Nonlinear Analysis 74 (2011) 3912–3925

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Nonlinear Analysis

journal homepage: www.elsevier.com/locate/na

Rotundity structure of local nature for generalized Orlicz–Lorentzfunction spaces

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

a r t i c l e i n f o

Article history:Received 22 December 2009Accepted 16 February 2011Communicated by: Ravi Agarwal

MSC:46E3046B2046A80

Keywords:Generalized Orlicz–Lorentz function spaceOrlicz–Lorentz function spaceExtreme pointRotund pointStrong U-pointRotundity

a b s t r a c t

Criteria for extremepoints and rotundpoints in generalizedOrlicz–Lorentz function spaces,which were introduced in Foralewski (2011) [27] are given. Some examples show that inthese spaces the notion of rotund point is essentially stronger than the notion of extremepoint. Finally, the criteria obtained in this paper are interpreted in the case of classicalOrlicz–Lorentz spaces. Results of this paper are related to the results from Carothers et al.(1992) [9], Kamińska (1990) [4], Foralewski et al. (2008) [26].

© 2011 Elsevier Ltd. All rights reserved.

In 1950 Lorentz introduced a class of function spaces later called Lorentz spaces (see [1,2]). These spaces play an importantrole in the theory of Banach spaces, in particular they are key objects in the interpolation theory of linear operators. At thebeginning of nineties Kamińska began to consider Orlicz–Lorentz spaces (see [3–5]). Both spaces, the Lorentz space and theOrlicz–Lorentz space, since the moment they have been defined are the objects of great interest to several mathematicians(see the papers [6–21] as well as the monographs [22–24]).

Recently generalized Orlicz–Lorentz sequence and function spaces have been defined and investigated (see [25–27]).In this paper we present results concerning the problem of existence of extreme points and rotund points in generalizedOrlicz–Lorentz function spaces.

Let (X, ‖ · ‖) be a real Banach space and let B(X) and S(X) be the closed unit ball and the unit sphere of X , respectively.A point x ∈ S(X) is called an extreme point of B(X) if for every y, z ∈ B(X) with x =

y+z2 , we have y = z. The notion of

extreme point plays an important role in some branches of mathematics. For example, the Krein–Milman theorem, Choquetintegral representation theorem, Rainwater theoremon the convergence in theweak topology, Bessaga–Pełczyński theoremand Elton test for unconditional convergence are formulated in terms of extreme points (see [28, Chapter IX]).

A point x ∈ S(X) is said to be a rotund point (strong U-point, SU-point) of B(X) if for any y ∈ S(X) with ‖x + y‖ = 2,we have x = y. Recall that the nature of a rotund point is such that a point x ∈ S(X) is a point of local uniform rotundity ifand only if x is a point of compact local uniform rotundity and a rotund point (see [29]). In [30] it has been shown that thenotion of the rotund point is important for the local best approximation problem.

E-mail address: katon@amu.edu.pl.

0362-546X/$ – see front matter© 2011 Elsevier Ltd. All rights reserved.doi:10.1016/j.na.2011.02.027

P. Foralewski / Nonlinear Analysis 74 (2011) 3912–3925 3913

It is obvious that a Banach space X is rotund if and only if every point of S(X) is an extreme point of B(X), as well as ifand only if any point of S(X) is a rotund point of B(X), but the notion of rotund point is essentially stronger than the notionof extreme point. Namely in l21 (two-dimensional l1 space) the points x = (1, 0) and y = (0, 1) are extreme points of B(l21).Since x = y and ‖x + y‖ = 2, neither x nor y is a rotund point of B(l21).

This paper is organized as follows. In Section 1 we recall the definition of generalized Orlicz–Lorentz function spaces.Next, in Section 2, we present few results that will be useful in the remaining part of this paper. Section 3 concerns criteriafor extreme points as well as for rotund points in generalized Orlicz–Lorentz function spaces. In Section 4 we will give fewexamples of extremepointswhich are not rotundpoint. Finally, in Section 5, the results obtained in this paper are interpretedin the case of classical Orlicz–Lorentz spaces.

1. Generalized Orlicz–Lorentz function spaces

Let L0 = L0([0, γ )) be the space of all (equivalence classes of) Lebesgue measurable real-valued functions defined onthe interval [0, γ ), where γ ≤ ∞. For any x, y ∈ L0, we write x ≤ y, if x(t) ≤ y(t) almost everywhere with respect to theLebesgue measurem on [0, γ ).

Given any x ∈ L0 we define its distribution function µx : [0,+∞) → [0, γ ] by

µx(λ) = m{t ∈ [0, γ ) : |x(t)| > λ}

(see [22–24]) and the nonincreasing rearrangement x∗: [0, γ ) → [0,∞] of x as

x∗(t) = inf{λ ≥ 0 : µx(λ) ≤ t}

(under the convention inf ∅ = ∞). We say that two functions x, y ∈ L0 are equimeasurable if µx(λ) = µy(λ) for all λ ≥ 0.Then we obviously have x∗

= y∗.A mapping σ : [0, γ ) → [0, γ ) is called a measure preserving transformation if for each measurable set A ⊂ [0, γ ), the

set σ−1(A) = {t ∈ [0, γ ) : σ(t) ∈ A} is measurable and m(σ−1(A)) = m(A) (see [22]). It is well known that a measurepreserving transformation induces equimeasurability, that is, if σ is a measure preserving transformation, then x and x ◦ σare equimeasurable functions. The converse is false (see [22]).

A Banach space E = (E,≤, ‖ · ‖), where E ⊂ L0, is said to be a Köthe space if the following conditions are satisfied:

(i) if x ∈ E, y ∈ L0 and |y| ≤ |x|, then y ∈ E and ‖y‖ ≤ ‖x‖,(ii) there exists a function x in E that is strictly positive on the whole [0, γ ).

Recall that the Köthe space E is called a symmetric space if E is rearrangement invariant in the sense that if x ∈ E, y ∈ L0and x∗

= y∗, then y ∈ E and ‖x‖ = ‖y‖ (see [12]). For basic properties of symmetric spaces we refer to [22–24].A map ϕ : [0, γ ) × [−∞,∞] → [0,∞] is said to be a Musielak–Orlicz function if for m-a.e. t ∈ [0, γ ) the function

ϕ(t, ·) : [−∞,∞] → [0,∞] (our definition is extended from R into Re:= [−∞,∞] by assuming that ϕ(t,−∞) =

ϕ(t,∞) = ∞) is a nonzero function that is convex, even, vanishing and continuous at zero, left continuous on (0,∞], andϕ(·, u) is anm-measurable function for any u ∈ R+ (see [31,32,35]).

In the definition of generalized Orlicz–Lorentz function spaces condition (L1), the definition of which is presented below,plays the key role.

Definition 1. Let us denote U∞ =

u ≥ 0 :

t0 ϕ(s, u)ds < ∞ for some t = t(u) > 0

and u∞ = supU∞. We say that a

Musielak–Orlicz function ϕ satisfies condition (L1), if there exists a measurable set A with m(A) = 0 such that for anys, t ∈ [0, γ ) \ A, s < t , and for any u ∈ [0, u∞), we have p(s, u) ≥ p(t, u), where by p(t, ·)we denote the right derivative ofϕ(t, ·) (cf. [33]) or equivalently, if for any s, t ∈ [0, γ ) \ A, s < t , the function Fs,t(u) := ϕ(s, u)− ϕ(t, u) is nondecreasingon the interval [0, u∞) (cf. [34] in the sequence case).

It is easy to show that condition (L1) yields that ϕ(s, u) ≥ ϕ(t, u) for all s, t ∈ [0, γ ) \ A, s < t , and u ∈ [0, u∞]. Hence,we have that

t0 ϕ(s, u)ds < ∞ for any u ∈ [0, u∞) (or u ∈ [0, u∞], if u∞ ∈ U∞) and any t ∈ [0, γ ).

In [27] it has been shown that the functional

ϱϕ (x) =

∫ γ

0ϕ(t, x∗(t))dt

is a convexmodular if and only if theMusielak–Orlicz functionϕ satisfies condition (L1). Then the generalizedOrlicz–Lorentzfunction space

Λϕ = {x ∈ L0 : ϱϕ (λx) < ∞ for some λ > 0}

equipped with the Luxemburg norm

‖x‖ = infλ > 0 : ϱϕ

xλ

≤ 1

is a symmetric Banach space. In [27] order continuity and separability, the Kadec–Klee property with respect to the localconvergence in measure as well as monotonicity and rotundity properties have been investigated inΛϕ among others.

3914 P. Foralewski / Nonlinear Analysis 74 (2011) 3912–3925

Remark 1. In this paper we will assume that the Musielak–Orlicz function ϕ satisfies conditions (L1). Without loss ofgenerality we can assume that the inequality p(s, u) ≥ p(t, u) is satisfied for any s, t ∈ [0, γ ) such that s ≤ t and anyu ∈ [0, u∞). We will also assume that the generalized Orlicz–Lorentz function spaceΛϕ is not trivial, that is, u∞ > 0.

2. Auxiliary results

In this part of paperwewill present few results thatwill be used in themain part of this paper.We startwith the following

Proposition 1. For any x, y ∈ Λϕ and any t ∈ (0, γ ), we have the inequality∫ t

0ϕ(s, (x + y)∗(s))ds ≤

∫ t

0ϕ(s, x∗(s)+ y∗(s))ds,

whence we get

ϱϕ (x + y) ≤ ϱϕx∗

+ y∗

for any x, y ∈ Λϕ .

Proof. Let x, y ∈ Λϕ be arbitrary fixed elements. Without loss of generality we can assume that∫ t0

0ϕ(s, x∗(s)+ y∗(s))ds < ∞

for some t0 ∈ (0, γ ). For fixed t ∈ (0, γ ) take arbitrary n ∈ N and divide the interval [0, t] into 2n not overlappingsubintervals of the same length by the points

0 = tn0 < tn1 < · · · < tn2n = t.

For any i = 1, 2, . . . , 2n one can find ani and bni such that

(x + y)∗(tni−1) ≥ ani ≥ (x + y)∗(tni ), x∗(tni−1)+ y∗(tni−1) ≥ bni ≥ x∗(tni )+ y∗(tni )

and ∫ tni

tni−1

(x + y)∗(s)ds = ani ·t2n,

∫ tni

tni−1

x∗(s)+ y∗(s)ds = bni ·t2n.

The sequences (ani )2ni=1 and (bni )

2ni=1 are nonincreasing sequences of nonnegative numbers. We also get (see [23, p. 65])

m−i=1

ani =2n

t

∫ tnm

0(x + y)∗(s)ds ≤

2n

t

∫ tnm

0x∗(s)+ y∗(s)ds =

m−i=1

bni (1)

form = 1, 2, . . . , 2n and ϕ(tn1 , bn1) < ∞. Now we will show that

m−i=1

ϕ(tni , ani ) ≤

m−i=1

ϕ(tni , bni ) (2)

form = 1, 2, . . . , 2n. By inequality (1) for m = 1, we get

ϕ(tn1 , an1) ≤ ϕ(tn1 , b

n1).

Now assume that inequality (2) is true for 1, 2, . . . ,m (1 ≤ m ≤ 2n− 1). We will show that it holds for m + 1 too. If

anm+1 ≤ bnm+1, then ϕ(tnm+1, a

nm+1) ≤ ϕ(tnm+1, b

nm+1), so on the basis of the induction assumption, inequality (2) for m + 1 is

obvious. Now assume that anm+1 > bnm+1. By virtue of Lemma 6.3 in [25], we get

ϕ(tnm+1, anm+1)− ϕ(tnm+1, b

nm+1)

anm+1 − bnm+1

m−i=1

(bni − ani ) ≤

m−i=1

(ϕ(tni , bni )− ϕ(tni , a

ni ))

and simultaneously, by inequality (1), we have anm+1 − bnm+1 ≤∑m

i=1(bni − ani ), whence

ϕ(tnm+1anm+1)− ϕ(tnm+1, b

nm+1) ≤

m−i=1

(ϕ(tni , bni )− ϕ(tni , a

ni )).

So, inequality (2) holds for m + 1.

P. Foralewski / Nonlinear Analysis 74 (2011) 3912–3925 3915

Since∫ t

0ϕ(s, (x + y)∗(s))ds = lim

n→∞

t2n

2n−i=1

ϕ(tni , ani ) ≤ lim

n→∞

t2n

2n−i=1

ϕ(tni , bni ) =

∫ t

0ϕ(s, x∗(s)+ y∗(s))ds,

the proof is complete. �

Remark 2. It is well known that for all x ∈ L0 and t ∈ [0, γ ) such that x∗(t) > lims→∞ x∗(s) = x∗(∞) one can find the setet(x) such thatm(et(x)) = t, |x(s)| ≥ x∗(t) form-a.e. s ∈ et(x) (in fact |x(s)| ≥ limw→t− x∗(w) form-a.e. s ∈ et(x)) and∫ t

0x∗(s)ds =

∫et (x)

|x(s)|ds. (3)

(see 7o on page 64 in [23]). It is also easy to show that if for any x ∈ L0 and t ∈ [0, γ ), x∗(t) > x∗(∞), there exists a set et(x)such that Eq. (3) is satisfied andm(et(x)) = t , then |x(s)| ≥ limw→t− x∗(w) form-a.e. s ∈ et(x).

The next lemma will be used in the proofs of the sufficiency of Theorems 1 and 2.

Lemma 1. (i) Let x ∈ L0 satisfy the condition: if γ = ∞ and limt→∞ x∗(t) = x∗(∞) > 0, then m({t : |x(t)| ≤ x∗(∞)}) = 0.Under these assumptions, for any y ∈ L0 such that y∗

= x∗ and (x + y)∗ = x∗+ y∗, we have x = y.

(ii) Let x ∈ L0 be such that m({t : |x(t)| < x∗(∞)}) = 0. Then for any y, z ∈ L0 satisfying x =y+z2 and x∗

= y∗= z∗, we have

y = z.

Proof. (i) Assume that the assumptions are satisfied. For any t ∈ [0, t0), where t0 = sup{t : x∗(t) > 0}, we have

(x + y)∗(t) = x∗(t)+ y∗(t) > limt→∞

(x∗(t)+ y∗(t)) = limt→∞

(x + y)∗(t) = (x + y)∗(∞);

obviously in the case when γ < ∞, we can assume that x(t) = y(t) = 0 for t ∈ [γ ,∞). Therefore, for any t ∈ [0, t0) thereexists a measurable set et(x + y) such thatm(et(x + y)) = t and∫ t

0(x + y)∗(s)ds =

∫et (x+y)

|x(s)+ y(s)|ds

(see Remark 2). Since (x + y)∗ = x∗+ y∗, analogously as in the proof of the necessity of 9o from [23, pages 65–66], we get

that the system of the sets (et(x + y))t∈[0,t0) can serve as the system (et(x))t∈[0,t0) and (et(y))t∈[0,t0), that is,∫ t

0x∗(s)ds =

∫et (x+y)

|x(s)|ds and∫ t

0y∗(s)ds =

∫et (x+y)

|y(s)|ds

for any t ∈ [0, t0) and |x(s)+ y(s)| = |x(s)| + |y(s)| form-a.e. s ∈

t∈[0,t0)et(x + y).

If x∗(∞) = y∗(∞) = (x + y)∗(∞) = 0, then by the Remark on page 65 in [23], we obtain that

t∈[0,t0)et(x + y) =

supp(x + y) = supp x = supp y, whence we have |x(s)+ y(s)| = |x(s)| + |y(s)| form-a.e. s ∈ [0, γ ).Nowwe will show that if x∗(∞) > 0, then

t∈[0,∞) et(x+ y) = [0,∞), whence we get |x(s)+ y(s)| = |x(s)| + |y(s)| for

m-a.e. s ∈ [0,∞). Suppose to the contrary that m

t ∈ ([0,∞) \

t∈[0,∞) et(x + y)) : |x(t)| ≥ x∗(∞)+1n

= a > 0 for

some n ∈ N . Denoting t1 = supt ∈ [0,∞) : x∗(t) ≥ x∗(∞)+

1n

, by equimeasurability of x and x∗, we obtain that there

exists a set A ⊂ et1(x + y) such that m(A) ≥ a and |x(t)| < x∗(∞) +1n ≤ limw→t1− x∗(w) for m-a.e. t ∈ A, which gives a

contradiction (see Remark 2).Now assume that x = y. We can assume, without loss of generality, that there exist n ∈ N and a set An such that

m(An) > 0 and |x(t)| ≥ |y(t)| +1n for m-a.e. t ∈ An. Next, we can find a positive number b > x∗(∞) such that the set

An,b :=t ∈ An : |x(t)| ∈

b, b +

12n

has a positive measure. For t2 := sup{t ∈ [0, γ ) : x∗(t) ≥ b}, we have t2 < ∞ and,

by Remark 2 and equimeasurability of x and x∗, An,b ⊂ et2(x) = et2(x + y). But et2(y) = et2(x + y), whence An,b ⊂ et2(y),which gives a contradiction (see Remark 2, we have |y(t)| ≤ b −

12n < limw→t2− x∗(w) = limw→t2− y∗(w) for m-a.e.

t ∈ An,b).(ii) If |x(t)| > x∗(∞) > 0 form-a.e. t ∈ [0,∞) or x∗(∞) = 0, proceeding analogously as above we can show that y = z.Now assume that m(A) = m({t : |x(t)| = x∗(∞) > 0}) > 0. For B = {t ∈ [0,∞) : |x(t)| > x∗(∞)} we have

m(B) = b ∈ [0,∞]. Since for any u, v ∈ L0 we have u∗≤ v∗ whenever u ≤ v (see Proposition 1.7 on page 41 in [22]) and∫ t

0(u + v)∗(s)ds ≤

∫ t

0(u)∗(s)ds +

∫ t

0(v)∗(s)ds

for all u, v ∈ L0 and t ∈ [0,∞) (see inequality (2.17) on page 65 in [23]), we get x∗χ[0,b) = (xχB)∗

= (yχB)∗

= (zχB)∗

whence, proceeding analogously as in (i), we conclude that yχB = zχB. Finally, suppose that yχA = zχA. By x =y+z2 , we

3916 P. Foralewski / Nonlinear Analysis 74 (2011) 3912–3925

can assume that there exists a natural number n such that the set C :=t ∈ A : |y(t)| > |x(t)| +

1n = x∗(∞)+

1n

satisfies

m(C) > 0. We have

m

t ∈ [0, γ ) : |y(t)| > x∗(∞)+1n

= m

t ∈ B : |y(t)| > x∗(∞)+

1n

+ m(C)

> m

t ∈ B : |x(t)| > x∗(∞)+1n

= m

t ∈ [0, γ ) : |x(t)| > x∗(∞)+

1n

,

which gives a contradiction since x and y are equimeasurable. �

The next two lemmas will be used in the proofs of the necessity of Theorems 1 and 2.

Lemma 2 ([4, Lemma 3.2]). Let x ∈ S(E), where E = (E,≤, ‖ · ‖) is a symmetric space (see page 2), be such that m({t : |x(t)| <limt→∞ x∗(t) = x∗(∞)}) = 0 (cf. condition (+) in [4]). Let x∗

=u+v2 , where u, v ∈ S(E), u = v and u, v are nonnegative.

(i) If m({t ∈ [0, γ ) : |x(t)| > x∗(∞)}) < ∞, then x is not an extreme point of the unit ball of E.(ii) If m({t ∈ [0, γ ) : |x(t)| > x∗(∞)}) = ∞ and µu(λ) = µv(λ) = ∞ for λ ∈ [0, x∗(∞)), then x is not an extreme point of

the unit ball of E either.

Lemma 3. Let x ∈ S(E), where E = (E,≤, ‖ · ‖) is a symmetric space, and let x satisfy the condition: if γ = ∞ andlimt→∞ x∗(t) = x∗(∞) > 0, then m({t : |x(t)| ≤ x∗(∞)}) = 0. If there exists v ∈ E such that v = v∗

= x∗ and‖v‖ = ‖(v + x∗)/2‖ = 1, then x is not a rotund point.

Proof. First assume that γ = ∞,m(supp x) = ∞ andm([0, γ )\supp x) > 0. By the assumptions on x, we have x∗(∞) = 0.Moreover, there exists a measure preserving transformation σ : supp x → [0, γ ) such that x∗(σ (t)) = |x(t)| for m-a.e.t ∈ supp x. Set v(t) = v(σ (t)) and u(t) =

v+x∗2 (σ (t)) for t ∈ supp x and v = u = 0 for t ∈ [0, γ ) \ supp x. Since for any

λ ≥ 0 we have

µv(λ) = m({t ∈ supp x : v(t) > λ})+ m({t ∈ [0, γ ) \ supp x : v(t) > λ})

= m({t ∈ supp x : v(σ (t)) > λ})+ m({t ∈ [0, γ ) \ supp x : 0 > λ})

= m({t ∈ [0, γ ) : v(t) > λ}) = µv(λ),

we conclude that v and v are equimeasurable. Analogously we can show equimeasurability of u and v+x∗2 . Hence we get

‖v‖ = ‖u‖ = 1. We also have u =v+|x|

2 . For the set A = {t ∈ [0, γ ) : v(t) = x∗(t)} we have m(A) > 0. DenotingB = σ−1(A) we get m(B) = m(A) and v(t) = v(σ (t)) = x∗(σ (t)) = |x(t)| for m-a.e. t ∈ B, so v = |x|. Therefore x is not arotund point.

Now let γ < ∞ or (γ = ∞ and supp x = [0, γ )). Then there exists a measure preserving transformation σ : [0, γ ) →

[0, γ ) such that x∗(σ (t)) = |x(t)| for m-a.e. t ∈ [0, γ ). Proceeding analogously as above we get that x is not a rotundpoint. �

3. Main results

Before a presentation of some criteria in order that x ∈ S(Λϕ) be an extremepoint of the unit ball B(Λϕ) for any x ∈ S(Λϕ)we must define two sets Ax and A0

x .

Definition 2. We will say that t0 (t0 ∈ [0, γ )) belongs to the set Ax if one of the following two conditions is satisfied:

(1) there exists t1 ∈ (t0, γ ) such that x∗(t1) < x∗(t) < x∗(t0) for all t ∈ (t0, t1) and functions ϕ(t, ·) of variable u are affineon the interval [x∗(t1), x∗(t0)] form-a.e. t ∈ [t0, t1],

(2) there exists t1 ∈ (t0, γ ] (in particular t1 = ∞) and ε > 0 such that x∗(t) ≥ x∗(t0) + ε for t < t0, x∗(t) = x∗(t0)for t ∈ [t0, t1), x∗(t) ≤ x∗(t0) − ε for t ∈ [t1, γ ) and functions ϕ(t, ·) of variable u are affine on the interval[x∗(t0)− ε, x∗(t0)+ ε] for m-a.e. t ∈ [t0, t1].

Additionally, if t0 satisfies condition (2), the functions ϕ(t, ·) have the form ϕ(t, u) = au + bt on the interval [x∗(t0 −

ε), x∗(t0 + ε)] for any t ∈ (t0, t1) (that is, p(t, u) is a constant function for all t ∈ (t0, t1) and all u ∈ [x∗(t0 − ε), x∗(t0 + ε)))and t1 < ∞, then also t1 ∈ Ax.

Definition 3. We will say that t0 (t0 ∈ [0, γ )) belongs to the set A0x if one of the following two conditions is satisfied:

(1) x∗(t) < x∗(t0) and ϕ(t, x∗(t0)) = 0 for any t ∈ (t0, γ ).(2) There exists ε > 0 such that x∗(t) ≥ x∗(t0)+ ε for t < t0 and ϕ(t, x∗(t0 + ε)) = 0 for t ∈ (t0, γ ).

P. Foralewski / Nonlinear Analysis 74 (2011) 3912–3925 3917

Remark 3. Note that if t0 satisfies condition (1) of Definition 2, then there exists at least countable set of points in the interval(t0, t1), which belong to the set Ax. In particular, if x∗ is strictly decreasing on the interval [t0, t1], then [t0, t1) ⊂ Ax.

In fact, by the right continuity of x∗, there exist s0 and s1 such that t0 < s0 < s1 < t1 and x∗(t0) > x∗(s0) > x∗(s1) >x∗(t1). If x∗(t) < x∗(s0) for t > s0, then s0 ∈ Ax. In the opposite case we denote s2 = sup{t > s0 : x∗(t) = x∗(s0)}. Obviouslys0 < s2 ≤ s1. If x∗(t) < x∗(s2) for t > s2, then s2 ∈ Ax. Now assume that there exists t > s2 such that x∗(t) = x∗(s2) anddefine s3 = sup{t > s2 : x∗(t) = x∗(s2)}. Obviously x∗(s2) < x∗(s0) and s3 ≤ t1. If x∗(s3) < x∗(s2), then s2 ∈ Ax (s2 satisfiescondition (2) of Definition 2). Whereas if x∗(s3) = x∗(s2), we get x∗(t) < x∗(s3) for t > s3 and s3 < t1, whence s3 ∈ Ax.

Now denote t ′1 = inf{t ≤ s0 : x∗(t) = x∗(s0)}. We have t0 < t ′1 ≤ s0. The above procedure can be repeated for t0 and t ′1.Therefore, we can find by induction a strictly decreasing sequence (tn) of elements from the interval (t0, t1) belonging to Ax.

Wewill now present few examples of elements of the unit sphere inΛϕ together with their sets fromDefinitions 2 and 3.

Example 1. It is easy to show that the Musielak–Orlicz function ϕ defined by the formula:if t ∈ [0, 1), then ϕ(t, u) = u2/4 for u ≥ 0;if t ∈ [1, 2), then ϕ(t, u) = u2/4 for u ∈ [0, 1) and ϕ(t, u) = u/2 − 1/4 for u ≥ 1;if t ∈ [2, γ = 3), then ϕ(t, u) = 0 for u ∈ [0, 1) and ϕ(t, u) = u/4 − 1/4 for u ≥ 1,satisfies condition (L1). We also have u∞ = ∞ (see Definition 1 on page 3).

Defining x1 = ((√33 − 3)/2)χ[0,γ ), x2 =

√2χ[0,1) + (4/3)χ[1,γ ) and x3 = (3/2)χ[0,1) + (11/8)χ[1,2) + 1χ[2,γ ), we get

Ax1 = ∅, Ax2 = {1}, Ax3 = {1, 2} and A0x1 = A0

x2 = A0x3 = ∅.

Now let ya =√5χ

0, 12

+ 1χ 12 ,2

+ aχ[2,γ ) for a ∈ [0, 1]. We have Ay1 = A0y1 = ∅ and Aya = {2, γ }, A0

ya = {2} for

a ∈ [0, 1).Finally, for z =

√2, 5χ[0,1) + (2 − t/2)χ[1,2) + 1χ[2,γ ), we get Az = [1, 2) and A0

z = ∅.

Theorem 1. An element x ∈ S(Λϕ) is an extreme point of B(Λϕ) if and only if |x| = u∞χ[0,γ ) (see Definition 1 on page 3) orthe following conditions hold:(i) m({t : |x(t)| < x∗(∞)}) = 0, where x∗(∞) = limt→∞ x∗(t) (cf. condition (+) in [4]),(ii) ϱϕ (x) = 1,(iii) the set Ax has at most one element,(iv) the set A0

x is empty.

The most important fact in the proof of this theorem is showing the equality x∗=

y∗+z∗

2 for x, y, z ∈ Λϕ such thatϱϕ (x) = ϱϕ (y) = ϱϕ (z) = 1 and x =

y+z2 . In papers [9,4] this equality has been shown under the additional assumption

that the weight function w is strictly decreasing. In this paper the above equality will be proved in Lemma 4 without anyadditional assumption.

Lemma 4. Let x satisfy conditions (ii)–(iv) of Theorem 1. If there exist y, z ∈ S(Λϕ) such that x =y+z2 , then

x∗=

12y∗

+12z∗. (4)

Proof. First assume that for any t ∈ [0, γ )which do not satisfy Eq. (4), we have

x∗(t) <12y∗(t)+

12z∗(t). (5)

Take arbitrary t0 for which inequality (5) holds and let [t1, t2] be the biggest interval such that t0 ∈ [t1, t2] and inequality(5) is satisfied for any t ∈ (t1, t2). By the right continuity of the rearrangement we get t1 < t2. Since ϱϕ (x) = 1 and themodular ϱϕ is convex, we have

0 ≤

∫ t2

t1

ϕ

t,

12y∗(t)+

12z∗(t)

− ϕ(t, x∗(t))

dt

≤

∫ t2

t1

12ϕ(t, y∗(t))+

12ϕ(t, z∗(t))− ϕ(t, x∗(t))

dt

≤12ϱϕ (y)+

12ϱϕ (z)− ϱϕ (x) = 0,

whence we get that ϕt, 1

2y∗(t)+

12 z

∗(t)

= 0 for any t ∈ (t1, t2] (see Remark 1). If 12y

∗(t1) +12 z

∗(t1) > x∗(t1), thenx∗(t) ≥

12y

∗(t1) +12 z

∗(t1) for any t < t1. Hence we get that t1 ∈ A0x (see Definition 3, condition (2)), which gives a

contradiction with condition (iv) of Theorem 1. Now let x∗(t1) =12y

∗(t1) +12 z

∗(t1). Then we have x∗(t) < x∗(t1) for anyt > t1 and ϕ(t, x∗(t1)) = 0 for the same t , whence we get again that t1 ∈ A0

x (see Definition 3, condition (1)) which againgives a contradiction with condition (iv) of Theorem 1.

3918 P. Foralewski / Nonlinear Analysis 74 (2011) 3912–3925

Now assume that the set

B =

t ∈ [0, γ ) : x∗(t) >

12y∗(t)+

12z∗(t)

is nonempty. By the right continuity of the rearrangement, we get that it consists of at most countable number of pairwisedisjoint intervals Bn. More precisely, for any n such that Bn is nonempty we have tn1 := inf{t ∈ Bn} < sup{t ∈ Bn} =: tn2 , theinequality

x∗(t) >12y∗(t)+

12z∗(t)

holds for all t ∈ (tn1 , tn2 ) and it does not hold for tn2 and for some t ∈ (t ′, tn1 ], where t ′ < tn1 is arbitrary. By inequality (2.17)

on page 65 in [23] we obtain that the set

C =

t ∈ [0, γ ) : x∗(t) <

12y∗(t)+

12z∗(t)

is also nonempty. Define the functions

b(t) =

∫[0,t)∩B

x∗(t)−

12y∗(t)+

12z∗(t)

dt

and

c(t) =

∫[0,t)∩C

12y∗(t)+

12z∗(t)

− x∗(t)

dt.

We have that b(t) ≤ c(t) < ∞ for any t ∈ [0, γ ) and both the functions are nondecreasing and continuous. Therefore, forany n such that the set Bn is nonempty, we can find sn1 and sn2 such that

sn1 = sup{t ∈ [0, γ ) : c(t) = b(tn1 )},sn2 = inf{t ∈ [0, γ ) : c(t) = b(tn2 )}.

We have 0 ≤ sn1 < sn2 ≤ tn1 . Since x∗(sn2) ≥ x∗(tn1 ), by condition (L1), we obtain that p(sn2, x∗(sn2)) ≥ p(sn2, x

∗(tn1 )) ≥

p(tn1 , x∗(tn1 )), where p(t, ·) denotes the right derivative of ϕ(t, ·). Hence, again applying condition (L1), we get∫

[sn1,sn2]∩C

12ϕ(y∗(t))+

12ϕ(z∗(t))

− ϕ(x∗(t))

dt ≥

∫[sn1,s

n2]∩C

ϕ

12y∗(t)+

12z∗(t)

− ϕ(x∗(t))

dt

=

∫[sn1,s

n2]∩C

∫ 12 y

∗(t)+ 12 z

∗(t)

x∗(t)p(t, u)du

dt

≥ p(sn2, x∗(sn2))(c(s

n2)− c(sn1))

≥ p(tn1 , x∗(tn1 ))(b(t

n2 )− b(tn1 ))

≥

∫Bn

∫ x∗(t)

12 y

∗(t)+ 12 z

∗(t)p(t, u)du

dt

=

∫Bn

ϕ(x∗(t))− ϕ

12y∗(t)+

12z∗(t)

dt

≥

∫Bn

ϕ(x∗(t))−

12ϕ(y∗(t))+

12ϕ(z∗(t))

dt. (6)

Since ϱϕ (x) = 1 and the modular ϱϕ is convex, we get the equality in (6) for any n such that the set Bn is nonempty, which

is possible if and only if p(t, u) is a constant function for any t ∈ (sn1, tn2 ) and u ∈

limt→tn2−

12y

∗(t)+12 z

∗(t), 1

2y∗(sn1) +

12 z

∗(sn1).

If there exists t ∈ [s11, t12 ) such that x∗(s) < x∗(t) for s > t , then t satisfies condition (1) of Definition 2. Hence, by

Remark 3, we obtain a contradiction with condition (iii) of Theorem 1. In the opposite case, there exist at least one interval[t0, t1] ⊂ [s11, t

12 ] and ε > 0 such that t1 < ∞, x∗(t) ≥ x∗(t0)+ε for t < t0, x∗(t) = x∗(t0) for t ∈ [t0, t1), x∗(t) ≤ x∗(t0)−ε

for t ∈ [t1, γ ) and functions ϕ(t, ·) of variable u have the form ϕ(t, u) = au + bt on the interval [x∗(t0)− ε, x∗(t0)+ ε] form-a.e. t ∈ [t0, t1], so t0, t1 ∈ Ax, which contradicts again to condition (iii) of Theorem 1. �

P. Foralewski / Nonlinear Analysis 74 (2011) 3912–3925 3919

Proof of Theorem 1. Sufficiency. First we assume that x ∈ S(Λϕ), |x| = u∞χ[0,γ ) and there exist y, z ∈ S(Λϕ) such thaty = z and x =

y+z2 . Without loss of generality we can assume thatm(B) > 0, where B = {t ∈ [0, γ ) : |y(t)| > |x(t)| = u∞}.

Denoting t1 = m(B), we have t1 ∈ (0, γ ] and y∗(t) > u∞ for t < t1. For t2 = min t1

2 , 1we have

ϱϕ (y) ≥

∫ t1

0ϕ(t, y∗(t))dt ≥

∫ t2

0ϕ(t, y∗(t2))dt = ∞,

whence we get y ∈ S(Λϕ)which is a contradiction.Now assume that x ∈ S(Λϕ) is not of the form |x| = u∞χ[0,γ ). Assume also that there exist y, z ∈ S(Λϕ) such that y = z

and x =y+z2 . By convexity of the modular ϱϕ , we have

1 = ϱϕ (x) ≤12ϱϕ (y)+

12ϱϕ (z) ≤ 1, (7)

whence we get

ϱϕ (y) = ϱϕ (z) = 1. (8)

By Lemma 4, we have x∗(t) =y∗(t)+z∗(t)

2 for t ∈ [0, γ ), whence by convexity of ϕ we obtain

ϕ(x∗(t)) ≤12ϕ(y∗(t))+

12ϕ(z∗(t))

form-a.e. t ∈ [0, γ ), so by (7), we get

ϕ(x∗(t)) =12ϕ(y∗(t))+

12ϕ(z∗(t)) (9)

for m-a.e. t ∈ [0, γ ). Since y = z, by Lemma 1(ii), we have y∗= z∗. Without loss of generality we can assume that there

exists t ′ ∈ [0, γ ) for which z∗(t ′) < y∗(t ′). For η =y∗(t ′)−x∗(t ′)

2 , we define

t0 = inf{t ≤ t ′ : x∗(t) < x∗(t ′)+ η}.

Since x∗(t0) ≤ x∗(t ′)+ η = y∗(t ′)− η and y∗(t) ≥ y∗(t ′) for t ≤ t ′, we get z∗(t) ≤ z∗(t0) ≤ x∗(t0)− η for t ≥ t0.First assume that x∗(t) < x∗(t0) for t > t0. By the right continuity of the rearrangement, we can find t1 ∈ (t0, γ ) such

that y∗(t1) ≥ x∗(t0), x∗(t1) ≥ x∗(t0) − η and x∗(t) > x∗(t1) for t < t1. By equality (9), the function ϕ(t, ·) is affine on theinterval [x∗(t1), x∗(t0)] for m-a.e. t ∈ [t0, t1]. Therefore t0 ∈ Ax. By Remark 3 there exists t2 ∈ (t0, t1) such that t2 ∈ Ax,which gives a contradiction with condition (iii) of Theorem 1.

Now let x∗(t) = x∗(t0) for some t > t0. By the definition of t0, we can find ε > 0 such that x∗(t0)+ ε ≤ x∗(t) for t < t0.Define t1 = sup{t ≥ t0 : x∗(t) = x∗(t0)}. If t1 < γ and x∗(t1) = x∗(t0), then proceeding analogously as above we obtainthat t1 satisfies condition (i) of Definition 2 and, in consequence, we get a contradiction. In the opposite case, it is easy toshow that t0 satisfies condition (ii) of Definition 2. By equality (8) and condition (iv) of Theorem 1, we get that there existss′ ∈ [0, γ ) for which y∗(s′) < z∗(s′). Proceeding analogously as for t ′, we can find s0 = t0 such that s0 ∈ Ax, which againgives a contradiction.

Necessity. Suppose that x ∈ S(Λϕ) is not of the form |x| = u∞χ[0,γ ). We will show that if not all conditions (i)–(iv) aresatisfied, then x is not an extreme point of B(Λϕ). First assume that (i) is not satisfied. Then we can find n ∈ N such that theset

An = {t ∈ [0, γ ) : |x(t)| ≤ x∗(∞)− 1/n}

has positive measure. Define y and z by formulae y(t) = x(t)+12n and z(t) = x(t)−

12n for t ∈ An and y(t) = z(t) = x(t)

for t ∈ [0, γ ) \ An. We have x = (y + z)/2 and y = z. Since x∗= y∗

= z∗, we get y, z ∈ S(Λϕ), so x is not an extreme pointof B(Λϕ).

Assume now that x satisfies condition (i) and ϱϕ (x) < 1.Wewill investigate two subcases. First assume that there exist t1and t2 such that 0 < t1 < t2 < γ and x∗(t1) > x∗(t2) > 0. For δ =

12 min(x∗(t1)− x∗(t2), x∗(t2))we can find η ∈ (0, γ − t2)

such that x∗(t2 + η)− δ ≥ 0 and∫ t1+η

t1ϕ(t, x∗(t2)+ δ)dt ≤ 1 − ϱϕ (x) .

Defining u = x∗+ δχ[t2,t2+η) and v = x∗

− δχ[t2,t2+η), we get that u and v are nonnegative, x∗=

u+v2 and u = v.

Simultaneously

ϱϕ (u) =

∫ γ

0ϕ(t, u∗(t))dt ≤

∫ γ

0ϕ(t, x∗(t))dt +

∫ t1+η

t1ϕ(t, x∗(t2)+ δ)dt ≤ 1.

3920 P. Foralewski / Nonlinear Analysis 74 (2011) 3912–3925

Since 0 ≤ v ≤ u, we have ϱϕ (v) ≤ 1. By convexity of the norm, we get ‖u‖ = ‖v‖ = 1. It is also easy to show thatµu(λ) = µv(λ) = ∞ for λ ∈ [0, x∗(∞)) whenever m({t ∈ [0, γ ) : |x(t)| > x∗(∞)}) = ∞. Therefore by Lemma 2, weconclude that x is not an extreme point.

Next assume that x∗(t) = x∗(0) for all t ∈ [0, b), where b = m(supp x). If b = γ , then we have 0 < x∗(0) < u∞ ≤ ∞.For δ =

12 min(u∞ − x∗(0), x∗(0))we can find η ∈ (0, γ ) such that∫ η

0ϕ(t, x∗(0)+ δ)dt ≤ 1 − ϱϕ (x) .

Defining u = x∗+ δχ[0,η) and v = x∗

− δχ[0,η), analogously as in first subcase, we get that x is not an extreme point. Inthe case when b < γ , we have x∗(0) = u∞. We can find a measurable set A ∈ [0, γ ) \ supp x and δ ∈ (0, u∞) such that0 < m(A) = a < ∞ and∫ b+a

bϕ(t, δ)dt ≤ 1 − ϱϕ (x) .

For y = x + δχA and z = x − δχA, we have y = z, x =y+z2 and ϱϕ (y) = ϱϕ (z) ≤ 1, so x is not an extreme point.

Now assume that conditions (i) and (ii) are satisfied and the set A0x is nonempty. First assume that t0 ∈ [0, γ ) satisfies

condition (1) of Definition 3. We can find η ∈ (0, γ − t0) for which x∗(t0 + η) ≥12x

∗(t0). It is easy to show thatu = x∗

+ (x∗(t0)− x∗)χ[t0,t0+η) and v = x∗− (x∗(t0)− x∗)χ[t0,t0+η) satisfy the assumptions of Lemma 2.

If t0 ∈ (0, γ ) satisfies condition (2) of Definition 3, then there exist ε > 0 such that x∗(t) ≥ x∗(t0) + ε for t < t0 andϕ(t, x∗(t0 + ε)) = 0 for t ∈ (t0, γ ). If x∗(t0) > 0, then we define u = x∗

+ δχ[t0,t0+η) and v = x∗− δχ[t0,t0+η), where

η ∈ (0, γ − t0) and δ =12 min(ε, x∗(t0 + η)) (we can find η ∈ (0, γ − t0) such that x∗(t0 + η) > 0). Again applying

Lemma 2, we have that x is not an extreme point. In the case when x∗(t0) = 0 we define y and z by the following formulae:y(t) = z(t) = x(t) for t ∈ supp x and y(t) = ε, z(t) = −ε for t ∈ [0, γ ) \ supp x. We have x =

y+z2 , y = z and

ϱϕ (x) = ϱϕ (y) = 1, whence we obtain that x is not an extreme point.Now assume that the set A0

x is empty but Ax has at least two elements. We will consider three cases. First assume thatt0 satisfies condition (2) of Definition 2, p(t, u) = p for all t ∈ (t0, t1) and all u ∈ [x∗(t0 − ε), x∗(t0 + ε)) and t1 < ∞,where t1 and ε are the constants from condition (2) of Definition 2. Therefore t1 ∈ Ax. For u = x∗

+ εχ[t0,t2) − εχ[t2,t1) andv = x∗

− εχ[t0,t2) + εχ[t2,t1), where t2 = (t0 + t1)/2 we have u = v and x∗=

u+v2 . Since u = u∗

= v∗ and∫ t2

t0ϕ(t, x∗(t0)+ ε)− ϕ(t, x∗(t0))dt =

t1 − t02

· ε · p =

∫ t1

t2ϕ(t, x∗(t0))− ϕ(t, x∗(t0)− ε)dt,

we get ϱϕ (u) = ϱϕ (v) = 1. Applying Lemma 2, we conclude that x is not an extreme point.Next suppose that there exist s0 and t0 (s0 < t0) which satisfy condition (2) of Definition 2. Let s1, t2 and ε > 0 be

constants with condition (2) of Definition 2. We have 0 ≤ s0 < s1 ≤ t0 < t1 ≤ γ and x∗(t0) − ε ≥ 0. For η ∈ [0, ε] wedefine functions

Fs0(η) =

∫ s1

s0(ϕ(t, x∗(s0)+ η)− ϕ(t, x∗(s0)))dt

Ft0(η) =

∫ t1

t0(ϕ(t, x∗(t0)+ η)− ϕ(t, x∗(t0)))dt.

Since the functions Fs0 and Ft0 are continuous on the interval [0, ε], we can find η1, η2 ∈ [0, ε] such that Fs0(η1) = Ft0(η2).For

u = x∗+ η1χ[s0,s1) − η2χ[t0,t1) and v = x∗

− η1χ[s0,s1) + η2χ[t0,t1)

wehave u = v and x∗=

u+v2 . Since the functionsϕ(t, ·) are affine on the interval [x∗(s0)−ε, x∗(s0)+ε] form-a.e. t ∈ [s0, s1]

and on the interval [x∗(t0)− ε, x∗(t0)+ ε] for m-a.e. t ∈ [t0, t1], we get ϱϕ (u) = ϱϕ (v) = 1. By Lemma 2, we again obtainthat x is not an extreme point.

Finally assume that there exists t0 which satisfies condition (1) of Definition 2. By Remark 3, we conclude that there existsat least countable set of elements of the set Ax belonging to the interval (t0, t1), where t1 is a constant from Definition 2(condition (1)). If at least two of them satisfy condition (2) of Definition 2, then we proceed in such a way as above. In theopposite case we can find t2 ∈ (t0, t1) satisfying condition (1) of the above definition. Denoting

t3 = inf{t ∈ (t0, t2] : x∗(t) = x∗(t2)},

we define

F(α) =

∫ t3

t0(ϕ(t, x∗(t)+ αmin(x∗(t0)− x∗(t), x∗(t)− x∗(t3)))− ϕ(t, x∗(t)))dt

P. Foralewski / Nonlinear Analysis 74 (2011) 3912–3925 3921

and

G(β) =

∫ t1

t2(ϕ(t, x∗(t)+ βmin(x∗(t2)− x∗(t), x∗(t)− x∗(t1)))− ϕ(t, x∗(t)))dt

for α, β ∈ [0, 1]. Since the functions F and G are continuous, we can find for α0, β0 ∈ (0, 1] for which F(α0) = G(β0).Defining

u = x∗+ α0 min(x∗(t0)− x∗(t), x∗(t)− x∗(t3))χ[t0,t3) − β0 min(x∗(t2)− x∗(t), x∗(t)− x∗(t1))χ[t2,t1),

v = x∗− α0 min(x∗(t0)− x∗(t), x∗(t)− x∗(t3))χ[t0,t3) + β0 min(x∗(t2)− x∗(t), x∗(t)− x∗(t1))χ[t2,t1)

we get u = v and x∗=

u+v2 . By affinity of the function ϕ(t, ·) on the interval [x∗(t1), x∗(t0)] form-a.e. t ∈ [t0, t1] we obtain

ϱϕ (u) = ϱϕ (v) = 1, which again leads to the consequence that x is not an extreme point. �

Now we will give criteria in order that x ∈ S(Λϕ) be a rotund point. Let us first for arbitrary x ∈ S(Λϕ) define two setsBx and Cx.

Definition 4. We will say that t0 (t0 ∈ [0, γ )) belongs to the set Bx if one of the following two conditions is satisfied:

(1) there exists t1 ∈ (t0, γ ) such that x∗(t) < x∗(t0) for t > t0 and the functions ϕ(t, ·) are affine on the interval[x∗(t), x∗(t0)] form-a.e. t ∈ [t0, t1],

(2) there exists t1 ∈ (t0, γ ) and ε > 0 such that x∗(t) ≥ x∗(t0)+ε for t < t0, x∗(t) = x∗(t0) for t ∈ [t0, t1) and the functionsϕ(t, ·) are affine on the interval [x∗(t0), x∗(t0)+ ε] for m-a.e. t ∈ [t0, t1].

Definition 5. Analogously we will say that t0 (t0 ∈ [0, γ )) belongs to the set Cx if one of the following two conditions issatisfied:

(1) there exists t1 ∈ (t0, γ ) such that x∗(t1) < x∗(t) < x∗(t0) for t ∈ (t0, t1) and the functions ϕ(t, ·) are affine on theinterval [x∗(t1), x∗(t)] for m-a.e. t ∈ [t0, t1],

(2) there exists t1 ∈ (t0, γ ) and ε > 0 such that x∗(t1) ≤ x∗(t0) − ε or (x∗(t0) > 0 and t1 = γ < ∞), x∗(t) = x∗(t0) fort ∈ [t0, t1) and the functions ϕ(t, ·) are affine on the interval [x∗(t0)− ε, x∗(t0)] for m-a.e. t ∈ [t0, t1].

Remark 4. Note that if t0 satisfies condition (1) of Definition 4, then for any t ∈ (t0, t1) such that x∗(s) < x∗(t) for any s > tor x∗(s) > x∗(t)+ ε for some ε > 0 and any s < t , we have t ∈ Bx.

It is also easy to show that if t0 satisfies condition (2) of Definition 5, then we have (t0, t1) ⊂ Cx.

Example 2. Let the function ϕ and the elements x1, x2, x3, y1 and z be as in Example 1. We have Bx1 = By1 = ∅, Bx2 = {1},Bx3 = {1, 2} and Bz = [1, 2). Simultaneously Cxi = Cz = [1, γ = 3) for i = 1, 2, 3 and Cy1 = [2, γ = 3).

Theorem 2. An element x ∈ S(Λϕ) is a rotund point of B(Λϕ) if and only if the following conditions are satisfied:

(i) if γ = ∞ and limt→∞ x∗(t) = x∗(∞) > 0, then m({t : |x(t)| ≤ x∗(∞)}) = 0,(ii) ϱϕ (x/λ) < ∞ for some λ ∈ (0, 1),(iii) at least one from the sets Bx and Cx is empty,(iv) the following implications hold:

(a) if γ < ∞, then for any t such that x∗(t) = lims→γ x∗(s) > 0, we have ϕ(t, x∗(t)) > 0,(b) if inf{t ∈ [0, γ ) : x∗(t) = 0} = a < γ and limt→a− x∗(t) > 0, then for any u > 0 there exists tu > a such that

ϕ(tu, u) > 0.

Proof. Sufficiency. Let x ∈ S(Λϕ) satisfy the assumptions of the theorem. First we will show that for any y ∈ S(Λϕ) suchthat x∗

= y∗, there exists a measurable set A (A ⊂ [0, γ )) such that m(A) > 0 and the inequality

ϕ

t,

x∗(t)+ y∗(t)2

<

12

ϕ(t, x∗(t))+ ϕ(t, y∗(t))

. (10)

holds for m-a.e. t ∈ A.Since x∗

= y∗, we can find t ′ ∈ [0, γ ) such that x∗(t ′) = y∗(t ′). Assume that x∗(t ′) < y∗(t ′) (the case x∗(t ′) > y∗(t ′) canbe proved analogously). Now, we define t0 and t1 by the formulae

t0 = inf{t ≤ t ′ : x∗(s) < y∗(s) for any s ∈ (t, t ′]},t1 = sup{t ≥ t ′ : x∗(s) < y∗(s) for any s ∈ [t ′, t)}

(if for any t < t ′ there exists s ∈ (t, t ′) such that x∗(s) ≥ y∗(s) we define t0 = t ′). By the right continuity of y∗, we havet0 < t1. Assume that ϕ(t, ·) is affine on the interval [x∗(t), y∗(t)] for m-a.e. t ∈ [t0, t1] (in the opposite case, there exists ameasurable set A ⊂ [t0, t1] such thatm(A) > 0 and inequality (10) holds for m-a.e. t ∈ A).

3922 P. Foralewski / Nonlinear Analysis 74 (2011) 3912–3925

If x∗(t0) < y∗(t0), by the right continuity of y∗, we get that t0 satisfies condition (1) of Definition 4 when x∗(t) < x∗(t0)for any t > t0 or t0 satisfies condition (2) of Definition 4 otherwise.

Now let x∗(t0) = y∗(t0). By the definition of t0, we have x∗(t) < x∗(t0) for any t > t0. If y∗(t) = y∗(t0) = x∗(t0) for somet > t0, then t0 satisfies condition (1) of Definition 4. Now assume that y∗(t) < y∗(t0) = x∗(t0) for any t > t0. By the rightcontinuity of x∗ there exist t and t such that t0 < t < t < t1 and x∗(t0) > x∗(t) > x∗(t). If x∗(t) > x∗(t) for any t > t , thent satisfies condition (1) of Definition 4 (we have y∗(t) > x∗(t)). In the opposite case we definet = sup{t ≥ t : x∗(t) = x∗(t)}.

We havet ≤ t < t1 and y∗(t) > x∗(t). Proceeding analogously as above we get thatt ∈ Bx.By condition (iii), we have ϕ(t, x∗(t)) > 0 whenever x∗(t) > limt→γ x∗(t). Hence by conditions (i) and (iv), we have∫ t1

t0ϕ(t, x∗(t))dt <

∫ t1

t0ϕ(t, y∗(t))dt,

whence, by condition (ii), we get that there exists t2 ∈ [0, t0) ∪ (t1, γ ) such that x∗(t2) > y∗(t2). Define

t3 = sup{t ≥ t2 : x∗(s) > y∗(s) for any s ∈ [t2, t)}.

Applying again right continuity of x∗ we have t2 < t3. Denoting t = sup{t : x∗(t) = x∗(t2)}, we have t2 ≤ t ≤ t3. Ifx∗(t) < x∗(t2) or t = t3 = γ < ∞, then there exists a measurable set A ⊂ [t2, t] such that m(A) > 0 and inequality(10) holds for m-a.e. t ∈ A (in the opposite case t2 satisfies condition (2) of Definition 5, which gives a contradiction withcondition (iii)). Now let x∗(t) = x∗(t2). Then t ∈ [t2, t3) and x∗(t) > x∗(t) for t > t , whence we get that there exists ameasurable set A ⊂ [t, t3] such that m(A) > 0 and inequality (10) holds for m-a.e. t ∈ A (in the opposite case, t satisfiescondition (1) of Definition 5, which would again give a contradiction with condition (iii)).

Therefore, for any y ∈ S(Λϕ) such that y∗= x∗, we have∫ γ

0ϕ

t,

x∗(t)+ y∗(t)2

dt <

12

∫ γ

0ϕ(t, x∗(t))dt +

∫ γ

0ϕ(t, y∗(t))dt

≤ 1,

whence, by Proposition 1, we get ϱϕ((x + y)/2) < 1.Now we will show that ‖(x + y)/2‖ < 1. Assume to the contrary that ‖(x + y)/2‖ = 1. Then ϱϕ

(1 + ε)

x+y2

= ∞ for

any ε > 0. By condition (ii), we have ϱϕ (x/β) < ∞ for some β ∈ (0, 1). Let ε = (1 − a)/(1 + a), where a ∈ (β, 1). Wehave ε > 0 and 1+ε

2 a +1+ε2 = 1. Consequently,

ϱϕ

(1 + ε)

x + y2

= ϱϕ

1 + ε

2axa

+1 + ε

2y

≤1 + ε

2aϱϕ

xa

+

1 + ε

2ϱϕ (y) < ∞,

which is impossible, so ‖(x + y)/2‖ < 1.Finally, take any y ∈ S(Λϕ) such that x = y and x∗

= y∗. Assume to the contrary that ‖z‖ = 1, where z =x+y2 . We have

z = x and by Lemma 1(i), we get z∗= x∗. By virtue of the first part of the proof we get

z+x2

< 1, and then

‖z‖ =

23x + z2

+

13y ≤

23

x + z2

+13‖y‖ < 1,

which is a contradiction, whence ‖(x + y)/2‖ < 1.Necessity. First assume that γ = ∞, limt→∞ x∗(t) = x∗(∞) > 0 and m({t ∈ [0, γ ) : |x(t)| = x∗(∞)}) > 0. Let

A ⊂ [0, γ ) be such that 0 < m(A) < ∞ and |x(t)| = x∗(∞) for m-a.e. t ∈ A. Define y = xχ[0,γ )\A. Then x = y andy∗

= ((x + y)/2)∗ = x∗, so x is not a rotund point.Now assume that x satisfies condition (i). We will prove the necessity of the remaining conditions. By virtue of Lemma 3,

it is enough to show that if x does not satisfy some of the conditions, then there exists v such that v = v∗= x∗ and

‖v‖ = ‖(v + x∗)/2‖ = 1.Let x be such that ϱϕ (x) = 1 and ϱϕ (x/λ) = ∞ for every λ ∈ (0, 1). First assume that∫ t

0ϕ(s, x∗(s)/λ)ds = ∞

for some t ∈ [0, γ ) and any λ ∈ (0, 1). We can find t ′ ∈ (0, t) such that t ′

0 ϕ(s, x∗(s)/λ)ds = ∞ for any λ ∈ (0, 1) and

x∗(t ′) > 0. For v = x∗χ[0,t ′) we have v = v∗= x∗, v ≤

v+x∗2 ≤ x∗ and ϱϕ (v) = ϱϕ ((v + x∗)/2) = ∞ for any λ ∈ (0, 1),

whence we get ‖v‖ = ‖(v + x∗)/2‖ = 1. Now let γ = ∞ and∫∞

tϕ(s, x∗(s)/λ)ds = ∞

P. Foralewski / Nonlinear Analysis 74 (2011) 3912–3925 3923

for some t ∈ [0, γ ) and anyλ ∈ (0, 1).Without loss of generalitywe can assume thatϕ(t, x∗(t)/λ) < ∞ for someλ ∈ (0, 1).Therefore, by condition (i), we can find t ′ ∈ (t, γ ) such that x∗(s) > x∗(t ′) for s < t ′ and

∞

t ′ ϕ(s, x∗(s)/λ)ds = ∞ for any

λ ∈ (0, 1). Define v = x∗(t ′)χ[0,t ′) + x∗χ[t ′,∞). Analogously as above, we get v = v∗= x∗ and ‖v‖ = ‖(v + x∗)/2‖ = 1.

Now assume that γ < ∞ and there exists t ∈ [0, γ ) such that x∗(t) = limt→γ x∗(t) > 0 and ϕ(t, x∗(t)) = 0. Obviouslywe have∫ t

0ϕ(s, x∗(s))ds = 1.

Therefore for v = x∗χ[0,t) we get v = v∗= x∗ and ‖v‖ = ‖(v + x∗)/2‖ = 1. Now let inf{t ∈ [0, γ ) : x∗(t) = 0} =

a < γ , limt→a− x∗(t) = b > 0 and there exists u > 0 such that for any t > a we have ϕ(t, u) = 0. It is easy to show thatv = x∗χ[0,a) + min( b2 , u)χ[a,γ ) satisfies the assumptions of Lemma 3.

Finally, wewill show the necessity of condition (iii). Suppose that both sets Bx and Cx are not empty. First we assume thatthere exists t0 ∈ [0, γ ) such that t0 ∈ Bx ∩ Cx. If there exists t > t0 such that x∗(t) = x∗(t0), then t0 satisfies condition (2)definition of Bx as well as condition (2) definition of Cx. Therefore we can find ε > 0 such that x∗(t) ≥ x∗(t0)+ ε for t < t0,the function ϕ(t, ·) is affine on the interval [x∗(t0), x∗(t0)+ ε] form-a.e. t ∈ [t0, t ′] where t ′ > t0 and the function ϕ(t, ·) isaffine on the interval [x∗(t0)−ε, x∗(t0)] form-a.e. t ∈ [t0, t1], where t1 = sup{t ≥ t0 : x∗(t) = x∗(t0)}. We have t0 < t ′ ≤ t1and x∗(t1) ≤ x∗(t0)− ε or t1 = γ < ∞. For t2 = (t ′ − t0)/2, we can find positive numbers ε1 and ε2 (ε1, ε2 ∈ (0, ε]) suchthat ∫ t2

t0(ϕ(t, x∗(t)+ ε1)− ϕ(t, x∗(t)))dt =

∫ t1

t2(ϕ(t, x∗(t))− ϕ(t, x∗(t)− ε2))dt.

Since

v = x∗χ[0,γ )\[t0,t1) + (x∗+ ε1)χ[t0,t2) + (x∗

− ε2)χ[t2,t1)

satisfies assumptions of Lemma 3, we get that x is not a rotund point.Now assume that x∗(t) < x∗(t0) for any t > t0. Then we can find t1 such that t1 > t0, x∗(t0) > x∗(t) > x∗(t1) for any

t ∈ (t0, t1) and the function ϕ(t, ·) is affine on the intervals [x∗(t1), x∗(t)] and [x∗(t), x∗(t0)] form-a.e. t ∈ [t0, t1]. Therefore,we can find t2 ∈ (t0, t1) such that∫ t1

t0ϕ(t, x∗(t))dt =

∫ t2

t0ϕ(t, x∗(t0))dt +

∫ t1

t2ϕ(t, x∗(t1))dt.

Obviously for the function

v = x∗χ[0,γ )\[t0,t1) + x∗(t0)χ[t0,t2) + x∗(t1)χ[t2,t1)

the assumptions of Lemma 3 are satisfied, so x is not a rotund point again.Now let there exist t0 ∈ Bx and s0 ∈ Cx such that t0 = s0. Proceeding analogously as above, we can find v satisfying the

assumptions of Lemma 3 (note that if s0 < t0, then, by Definitions 4 and 5, x∗(s0) > x∗(t0)). �

4. Some examples

It is easy to observe that the conditions of Theorem 2 imply the corresponding conditions of Theorem 1. We will showthat they are stronger, that is, we will find examples of extreme points which are not rotund points. Example 3 shows thatthe element x ∈ S(Λϕ) of the form |x| = u∞χ[0,γ ) (see Definition 1 on page 3) is not a rotund point.

Example 3. For any t ∈ [0, γ =12 ) let ϕ(t, u) = u2 for u ∈ [0, 1] and ϕ(t, u) =

1t (u

2−1)+1 for u > 1 (for t = 0we define

ϕ(0, u) = ∞ for any u > 1). The function ϕ satisfies condition (L1) and u∞ = 1. By Theorem 1, the element x = 1χ[0,γ ) isan extreme point. Since for y = 1χ

0, 14

we have ‖y‖ = ‖(x + y)/2‖ = 1, x is not a rotund point.

The next example shows that condition (i) from Theorem 2 is stronger than condition (i) from Theorem 1.

Example 4. Assume that ϕ(t, u) = 0 for t ∈ [0, γ = ∞) and u ∈ [0, 1], ϕ(t, u) =12 (u − 1)2 for t ∈ [0, 1] and u ∈ (1,∞)

and ϕ(t, u) =12t (u − 1)2 for t ∈ (1,∞) and u > 1. Obviously the function ϕ satisfies condition (L1). For x given by the

formulae: x(t) = 1 for t = [0, 1), x(t) = 2 for t = [1, 2) and x(t) = 1 +1

√t−1

for t = [2,∞), we have x∗(t) = 2 for

t = [0, 1) and x∗(t) = 1 +1

√tfor t = [1,∞). Since x does not satisfy condition (i) from Theorem 2, we get that x is not a

rotund point. At the same time, by Theorem 1, x is an extreme point.

Examples 5 and 6 will show that for x ∈ S(Λϕ) the assumption that ϱϕ (x/λ) < ∞ for some λ ∈ (0, 1) is essentiallystronger than ϱϕ (x) = 1.

3924 P. Foralewski / Nonlinear Analysis 74 (2011) 3912–3925

Example 5. Let ϕ(t, u) =1

t+1u2 for t ∈ [0,∞) and u ∈ [0, 1] and ϕ(t, u) = ∞ for t ∈ [0,∞) and u > 1. The function ϕ is

defined in such a way that it satisfies condition (L1) (u∞ = 1). It is easy to show that x = x(t), where x(t) =1

√t+1

satisfiesthe conditions of Theorem 1. At the same time ϱϕ (x/λ) = ∞ for any λ ∈ (0, 1), so x is not a rotund point.

Example 6. Assume that γ = b, where b =∑

∞

n=11

n!(n2n−1)!2n and let

p(t, u) =

(n2n

− 1)! for u ∈ [0, n!),(n2n)! for u ∈ [n!,∞)

for t ∈ [bn, bn−1), where b0 = γ and bn = b −∑n

k=11

k!(k2k−1)!2kfor n ∈ N . The Musielak–Orlicz function ϕ, where

ϕ(t, u) =

|u|0 p(t, v)dv for t ∈ (0, γ ) and u ∈ R, satisfies condition (L1). Simultaneously

ϕ

t,1 +

1n

n!

≥

∫ (1+1/n)n!

n!(n2n)!du =

1nn!(n2n)! = 2n

∫ n!

0(n2n

− 1)!du = 2nϕ(t, n!)

for t ∈ [bn, bn−1), n ∈ N , and∫ bn−1

bnϕ(t, n!)dt =

∫ bn−1

bn

∫ n!

0p(t, u)du

dt =

∫ bn−1

bnn!(n2n

− 1)!dt =12n

for n ∈ N . Therefore for x such that x(t) = n! for t ∈ [bn, bn−1), n ≥ 1, we have x = x∗ and in consequence, ϱϕ (x) = 1 andϱϕ (λx) = ∞ for any λ > 1. We also have that Ax = A0

x = ∅. Hence we get that x is an extreme point and it is not a rotundpoint.

Finally in Example 7 we will show that conditions (iii) and (iv) from Theorem 1 are essentially weaker than thecorresponding conditions (iii) and (iv) from Theorem 2.

Example 7. Assume that the function ϕ and the elements x2 and y1 are the same as in Example 1.We have ϱϕ (λx2) < ∞ for any λ > 0, Ax2 = {1} and A0

x2 = ∅, whence we obtain that x2 is an extreme point. SinceBx2 = {1} and Cz = [1, γ ) (see Example 2), x2 does not satisfy condition (iii) of Theorem 2, so x2 is not a rotund point.

Since ϱϕ (λy1) < ∞ for any λ > 0, Ay1 = A0y1 = ∅, we have that y1 is an extreme point. The element y1 also satisfies

condition (iii) of Theorem 2 (see Example 2), but it does not satisfy condition (iv)(a) of this theorem (for t ∈ [2, γ )we havex∗(t) = lims→γ=3 x∗(s) = 1 and ϕ(t, x∗(t)) = 0), so x2 is not a rotund point.

5. Applications to Orlicz–Lorentz function spaces

Suppose that ϕψ,ω(t, u) = ψ(u)ω(t), where ψ is an Orlicz function and ω is a nonnegative function on [0, γ ). Withoutloss of generality we can assume that the number u∞ from Definition 1 is strictly positive. Then ψ(u) < ∞ for anyu ∈ [0, u∞) and

t0 ω(s)ds < ∞ for some t ∈ (0, γ ), whenever ψ(u∞) > 0. We have pϕ(t, u) = pψ (u)ω(t), where pϕ(t, ·)

and pψ (·) denote the right derivatives of ϕψ,ω(t, ·) and ψ(·), respectively. Therefore the Musielak–Orlicz function ϕψ,ωsatisfies condition (L1) if and only if ψ(u) = 0 for any u ∈ [0, u∞] or ω is a nonincreasing function on [0, γ ).

In the remaining part of this section we will assume that ϕψ,ω satisfies condition (L1). Note that in the case whenψ(u∞) > 0, we get that ω is a nonincreasing and locally integrable function on [0, γ ).

It is well known that Musielak–Orlicz functions ϕψ,ω satisfying condition (L1) generates Orlicz–Lorentz spaces, which isdenoted byΛψ,ω (see [3]).

Before formulating corollaries concerning extreme and rotund points in Orlicz–Lorentz spaces we introduce somedenotations and notions.

By Sψ ⊂ [0,∞) we will denote the set of all points of strict convexity of ψ , that is, u ∈ Sψ if u ≥ 0 and for any v and wsuch that u =

v+w2 and v = w, we have ψ(u) < 1

2ψ(v)+12ψ(w). We also denote

S+

ψ = {u ∈ Sψ : there exists ε > 0 such that ψ is affine on [u, u + ε]}

and

S−

ψ = {u ∈ Sψ \ {0} : there exists ε > 0 such that ψ is affine on [u − ε, u]}.

For any Orlicz function ψ we also define aψ = sup{u ≥ 0 : ψ(u) = 0} and bψ = sup{u ≥ 0 : ψ(u) < ∞}.By Theorem 1, we get the following

Corollary 1. An element x ∈ S(Λψ,ω) is an extreme point of B(Λψ,ω) if and only if |x| = u∞χ[0,γ ) or the following conditionshold:(i) m({t : |x(t)| < x∗(∞)}) = 0, where x∗(∞) = limt→∞ x∗(t),(ii) ϱϕ (x) = 1,

P. Foralewski / Nonlinear Analysis 74 (2011) 3912–3925 3925

(iii) there exist α > aψ , 0 ≤ t1 ≤ t2 ≤ γ and a measurable function y such that x∗= αχ[t1,t2) + y, where [t1, t2) ∩ supp y =

∅, α ∈ Sψ and y(t) ∈ Sψ for any t ∈ [0.γ ) \ [t1, t2), and the weight function ω is not a constant function on the interval(t1, t2).

If we additionally suppose that aψ = 0, u∞ = bψ = ∞ and the weight function ω is strictly decreasing on [0, γ ), then wehave

Corollary 1A ([4, Theorem 7]). An element x ∈ S(Λψ,ω) is an extreme point of B(Λψ,ω) if and only if the following conditionsare satisfied:

(i) m({t : |x(t)| < x∗(∞)}) = 0, where x∗(∞) = limt→∞ x∗(t) (cf. condition (+) in [4]),(ii) ϱϕ (x) = 1,(iii) there exist α > 0, A ⊂ [0, γ ) and a measurable function y such that |x| = αχA + y, where A ∩ supp y = ∅, α ∈ Sψ and

y(t) ∈ Sψ for m-a.e. t ∈ [0, γ ).

For any x ∈ Λψ,ω , we define Fx = {u ∈ [0,∞) : there exists t ∈ [0, γ ) such that u = x∗(t)}. Now applying Theorem 2,we obtain the following

Corollary 2. An element x ∈ S(Λψ,ω) is a rotund point of B(Λψ,ω) if and only if the following conditions are satisfied:

(i) if γ = ∞ and limt→∞ x∗(t) = x∗(∞) > 0, then m({t : |x(t)| ≤ x∗(∞)}) = 0,(ii) ϱϕ (x/λ) < ∞ for some λ ∈ (0, 1),(iii) Fx ⊂ Sψ and if Fx ∩ S−

ψ = ∅, then for any u ∈ S+

ψ we have m({t : x∗(t) = u}) = 0,(iv) if γ < ∞, and there exists t < γ such that x∗(t) = lims→γ x∗(s) > 0, then aψ < lims→γ x∗(s).

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