where (x ∧ y)(t) = min{x(t), y(t)} (see and .

P r o o f. The lemma results from obvious fact that the mapping y → |x| ∧ |y| is weak-norm continuous for every fixed x ∈ h_{Φ}. In fact weak convergence of (y_{n}) to zero implies that y_{n} → 0 coordinatevise. So, if x ∈ h_{Φ} then |y_{n}| ∧ |x| ≤ |x| and |y_{n}| ∧ |x| → 0 coordinatevise. By the dominated Lebesgue convergence theorem, we get ∥ |y_{n}| ∧ |x| ∥ → 0.

Lemma 1.9. The Riesz angle α(l_{Φ}) < 2 if and only if Φ ∈ ∇_{2}, where

α(l_{Φ}) = sup{∥ |x| ∨ |y| ∥ : ∥x∥ ≤ 1, ∥y∥ ≤ 1}.

P r o o f. If Φ ∉ ∇_{2}, then there exist u_{n} ↓ 0 such that

(1.31)  2Φ(\frac{u_{n}}{2} ) > (1 − \frac{1}{n})Φ(u_{n}) (n ∈ N).

Let m_{n} be an integer satisfying m_{n}Φ(u_{n}) ≤ 1 and (m_{n} + 1)Φ(u_{n}) > 1. Define

x_{n} = u_{n} ∑_{i=1}^{m_{n}} e_{i}, y_{n} = u_{n} ∑_{i=m_{n}+1}^{2m_{n}} e_{i}

.

Then it is easy to check that 1 ≥ I_{Φ}(y_{n}) → 1 and by (1.31),

I_{Φ}(\frac{x_{n} ∨ y_{n}}{2} ) = 2m_{n}Φ(\frac{u_{n}}{2} ) > (1 − \frac{1}{n})m_{n}Φ(u_{n}) → 1.

This shows that ∥x_{n} ∨ y_{n}∥ → 2. Next we assume Φ ∈ ∇_{2}, i.e., there exist δ > 0 such that

Φ((2 − δ)u) ≥ 2Φ(u) (|u| ≤ Φ^{−1}(1)).

Given x, y ∈ B(l_{Φ}), we have |x(i)|, |y(i)| ≤ Φ^{−1}(1), whence

\begin{aligned}I_{Φ}(\frac{|x| ∨ |y|}{2 − δ}) & = I_{Φ}(\frac{x}{2 − δ}) + I_{Φ}(\frac{y}{2 − δ}) ≤ \frac{1}{2}[I_{Φ}(x) + I_{Φ}(y)] ≤ 1, \\ & , ∥|x| ∨ |y|∥ ≤ 2 − δ. \end{aligned}

Applying a result of J.M. Borwein and B. Sims [6] stating that every weakly orthogonal Banach lattice X with Riesz angle α(X) < 2 has the weakly fixed point property, from Lemmas 1.7 and 1.8 we deduce the following

Theorem 1.5. If Φ ∈ ∇_{2}, then h_{Φ} has the weak fixed point property.

Remark 1.1. Theorems 1.4 and 1.5 furnish a natural example of a space with the weakly fixed point property but without WNS.

Remark 1.2. P.N. Dowling, C.J. Lennard and B. Turett [28] investigated Orlicz spaces for which every nonexpansive self-mapping of a nonempty, closed, bounded, convex subset has a fixed point. This property is called the fixed point property (FPP). They proved that L_{Φ}^{0} has FPP if and only if it is reflexive. In fact, this can be obtained immediately from Theorem 1.1 presented above, Theorem 1.90 in [10], Theorem 1.1 and the following two results given by P.N. Dowling and C.J. Lennard [27]: (a) A Banach space X fails FPP if it contains an asymptotically isometric copy of l^{1} i.e., for every positive sequence (ε_{n}) decreasing to 0, there exists a sequences {x_{n}} of norm-one elements in X such that ∑_{n}(1 − ε_{n})|α_{n}| ≤ ∥ ∑_{n} α_{n}x_{n}∥ for all sequences (α_{n}) of real numbers. (b) If the dual of X contains an isometric copy of l^{∞}, then X contains an asymptotically isometric copy of l^{1}. It is still an open problem whether the above conclusion is true or not for the Orlicz space L_{Φ} equipped with the Luxemburg norm. The only trouble is that one cannot prove the necessity of Φ ∈ △_{2} in the same way as for the Orlicz norm.

Notes. Criteria for normal structure and uniform normal structure of Musielak-Orlicz spaces were given by E. Katirtzoglou [53]. In Orlicz-Lorentz spaces the criteria were presented by A. Kamińska, Pei-Kee Lin and H. Sun [52].

2. Uniform rotundity in every direction

Recall that a Banach space X is said to be uniformly rotund in every direction (URED) if for any z ∈ S(X) and every two sequences {x_{n}} and {y_{n}} in S(X) such that x_{n} − y_{n} = ε_{n}z, where {ε_{n}} is a sequence of reals, and ∥x_{n} + y_{n}∥ → 2 there holds ∥x_{n} − y_{n}∥ → 0. If we change S(X) into B(X) in the above definition, we get the same property. In the fixed point theory this geometric property is important because of the following

Theorem 2.1. Any Banach space X which is uniformly rotund in every direction has normal structure.

P r o o f. Let A be a bounded and convex set with d(A) > 0. We need to show that A is not diametral, that is, there exists w ∈ A such that d_{w}(A) := sup{∥w −x∥: x ∈ A} < d(A) := sup{∥x−y∥: x, y ∈ A}. Take arbitrary x, y ∈ A such that x ≠ y and define w = (x + y)/2. We will show that w has the desired property. Since A is convex, w ∈ A. Assume for the contrary that A is a diametral point for A, i.e. d_{w}(A) = d(A). Then there is a sequence {y_{n}}_{n=1}^{∞} in A such that

lim_{n→∞} ∥w − y_{n}∥ = d(A).

We can write this in the form

\begin{aligned}∥\frac{x + y}{2} − y_{n}∥ & = ∥\frac{x + y}{2} − \frac{y_{n} + y_{n}}{2} ∥ \\ & = ∥\frac{1}{2}(x − y_{n}) + \frac{1}{2}(y − y_{n})∥ → d(A). \end{aligned}

Moreover, ∥(x−y_{n})−(y −y_{n})∥ = ∥x−y∥ > 0, ∥x−y_{n}∥ ≤ d(A) and ∥y −y_{n}∥ ≤ d(A). Defining

\overline{x} = x/d(A), \overline{y} = y/d(A) \overline{y}_{n} = y_{n}/d(A),

we have

∥\overline{x} − \overline{y}_{n}∥ ≤ 1 ∥\overline{y} − \overline{y}_{n}∥ ≤ 1 (n ∈ N)

and

∥\frac{(\overline{x} − \overline{y}_{n}) + (\overline{y} − \overline{y}_{n})}{2}∥ → 1.

By the assumption that X is URED, we conclude that

∥\overline{x} − \overline{y}∥ = ∥(\overline{x} − \overline{y}_{n}) − (\overline{y} − \overline{y}_{n})∥ → 0 n → ∞,

which gives ∥\overline{x}−\overline{y}∥ = 0, that is, \overline{x} = \overline{y}, a contradiction. So, w is not a diametral point of A. Now we will present criteria for URED of Orlicz spaces. We do not assume in this section that Orlicz functions Φ satisfy conditions (ii) from the definition (see page 1).

Theorem 2.2. Let (T, Σ, μ) be a nonatomic complete and σ-finite measure space and Φ be an Orlicz function. Then the Orlicz space L_{Φ} equipped with the Luxemburg norm is uniformly rotund in every direction if and only if Φ is strictly convex and Φ satisfies the Δ_{2}-condition on ℝ_{+} if μ is infinite and the Δ_{2}-condition at infinity if it μ is finite.

P r o o f. Sufficiency. Let {x_{n}}, {y_{n}} be sequences in S(L_{Φ}), z ∈ S(L_{Φ}), x_{n} − y_{n} = ε_{n}z, where \{ε_{n}\} is a sequence of reals and ∥\frac{x_{n}+y_{n}}{2} ∥ → 1. Then by the Δ_{2}-

condition, we have I_{Φ}(\frac{x_{n}+y_{n}}{2} ) → 1 (see [10]). We will show that x_{n} − y_{n} → 0 in measure. Assume for the contrary that it is not true. Then we can assume (passing to a subsequence if necessary that for some ε, σ > 0 there holds μE_{n} ≥ ε for all n ∈ ℕ, where

E_{n} = {t ∈ T: |x_{n}(t) − y_{n}(t)| ≥ σ}.

Choose k > 1 such that F ∈ Σ and μF = ε/4 implies that ∥χ_{F} ∥ = 1/k and define

A_{n} = {t ∈ T: |x_{n}(t)| > k}, B_{n} = {t ∈ T: |y_{n}(t)| > k}.

Then we have 1 = ∥x_{n}∥_{Φ} ≥ ∥x_{n}χ_{A_{n}}∥_{Φ} > k∥χ_{A_{n}}∥_{Φ}, whence ∥χ_{A_{n}}∥_{Φ} < \frac{1}{k} and consequently, μ(A_{n}) < ε/4. Similarly, μ(B_{n}) < ε/4 (n ∈ N). By strict convexity of Φ there is δ > 0 such that if u, v ∈ [0, k] and |u − v| ≥ σ, then

Φ(\frac{u + v}{2} ) ≤ \frac{1 − δ}{2} \{Φ(u) + Φ(v)\}.

Denote C_{n} = E_{n}\(A_{n}∪B_{n}). Then we have μ(C_{n}) ≥ μE_{n}−(μ(A_{n})+μ(B_{n})) = ε− \frac{ε}{2} = \frac{ε}{2}. Moreover for any t ∈ C_{n}, we have |x_{n}(t) − y_{n}(t)| ≥ σ and |x_{n}(t)| ≤ k, |y_{n}(t)| ≤ k for any k ∈ N. Consequently

Φ(\frac{x_{n}(t) + y_{n}(t)}{2}) ≤ \frac{1 − δ}{2} \{Φ(x_{n}(t)) + Φ(y_{n}(t))\} (t ∈ C_{n}),

n = 1, 2, …. Consequently

0 ← 1 − I_{Φ}(\frac{x_{n} + y_{n}}{2} ) = \frac{1}{2}\{I_{Φ}(x_{n}) + I_{Φ}(y_{n})\} − I_{Φ}(\frac{x_{n} + y_{n}}{2} )

≥ \frac{1}{2}\{I_{Φ}(x_{n})χ_{Cn}) + I_{Φ}(y_{n})χ_{Cn})\} − I_{Φ}(\frac{x_{n} + y_{n}}{2} χ_{Cn} )

≥ \frac{1}{2}\{I_{Φ}(x_{n}χ_{Cn}) + I_{Φ}(y_{n}χ_{Cn})\} − \frac{1 − δ}{2} \{I_{Φ}(x_{n}χ_{Cn}) + I_{Φ}(y_{n}χ_{Cn})\}

= \frac{δ}{2}\{I_{Φ}(x_{n}χ_{Cn}) + I_{Φ}(y_{n}χ_{Cn})\} ≥ δI_{Φ}(\frac{x_{n} − y_{n}}{2} χ_{Cn} )

≥ δΦ(\frac{σ}{2} )μ(C_{n}) ≥ δΦ(\frac{σ}{2} )\frac{ε}{2},

a contradiction. Therefore x_{n} − y_{n} → 0 in measure. Since z ∈ S(L_{Φ}) and so z ≠ 0, we conclude from the equality x_{n} − y_{n} = ε_{n}z (n ∈ N) that ε_{n} → 0. Consequently, there is n_{0} ∈ N such that |ε_{n}| ≤ 1 for n ≥ n_{0} and so |x_{n} − y_{n}| ≤ |z| for n ≥ n_{0}. By the Lebesgue dominated convergence theorem, we get (x_{n} − y_{n}) → 0 and by the suitable Δ_{2}-condition for Φ, we get I_{Φ}(λ(x_{n} − y_{n})) → 0 as n → ∞ for any λ > 0, which means that ∥x_{n} − y_{n}∥ → 0 as n → ∞.

Necessity. Assume that Φ does not satisfy the suitable Δ_{2}-condition. Then L^{Φ} contains an order isometric copy of l_{∞} (see [10], [69] and [74]). Since l_{∞} is not URED, L_{Φ} is not URED. Assume now that Φ is not strictly convex. Then there exist u > v > 0 such that Φ(\frac{u+v}{2} ) = \frac{1}{2}\{Φ(u) + Φ(v)\}. We will show that L_{Φ} is not rotund, so it is not URED Choose two disjoint sets A, B ∈ Σ and a > 0 such that μ(A) > 0, μ(B) > 0 and

\frac{1}{2}(Φ(u) + Φ(v))μ(A) + Φ(a)μ(B) = 1.

Let C, D ⊂ A be measurable sets such that \frac{μ(C)=μ(D)=1}{2μ(A)} and C ∩ D = ∅. Define

\begin{aligned}x & = uχ_{C} + vχ_{D} + aχ_{B}, \\ y & = vχ_{C} + uχ_{D} + aχ_{.} \end{aligned}

Then

I_{Φ}(x) = Φ(u)μ(C) + Φ(v)μ(D) + Φ(a)μ(B) = \frac{1}{2}(Φ(u) + Φ(v))μ(A) + Φ(a)μ(B) = 1

. In the same way we can prove that I_{Φ}(y) = 1. Moreover,

\begin{aligned}I_{Φ}(\frac{x + y}{2} ) & = Φ(\frac{u + v}{2} )μ(A) + Φ(a)μ(B) \\ & = \frac{1}{2}(Φ(u) + Φ(v))μ(A)Φ(a)μ(B) = 1. \end{aligned}

Consequently, ∥x∥_{Φ} = ∥y∥_{Φ} = ∥\frac{x+y}{2} ∥_{Φ} = 1. Since, evidently, x ≠ y, L_{Φ} is not rotund. This finishes the proof.

Notes. Kamińska [50] first gave criteria for URED of Musielak-Orlicz spaces of Bochner type. Theorem 2.2 can be easily deduced from her paper. The proof that we presented here is different.

3. B-convexity and uniform monotonicity

These properties are related to the fixed point theory by the following

Theorem (see [3]) If a Köthe function space X is B-convex and uniformly monotone, then it has the fixed point property.

Recall that a Banach space X is said to be B-convex if no nonreflexive space Y is finitely represented in X (see [3] and [10]). Since UR implies nonsquareness, nonsquareness implies B-convexity and B-convexity is preserved by equivalent norms, we know that uniformly convexitiable Banach spaces are B-convex. The converse is also true. Now, we will present criteria for B-convexity and uniform monotonicity of Orlicz spaces. We do not assume generally in this section that Orlicz functions that they must satisfy condition (ii) from the definition on page 1. First we will prove the following

Lemma 3.1. Let Φ be an Orlicz function such that its right derivative p on ℝ_{+} satisfies the condition: for any ε > 0, there exists K > 1 such that

p((1 + ε)t) ≥ Kp(t) (t ≥ 0).

Then Φ is uniformly convex.

P r o o f. Given ε ∈ (0, 1), take K > 1 such that

p((1 + ε/2)t) ≥ Kp(t) (t ≥ 0).

We shall show that for any u, v ∈ ℝ satisfying |u − v| ≥ ε max{|u|, |v|}, the inequality

(3.1)  Φ(\frac{u + v}{2} ) ≤ (1 − δ)\frac{Φ(u) + Φ(v)}{2}

holds for δ = ε(1 − 1/K)/4 > 0. We may assume without loss of generality that u − v ≥ εu > εv > 0, that is (1 − ε)u ≥ v > 0. Define

φ(t) = Φ(u) + Φ(t) − 2Φ(\frac{u + t}{2}) (t ≥ 0).

Then for almost all t ∈ [0, u],

φ′(t) = p(t) − p(\frac{u + t}{2}) ≤ 0.

Hence φ(t) is nonincreasing on [0, u]. Therefore,

φ(v) = Φ(u) + Φ(v) − 2Φ(\frac{u + v}{2} )

≥ Φ(u) + Φ((1 − ε)u) − 2Φ((1 − ε/2)u)

= ∫_{(1−ε/2)u}^{u} p(t)dt − ∫_{(1−ε)u}^{(1−ε/2)u}p(t)dt = ∫_{(1−ε/2)u}^{u} [p(t) − p((1 − ε/2)t)]dt

≥ ∫_{(1−ε/2)u}^{u} (1 − 1/K)p(t)dt ≥ (1 − 1/K)[Φ(u) − (1 − ε/2)Φ(u)]

> \frac{ε}{4}(1 − 1/K)[Φ(u) + Φ(v)]

for u and v as above, that is condition (+) holds with δ = (ε/4)(1 − 1/K) > 0.

Theorem 3.1. For any Orlicz space L_{Φ} the following are equivalent:

L_{Φ} is reflexive,

Φ ∈ Δ_{2} and P si ∈ Δ_{2}, where Ψ is the compementary function of Φ, L_{Φ} is uniformly convexitiable, L_{Φ} is B-convex.

P r o o f. It is well known that (a) ⇔ (b) (see [69] and [74]). Let us prove that (b) ⇒ (c). We consider only the case of a nonatomic finite measure space, when Φ ∈ Δ_{2} means that Φ satisfies the Δ_{2}-condition at ∞. By (b) there exist u_{0} > 0, K > 2 and δ > 0 such that

(2 + δ)Φ(u) ≤ Φ(2u) ≤ KΦ(u) (u ≥ u_{0}).

Since changing the value of Φ on [0, u_{0}] does not affect the equivalence, we may assume that the above inequalities hold for all u ∈ ℝ. Let

Φ_{0}(u) = ∫_{0}^{|u|} \frac{Φ(t)}{t}dt, Φ_{1} = ∫_{0}^{|u|} \frac{Φ_{0}(t)}{t}dt.

We claim that Φ_{0} ∼ Φ_{1} ∼ Φ. Indeed, denoting by p the right derivative of Φ on ℝ_{+}, we have

p(u) ≥ \frac{Φ(u)}{u} ≥ \frac{Φ(2u)}{Ku} ≥ \frac{1}{K} p(u) (u > 0).

Integrating each term of the last inequalities from zero to u, we get KΦ(u) ≥ KΦ_{0}(u) ≥ Φ(u), i.e. Φ ∼ Φ_{0}. Similarly Φ_{0} ∼ Φ_{1}. Next, we will show that Φ_{1} is uniformly convex. Since

K^{−1}Φ(u) ≤ Φ(\frac{u}{2}) = ∫_{0}^{|u|/2}\frac{Φ(u/2)}{|u/2|}dt ≤ ∫_{|u|/2}^{|u|} \frac{Φ(t)}{t} ≤ Φ_{0}(u)

= ∫_{0}^{|u|/2}\frac{Φ(t)}{t}dt + ∫_{|u|/2}^{|u|} \frac{Φ(t)}{t}dt

≤ Φ(\frac{u}{2}) + \frac{Φ(u)}{2} ≤ (\frac{1}{2 + δ} + \frac{1}{2})Φ(u) = \frac{1}{L}Φ(u),

where L = \frac{4+2δ}{4+δ} > 1, we obtain for t > 0,

L ≤ \frac{Φ(t)}{Φ_{0}(t)} = \frac{tΦ′_{0}(t)}{Φ_{0}(t)} ≤ K.

Hence we get

\frac{L}{t} ≤ \frac{Φ′_{0}(t)}{Φ_{0}(t)} ≤ \frac{K}{t} .

Integrating this inequality with respect to t from u to θu, where θ ≥ 1, we have

θ^{L}Φ_{0}(u) ≤ Φ_{0}(θu) ≤ θ^{K}Φ(u) (θ ≥ 1, u ∈ ℝ).

Set p_{1}(t) = M′_{1}(t) = Φ_{0}(t)/t for t > 0. We have for any ε > 0 and u > 0,

\begin{aligned}p_{1}((1 + ε)u) & = \frac{Φ_{0}((1 + ε)u)}{(1 + ε)u} ≥ \frac{(1 + ε)^{L}Φ_{0}(u)}{(1 + ε)u} \\ & = (1 + ε)^{L−1}Φ_{0}(u)/u = (1 + ε)^{L−1}p_{1}(u), \end{aligned}

whence, by Lemma 3.1, Φ_{1} is uniformly convex. Finally we will show that Ψ_{1} is also uniformly convex, where Ψ_{1} denotes the function complementary to Φ_{1} in the sens of Young. Let (Ψ_{1})′(u) =: q_{1}(u) and

q_{1}((1 + ε)v) = α(v)q_{1}(v) (v > 0).

Then α(v) > 1 (v > 0). Replacing v by p_{1}(u), we get

\begin{aligned}(1 + ε)p_{1}(u) & = p_{1}(α(v)u) = \frac{Φ_{0}(α(v)u)}{α(v)u} \\ & ≤ \frac{α^{K}(v)Φ_{0}(u)}{α(v)u} = α^{K−1}(v)p_{1}(u). \end{aligned}

Hence, α(v) ≥ (1 + ε)^{1/(K−1)}, and so q_{1}((1 + ε)v) ≥ (1 + ε)^{1/(K−1)}q_{1}(v), i.e. Ψ_{1} is uniformly convex. Therefore, applying the result of Kamińska [51] and the fact that a Banach space X is uniformly rotund iff X^{∗} is uniformly smooth, we get that both spaces L_{Φ_{1}} and L_{Ψ} are uniformly rotund and uniformly smooth for both, the Luxemburg and the Orlicz norm. Since Φ ∼ Φ_{1}, the space L_{Φ} is B-convex, which finishes the proof of the implications (b) ⇒ (c) ⇒ (d). Let us prove that (d) ⇒ (a). We will show that if L_{Φ} is not reflexive, then L_{Φ} is not B-complex. L_{Φ} is nonreflexive iff Φ ∉ Δ_{2} or Ψ ∉ Δ_{2}. If Φ ∉ Δ_{2}, then L_{Φ} equipped with the Luxemburg norm contains an order isometric copy of l_{∞} (see [74]). Therefore L_{Φ} is not B-complex. Assume now that Ψ ∉ Δ_{2} and Φ ∈ Δ_{2}. Then the dual space of L_{Φ}^{0} is L_{Ψ}, which contains an order isometric copy of l_{∞} and consequently L_{Φ}^{0} contains an asympthotically isometric copy of l_{1} (see [12]), so L_{Φ}^{0} (and consequently L_{Φ}) is not B-convex. This completes the proof.

Let us denote by Φ ∈ Δ_{2} the fact that the Orlicz function Φ satisfies the suitable Δ_{2}-condition, which means the Δ_{2}-condition on ℝ_{+} if μ is nonatomic and infinite, the Δ_{2}-condition at infinity if μ is non-atomic and finite and the Δ_{2}-condition at zero if μ is the counting measure.

Theorem 3.2. An Orlicz space L_{Φ} equipped with the Luxemburg norm is uniformly monotone if and only if Φ vanishes only at zero and Φ ∈ Δ_{2}.

P r o o f. Sufficiency. Assume that a(Φ) := sup{u ≥ 0: Φ(u) = 0} = 0 and Φ ∈ Δ_{2}. Let x ∈ S(L_{Φ}), 0 ≤ y ≤ x and ∥y∥_{Φ} ≥ ε, where ε ∈ (0, 1). Then, by Φ ∈ Δ_{2}, I_{Φ}(x) = 1, and there is δ(ε) ∈ (0, ε] such that I_{Φ}(y) ≥ δ(ε) (see [10]). Since Φ is superadditive on ℝ_{+}, we have

I_{Φ}(x) = I_{Φ}((x − y) + y) ≥ I_{Φ}(x − y) + I_{Φ}(y),

whence I_{Φ}(x − y) ≤ 1 − I_{Φ}(y) ≤ 1 − δ(ε). Again by Φ ∈ Δ_{2}, there is a function σ: (0, 1) → (0, 1) such that ∥x∥ ≤ 1 − σ(ε) whenever I_{Φ}(x) ≤ 1 − ε (see [10]). Consequently, ∥x − y∥ ≤ 1 − σ(δ(ε)), i.e. L_{Φ} is uniformly monotone. Necessity. Assume that a(Φ) > 0. First assume that the measure space is infinite. Then x = χ_{T} ∈ S(L_{Φ}). Let A ∈ Σ be such that μ(A) = μ(T \ A) = ∞ and define y = χ_{A}.

Then 0 ≤ y ≤ x and ∥x−y∥_{Φ} = ∥χ_{T \A}∥_{Φ} = 1, which means that L_{Φ} is not uniformly monotone. If μ is finite take y ≥ 0 such that ∥y∥_{Φ} = 1 and μ(T \ y) > 0. Define x = y + a(Φ)χ_{T \ supp y}. Then 0 ≤ y ≤ x and ∥x∥_{Φ} = 1, whence we get that L_{Φ} is not uniformly monotone. Assume now that Φ ∉ Δ_{2}. Then there exists x ∈ S(L_{Φ}), x ≥ 0, such that I_{Φ}(λx) = +∞ for any λ > 1 (see [10] and [69]). Consequently, there exists A ∈ Σ such that ∥xχ_{A}∥_{Φ} = ∥xχ_{T \A}∥_{Φ} = ∥x−xχ_{A}∥_{Φ} = 1, that is L_{Φ} is not uniformly monotone. This finishes the proof of the theorem.

Thoerem 3.3. Let Φ be an Orlicz function such that Φ(u)/u → ∞ as u → ∞. The Orlicz space L_{Φ}^{0} equipped with the Orlicz norm is uniformly monotone if and only if Φ ∈ Δ_{2} and Φ vanishes only at zero.

P r o o f. Sufficiency. Let a(Φ) = 0, Φ ∈ Δ_{2}, x ∈ S(L_{Φ}^{0}), 0 ≤ y ≤ x and ∥y∥_{Φ}^{0} ≥ ε. By Φ(u)/u → ∞ as u → ∞, there is k > 1 such that

1 = ∥x∥_{Φ}^{0} = \frac{1}{k} (1 + I_{Φ}(kx)).

By Φ ∈ Δ_{2} and ∥y∥_{Φ}^{0} ≥ ε, we have I_{Φ}(y) ≥ δ(ε) with some δ(ε) ∈ (0, ε] with δ(ε) independent of x and y. By k > 1, we have \frac{1}{k} I_{Φ}(ky) ≥ I_{Φ}(y) ≥ δ(ε). Consequently

∥x − y∥_{Φ}^{0} ≤ \frac{1}{k} (1 + I_{Φ}(k(x − y)))

≤ \frac{1}{k} (1 + I_{Φ}(kx) − I_{Φ}(ky))

= \frac{1}{k} (1 + I_{Φ}(kx)) − \frac{1}{k} I_{Φ}(ky) ≤ 1 − δ(ε),

i.e. L_{Φ}^{0} is uniformly monotone. Necessity. Assume first that a(Φ) = 0. Choose y ≥ 0 in S(L_{Φ}^{0}) such that μ(T \ y) > 0. Define x = y + \frac{1}{k} a(Φ)χ_{T \ supp y}, where \frac{1}{k} (1 + I_{Φ}(ky)) = ∥x∥_{Φ}^{0} = 1. Then 0 ≤ y ≤ x, so ∥x∥_{Φ}^{0} ≥ ∥y∥_{Φ}^{0} ≥ 1. Moreover

\begin{aligned}∥x∥_{Φ}^{0} ≤ \frac{1}{k} (1 + I_{Φ}(kx)) & = \frac{1}{k} (1 + I_{Φ}(ky)) + \frac{1}{k} Φ(a(Φ))μ(T \ y) \\ & = \frac{1}{k} (1 + I_{Φ}(ky)) = ∥y∥_{Φ}^{0} = 1. \end{aligned}

Consequently ∥x∥_{Φ}^{0} = 1. Moreover,

I_{Φ}(2(x − y)) = Φ(2a(Φ))μ(T \ y) =: ε > 0.

Hence

∥2(x − y)∥_{Φ}^{0} ≥ min(1, ε), ∥x − y∥_{Φ}^{0} ≥ \frac{1}{2} min(1, ε).

This shows that L_{Φ}^{0} is not uniformly monotone. Assume now that Φ ∉ Δ_{2}. Let A ∈ Σ be such that μ(A) > 0 and μ(T \ A) > 0. By Proposition 2.1 in [40] there is a sequence {x_{n}} in L_{Φ}^{0} such that I_{Φ}(x_{n}) ≤ 2^{−n}, ∥x_{n}∥_{Φ}^{0} ≥ 1/(1 + 2^{−n}) ≥ 2/3 and x_{n} ⊂ T \ A for n = 1, 2, …. Consequently, if x ∈ L_{Φ}^{0} is such that x ⊂ A, ∥x∥_{Φ}^{0} = 1 and ∥x∥_{Φ}^{0} = \frac{1}{k_{0}} (1 + I_{Φ}(k_{0}x)) with some k_{0} ≥ 1, defining y_{n} = \frac{1}{k_{0}} x_{n}, we have I_{Φ}(k_{0}y_{n}) ≤ 2^{−n} and ∥y_{n}∥_{Φ}^{0} ≥ 2/3k_{0}. Moreover, ∥x+y_{n}∥_{Φ}^{0} ≤ \frac{1}{k_{0}} (1+I_{Φ}(k_{0}(x+y_{n}))) = \frac{1}{k_{0}} (1 + I_{Φ}(k_{0}x)) + \frac{1}{k_{0}} I_{Φ}(k_{0}y_{n}) ≤ 1 + 2^{−n}. Combining this with ∥y_{n}∥_{Φ}^{0} ≥ 2/3k_{0} for any n ∈ N, we conclude that L_{Φ}^{0} is not UM, which finishes the proof.

Notes. B-convexity for Musielak-Orlicz spaces was characterized in [41’] and for Orlicz-Lorentz spaces in [43]. Uniform monotonicity of Musielak-Orlicz spaces was characterized in [42], [26] and [55”]. For Orlicz-Lorentz spaces it was done in [41]. In some Calderon-Lozanowsky spaces and Banach lattices monotonicity properties were considered in [8], [29], [30] and [44].

4. Nearly uniform convexity and nearly uniform smoothness

First we introduce the notions of nearly uniform convexity, k-nearly uniform convexity and nearly uniform smoothness. For a given ε > 0 a sequence {x_{n}} in a Banach space X is said to be ε-separated if

sep({x_{n}}) := inf_{m≠n} ∥x_{m} − x_{n}∥ > ε.

A Banach space X is called nearly uniformly convex (NUC) if for any ε > 0 there is δ > 0 such that for every sequence {x_{n}} in B(X) with sep({x_{n}}) > ε there is an element x ∈ co({x_{n}}), the convex hull of {x_{n}}, such that ∥x∥ < 1 − δ. This notion has been introduced by Huff [48], where it has also been proved that a Banach space X is NUC if and only if it is refelxive and it has the uniform Kadec-Klee property (UKK). Recall that a Banach space X is said to have the UKK - property if for any ε > 0 there is δ = δ(ε) > 0 such that for any sequence {x_{n}} with sep({x_{n}}) > ε and any x ∈ B(X), we have ∥x∥ ≤ 1 − δ whenever x_{n} → x weakly. It is well known that NUC Banach spaces have the FPP (see [27] and [33]). The property NUC has also been defined by using the measure of noncompactness by Goebel and Sȩkowski [34]. Kutzarova [55’] introduced the notion of k-nearly uniform convexity of Banach spaces (K-NUC). Let k be an integer, k ≥ 2. A Banach space X is said to be k-NUC if for any ε > 0 there exists δ > 0 such that for every sequence {x_{n}} in B(X) with sep(\{x_{n}\}) > ε, there are n_{1}, n_{2}, …, n_{k} ∈ N such that ∥\frac{1}{k} (x_{n_{1}} + x_{n_{2}} + ⋯ + x_{n_{k}} ∥ < 1 − δ. Clearly k-NUC Banach spaces are NUC but the opposite implication does not hold in general (see [22’]). The notion of nearly uniform smoothness (NUS) has been introduced by Sȩkowski and Stachura [69’]. The definition uses the notion of the measure of noncompactness. Prus [67], [68] used another (equivalent) definition of this property which is easier to formulate. Namely a Banach space X is said to be NUS if for any ε > 0 there exists δ > 0 such that for each basic sequence {x_{n}} in B(X) there is k > 1 such that

∥x_{1} + tx_{k}∥ ≤ 1 + tε

for each t ∈ [0, δ]. Prus [67] showed that a Banach space X is NUC if and only if its dual space X^{∗} is NUS. Let A be a bounded set of X. Its Kuratowski measure of non-compactness α(A) is defined as the infimum of all numbers d > 0 such that A may be covered by finitely many sets of diameter smaller than d. A Banach space X is said to be nearly uniformly ∗-smooth provided that for every ε > 0 there exists δ > 0 such that if x ∈ S(X), then

α(S^{∗}(x, δ)) ≤ ε,

where S^{∗}(x, δ) = {x^{∗} ∈ B(X^{∗}): x^{∗}(x) ≥ 1 − δ}. NUC and NUS have been also studied by Banaś [4], [5]. Also nearly uniform smoothness and weakly nearly uniform smoothness are related to the fixed point theory as it follows from the following

Theorem 4.1 (see [31]) If X is a W NUS Banach space, then X has the F P P. In particular, NUS Banach space have the F P P.

In order to get criteria for NUS of Orlicz spaces it is natural to present first criteria for NUC of these spaces because NUC and NUS are dual properties (see [67]). Since k-NUC implies NUC we present first criteria for k-NUC of Orlicz spaces given in [22’].

Theorem 4.2. Let (T, Σ, μ) be a nonatomic and finite measure spaces and Φ be a Orlicz function satisfying (Φ(u)/u) → ∞ as u → ∞ and X be equal to L_{Φ} or L_{Φ}^{0}. Then X is (kNUC) if and only if Φ is a strictly convex and satisfies the Δ_{2}-condition at infinity and Φ is uniformly convex outside a neighbourhood of zero.

P r o o f. The sufficiency follows from the fact that L_{Φ} ( or L_{Φ}^{0} ) is (UC) under the assumption that Φ is a strictly convex Orlicz function satisfying the Δ_{2}- condition at infinity and Φ is uniformly convex outside a neighbourhood of zero. We only need to prove the necessity of the theorem. Obviously, every Banach space X with property (kNUC) has the Kadec-Klee property. In case of L_{Φ}, the proof of the necessity of strict convexity of Φ for the Kadec-Klee property, can be found in [19]. Next, we will show that Φ must be strictly convex in case when L_{Φ}^{0} has the Kadec-Klee property. Assume the contrary, i.e. that there is an interval [a, b] such that the right-hand side derivative p of Φ is constant in [a, b]. Take G^{0} ⊂ G such that 0 < μ(G \ G^{0}) < μ(G) and choose c > 0 and G′ ⊂ G \ G^{0} satisfying

Ψ (p(a)) μ(G^{0}) + Ψ (p(c)) μ(G′) = 1.

Such a constant c exists because by the assumption (Φ(u)/u) → ∞ as u → ∞ the function Ψ has only finite values. Divide G^{0} into two subsets G_{1}^{1} and G_{2}^{1} such that G^{0} = G_{1}^{1}∪G_{2}^{1} and μ(G_{1}^{1}) = μ(G_{2}^{1}). Suppose that the sequence (G_{1}^{n−1}, G_{2}^{n−1}, …, G_{2^{n−1}}^{n−1}) of pairwise disjoint subsets of G^{0} all being of the same measure equal to 2^{−n+1}μ(G^{0}) was already defined. We devide every set G_{i}^{n−1} into two subsets G_{2i−1}^{n}, G_{2i}^{n} such that G_{i}^{n−1} = G_{2i−1}^{n} ∪ G_{2i}^{n} and μ(G_{2i−1}^{n}) = μ(G_{2i}^{n}) (i = 1, 2, …, 2^{n−1}). In such a way, we obtain a partition (G_{1}^{n}, G_{2}^{n}, …, G_{2^{n}}^{n}) of G^{0} such that

μ(G_{i}^{n}) = 2^{−n}μ(G^{0}) (i = 1, 2, …, 2^{n}).

Denote

k = 1 + Φ (\frac{a + b}{2}) μ(G^{0}) + Φ(c)μ(G′)

and put

x_{n} = \frac{1}{k} (aχ_{E_{1,n}} + bχ_{E_{2,n}}) ,

where E_{1,n} = ⋃_{k=1}^{2^{n−1}} G_{2k−1}^{n}, E_{2,n} = ⋃_{k=1}^{2^{n−1}} G_{2k}^{n} (n = 1, 2, …). Since I_{Ψ}(p(kx_{n}))

= 1, we have

∥x_{n}∥_{0} = \frac{1}{k} (1 + I_{Φ}(kx_{n})) = 1 (n = 1, 2, …).

By the reflexivity of L_{Φ}^{0}, we can assume that there exists x ∈ B(L_{Φ}^{0}) such that x_{n}→^{w}x. Since v = p (\frac{a+b}{2}) χ_{G^{0}} + cχ_{G′} defines a functional which supports x_{n} (n = 1, 2, …), so x ∈ S(L_{Φ}^{0}). But

∥x_{n} − x_{m}∥_{0} = \frac{b − a}{k} · \frac{μ(G^{0})}{2}Ψ^{−1} (\frac{2}{μ(G^{0})}) (n, m = 1, 2, …, n ≠ m).

This contradicts the Kadec-Klee property. The necessity of the uniform convexity of Φ outside a neighbourhood of zero and of the Δ_{2}-condition at infinity is proved in [2], Theorem 3.15.

Corollary 4.1. Under the assumptions of Theorem 4.2 on μ and Φ, the spaces L_{Φ} and L_{Φ}^{0} are NUC if and only if both Φ and Ψ (where Ψ is the Young conjugate satisfy the condition Δ_{2} at infinity.

P r o o f. It follows directly from the facts that k-NUC implies NUC, NUC implies reflexivity and reflexivity of L_{Φ} (resp. L_{Φ}^{0}) is equivalent to the fact that both Φ and Ψ satisfy the suitable Δ_{2}-condition.

Theorem 4.3. The Orlicz sequence space l_{Φ} is (kNUC) if and only if both Φ and Ψ satisfy both the Δ_{2}-condition at zero, i.e. l_{Φ} is reflexive.

P r o o f. We need only to prove the sufficiency of the theorem . Suppose that the implication is not true. Let any ε > 0 and {x_{n}} ⊂ B(l_{Φ}) with sep(x_{n})> ε be given. By Φ ∈ Δ_{2}(0), there exists δ = δ(ε) > 0 such that

inf \{I_{Φ}(\frac{x_{n} − x_{m}}{2} ) : n ≠ m\} ≥ δ.

Next, we will show that for any j ∈ N there exists n_{j} ∈ N such that

(4.1)  ∑_{i=j}^{∞} Φ(x_{n_{j}}(i)) ≥ \frac{δ}{3}.

Otherwise, there exists j_{0} ∈ N such that

∑_{i=j}^{∞} Φ(x_{n_{j}}(i)) < \frac{δ}{3}

for any j ∈ N. Defining \overline{x}_{n} = (x_{n}(1), x_{n}(2), ⋯, x_{n}(j_{0}), 0, 0, …) for n ∈ N, we easily get that there exists a subsequence \{\overline{x}_{n_{k}}\} of \{\overline{x}_{n}\} such that

I_{Φ} (\frac{\overline{x}_{n_{i}} − \overline{x}_{n_{j}}}{2} ) < \frac{δ}{3}

for any i ≠ j. Hence

I_{Φ} (\frac{x_{n_{i}} − x_{n_{j}}}{2} )

= I_{Φ} (\frac{∑_{k=1}^{j_{0}}(x_{n_{i}}(k) − x_{n_{j}}(k))e_{k}}{2} ) + I_{Φ} (\frac{∑_{k=j_{0}+1}^{∞}(x_{n_{i}}(k) − x_{n_{j}}(k))e_{k}}{2} )

≤ I_{Φ} (\frac{∑_{k=1}^{j_{0}}(x_{n_{i}}(k) − x_{n_{j}}(k))e_{k}}{2} ) + \frac{1}{2} ∑_{k=j_{0}+1}^{∞} Φ(x_{n_{i}}(k)) + \frac{1}{2} ∑_{k=j_{0}+1}^{∞} Φ(x_{n_{j}}(k))

= I_{Φ} (\frac{\overline{x}_{n_{i}} − \overline{x}_{n_{j}}}{2} ) + \frac{1}{2} ∑_{k=j_{0}+1}^{∞} Φ(x_{n_{i}}(k)) + \frac{1}{2} ∑_{k=j_{0}+1}^{∞} Φ(x_{n_{j}}(k))

< \frac{δ}{3} + \frac{δ}{6} + \frac{δ}{6} = \frac{2}{3}δ < δ.

This contradiction shows that (4.1) holds. Since Ψ satisfies the Δ_{2}-condition at zero, there is Θ ∈ (0, 1) such that

(4.2)  Φ (\frac{u}{k}) ≤ (1 − Θ)\frac{Φ(u)}{k}  (∀0 ≤ u ≤ Φ^{−1}(1))

(see [2] and [8]). By Φ ∈ Δ_{2}(0), there exists σ > 0 such that

(4.3)  |I_{Φ}(x + y) − I_{Φ}(x)| < \frac{Θδ}{6k}

whenever I_{Φ}(x) ≤ 1, I_{Φ}(y) ≤ σ (see [2], [10]). Take n_{1} < n_{2} < ⋯ < n_{k−1}, n_{1}, n_{2}, ⋯ n_{k−1} ∈ N. Notice that

I_{Φ} (\frac{x_{n_{1}} + x_{n_{2}} + ⋯ + x_{n_{k−1}}}{k} ) ≤ 1

and I_{Φ}(x_{n_{i}}) ≤ 1 for i = 1, 2, ⋯, k − 1. There exists j_{0} ∈ N such that

(4.4)  ∑_{i=j_{0}+1}^{∞} Φ (\frac{x_{n_{1}}(i) + x_{n_{2}}(i) + ⋯ + x_{n_{k−1}}(i)}{k}) < σ

and

∑_{i=j_{0}+1}^{∞} Φ (x_{n_{j}}(i)) < \frac{δ}{3} (j = 1, 2, …, k − 1).

By (4.1), there exists n_{k} ∈ N such that

(4.5)  ∑_{i=j_{0}+1}^{∞} Φ(x_{n_{k}}(i)) ≥ \frac{δ}{3}.

So, in virtue of (4.2), (4.3), (4.4) and (4.5), we get

I_{Φ} (\frac{x_{n_{1}} + ⋯ + x_{n_{k}}}{k} )

= ∑_{i=1}^{j_{0}} Φ (\frac{x_{n_{1}}(i) + ⋯ + x_{n_{k}}(i)}{k}) + ∑_{i=j_{0}+1}^{∞} Φ (\frac{x_{n_{1}}(i) + ⋯ + x_{n_{k}}(i)}{k})

≤ \frac{1}{k} ∑_{j=1}^{k} ∑_{i=1}^{j_{0}} Φ(x_{n_{j}}(i)) + ∑_{i=j_{0}+1}^{∞} Φ (\frac{x_{n_{k}}(i)}{k}) + \frac{Θδ}{6k}

≤ \frac{1}{k} ∑_{j=1}^{k} ∑_{i=1}^{j_{0}} Φ(x_{n_{j}}(i)) + \frac{1 − Θ}{k} ∑_{i=j_{0}+1}^{∞} Φ (x_{n_{k}}(i)) + \frac{Θδ}{6k}

= \frac{1}{k} ∑_{j=1}^{k} ∑_{i=1}^{∞} Φ(x_{n_{j}}(i)) − \frac{Θ}{k} ∑_{i=j_{0}+1}^{∞} Φ (x_{n_{k}}(i)) + \frac{Θδ}{6k}

≤ 1 − \frac{Θδ}{3k} + \frac{Θδ}{6k} = 1 − \frac{Θδ}{3k} ,

which completes the proof.

Theorem 4.4. For any Orlicz function Φ satisfying (Φ(u)/u) → ∞ as u → ∞ the Orlicz sequence space l_{Φ}^{0} is (kNUC) if and only if both Φ and Ψ satisfy the Δ_{2}-condition at zero, i.e. l_{Φ}^{0} is reflexive.

P r o o f. We only need to prove the sufficiency. Let any ε > 0 and {x_{n}} ⊂ B(l_{Φ}^{0}) with sep({x_{n}}) > ε be given. By Φ ∈ Δ_{2}(0), there exists δ > 0 such that

inf \{I_{Φ}(\frac{x_{n} − x_{m}}{2} ): n ≠ m\} ≥ δ.

By the same argumentation as in Theorem 4.3, we have that for any j ∈ N there exists n_{j} ∈ N such that

(4.6)  ∑_{i=j}^{∞} Φ(x_{n_{j}}(i)) ≥ \frac{δ}{3}.

Take k_{n} ≥ 1 satisfying

∥x_{n}∥_{0} = \frac{1}{k_{n}} (1 + I_{Φ}(k_{n}x_{n})) (n = 1, 2, …).

Such numbers k_{n} ≥ 1 exist by the assumption (Φ(u)/u) → ∞ as u → ∞ (see [10]). Since Ψ satisfies the Δ_{2}-condition at zero, the number

k_{0} = sup{k_{n} : n = 1, 2, … }

is finite (see [2]). Fix n_{1} < n_{2} < ⋯ < n_{k−1}, n_{1}, n_{2}, … n_{k−1} ∈ N. For any n_{k} ∈ N ,

put H = ∏_{i=1}^{k} k_{n_{i}}, h_{j} = ∏_{i≠j} k_{n_{i}}, h = ∏_{i=1}^{k} _{\frac{k_{n_{i}}}{∑_{j=1}^{k} h_{j}}} and λ = \frac{k_{0}^{k−1}}{k_{0}^{k−1}+1}. By Φ ∈ Δ_{2}(0), there

exists Θ ∈ (0, 1) such that

Φ(λu) ≤ (1 − Θ)λΦ(u) 0 ≤ u ≤ Φ^{−1}(k_{0}).

Since Φ is convex, for any l ∈ [0, λ] and u ∈ [0, Φ^{−1}(k_{0})], we have

Φ(lu) = Φ(λ\frac{l}{λ}u) ≤ (1 − Θ)λΦ(\frac{l}{λ}u)

≤ λ(1 − Θ)\frac{l}{λ}Φ(u) ≤ (1 − Θ)lΦ(u).

Since _{\frac{h_{k}}{∑_{i=1}^{k} h_{i}}} = _{\frac{h_{k}}{h_{k}+∑_{i=1}^{k−1} h_{i}}} ≤ \frac{k_{0}^{k−1}}{1+k_{0}^{k−1}} = λ, there holds

(4.7)  (\frac{h_{k}}{∑_{i=1}^{k} h_{i}} u) ≤ (1 − Θ)\frac{h_{k}}{∑_{i=1}^{k} h_{i}} Φ(u)

whenever 0 ≤ u ≤ Φ^{−1}(k_{0}). By Φ ∈ Δ_{2}(0), there exists σ > 0 such that

|I_{Φ}(x + y) − I_{Φ}(x)| < \frac{Θk_{0}^{k}}{1 + k_{0}^{k−1}} · \frac{δ}{6}

if I_{Φ}(x) ≤ k_{0} and I_{Φ}(y) ≤ σ (see [2] and [10]). Notice that I_{Φ} (\frac{x_{n_{1}} +x_{n_{2}} +···+x_{n_{k−1}}}{k} ) < ∞ and I_{Φ}(x_{n_{i}}) < ∞ for i = 1, 2, …, k − 1. So, there exists j_{0} ∈ N such that

∑_{i=j_{0}+1}^{∞} Φ (\frac{x_{n_{1}}(i) + x_{n_{2}}(i) + ⋯ + x_{n_{k−1}}(i)}{k}) < σ

and

∑_{i=j_{0}+1}^{∞} Φ (x_{n_{j}}(i)) < \frac{δ}{3} (j = 1, 2, …, k − 1).

By (4.6), there exists n_{k} ∈ N such that

∑_{i=j_{0}+1}^{∞} Φ(x_{n_{k}}(i)) ≥ \frac{δ}{3}.

Hence

∥∑_{i=1}^{k} x_{n_{i}} ∥_{0} ≤ \frac{∑_{i=1}^{k} h_{i}}{H} (1 + I_{Φ}(\frac{H}{∑_{i=1}^{k} h_{i}} (x_{n_{1}} + x_{n_{2}} + ⋯ + x_{n_{k−1}} + x_{n_{k}})))

= \frac{∑_{i=1}^{k} h_{i}}{H} (1 + ∑_{i=1}^{j_{0}} Φ(\frac{H}{∑_{i=1}^{k} h_{i}} (x_{n_{1}}(i) + ⋯ + x_{n_{k}}(i)))

+ ∑_{i=j_{0}+1}^{∞} Φ(\frac{H}{∑_{i=1}^{k} h_{i}} (x_{n_{1}}(i) + ⋯ + x_{n_{k}}(i))))

= \frac{∑_{i=1}^{k} h_{i}}{H} (1 + ∑_{i=1}^{j_{0}} Φ(\frac{h_{1}}{∑_{i=1}^{k} h_{i}} k_{n_{1}}x_{n_{1}}(i) + ⋯ + \frac{h_{k}}{∑_{i=1}^{k} h_{i}} k_{n_{k}}x_{n_{k}}(i))

+ ∑_{i=j_{0}+1}^{∞} Φ(\frac{H}{∑_{i=1}^{k} h_{i}} (x_{n_{1}}(i) + ⋯ + x_{n_{k−1}}(i)) + \frac{H}{∑_{i=1}^{k} h_{i}} x_{n_{k}}(i)))

≤ \frac{∑_{i=1}^{k} h_{i}}{H} (1 + ∑_{i=1}^{j_{0}}(\frac{h_{1}}{∑_{i=1}^{k} h_{i}} Φ(k_{n_{1}}x_{n_{1}}(i)) + ⋯ + \frac{h_{k}}{∑_{i=1}^{k} h_{i}} Φ(k_{n_{k}}x_{n_{k}}(i)))

+ ∑_{i=j_{0}+1}^{∞} Φ(\frac{h_{k}}{∑_{i=1}^{k} h_{i}} k_{n_{k}}x_{n_{k}}(i)) + \frac{δΘk_{0}^{k}}{6(1 + k_{0}^{k−1})})

≤ \frac{∑_{i=1}^{k} h_{i}}{H} (1 + ∑_{i=1}^{j_{0}}(\frac{h_{1}}{∑_{i=1}^{k} h_{i}} Φ(k_{n_{1}}x_{n_{1}}(i)) + ⋯ + \frac{h_{k}}{∑_{i=1}^{k} h_{i}} Φ(k_{n_{k}}x_{n_{k}}(i)))

+ (1 − Θ)\frac{h_{k}}{∑_{i=1}^{k} h_{i}} ∑_{i=j_{0}+1}^{∞} Φ(k_{n_{k}}x_{n_{k}}(i)) + \frac{δΘk_{0}^{k}}{6(1 + k_{0}^{k−1})})

≤ ∑_{i=1}^{k} \frac{1}{k_{n_{i}}} (1 + I_{Φ}(k_{n_{i}}x_{n_{i}})) − Θ\frac{h_{k}}{∑_{i=1}^{k} h_{i}} ∑_{i=j_{0}+1}^{∞} Φ(k_{n_{k}}x_{n_{k}}(i)) + \frac{δΘk_{0}^{k}}{6(1 + k_{0}^{k−1})}

≤ k − \frac{Θk_{0}^{k−1}}{1 + k_{0}^{k−1}} ∑_{i=j_{0}+1}^{∞} Φ(k_{0}x_{n_{k}}(i)) + \frac{δΘk_{0}^{k}}{6(1 + k_{0}^{k−1})}

≤ k − \frac{Θk_{0}^{k}}{1 + k_{0}^{k−1}} ∑_{i=j_{0}+1}^{∞} Φ(x_{n_{k}}(i)) + \frac{δΘk_{0}^{k}}{6(1 + k_{0}^{k−1})}

≤ k − \frac{Θk_{0}^{k}}{1 + k_{0}^{k−1}} \frac{δ}{3} + \frac{δΘk_{0}^{k}}{6(1 + k_{0}^{k−1})} = k − \frac{δΘk_{0}^{k}}{6(1 + k_{0}^{k−1})}.

This completes the proof.

Corollary 4.2. If Φ is an Orlicz function satisfying (Φ(u)/u) → ∞ as u → ∞, then the Orlicz sequence spaces l_{Φ} and l_{Φ}^{0} are NUC if and only if both Φ and Ψ satisfy the Δ_{2}-condition at zero.

P r o o f. It can be proceeded analogously to the proof of Corollary 4.1.

Theorem 4.5. For any Orlicz function Φ satisfying Φ(u)/u → ∞ as u → ∞ the following statements are equivalent:

l_{Φ} is nearly uniformly smooth. l_{Φ} is nearly uniformly ∗-smooth. Φ and Ψ satisfy the δ_{2}-condition at zero.

P r o o f. (c) ⇒ (a). By the last theorem, l_{Ψ}^{0} is NUC. Therefore, its dual l_{Φ} is NUS (see [67]). Consequently, we need only to prove that (b) ⇒ (c). We will show that (b) implies the Δ_{2}-condition at zero for Φ. If Φ does not satisfy the Δ_{2}-condition at zero, we can construct x ∈ S(l_{Φ}) such that I_{Φ}(x) ≤ 1 and I_{Φ}((1 + \frac{1}{n})x) = ∞ for every n ∈ ℕ (see [10] and [50’]). Take a sequence {i_{k}} of natural numbers such that i_{k} ↑ and

∑_{i=i_{k}+1}^{i_{k+1}} Φ((1 + \frac{1}{k} )x(i)) ≥ 1 (k ∈ ℕ).

Put

x_{k} = (0, 0, …, 0, x(i_{k} + 1), x(i_{k} + 2), …, x(i_{k+1}), 0, 0, …) (k ∈ ℕ).

Then it is obvious that

\frac{k}{k + 1} ≤ ∥x_{k}∥ ≤ 1 (k ∈ ℕ).

Moreover,

(4.8)  x_{k} → 0 .

Indeed, for every y^{∗} ∈ (l_{Φ})^{∗} we have y^{∗} = y_{0}^{∗} + y_{1}^{∗} uniquely, where y_{0}^{∗} is the order continuous part of y^{∗} and y_{1}^{∗} is the singular part of y^{∗}, i.e. y_{1}^{∗}(x) = 0 for any x ∈ h_{Φ} (see [10]). The functional y_{0}^{∗} is generated by some y_{0} ∈ l_{Ψ} by the formmula

y_{0}^{∗}(x) = ⟨x, y_{0}⟩ = ∑_{i=1}^{∞} x(i)y_{0}(i) (x ∈ l_{Φ}).

Let λ > 0 be such that ∑_{i=1}^{∞} Ψ(λy_{0}(i)) < ∞. Since x_{k} ∈ h_{Φ} for any k ∈ ℕ, we have

⟨x_{k}, y^{∗}⟩ = ⟨x_{k}, y_{0}^{∗}⟩ = ∑_{i=i_{k}+1}^{i_{k+_{1}}} x(i)y_{0}(i)

≤ \frac{1}{λ}( ∑_{i=i_{k}+1}^{i_{k+1}} Φ(x(i)) + ∑_{i=i_{k}+1}^{i_{k+1}} Ψ(λy_{0}(i)))

→ 0 k → ∞,

i.e. (4.8) holds. Since the space l_{Φ} is nearly uniformly ∗-smooth, it has property A_{2}^{ε}, i.e. for any ε > 0 there exists δ ∈ (0, 1) such that for each weakly null sequence (z_{n}) in B(l_{Φ}) there is m > 1 such that

∥z_{1} + tz_{m}∥ ≤ 1 + tε

whenever t ∈ [0, δ] (see [67] and [68]). Take k_{0} ∈ ℕ such that \frac{2}{k+1} < (1−ε)δ if k ≥ k_{0}. We have for k ≥ k_{0},

\begin{aligned}1 + δε & ≥ ∥x + δx_{k}∥ ≥ ∥(1 + δ)x_{k}∥ ≥ (1 + δ)\frac{k}{k + 1} \\ & = (1 + δ)(1 − \frac{1}{k + 1}) > 1 + δ − \frac{2}{k + 1}, \end{aligned}

whence \frac{2}{k+1} > (1 − ε)δ. This is a contradiction which finishes the proof of the fact that (b) implies the Δ_{2}-condition at zero for Φ. Next, we will show that (b) implies the δ_{2}-condition for Ψ. By the above part of the proof, we can assume that l_{Ψ} is nearly uniformly ∗-smooth and Φ satisfies the Δ_{2}-condition at zero. So, l_{Φ} is order continuous. Moreover, any Orlicz space l_{Φ} has the Fatou property and consequently, it is weakly sequentially complete. So, in view of Corollary 4.1, l_{Φ} is nearly uniformly smooth and consequently reflexive. This yields the Δ_{2}-condition at zero for Ψ.

5. Weak orthogonality and uniform non-squareness

Garcia-Falset [30’] has proved that if a Banach space X is weakly orthogonal and uniformly nonsquare, then X has the FPP. So, we will present now criteria for uniform non-squareness in Orlicz spaces and criteria for the weak orthogonality of Köthe sequence spaces. We say following Sims [72] that a Banach space X is weakly orthogonal if for any x ∈ S(X) and any weakly null sequence in X, we have

lim_{n→∞}| ∥x_{n} + x∥ − ∥x_{n} − x∥ | → 0.

Theorem 5.1 (see [25]). A Köthe sequence space X is weakly orthogonal if and only if it is order continuous.

P r o o f. Necessity. If X is not order continuous, then X_{a}, the subspace of all absolutely continuous elements from X, is a closed proper subspace of X. By Riesz’s Lemma, for any θ ∈ (0, 1) there is x_{0} ∈ S(X) such that ∥x_{0} − x∥ ≥ θ for any x ∈ X_{a}. Take a sequence {n_{i}} of natural numbers such that n_{i} ↑ ∞ and

∥ ∑_{j=n_{i}+1}^{n_{i+1}} x_{θ}(j)e_{j}∥ ≥ (1 − \frac{1}{i})θ,

where θ ∈ (\frac{2}{3}, 1). Then, setting

x_{i} = ∑_{j=n_{i}+1}^{n_{i+1}} x_{θ}(j)e_{j}

for i = 1, 2, …, we immediately get

(5.1)  (1 − \frac{1}{i})θ ≤ ∥x_{i}∥ ≤ 1

for i = 1, 2, …. Moreover,

(5.2)  x_{i} → 0 i → ∞.

Really, it is easy to see that for any f = g + φ ∈ X^{∗} with g ∈ X′ (the Köthe dual of X) and φ ∈ (X_{a})^{⊥}, we have ⟨x_{i}, f⟩ = ⟨x_{i}, g⟩. Since ∑_{j=1}^{∞} x_{θ}(j)g(j), we get

⟨x_{i}, g⟩ = ∑_{j=n_{i}+1}^{n_{i+1}} x_{θ}(j)g(j) → 0 i → ∞.

Moreover, by (5.1) we have

∥x_{θ} + x_{i}∥ ≥ 2∥x_{i}∥ ≥ 2(1 − \frac{1}{i})θ

for i = 1, 2, …. However, ∥x_{θ} − x_{i}∥ ≤ 1, so by

2(1 − \frac{1}{i})θ > \frac{4}{3}(1 − \frac{1}{i}) → \frac{4}{3}

we have

lim_{i→∞} | ∥x_{θ} + x_{i}∥ − ∥x_{θ} − x_{i}∥ | ≥ \frac{1}{3},

i.e. X is not weakly orthogonal. Sufficiency. For any ε > 0, any x ∈ S(X) and any weakly null sequence {x_{n}} in X, there are i_{0} and n_{0} ∈ ℕ such that

∥ ∑_{i=i_{0}+1}^{∞} x(i)e_{i}∥ < \frac{ε}{4}  ∥∑_{i=1}^{i_{0}} x_{n}(i)e_{i}∥ < \frac{ε}{4}

for n ≥ n_{0}. Put

\overline{x}_{n} = ∑_{i=1}^{i_{0}} x(i)e_{i} + ∑_{i=i_{0}+1}^{∞} x_{n}(i)e_{i} \overline{y}_{n} = ∑_{i=1}^{i_{0}} x(i)e_{i} − ∑_{i=i_{0}+1}^{∞} x_{n}(i)e_{i}

for n = 1, 2, … Then ∥\overline{x}_{n}∥ = ∥\overline{y}_{n}∥ for every n ∈ ℕ and

\begin{aligned}∥(x + x_{n}) − \overline{x}_{n}∥ & = ∥∑_{i=1}^{i_{0}} x_{n}(i)e_{i} + ∑_{i=i_{0}+1}^{∞} x(i)e_{i}∥ \\ & ≤ ∥∑_{i=1}^{i_{0}} x_{n}(i)e_{i}∥ + ∥ ∑_{i=i_{0}+1}^{∞} x(i)e_{i}∥ ≤ q \frac{ε}{4} + \frac{ε}{4} = \frac{ε}{2} \end{aligned}

for n ≥ n_{0}. Moreover,

\begin{aligned}∥(x − x_{n}) − \overline{y}_{n}∥ & = ∥ ∑_{i=i_{0}+1}^{∞} x(i)e_{i} − ∑_{i=1}^{i_{0}} x_{n}(i)e_{i}∥ \\ & ≤ ∥∑_{i=1}^{i_{0}} x_{n}(i)e_{i}∥ + ∥ ∑_{i=i_{0}+1}^{∞} x(i)e_{i}∥ ≤ \frac{ε}{4} + \frac{ε}{4} = \frac{ε}{2} \end{aligned}

for n ≥ n_{0}. Hence, we have

| ∥x + x_{n}∥ − ∥x − x_{n}∥ | = | ∥x + x_{n}∥ − ∥\overline{x}_{n}∥ + ∥x − x_{n}∥ − ∥\overline{y}_{n}∥ |

≤ | ∥x + x_{n}∥ − ∥\overline{x}_{n}∥ | + | ∥x − x_{n}∥ − ∥\overline{y}∥ |

≤ \frac{ε}{2} + \frac{ε}{2} = ε

for n ≥ n_{0}. This means that lim_{n→∞} | ∥x + x_{n}∥ − ∥x − x_{n}∥ | = 0.

In this section we do not assume that Φ satisfies condition (ii) from the definition of an Orlicz function.

Corollary 5.1. Orlicz sequence spaces l_{Φ} equipped with the Luxemburg norm or with the Orlicz norm are weakly orthogonal if and only if Φ ∈ Δ_{2}(0).

P r o o f. Since OC of l_{Φ} and l_{Φ}^{0} is equivalent to Φ ∈ Δ_{2}(0), the corollary follows immediately by Theorem 5.1.

The notion of inform non-squarences of a Banach space has been introduced by R.S. James [49]. Recall that a Banach space X is said to be uniformly non-square (UNSQ for short) if there is ε ∈ (0, 1) such that for every x, y ∈ B(X) there holds

min(∥\frac{x + y}{2} ∥, ∥\frac{x − y}{2} ∥) ≤ 1 − ε.

Theorem 5.2 (see [39], [46] and [73]). (a) In the case of a nonatomic infinite and σ-finite measure space as well as in the case of the counting measure space the Orlicz space L_{Φ} equipped with the Luxemburg norm is uniformly non-square if and only if it is reflexive. In the case of any finite nonatomic measure space the Orlicz space L_{Φ} equipped with the Luxemburg norm is uniformly non-square if and only if L_{Φ} is reflexive and Φ(b(Φ))μ(T) < 2, where b(Φ) = sup{u ≥ 0: Φ is linear on the interval [0, u]}.

P r o o f. We only prove (b) beacuse in case (a) the proof is similar and easier. Necessity. James [49] has proved that UNSQ implies reflexivity. Assume that Φ(b(Φ))μ(T) ≥ 2. Since μ is nonatomic, there are measurable disjoint sets A, B in T such that Φ(b(Φ))μ(A) = 1 and μ(A) = μ(B). Define

x = b(Φ)χ_{A}, y = b(Φ)χ_{B}.

Then I_{Φ}(x) = I_{Φ}(y) = 1, whence ∥x∥_{Φ} = ∥y∥_{Φ} = 1. Since Φ is linear on [0, b(Φ)], we have Φ(\frac{b(Φ)}{2}) = \frac{1}{2}Φ(b(Φ)), whence I_{Φ}(\frac{x±y}{2} ) = \frac{1}{2}\{I_{Φ}(x)+I_{Φ}(y)\} = 1. Consequently, ∥\frac{x+y}{2} ∥_{Φ} = 1, and L_{Φ} is not UNSQ. Sufficiency. By Lemma 0.1 there is u_{0} > b(Φ) and θ ∈ (0, 1) such that

Φ(\frac{u}{2}) ≤ \frac{1 − θ}{2} Φ(u) (u ≥ u_{0}).

Let c ∈ (b(Φ), u_{0}) satisfy Φ(c)μ(T) < 2. Since Φ(\frac{u}{2}) < \frac{1}{2}Φ(u) for any u ≥ c and the interval [c, u_{0}] is compact, we get

η := sup\{\frac{2Φ(\frac{u}{2})}{Φ(u)}: u ∈ [c, u_{0}]\} < 1,

by continuity of the function 2Φ(u/2)/Φ(u). Denoting δ = min(θ, 1 − η), we have

(5.3)  Φ(\frac{u}{2}) ≤ \frac{1 − δ}{2} Φ(u) (u ≥ c).

There is σ ∈ (0, 1) such that Φ(c)μ(T) = 2σ. Denote α = 1/\sqrt{σ}. We will show that Φ(u) + Φ(v) ≥ αΦ(c) implies Φ(\frac{u+v}{2} ) + Φ(\frac{u−v}{2} ) ≤ (1 − β)[Φ(u) + Φ(v)] for some β ∈ (0, 1) inependent of u and v. We will consider two cases.

I. max(|u|, |v|) ≥ c. Then we have min(|u + v|, |u − v|) ≤ max(|u|, |v|). Consequently, by (5.3), we get

(5.4)

Φ(\frac{u + v}{2} ) + Φ(\frac{u − v}{2} ) ≤ \frac{1}{2}\{Φ(u) + Φ(v)\} + \frac{1 − θ}{2} Φ(max(|u|, |v|))

≤ \frac{1}{2}\{Φ(u) + Φ(v)\} + \frac{1 − θ}{2} \{Φ(u) + Φ(v)\}

= (1 − \frac{θ}{2})\{Φ(u) + Φ(v)\}.

II. max(|u|, |v|) < c. Assume without loss of generality that |u| ≥ |v|. We have Φ(v) ≥ (α−1)Φ(c), whenever Φ(u)+Φ(v) ≥ αΦ(c), because otherwise Φ(u)+Φ(v) < Φ(c) + (α − 1)Φ(c) = αΦ(c), a contradiction. Therefore

\frac{Φ(u)}{Φ(u) + Φ(v)} < \frac{Φ(c)}{Φ(c) + Φ(v)} ≤ \frac{Φ(c)}{Φ(c) + (α − 1)Φ(c)} = \frac{1}{α} = \sqrt{σ} < 1.

Consequently, assuming without loss of generality that min(|u + v|, |u − v|) = |u + v|, we have |u + v| ≤ max(|u|, |v|) and so

Φ(\frac{u + v}{2} ) ≤ Φ(\frac{max(|u|, |v|)}{2}) = Φ(\frac{u}{2}) ≤ \frac{1}{2}Φ(u)

≤ \frac{\sqrt{σ}}{2} \{Φ(u) + Φ(v)\}.

Therefore

Φ(\frac{u + v}{2} ) + Φ(\frac{u − v}{2} ) ≤ \frac{1 + \sqrt{σ}}{2} \{Φ(u) + Φ(v)\}.

Combining the last inequality and inequality (5.4), we get

(5.5) Φ(u) + Φ(v) ≥ αΦ(c) ⇒ Φ(\frac{u + v}{2} ) + Φ(\frac{u − v}{2} ) ≤ (1 − β)[Φ(u) + Φ(v)],

where β = min(\frac{θ}{2}, \frac{1−\sqrt{σ}}{2} ). Assume that x, y ∈ S(L_{Φ}) and define

A = {t ∈ T: Φ(x(t)) + Φ(y(t)) ≥ αΦ(c)}.

We have I_{Φ}(xχ_{T \A}) + I_{Φ}(yχ_{T \A}) < αΦ(c)μ(T) = 2\sqrt{σ}. Since, by the Δ_{2}-condition at infinity for Φ, there holds I_{Φ}(x) = I_{Φ}(y) = 1, we have

I_{Φ}(xχ_{A}) + I_{Φ}(yχ_{A}) > 2(1 − \sqrt{σ}).

Therefore,

2−\{I_{Φ}(\frac{x + y}{2} ) + I_{Φ}(\frac{x − y}{2} ) = I_{Φ}(x) + I_{Φ}(y) − \{I_{Φ}(\frac{x + y}{2} ) + I_{Φ}(\frac{x − y}{2} )\}

≥ I_{Φ}(xχ_{A}) + I_{Φ}(yχ_{A}) − \{I_{Φ}(\frac{x + y}{2} χ_{A}) + I_{Φ}(\frac{x − y}{2} χ_{A})\}

≥ β(I_{Φ}(xχ_{A}) + I_{Φ}(yχ_{A})\} ≥ 2β(1 − \sqrt{σ}).

Consequently

I_{Φ}(\frac{x + y}{2} ) + I_{Φ}(\frac{x − y}{2} ) ≤ 2 − 2β(1 − \sqrt{σ}) = 2[1 − β(1 − \sqrt{σ})]

and so

min(I_{Φ}(\frac{x + y}{2} , I_{Φ}(\frac{x + y}{2} )) ≤ 1 − β(1 − \sqrt{σ}).

Since Φ satisfies the Δ_{2}-condition at ∞, there is q ∈ (0, 1) independent of x and y such that

min(∥\frac{x + y}{2} ∥_{Φ}, ∥\frac{x + y}{2} ∥_{Φ}) ≤ 1 − q,

which finishes the proof.

Notes. Uniform non-squareness of Musielak-Orlicz spaces was characterized in [42]. For Orlicz-Lorentz spaces it was done in [43], where uniform non-squareness of some Calderon-Lozanowsky spaces was also considered.

Notes. The characteristic of convexity of Orlicz function spaces equipped with the Luxemburg norm was calculated in [46] in the case when the measure spaces is nonatomic and infinite. Recall that this coefficient for a Banach space X is defined by

ε_{0}(X) = inf{ε ∈ (0, 2]: δ_{X}(ε) > 0},

where δ_{X} denotes the modulus of convexity of X. In the case of nonatomic and finite measure space ε_{0}(L^{Φ}) was calculated in [46’]. Lower and upper estimates for the characteristic of convexity of Köthe-Bochner spaces were given in [46^{∗}].

6. Opial property and uniform Opial property in modular sequence spaces

In this section we will present some results on uniform Opial property of modular sequence spaces. As a corollary we will obtain criteria for Opial property and uniform Opial property of Orlicz sequence spaces for both (the Luxemburg and the Orlicz) norms.

Let X be a real vector space. A functional m: X → [0, ∞] is called a modular if (see [63] and [61]):

m(x) = 0 if and only if x = 0, m(−x) = m(x) for all x ∈ X, m(αx + βy) ≤ αm(x) + βm(y) for all x, y ∈ X and α, β ≥ 0 such that α + β = 1 (i.e. m is convex). For any modular m on X, the space

X_{m} = {x ∈ X: m(λx) < ∞ λ > 0}

is called the modular space (generated by m). It is obvious that X_{m} is a vector space. The functional

∥x∥ = inf{λ > 0: m(x/λ) ≤ 1}

is a norm in X_{m}, which is called the Luxemburg norm (see [14] and [15]). A modular m is said to satisfy the Δ_{2}-condition (m ∈ Δ_{2} for short) if for any ε > 0 there exist constants K ≥ 2 and a > 0 such that

m(2x) ≤ Km(x) + ε

for all x ∈ X_{m} with m(x) ≤ a. If m satisfies the Δ_{2}-condition for any a > 0 with K ≥ 2 dependent on a, we say that m satisfies the strong Δ_{2}-condition (m ∈ Δ_{2}^{s} for short). In this section a function Φ: (−∞, ∞) → [0, ∞) is said to be an Orlicz function if it is convex, even and Φ(0) = 0 (see [10], [60], [61], [63], [55] and [69]). For a given Orlicz function Φ one can define on the the space l^{0} of all real sequences x = (x(i)) the modular

m_{Φ}(x) = ∑_{i=1}^{∞} Φ(x(i)).

The modular space (ℓ^{0})_{m_{Φ}} is called an Orlicz sequence space (see [10], [55], [60], [61], [63] and [69]). An Orlicz function Φ is said to satisfy the Δ_{2}-condition at zero (Φ ∈ Δ_{2} for short) if there are constants K ≥ 2 and a > 0 such that Φ(a) > 0 and Φ(2u) ≤ KΦ(u) for all real u with |u| < a. It is easy to see that if Φ vanishes only at zero, then m_{Φ} ∈ Δ_{2}^{s} whenever Φ ∈ Δ_{2}(0). Let ℓ^{0} be defined as above and ℕ denote the set of natural numbers. A Banach space X ⊂ ℓ^{0} is said to be a Köthe sequence space (or a Banach sequence lattice) if there is a sequence x = (x(i))_{i=1}^{∞} ∈ X with all x(i) ≠ 0 and for every x ∈ ℓ^{0} and y ∈ X with |x(i)| ≤ |y(i)| for all i ∈ ℕ it follows that x ∈ X and ∥x∥ ≤ ∥y∥. An element x ∈ X is said to be absolutely continuous if

lim_{n→∞} ∥(0, …, 0, x(n + 1), x(n + 2), …)∥ = 0.

The set of all absolutely continuous elements in X is denoted by X_{a} and it is a subspace of X. X is called absolutely continuous if X_{a} = X. We say that a Banach sequence lattice X has the Fatou property if for any x ∈ X and a sequence {x_{n}} in X such that 0 ≤ x_{n} ≤ x and x_{n} ↑ x, there holds ∥x_{n}∥ ↑ ∥x∥ (for the theory of Köthe sequence spaces we refer to [52’]). A Banach space X is said to have the Opial property (see [64] and [2]) if for every weakly null seuqence {x_{n}} and every x ≠ 0 in X there holds

lim inf_{n→∞} ∥x_{n}∥ < lim inf_{n→∞} ∥x_{n} + x∥.

The Opial property is important because Banach spaces with this property have the weak fixed point property (see [66]). Opial has proved in [64’] that the Lebesgue sequence space ℓ_{p} (1 < p < ∞) have this property but L_{p}[0, 2π] (p ≠ 2, 1 < p < ∞) have not. Franchetti [30^{∗}] has shown that any infinite-dimensional Banach space admits an equivalent norm under which it has the Opial property. A Banach space X is said to have the uniform Opial property (see [66]) if for every ε > 0 there exists τ > 0 such that for any weakly null sequence {x_{n}} in S(X) (the unit sphere of X) and x ∈ X with ∥x∥ ≥ ε there holds

1 + τ ≤ lim inf_{n→∞} ∥x_{n} + x∥.

Let β be the ball-measure (that is the Hausdorff measure) of noncompactness in X, i.e.

\begin{matrix}β(A) = inf\{ & ε > 0: \\ ≤ ε\} \end{matrix}

for any A ⊂ X. A Banach space X is said to have property (L) if lim_{ε→1−} Δ(ε) = 1, where (see [33] and [34])

Δ(ε) = inf{1 − inf[∥x∥: x ∈ A]},

and the first infimum is taken over all closed sets A in the unit ball B(X) of X with β(A) ≥ ε. The function Δ is called the modulus of noncompact convexity (see [34]). It has been proved in [66] that property (L) is useful to study the fixed point property and that a Banach space X has property (L) if and only if it is reflexive and has the uniform Opial property.

We start with the following auxiliary lemma.

Lemma 6.1. Assume that m ∈ Δ_{2}^{s}. Then for every L > 0 and ε > 0 there exists δ = δ(L, ε) > 0 such that for all x, y in X_{m} with m(x) ≤ L and m(y) ≤ δ, there holds |m(x + y) − m(y)| < ε.

P r o o f. Let L > 0 and ε > 0 be given. By m ∈ Δ_{2}^{s}, we conclude that there is K_{0} ≥ 2 such that

m(2x) ≤ K_{0}m(x) + ε/8

for all x ∈ X_{m} with m(x) ≤ L. Set β = ε/2K_{0}L. Using again m ∈ Δ_{2}^{s}, one can find K_{1} ≥ 2 such that

m (\frac{2}{β} x) ≤ K_{1}m(x) + ε/8

for all x ∈ X_{m} with m(x) ≤ L. Set δ = ε/2βK_{1} and assume that m(x) ≤ L and m(y) ≤ δ. Then

m(x + y) = m((1 − β)x + β(x + β^{−1}y))

≤ (1 − β)m(x) + βm(x + β^{−1}y) ≤ m(x) + βm(2^{−1}(2x + 2β^{−1}y)) ≤ m(x) + 2^{−1}βm(2x) + 2^{−1}βm(2β^{−1}y)

≤ m(x) + 2^{−1}βK_{0}L + \frac{ε}{8} + 2^{−1}βm(y) + \frac{ε}{8}

< m(x) + ε.

In a similar way we can show that m(x)−ε < m(x+y). Hence |m(x+y)−m(x)| < ε whenever m(x) ≤ L and m(y) ≤ δ, which finishes the proof.

Corollary 6.1. If m ∈ Δ_{2}^{s}, then for any x ∈ X_{m}, ∥x∥ = 1 if and only if m(x) = 1.

P r o o f. We only need to show that ∥x∥ = 1 implies m(x) = 1 because the opposite implication is obvious. Assume that m ∈ Δ_{2}^{s}. We can easily get from Lemma 6.1 that the function f defined on ℝ := (−∞, ∞) by f(λ) = m(λx) is continuous. Namely, it easily follows by m ∈ Δ_{2}^{s} that f is finitely valued, which yields that f is continuous. We can also get this fact on the base of Lemma 6.1. Take any ε > 0 and λ_{0} ∈ ℝ \ {0}, and apply Lemma 6.1 which L = m(λ_{0}x) and δ = δ(L, ε). We have |m(x + y) − m(x)| < ε whenever m(x) ≤ L and m(y) ≤ δ. Hence

|m(λx) − m(λ_{0}x)| = |m((λ − λ_{0})x + λ_{0}x) − m(λ_{0}x)| < ε

whenever |λ − λ_{0}| < δ. So, we easily get that m(x) = 1 whenever ∥x∥ = 1.

Lemma 6.2. If m ∈ Δ_{2}, then for any sequence (x_{n}) in X_{m} the condition ∥x_{n}∥ → 0 holds if and only if m(x_{n}) → 0.

P r o o f. It is easy to see that ∥x_{n}∥ → 0 if and only if m(λx_{n}) → 0 for each λ > 0. By m ∈ Δ_{2} it follows that the property from Lemma 6.1 holds for sufficiently small positive L (say L ≤ L_{0}). Assume that m(x_{n}) → 0. There is m ∈ ℕ such that m(x_{n}) ≤ L_{0} for all n ≥ m. So, for any ε ∈ (0, 1) there is K_{ε} > 0 such that

m (\frac{x_{n}}{ε} ) ≤ K_{ε}m(x_{n}) + \frac{ε}{2}

for n sufficiently large. Let n_{0} ∈ ℕ be so large that m(x_{n}) ≤ \frac{1}{2K_{ε}} for n > n_{0}. Hence m(x_{n}/ε) ≤ K_{ε}m(x_{n}) + \frac{1}{2} ≤ 1 for n > n_{0} sufficiently large. This yields ∥x_{n}∥ ≤ ε for n sufficiently large. The opposite implication follows by the inequality m(x) ≤ ∥x∥ for x with ∥x∥ ≤ 1.

Lemma 6.3. If m ∈ Δ_{2}^{s}, then for any ε > 0 there exists δ = δ(ε) > 0 such that ∥x∥ ≥ 1 + δ whenever m(x) ≥ 1 + ε.

P r o o f. Suppose that there exist ε_{0} > 0 and a sequence {x_{n}} in X_{m} such that ∥x_{n}∥ ↓ 1 and m(x_{n}) ≥ 1 + ε_{0}. Since m ∈ Δ_{2}^{s}, for any ε > 0 there exists δ > 0 such that

|m(x + y) − m(ε)| < ε

whenever m(x) ≤ 1 and m(y) ≤ δ (see Lemma 6.1). We may assume without loss of generality that 1 − 1/∥x_{n}∥ < δ. Hence, applying the fact that m(x_{n}/∥x_{n}∥) = 1 for any n ∈ N (see the proof of Lemma 6.1), we get

|m((1 − 1/∥x_{n}∥)x_{n} + x_{n}/∥x_{n}∥) − m(x_{n}/∥x_{n}∥)| < ε,

i.e. |m(x_{n}) − 1| < ε. This contradiction shows that Lemma 6.3 is true.

Theorem 6.1. Suppose that a Köthe sequence space X has the Fatou property. Then X is absolutely continuous whenever it has the uniform Opial property.

P r o o f. Assume that X is not order continuous. Take ε = 1/2 and an arbitrary τ > 0. Let θ = 1/(1 + τ/2), whence θ > 1/(1 + τ). By Riesz’s lemma (see [73’], p. 64), for any θ ∈ (0, 1) there is x_{0} ∈ S(X) such that ∥x_{0} − x∥ > θ for any x ∈ X_{a}. Let x_{0} corresponds to θ = 1/(1 + τ/2). By the Fatou property of X, ∥∑_{i=1}^{n} x_{0}(i)e_{i}∥ ↑ ∥x_{0}∥ = 1. Let n_{1} = 0. There is n_{2} ∈ ℕ such that

∥∑_{i=1}^{n_{2}} x_{0}(i)e_{i}∥ ≥ (1 − \frac{1}{2}) θ.

Since ∑_{i=1}^{n_{2}} x_{0}(i)e_{i} ∈ X_{a}, it follows that

∥ ∑_{i=n_{2}+1}^{∞} x_{0}(i)e_{i}∥ ≥ θ.

So, there is n_{3} ∈ ℕ, n_{3} > n_{2}, such that

∥ ∑_{i=n_{2}+1}^{n_{3}} x_{0}(i)e_{i}∥ ≥ (1 − \frac{1}{3}) θ.

One can find by induction a sequence (n_{j})_{j=1}^{∞} in ℕ such that n_{1} = 0, n_{1} < n_{2} < ⋯, and

∥ ∑_{i=n_{j}+1}^{n_{j+1}} x_{0}(i)e_{i}∥ ≥ (1 − \frac{1}{j + 1}) θ > (1 − \frac{1}{j}) θ (j = 1, 2, …).

Define

x_{j} = ∑_{i=n_{j}+1}^{n_{j+1}} x_{0}(i)e_{i} (i = 1, 2, …).

It is obvious that x^{∗}(x_{j}) = 0, j = 1, 2, …, for any singular functional x^{∗} ∈ X^{∗}. If x^{∗} ∈ X^{∗} is order continuous, there is y = (y(i))_{i=1}^{∞} ∈ X′ (= the Köthe dual of X, see [52’]) such that

x^{∗}(z) = ∑_{i=1}^{∞} y(i)z(i) (∀z = (z(i))_{i=1}^{∞} ∈ X).

Since ∑_{i=1}^{∞} y(i)x_{0}(i) is convergent it follows that

x^{∗}(x_{j}) = x^{∗}( ∑_{i=n_{j}+1}^{n_{j+1}} x_{0}(i)e_{i}) = ∑_{i=n_{j}+1}^{n_{j+1}} y(i)x_{0}(i) → 0

as j → ∞. Therefore, x_{j} → 0 weakly as j → ∞. Moreover

(6.1)  (1 − \frac{1}{j}) θ ≤ ∥x_{j}∥ ≤ 1 (j = 1, 2, …).

Define y_{j} = x_{j}/∥x_{j}∥. Then ∥y_{j}∥ = 1 for all j ∈ ℕ and

∥x_{0} − y_{j}∥ ≤ ∥\frac{x_{0} − x_{j}}{∥x_{j}∥} ∥ ≤ \frac{∥x_{0}∥}{∥x_{j}∥} ≤ \frac{1}{(1 − \frac{1}{j})θ} → \frac{1}{θ}.

Since 1/θ < 1 + τ, there is j_{0} ∈ ℕ such that

(6.2)  ∥x_{0} − y_{j}∥ < 1 + τ (∀j > j_{0}).

Since x_{j} → 0 weakly, inequalities (6.1) yield that y_{j} → 0 weakly. Hence and by (6.2) it follows that X has not the uniform Opial property. This finishes the proof.

Theorem 6.2. Assume that a modular m ∈ Δ_{2}^{s} and m is countably orthogonally additive and that the modular sequence space X_{m} is a Banach space. Then X_{m} has the uniform Opial property.

P r o o f. Let ε > 0 be given. There is ε_{1} > 0 such that (see Lemma 6.2) m(x) ≥ ε_{1} whenever ∥x∥ ≥ ε. Since m ∈ Δ_{2}^{s}, by Lemma 6.1, there is δ ∈ (0, ε_{1}/4) such that

|m(x + y) − m(x)| < ε_{1}/8

whenever m(x) ≤ 1 and m(y) ≤ δ. By countable orthogonal additivity of m, there is i_{0} ∈ ℕ such that

m( ∑_{i=i_{0}+1}^{∞} x(i)e_{i}) ≤ δ.

Let {x_{n}} be a weakly null sequence in S(X). It is obvious that x_{n} → 0 coordinatewise. Hence, there is n_{0} ∈ ℕ such that

m(∑_{i=1}^{i_{0}} x_{n}(i)e_{i}) ≤ δ (∀n ≥ n_{0}).

Therefore

m(x_{n} + x) = m(∑_{i=1}^{i_{i}}(x_{n}(i) + x(i))e_{i}) + m(∑_{i=1}^{∞}(x_{n}(i) + x(i))e_{i})

≥ m(∑_{i=1}^{i_{0}} x(i)e_{i}) − \frac{ε_{1}}{8} + m( ∑_{i=i_{0}+1}^{∞} x_{n}(i)e_{i}) − \frac{ε_{1}}{8}

≥ \frac{3}{4}ε_{1} − \frac{ε_{1}}{8} − \frac{ε_{1}}{8} + m(x_{n}) − \frac{ε_{1}}{8} − \frac{ε_{1}}{8}

= 1 + \frac{ε_{1}}{4}

for n ≥ n_{0}. By Lemma 6.3, there is ε_{2} > 0 that depends only on ε_{1} and such that ∥x_{n} + x∥ > 1 + ε_{2} whenever n ≥ n_{0}. This means that X_{m} has the uniform Opial property.

In this section we write m in place of I_{Φ} in Orlicz spaces.

Corollary 6.2. Orlicz sequence spaces ℓ^{Φ} equipped with the Luxemburg norm have the uniform Opial property if and only if Φ ∈ Δ_{2}(0).

P r o o f. Sufficiency. Orlicz spaces ℓ^{Φ} are Banach spaces and they are modular spaces (ℓ^{0})_{m_{Φ}}, where m_{Φ}(x) = ∑_{i=1}^{∞} Φ(x(i)) for x = (x(i)) ∈ ℓ^{0}. If Φ ∈ Δ_{2}(0), then

m_{Φ} is countably orthogonally additive and m_{Φ} ∈ Δ_{2}^{s}. Therefore, the uniform Opial property of ℓ^{Φ} follows by Theorem 6.2. Necessity. If Φ ∉ Δ_{2}(0), then ℓ^{Φ} contains an order isometric copy of ℓ^{∞} (see [50’]), so ℓ^{Φ} is not absolutely continuous. Since ℓ^{Φ} has the Fatou property (see [60]), by Theorem 6.1, ℓ^{Φ} has not the uniform Opial property.

Corollary 6.3. Orlicz sequence spaces l^{Φ} equipped with the Luxemburg norm have the Opial property if and only if Φ ∈ Δ_{2}(0).

P r o o f. If Φ ∈ Δ_{2}(0), then by Corollary 6.3, l^{Φ} has the uniform Opial property, so Opial property as well. Assuming that Φ ∉ Δ_{2}(0), one can find a sequence {x_{n}} in s(l^{Φ}) such that x_{m}⊥x_{n} for m ≠ n and x = ∑_{n=1}^{∞} x_{n} ∈ S(l^{Φ}) (see [50’]). Then we

easily get that x_{n} → 0 weakly. However, ∥x_{1} + x_{n}∥ = 1 for any n > 2. Consequently l^{Φ} fails to have the Opial property.

Corollary 6.4. The Nakano sequence spaces ℓ^{(p_{i})} with 1 ≤ p_{i} < ∞ for all i ∈ ℕ has the uniform Opial property if and only if lim sup_{i→∞} p_{i} < ∞.

P r o o f. The Nakano space ℓ^{(p_{i})} is a Banach space and it is generated by the modular m(x) = ∑_{i=1}^{∞} |x(i)|^{p_{i}} defined on ℓ^{0} (see [63’] and [63]). If lim sup_{i→∞} p_{i} < ∞, then

m ∈ Δ_{2}^{s} and m is countably orthogonally additive. Therefore, by Theorem 6.2, ℓ^{(p_{i})} has the uniform Opial property. If lim sup_{i→∞} p_{i} = ∞, then the Musielak-Orlicz function

Φ = (Φ)_{i=1}^{∞}, where Φ_{i}(u) = |u|^{p_{i}}, does not satisfy the δ_{2}-condition (for the definition of Φ ∈ δ_{2} see [50’] and [30]). Therefore (see [50’], [30] and [37]) ℓ^{(p_{i})} contains an order isometric copy of ℓ^{∞}, whence ℓ^{(p_{i})} is not absolutely continuous. Moreover, ℓ^{(p_{i})} has the Fatou property (see [6]) whence, by Theorem 6.1, it follows that ℓ^{(p_{i})} has not the uniform Opial property, which finishes the proof. For some other properties of ℓ^{(p_{i})} we refer to [47]. In the following we will consider the uniform Opial property for Orlicz spaces equipped with the Amemiya norm

∥x∥_{Φ}^{A} = inf_{k>0} \frac{1}{k} (1 + m_{Φ}(kx)).

We write shortly ℓ_{A}^{Φ} in place of (ℓ^{Φ}, ∥ · ∥_{Φ}^{A}). Denote by K(x) the set of all k > 0 such that ∥x∥_{Φ}^{A} = \frac{1}{k} (1+m_{Φ}(kx)). It is known (see [10], [69] and [75]) that K(x) = [k_{x}^{∗}, k_{x}^{∗∗}], where k_{x}^{∗} = inf{k > 0: m_{Φ}(p ∘ kx) ≥ 1} and k_{x}^{∗∗} = sup{k > 0: m_{Φ}(p ∘ kx) ≤ 1} whenever K(x) ≠ ∅ (that is k_{x}^{∗} < ∞, where p denotes the right hand side derivative of Φ on ℝ_{+} = [0, ∞) and p ∘ kx denotes the composition of p and kx. It is also well known that K(x) ≠ ∅ for all x ∈ ℓ_{A}^{Φ} whenever (Φ(u)/u) → ∞ as u → ∞ (see [10] and [25’]). The following lemma from [25’] will be useful to get criteria for the uniform Opial property of ℓ_{A}^{Φ}.

Lemma 6.4. If x ∈ ℓ^{Φ} and K(x) = ∅, then A := lim_{u→∞}(Φ(u)/u) < ∞ and

∥x∥_{Φ}^{A} = A ∑_{i=1}^{∞} |x(i)|.

Theorem 6.3. The Orlicz space ℓ_{A}^{Φ} has the uniform Opial property if and only if Φ ∈ Δ_{2}(0).

P r o o f. Since ℓ_{A}^{Φ} is not absolutely continuous whenever Φ ∉ Δ_{2}(0), so by Theorem 6.1, the necessity is obvious. Sufficiency. Take any ε > 0 and x ∈ ℓ_{A}^{Φ} with ∥x∥_{Φ}^{A} ≥ ε. Let (x_{n}) be a weakly null sequence in S(ℓ_{A}^{Φ}). By Φ ∈ Δ_{2}(0) there is δ ∈ (0, ε) independent of x such that m_{Φ} (\frac{x}{2}) ≥ δ. Take j ∈ ℕ such that

∥ ∑_{i=j+1}^{∞} x(i)e_{i}∥_{Φ}^{A} < \frac{δ}{8}.

By x_{n}→^{w}0 it follows that x_{n}(i) → 0 for any i ∈ ℕ. So, there exists n_{0} ∈ ℕ such that

∥∑_{i=1}^{j} x_{n}(i)e_{i}∥_{Φ}^{A} < \frac{δ}{8} (∀n > n_{0}).

Hence

(6.3)  ∥x + x_{n}∥_{Φ}^{A} = ∥∑_{i=1}^{j}(x(i) + x_{n}(i))e_{i} + ∑_{j=j+1}^{∞}(x(i) + x_{n}(i))e_{i}∥_{Φ}^{A}

≥ ∥∑_{i=1}^{j}(x(i) + x_{n}(i))e_{i} + ∑_{i=j+1}^{∞} x_{n}(i)e_{i}∥_{Φ}^{A} − \frac{δ}{8}

≥ ∥∑_{i=1}^{j} x(i)e_{i} + ∑_{i=j+1}^{∞} x_{n}(i)e_{i}∥_{Φ}^{A} − \frac{δ}{4}

whenever n > n_{0}. We will consider now two cases for n > n_{0}.

Case I. K(∑_{i=1}^{j} x(i)e_{i} + ∑_{i=j+1}^{∞} x_{n}(i)e_{i}) ≠ ∅. Then there exists k_{n} > 0 such that

∥x_{n} + x∥_{Φ}^{A} = \frac{1}{k_{n}} (1 + m(k_{n}(∑_{i=1}^{j} x(i)e_{i} + ∑_{i=j+1}^{∞} x_{n}(i)e_{i}))).

Combining this with (6.3), we get

(6.4)  ∥x_{n} + x∥_{Φ}^{A} ≥ \frac{1}{k_{n}} (∑_{i=1}^{j} Φ(k_{n}x(i)) + ∑_{i=j+1}^{∞} Φ(k_{n}x_{n}(i))) − \frac{δ}{4}

= \frac{1}{k_{n}} (∑_{i=1}^{∞} Φ(k_{n}x(i))) + \frac{1}{k_{n}} ∑_{i=j+1}^{j} Φ(k_{n}x_{n}(i)) − \frac{δ}{4}.

Moreover, by the inequalities

m( ∑_{i=j+1}^{∞} x(i)e_{i}) ≤ ∥ ∑_{i=j+1}^{∞} x(i)e_{i}∥_{Φ}^{A} < \frac{δ}{8} m (\frac{x}{2}) ≥ δ,

it follows that

(6.5)  m(∑_{i=1}^{j} x(i)e_{i}) > \frac{7δ}{8} .

We may assume without loss of generality that k_{n} ≥ \frac{1}{2}. Hence, inequalities (6.4) and (6.5) yield

\begin{aligned}∥x_{n} + x∥_{Φ}^{A} & ≥ ∥ ∑_{i=j+1}^{∞} x_{n}(i)e_{i}∥_{Φ}^{A} + 2 ∑_{i=1}^{j} Φ (\frac{1}{2}x(i)) − \frac{δ}{4} \\ & ≥ 1 − \frac{δ}{8} + \frac{7δ}{8} − \frac{δ}{4} = 1 + \frac{δ}{2}. \end{aligned}

Case II. K(∑_{i=1}^{m} x(i)e_{i} + ∑_{i=j+1}^{∞} x_{n}(i)e_{i}) = ∅. Then

∥∑_{i=1}^{j} x(i)e_{i} + ∑_{i=j+1}^{∞} x_{n}(i)e_{i}∥_{Φ}^{A} = A ∑_{i=1}^{j} |x(i)| + A ∑_{i=j+1}^{∞} |x_{n}(i)|

≥ ∥∑_{i=1}^{j} x(i)e_{i}∥_{Φ}^{A} + ∥ ∑_{i=j+1}^{∞} x_{n}(i)e_{i}∥

> \frac{7δ}{8} + 1 − \frac{δ}{8} = 1 + \frac{3δ}{4} .

Therefore, by inequality (6.3) we get

∥x_{n} + x∥_{Φ}^{A} ≥ 1 + \frac{δ}{2}.

So, in any case, ∥x_{n} + x∥_{Φ}^{A} ≥ 1 + \frac{δ}{2} for n > n_{0}, which finishes the proof.

Corollary 6.5. Orlicz spaces l_{Φ} generated by Orlicz functions Φ satisfying (Φ(u)/u) → ∞ as u → ∞ have normal structure if and only if Φ ∈ Δ_{2}(0).

P r o o f. If Φ ∈ Δ_{2}(0), then l_{Φ}^{A} has uniform normal structure, so normal structure as well. Assume that Φ ∉ Δ_{2}(0). Then there exist x ∈ S(l_{Φ}^{0}) and a sequence {x_{n}} in (l_{Φ}^{A})^{+} such that x_{m}⊥x_{n} for m ≠ n, x_{n}⊥x, {x_{n}} has a majorant in (l_{Φ}^{A})^{+} and I_{Φ}(k_{0}x_{n}) ≤ 2^{−n} for any n, where k_{0} ≥ 1 satisfies \frac{1}{k_{0}} (1 + m(k_{0}x)) = ∥x∥_{Φ}^{A} = 1, and ∥x_{n}∥_{Φ}^{A} → 1 as n → ∞. Then x_{n} → 0 weakly. Moreover,

\frac{1}{k_{0}} (1 + m(k_{0}(x + x_{n})) = \frac{1}{k_{0}} (1 + m(k_{0}x)) + \frac{1}{k_{0}} m(k_{0}x_{n}) ≤ 1 + 2^{−n}.

Therefore, lim_{n→∞} ∥x_{n} + x_{n}∥_{Φ}^{A} = lim_{n→∞} ∥x_{n}∥_{Φ}^{A} = 1, that is, l_{Φ}^{0} fails to have the Opial property.

7. Garcia – Falset coefficient

First we need to fix some notations and definitions. Let X be a Banach space with the unit sphere S(X) and the unit ball B(X). Garcia-Falset [30’] defined the coefficient

R(X) = sup{lim inf_{n→∞} ∥x_{n} − x∥: x ∈ B(X), {x_{n}} ⊂ B(X), x_{n}→^{w}0}

and proeved that any Banach space X with R(X) < 2 has the weak fixed point property. Let l^{0} denote the space of all real sequences. If a Banach space X is a subspace of l^{0} such that there exists x ∈ X such that x(i) ≠ 0 for all i ∈ N and if x ∈ l^{0}, y ∈ X and |x(i)| ≤ |y(i()| for all i ∈ N implies that x ∈ X and ∥x∥ ≤ ∥y∥, then X is said to be a Köthe sequence space. A Köthe sequence space X is said to have semi-Fatou property (X ∈ SF P for short) if for every sequence {x_{n}} in X and x ∈ X such that 0 ≤ x_{n} ↑ x, there holds ∥x_{n}∥ → ∥x∥.

Theorem 7.1 (see [43’]). If X is a Köthe sequence space with the semi-Fatou property and with the norm not being absolutely continuous, then X contains almost isometric copy of l_{∞}, i.e. for any ε > 0 the exists a closed subspace Y of X and an isomorphism of l_{∞} onto Y such that 1 ≥ ∥P ∥ ≤ 1 + ε.

Corollary 7.1. If a Köthe sequence space with the semi-Fatou property is not absolutely continuous, then R(X) = 2.

P r o o f. It is easy to see that R(l_{∞}) = 2. Moreover, by Theorem 7.1, R(X) = R(l_{∞}).

Corollary 7.2. If Φ does not satisfy the Δ_{2}-condition at zero, then R(l_{Φ}) = R(l_{Φ}^{0}) = 2.

P r o o f. Both norms ∥ ∥ and ∥ ∥^{0} have the semi-Fatou property (in fact they even have the Fatou property). Moreover, if Φ ∉ Δ_{2}(0), then l_{Φ} and l_{Φ}^{0} are not absolutely continuous (see [10] and [68]). So, by Corollary 7.1, we get the desired conclusion.

Theorem 7.2 (see [22]). For any Orlicz function Φ, the equality R(h_{Φ}) = sup{c_{x}: x = ∑_{i=1}^{m} x(i)e_{i} ∈ S(l_{Φ}) for some m ∈ N} holds, where c_{x} is positive number

satisfying the equality I_{Φ}(x/c_{x}) = 1/2.

P r o o f. Write ∥ ∥ for the Luxemburg norm ∥ · ∥_{Φ}. First, we will show that R(h_{Φ}) = R_{1}(h_{Φ}), where

R_{1}(h_{Φ}) = sup{lim inf_{n→∞} ∥x_{n} − x∥: x_{n} = ∑_{i=I_{n−1}+1}^{I_{n}} x_{n}(i)e_{i} ∈ S(h_{Φ}), x_{n}→^{w}0,

x = ∑_{i=1}^{I_{1}} x(i)e_{i} ∈ S)h_{Φ}), I_{1} < I_{2} < ⋯, n = 2, 3, …, }.

We only need to show that R(h_{Φ}) ≤ R_{1}(h_{Φ}). For any ε > 0 there exist a weakly null sequence {x_{n}} ⊂ B(h_{Φ}) and x_{0} ∈ B(h_{Φ}) such that

lim inf_{n→∞} ∥x_{n} − x_{0}∥ ≥ R)h_{Φ}) − ε.

We may assume for the sake of convenience that ∥x_{n} − x_{0}∥ ≥ R(h_{Φ}) − 2ε for all n ∈ 𝒩. For the above ε > 0 there exists i_{0} ∈ 𝒩 such that

∥ ∑_{i=i_{0}+1}^{∞} x_{0}(i)e_{i}∥ < ε.

By x_{n}→^{w}0, we have x_{n}(i) → 0 as n → ∞ for all i ∈ 𝒩. So, there exists n_{1} ∈ 𝒩 such that

∥∑_{i=1}^{i_{0}} x_{n}(i)e_{i}∥ < ε n ≥ n_{1}.

Define z_{1} = ∑_{i=i_{0}+1}^{∞} x_{n_{1}}(i)e_{i}. Then there exists i_{1} ∈ 𝒩 such that i_{1} > i_{0} and

∥ ∑_{i=i_{1}+1}^{∞} z_{1}(i)e_{i}∥ < ε.

Again by the coordinatewise convergence of x_{n} to 0, there exists n_{2} > n_{1} such that

∥∑_{i=1}^{i_{1}} x_{n}(i)e_{i}∥ < ε n ≥ n_{2}.

Define z_{2} = ∑_{i=i_{1}+1}^{∞} x_{n_{2}}(i)e_{i}. Proceeding by induction in such a way, we get a sequence

{x_{k}} ⊂ B(h_{Φ}) such that z_{k} = ∑_{i=i_{k−1}+1}^{∞} x_{n_{k}}(i)e_{i} and

∥x_{n_{k}}∥ − ε ≤ ∥z_{k}∥ ≤ ∥x_{n_{k}}∥

for all k ∈ 𝒩. Put w_{0} = ∑_{i=1}^{i_{0}} x_{0}(i)e_{i} and w_{k} = ∑_{i=i_{k−1}+1}^{i_{k}} z_{k}(i)e_{i} for all k ∈ 𝒩. We have

{w_{k}} ⊂ B(h_{Φ}) and

∥x_{n_{k}}∥ − 2ε ≤ ∥w_{k}∥ ≤ ∥x_{n_{k}}∥

for any k ∈ 𝒩. Next, we will show that w_{k}→^{w} 0. Since

(7.1)  lim sup_{λ→0} I_{Φ}(λw_{k})λ = lim sup_{λ→0} I_{Φ}(λx_{n_{k}})λ = 0

and every singular functional vanishes on any w_{k}, we only need to show that w_{k}→^{l_{Ψ}} 0 as k → ∞. Let y ∈ L_{Ψ} and λ_{1} > 0 satisfy I_{Ψ}(λ_{1}y) < ∞. Then, by the Young inequality, we have

⟨w_{k}, y⟩ = ∑_{i=1}^{∞} w_{k}(i)y(i) ≤ \frac{1}{λ_{1}λ}(I_{Φ}(λw_{k}) + ∑_{i=i_{k−1}+1}^{i_{k}} Ψ(λ_{1}y(i))),

which holds for all k ∈ 𝒩 and λ > 0. So, by the fact that ∑_{i=i_{k−1}+1}^{i_{k}} Ψ(λ_{1}y(i)) → 0 as

k → ∞ as well as by condition (7.1), we conclude that w_{k}→^{w} 0. Notice that

lim inf_{k→∞} ∥w_{k} − w_{0}∥

= lim inf_{k→∞}∥x_{n_{k}} − x_{0} − ∑_{i=1}^{i_{k−1}} x_{n_{k}}(i)e_{i} − ∑_{i=i_{k}+1}^{∞} x_{0}(i)e_{i} + ∑_{i=i_{0}+1}^{∞} x_{n_{k}}(i)e_{i}∥
≤ lim inf_{k→∞} ∥x_{n_{k}} − x_{0}∥ − 3ε ≥ R(h_{Φ}) − 4ε.

By the arbitrariness of ε > 0, we obtain R(h_{Φ}) = R_{1}(h_{Φ}). Finally, we put d = sup{c_{x}: x ∈ S(h_{Φ}) and x(n) = 0 for n > m with some m ∈ 𝒩}, where for any x ∈ h_{Φ} with finitely many coordinates different from 0, c_{x} denotes the positive number such that I_{Φ}(x/c_{x}) = 1/2. Such a positive number c_{x} exists by the continuity of Φ. Then

for any x_{n} = ∑_{i=I_{n−1}+1}^{I_{n}} x_{n}(i)e_{i} ∈ S(h_{Φ}) (n = 2, 3, …), x_{n}→^{w}0, x = ∑_{i=1}^{I_{1}} x(i)e_{i} ∈ S(h_{Φ}),

with I_{1} < I_{2} < ⋯, there holds I_{Φ}((x_{n}−x)/d) = ∑_{i=1}^{I_{1}} Φ(x(i)/d)+∑_{i=I_{n−1}+1}^{I_{n}} Φ(x_{n}(i)/d) ≤ 1/2 + 1/2 = 1. So ∥(x_{n} − x)/d∥ ≤ 1, i.e. ∥x_{n} − x∥ ≤ d, whence R(h_{Φ}) ≤ d. On the other hand, for any ε > 0 there exists y = ∑_{i=1}^{I_{1}} y(i)e_{i} ∈ S(h_{Φ}) such that c_{y} + ε > d.

Put y_{n} = (\overbrace{0, 0, ⋯, 0}^{(n+1)I_{1}}, y(1), y(2), …, y(I_{1}), 0, 0, …) for all n ∈ 𝒩. Since

lim sup_{λ→0}(I_{Φ}(λx_{n})/λ) = lim_{λ→0}(I_{Φ}(λx)/λ) = 0,

we obtain as above that x_{n}→_{l_{Ψ}} 0, whence x_{n}→^{w} 0. Since

I_{Φ}((y_{n} − y)/c_{y}) = 2 ∑_{i=1}^{I_{1}} Φ(y(i)/c_{y}) = 1,

we get ∥y_{n} − y∥ = c_{y} > d − ε. By the arbitrariness of ε > 0, R(h_{Φ}) ≥ d. Hence R(X) = d and the proof is finished.

Remark 7.2. Note that R(X) = R_{1}(X) for any Köthe sequence space with the semi-Fatou property and an absolutely continuous norm.

P r o o f. We can repeat the proof for h_{Φ} in Theorem 7.2.

Corollary 7.3. For any Lebesgue sequence space l_{p} (1 < p < ∞), we have R(l_{p}) = 2^{1/p}.

P r o o f. For any x ∈ S(l_{p}) we have c_{x} = 2^{1/p}, which follows by the equalities

I_{p} (\frac{x}{c_{x}} ) = ∑_{i=1}^{∞} |\frac{x(i)}{c_{x}} |^{p} = \frac{1}{c_{x}^{p}} ∑_{i=1}^{∞} |x(i)|^{p} = \frac{1}{c_{x}^{p}} = \frac{1}{2}.

To formulate the next corollary we need to define the Riesz angle for any Banach lattice X. It is defined by

α(X) = sup{∥|x| ∨ |y|∥ : x, y ∈ B(X), |x| ∧ |y| = 0}.

Corollary 7.4. For any Orlicz function Φ, the equality R(h_{Φ}) = α(h_{Φ}) holds.

P r o o f. By the equality R(h_{Φ}) = d that was obtained in the proof of Theorem 7.2, we can easily get the inequality R(h_{Φ}) ≤ α(h_{Φ}). On the other hand, for any ε > 0 there exist x ∈ S(h_{Φ}) and y ∈ S(h_{Φ}) such that |x| ∧ |y| = 0 and

∥|x| ∨ |y|∥ > α(l_{Φ}) − ε.

For the sake of convenience, we may assume that x ∨ y = (x(1), y(1), x(2), y(2), ⋯). By the fact that h_{Φ} has an absolutely continuous norm, there exists i_{0} ∈ 𝒩 such that

∥(|x| (1), |y| (1), ⋯; |x| (i_{0}), |y| (i_{0}), 0, 0, ⋯)∥ ≥ α(h_{Φ}) − ε.

Defining x_{0} = (x(1), x(2), ⋯, x(i_{0}), 0, 0, ⋯) and y_{n} = (\overbrace{0, ⋯, 0}^{(i_{0}+n)th}, |y| (1), |y| (2), ⋯,

|y| (i_{0}), 0, 0, ⋯) for all n ∈ 𝒩, we get y_{n}→^{w} 0 and

lim inf_{n→∞} ∥y_{n} + x_{0}∥ ≥ α(h_{Φ}) − ε.

Hence R(X) ≥ α(X) − ε. By the arbitrariness of ε > 0, we get R(h_{Φ}) ≥ α(h_{Φ}) and consequently R(h_{Φ}) = α(h_{Φ}). For Orlicz function Φ we write as usual Φ ∈ Δ_{2}(0) if Φ satisfies the Δ_{2}-condition in a neighbourhood of zero.

Corollary 7.5. For any Orlicz sequence space l_{Φ}, R(l_{Φ}) < 2 if and only if Φ ∈ Δ_{2}(0) and Ψ ∈ Δ_{2}(0).

Corollary 7.6. For any Orlicz function Φ, R(h_{Φ}) < 2 if and only if Ψ ∈ Δ_{2}(0), where Ψ denotes the Young conjugate of Φ.

P r o o f. By Corollary 7.1 and Theorem 3.11 in [10], which says that if Ψ ∈ Δ_{2}(0), then h_{Φ} has the w − F P P, both Corollaries 7.5 and 7.6 follow.

Notes. It is known that property (β) which has been introducent by Rolewicz [69^{+}] is stronger than NUC and it implies normal structure of the dual space (see [56]). Property (β) has been considerated in Orlicz-Bochner spaces, Musielak-Orlicz sequence spaces of Bochner type, Orlicz-Lorentz spaces and Calderón-Lozanovskiǐ spaces in [54’], [54^{+}] and [54^{∗}]. Properties UKK and NUC in Köthe-Bochner spaces have been considered in [54^{#}].

Cesaro Sequence Spaces

8. Garcia-Falset coefficient

For 1 ≤ p < ∞ the Cesaro sequence space ces_{p} is defined by

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

(see [59] and [70]).

Lemma 8.1 (see [20]). For any ε > 0 and L > 0, there exists δ > 0 such that

|∑_{n=1}^{∞}(\frac{1}{n} ∑_{i=1}^{n} |x(i) + y(i)|)^{p} − ∑_{n=1}^{∞}(\frac{1}{n} ∑_{i=1}^{n} |x(i)|)^{p}| < ε

whenever 1 ≤ p < ∞, ∑_{n=1}^{∞}(\frac{1}{n} ∑_{i=1}^{n} |x(i)|)^{p} ≤ L and ∑_{n=1}^{∞}(\frac{1}{n} ∑_{i=1}^{n} |y(i)|)^{p} ≤ δ.

P r o o f. It follows by the uniform continuity of the function f(u) = u^{p} on any compact interval [0, l].

Theorem 8.1. For the Cesaro sequence space ces_{p} (1 < p < ∞) there holds R(ces_{p}) = 2^{1/p}.

P r o o f. We can apply Remark 6.2.Let ε > 0 be given. For any {x_{n} =

∑_{i=I_{n−1}+1}^{I_{n}} x_{n}(i)e_{i}}_{n=2}^{∞} ⊂ S(ces_{p}), x_{n}→^{w} 0, x = ∑_{i=1}^{I_{1}} x(i)e_{i} ∈ S(ces_{p}), I_{1} < I_{2} < ⋯,

there exists n_{0} ∈ N such that ∑_{k=i_{n_{0}} +1}^{∞} (a/k)^{p} < min(ε, δ), where a = ∑_{i=1}^{I_{1}} |x(i)| and

δ > 0 is the number corresponding to our ε > 0 and L = 1 in Lemma 7.1. Hence for any m > n_{0} there holds

∥x_{m} − x∥^{p} = ∑_{k=1}^{I_{m−1}}(\frac{1}{k} ∑_{i=1}^{k}|x(i)|)^{p} + ∑_{k=I_{m−1}+1}^{∞}(\frac{1}{k} (a + ∑_{i=1}^{k}|x_{m}(i)|))^{p}

≥ ∑_{k=1}^{I_{m−1}}(\frac{1}{k} ∑_{i=1}^{k}|x(i)|)^{p} + ∑_{k=I_{m−1}+1}^{∞}(\frac{1}{k} ∑_{i=1}^{k}|x_{m}(i)|))^{p}

= ∑_{k=1}^{∞}(\frac{1}{k} ∑_{i=1}^{k}|x(i)|)^{p} − ∑_{k=I_{m−1}+1}^{∞}(\frac{a}{k} )^{p} + ∑_{k=I_{m−1}+1}^{∞}(\frac{1}{k} ∑_{i=1}^{k}|x_{m}(i)|))^{p}

> 1 − ε + 1 = 2 − ε,

i.e. lim inf_{n→∞} ∥x_{n} − x∥ ≥ (2 − ε)^{\frac{1}{p}}. On the other hand, for any m > n_{0},

\begin{aligned}∥x_{m} − x∥^{p} & = ∑_{k=1}^{I_{m−1}}(\frac{1}{k} ∑_{i=1}^{k}|x(i)|)^{p} + ∑_{k=i_{m−1}+1}^{∞}(\frac{1}{k} (a + ∑_{i=1}^{k}|x_{m}(i)|))^{p} \\ & ≤ ∑_{k=1}^{∞}(\frac{1}{k} ∑_{i=1}^{k}|x(i)|)^{p} + ∑_{k=I_{m−1}+1}^{∞}(\frac{1}{k} (∑_{i=1}^{k}|x_{m}(i)|))^{p} + ε = 2 + ε, \end{aligned}

i.e. lim inf_{n→∞}∥x_{n} − x∥ ≤ (2 + ε)^{\frac{1}{p}}. By the arbitrariness of ε > 0 and by Remark 7.2, we

get R(ces_{p}) = 2^{\frac{1}{p}}.

Corollary 8.1. Cesaro sequence spaces ces_{p} (1 < p < ∞) have the fixed point property.

P r o o f. For 1 < p < ∞, ces_{p} is a reflexive space and since R(ces_{p}) = 2^{1/p} < 2, ces_{p} has the weakly fixed point property. Therefore, ces_{p} has the fixed point property.

9. Weakly convergence sequence coefficient, uniform Opial property, k-NUC and uniform normall structure

Our main aim in this section is to calculate the weakly convergence sequence coefficient for Cesáro sequence space ces_{p} and to prove that for any p ∈ (1, ∞), ces_{p} is (kNUC) for any integer k ≥ 2 and has the uniform Opial property and property (L). The weakly convergence sequence coefficient which is connected with normal structure is an important geometric constant. It was introduced by Bynum [22’]. In the following, X denotes a Banach space and S(X) denotes the unit sphere of X. For a sequence {x_{n}} ⊂ X, we consider

A({x_{n}}) = lim_{n→∞}{sup{∥x_{i} − x_{j}∥: i, j ≥ n, i ≠ j}}
A_{1}({x_{n}}) = lim_{n→∞}{inf{∥x_{i} − x_{j}∥: i, j ≥ n, i ≠ j}}.

The weakly convergence sequence coefficient of X, denoted by W CS(X), is defined as follows: W CS(X) = sup{k > 0: for each weakly convergent sequence {x_{n}}, there is y ∈ co({x_{n}}) such that k·lim sup_{n→∞} ∥x_{n}−y∥ ≤ A({x_{n}})}, where co({x_{n}}) denotes the

convex hull of the elements of {x_{n}} (see [4]). The number M(X) = 1/W CS(X) for a reflexive Banach space is called the Maluta coefficient and it is known that M(X) = 1 for every non-reflexive Banach space X (see [62]). It is also well known that a Banach space X with W CS(X) > 1 has weak normal structure (see [22’]). A sequence {x_{n}} is said to be an asymptotic equidistant sequence if A({x_{n}}) = A_{1}({x_{n}}) (see [76]). The formula W CS(X) = inf{A({x_{n}}): {x_{n}} ⊂ S(X) and x_{n}→^{w}0} = inf{A({x_{n}}): {x_{n}} an asymptotic equidistant sequence in S(X) and x_{n}→^{w}0} was obtained in [76]. A Banach space X is said to have weak uniform normal structure if W CS(X) > 1. For a bounded subset A ⊂ X, the set-measure of noncompactness was defined in [52’] by

\begin{aligned}α(A) & = inf\{ε > 0: A \\ & ≤ ε\}. \end{aligned}

The ball-measure of noncompactness is defined by (see [9] and [11])

\begin{aligned}β(A) & = inf\{ε > 0: A \\ & ≤ ε\}. \end{aligned}

The functions α and β are called the Kuratowski measure of noncompactness and the Hausdorff measure of noncompactness in X, respectively. We can associate these functions with the notions of the set-contraction and the ball-contraction (see [26’]). These notions are very useful tools to study nonlinear operator problems (see [26’] and [66]).

The packing rate of a Banach space X is denoted by γ(X) and it is defined by the formula

γ(X) = δ(X)/σ(X),

where δ(X) and σ(X) are defined as the supremum and the infimum, respectively, of the set

\{\frac{β(A)}{α(A)}: A ⊂ X, A αα(A) > 0\}.

Recall that A ⊂ X is said to be α-minimal if α(B) = α(A) for any infinite subset of A. For those definitions and for results concerning the existence of α-minimal and β-minimal sets we refer to [2, Chapter X].

For each ε > 0 define Δ(ε) = inf{1−inf[∥x∥: x ∈ A]: A is a closed convex subset of B(X) with β(A) ≥ ε}. The function Δ is called the modulus of noncompact convexity (see [34]). A Banach space X is said to have property (L) if lim_{ε→1−} Δ(ε) = 1. It has been proved in [66] that property (L) is a useful tool in the fixed point theory and that a Banach space X has property (L) if and only if it is reflexive and has the uniform Opial property.

Theorem 9.1. If 1 < p < ∞, then the space ces_{p} is (k − NUC) for any integer k ≥ 2.

P r o o f. Let ε > 0 be given. For every sequence {x_{n}} ⊂ B(X) with sep({x_{n}}) > ε, we put x_{n}^{m} = (0, 0, …, 0, x_{n}(m), x_{n}(m + 1), …). For each i ∈ N, the sequence {x_{n}(i)}_{i=1}^{∞} is bounded. Therefore, using the diagonal method one can find a subsequence {x_{n_{k}}} of {x_{n}} such that the sequence {x_{n_{k}}(i)} converges for each i ∈ N. Therefore, for any m ∈ N there exists k_{m} such that sep({x_{n_{k}}^{m}}_{k≥k_{m}}) ≥ ε. Hence for each m ∈ N there exists n_{m} ∈ N such that

(9.1)  ∥x_{n_{m}}^{m}∥ ≥ \frac{ε}{2}.

Write I_{p}(x) = ∑_{n=1}^{∞}(\frac{1}{n} ∑_{i=1}^{n} |x(i)|)^{p} and put ε_{1} = \frac{k^{p−1}−1}{2k^{p}(k−1)}(\frac{ε}{2})^{p}. Then there exists δ > 0

such that

(9.2)  |I_{p}(x + y) − I_{p}(x)| < ε_{1}

whenever I_{p}(x) ≤ 1 and I_{p}(y) ≤ δ (see Lemma 7.1). There exists m_{1} ∈ 𝒩 such that I_{p}(x_{1}^{m_{1}}) ≤ δ. Next, there exists m_{2} > m_{1} such that I_{p}(x_{2}^{m_{2}}) ≤ δ. In such a way, there exists m_{2} < m_{3} < ⋯ < m_{k−1} such that I_{p}(x_{j}^{m_{j}}) ≤ δ for all j = 1, 2, …, k−1. Define m_{k} = m_{k−1} + 1. By condition (9.1), there exists n_{k} ∈ 𝒩 such that I_{p}(x_{n_{k}}^{m_{k}}) ≥ (\frac{ε}{2})^{p}. Put n_{i} = i for 1 ≤ i ≤ k − 1. Then in virtue of (9.1), (9.2) and convexity of the function f(u) = |u|^{p}, we get

I_{p}(\frac{x_{n_{1}} + x_{n_{2}} + ⋯ + x_{n_{k−1}} + x_{n_{k}}}{k} ) = ∑_{n=1}^{m_{1}}(\frac{1}{n} ∑_{i=1}^{n}|\frac{x_{n_{1}}(i) + ⋯ + x_{n_{k}}(i)}{k}|)^{p}

+ ∑_{n=m_{1}+1}^{∞}(\frac{1}{n} ∑_{i=1}^{n}|\frac{x_{n_{1}}(i) + x_{n_{2}}(i) + ⋯ + x_{n_{k−1}}(i) + x_{n_{k}}(i)}{k}|)^{p}

≤ ∑_{n=1}^{m_{1}} \frac{1}{k} ∑_{j=1}^{k}(\frac{1}{n} ∑_{i=1}^{n} |x_{n_{j}}(i)|)^{p} + ∑_{n=m_{1}+1}^{∞}(\frac{1}{n} ∑_{i=1}^{n}|\frac{x_{n_{2}}(i) + ⋯ + x_{n_{k}}(i)}{k}|)^{p} + ε_{1}

= ∑_{n=1}^{m_{1}} \frac{1}{k} ∑_{j=1}^{k}(\frac{1}{n} ∑_{i=1}^{n} |x_{n_{j}}(i)|)^{p} + ∑_{n=m_{1}+1}^{m_{2}}(\frac{1}{n} ∑_{i=1}^{n}|\frac{x_{n_{2}}(i) + ⋯ + x_{n_{k}}(i)}{k}|)^{p}

+ ∑_{n=m_{2}+1}^{∞}(\frac{1}{n} ∑_{i=1}^{n}|\frac{x_{n_{2}}(i) + x_{n_{3}}(i) + ⋯ + x_{n_{k−1}}(i) + x_{n_{k}}(i)}{k}|)^{p} + ε_{1}

≤ ∑_{n=1}^{m_{1}} \frac{1}{k} ∑_{j=1}^{k}(\frac{1}{n} ∑_{i=1}^{n} |x_{n_{j}}(i)|)^{p} + ∑_{n=m_{1}+1}^{m_{2}} \frac{1}{k} ∑_{j=2}^{k}(\frac{1}{n} ∑_{i=1}^{n} |x_{n_{j}}(i)|)^{p}

+ ∑_{n=m_{3}+1}^{∞}(\frac{1}{n} ∑_{i=1}^{n}|\frac{x_{n_{3}}(i) + x_{n_{4}}(i) + ⋯ + x_{n_{k−1}}(i) + x_{n_{k}}(i)}{k}|)^{p} + 2ε_{1}

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

≤ \frac{I_{p}(x_{n_{1}}) + ⋯ + I_{p}(x_{n_{k−1}})}{k} + \frac{1}{k} ∑_{n=1}^{m_{k}−1}(\frac{1}{n}(∑_{i=1}^{n} |x_{n_{k}}(i)|))^{p}

+ ∑_{n=m_{k−1}+1}^{∞}(\frac{1}{n}(∑_{i=1}^{n} \frac{|x_{n_{k}}(i)|}{k}))^{p} + (k − 1)ε_{1}

≤ \frac{k − 1}{k} + \frac{1}{k} ∑_{n=1}^{m_{k}−1}(\frac{1}{n}(∑_{i=1}^{n} |x_{n_{k}}(i)|))^{p}

+ \frac{1}{k^{p}} ∑_{n=m_{k−1}+1}^{∞}(\frac{1}{n}(∑_{i=1}^{n} |x_{n_{k}}(i)|))^{p} + (k − 1)ε_{1}

\begin{matrix}= 1 − \frac{1}{k} + \frac{1}{k} (1 − ∑_{n=m_{k−1}+1}^{∞}(\frac{1}{n} ∑_{i=1}^{n} |x_{n_{k}}(i)|))^{p} \\ + \frac{1}{k^{p}} ∑_{n=m_{k−1}+1}^{∞}(\frac{1}{n}(∑_{i=1}^{n} |x_{n_{k}}(i)|))^{p} + (k − 1)ε_{1} \end{matrix}

\begin{matrix}≤ 1 + (k − 1)ε_{1} − (\frac{k^{p−1} − 1}{k^{p}}) ∑_{n=m_{k−1}+1}^{∞}(\frac{1}{n}(∑_{i=1}^{n} |x_{n_{k}}(i)|))^{p} \\ ≤ 1 + (k − 1)ε_{1} − (\frac{k^{p−1} − 1}{k^{p}})(\frac{ε}{2})^{p} = 1 − \frac{1}{2}(\frac{k^{p−1} − 1}{k^{p}})(\frac{ε}{2})^{p}. \end{matrix}

Therefore, ces_{p} is (k − NUC) for any integer k ≥ 2.

Theorem 9.2. For any 1 < p < ∞, the space ces_{p} has the uniform Opial property.

P r o o f. For any ε > 0 we can find a positive number ε_{0} ∈ (0, ε) such that

1 + \frac{ε^{p}}{2} > (1 + ε_{0})^{p}.

Let x ∈ X and ∥x∥ ≥ ε. There exists n_{1} ∈ 𝒩 such that

∑_{i=n_{1}+1}^{∞}(\frac{1}{n} ∑_{i=1}^{n} |x(i)|)^{p} < (\frac{ε_{0}}{4} )^{p}.

Hence we have

∥ ∑_{i=n_{1}+1}^{∞} x(i)e_{i}∥ < \frac{ε_{0}}{4} < \frac{ε}{4},

where e_{i} = (0, …, 1^{ith}, 0, 0, …). Furthermore, we have

\begin{aligned}ε^{p} & ≤ ∑_{n=1}^{n_{1}}(\frac{1}{n} ∑_{i=1}^{n} |x(i)|)^{p} + ∑_{n=n_{1}+1}^{∞}(\frac{1}{n} ∑_{i=1}^{n} |x(i)|)^{p} \\ & < ∑_{n=1}^{n_{1}}(\frac{1}{n} ∑_{i=1}^{n} |x(i)|)^{p} + (\frac{ε_{0}}{4} )^{p} < ∑_{n=1}^{n_{1}}(\frac{1}{n} ∑_{i=1}^{n} |x(i)|)^{p} + \frac{ε^{p}}{4} , \end{aligned}

whence

\frac{3ε^{p}}{4} ≤ ∑_{n=1}^{n_{1}}(\frac{1}{n} ∑_{i=1}^{n} |x(i)|)^{p}.

For any weakly null sequence {x_{m}} ⊂ S(X), in virtue of x_{m}(i) → 0 for i = 1, 2, …, there exists m_{0} ∈ N such that

∥∑_{n=1}^{n_{1}} x_{m}(i)e_{i}∥ < \frac{ε_{0}}{4}

when m > m_{0}. Therefore,

∥x_{m} + x∥ = ∥∑_{i=1}^{n_{1}}(x_{m}(i) + x(i))e_{i} + ∑_{i=n_{1}+1}^{∞}(x_{m}(i) + x(i))e_{i}∥

≥ ∥∑_{i=1}^{n_{1}} x(i)e_{i} + ∑_{i=n_{1}+1}^{∞} x_{m}(i)e_{i}∥ − ∥∑_{i=1}^{n_{1}} x_{m}(i)e_{i}∥ − ∥ ∑_{i=n_{1}+1}^{∞} x(i)e_{i}∥

≥ ∥∑_{i=1}^{n_{1}} x(i)e_{i} + ∑_{i=n_{1}+1}^{∞} x_{m}(i)e_{i}∥ − \frac{ε_{0}}{2}

when m > m_{0}. Moreover for a := ∑_{i=1}^{n_{1}} |x(i)| there holds

∥∑_{i=1}^{n_{1}} x(i)e_{i} + ∑_{i=n_{1}+1}^{∞} x_{m}(i)e_{i}∥^{p}

= ∑_{n=1}^{n_{1}}(\frac{1}{n} ∑_{i=1}^{n} |x(i)e_{i}|)^{p} + ∑_{n=n_{1}+1}^{∞}(\frac{1}{n} ∑_{i=1}^{n}(a + |x_{m}(i)|))^{p}

≥ ∑_{n=1}^{n_{1}}(\frac{1}{n} ∑_{i=1}^{n} |x(i)e_{i}|)^{p} + ∑_{n=n_{1}+1}^{∞}(\frac{1}{n} ∑_{i=1}^{n} |x_{m}(i)|)^{p}

≥ \frac{3ε^{p}}{4} + (1 − \frac{ε^{p}}{4} ) = 1 + \frac{ε^{p}}{2} > (1 + ε_{0})^{p}.

Therefore, combining this with the previous inequality, we get

\begin{aligned}∥x_{m} + x∥ & ≥ ∥∑_{i=1}^{n_{1}} x(i)e_{i} + ∑_{i=n_{1}+1}^{∞} x_{m}(i)e_{i}∥ − \frac{ε_{0}}{2} \\ & ≥ 1 + ε_{0} − \frac{ε_{0}}{2} = 1 + \frac{ε_{0}}{2} . \end{aligned}

This means that ces_{p} has the uniform Opial property.

By the facts presented in the introduction and the reflexity of ces_{p} for 1 < p < ∞, we get the following.

Corollary 9.1. For 1 < p < ∞ the space ces_{p} has property and the fixed point property.

Now, we will calculate the weakly convergence sequence coefficient of ces_{p}. Theorem 9.3. There holds the equality W CS(ces_{p}) = 2^{\frac{1}{p}} whenever 1 < p < ∞.

P r o o f. Take any ε > 0 and an asymptotic equidistant sequence {x_{n}} ⊂ S(X) with x_{n}→^{w}0 and put v_{1} = x_{1}. There exists i_{1} ∈ 𝒩 such that

∥ ∑_{i=i_{1}+1}^{∞} v_{1}(i)e_{i}∥ < ε.

Since x_{n} → 0 coordinatewise, there exists n_{2} ∈ 𝒩 such that

∥∑_{i=1}^{i_{1}} x_{n}(i)e_{i}∥ < ε

whenever n ≥ n_{2}. Take v_{2} = x_{n_{2}}. Then there is i_{2} > i_{1} such that

∥ ∑_{i=i_{2}+1}^{∞} v_{2}(i)e_{i}∥ < ε.

Since x_{n}(i) → 0 coordinatewise, there exists n_{3} ∈ 𝒩 such that

∥∑_{i=1}^{i_{2}} x_{n}(i)e_{i}∥ < ε

whenever n ≥ n_{3}. Continuing this process in such a way by induction, we get a subsequence {v_{n}} of {x_{n}} such that

∥ ∑_{i=i_{n}+1}^{∞} v_{n}(i)e_{i}∥ < ε

and

∥∑_{i=1}^{i_{n}} v_{n+1}(i)e_{i}∥ < ε.

Put

z_{n} = ∑_{i=i_{n−1}+1}^{i_{n}} v_{n}(i)e_{i}

for n = 2, 3, …. Then

1 ≥ ∥z_{n}∥ = ∥∑_{i=1}^{∞} v_{n}(i)e_{i} − ∑_{i=1}^{i_{n−1}} v_{n}(i)e_{i} − ∑_{i=i_{n}+1}^{∞} v_{n}(i)e_{i}∥ (9.3)

≥ ∥∑_{i=1}^{∞} v_{n}(i)e_{i}∥ − ∥∑_{i=1}^{i_{n−1}} v_{n}(i)e_{i}∥ − ∥ ∑_{i=i_{n}+1}^{∞} u_{n}(i)e_{i}∥ > 1 − 2ε.

Moreover, for any n, m ∈ 𝒩 with n ≠ m, we have

∥v_{n} − v_{m}∥ = ∥∑_{i=1}^{∞} u_{n}(i)e_{i} − ∑_{i=1}^{∞} v_{m}(i)e_{i}∥ (9.4)

≥ ∥ ∑_{i=i_{n−1}+1}^{i_{n}} v_{n}(i)e_{i} − ∑_{i=i_{m−1}+1}^{i_{m}} v_{m}(i)e_{i}∥ − ∥∑_{i=1}^{i_{n−1}} v_{n}(i)e_{i}∥

− ∥ ∑_{i=i_{n}+1}^{∞} v_{n}(i)e_{i}∥ − ∥ ∑_{i=1}^{i_{m−1}} v_{m}(i)e_{i}∥ − ∥ ∑_{i=i_{m}+1}^{∞} u_{m}(i)e_{i}∥

≥ ∥z_{n} − z_{m}∥ − 4ε.

This means that A(\{x_{n}\}) = A(\{v_{n}\}) ≥ A(\{z_{n}\})−4ε. Put u_{n} = \frac{z_{n}}{∥z_{n}∥} for n = 2, 3, …. Then

(9.5)  u_{n} ∈ S(ces_{p});
(9.6)  A({x_{n}}) ≥ 1 − εA({u_{n}}) − 4ε.

On the other hand

∥v_{n} − v_{m}∥ ≤ ∥z_{n} − z_{m}∥ + 4ε ≤ ∥u_{n} − u_{m}∥ + 4ε

for every m, n ∈ 𝒩, m ≠ n. Therefore

(9.7)  A({u_{n}}) ≥ A({x_{n}}) − 4ε.

By the arbitrariness of ε > 0, we have from (9.5), (9.6) and (9.7) that W CS(ces_{p}) =

inf{A({u_{n}}): u_{n} = ∑_{i=i_{n−1}+1}^{i_{n}} u_{n}(i)e_{i} ∈ S(ces_{p}), 0 = i_{0} < i_{1} < i_{2} < …, u_{n}→^{w}0}.

Using Lemma 2 in [76], we have W CS(ces_{p}) = inf{A({u_{n}}): u_{n} = ∑_{i=i_{n−1}+1}^{i_{n}} u_{n}(i)e_{i} ∈ S(ces_{p}), 0 = i_{0} < i_{1} < i_{2} < ⋯, u_{n}→^{w}0 and {u_{n}} is asymptotic equidistant}. Take m ∈ 𝒩 large enough such that

∑_{k=i_{m−1}+1}^{∞}(\frac{b}{k} )^{p} < ε,

where b := ∑_{i=i_{n−1}+1}^{i_{n}} |u_{n}(i)|. We have for n < m

∥u_{n} − u_{m}∥^{p}

= ∑_{k=i_{n−1}+1}^{i_{m−1}}(\frac{1}{k} ∑_{i=1}^{k} |u_{n}(i)|)^{p} + ∑_{k=i_{m−1}+1}^{∞}(\frac{1}{k} (b + ∑_{i=1}^{k} |u_{m}(i)|))^{p}

≥ ∑_{k=i_{n−1}+1}^{i_{m−1}}(\frac{1}{k} ∑_{i=1}^{k} |u_{n}(i)|)^{p} + ∑_{k=i_{m−1}+1}^{∞}(\frac{1}{k} ∑_{i=1}^{k} |u_{m}(i)|)^{p}

= ∑_{k=i_{n−1}+1}^{∞}(\frac{1}{k} ∑_{i=1}^{k} |u_{n}(i)|)^{p} − ∑_{k=i_{m−1}+1}^{∞}(\frac{b}{k} )^{p} + ∑_{k=i_{m−1}+1}^{∞}(\frac{1}{k} ∑_{i=1}^{k} |u_{m}(i)|)^{p}

> 1 − ε + 1 = 2 − ε, A_{1}(\{u_{1}\}) ≥ (2 − ε)^{\frac{1}{p}} . Note that

\begin{matrix}[ ∑_{k=i_{m−1}+1}^{∞}(\frac{1}{k} (b + ∑_{i=1}^{k} |u_{m}(i)|))^{p}]^{\frac{1}{p}} = [ ∑_{k=i_{m−1}+1}^{∞}(\frac{b}{k} + \frac{1}{k} ∑_{i=1}^{k} |u_{m}(i)|)^{p}]^{\frac{1}{p}} \\ ≤[ ∑_{k=i_{m−1}+1}^{∞}(\frac{b}{k} )^{p}]^{\frac{1}{p}} + [ ∑_{k=i_{m−1}+1}^{∞}(\frac{1}{k} ∑_{i=1}^{k} |u_{m}(i)|)^{p}]^{\frac{1}{p}} < ε^{\frac{1}{p}} + 1. \end{matrix}

Therefore ∥u_{n} − u_{m}∥^{p}

= ∑_{k=i_{n−1}+1}^{i_{m−1}}(\frac{1}{k} ∑_{i=1}^{k} |u_{m}(i)|)^{p} + ∑_{k=i_{m−1}+1}^{∞}(\frac{1}{k} (b + ∑_{i=1}^{k} |u_{m}(i)|))^{p}

≤ ∑_{k=i_{n−1}+1}^{∞}(\frac{1}{k} ∑_{i=1}^{k} |u_{m}(i)|)^{p} + ∑_{k=i_{m−1}+1}^{∞}(\frac{1}{k} (b + ∑_{i=1}^{k} |u_{m}(i)|))^{p}

≤ 1 + (1 + ε^{\frac{1}{p}})^{p}

for any n, m ∈ 𝒩, m ≠ n. This yields A(\{u_{n}\}) ≤ [1 + (1 + ε^{\frac{1}{p}} )^{p}]^{\frac{1}{p}} and, by the arbitrariness of ε > 0, we obtain W CS(ces_{p}) = 2^{\frac{1}{p}}.

Corollary 9.2. For 1 < p < ∞, ces_{p} has the weak uniform normal structure and normal structure.

Corollary 9.. For any 1 < p < ∞ there holds the equality γ(ces_{p}) = 2^{(p−1)/p}.

P r o o f. By [1’], if X is reflexive Banach space with the uniform Opial property, then γ(X) = \frac{2}{W CS(X)}. Since, by Theorem 9.1, ces_{p} is (NUC) for 1 < p < ∞ and property (NUC) implies reflexivity, Theorem 9.3 yields γ(ces_{p}) = 22^{1p} = 2^{(p−1)p}.

Note. Banach-Saks and weak Banach-Saks properties in Cesaro sequence spaces has been characterized in [17].

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