Characteristic of monotonicity of Orlicz function spaces equipped with the Orlicz norm

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P. Foralewski, H. Hudzik, R. Kaczmarek, M. Krbec

Characteristic of monotonicity of Orlicz function spaces equipped with the Orlicz norm

Dedicated to Professor Julian Musielak on his 85^th birthday

Abstract. First we prove that the property of strict monotonicity of a Köthe space (E,k.kE)and/or of its Köthe dual (E⁰,k.kE0)can be used successfully to compare the supports of x ∈ E\{θ} and y ∈ S(E⁰), where < x, y >= kxkE. Next we prove that any element x ∈ S+(E)with µ(T \ supp x) = 0 is a point of order smoothness in E, whenever E is an order continuous Köthe space. Next, the characteristic of monotonicity of Orlicz function spaces equipped with the Orlicz norm is considered. In the case when the measure is finite, the Orlicz function Φ vanishes outside zero, supu>0[A|u| − Φ(u)] = ∞, where A = limu→∞Φ(u)

u and Φ satisfies the ∆2- condition at ∞, we give only some lower and upper estimates for this characteristic.

In all others (easier) cases the characteristic is calculated.

2010 Mathematics Subject Classification: 46B42, 46E30.

Key words and phrases: Orlicz space, Orlicz norm, Köthe space, Köthe dual, characteristic of monotonicity, strict monotonicity, point of order smoothness.

1. Introduction. The monotonicity properties of normed lattices have been introduced and studied by Birkhoff [3] in the context of their geometric structure.

Next, Akcoglu and Sucheston [1] discovered the connection of strict and uniform monotonicity to the ergodic theory.

It is well known that monotonicity properties of normed lattices are restrictions of respective rotundity properties to the couples of comparable elements in the positive cone X+of X (see [13]). In consequence, the role of monotonicity properties in the dominated best approximation problems is similar to the role of rotundity properties, when the more general (not dominated) best approximation problems are considered in Banach spaces (see [14], [24], [6]). The problem of calculating the characteristic of monotonicity of a Banach lattice seems to be of great interest because of the result of Betiuk-Pilarska and Prus [2] stating that if a Banach lattice

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X has this characteristic strictly smaller then 1 and X is weakly orthogonal, then X has the weak normal structure. Consequently, X has the weak fixed point property (for the definition we refer to [19]). Therefore, it was quite natural that several papers were devoted to the problem of calculating the exact value or getting good estimates of the characteristic of monotonicity in a concrete specific class of Banach lattices (see [27], [9] or [10]).

It is also worth noticing that monotonicity properties have close relationships to complex rotundities and their applications. Namely, for any real Köthe space E, strict monotonicity and uniform monotonicity of E coincide, respectively, with complex rotundity and complex uniform rotundity of the complexification E^cof E.

This result was obtained independently by Hudzik and Narloch [18] and Lee [25].

In [14] the concepts of lower and upper local uniform monotonicities were inve- stigated and it was shown that these monotonicity properties are different. In turn, Kolwicz and Płuciennik [20] introduced a new monotonicity property of normed lattices, called uniform monotonicity in every order interval and they discovered that this property is useful in order to give a criterion for uniform rotundity in every direction of Calder´on-Lozanovski˘ı spaces.

Recently, Wójtowicz [32] proved that if X is a Dedekind σ-complete Banach lattice and its dual X^∗ is strictly monotone, then for any x ∈ X\{0} all supporting functionals at |x| are positive.

For more results connected with monotonicity properties we refer to the papers [7, 8, 12, 15].

Let us denote S+(X) = S(X)∩ X⁺, where X+is the positive cone of a Banach lattice X = (X, ≤, k.k) and S(X) is the unit sphere of X.

A Banach lattice X is said to be strictly monotone, if kyk < kxk for all x, y ∈ X⁺ such that y ≤ x and y 6= x. A Banach lattice X is said to be uniformly monotone if for any ε ∈ (0, 1) there is δ(ε) ∈ (0, 1) such that kx − yk ≤ 1 − δ(ε) whenever 0≤ y ≤ x, kxk = 1 and kyk ≥ ε.

For a given Banach lattice X, the function δm,X: [0, 1]→ [0, 1] defined by δm,X(ε) = inf{1 − kx − yk : 0 ≤ y ≤ x, kxk = 1, kyk ≥ ε}

is called the lower modulus of monotonicity of X. It is known (see [11]) that δm,X(ε) = inf{1 − kx − yk : 0 ≤ y ≤ x, kxk = 1, kyk ≥ ε}

= inf{1 − kx − yk : 0 ≤ y ≤ x, kxk = 1, kyk = ε}

= inf{1 − kx − yk : 0 ≤ y ≤ x, kxk ≤ 1, kyk ≥ ε}

= inf{1 − kx − yk : 0 ≤ y ≤ x, kxk ≤ 1, kyk = ε}

for any ε ∈ (0, 1). The lower modulus of monotonicity δm,X is a non-decreasing and convex function on the interval [0, 1] (see [23]), so δm,X is continuous on the interval [0, 1). It is also clear that δm,X(ε)≤ ε for any ε ∈ [0, 1]. Obviously, X is uniformly monotone if and only if δm,X(ε) > 0for every ε ∈ (0, 1]. It is also easy to see that a Banach lattice X is strictly monotone if and only if δm,X(1) = 1.

The number ε0,m(X)∈ [0, 1] defined by

ε0,m(X) = sup{ε ∈ [0, 1): δ^m,X(ε) = 0} = inf{ε ∈ (0, 1): δ^m,X(ε) > 0},

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(where inf ∅ := 1) is said to be the characteristic of monotonicity of X. A Banach lattice X is uniformly monotone if and only if ε0,m(X) = 0.

Let us recall some useful facts.

Theorem 1.1 ([11, Theorem 5]) For any Banach lattice X the following formula for the characteristic of monotonicity holds true

ε0,m(X) = sup{lim sup

n→∞ kxⁿ− yⁿk : kxⁿk = 1 ∧ 0 ≤ yⁿ ≤ xⁿ∀ n ∈ N ∧ kyⁿk → 1}.

The relation between the characteristic of monotonicity and the modulus of monotonicity reveals the following

Theorem 1.2 ([9, Theorem 2.1]) For any Banach lattice X the following equality is true

ε0,m(X) = 1− δ^m,X(1⁻), where δm,X(1⁻) = lim

ε→1⁻δm,X(ε). Moreover,

δm,X(1− δ^m,X(ε)) = 1− ε

for arbitrary ε ∈ (ε0,m(X), 1] if ε0,m(X) < 1 as well as also in the case when ε = ε0,m(X) = 1.

2. Some remarks on strict monotonicity and order smoothness in Köthe spaces. Let us denote by (T, Σ, µ) a positive, complete and σ−finite measure space and by L⁰ = L⁰(T, Σ, µ) the space of all (equivalence classes of) real-valued and Σ−measurable functions defined on T . For two functions x, y ∈ L⁰ we write y ≤ x if and only if y(t) ≤ x(t) µ−a.e. in T .

By E = (E, ≤, k · kE)we denote a Köthe space over the measure space (T, Σ, µ), that is, E is a Banach subspace of L⁰ which satisfies the following conditions:

(i) if |y| ≤ |x|, x ∈ E and y ∈ L⁰, then y ∈ E and kykE ≤ kxkE, (ii) there exists a function x ∈ E which is strictly positive µ−a.e. in T .

Let E⁰ be the Köthe dual of E, that is, E⁰ =

y∈ L⁰(T, Σ, µ) : x· y ∈ L¹(T, Σ, µ)for any x ∈ E , and let E⁰ be equipped with the norm

kykE⁰ = sup

kxkE≤1

Z

T

x(t)y(t)dµ

= sup

kxkE≤1

Z

T

|x(t)y(t)| dµ = sup

kxkE≤1kx · yk^L¹ for any y ∈ E⁰.

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Remark 2.1 Let us assume that x ∈ E\ {0}, y ∈ E⁰, kykE⁰ = 1andR

T

x(t)y(t)dµ = kxkE. Then:

1⁰ supp x⊂ supp y up to a set of measure zero whenever E is strictly monotone, 2⁰ supp y⊂ supp x up to a set of measure zero whenever E⁰ is strictly monotone, 3⁰ supp x = supp y up to a set of measure zero if both E and E⁰ are strictly

monotone.

Proof We start from 1⁰. Let us define the functional ξy: E → R by the formula

ξy(z) =< z, y >=

Z

T

z(t)y(t)dµfor any z ∈ E.

It is obvious that kξ^yk = kykE⁰ by the definitions of these two norms. Moreover, we have ξy(x) = kxkE, whence ξy

x kxkE

= 1 = kξ^yk = kykE⁰. This means that the supremum in the definition of the norm kykE⁰ is attained at the point

x

kxkE. Therefore, it must be x(t)y(t) ≥ 0 µ-a.e. in T because otherwise defining

˜

x(t) = _kxk^x(t)_Esgn y(t) µ−a.e. in T , we get ξ^y(˜x) > ξy

x kxkE

, a contradiction. It is obvious that

ξy(xχsupp x∩supp y) = ξy(x) =kxkE,

whence by the inequality ξy(z) ≤ kξ^yk kzkE for any z ∈ E, we conclude that kxχ^{supp x}∩supp ykE =kxkE. By the assumption that E is strictly monotone, we get that supp x ⊂ supp y up to a set of measure zero.

2⁰. By the equality ξyχsupp x∩supp y(x) =kxkE, we get kyχsupp x∩supp ykE⁰ = 1 =kykE⁰.

Therefore, the assumption that E⁰ is strictly monotone gives that supp y ⊂ supp x up to a set of measure zero.

3⁰. This follows by 1⁰ and 2⁰.

A point x ∈ S⁺(E), where E is a Köthe space, is said to be order smooth if Grad(x) contains no order interval which is not a singleton. Let us recall that Grad(x) consists of all functionals x^∗ ∈ S(E^∗)(the unit sphere of the topological dual E^∗of E) such that x^∗(x) =kxk^E. A Köthe space E is said to be order smooth if all points x ∈ S⁺(E)are points of order smoothness.

Recall that a Köthe space E is called order continuous if for any element x ∈ E and any sequence (xn) in E+ (the positive cone in E) with 0 ≤ xⁿ ≤ |x| for all n∈ N and xⁿ→ 0 µ-a.e., there holds kxⁿk^E → 0.

Remark 2.2 If E is a Köthe space with an order continuous norm k · k^E, then any x∈ S⁺(E)with µ(T \ supp x) = 0 is a point of order smoothness.

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Proof Let the assumptions about x be satisfied and let y^∗, z^∗ ∈ Grad(x) and 0 ≤ y^∗ ≤ z^∗. Since E is order continuous, so E^∗= E⁰ isometrically, whence there exist y, z ∈ S(E⁰)such that

y^∗(w) = Z

T

w(t)y(t)dµand z^∗(w) = Z

T

w(t)z(t)dµ

for any w ∈ E. Since y^∗, z^∗∈ Grad(x), we have ky^∗k = kz^∗k = y^∗(x) = z^∗(x) = 1.

This yields that x(t)y(t) ≥ 0 and x(t)z(t) ≥ 0 µ−a.e. in T . Since y(t) ≤ z(t) µ−a.e.

in T , we also have x(t)y(t) ≤ x(t)z(t) µ−a.e. in T . Since µ(T \ supp x) = 0, it must be y(t) = z(t) µ−a.e. in T because otherwise y(t) < z(t) and so x(t)y(t) < x(t)z(t) on a set of positive measure, whence

kxk^E= y^∗(x) = Z

T

x(t)y(t)dµ <

Z

T

x(t)z(t)dµ = z^∗(x) =kxk^E,

a contradiction, which finishes the proof.

3. Characteristic of monotonicity in Orlicz spaces equipped with the Orlicz norm. A map Φ : R → [0, ∞] is said to be an Orlicz function if Φ is a nonzero function which is convex, even, continuous and vanishing at zero and left continuous on R⁺. Let us note that the left continuity of an Orlicz function Φ on R⁺ is equivalent to the fact that limu→b(Φ)⁻Φ(u) = Φ(b(Φ)), where b(Φ) := sup{u >

0 : Φ(u) <∞}. We will also use another parameter a(Φ) := sup{u > 0 : Φ(u) = 0}

of the Orlicz function Φ.

For any Orlicz function Φ we define its complementary function in the sense of Young by the formula

Ψ(u) = sup

v>0{|u|v − Φ(v)}

for all u ∈ R. It is easy to show that Ψ is also an Orlicz function.

Given any Orlicz function Φ we define on L⁰ = L⁰(T, Σ, µ) a convex modular IΦby

IΦ(x) = Z

T

Φ(x(t))dµ

(see [4, 22, 26, 28, 29, 31]). The Orlicz space L^Φ = L^Φ(T, Σ, µ) generated by an Orlicz function Φ is defined as

L^Φ={x ∈ L⁰: IΦ(λx) < +∞ for some λ > 0}.

Orlicz spaces L^Φ are usually considered as Banach spaces equipped with the Lu- xemburg norm

kxk^Φ= infn

λ > 0 : IΦ x λ

≤ 1o

or with the Orlicz norm

kxk^oΦ= sup Z

T

x(t)y(t)dµ

: I^Ψ(y)≤ 1

.

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It is known (see [22] for so-called N-functions and [16] in general) that for any Orlicz function Φ the following Amemiya formula

kxk^oΦ= inf

k>0

1

k(1 + IΦ(kx))

for the Orlicz norm holds. Moreover, the Amemiya formula for the Luxemburg norm (see [30] or [16]) is the following

kxkΦ= inf

k>0

max 1

k{1, I^Φ(kx)}

. Therefore, the inequalities

(1) kxkΦ≤ kxk^oΦ≤ 2 kxkΦ,

which are true for any x ∈ L^Φ are obvious. Orlicz spaces are Köthe spaces under both (the Luxemburg and the Orlicz) norms when they are considered with the natural semi-order ≤.

We say that an Orlicz function Φ satisfies condition ∆2for all u ∈ R⁺(at infinity) if there is K > 0 such that the inequality Φ(2u) ≤ KΦ(u) holds for all u ∈ R (for all u ∈ R satisfying |u| ≥ u⁰ with some u0 > 0 such that Φ(u0) <∞). We write then Φ ∈ ∆²(R⁺) (Φ∈ ∆²(∞)), respectively. Let us note that Φ ∈ ∆²(R⁺)implies that a(Φ) = 0 and b(Φ) = ∞. Analogously, Φ ∈ ∆²(∞) implies b(Φ) = ∞.

Given any x ∈ L^Φ, let K(x) be the set of all positive numbers k such that kxk^oΦ= 1

k(1 + IΦ(kx)) and let

k^∗(x) = inf{k > 0: I^Ψ(p(k|x|)) ≥ 1} , k^∗∗(x) = sup{k > 0: I^Ψ(p(k|x|)) ≤ 1} ,

where p is the right hand-side derivative of Φ. It is known that K(x) = [k^∗(x), k^∗∗(x)], whenever k^∗∗(x) <∞ and K(x) = [k^∗(x),∞) whenever k^∗(x) <∞ and k^∗∗(x) =

∞. In the second case we also have that kxk^oΦ = lim_k→∞¹_kIΦ(kx). It is obvious that K(x) 6= ∅ if and only if k^∗(x) <∞. It is known that k^∗(x) <∞ (equivalen- tly K(x) 6= ∅) for any x ∈ S(L^Φ)if and only if supu>0[A|u| − Φ(u)] = ∞, where A = lim_u→∞^Φ(u)_u (see [5]). It is obvious that this condition is satisfied if A = ∞.

However, there are Orlicz functions with A < ∞ and supu>0[A|u| − Φ(u)] = ∞ (see [5]).

Now, we will present some results concerning the characteristic of monotonicity in Orlicz spaces L^Φo = (L^Φ,k · k^oΦ). Recall that analogous results for the spaces (L^Φ,k·kΦ)have been presented in [9].

Lemma 3.1 Assume that µ(T ) < ∞, Φ is an Orlicz function with a(Φ) > 0 and satisfying the condition ∆2(∞) and let c ∈ (a(Φ), ∞). Then for any ε ∈ (0, 1) there exists δ(ε) ∈ (0, 1) such that if x ∈ L^Φ, |x(t)| ≥ c for µ−a.e. t ∈ supp x and IΦ(x)≤ δ(ε), then kxk^oΦ≤ ε.

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This lemma follows immediately from [9, Lemma 3.1] and inequalities (1).

Lemma 3.2 Let µ(T ) < ∞, supu>0[A|u| − Φ(u)] = ∞, where A = limu→∞Φ(u) u , a(Φ) > 0 and Φ ∈ ∆2(∞). Then

1−a(Φ)

l(Φ)kχ^Tk^oΦ≤ δm,L^Φ_o(1)≤ 1 −a(Φ) k(Φ)kχ^Tk^oΦ, where

k(Φ) := inf{lim inf_n→∞ k^∗(yn) : yn∈ S⁺(L^Φ_o)for any n ∈ N and lim_n→∞µ(supp yn) = 0} and

l(Φ) := inf{k^∗(x) : x∈ S(L^Φo)}.

Proof Let us note that δm,L^Φ_o(1) = 1−sup {kx − ykôΦ: 0≤ y ≤ x, kykôΦ=kxkôΦ= 1}.

Assume now that 0 ≤ y ≤ x, kyk^oΦ=kxk^oΦ= 1and let k ∈ K(x). Then k ≥ 1 and, by superadditivity of the modular IΦon L⁰

+, 1 = kxk^oΦ= 1

k(1 + IΦ(kx))

= 1

k(1 + IΦ(k(x− y) + ky))

≥ 1

k(1 + IΦ(k(x− y)) + I^Φ(ky))

= 1

k(1 + IΦ(ky)) + 1

kIΦ(k (x− y))

≥ kyk^oΦ+ 1

kIΦ(k(x− y)) = 1 +1

kIΦ(k(x− y)),

whence IΦ(k(x− y)) = 0, and consequently x − y ≤ ^a(Φ)k χT. Therefore, kx − yk^oΦ≤

a(Φ)

k kχ^Tk^oΦ. By the arbitrariness of the functions x, y ∈ S(L^Φo)such that 0 ≤ y ≤ x, we get the inequality

sup{kx − ykôΦ: 0≤ y ≤ x, kykôΦ=kxkôΦ= 1} ≤ a(Φ)

l(Φ) kχ^Tk^oΦ, i.e.

(2) δm,L^Φ_o(1)≥ 1 −a(Φ)

l(Φ) kχ^Tk^oΦ.

By the definition of the number k(Φ) there exists a sequence (yn)^∞_n=1in S+(L^Φ_o)such that µ(supp yn)→ 0 and k^∗(yn)→ k(Φ). Let us denote for simplicity k^∗(yn)by kn

and supp yn by An, and define the sequence of functions (xn)^∞_n=1 in the following way

xn = yn+a(Φ) kn

χT\An

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for any n ∈ N. It is clear that 1 = kynk^oΦ≤ kxⁿk^oΦ for any n ∈ N. Moreover,

kxⁿk^oΦ≤ 1 kn

(1 + IΦ(knxn)) = 1 kn

(1 + IΦ(knyn)) =kyⁿkôΦ, so kxⁿkôΦ=kyⁿkôΦ= 1. Moreover,

kxⁿ− yⁿk^oΦ= a(Φ)

kn

χT\An

o

Φ

= a(Φ) kn

χT\An

^o

Φ→ a(Φ) k(Φ)kχ^Tk^oΦ

as n → N, which follows by the facts that kn → k(Φ), µ(Aⁿ)→ 0 as n → ∞ and the element χT is order continuous (because of the assumptions that µ(T ) < ∞ and Φ∈ ∆²(∞), whence b(Φ) = ∞). Consequently, we have

sup{kx − ykôΦ: 0≤ y ≤ x, kykôΦ=kxkôΦ= 1} ≥ lim_n→∞kxⁿ− yⁿkôΦ= a(Φ)

k(Φ)kχ^Tk^oΦ, i.e.

(3) δ_m,L^Φ_o(1)≤ 1 −a(Φ)

k(Φ)kχ^Tk^oΦ.

Combining inequalities (2) and (3), we get our thesis.

Remark 3.3 The formula for kχ^Ak^oΦ is the following:

kχ^Tk^oΦ= µ(T )Ψ⁻¹

1 µ(T )

.

Really, it is known (see [21], p. 107) that for any symmetric space E and its Köthe dual E⁰ the equality kχÂkÊkχÂkÊ⁰ = kχÂkL¹ = µ(A) holds for any A ∈ Σ with 0 < µ(A) < ∞. It is also known that the Köthe dual of L^Φo = (L^Φ,k · kôΦ) is (L^Ψ,k·k^Ψ). Denoting λ(A) = 1/Ψ⁻¹(1/µ(A))for any A ∈ Σ with 0 < µ(A) < ∞, we have IΨ

χA

λ(A)

= 1, whence kχAk^Ψ= λ(A). In consequence, kχAkôΦ· λ(A) = µ(A), whence kχÂkôΦ= ^µ(A)_λ(A) = µ(A)Ψ⁻¹(1/µ(A)).

Remark 3.4 Let us note that if µ(T ) < ∞, supu>0[A|u| − Φ(u)] = ∞, where A = limu→∞Φ(u)

u , Φ ∈ ∆²(∞) and Ψ ∈ ∆²(∞), then the constant k(Φ) from Lemma 3.2 is finite. Indeed, in [4, Theorem 1.35(2)] it has been shown that if µ(T ) < ∞, the Orlicz function Φ vanishes only at 0, Φ has finite values on the whole R, lim^u→0Φ(u)

u = 0, limu→∞Φ(u)

u = ∞ and Ψ ∈ ∆²(∞), then sup{k >

0 : k ∈ K(x), x ∈ S(L^Φo)} < ∞. It is easy to show that Theorem 1.35(2) from [4]

remains also true if Φ has finite values on the whole R and supu>0[A|u| − Φ(u)] = ∞ (also if a(Φ) > 0). In consequence, taking any sequence (yn)^∞_n=1from S+(L^Φ_o)with limn→∞µ(supp yn) = 0, we get that the sequence (k^∗(yn))^∞_n=1 is bounded, whence lim infn→∞ k^∗(yn) <∞ and k(Φ) ≤ lim infⁿ→∞ k^∗(yn).

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Theorem 3.5 Let µ(T ) < ∞ and Φ be an Orlicz function. Then the following statements hold true:

a) If supu>0[A|u| − Φ(u)] = ∞, where A = lim^u→∞Φ(u)

u , Φ ∈ ∆²(∞) and a(Φ) = 0, then ε0,m(L^Φ_o) = 0.

b) If supu>0[A|u| − Φ(u)] = ∞, Φ ∈ ∆²(∞), Ψ ∈ ∆²(∞) and a(Φ) > 0, then

(4) a(Φ)

k(Φ)kχ^Tk^oΦ≤ ε^0,m(L^Φ_o)≤a(Φ)

l(Φ) kχ^Tk^oΦ,

where the constants k(Φ) and l(Φ) are defined as in Lemma 3.2. In the case when Ψ /∈ ∆²(∞), we have

(5) a(Φ)

k(Φ)kχ^Tk^oΦ≤ ε^0,m(L^Φ_o)≤ a(Φ) kχ^Tk^oΦ, c) If Φ /∈ ∆2(∞), then ε^0,m(L^Φ_o) = 1.

Proof a) Proceeding analogously as in [14] we get that the space L^Φo is uniformly monotone (since the condition limu→∞Φ(u)

u = ∞ that was used in [14], can be replaced successfully by supu>0[A|u| − Φ(u)] = ∞), whence we have immediately ε0,m(L^Φ_o) = 0.

b)Now we will show formula (4). Assume that 0 ≤ yⁿ ≤ xⁿ, kxⁿk^oΦ= 1for all n∈ N and kyⁿk^oΦ→ 1 as n → ∞. Let kⁿ∈ K(xⁿ)for any n ∈ N. By superadditivity of the modular IΦon L⁰

+, we have 1 = kxⁿk^oΦ= 1

kn

(1 + IΦ(knxn))

= 1

kn

(1 + IΦ(kn(xn− yⁿ) + knyn))

≥ 1

kn

(1 + IΦ(kn(xn− yⁿ)) + IΦ(knyn))

= 1

kn

(1 + IΦ(knyn)) + 1 kn

IΦ(kn(xn− yⁿ))

≥ kyⁿk^oΦ+ 1 kn

IΦ(kn(xn− yⁿ)) .

The assumption that kyⁿk^oΦ → 1 as n → ∞ and the above inequality yield that

1

knIΦ(kn(xn− yⁿ))→ 0 as n → ∞. Since sup{kⁿ: n∈ N} < ∞ (see Remark 3.4), we have IΦ(kn(xn− yⁿ))→ 0.

Take arbitrary ε > 0 and define the sets

An={t ∈ T : |kⁿ(xn(t)− yⁿ(t))| ≥ (1 + ε)a(Φ)} (∀ n ∈ N).

Obviously, IΦ(kn(xn− yⁿ) χAn)→ 0 as n → ∞, whence by virtue of Lemma 3.1,

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we obtain that kkn(xn− yⁿ)χAnk^oΦ→ 0 as n → ∞. Hence, we have l(Φ)· lim sup

n→∞ k(xⁿ− yⁿ)k^oΦ ≤ lim sup

n→∞ kkⁿ(xn− yⁿ)k^oΦ

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≤ lim sup

n→∞

kn(xn− yⁿ) χT\An

^o

Φ

+ lim sup

n→∞ kkⁿ(xn− yⁿ) χAnk^oΦ

= lim sup

n→∞

kn(xn− yⁿ) χT\An

^o

Φ

≤ lim sup

n→∞ (1 + ε)a(Φ) χT\An

^o

Φ

≤ (1 + ε)a(Φ) kχ^Tk^oΦ. Since ε > 0 was arbitrary, we get the inequality

lim sup

n→∞ kxⁿ− yⁿk^oΦ≤ a(Φ)

l(Φ) kχ^Tk^oΦ.

By the arbitrariness of (xn)^∞_n=1 and (yn)^∞_n=1with the desired properties and The- orem 1.1, we obtain ε0,m L^Φ_o

≤ ^a(Φ)l(Φ)kχ^Tk^oΦ.

Moreover, by Theorem 1.2 and Lemma 3.2, we have ε0,m L^Φ_o

= 1− δm,L^Φ_o(1⁻)≥ 1 − δm,L^Φ_o(1)

≥ 1 −

1−a(Φ) k(Φ)kχ^Tk^oΦ

=a(Φ)

k(Φ)kχ^Tk^oΦ.

If we do not assume that Ψ ∈ ∆²(∞), then analogously as in the proof of (4), we get _k¹_nIΦ(kn(xn− yⁿ))→ 0 as n → ∞. Since ^Φ(u)u is a nondecreasing function with respect to u, we obtain that for any x ∈ L⁰(T, Σ, µ)the function f(k) := ¹_kIΦ(kx) is nondecreasing with respect to k. Hence

0≤ I^Φ(xn− yⁿ)≤ 1 kn

IΦ(kn(xn− yⁿ))

for any n ∈ N, whence IΦ(xn− yⁿ)→ 0 as n → ∞. Proceeding analogously as in (6) and defining An = {t ∈ T : |(xⁿ(t)− yⁿ(t))| ≥ (1 + ε)a(Φ)} for any n ∈ N, we get ε0,m L^Φ_o

≤ a(Φ) kχ^Tk^oΦ.

In order to prove c) let us notice that if an Orlicz function Φ does not satisfy condition ∆2(∞), then the Orlicz space L^Φo contains an order almost isometric copy of `^∞(see [17]). Consequently, ε0,m L^Φ_o= 1.

Theorem 3.6 Let µ(T ) = ∞, then ε0,m(L^Φ_o) = 0whenever supu>0[A|u| − Φ(u)] =

∞, where A = limu→∞Φ(u)

u and Φ ∈ ∆2(R⁺), and ε0,m(L^Φ_o) = 1 if Φ /∈ ∆2(R⁺).

Proof Condition Φ ∈ ∆2(R⁺)implies that a(Φ) = 0. Therefore, using the condition supu>0[A|u| − Φ(u)] = ∞ instead of lim^u→∞Φ(u)

u =∞, we can prove in the same way as in [14] that the space L^Φo is uniformly monotone and, in consequence, ε0,m(L^Φ_o) = 0. If Φ /∈ ∆²(R⁺), then L^Φo contains an order almost isometric copy of

`^∞(see [17]), whence ε0,m(L^Φ_o) = 1.

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P. Foralewski

Adam Mickiewicz University, Faculty of Mathematics and Computer Science Umultowska 87, 61-614 Poznań, Poland

E-mail: katon@amu.edu.pl H. Hudzik

E-mail: hudzik@amu.edu.pl R. Kaczmarek

E-mail: radekk@amu.edu.pl M. Krbec

Institute of Mathematics, Academy of Sciences of the Czech Republic Zitná 25, CZ-115 67 Prague 1, Czech Republic

(Received: 20.11.2013)