Extreme Points and Rotundity in Musielak-Orlicz-Bochner Function Spaces Endowed with Orlicz Norm

Abstract

The criteria for extreme point and rotundity of Musielak-Orlicz-Bochner function spaces equipped with Orlicz norm are given. Although criteria for extreme point of Musielak-Orlicz function spaces equipped with the Orlicz norm were known, we can easily deduce them from our main results.

1. Introduction

Let (𝑋,‖⋅‖) be a real Banach space. 𝑆(𝑋) and 𝐵(𝑋) denote the unit sphere and unit ball, respectively. By 𝑋^∗ denote the dual space of 𝑋. Let 𝑁,𝑅, and 𝑅⁺ denote the set natural number, reals, and nonnegative reals, respectively.

A point 𝑥∈𝐴 is said to be extreme point of 𝐴 if 2𝑥=𝑦+𝑧 and 𝑦,𝑧∈𝐴 imply 𝑦=𝑧. The set of all extreme points of 𝐴 is denoted by Ext𝐴. If Ext𝐵(𝑋)=𝑆(𝑋), then 𝑋 is said to be rotund. A point 𝑥∈𝑆(𝑋) is said to be strongly extreme point if for any {𝑥_𝑛}^∞_𝑛=1,{𝑦_𝑛}^∞_𝑛=1∈𝑋 with ‖𝑥_𝑛‖→1,‖𝑦_𝑛‖→1, and 𝑥=(1/2)(𝑥_𝑛+𝑦_𝑛), there holds ‖𝑥_𝑛−𝑦_𝑛‖→0(𝑛→∞). If the set of all strongly extreme points of 𝐵(𝑋) is equal to 𝑆(𝑋), then 𝑋 is said to be midpoint local uniform rotund.

The notion of extreme point plays an important role in some branches of mathematics. For example, the Krein-Milman theorem, Choquet integral representation theorem, Rainwater theorem on convergence in weak topology, Bessaga-Pelczynski theorem, and Elton test unconditional convergence are strongly connected with this notion. In [1], using the principle of locally reflexivity, a remarkable theorem describing connections between extreme points of 𝑆(𝑋) and strongly extreme points of 𝑆(𝑋) is proved. Namely, a Banach space 𝑋 is midpoint local uniformly rotund if and only if every point of 𝑆(𝑋) is an extreme point in 𝑋^∗∗. Another proof of this theorem based on Goldstein's theorem is given in [2]. Analyzing the proof of this fact one can easily see its local version, namely, if 𝑥∈𝑆(𝑋) is a strongly extreme point in 𝑋, then 𝜅(𝑥) is an extreme point in 𝑋^∗∗, where 𝜅 is the mapping of canonical embedding of 𝑋 into 𝑋^∗∗.

The criteria for extreme point and rotundity in the classical Musielak-Orlicz function spaces have been given in [3] already. However, because of the complication of Musielak-Orlicz-Bochner function spaces equipped with Orlicz norm, at present, the criteria for extreme point and rotundity have not been discussed yet. The aim of this paper is to give criteria for extreme point and rotundity of Musielak-Orlicz-Bochner function spaces equipped with Orlicz norm. By the result of this paper, it is easy to see that the result of [3] is true.

Let (𝑇,∑,𝜇) be nonatomic measurable space. Suppose that a function 𝑀∶𝑇×[0,∞)→[0,∞] satisfies the following conditions: (1)for 𝜇-𝑎.𝑒, 𝑡∈𝑇, 𝑀(𝑡,0)=0, lim_𝑢→∞𝑀(𝑡,𝑢)=∞ and 𝑀(𝑡,𝑢^′)<∞ for some 𝑢^′>0; (2)for 𝜇-𝑎.𝑒, 𝑡∈𝑇, 𝑀(𝑡,𝑢) is convex on [0,∞) with respect to 𝑢; (3)for each 𝑢∈[0,∞),𝑀(𝑡,𝑢) is a 𝜇-measurable function of 𝑡 on 𝑇.

Let 𝑝(𝑡,𝑢) denote the right derivative of 𝑀(𝑡,⋅) at 𝑢∈𝑅⁺ (where if 𝑀(𝑡,𝑢)=∞, let 𝑝(𝑡,𝑢)=∞) and let 𝑞(𝑡,⋅) be the generalized inverse function of 𝑝(𝑡,⋅) defined on 𝑅⁺ by𝑞(𝑡,𝑣)=sup_𝑢≥0{𝑢≥0∶𝑝(𝑡,𝑢)≤𝑣}. (1.1) Then 𝑁(𝑡,𝑣)=∫^𝑣₀𝑞(𝑡,𝑠)𝑑𝑠 for any 𝑣∈𝑅 and 𝜇-a.e. 𝑡∈𝑇. It is well known that there holds the Young inequality 𝑢𝑣≤𝑀(𝑡,𝑢)+𝑁(𝑡,𝑣) for 𝜇-a.e. 𝑡∈𝑇. And 𝑢𝑣=𝑀(𝑡,𝑢)+𝑁(𝑡,𝑢)⇔𝑢=𝑞(𝑡,𝑣) or 𝑣=𝑞(𝑡,𝑢). Let 𝑒(𝑡)=sup{𝑢>0∶𝑀(𝑡,𝑢)=0},𝐸(𝑡)=sup{𝑢>0∶𝑀(𝑡,𝑢)<∞}. (1.2)

For fixed 𝑡∈𝑇 and 𝑣≥0, if there exists 𝜖∈(0,1) such that 1𝑀(𝑡,𝑣)=21𝑀(𝑡,𝑣+𝜀)+2𝑀(𝑡,𝑣−𝜀)<∞, (1.3) then we call 𝑣 a nonstrictly convex points of 𝑀 with respects to 𝑡. The set of all nonstrictly convex point of 𝑀 with respect to 𝑡 is denoted by 𝐾_𝑡.

For fixed 𝑡∈𝑇, if 𝐾_𝑡=Φ, then we call that 𝑀(𝑡,𝑢) is strictly convex with respect 𝑢 for 𝑡.

Moreover, for a given Banach space (𝑋,‖⋅‖), we denote by 𝑋_𝑇 the set of all strongly 𝜇-measurable functions from 𝑇 to 𝑋, and for each 𝑢∈𝑋_𝑇, define the modular of 𝑢 by 𝜌_𝑀(𝑢)=_𝑇𝑀(𝑡,‖𝑢(𝑡)‖)𝑑𝑡. (1.4) Put𝐿⁰_𝑀(𝑋)={𝑢∈𝑋_𝑇∶𝜌_𝑀(𝜆𝑢)<∞forsome𝜆>0}. (1.5) Then the Musielak-Orlicz-Bochner function space‖𝑢‖⁰=inf_𝑘>01𝑘[1+𝜌_𝑀(𝑘𝑢)] (1.6) is Banach space. If 𝑋=𝑅, 𝐿⁰_𝑀(𝑅) is said to be Musielak-Orlicz function space. Set1𝐾(𝑢)={𝑘>0∶𝑘(1+𝜌_𝑀(𝑘𝑢))=‖𝑢‖⁰}. (1.7) In particular, the set 𝐾(𝑢) can be nonempty. To show that, we give a proposition.

Proposition 1.1. If  lim_𝑢→∞(𝑀(𝑡,𝑢)/𝑢)=∞𝜇-a.e. 𝑡∈𝑇, then 𝐾(𝑢)≠𝜙  for any 𝑢∈𝐿⁰_𝑀(𝑋).

Proof. For any 𝑢∈𝐿⁰_𝑀(𝑋), there exists 𝑎>0 such that 𝜇𝑇₀>0, where 𝑇₀={𝑡∈𝑇∶‖𝑢(𝑡)‖≥𝑎}. It is easy to see that 𝑇₀=∪^∞_𝑛=1𝐺_𝑛, where 𝐺_𝑛={𝑡∈𝑇₀∶𝑀(𝑡,𝑣)𝑣≥3‖𝑢‖⁰𝑎⋅𝜇𝑇₀,𝑣≥𝑛}. (1.8) Noticing that 𝐺₁⊂𝐺₂⊂⋯⊂𝐺_𝑛⊂⋯, then lim_𝑛→∞𝜇𝐺_𝑛=𝜇𝑇₀. Hence there exists 𝑛₁ such that 𝜇𝐺_𝑛1>(1/2)𝜇𝑇₀. This means that if 𝑘>𝑛₁/𝑎, we have 1𝑘[1+_𝑇𝑀(𝑡,𝑘‖𝑢(𝑡)‖)𝑑𝑡]≥_𝐺𝑛1𝑀(𝑡,𝑘‖𝑢(𝑡)‖)𝑘𝑑𝑡≥_𝐺𝑛1𝑀(𝑡,𝑘𝑎)𝑘𝑑𝑡=𝑎_𝐺𝑛1𝑀(𝑡,𝑘𝑎)𝑘𝑎𝑑𝑡≥𝑎_𝐺𝑛1𝑀(𝑡,𝑛₁)𝑛₁𝑑𝑡≥𝑎3‖𝑢‖⁰𝑎⋅𝜇𝑇₀⋅𝜇𝐺_𝑛1>32‖𝑢‖⁰. (1.9) This implies that if (1/𝑘_𝑛)(1+𝜌_𝑀(𝑘_𝑛𝑢))→‖𝑢‖⁰(𝑛→∞), then sequence {𝑘_𝑛}^∞_𝑛=1 is bounded. Without loss of generality, we may assume that 𝑘_𝑛→𝑘₀. Without loss of generality, we may assume that 𝑘₁≤𝑘₂≤⋯𝑘_𝑛≤⋯≤𝑘₀ or 𝑘₁≥𝑘₂≥⋯𝑘_𝑛≥⋯≥𝑘₀. If 𝑘₁≤𝑘₂≤⋯𝑘_𝑛≤⋯≤𝑘₀, by Levi theorem, we have lim_𝑛→∞1𝑘_𝑛[1+𝜌_𝑀(𝑘_𝑛𝑢)]=lim_𝑛→∞1𝑘_𝑛+lim_{𝑛→∞𝑇}𝑀(𝑡,‖𝑘_𝑛𝑢(𝑡)‖)𝑘_𝑛=1𝑑𝑡𝑘₀+_𝑇𝑀(𝑡,‖𝑘₀𝑢(𝑡)‖)𝑘₀=1𝑑𝑡𝑘₀[1+𝜌_𝑀(𝑘₀𝑢)]. (1.10) If 𝑘₁≥𝑘₂≥⋯𝑘_𝑛≥⋯≥𝑘₀, by dominated convergence theorem, we have lim_𝑛→∞1𝑘_𝑛[1+𝜌_𝑀(𝑘_𝑛𝑢)]=lim_𝑛→∞1𝑘_𝑛+lim_{𝑛→∞𝑇}𝑀(𝑡,‖𝑘_𝑛𝑢(𝑡)‖)𝑘_𝑛=1𝑑𝑡𝑘₀+_𝑇𝑀(𝑡,‖𝑘₀𝑢(𝑡)‖)𝑘₀=1𝑑𝑡𝑘₀[1+𝜌_𝑀(𝑘₀𝑢)]. (1.11) Therefore (1/𝑘_𝑛)[1+𝜌_𝑀(𝑘_𝑛𝑢)]→(1/𝑘₀)[1+𝜌_𝑀(𝑘₀𝑢)](𝑛→∞), namely, (1/𝑘₀)[1+𝜌_𝑀(𝑘₀𝑢)]=‖𝑢‖⁰. This implies 𝑘₀∈𝐾(𝑢).

2. Main Results

In order to obtain the main theorems of this paper, we first give some lemmas.

Lemma 2.1. If 𝐾(𝑢)=𝜙, then ‖𝑢‖⁰=∫_𝑇𝐴(𝑡)⋅‖𝑢(𝑡)‖𝑑𝑡, where 𝐴(𝑡)=lim_𝑢→∞(𝑀(𝑡,𝑢)/𝑢).

Proof. By proof of Proposition 1.1, we know that if 𝐾(𝑢)=𝜙, then there exists {𝑘_𝑛}^∞_𝑛=1 such that (1/𝑘_𝑛)(1+𝜌_𝑀(𝑘_𝑛𝑢))→‖𝑢‖⁰ and 𝑘_𝑛→∞ as 𝑛→∞. Without loss of generality, we may assume that 𝑘₁≤𝑘₂≤⋯𝑘_𝑛≤⋯. By Levi theorem, we have ‖𝑢‖⁰=lim_𝑛→∞1𝑘_𝑛(1+_𝑇𝑀(𝑡,𝑘_𝑛‖𝑢(𝑡)‖)𝑑𝑡)=lim_𝑛→∞1𝑘_𝑛(1+_{{𝑡∈𝑇∶‖𝑢(𝑡)‖≠0}}𝑀(𝑡,𝑘_𝑛‖𝑢(𝑡)‖)𝑑𝑡)=lim_𝑛→∞1𝑘_𝑛+lim_{𝑛→∞{𝑡∈𝑇∶‖𝑢(𝑡)‖≠0}}𝑀(𝑡,𝑘_𝑛‖𝑢(𝑡)‖)𝑘_𝑛=‖𝑢(𝑡)‖⋅‖𝑢(𝑡)‖𝑑𝑡_{{𝑡∈𝑇∶‖𝑢(𝑡)‖≠0}}lim_𝑛→∞𝑀(𝑡,𝑘_𝑛‖𝑢(𝑡)‖)𝑘_𝑛=‖𝑢(𝑡)‖⋅‖𝑢(𝑡)‖𝑑𝑡_{{𝑡∈𝑇∶‖𝑢(𝑡)‖≠0}}𝐴(𝑡)⋅‖𝑢(𝑡)‖𝑑𝑡=_𝑇𝐴(𝑡)⋅‖𝑢(𝑡)‖𝑑𝑡. (2.1) Hence the conclusion of the lemma is true.

Lemma 2.2. If the set 𝐾(𝑢) consists of one element from (0,+∞), then ‖𝑢‖⁰<∫_𝑇𝐴(𝑡)⋅‖𝑢(𝑡)‖𝑑𝑡.

Proof. Pick 𝑘₁>𝑘₂>0; then we have 𝜌_𝑀(𝑘₁𝑢𝜒_𝐸𝑛)≥_𝑇𝑘₁‖𝑢(𝑡)𝜒_𝐸𝑛(𝑡)‖⋅𝑝(𝑡,‖𝑘₂𝑢(𝑡)𝜒_𝐸𝑛(𝑡)‖)𝑑𝑡−𝜌_𝑁(𝑝(𝑘₂𝑢𝜒_𝐸𝑛𝜌)),_𝑀(𝑘₂𝑢𝜒_𝐸𝑛)=_𝑇𝑘₂‖𝑢(𝑡)𝜒_𝐸𝑛(𝑡)‖⋅𝑝(𝑡,‖𝑘₂𝑢(𝑡)𝜒_𝐸𝑛(𝑡)‖)𝑑𝑡−𝜌_𝑁(𝑝(𝑘₂𝑢𝜒_𝐸𝑛)), (2.2) where 𝐸_𝑛={𝑡∈𝑇∶𝑘₂‖𝑢(𝑡)‖≤𝑛,𝑝(𝑡,‖𝑘₂𝑢(𝑡)‖)≤𝑛}. (2.3) It follows that 1𝑘₁[1+𝜌_𝑀(𝑘₁𝑢𝜒_𝐸𝑛1)]−𝑘₂[1+𝜌_𝑀(𝑘₂𝑢𝜒_𝐸𝑛=𝑘)]₁−𝑘₂𝑘₁𝑘₂𝑘(−1+²𝑘₁−𝑘₂[𝜌_𝑀(𝑘₁𝑢𝜒_𝐸𝑛)−𝜌_𝑀(𝑘₂𝑢𝜒_𝐸𝑛)]−𝜌_𝑀(𝑘₂𝑢𝜒_𝐸𝑛≥𝑘))₁−𝑘₂𝑘₁𝑘₂𝑘(−1+²𝑘₁−𝑘_2𝑇(𝑘₁−𝑘₂)‖𝑢(𝑡)𝜒_𝐸𝑛(𝑡)‖⋅𝑝(𝑡,‖𝑘₂𝑢(𝑡)𝜒_𝐸𝑛(𝑡)‖)𝑑𝑡−𝜌_𝑀(𝑘₂𝑢𝜒_𝐸𝑛=𝑘))₁−𝑘₂𝑘₁𝑘₂(𝜌_𝑁(𝑝(𝑘₂𝑢𝜒_𝐸𝑛))−1). (2.4) Let 𝑛→∞; then we obtain 1𝑘₁[1+𝜌_𝑀(𝑘₁𝑢𝜒_𝐸𝑘)]≥¹−𝑘₂𝑘₁𝑘₂(𝜌_𝑁(𝑝(𝑘₂𝑢𝜒_𝐸1))−1)+𝑘₂[1+𝜌_𝑀(𝑘₂𝑢𝜒_𝐸)], (2.5) where 𝐸={𝑡∈𝑇∶𝑝(𝑡,‖𝑘₂𝑢(𝑡)‖)<∞}. (2.6) If 𝑝(𝑡,‖𝑘₂𝑢(𝑡)‖)=∞, then 𝑀(𝑡,𝑘₁‖𝑢(𝑡)‖)≥𝑀(𝑡,𝑘₂‖𝑢(𝑡)‖)=∞. Hence we have 1𝑘₁[1+𝜌_𝑀(𝑘₁𝑘𝑢)]≥¹−𝑘₂𝑘₁𝑘₂(𝜌_𝑁(𝑝(𝑘₂1𝑢))−1)+𝑘₂[1+𝜌_𝑀(𝑘₂𝑢)]. (2.7) Moreover, there exists 𝑘₀∈𝑅⁺ such that 𝜌_𝑁(𝑝(𝑘𝑢))≥1, whenever 𝑘≥𝑘₀. This means that function 𝐹(𝑘)=(1/𝑘)[1+𝜌_𝑀(𝑘𝑢)] is nondecreasing, when 𝑛 is large enough. Pick sequence {𝑘_𝑛}^∞_𝑛=1 such that 0<𝑘₁<𝑘₂⋯<𝑘_𝑛<⋯. By Levi theorem, we have _𝑇𝐴(𝑡)⋅‖𝑢(𝑡)‖𝑑𝑡=_𝑇lim_𝑛→∞𝑀(𝑡,𝑘_𝑛‖𝑢(𝑡)‖)𝑘_𝑛‖𝑢(𝑡)‖⋅‖𝑢(𝑡)‖𝑑𝑡=lim_𝑛→∞T𝑀(𝑡,𝑘_𝑛‖𝑢(𝑡)‖)𝑘_𝑛‖𝑢(𝑡)‖⋅‖𝑢(𝑡)‖𝑑𝑡=lim_𝑛→∞1𝑘_𝑛𝑇𝑀(𝑡,𝑘_𝑛‖𝑢(𝑡)‖)𝑑𝑡=lim_𝑛→∞1𝑘_𝑛[1+_𝑇𝑀(𝑡,𝑘_𝑛>1‖𝑢(𝑡)‖)𝑑𝑡]𝑙[1+_𝑇𝑀(𝑡,𝑙‖𝑢(𝑡)‖)𝑑𝑡]=‖𝑢‖⁰, (2.8) where {𝑙}=𝐾(𝑢). Hence the conclusion of the lemma is true.

Lemma 2.3 (see [3]). Let 𝐿⁰_𝑀(𝑅) be rotund, then 𝑀(𝑡,𝑢) is strictly convex with respect to 𝑢 for almost all 𝑡∈𝑇.

Theorem 2.4. Let 𝐿⁰_𝑀(𝑋) be Musielak-Orlicz-Bochner function spaces, then 𝑢∈𝑆(𝐿⁰_𝑀(𝑋)) is an extreme point of 𝐵(𝐿⁰_𝑀(𝑋)) if and only if (a)the set 𝐾(𝑢) consists of one element from (0,+∞); (b)𝑣,𝑤∈𝑋_𝑇 with 𝑢=𝜆𝑣+(1−𝜆)𝑤 and ‖𝑢(𝑡)‖=‖𝑣(t)‖=‖𝑤(𝑡)‖𝜇-𝑎.𝑒 that on 𝑇 implies 𝑣=𝑤, where 𝜆∈(0,1); (c)𝜇{𝑡∈𝑇∶𝑘‖𝑢(𝑡)‖∈𝐾_𝑡}=0, where 𝑘∈𝐾(𝑢).

Proof. Necessity. (a1) Suppose that 𝑢 is an extreme point of the unit ball 𝐵(𝐿⁰_𝑀) and 𝐾(𝑢)=𝜙, then ‖𝑢‖⁰=∫_𝑇𝐴(𝑡)⋅‖𝑢(𝑡)‖𝑑𝑡 by Lemma 2.1. Decompose 𝑇 into 𝑇₁ and 𝑇₂ such that ∫_𝑇1𝐴(𝑡)⋅‖𝑢(𝑡)‖𝑑𝑡=∫_𝑇2𝐴(𝑡)⋅‖𝑢(𝑡)‖𝑑𝑡. Pick 𝜀∈(0,1). Put 𝑢₁(𝑡)=(𝑢(𝑡)+𝜀𝑢(𝑡))𝜒_𝑇1+(𝑢(𝑡)−𝜀𝑢(𝑡))𝜒_𝑇2,𝑢₂(𝑡)=(𝑢(𝑡)−𝜀𝑢(𝑡))𝜒_𝑇1+(𝑢(𝑡)+𝜀𝑢(𝑡))𝜒_𝑇2. (2.9) Obviously, 𝑢=(1/2)(𝑢₁+𝑢₂) and 𝑢₁≠𝑢₂. Moreover, we have ‖𝑢₁‖⁰≤_𝑇𝐴(𝑡)⋅‖𝑢₁=(𝑡)‖𝑑𝑡_𝑇1𝐴(𝑡)⋅‖𝑢(𝑡)+𝜀𝑢(𝑡)‖𝑑𝑡+_𝑇2=𝐴(𝑡)⋅‖𝑢(𝑡)−𝜀𝑢(𝑡)‖𝑑𝑡_𝑇1𝐴(𝑡)⋅‖𝑢(𝑡)‖𝑑𝑡+_𝑇1𝜀𝐴(𝑡)⋅‖𝑢(𝑡)‖𝑑𝑡+_𝑇2−𝐴(𝑡)⋅‖𝑢(𝑡)‖𝑑𝑡_𝑇2𝜀𝐴(𝑡)⋅‖𝑢(𝑡)‖𝑑𝑡=_𝑇1𝐴(𝑡)⋅‖𝑢(𝑡)‖𝑑𝑡+_𝑇2=𝐴(𝑡)⋅‖𝑢(𝑡)‖𝑑𝑡_𝑇𝐴(𝑡)⋅‖𝑢(𝑡)‖𝑑𝑡=1. (2.10) Similarly, we have ‖𝑢₂‖⁰≤1. Hence 𝑢₁,𝑢₁∈𝐵(𝐿⁰_𝑀(𝑋)). Therefore 𝑢∈𝑆(𝐿⁰_𝑀(𝑋)) is not an extreme point of 𝐵(𝐿⁰_𝑀(𝑋)), a contradiction. Hence 𝐾(𝑢)≠𝜙. Suppose that ‖𝑢‖⁰=∫_𝑇𝐴(𝑡)⋅‖𝑢(𝑡)‖𝑑𝑡. Similarly, we get a contradiction.
The necessity of (b) is obvious.
(c) Set𝐻₁𝐻={𝑡∈𝑇∶𝑢(𝑡)=0,𝑒(𝑡)>0},₂={𝑡∈𝑇∶2𝑀(𝑡,𝑘‖𝑢(𝑡)‖)=𝑀(𝑡,𝑘‖𝑢(𝑡)‖+𝜀)+𝑀(𝑡,𝑘‖𝑢(𝑡)‖−𝜀),‖𝑢(𝑡)‖≠0}, (2.11) where {𝑘}∈𝐾(𝑢). Suppose that (c) does not hold. Then 𝜇𝐻₁>0 or 𝜇𝐻₂>0.
If 𝜇𝐻₁>0, then for any 𝑥∈𝑆(𝑋), by setting (1(𝑣(𝑡),𝑤(𝑡))=12𝑘𝑒(𝑡)𝑥,−2𝑘𝑒(𝑡)𝑥),𝑡∈𝐻₁,(𝑢(𝑡),𝑢(𝑡)),𝑡∈𝑇⧵𝐻₁ (2.12) we have 𝑢≠𝑤 and 𝑢=(1/2)(𝑣+𝑤). Moreover, we have ‖𝑣‖⁰≤1𝑘(1+𝜌_𝑀=1(𝑘𝑢))𝑘(1+_𝐻11𝑀(𝑡,𝑘‖2𝑘𝑒(𝑡)‖)𝑑𝑡+_𝑇⧵𝐻1=1𝑀(𝑡,𝑘‖𝑢(𝑡)‖)𝑑𝑡)𝑘(1+_𝑇⧵𝐻1𝑀(𝑡,𝑘‖𝑢(𝑡)‖)𝑑𝑡)≤‖𝑢‖⁰. (2.13)
Similarly, we have ‖𝑤‖⁰≤1. Hence 𝑣,𝑤∈𝐵(𝐿⁰_𝑀(𝑋)). Therefore 𝑢∈𝑆(𝐿⁰_𝑀(𝑋)) is not an extreme point of 𝐵(𝐿⁰_𝑀(𝑋)), a contradiction.
If 𝜇𝐻₂>0, it is easy to see that 𝐻₂⊂∪^∞_𝑛=1{𝑡∈𝑇∶‖𝑢(𝑡)‖≠0,𝑀(𝑡,𝑘‖𝑢(𝑡)‖)=(1/2)𝑀(𝑡,(1+1/𝑛)𝑘‖𝑢(𝑡)‖)+(1/2)𝑀(𝑡,(1−1/𝑛)𝑘‖𝑢(𝑡)‖)}, where {𝑘}∈𝐾(𝑢). Then there exists 𝑛₀∈𝑁 such that 1𝐻={𝑡∈𝑇∶𝑢(𝑡)≠0,𝑀(𝑡,𝑘‖𝑢(𝑡)‖)=21𝑀(𝑡,(1+𝑛₀1)𝑘‖𝑢(𝑡)‖)+21𝑀(𝑡,(1−𝑛₀)𝑘‖𝑢(𝑡)‖)<∞} (2.14) is not a noll set. Decompose 𝐻 into 𝐸 and 𝐹 such that ∫_𝐸𝑝(𝑡,(1/𝑛₀)𝑘‖𝑢(𝑡)‖)𝑑𝑡=∫_𝐹𝑝(𝑡,(1/𝑛₀)𝑘‖𝑢(𝑡)‖)𝑑𝑡. Define 1(𝑣(𝑡),𝑤(𝑡))=((1+𝑛₀1)𝑢(𝑡),(1−𝑛₀1)𝑢(𝑡)),𝑡∈𝐸,((1−𝑛₀1)𝑢(𝑡),(1+𝑛₀)𝑢(𝑡)),𝑡∈𝐹,(𝑢(𝑡),𝑢(𝑡)),𝑡∈𝑇⧵(𝐸∪𝐹). (2.15) Then 𝑢≠𝑤 and 𝑢=(1/2)(𝑣+𝑤). Furthermore, we have ‖𝑣‖⁰≤1𝑘(1+𝜌_𝑀=1(𝑘𝑣))𝑘(1+_{𝑇⧵(𝐸∪𝐹)}𝑀(𝑡,𝑘‖𝑢(𝑡)‖)𝑑𝑡+_𝐸1𝑀(𝑡,(1+𝑛₀+)𝑘‖𝑢(𝑡)‖)𝑑𝑡_𝐹1𝑀(𝑡,(1−𝑛₀=1)𝑘‖𝑢(𝑡)‖)𝑑𝑡)𝑘(1+_{𝑇⧵(𝐸∪𝐹)}𝑀(𝑡,𝑘‖𝑢(𝑡)‖)𝑑𝑡+_𝐸𝑀(𝑡,𝑘‖𝑢(𝑡)‖)𝑑𝑡+_𝐸1𝑝(𝑡,𝑛₀𝑘‖𝑢(𝑡)‖)𝑑𝑡+_𝐹𝑀(𝑡,𝑘‖𝑢(𝑡)‖)𝑑𝑡−_𝐹1𝑝(𝑡,𝑛₀=1𝑘‖𝑢(𝑡)‖)𝑑𝑡)𝑘(1+_{𝑇⧵(𝐸∪F)}𝑀(𝑡,𝑘‖𝑢(𝑡)‖)𝑑𝑡+_𝐸𝑀(𝑡,𝑘‖𝑢(𝑡)‖)𝑑𝑡+_𝐹=1𝑀(𝑡,𝑘‖𝑢(𝑡)‖)𝑑𝑡)𝑘(1+𝜌_𝑀(𝑘𝑢))=‖𝑢‖⁰=1. (2.16) Similarly, we have ‖𝑤‖⁰≤1. Hence 𝑣,𝑤∈𝐵(𝐿⁰_𝑀(𝑋)). Therefore 𝑢∈𝑆(𝐿⁰_𝑀(𝑋)) is not an extreme point, a contradiction. Hence (c) is true.
(a2) If 𝐾(𝑢)≠𝜙 and 𝑢∈𝑆(𝐿⁰_𝑀(𝑋)) is an extreme point, suppose that there exists 𝑘₁,𝑘₂∈𝐾(𝑢) satisfying 𝑘₁≠𝑘₂. Define𝑘=𝑘₁𝑘₂/(𝑘₁+𝑘₂), 2=‖𝑢‖⁰+‖𝑢‖⁰=𝑘₁+𝑘₂𝑘₁𝑘₂𝑘[1+²𝑘₁+𝑘₂𝜌_𝑀(𝑘₁𝑘𝑢)+¹𝑘₁+𝑘₂𝜌_𝑀(𝑘₂=𝑘𝑢)]₁+𝑘₂𝑘₁𝑘₂𝑘[1+²𝑘₁+𝑘_2𝑇𝑀(𝑡,‖𝑘₁𝑘𝑢(𝑡)‖)𝑑𝑡+¹𝑘₁+𝑘_2𝑇𝑀(𝑡,‖𝑘₂≥𝑘𝑢(𝑡)‖)𝑑𝑡]₁+𝑘₂𝑘₁𝑘₂[1+_𝑇𝑘𝑀(𝑡,²𝑘₁+𝑘₂‖𝑘₁𝑘𝑢(𝑡)‖+¹𝑘₁+𝑘₂‖𝑘₂=𝑘𝑢(𝑡)‖)𝑑𝑡]₁+𝑘₂𝑘₁𝑘₂[1+_𝑇𝑀(𝑡,‖2𝑘₁𝑘₂𝑘₁+𝑘₂1𝑢(𝑡)‖)𝑑𝑡]=22𝑘[1+𝜌_𝑀(2𝑘𝑢)]≥2‖𝑢‖⁰=2. (2.17) This implies that ‖𝑢‖⁰=12𝑘[1+𝜌_𝑀(2𝑘𝑢)] (2.18) (i.e., 2𝑘∈𝐾(𝑢)), 𝑘₂𝑘₁+𝑘₂𝑀(𝑡,𝑘₁𝑘‖𝑢(𝑡)‖)+¹𝑘₁+𝑘₂𝑀(𝑡,𝑘₂‖𝑢(𝑡)‖)=𝑀(𝑡,2𝑘‖𝑢(𝑡)‖). (2.19) Since 𝑘₁‖𝑢(𝑡)‖≠𝑘₂‖𝑢(𝑡)‖ on {𝑡∈𝑇∶‖𝑢(𝑡)‖≠0}, then 2𝑘‖𝑢(𝑡)‖∈𝐾_𝑡 on {𝑡∈𝑇∶‖𝑢(𝑡)‖≠0}, a contradiction. Therefore (a) is true.
Sufficiency. We first prove that for 𝑢,𝑢₁,𝑢₂∈𝑆(𝐿⁰_𝑀(𝑋)) with 𝑢=(1/2)(𝑢₁+𝑢₂) at least one of the sets 𝐾(𝑢₁) or 𝐾(𝑢₂) is nonempty. Suppose that 𝐾(𝑢₁)=𝜙 and 𝐾(𝑢₂)=𝜙. Hence we have 11=‖2(𝑢₁+𝑢₂)‖⁰<_𝑇1𝐴(𝑡)⋅‖2(𝑢₁(𝑡)+𝑢₂≤1(𝑡))‖𝑑𝑡2_𝑇𝐴(𝑡)⋅‖𝑢₁1(𝑡)‖𝑑𝑡+2_𝑇𝐴(𝑡)⋅‖𝑢₂=1(𝑡)‖𝑑𝑡2‖𝑢₁‖⁰+12‖𝑢₂‖⁰=1, (2.20) a contradiction. This contradiction shows that 𝐾(𝑢₁)≠𝜙 or 𝐾(𝑢₂)≠𝜙.
Now we will prove that 𝐾(𝑢₁)≠𝜙 and 𝐾(𝑢₂)≠𝜙. Otherwise, we can assume without loss of generality that 𝐾(𝑢₁)≠𝜙 and 𝐾(𝑢₂)=𝜙. Put [𝑢₁,𝑢)={(1−𝜆)𝑢₁+𝜆𝑢∶0<𝜆<1},(𝑢,𝑢₂]={(1−𝜆)𝑢+𝜆𝑢₂∶0<𝜆<1}. (2.21) Next we will prove that 𝐾(𝑦)≠𝜙 for all 𝑦∈[𝑢₁,𝑢) and 𝐾(𝑦)=𝜙 for all 𝑦∈(𝑢,𝑢₂]. Assume first for the contrary that this is 𝑢₃∈[𝑢₁,𝑢) such that 𝐾(𝑢₃)=𝜙. Then there exists 𝜆₃∈[0,1) such that 𝑢₃=(1−𝜆₃)𝑢₁+𝜆₃𝑢. Since 𝑢₁=2𝑢−𝑢₂, we have 𝑢₃=(1−𝜆₃)(2𝑢−𝑢₂)+𝜆₃𝑢=(2−𝜆₃)𝑢−(1−𝜆₃)𝑢₂. (2.22) Hence 𝑢=(1/(2−𝜆₃))𝑢₃+((1−𝜆₃)/(2−𝜆₃))𝑢₂. Therefore 1=‖𝑢‖⁰<_𝑇1𝐴(𝑡)⋅‖2−𝜆₃𝑢₃(𝑡)+1−𝜆₃2−𝜆₃𝑢₂≤1(𝑡)‖𝑑𝑡2−𝜆_3𝑇𝐴(𝑡)⋅‖𝑢₃(𝑡)‖𝑑𝑡+1−𝜆₃2−𝜆_3𝑇𝐴(𝑡)⋅‖𝑢₂=1(𝑡)‖𝑑𝑡2−𝜆₃‖𝑢₃‖⁰+1−𝜆₃2−𝜆₃‖𝑢₂‖⁰=1, (2.23) a contradiction.
Assume now for the contrary that this is 𝑢₄∈(𝑢,𝑢₂] such that 𝐾(𝑢₄)≠𝜙. We can find 𝑢₅∈(𝑢,𝑢₂] such that 𝑢=(1/2)(𝑢₄+𝑢₅) and 𝑢₄≠𝑢₅. Therefore there are 𝑘₄≥1 and 𝑘₅≥1 such that‖𝑢₄‖⁰=1𝑘₄[1+𝜌_𝑀(𝑘₄𝑢₄)],‖𝑢₅‖⁰=1𝑘₅[1+𝜌_𝑀(𝑘₅𝑢₅)]. (2.24)
By the convexity of the modular 𝜌_𝑀 we have 𝜌_𝑀(2𝑘₄𝑘₅𝑘₄+𝑘₅𝑢)=𝜌_𝑀(𝑘₄𝑘₅𝑘₄+𝑘₅(𝑢₄+𝑢₅))=𝜌_𝑀(𝑘₅𝑘₄+𝑘₅𝑘₄𝑢₄+𝑘₄𝑘₄+𝑘₅𝑘₅𝑢₅)=_𝑇𝑘𝑀(𝑡,‖⁵𝑘₄+𝑘₅𝑘₄𝑢₄𝑘(𝑡)+⁴𝑘₄+𝑘₅𝑘₅𝑢₅≤(𝑡)‖)𝑑𝑡_𝑇𝑘𝑀(𝑡,⁵𝑘₄+𝑘₅‖𝑘₄𝑢₄𝑘(𝑡)‖+⁴𝑘₄+𝑘₅‖𝑘₅𝑢₅≤𝑘(𝑡)‖)𝑑𝑡₅𝑘₄+𝑘_5𝑇𝑀(𝑡,‖𝑘₄𝑢₄𝑘(𝑡)‖)𝑑𝑡+⁴𝑘₄+𝑘_5𝑇𝑀(𝑡,‖𝑘₅𝑢₅=𝑘(𝑡)‖)𝑑𝑡₅𝑘₄+𝑘₅𝜌_𝑀(𝑘₄𝑢₄𝑘)+⁴𝑘₄+𝑘₅𝜌_𝑀(𝑘₅𝑢₅). (2.25) Hence 2=2‖𝑢‖⁰≤𝑘₄+𝑘₅𝑘₄𝑘₅(1+𝜌_𝑀(𝑘₄𝑘₅𝑘₄+𝑘₅≤𝑘2𝑢))₄+𝑘₅𝑘₄𝑘₅𝑘(1+⁵𝑘₄+𝑘₅𝜌_𝑀(𝑘₄𝑢₄𝑘)+⁴𝑘₄+𝑘₅𝜌_𝑀(𝑘₅𝑢₅≤1))𝑘₄(1+𝜌_𝑀(𝑘₄𝑢₄1))+𝑘₅(1+𝜌_𝑀(𝑘₅𝑢₅))=2. (2.26) Consequently, all inequalities from the last three lines are equalities in fact. Therefore 2(𝑘₄k₅/(𝑘₄+𝑘₅))=𝑘 and 𝑘𝑀(𝑡,‖𝑘𝑢(𝑡)‖)=⁵𝑘₄+𝑘₅𝑀(𝑡,‖𝑘₄𝑢₄𝑘(𝑡)‖)+⁴𝑘₄+𝑘₅𝑀(𝑡,‖𝑘₅𝑢₅(𝑡)‖) (2.27) for 𝜇-𝑎.𝑒  𝑡∈𝑇. By 𝜇{𝑡∈𝑇∶𝑘‖𝑢(𝑡)‖∈𝐾_𝑡}=0, it follows that ‖𝑘𝑢(𝑡)‖=‖𝑘₄𝑢₄(𝑡)‖=‖𝑘₅𝑢₅(𝑡)‖ for 𝜇-𝑎.𝑒  𝑡∈𝑇. And we have 𝑘𝑢(𝑡)=(𝑘₅/(𝑘₄+𝑘₅))𝑘₄𝑢₄(𝑡)+(𝑘₄/(𝑘₄+𝑘₅))𝑘₅𝑢₅(𝑡) for 𝜇-𝑎.𝑒  𝑡∈𝑇. By (b), we have 𝑘𝑢=𝑘₄𝑢₄=𝑘₅𝑢₅. Since 𝑢₄,𝑢₅,𝑢∈𝑆(𝐿⁰_𝑀(𝑋)), we get 𝑘₄=𝑘₅=𝑘, which gives 𝑢₄=𝑢₅=𝑢. This contradicts the inequality 𝑢₄≠𝑢₅. Thus 𝐾(𝑦)=𝜙 for any 𝑦∈(𝑢,𝑢₂]. Take 𝑢_𝑛=(1−1/𝑛)𝑢+(1/𝑛)𝑢₂ for all 𝑛∈𝑁. Then 𝑢_𝑛∈(𝑢,𝑢₂] for all 𝑛∈𝑁. Hence 𝐾(𝑢_𝑛)=𝜙, and consequently ‖𝑢_𝑛‖⁰=∫_𝑇𝐴(𝑡)⋅‖𝑢_𝑛(𝑡)‖𝑑𝑡 for all 𝑛∈𝑁. Note that ‖𝑢_𝑛−𝑢‖⁰→0(𝑛→∞) and lim_𝑛→∞‖𝑢_𝑛(𝑡)‖=‖𝑢(𝑡)‖ for 𝜇-𝑎.𝑒,𝑡∈𝑇. Since 𝐾(𝑢)={𝑘}, with 0<𝑘<∞, we have lim_𝑛→∞‖𝑢_𝑛‖⁰=‖𝑢‖⁰=lim_{𝑛→∞𝑇}𝐴(𝑡)⋅‖𝑢_𝑛(𝑡)‖𝑑𝑡≥_𝑇𝐴(𝑡)⋅‖𝑢(𝑡)‖𝑑𝑡>‖𝑢‖⁰, (2.28) a contradiction. Therefore 𝐾(𝑢₁)≠𝜙 and 𝐾(𝑢₂)≠𝜙. Now repeating the same procedure as above, putting 𝑢₁ and 𝑢₂ instead of 𝑢₄ and 𝑢₅, respectively, we get 𝑘₁𝑢₁(𝑡)=𝑘₂𝑢₂(𝑡)=𝑘𝑢(𝑡) (2.29) for 𝜇-𝑎.𝑒,𝑡∈𝑇. Hence, by the fact that 𝑢₁,𝑢₂,𝑢∈𝑆(𝐿⁰_𝑀(𝑋)), we have 𝑘₁=𝑘₂=𝑘, and consequently, 𝑢₁=𝑢₂=𝑢. Thus 𝑢 is an extreme point of 𝐵(𝐿⁰_𝑀(𝑋)).

Corollary 2.5 (see [3]). 𝑢∈𝑆(𝐿⁰_𝑀(𝑅)) is an extreme point of 𝐵(𝐿⁰_𝑀(𝑅)) if and only if (a)the set 𝐾(𝑢) consists of one element from (0,+∞); (b)𝜇{𝑡∈𝑇∶𝑘|𝑢(𝑡)|∈𝐾_𝑡}=0, where 𝑘∈𝐾(𝑢).

Finally, we investigate the rotundity of 𝐿⁰_𝑀(𝑋).

Theorem 2.6. 𝐿⁰_𝑀(𝑋) is rotund if and only if: (a)for any 𝑢∈𝑆(𝐿⁰_𝑀(𝑋)), the set 𝐾(𝑢) consists of one element from (0,+∞); (b)𝑋 is rotund; (c)𝑀(𝑡,𝑢) is strictly convex with respect to 𝑢 for almost all 𝑡∈𝑇.

Proof. Sufficiency is obvious by Theorem 2.4.
Necessity. (a) is obvious by (a) of Theorem 2.4. 𝐿⁰_𝑀(𝑅) is isometrically isomorphic to closed subspace of 𝐿⁰_𝑀(𝑋), thus 𝐿⁰_𝑀(𝑅) is rotund. By Lemma 2.3, (c) is obvious.
If (b) is not true, then there exist 𝑥,𝑦,𝑧∈𝑆(𝑋) with 2𝑥=𝑦+𝑧 and 𝑦≠𝑧. Pick ℎ(𝑡)∈𝑆(𝐿⁰_𝑀(𝑋)), then there exists 𝑑>0 such that 𝜇𝐻>0, where 𝐻={𝑡∈𝑇∶‖ℎ(𝑡)‖≥𝑑}. Since ℎ(𝑡)∈𝑆(𝐿⁰_𝑀(𝑋)), then there exists 𝑘^′>0 such that_𝐻𝑀(𝑡,𝑘^′‖ℎ(𝑡)‖)𝑑𝑡≤_𝑇𝑀(𝑡,𝑘^′‖ℎ(𝑡)‖)𝑑𝑡<∞. (2.30) Set 𝑢(𝑡)=𝑑⋅𝑥⋅𝜒_𝐻(𝑡),𝑣(𝑡)=𝑑⋅𝑦⋅𝜒_𝐻(𝑡),𝑤(𝑡)=𝑑⋅𝑧⋅𝜒_𝐻(𝑡). (2.31) We have _𝑇𝑀(𝑡,𝑘^′‖𝑢(𝑡)‖)𝑑𝑡=_𝐻𝑀(𝑡,𝑘^′𝑑)𝑑𝑡≤_𝐻𝑀(𝑡,𝑘‖ℎ(𝑡)‖)𝑑𝑡<∞. (2.32) This implies that 𝑢(𝑡)∈𝐿⁰_𝑀(𝑋). Similarly, we have 𝑣(𝑡),𝑤(𝑡)∈𝐿⁰_𝑀(𝑋). It is easy to see that ‖𝑢‖⁰=‖𝑣‖⁰=‖𝑤‖⁰. Then ‖𝑢‖𝑢‖⁰‖⁰𝑣=‖‖𝑢‖⁰‖⁰𝑤=‖‖𝑢‖⁰‖⁰𝑢=1,‖𝑢‖⁰=12⋅𝑣‖𝑢‖⁰+12⋅𝑤‖𝑢‖⁰. (2.33) However, ‖𝑢(𝑡)‖=‖𝑣(𝑡)‖=‖𝑤(𝑡)‖ for 𝑡∈𝑇. By (b) of Theorem 2.4, we have 𝑢=𝑤. Hence 𝑣/‖𝑢‖⁰=𝑤/‖𝑢‖⁰. So 𝑢 is not an extreme point of 𝐵(𝐿⁰_𝑀(𝑋)). Contradicting the rotundity of 𝐿⁰_𝑀(𝑋).

Acknowledgment

The authors would like to thank the anonymous referees for some suggestions to improve the paper. This work was supported by Heilong Jiang Natural Science Fund A200902.