# Köthe-Bochner Spaces and Some Geometric Properties Related to Rotundity and Smoothness

Authors: Jan-David Hardtke
Journal: Journal of Function Spaces and Applications
 (2013)

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## Abstract

In 2000, Kadets et al. introduced the notions of acs, luacs, and uacs spaces, which form common generalisations of well-known
rotundity and smoothness properties of Banach spaces. In a recent paper, the author introduced some further related notions and investigated the behaviour of these geometric properties under the formation of absolute sums. The present paper is in a sense a continuation of this work. Here we will study the behaviour of the said properties under the formation of Köthe-Bochner spaces, thereby generalising some results of G. Sirotkin on the acs, luacs, and uacs properties ofLp-Bochner spaces.


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## Body

## 1. Introduction

We begin with some notation and definitions. Throughout this paper,X denotes a real Banach space, X* its dual, BX its unit ball and SX its unit sphere.In the next definition, we summarise the most important rotundity properties.Definition 1.
A Banach spaceX is called(i)
rotund (R) if for any two elements x,y∈SX the equality ∥x+y∥=2 implies x=y,(ii)
locally uniformly rotund (LUR) if for every x∈SX, the implication
(1)∥xn+x∥⟶2⟹∥xn-x∥⟶0
holds for every sequence (xn)n∈ℕ in SX,(iii)
weakly locally uniformly rotund (WLUR) if for every x∈SX and every sequence (xn)n∈ℕ in SX one has
(2)∥xn+x∥⟶2⟹xn⟶xweakly,(iv)
uniformly rotund (UR) if for any two sequences (xn)n∈ℕ and (yn)n∈ℕ in SX the implication
(3)∥xn+yn∥⟶2⟹∥xn-yn∥⟶0
holds,(v)
weakly uniformly rotund (WUR) if for any two sequences (xn)n∈ℕ and (yn)n∈ℕ the following implication holds:
(4)∥xn+yn∥⟶2⟹xn-yn⟶0weakly.Figure1 shows the obvious implications between these notions. No other implications are valid in general (see the examples in [1]). Note, however, that all these notions coincide in finite-dimensional spaces, by the compactness of BX.Figure 1The modulus of convexity of the spaceX is defined by
(5)δX(ε)=inf{1-12∥x+y∥:x,y∈BX,∥x-y∥≥ε}
for every ε in the interval ]0,2]. Then X is UR if and only if δX(ε)>0 for all 0<ε≤2.For the local version one defines(6)δX(x,ε)=inf{1-12∥x+y∥:y∈BX,∥x-y∥≥ε}
for every x∈SX and each ε∈]0,2]. Then X is LUR if and only if δX(x,ε)>0 for all x∈SX and all 0<ε≤2.Let us also recall some notions of smoothness. The spaceX is called smooth (S) if its norm is Gâteaux-differentiable at every nonzero point (equivalently at every point of SX), which is the case if and only if for every x∈SX there is a unique functional x*∈SX* with x*(x)=1 (cf. [2, Lemma 8.4 (ii)]). X is called Fréchet-smooth (FS) if the norm is Fréchet-differentiable at every nonzero point. The norm of the space X is said to be uniformly Gâteaux-differentiable (UG) if for each y∈SX the limit limτ→0(∥x+τy∥-1)/τ exists uniformly in x∈SX. Finally, X is called uniformly smooth (US) if limτ→0ρX(τ)/τ=0, where ρX denotes the modulus of smoothness of X defined by ρX(τ)=sup{1/2(∥x+τy∥+∥x-τy∥-2):x,y∈SX} for every τ>0.In [3] the following notions were introduced (in connection with the so called Anti-Daugavet property).Definition 2.
A Banach spaceX is called (i)
alternatively convex or smooth (acs) if for every x,y∈SX with ∥x+y∥=2 and every x*∈SX* with x*(x)=1 one has x*(y)=1 as well,(ii)
locally uniformly alternatively convex or smooth (luacs) if for every x∈SX, every sequence (xn)n∈ℕ in SX and every functional x*∈SX* one has
(7)∥xn+x∥⟶2,x*(xn)⟶1⟹x*(x)=1,(iii)
uniformly alternatively convex or smooth (uacs) if for all sequences (xn)n∈ℕ, (yn)n∈ℕ in SX and (xn*)n∈ℕ in SX*, one has
(8)∥xn+yn∥⟶2,xn*(xn)⟶1⟹xn*(yn)⟶1.
The author introduced the following related notions in [4].Definition 3.
A Banach spaceX is called(i)
strongly locally uniformly alternatively convex or smooth (sluacs) if for every x∈SX and all sequences (xn)n∈ℕ in SX and (xn*)n∈ℕ in SX*, one has
(9)∥xn+x∥⟶2,xn*(xn)⟶1⟹xn*(x)⟶1,(ii)
weakly uniformly alternatively convex or smooth (wuacs) if for any two sequences (xn)n∈ℕ, (yn)n∈ℕ in SX and every functional x*∈SX*, one has
(10)∥xn+yn∥⟶2,x*(xn)⟶1⟹x*(yn)⟶1.The obvious implications between the acs properties and the rotundity properties are indicated in Figure2. No other implications are generally valid (see the examples in [4]), but note again that the properties acs, luacs, sluacs, wuacs and uacs coincide in finite-dimensional spaces, by compactness.Figure 2The connection between some of the acs properties to smoothness properties is illustrated in Figure3.Figure 3Let us mention that if we replace the conditionxn*(xn)→1 by xn*(xn)=1 for every n∈ℕ in the definitions of the properties uacs and sluacs, respectively we still obtain the same classes of spaces. For uacs spaces this was first proved by Sirotkin in [5] using the fact that uacs spaces are reflexive (see below). For sluacs spaces this characterisation can be proved by means of the Bishop-Phelps-Bollobás theorem (see [4, Proposition 2.1]).This characterisation enables us to define the following “uacs-modulus” of a given Banach space (cf. [4, Definition 1.4]).Definition 4.
For a Banach spaceX one defines
(11)DX(ε)={(x,y)∈SX×SX:∃x*∈SX*x*(x)=1,x*(y)≤1-ε},δuacsX(ε)=inf{1-∥x+y2∥:(x,y)∈DX(ε)}∀ε∈]0,2].
Then X is uacs if and only if δuacsX(ε)>0 for every ε∈]0,2] and one clearly has δX(ε)≤δuacsX(ε) for each ε∈]0,2].The above characterisation shows that the class of uacs spaces coincides with the class ofu-spaces introduced by Lau in [6] and our modulus δuacsX is the same as the modulus of u-convexity from [7]. Also, the notion of u-spaces which was introduced in [8] coincides with the notion of acs spaces.Recall that a Banach spaceX is said to be uniformly nonsquare if there is some δ>0 such that for all x,y∈BX we have ∥x+y∥≤2(1-δ) or ∥x-y∥≤2(1-δ). It is easily seen that uacs spaces are uniformly nonsquare and hence by a well-known theorem of James (cf. [9, page 261]) they are superreflexive, as was observed in [3, Lemma 4.4]. For a proof of the superreflexivity of uacs spaces that does not rely on James’ result on uniformly nonsquare spaces, see [4, Proposition 2.8].Let us also restate here the following auxiliary result [4, Lemma 2.30] (it is the generalisation of [10, Lemma 2.1] to sequences, with a completely analogous proof).Lemma 5.
Let(xn)n∈ℕ and (yn)n∈ℕ be sequences in the (real or complex) normed space X such that ∥xn+yn∥-∥xn∥-∥yn∥→0.
Then for any two bounded sequences(αn)n∈ℕ, (βn)n∈ℕ of nonnegative real numbers one also has ∥αnxn+βnyn∥-αn∥xn∥-βn∥yn∥→0.Finally, we will need two more definitions from [4].Definition 6.
A Banach spaceX is called (i)
aluacs+ space if for every x∈SX, every sequence (xn)n∈ℕ in SX with ∥xn+x∥→2, and all x*∈SX* one has
(12)x*(xn)⟶1⟺x*(x)=1,(ii)
asluacs+ space if for every x∈SX, every sequence (xn)n∈ℕ in SX with ∥xn+x∥→2, and all sequences (xn*)n∈ℕ in SX*, one has
(13)xn*(xn)⟶1⟺xn*(x)⟶1.
Obviously, every WLUR space is luacs+, and every LUR space is sluacs+.In the next section we will recall some facts on Köthe-Bochner spaces.


## 2. Preliminaries on Köthe-Bochner Spaces

If not otherwise stated,(S,𝒜,μ) will denote a complete, σ-finite measure space. For A∈𝒜 we denote by χA the characteristic function of A.A Köthe function space over(S,𝒜,μ) is a Banach space (E,∥·∥E) of real-valued measurable (i.e., 𝒜-Borel-measurable) functions on S modulo equality μ-almost everywhere (we will henceforth abbreviate this by μ-a.e. or simply a.e. if μ is tacitly understood) such that (i)
χA∈E for every A∈𝒜 with μ(A)<∞,(ii)
for everyf∈E and every set A∈𝒜 with μ(A)<∞f is μ-integrable over A,(iii)
ifg is measurable and f∈E such that |g(t)|≤|f(t)|μ-a.e., then g∈E and ∥g∥E≤∥f∥E.The standard examples are of course the spacesLp(μ) for 1≤p≤∞.Every Köthe function spaceE is a Banach lattice when endowed with the natural order f≤g if and only if f(t)≤g(t)μ-a.e.Recall that a Banach latticeE is said to be order complete (σ-order complete) if for every net (sequence) in E which is order bounded, the supremum of said net (sequence) in E exists. A Banach lattice E is called order continuous (σ-order continuous) provided that every decreasing net (sequence) in E whose infimum is zero is norm-convergent to zero.It is easy to see that a Köthe function spaceE is always σ-order complete, and thus by [11, Proposition 3.1.5] E is order continuous if and only if E is σ-order continuous if and only if E is order complete and order continuous. Also, reflexivity of E implies order continuity, for any σ-order complete Banach lattice which is not σ-order continuous contains an isomorphic copy of ℓ∞ (cf. [11, Proposition  3.1.4]).Let us also mention the following well-known fact that will be needed later.Lemma 7.
IfE is a Köthe function space, (fn)n∈ℕ a sequence in E, and f∈E such that ∥fn-f∥E→0, then there is a subsequence of (fn)n∈ℕ which converges pointwise almost everywhere to f.For a Köthe function spaceE we denote by E′ the space of all measurable functions g:S→ℝ (modulo equality μ-a.e.) such that
(14)∥g∥E′:=sup{∫S‍|fg|dμ:f∈BE}<∞.
Then (E′,∥·∥E′) is again a Köthe function space, the so called Köthe dual of E. The operator T:E′→E* defined by
(15)(Tg)(f)=∫S‍fgdμ∀f∈E,∀g∈E′
is well-defined, linear and isometric. Moreover, T is surjective if and only if E is order continuous (cf. [11, page 149]), thus for order continuous E we have E*=E′.We refer the reader to [12] or [11] for more information on Banach lattices in general and Köthe function spaces in particular.Now recall that ifX is a Banach space, a function f:S→X is called simple if there are finitely many disjoint measurable sets A1,…,An∈𝒜 such that μ(Ai)<∞ for all i=1,…,n, f is constant on each Ai and f(t)=0 for every t∈S∖⋃i=1n‍Ai. The function f is said to be Bochner-measurable if there exists a sequence (fn)n∈ℕ of simple functions such that limn→∞∥fn(t)-f(t)∥=0μ-a.e. and weakly measurable if x*∘f is measurable for every functional x*∈X*. According to Pettis’ measurability theorem (cf. [11, Theorem  3.2.2]) f is Bochner-measurable if and only if f is weakly measurable and almost everywhere separably valued (i.e., there is a separable subspace Y⊆X such that f(t)∈Yμ-a.e.).For a Köthe function spaceE and a Banach space X we denote by E(X) the space of all Bochner-measurable functions f:S→X (modulo equality a.e.) such that ∥f(·)∥∈E. Endowed with the norm ∥f∥E(X)=∥∥f(·)∥∥EE(X) becomes a Banach space, the so called Köthe Bochner space induced by E and X. The most prominent examples are again the Lebesgue-Bochner spaces Lp(X) for 1≤p≤∞.Next we recall how the dual ofE(X) can be described provided that E is order continuous. A function F:S→X* is called weak*-measurable if F(·)(x) is measurable for every x∈X. We define an equivalence relation on the set of all weak*-measurable functions by setting F~G if and only if for every x∈XF(t)(x)=G(t)(x) a.e., and we write E′(X*,w*) for the space of all (equivalence classes of) weak*-measurable functions F such that there is some g∈E′ with ∥F(t)∥≤g(t) a.e.A norm onE′(X*,w*) can be defined by
(16)∥[F]∥E′(X*,w*):=inf{∥g∥E′:g∈E′,F(t)≤g(t)a.e.}.
Then the following deep theorem holds.Theorem 8 (cf. [13]).
LetE be an order-continuous Köthe function space over the complete, σ-finite measure space (S,𝒜,μ), and let X be a Banach space. Then the map V:E′(X*,w*)→E(X)* defined by
(17)V([F])(f)≔∫S‍F(t)(f(t))dμ(t),∀f∈E(X),∀[F]∈E′(X*,w*)
is an isometric isomorphism, and, moreover, every equivalence class L in E′(X*,w*) has a representative F such that ∥F(·)∥∈E′ and ∥L∥E′(X*,w*)=∥∥F(·)∥∥E′.There are a number of papers on various rotundity and smoothness properties in Köthe-Bochner spaces in general and Lebesgues-Bochner spaces in particular, see for example [14–17] and references therein.Sirotkin proved in [5] that for 1<p<∞ the Lebesgue-Bochner space Lp(X) is acs, respectively luacs, and uacs, whenever X has the respective property. In the next section we will study the more general case of Köthe-Bochner spaces.


## 3. Results and Proofs

We begin with the acs spaces, for which we have the following result.Proposition 9.
IfE is an order-continuous acs Köthe function space and X is an acs Banach space, then E(X) is acs as well.Proof.
The proof is similar to that of [4, Proposition 3.3]. First we fix two elements f,g∈SE(X) such that ∥f+g∥E(X)=2 and a functional l∈SE(X)* with l(f)=1.
SinceE is order continuous, by Theorem 8,  l can be represented via an element [F]∈E′(X*,w*) such that ∥F(·)∥∈E′ and ∥∥F(·)∥∥E′=∥[F]∥E′(X*,w*)=∥l∥=1. It follows that
(18)1=l(f)=∫S‍F(t)(f(t))dμ(t)≤∫S‍∥F(t)∥∥f(t)∥dμ(t)≤∥∥F(·)∥∥E′∥∥f(·)∥∥E=∥l∥∥f∥E(X)=1
and hence
(19)∫S‍∥F(t)∥∥f(t)∥dμ(t)=1,(20)F(t)(f(t))=∥F(t)∥∥f(t)∥a.e.
We also have
(21)2=∥f+g∥E(X)=∥∥f(·)+g(·)∥∥E≤∥∥f(·)∥+∥g(·)∥∥E≤∥f∥E(X)+∥g∥E(X)=2
and thus
(22)∥∥f(·)∥+∥g(·)∥∥E=2.
Since E is acs it follows from (19) and (22) that
(23)∫S‍∥F(t)∥∥g(t)∥dμ(t)=1.
In a similar way as we have obtained (22) we can also show
(24)∥∥f(·)+g(·)∥+∥f(·)∥+∥g(·)∥∥E=4.
Because E is acs this together with (19), (22) and (23) implies
(25)∫S‍∥F(t)∥∥f(t)+g(t)∥dμ(t)=2.
From (19), (23) and (25) we get
(26)∥F(t)∥(∥f(t)∥+∥g(t)∥-∥f(t)+g(t)∥)=0a.e.
Now we will show that
(27)F(t)(g(t))=∥F(t)∥∥g(t)∥a.e.
To this end, let us denote by N1 and N2 the null sets on which the equality from (20) and (26), respectively does not hold. Let N=N1∪N2.
PutB={t∈S∖N:F(t)≠0andg(t)≠0} and C={t∈B:f(t)=0}. We claim that C is a null set.
To see this, defineh:S→ℝ by h(t)=∥F(t)∥ for t∈S∖C and h(t)=0 for t∈C. Then h is measurable and since h(t)≤∥F(t)∥ for all t∈S we have h∈E′ with ∥h∥E′≤1. We also have h(t)∥f(t)∥=∥F(t)∥∥f(t)∥ for every t∈S and hence by (19)
(28)∫S‍h(t)∥f(t)∥dμ(t)=1
which also implies ∥h∥E′=1. Together with (22) we now get
(29)∫S‍h(t)∥g(t)∥dμ(t)=1,
since E is acs. Taking into account (23) we arrive at
(30)∫S‍(∥F(t)∥-h(t))∥g(t)∥dμ(t)=0.
Hence (∥F(t)∥-h(t))∥g(t)∥=0 a.e. and thus C must be a null set.
Now ift∈(S∖C)∩B then F(t)≠0, f(t)≠0 and g(t)≠0 and ∥F(t)∥∥f(t)∥=F(t)(f(t)) as well as
(31)∥f(t)+g(t)∥=∥f(t)∥+∥g(t)∥.
By [10, Lemma 2.1] this implies
(32)∥f(t)∥f(t)∥+g(t)∥g(t)∥∥=2.
Since X is acs it follows that ∥F(t)∥∥g(t)∥=F(t)(g(t)).
SoM:=N∪C is a null set with ∥F(t)∥∥g(t)∥=F(t)(g(t)) for every t∈S∖M and (27) is proved.
Now combining (23) and (27) we obtain
(33)l(g)=∫S‍F(t)(g(t))dμ(t)=1,
which finishes the proof.Before we turn to the case of luacs spaces, let us recall Egorov’s theorem (cf. [18, Theorem A, page 88]), which states that for any finite measure space (S,𝒜,μ) and every sequence (fn)n∈ℕ of measurable functions on S which converges to zero pointwise μ-a.e. and each ε>0 there is a set A∈𝒜 with μ(S∖A)≤ε such that (fn)n∈ℕ is uniformly convergent to zero on A.Now we are ready to prove the following theorem.Theorem 10.
LetE be an order-continuous Köthe function space over the complete σ-finite measure space (S,𝒜,μ) and X an luacs Banach space. If (a)
E is WLUR or(b)
E is luacs+ and E′ is also order continuous,then E(X) is also luacs.Proof.
Suppose that we are given a sequence(fn)n∈ℕ in SE(X) and an element f∈SE(X) such that ∥fn+f∥E(X)→2 as well as a functional l∈SE(X)* such that l(fn)→1. As before, we can represent l by an element [F]∈E′(X*,w*). We then have
(34)l(fn)=∫S‍F(t)(fn(t))dμ(t)≤∫S‍∥F(t)∥∥fn(t)∥dμ(t)≤1
and hence
(35)limn→∞∫S‍∥F(t)∥∥fn(t)∥dμ(t)=1.
By passing to a subsequence we may also assume that
(36)limn→∞(∥F(t)∥∥fn(t)∥-F(t)(fn(t)))=0a.e.
We further have
(37)∥fn+f∥E(X)=∥∥fn(·)+f(·)∥∥E≤∥∥fn(·)∥+∥f(·)∥∥E≤2
and thus
(38)limn→∞∥∥fn(·)∥+∥f(·)∥∥E=2.
An analogous argument also shows
(39)limn→∞∥∥fn(·)+f(·)∥+∥fn(·)∥+∥f(·)∥∥E=4.
Moreover, the inequality
(40)∥fn+f∥E(X)+1≥∥∥fn(·)+f(·)∥+∥fn(·)∥∥E≥∥∥fn(·)+f(·)∥+∥fn(·)∥+∥f(·)∥∥E-1
holds for every n∈ℕ. It follows that
(41)limn→∞∥∥fn(·)+f(·)∥+∥fn(·)∥∥E=3.
Analogously one can see that
(42)limn→∞∥∥fn(·)+f(·)∥+∥f(·)∥∥E=3.
Finally, we have
(43)∥∥fn(·)+f(·)∥+∥fn(·)∥∥E+3≥∥∥fn(·)+f(·)∥+∥fn(·)∥+3∥f(·)∥∥E≥2∥∥fn(·)+f(·)∥+∥f(·)∥∥E.
Consequently
(44)limn→∞∥∥fn(·)+f(·)∥+∥fn(·)∥+3∥f(·)∥∥E=6.
Since E is in particular luacs we get from (35) and (38) that
(45)∫S‍∥F(t)∥∥f(t)∥dμ(t)=1.
Because E is in any case luacs+, it follows from (41), (44), and (45) that
(46)limn→∞∫S‍∥F(t)∥(∥fn(t)∥+∥fn(t)+f(t)∥)dμ(t)=3
and thus
(47)limn→∞∫S‍∥F(t)∥(∥fn(t)∥+∥f(t)∥-∥fn(t)+f(t)∥)dμ(t)=0.
So by passing to a further subsequence we may assume
(48)limn→∞∥F(t)∥(∥fn(t)∥+∥f(t)∥-∥fn(t)+f(t)∥)=0a.e.
Next we will show that
(49)F(t)(f(t))=∥F(t)∥∥f(t)∥a.e.
Since (S,𝒜,μ) is σ-finite there is an increasing sequence (Am)m∈ℕ in 𝒜 such that μ(Am)<∞ for every m∈ℕ and ⋃m=1∞‍Am=S.
Denote byN1 and N2 the null sets on which the convergence statement from (36) and (48), respectively does not hold, and let N=N1∪N2. Put B={t∈S∖N:F(t)≠0andf(t)≠0} and C={t∈B:fn(t)→0}. We will see that C is a null set.
First we define for everym∈ℕ a function am:S→ℝ by setting am(t)=∥F(t)∥ for t∈S∖(C∩Am) and am(t)=0 for t∈C∩Am. Note that each am is measurable and since |am(t)|≤∥F(t)∥ for every t∈S we have am∈BE′.
We havelimk→∞∥F(t)∥∥fk(t)∥χC∩Am(t)=0 for every t∈S and every m∈ℕ, so by Egorov’s theorem we can find for every m∈ℕ an increasing sequence (Bn,m)n∈ℕ in 𝒜|Am with μ(Am∖Bn,m)≤1/n such that (∥F(·)∥∥fk(·)∥χC∩Am)k∈ℕ converges uniformly to zero on each Bn,m.
It follows thatMm:=⋂n=1∞‍Am∖Bn,m is a null set for every m∈ℕ.
Let us now first suppose that (b) holds, soE′ is order continuous. We have
(50)limn→∞∥F(t)∥χC∩(Am∖Bn,m)(t)=0∀t∈S∖Mm
and moreover this sequence is decreasing, so the order continuity of E′ implies
(51)limn→∞∥∥F(·)∥χC∩(Am∖Bn,m)∥E′=0.
So if m∈ℕ and ε>0 are given, we can find an index n∈ℕ such that ∥∥F(·)∥χC∩(Am∖Bn,m)∥E′≤ε, and then, by uniform convergence, an index k0∈ℕ such that ∥F(t)∥∥fk(t)∥χC∩Bn,m(t)≤εμ(Am)-1 for every t∈S and every k≥k0.
Then we have(52)∫C∩Am‍∥F(t)∥∥fk(t)∥dμ(t)=∫C∩Bn,m‍∥F(t)∥∥fk(t)∥dμ(t)+∫C∩(Am∖Bn,m)‍∥F(t)∥∥fk(t)∥dμ(t)≤∫C∩Bn,m‍εμ(Am)dμ(t)+∥∥F(·)∥χC∩(Am∖Bn,m)∥E′≤2ε
for each k≥k0.
In conclusion we have(+)limk→∞∫C∩Am‍∥F(t)∥∥fk(t)∥dμ(t)=0∀m∈ℕ.
Now if (a) holds, that is, if E is WLUR, then by (38) the sequence (∥fk(·)∥)k∈ℕ must be weakly convergent to ∥f(·)∥ in E, and hence
(53)limk→∞∫C∩(Am∖Bn,m)‍∥F(t)∥∥fk(t)∥dμ(t)=∫C∩(Am∖Bn,m)‍∥F(t)∥∥f(t)∥dμ(t)
for all n,m∈ℕ. Since (∥f(·)∥χC∩(Am∖Bn,m))n∈ℕ decreases to zero a.e., the order continuity of E gives us limn→∞∥∥f(·)∥χC∩(Am∖Bn,m)∥E=0 for every m∈ℕ.
A similiar argument as before now easily yields that(+) also holds in case (a). But (+) is nothing else than
(54)limn→∞∫S‍(∥F(t)∥-am(t))∥fn(t)∥dμ(t)=0∀m∈ℕ.
Combining this with (35) leaves us with
(55)limn→∞∫S‍am(t)∥fn(t)∥dμ(t)=1∀m∈ℕ.
Since E is luacs and because of (38), it follows that
(56)∫S‍am(t)∥f(t)∥dμ(t)=1∀m∈ℕ.
Taking into account (45) we get
(57)∫S‍(∥F(t)∥-am(t))∥f(t)∥dμ(t)=0∀m∈ℕ,
and hence for every m∈ℕ we have (∥F(t)∥-am(t))∥f(t)∥=0 a.e. Consequently, C∩Am is a null set for every m, and thus C=⋃m=1∞‍C∩Am is also a null set.
Now suppose thatt∈(S∖C)∩B. Then we have F(t)≠0, f(t)≠0 and ∥fn(t)∥↛0, as well as ∥F(t)∥∥fn(t)∥-F(t)(fn(t))→0 and
(58)limn→∞(∥fn(t)∥+∥f(t)∥-∥fn(t)+f(t)∥)=0.
By passing to a subsequence we may assume that (∥fn(t)∥)n∈ℕ is bounded away from zero. Then it follows from Lemma 5 that
(59)limn→∞∥fn(t)∥fn(t)∥+f(t)∥f(t)∥∥=2.
Also, we have
(60)limn→∞F(t)∥F(t)∥(fn(t)∥fn(t)∥)=1.
Since X is luacs we can conclude that F(t)(f(t))=∥F(t)∥∥f(t)∥.
SoM:=N∪C is a null set with F(t)(f(t))=∥F(t)∥∥f(t)∥ for every t∈S∖M and (49) is proved.
From (45) and (49) it follows that
(61)l(f)=∫S‍F(t)(f(t))dμ(t)=1
and we are done.Recall that a subsetA⊆L1(μ) is said to be equi-integrable if for every ε>0 there is some δ>0 such that
(62)B∈𝒜withμ(B)≤δ⟹|∫B‍fdμ|≤ε∀f∈A.
It is well known that for a finite measure μ a bounded subset A⊆L1(μ) is relatively weakly compact in L1(μ) if and only if A is equi-integrable (see, e.g., [19, Theorem 13.6]). One ingredient for the usual proof of this fact is the following lemma (see [19, Proposition 13.4]), which we will also need in the sequel.Lemma 11.
For a finite measure space(S,𝒜,μ), a sequence (fn)n∈ℕ in L1(μ) is equi-integrable whenever the sequence (∫B‍fndμ)n∈ℕ is convergent for each B∈𝒜.We will also need Vitali’s lemma, which reads as follows (see, e.g., [11, Lemma  3.1.13] for an even more general version).Lemma 12.
Let(S,𝒜,μ) be a finite measure space, and let (fn)n∈ℕ be a sequence in L1(μ) such that {|fn|:n∈ℕ} is equi-integrable. Let f be a measurable function on S such that fn(t)→f(t)μ-a.e. Then f∈L1(μ) and ∥fn-f∥1→0.Finally, let us recall that a Banach spaceX is said to have the Kadets-Klee property (also known as property (H)) if for every sequence (xn)n∈ℕ in X and each x∈X the implication
(63)xn⟶σx,∥xn∥⟶∥x∥⟹∥xn-x∥⟶0
holds. For example, every LUR space and every dual of a reflexive, FS space has the Kadets-Klee property.It is known that every Banach lattice with the Kadets-Klee property is order continuous, (cf. [12, page 28]). With this in mind we can prove the following result concerning luacs+ spaces.Theorem 13.
If the measureμ is finite and E is LUR, then E(X) is a luacs+ space whenever X is luacs+. If in addition E′ is order continuous then the assertion also holds if μ is merely σ-finite.Proof.
By the previous theorem,E(X) is luacs, so we only have to show the implication “⇐” in Definition 6 (i). To this end, let (fn)n∈ℕ be a sequence in SE(X) and f∈SE(X) such that ∥fn+f∥E(X)→2, and let l∈SE(X)* such that l(f)=1. It will be enough to show that a subsequence of (l(fn))n∈ℕ converges to one.
SinceE is order continuous, we can as before represent l by some [F]∈E′(X*,w*) and conclude
(64)∫S‍∥F(t)∥∥f(t)∥dμ(t)=1,(65)∥F(t)∥∥f(t)∥=F(t)(f(t))a.e.
Also, just as we have done in the previous proof, we find that
(66)limn→∞∥∥fn(·)∥+∥f(·)∥∥E=2,limn→∞∥∥fn(·)+f(·)∥+∥fn(·)∥+∥f(·)∥∥E=4,limn→∞∥∥fn(·)+f(·)∥+∥fn(·)∥∥E=3,limn→∞∥∥fn(·)+f(·)∥+∥f(·)∥∥E=3,limn→∞∥∥fn(·)+f(·)∥+∥fn(·)∥+3∥f(·)∥∥E=6.
Since E is LUR, it follows that
(67)limn→∞∥∥fn(·)∥-∥f(·)∥∥E=0,(68)limn→∞∥∥fn(·)+f(·)∥+∥fn(·)∥-3∥f(·)∥∥E=0.
Hence, by passing to a subsequence, we may assume that (cf. Lemma 7)
(69)limn→∞∥fn(t)∥=∥f(t)∥a.e.,limn→∞∥fn(t)+f(t)∥=2∥f(t)∥a.e.
By (67) and (64) we also have
(70)limn→∞∫S‍∥F(t)∥∥fn(t)∥dμ(t)=1.
Since X is luacs+, it follows from (65), (69) that
(71)limn→∞(∥F(t)∥∥fn(t)∥-F(t)(fn(t)))=0a.e.
From (67) we also get
(72)limn→∞∫A‍∥F(t)∥∥fn(t)∥dμ(t)=∫A‍∥F(t)∥∥f(t)∥dμ(t)∀A∈𝒜.
Thus by Lemma 11 the sequence (∥F(·)∥∥fn(·)∥χB)n∈ℕ and hence also the sequence (∥(F(·)∥∥fn(·)∥-F(·)(fn(·)))χB)n∈ℕ are equi-integrable with respect to (B,𝒜|B,μ𝒜|B) for every B∈𝒜 with μ(B)<∞. This combined with Vitali’s lemma and (71) implies
(73)Limn→∞∫B‍(∥F(t)∥∥fn(t)∥-F(t)(fn(t)))dμ(t)=0∀B∈𝒜withμ(B)<∞.
So if μ(S)<∞, we immediately get
(74)l(fn)=∫S‍F(t)(fn(t))dμ(t)⟶1,
because of (70).
Ifμ is merely σ-finite but E′ is order continuous, we can fix an increasing sequence (Am)m∈ℕ in 𝒜 such that ⋃m=1∞‍Am=S and μ(Am)<∞ for every m∈ℕ. Then the sequence (∥F(·)∥χS∖Am)m∈ℕ decreases pointwise to zero, and, by the order continuity of E′, we can conclude that ∥∥F(·)∥χS∖Am∥E′→0.
Thus given anyε>0, we find an m0∈ℕ such that ∥∥F(·)∥χS∖Am0∥E′≤ε/3. Since μ(Am0)<∞, there exists N∈ℕ such that
(75)∫Am0‍(∥F(t)∥∥fn(t)∥-F(t)(fn(t)))dμ(t)≤ε3∀n≥N.
It follows that for every n≥N(76)∫S‍(∥F(t)∥∥fn(t)∥-F(t)(fn(t)))dμ(t)≤ε3+2∫S∖Am0‍∥F(t)∥∥fn(t)∥dμ(t)≤ε3+2∥∥F(·)∥χS∖Am0∥E′≤ε.
So we have
(77)limn→∞∫S‍(∥F(t)∥∥fn(t)∥-F(t)(fn(t)))dμ(t)=0,
and because of (70) it follows as before that
(78)l(fn)=∫S‍F(t)(fn(t))dμ(t)⟶1,
finishing the proof.Now we turn to the sluacs spaces. An easy normalisation argument shows that a Banach spaceX is sluacs if and only if for every x∈SX, every sequence (xn*)n∈ℕ in SX*, and all sequences (xn)n∈ℕ in X with ∥xn+x∥→2, ∥xn∥→1 and xn*(xn)→1, we have xn*(x)→1. In view of this characterisation, X is sluacs if and only if for every x∈SX and every 0<ε≤2 the number
(79)βX(x,ε):=inf{max{1-∥x+y2∥,|∥y∥-1|,|x*(y)-1|∥x+y2∥}:(y,x*)∈Vx,ε}
is strictly positive, where
(80)Vx,ε:={(y,x*)∈X×SX*:x*(y-x)≥ε}.
Next we will prove an easy lemma on the continuity of βX.Lemma 14.
For all0<ε, ε~,≤2 and all x,x~∈SX, one has
(81)|βX(x,ε)-βX(x~,ε~)|≤∥x-x~∥+|ε-ε~|;
that is βX is 1-Lipschitz continuous with respect to the norm of X⊕1ℝ.Proof.
First we fix0<ε≤2 and x,x~∈SX. Put δ=∥x-x~∥ and take y∈X, x*∈SX* such that x*(y-x)≥ε. It follows that x*(y-x~)≥ε-δ.
Now let0<τ<1 be arbitrary. We can find z∈SX with x*(z)≥1-τ. Define y~=y+δ(1-τ)-1z. Then
(82)x*(y~-x~)=δ1-τx*(z)+x*(y-x~)≥δ+x*(y-x~)=ε
and hence
(83)max{1-∥x~+y~2∥,|∥y~∥-1|,|x*(y~)-1|}≥βX(x~,ε).
But we have |∥y~∥-∥y∥|≤∥y-y~∥=δ(1-τ)-1 and |x*(y~)-x*(y)|≤∥y-y~∥=δ(1-τ)-1 as well as
(84)|∥x+y2∥-∥x~+y~2∥|≤12(∥x-x~∥+∥y-y~∥)=12(δ+δ1-τ)≤δ1-τ.
Thus we get
(85)max{1-∥x+y2∥,|∥y∥-1|,|x*(y)-1|}≥βX(x~,ε)-δ1-τ,
and since 0<τ<1 was arbitrary, it follows that
(86)max{1-∥x+y2∥,|∥y∥-1|,|x*(y)-1|}≥βX(x~,ε)-δ.
Again, since (y,x*)∈Vx,ε was arbitrary we can conclude that
(87)βX(x~,ε)-βX(x,ε)≤δ=∥x-x~∥
and by symmetry it follows that
(88)|βX(x~,ε)-βX(x,ε)|≤∥x-x~∥.
Analogously one can prove that
(89)|βX(x,ε~)-βX(x,ε)|≤|ε-ε~|
for all x∈SX and all 0<ε, ε~,≤2. An application of the triangle inequality then yields the result.In the paper [16] Kamińska and Turett proved various theorems concerning different rotundity properties of Köthe-Bochner spaces. For example, by [16, Theorem 5] if E has the so-called Fatou property and is LUR, then E(X) is LUR whenever X is LUR. We will adopt the technique of proof from [16, Theorem 5] to show the following result.Theorem 15.
IfE is LUR and X is sluacs smooth, then E(X) is also sluacs.Proof.
SinceE is LUR it is order continuous.
Let0<ε≤2 and f∈SE(X) be arbitrary and let
(90)An:={t∈S:f(t)≠0,βX(f(t)∥f(t)∥,ε8)≥1n}
for every n∈ℕ. Since by the previous lemma βX(·,ε/8) is continuous, it follows that the sets An are measurable. Also, the sequence (An)n∈ℕ is increasing and because X is sluacs, we have ⋃n=1∞‍An={t∈S:f(t)≠0}; hence (∥f(·)∥χS∖An)n∈ℕ decreases pointwise to zero. The order continuity of E implies ∥∥f(·)∥χS∖An∥E→0, and thus we can find n0∈ℕ with
(91)∥∥f(·)∥χS∖An0∥E≤ε64.
Now let us take g∈SE(X) and l∈SE(X)* with l(g)=1 and l(f)≤1-ε. Let l be represented by [F]∈E′(X*,w*). As in the proof of Proposition 9 we can conclude that
(92)∫S‍∥F(t)∥∥g(t)∥dμ(t)=1,(93)∥F(t)∥∥g(t)∥=F(t)(g(t))a.e.
Next we define
(94)C:={t∈S:F(t)≠0},B≔{ε4t∈C:F(t)(g(t)-f(t))≥ε4∥F(t)∥max{∥f(t)∥,∥g(t)∥}}.
Then B is measurable, and
(95)∫S∖B‍F(t)(g(t)-f(t))dμ(t)≤ε4∫S∖B‍∥F(t)∥max{∥f(t)∥,∥g(t)∥}dμ(t)≤ε4∫S∖B‍∥F(t)∥(∥f(t)∥+∥g(t)∥)dμ(t)≤ε42=ε2.
Since l(g-f)≥ε, it follows that
(96)∫B‍F(t)(g(t)-f(t))dμ(t)≥ε2.
Let us fix 0<η<min{ε/16,1/2n0} such that
(97)η1-η<2n0.
Now consider the following sets:
(98)B1:={t∈B:∥g(t)∥<(1-η)∥f(t)∥},B2:={t∈B:(1-η)∥f(t)∥≤∥g(t)∥≤∥f(t)∥},B3:={t∈B:(1-η)∥g(t)∥≤∥f(t)∥<∥g(t)∥},B4:={t∈B:(1-η)∥g(t)∥>∥f(t)∥}.
Then B1,…,B4 are measurable, pairwise disjoint, and ⋃i=14‍Bi=B. Thus by (96) there exists some i∈{1,…,4} such that
(99)∫Bi‍F(t)(g(t)-f(t))dμ(t)≥ε8.
If i=1, then, since ∥g(t)∥≤∥f(t)∥ for t∈B1, it follows that
(100)∫B1‍∥F(t)∥∥f(t)∥dμ(t)≥ε16,
and again by definition of B1 we obtain
(101)∥∥g(·)∥-∥f(·)∥∥E=∥∥g(·)∥-∥f(·)∥∥E≥∫B1‍∥F(t)∥(∥f(t)∥-∥g(t)∥)dμ(t)≥η∫B1‍∥F(t)∥∥f(t)∥dμ(t)≥ηε16
and hence
(102)∥f+g2∥E(X)≤∥∥f(·)∥+∥g(·)∥2∥E≤1-δE(∥f(·)∥,ηε16).
In the case of i=4 one can obtain the same statement by an analogous argument. To treat the remaining cases we need some preliminary considerations.
Let us denote byN the null set on which the equality from (93) does not hold and suppose that t∈B2∩An0∩(S∖N). Then in particular t∈B and ∥f(t)∥≥∥g(t)∥, and hence
(103)F(t)∥F(t)∥(g(t)∥f(t)∥-f(t)∥f(t)∥)≥ε4.
Moreover, by the definitions of B2 and An0 and the choice of η we have
(104)|∥g(t)∥f(t)∥∥-1|=|∥g(t)∥∥f(t)∥-1|≤η<1n0≤βX(f(t)∥f(t)∥,ε8)≤βX(f(t)∥f(t)∥,ε4).
Since t∈(S∖N), we also have
(105)|F(t)∥F(t)∥(g(t)∥f(t)∥)-1|=|∥g(t)∥∥f(t)∥-1|<βX(f(t)∥f(t)∥,ε4).
So by definition of βX we must have
(106)12∥f(t)∥f(t)∥+g(t)∥f(t)∥∥≤1-βX(f(t)∥f(t)∥,ε4)≤1-1n0.
Once more by the definition of B1 this implies
(107)∥f(t)+g(t)2∥≤(1-1n0)∥f(t)∥≤1-1/n02(1-η)(∥f(t)∥+∥g(t)∥)=12(1-α1)(∥f(t)∥+∥g(t)∥),
where α1:=(1/n0-η)(1-η)-1>0.
Now suppose thatt∈B3∩An0∩(S∖N). Then
(108)F(t)∥F(t)∥(g(t)∥g(t)∥-f(t)∥g(t)∥)≥ε4.
Consequently
(109)F(t)∥F(t)∥(g(t)∥g(t)∥-f(t)∥f(t)∥)≥ε4+F(t)∥F(t)∥(f(t)∥g(t)∥-f(t)∥f(t)∥)≥ε4-∥f(t)∥g(t)∥-f(t)∥f(t)∥∥=ε4-|∥f(t)∥∥g(t)∥-1|≥ε4-η≥ε8.
Since ∥F(t)∥∥g(t)∥=F(t)(g(t)), the definition of βX implies that
(110)12∥f(t)∥f(t)∥+g(t)∥g(t)∥∥≤1-βX(f(t)∥f(t)∥,ε8)≤1-1n0,
where the latter inequality holds because of t∈An0. It follows that
(111)12∥f(t)∥f(t)∥+g(t)∥f(t)∥∥≤1-1n0+12∥g(t)∥f(t)∥-g(t)∥g(t)∥∥=1-1n0+12|∥g(t)∥∥f(t)∥-1|≤1-1n0+12(11-η-1)=1-α2,
where α2:=1/n0-η(2-2η)-1 which by (97) is greater than zero. Because of ∥f(t)∥≤∥g(t)∥, it follows that
(112)∥f(t)+g(t)∥≤(1-α2)(∥f(t)∥+∥g(t)∥).
So if we put α=min{α1,α2} and P=B2∩An0∩(S∖N), Q=B3∩An0∩(S∖N) then
(113)∥f(t)+g(t)∥≤(1-α)(∥f(t)∥+∥g(t)∥),∀t∈P∪Q.
Now we will show that if i=2 and i=3, respectively then
(114)∫P‍∥F(t)∥∥f(t)∥dμ(t)≥ε64resp.∫Q‍∥F(t)∥∥f(t)∥dμ(t)≥ε64.
Let us first assume that i=2, that is,
(115)∫B2‍F(t)(g(t)-f(t))dμ(t)≥ε8.
Since ∥f(t)∥≥∥g(t)∥ for t∈B2 it follows that
(116)∫B2‍∥F(t)∥∥f(t)∥dμ(t)≥ε16.
Because N is a null set, we have
(117)∫P‍∥F(t)∥∥f(t)∥dμ(t)=∫B2∩An0‍∥F(t)∥∥f(t)∥dμ(t)=∫B2‍∥F(t)∥∥f(t)∥dμ(t)-∫B2∖An0‍∥F(t)∥∥f(t)∥dμ(t)≥ε16-∫S∖An0‍∥F(t)∥∥f(t)∥dμ(t)≥ε16-∥∥f(·)∥χS∖An0∥E≥ε16-ε64≥ε64,
where the second last inequality holds because of (91).
Now assume thati=3; that is,
(118)∫B3‍F(t)(g(t)-f(t))dμ(t)≥ε8.
It follows that
(119)ε8≤∫B3‍∥F(t)∥(∥g(t)∥+∥f(t)∥)dμ(t)≤∫B3‍∥F(t)∥(1+11-η)∥f(t)∥dμ(t)≤4∫B3‍∥F(t)∥∥f(t)∥dμ(t),
and hence as before we get
(120)∫Q‍∥F(t)∥∥f(t)∥dμ(t)≥ε32-ε64=ε64.
So if i=2, or i=3 then there is R∈{P,Q} such that
(121)∥∥f(·)∥χR∥E≥∫R‍∥F(t)∥∥f(t)∥dμ(t)≥ε64.
Put h=∥f(·)∥(1-2αχR). Then h∈BE, and moreover ∥∥f(·)∥-h∥E=2α∥∥f(·)∥χR∥E≥αε/32; hence
(122)∥∥f(·)∥(1-αχR)∥E=12∥∥f(·)∥+h∥E≤1-δE(∥f(·)∥,εα32).
We further have
(123)∥f+g2∥E(X)≤12∥(∥f(·)∥+∥g(·)∥)χS∖R+∥f(·)+g(·)∥χR∥E≤(113)12∥(∥f(·)∥+∥g(·)∥)χS∖R+(1-α)(∥f(·)∥+∥g(·)∥)χR∥E≤12∥∥g(·)∥+∥f(·)∥-α∥f(·)∥χR∥E≤12+12∥∥f(·)∥(1-αχR)∥E≤12+12(1-δE(∥f(·)∥,εα32))=1-12δE(∥f(·)∥,εα32).
Altogether we have shown that, for
(124)δ:=min{12δE(∥f(·)∥,εα32),δE(∥f(·)∥,εη16)}>0,
we have for every g∈SE(X) and every l∈SE(X)* with l(g)=1 and l(f)≤1-ε(125)∥f+g2∥E(X)≤1-δ.
By the aforementioned characterisation of sluacs spaces ([4, Proposition 2.1]), this implies that E(X) is sluacs.Next we will have a look at the case of wuacs spaces.Theorem 16.
Ifμ is a σ-finite measure and E is wuacs, reflexive, and has the Kadets-Klee property, then E(X) is wuacs whenever X is wuacs.Proof.
Note that sinceE is reflexive (or since it has the Kadets-Klee property), it is order continuous.
Let us take two sequences(fn)n∈ℕ and (gn)n∈ℕ in the unit sphere of E(X) such that ∥fn+gn∥E(X)→2 and a functional l∈SE(X)*, as usual represented by [F]∈E′(X*,w*), with l(fn)→1.
As in the proof of Theorem10 we find
(126)limn→∞∫S‍∥F(t)∥∥fn(t)∥dμ(t)=1,
and by passing to a subsequence also
(127)limn→∞(∥F(t)∥∥fn(t)∥-F(t)(fn(t)))=0a.e.
It is also easy to see that
(128)limn→∞∥∥fn(·)∥+∥gn(·)∥∥E=2,(129)limn→∞∥∥fn(·)∥+∥gn(·)∥+∥fn(·)+gn(·)∥∥E=4.
Since E is wuacs, it follows from (126) and (128) that
(130)limn→∞∫S‍∥F(t)∥∥gn(t)∥dμ(t)=1.
Again since E is wuacs and because of (126), (128), (129), and (130) we can deduce that
(131)limn→∞∫S‍∥F(t)∥(∥fn(t)∥+∥gn(t)∥-∥fn(t)+gn(t)∥)dμ(t)=0.
Hence we can pass to a further subsequence such that
(132)limn→∞∥F(t)∥(∥fn(t)∥+∥gn(t)∥-∥fn(t)+gn(t)∥)=0a.e.
By the reflexivity of E we can pass once more to a subsequence such that (∥fn(·)∥)n∈ℕ and (∥gn(·)∥)n∈ℕ are weakly convergent to h1∈BE and h2∈BE, respectively. In view of (126) and (130) it follows that
(133)∫S‍∥F(t)∥hi(t)dμ(t)=1∀i∈{1,2},
hence ∥h1∥E=∥h2∥E=1, and moreover
(134)∥h1+h2∥E=2.
The fact that E has the Kadets-Klee property implies that
(135)∥∥fn(·)∥-h1∥E⟶0,∥∥gn(·)∥-h2∥E⟶0,
and thus by Lemma 7 we can, for the last time, pass to a subsequence such that
(136)limn→∞∥fn(t)∥=h1(t),limn→∞∥gn(t)∥=h2(t)a.e.
Let N1, N2, and N3 denote the null sets on which the convergence statement from (127), (132), and (136), respectively, does not hold and put N=N1∪N2∪N3 as well as B={t∈S∖N:F(t)≠0andh2(t)≠0} and C={t∈B:h1(t)=0}.
Because of (134) and since E is in particular acs, we can show just as in the proof of Proposition 9 that C is a null set.
The fact thatX is wuacs together with Lemma 5 easily implies that
(137)limn→∞(∥F(t)∥∥gn(t)∥-F(t)(gn(t)))=0,∀t∈S∖(N∪C).
By the weak convergence of (∥gn(·)∥)n∈ℕ to h2 we have
(138)limn→∞∫A‍∥F(t)∥∥gn(t)∥dμ(t)=∫A‍∥F(t)∥h2(t)dμ(t)∀A∈𝒜.
Since E is reflexive, E′ is order continuous, and thus we can deduce as in the proof of Theorem 13, with the aid of Vitali’s lemma, (138), (137), and the fact that N∪C is a null set, that
(139)limn→∞∫S‍(∥F(t)∥∥gn(t)∥-F(t)(gn(t)))dμ(t)=0.
Because of (130), it follows that
(140)limn→∞l(gn)=limn→∞∫S‍F(t)(gn(t))dμ(t)=1,
and we are done.If we combine the techniques of the proofs of Theorems16 and 13, we can also obtain another result concerning luacs+ spaces (we omit the details).Theorem 17.
Ifμ is a σ-finite measure and E is luacs+, reflexive and has the Kadets-Klee property, then E(X) is luacs+ whenever X is luacs+.It is further possible to obtain another sufficient condition forE(X) to be sluacs.Theorem 18.
Ifμ is a σ-finite measure and E is sluacs+ and reflexive and both E and E* have the Kadets-Klee property, then E(X) is sluacs whenever X is sluacs.Proof.
Let(fn)n∈ℕ be a sequence in SE(X) and f∈SE(X) such that we have ∥fn+f∥E(X)→2. Also, let (ln)n∈ℕ be a sequence in SE(X)* such that ln(fn)→1. If we represent each ln by [Fn]∈E′(X*,w*), we can obtain as usual
(141)limn→∞∫S‍∥Fn(t)∥∥fn(t)∥dμ(t)=1,
and by passing to a subsequence also
(142)limn→∞(∥Fn(t)∥∥fn(t)∥-Fn(t)(fn(t)))=0a.e.
as well as
(143)limn→∞∥∥fn(·)∥+∥f(·)∥∥E=2,(144)limn→∞∥∥fn(·)+f(·)∥+∥fn(·)∥+∥f(·)∥∥E=4,(145)limn→∞∥∥fn(·)+f(·)∥+∥fn(·)∥∥E=3,(146)limn→∞∥∥fn(·)+f(·)∥+∥f(·)∥∥E=3,(147)limn→∞∥∥fn(·)+f(·)∥+∥fn(·)∥+3∥f(·)∥∥E=6.
Using the fact that E is sluacs+ we can conclude that
(148)limn→∞∫S‍∥Fn(t)∥∥f(t)∥dμ(t)=1,(149)limn→∞∫S‍∥Fn(t)∥(∥fn(t)∥+∥f(t)∥-∥fn(t)+f(t)∥)dμ(t)=0.
So we can pass to another subsequence such that
(150)limn→∞∥Fn(t)∥(∥fn(t)∥+∥f(t)∥-∥fn(t)+f(t)∥)=0a.e.
Since E (and hence also E*) is reflexive, we may assume without loss of generality that (∥fn(·)∥)n∈ℕ is weakly convergent to some h∈BE and that (∥Fn(·)∥)n∈ℕ is weakly convergent to some g∈BE*=BE′.
It follows from (148) that
(151)∫S‍g(t)∥f(t)∥dμ(t)=1
and hence g∈SE*. Because of (151), (143), and the fact that E is sluacs+, we get that
(152)limn→∞∫S‍g(t)∥fn(t)∥dμ(t)=1,
and consequently
(153)∫S‍g(t)h(t)dμ(t)=1,
whence h∈SE. Since both E and E* have the Kadets-Klee property, it follows that
(154)∥∥fn(·)∥-h∥E⟶0,∥∥Fn(·)∥-g∥E′⟶0.
Thus we can pass once more to a subsequence such that
(155)limn→∞∥fn(t)∥=h(t),limn→∞∥Fn(t)∥=g(t)a.e.
Combining (153) and (151) we also obtain
(156)∥h+∥f(·)∥∥E=2.
Let N be a null set such that the convergence statements of (142), (150), and (155) hold for every t∈S∖N, and put B={t∈S∖N:g(t)≠0andf(t)≠0} as well as C={t∈B:h(t)=0}.
Similar to the arguments in the proof of Theorem16 one can see that C is a null set and then, using the fact that X is sluacs, deduce that
(157)limn→∞(∥Fn(t)∥∥f(t)∥-Fn(t)(f(t)))=0a.e.
By our usual method based on Vitali’s lemma we can conclude that for every A∈𝒜 with μ(A)<∞ we have
(158)limn→∞∫A‍(∥Fn(t)∥∥f(t)∥-Fn(t)(f(t)))dμ(t)=0.
Now we fix an increasing sequence (Am)m∈ℕ in 𝒜 such that μ(Am)<∞ for all m∈ℕ and ⋃m=1∞‍Am=S. The order continuity of E implies ∥∥f(·)∥χS∖Am∥E→0. Analogous to the argument at the end of the proof of Theorem 13, this together with (158) leads to
(159)limn→∞∫S‍(∥Fn(t)∥∥f(t)∥-Fn(t)(f(t)))dμ(t)=0.
Taking into account (148) we arrive at
(160)limn→∞ln(f)=limn→∞∫S‍Fn(t)(f(t))dμ(t)=1,
and the proof is finished.Next we will consider sufficient conditions for a Köthe-Bochner space to besluacs+ (recall that a dual Banach space X* is said to have the Kadets-Klee* property if it fulfils the definition of the Kadets-Klee property with weak-replaced by weak*-convergence).Theorem 19.
LetE be a Köthe function space over the complete σ-finite measure space (S,𝒜,μ), and let X be an sluacs+ Banach space. If E* has the Kadets-Klee* property and in addition (a)
E is sluacs+, reflexive, and has the Kadets-Klee property or(b)
E is LUR and BE* is weak*-sequentially compact,then E(X) is sluacs+.Proof.
By Theorems15 and 18 we already know that E(X) is in both cases sluacs. Note also that in both cases E is order continuous. Now take a sequence (fn)n∈ℕ in SE(X) and f∈SE(X) such that ∥fn+f∥E(X)→2, and let (ln)n∈ℕ be a sequence in SE(X)* such that ln(f)→1. If we represent each ln by [Fn]∈E′(X*,w*) we can obtain as usual
(161)limn→∞∫S‍∥Fn(t)∥∥f(t)∥dμ(t)=1
and by passing to a subsequence also
(162)limn→∞(∥Fn(t)∥∥f(t)∥-Fn(t)(f(t)))=0a.e.
as well as
(163)limn→∞∥∥fn(·)∥+∥f(·)∥∥E=2,(164)limn→∞∥∥fn(·)+f(·)∥+∥fn(·)∥+∥f(·)∥∥E=4,(165)limn→∞∥∥fn(·)+f(·)∥+∥fn(·)∥∥E=3,(166)limn→∞∥∥fn(·)+f(·)∥+∥f(·)∥∥E=3,(167)limn→∞∥∥fn(·)+f(·)∥+∥fn(·)∥+3∥f(·)∥∥E=6.
Since E is sluacs+, it follows that
(168)limn→∞∫S‍∥Fn(t)∥∥fn(t)∥dμ(t)=1,(169)limn→∞∫S‍∥Fn(t)∥(∥fn(t)∥+∥f(t)∥-∥fn(t)+f(t)∥)dμ(t)=0,
so that by passing to another subsequence we can assume
(170)limn→∞∥Fn(t)∥(∥fn(t)∥+∥f(t)∥-∥fn(t)+f(t)∥)=0a.e.
In both cases (a) and (b) the dual unit ball BE* is weak*-sequentially compact, so that we can also assume the weak*-convergence of (∥Fn(·)∥)n∈ℕ to some g∈BE*. It follows from (161) that
(171)∫S‍g(t)∥f(t)∥dμ(t)=1,
and hence ∥g∥E′=1. Since E* has the Kadets-Klee* property, we get
(172)∥∥Fn(·)∥-g∥E′⟶0,
and thus we can, by passing to yet another subsequence, assume that
(173)limn→∞∥Fn(t)∥=g(t)a.e.
Next we claim that there is an h∈SE such that
(174)∫S‍g(t)h(t)dμ(t)=1,
and, after passing to a subsequence once more,
(175)∥∥fn(·)∥-h∥E⟶0.
In case (b) E is LUR, and thus by (163) and (171) we can take h=∥f(·)∥. In case (a) E is reflexive, and hence we can assume that (∥fn(·)∥)n∈ℕ is weakly convergent to some h∈BE. Then (174) follows from (172) and (168). This also implies ∥h∥E=1, and by the Kadets-Klee property of E we have (175).
By (175) we may assume that
(176)limhn→∞∥fn(t)∥=h(t)a.e.
Note that (171) and (174) imply that ∥∥f(·)∥+h∥E=2. Using all this and the fact that X is sluacs+ one can first prove, analogously to the arguments in the proof of Theorem 18, that
(177)limn→∞(∥Fn(t)∥∥fn(t)∥-Fn(t)(fn(t)))=0a.e.
and then
(178)limn→∞∫A‍(∥Fn(t)∥∥fn(t)∥-Fn(t)(fn(t)))dμ(t)=0,
for every A∈𝒜 with μ(A)<∞.
Let us now fix a sequence(Am)m∈ℕ in 𝒜 as in the proof of Theorem 18. The order continuity of E implies ∥∥f(·)∥χS∖Am∥E→0.
Letε>0 be arbitrary. Since E is sluacs+ there exists a δ>0 such that for all b∈SE and all l∈BE* with ∥b+∥f(·)∥∥E≥2(1-δ) and l(∥f(·)∥)≥1-δ one has l(b)≥1-ε.
Fixm0∈ℕ with ∥∥f(·)∥χS∖Am0∥E≤δ/2. Because of (163), (161) and (178) there is an N∈ℕ such that for all n≥N the inequalities
(179)∥∥fn(·)∥+∥f(·)∥∥E≥2(1-δ),∫S‍∥Fn(t)∥∥f(t)∥dμ(t)≥1-δ2,∫Am0‍(∥Fn(t)∥∥fn(t)∥-Fn(t)(fn(t)))dμ(t)≤ε
hold.
It follows that for everyn≥N we have
(180)|∫Am0‍∥Fn(t)∥∥f(t)∥dμ(t)-1|≤∥∥f(·)∥χS∖Am0∥E+|∫S‍∥Fn(t)∥∥f(t)∥dμ(t)-1|≤δ,
and hence by the choice of δ,
(181)∫Am0‍∥Fn(t)∥∥fn(t)∥dμ(t)≥1-ε.
Consequently, for every n≥N we have
(182)∫S‍(∥Fn(t)∥∥fn(t)∥-Fn(t)(fn(t)))dμ(t)≤ε+∫S∖Am0‍(∥Fn(t)∥∥fn(t)∥-Fn(t)(fn(t)))dμ(t)≤ε+2∫S∖Am0‍∥Fn(t)∥∥fn(t)∥dμ(t)≤ε+2(1-(1-ε))=3ε.
Thus we have shown
(183)limn→∞∫S‍(∥Fn(t)∥∥fn(t)∥-Fn(t)(fn(t)))dμ(t)=0.
Together with (168) it follows ln(fn)→1, as desired.Now let us treat the case of uacs spaces. In analogy to [4, Definition 3.15] we say that an order continuous Köthe function space E has property (u+) if for every ε>0 there is some δ>0 such that for all f,g∈SE and every h∈SE′ we have
(184)∥f+g∥E≥2(1-δ),∫S‍fhdμ=1⟹∫S‍|h||f-g|dμ≤ε.
This property certainly implies that E is uacs, while every UR space has property (u+), but the author does not know whether these implications are strict.The following theorem holds. Its proof is completely analogous to the one of [4, Theorem 3.16] (which is a modification of the proof of [20, Theorem 3]), but we will explicitly give it here, for the readers convenience.Theorem 20.
IfE is an order continuous Köthe function space with the property (u+) (in particular, if E is UR) and X is a uacs Banach space, then E(X) is also uacs.Proof.
Let0<ε≤2 be arbitrary. Since E is in particular uacs there is a number η>0 such that for all functions a,b∈BE and every functional l∈BE* with l(a)=1 one has
(185)l(b)<1-ε4δuacsX(ε2)⟹∥a+b∥E≤2(1-η).
Now let f,g∈SE(X) such that ∥f(t)∥=∥g(t)∥ a.e., and let L∈E(X)* such that L(f)=1 and L(g)<1-ε. We claim that ∥f+g∥E(X)≤2(1-η).
LetL be represented by [F]∈E′(X*,w*), and put β=∥g(·)∥, ν=∥F(·)∥. Define γ by γ(t)=ν(t)β(t)-F(t)(g(t)). Note that γ is measurable and
(186)0≤γ(t)≤2ν(t)β(t)∀t∈S.
As before we can deduce from L(f)=1 that
(187)∫S‍∥F(t)∥∥f(t)∥dμ(t)=1
and F(t)(f(t))=∥F(t)∥∥f(t)∥ a.e., hence
(188)F(t)(f(t))=ν(t)β(t)a.e.
Next we define
(189)α(t)={12δuacsX(γ(t)ν(t)β(t))if0<γ(t)<ν(t)β(t),0ifγ(t)=0,12δuacsX(1)otherwise.
Note that since δuacsX is continuous on (0,1) (see [21, Lemma 3.10] or [4, Lemma 2.11]), the function α is measurable. Using (188) it is easy to see that
(190)∥f(t)+g(t)∥≤2(1-α(t))β(t)a.e.
By (186) and (187) we have ∫S‍γ(t)dμ(t)≤2. Furthermore, we also have
(191)ε<1-L(g)=L(f-g)=∫S‍F(t)(f(t)-g(t))dμ(t)≤∫S‍γ(t)dμ(t);
thus
(192)ε<∫S‍γ(t)dμ(t)≤2.
Now put A={t∈S:2γ(t)>εν(t)β(t)} and B=S∖A. We then have (because of (187))
(193)∫B‍γ(t)dμ(t)≤ε2∫B‍ν(t)β(t)dμ(t)≤ε2∫S‍ν(t)β(t)dμ(t)=ε2.
Together with (192) it follows that
(194)∫A‍γ(t)dμ(t)>ε-ε2=ε2.
Taking into account (186) we get
(195)∫A‍ν(t)β(t)dμ(t)>ε4.
Next we define h=βχB and h′=βχA as well as h′′=(1-δuacsX(ε/2))h′. Then ∥h+h′′∥E≤∥h+h′∥E=∥β∥E=1. Let l be the functional on E represented by ν=∥F(·)∥. We have l(h+h′)=l(β)=1 (by (187)) and further, by (195),
(196)l(h+h′′)=1-δuacsX(ε2)l(h′)=1-∫A‍ν(t)β(t)dμ(t)<1-ε4δuacsX(ε2).
So by our choice of η we get ∥2h+h′+h′′∥E≤2(1-η); that is,
(197)∥h+(1-12δuacsX(ε2))h′∥E≤1-η.
By monotonicity of δuacsX we have
(198)α(t)≥12δuacsX(ε2)∀t∈A.
Using (190), (198), and (197) we obtain
(199)∥f+g∥E(X)=∥∥f(·)+g(·)∥∥E≤2∥(1-α)β∥E≤2∥(1-2-1δuacsX(ε2))h′+h∥E≤2(1-η).
The first step of the proof is completed. Next we wish to remove the restriction ∥f(·)∥=∥g(·)∥ a.e. So let again 0<ε≤2 be arbitrary, and choose η as above but corresponding to the value ε/2. Take 0<ω<2η/3.
SinceE is uacs, we may find τ>0 such that for all a,b∈BE and every l∈BE* we have
(200)l(a)≥1-τ,∥a+b∥E≥2(1-τ)⟹l(b)≥1-ω.
Next we fix 0<ρ<min{ε/2,2τ,ω} and find a number τ~ to the value ρ according to the definition of the property (u+) of E. Finally, let 0<ξ<min{τ,τ~}.
Letf,g∈SE(X) be arbitrary and L∈SE(X)* (as usually represented by F) such that L(f)=1 and ∥f+g∥E(X)≥2(1-ξ). We are going to prove that L(g)>1-ε, thus showing that E(X) is uacs.
To this end, we definez:S→X by
(201)z(t)={∥f(t)∥∥g(t)∥g(t)ifg(t)≠0,f(t)ifg(t)=0.
Then z is Bochner-measurable, and ∥z(t)∥=∥f(t)∥ for all t∈S (hence z∈E(X)). Furthermore,
(202)∥z(t)-g(t)∥=|∥f(t)∥-∥g(t)∥|∀t∈S.
As before we have
(203)∫S‍∥F(t)∥∥f(t)∥dμ(t)=1.
Also,
(204)2(1-τ~)≤2(1-ξ)≤∥f+g∥E(X)≤∥∥f(·)∥+∥g(·)∥∥E,
so the choice of τ~ together with (202) implies
(205)∫S‍∥F(t)∥∥z(t)-g(t)∥dμ(t)≤ρ.
Next we observe that
(206)∥∥f(·)∥+∥g(·)∥+∥f(·)+g(·)∥∥E≥2∥f+g∥E(X)≥4(1-ξ)≥4(1-τ),
and (because of (203) and (205))
(207)∫S‍∥F(t)∥(∥f(t)∥+∥g(t)∥)dμ(t)=1+∫S‍∥F(t)∥∥g(t)∥dμ(t)≥1+∫S‍∥F(t)∥∥f(t)∥dμ(t)-∫S‍∥F(t)∥|∥f(t)∥-∥g(t)∥|dμ(t)=2-∫S‍∥F(t)∥|∥f(t)∥-∥g(t)∥|dμ(t)≥2-ρ≥2(1-τ).
So (200) implies
(208)∫S‍∥F(t)∥∥f(t)+g(t)∥dμ(t)≥2(1-ω).
Using (205) and (208) we can conclude
(209)∥f+z∥E(X)≥∫S‍∥F(t)∥∥f(t)+z(t)∥dμ(t)≥∫S‍∥F(t)∥∥f(t)+g(t)∥dμ(t)-∫S‍∥F(t)∥∥g(t)-z(t)∥dμ(t)≥2(1-ω)-ρ>2(1-η).
By the choice of η this implies L(z)≥1-ε/2. But by (205) we also have |L(g)-L(z)|≤ρ; hence L(g)≥L(z)-ρ≥1-ε/2-ρ>1-ε.The above theorem admits the following corollary.Corollary 21.
IfE is a US Köthe function space and X is a uacs Banach space, and hence also E(X) is uacs.Proof.
Since uacs is a self-dual property (cf. [4, Corollary 2.13]) X* is also uacs, and since E is US, we have that E*=E′ is UR (cf. [2, Theorem 9.10]). So by the previous theorem E′(X*) is uacs. But as a uacs space, X* is reflexive, and hence it has the Radon-Nikodým property. It follows from the general theory in [13] that in this case E(X)* is isometrically isomorphic to E′(X*), so E(X)* and hence also E(X) is uacs.Finally, we consider some midpoint version of luacs and sluacs spaces. Let us first recall the following well-known notions: a Banach spaceX is said to be midpoint locally uniformly rotund (MLUR) if for any two sequences (xn)n∈ℕ and (yn)n∈ℕ in SX and every x∈SX we have
(210)∥x-xn+yn2∥⟶0⟹∥xn-yn∥⟶0.X is called weakly midpoint locally uniformly rotund (WMLUR) if it satisfies the above condition with ∥xn-yn∥→0 replaced by xn-yn→σ0, where the symbol →σ denotes the convergence in the weak topology of X. The notion of MLUR spaces was originally introduced by Anderson in [22].In [4] the author introduced the following analogous definitions.Definition 22.
LetX be a Banach space.(i)
The spaceX is said to be midpoint locally uniformly alternatively convex or smooth (mluacs) if for any two sequences (xn)n∈ℕ and (yn)n∈ℕ in SX, every x∈SX, and every x*∈SX*, one has
(211)∥x-xn+yn2∥⟶0,x*(xn)⟶1⟹x*(yn)⟶1.(ii)
The spaceX is called midpoint strongly locally uniformly alternatively convex or smooth (msluacs) if for any two sequences (xn)n∈ℕ and (yn)n∈ℕ in SX, every x∈SX and every sequence (xn*)n∈ℕ in SX*, one has
(212)∥x-xn+yn2∥⟶0,xn*(xn)⟶1⟹xn*(yn)⟶1.
Figure4 summarises the obvious implications. No other implications are true in general (see the examples in [4]).Figure 4Concerning the properties msluacs and mluacs for Köthe-Bochner spaces we have the following result.Theorem 23.
LetE be an MLUR Köthe function space over a complete σ-finite measure space and X a Banach space. If X is mluacs, then so is E(X). If X is msluacs and in addition E* has the Kadets-Klee* property and BE* is weak*-sequentially compact, then E(X) is also msluacs.Proof.
Let us first recall thatℓ∞ has no equivalent MLUR norm (cf. [11, Theorem 2.1.5]), and so by [11, Propositions 3.1.4 and 3.1.5] (and since every Köthe function space is σ-order complete), E must be order continuous.
Now let us assume thatX is msluacs and E* has the Kadets-Klee* property and weak*-sequentially compact unit ball. To show that E(X) is msluacs we will proceed in an analogous way to the proof of [4, Proposition 4.7], which in turn uses techniques from the proof of [23, Proposition 4].
So let us take two sequences(fn)n∈ℕ, (gn)n∈ℕ in SE(X) and f∈SE(X) such that ∥fn+gn-2f∥E(X)→0. Also, take a sequence (ln)n∈ℕ of norm-one functionals on E(X) such that ln(fn)→1. As usual, ln will be represented by [Fn]∈E′(X*,w*), and we conclude
(213)limn→∞∫S‍∥Fn(t)∥∥fn(t)∥dμ(t)=1,
and, after passing to an appropriate subsequence,
(214)limn→∞(∥Fn(t)∥∥fn(t)∥-Fn(t)(fn(t)))=0a.e.
We also have
(215)∥2∥f(·)∥-∥fn(·)+gn(·)∥∥E=∥|2∥f(·)∥-∥fn(·)+gn(·)∥|∥E≤∥∥2f(·)-fn(·)-gn(·)∥∥E=∥2f-fn-gn∥E(X);
hence
(216)∥2∥f(·)∥-∥fn(·)+gn(·)∥∥E⟶0.
As before we can also show
(217)∥∥fn(·)∥+∥gn(·)∥∥E⟶2.
Also, because of ∥fn+gn-2f∥E(X)→0, we may pass to a further subsequence such that
(218)limn→∞∥fn(t)+gn(t)-2f(t)∥=0a.e.
Let us define for every n∈ℕ(219)an(t):=2∥f(t)∥-12(∥fn(t)∥+∥gn(t)∥),bn(t):=∥f(t)∥-12∥fn(t)+gn(t)∥.
Note that
(220)∥f(t)∥≤bn(t)+12(∥fn(t)∥+∥gn(t)∥).
So if an(t)≥0, then
(221)|an(t)|=2∥f(t)∥-12(∥fn(t)∥+∥gn(t)∥)≤2|bn(t)|+12(∥fn(t)∥+∥gn(t)∥).
If an(t)<0, then
(222)|an(t)|=12(∥fn(t)∥+∥gn(t)∥)-2∥f(t)∥≤2|bn(t)|+12(∥fn(t)∥+∥gn(t)∥).
So we always have
(223)|an(t)|≤2|bn(t)|+12(∥fn(t)∥+∥gn(t)∥).
It follows that
(224)12∥∥fn(·)∥+∥gn(·)∥∥E+2∥bn∥E≥∥2|bn|+12(∥fn(·)∥+∥gn(·)∥)∥E≥∥an∥E≥2-12∥∥fn(·)∥+∥gn(·)∥∥E,
and we can conclude with (216) and (217) that ∥an∥E→1.
Using this together with (216), ∥fn(·)∥+∥gn(·)∥+2an=4∥f(·)∥, and the fact that E is MLUR we get
(225)limn→∞∥2∥f(·)∥-∥fn(·)∥-∥gn(·)∥∥E=0.
Again, since E is MLUR, this implies
(226)limn→∞∥∥fn(·)∥-∥gn(·)∥∥E=0.
Because of (225) and (226), we can pass to a further subsequence such that
(227)limn→∞∥fn(t)∥=∥f(t)∥=limn→∞∥gn(t)∥a.e.
Since BE* is weak*-sequentially compact we may also asssume that (∥Fn(·)∥)n∈ℕ weak*-converges to some g∈BE′.
Equations (225) and (226) imply ∥∥fn(·)∥-∥f(·)∥∥E→0. Together with (213) this gives us
(228)limn→∞∫S‍∥Fn(t)∥∥f(t)∥dμ(t)=1;
hence we also have
(229)∫S‍∥g(t)∥∥f(t)∥dμ(t)=1;
thus ∥g∥E′=1. Since E* has the Kadets-Klee* property, it follows that ∥∥Fn(·)∥-g∥E′→0, so if we pass again to a subsequence, we may assume
(230)limn→∞∥Fn(t)∥=g(t)a.e.
Now if we combine (214), (216), (227), and (230) we obtain
(231)limn→∞(∥Fn(t)∥∥f(t)∥-Fn(t)(f(t)))=0a.e.,
since X is msluacs.
Using our usual argument via equi-integrability and Vitali’s lemma this leads to(232)limn→∞∫A‍(∥Fn(t)∥∥f(t)∥-Fn(t)(f(t)))dμ(t)=0
for every A∈𝒜 with μ(A)<∞.
By the order continuity ofE we can derive from this
(233)limn→∞∫S‍(∥Fn(t)∥∥f(t)∥-Fn(t)(f(t)))dμ(t)=0
also in the σ-finite case (cf. the proof of Theorem 18).
Combining (233) and (228) gives us ln(f)→1, and we are done.
The statement about mluacs spaces can be proved similarly.We remark that the results proved in this section especially apply toLp spaces for 1<p<∞ (as we said before, for the properties acs/luacs/uacs this was already proved by Sirotkin in [5]).Corollary 24.
IfX is acs/luacs/luacs+/sluacs/sluacs+/mluacs/msluacs/wuacs/uacs, then for any complete, σ-finite measure space (S,𝒜,μ) and any 1<p<∞ the Lebesgue-Bochner space Lp(μ)(X) has the same property.In the last section we will establish some further connections between the various properties that we considered in this paper.


## 4. Miscellaneous

In [24] Lovaglia called a Banach space X weakly locally uniformly rotund if for every sequence (xn)n∈ℕ in SX, every x∈SX and each x*∈SX*, the implication
(234)∥xn+x∥⟶2,x*(x)=1⟹x*(xn)⟶1
holds. Since this notion of weak local uniform rotundity is strictly weaker than the notion of WLUR spaces that is nowadays commonly used, we will call such spaces WLUR in the sense of Lovaglia. (A dual Banach space will be called WLUR* in the sense of Lovaglia, if it fulfils Lovaglia’s definition for all evaluation functionals.) By definition, a Banach space is luacs+ if and only if it is luacs and WLUR in the sense of Lovaglia. Also, the following is valid.Proposition 25.
A Banach spaceX is luacs+ if and only if X is WLUR in the sense of Lovaglia and for all x*,y*∈SX* with ∥x*+y*∥=2 and every x∈SX with x*(x)=1 one also has y*(x)=1.Proof.
The necessity is clear because of [4, Proposition 2.16 (i)]. For the sufficiency we only have to prove that X is luacs, so let us take a sequence (xn)n∈ℕ in SX and x∈SX such that ∥xn+x∥→2 as well as x*∈SX* with x*(xn)→1. Since BX** is weak*-compact, we can find x**∈BX** and a subnet (xφ(i))i∈I which is weak*-convergent to x**. It follows that x**(x*)=1=∥x**∥.
Now fix a sequence(yn*)n∈ℕ in SX* such that yn*(xn+x)→2. Then yn*(xn)→1 and yn*(x)→1. There are y*∈BX* and a subnet (yψ(j)*)j∈J which is weak*-convergent to y*. It follows that y*(x)=1=∥y*∥. Since X is WLUR in the sense of Lovaglia, we conclude y*(xn)→1. It follows that x**(y*)=1=x**(x*); hence ∥x*+y*∥=2.
Because ofy*(x)=1, our assumption implies x*(x)=1 and we are done.The following assertion is also easy to prove (we omit the details).Proposition 26.
IfX is a Banach space which is WLUR in the sense of Lovaglia, and such that X* is WLUR* in the sense of Lovaglia then X is sluacs.Under additional assumptions on the spaceX it is possible to prove some more results.Proposition 27.
LetX be a reflexive Banach space. (i)
IfXis WLUR in the sense of Lovaglia, thenXisluacs+.(ii)
IfXis sluacs andluacs+,
thenXis wuacs.(iii)
IfXis wuacs andR, thenXis WLUR.Proof.
(i) follows directly from the Proposition25 and [4, Proposition 2.15]. Of the remaining assertions we will only prove (iii) explicitly.
Let(xn)n∈ℕ be a sequence in SX and x∈SX such that ∥xn+x∥→2. We can find a sequence (xn*)n∈ℕ in SX* such that xn*(xn+x)→2, and hence xn*(xn)→1 and xn*(x)→1.
SinceX is reflexive, we may assume that (xn*)n∈ℕ is weak*-convergent to some y*∈BX* and (xn)n∈ℕ is weakly convergent to some y∈BX. It follows that y*(x)=1 and hence ∥xn*+y*∥→2.
SinceX is wuacs, the dual space X* is sluacs (cf. [4, Proposition 2.16]), and thus (because of xn*(xn)→1) we can conclude that y*(xn)→1, whence y*(y)=1=y*(x), which implies ∥x+y∥=2, which by the rotundity of X implies x=y.Proposition 28.
LetX be a reflexive Banach space with the Kadets-Klee property. (i)
IfXis acs thenX, is luacs.(ii)
IfXis WLUR in the sense of Lovaglia, thenXis wuacs andsluacs+.(iii)
IfXis WLUR in the sense of Lovaglia andR, thenXis wuacs and LUR.Proof.
(i) Let(xn)n∈ℕ, x, and y be as in the proof of (iii) of the previous proposition and let x*∈SX* with x*(xn)→1. Then x*(y)=1 and hence ∥y∥=1. Since X has the Kadets-Klee property, it follows that ∥xn-y∥→0 and thus ∥x+y∥=2. Because X is acs, we obtain x*(x)=1, as desired.
(ii) We first show thatX is wuacs. Take two sequences (xn)n∈ℕ and (yn)n∈ℕ in SX such that ∥xn+yn∥→2 and a functional x*∈SX* with x*(xn)→1. By the reflexivity of X we may assume that (xn)n∈ℕ is weakly convergent to some x∈BX. Then x*(x)=1; hence ∥x∥=1.
ButX has the Kadets-Klee property, so this implies ∥xn-x∥→0.
Now fix a sequence(yn*)n∈ℕ in SX* such that yn*(xn)→1, and yn*(yn)→1. It follows that yn*(x)→1 and consequently ∥yn+x∥→2.
Sincex*(x)=1 and X is WLUR in the sense of Lovaglia, we get x*(yn)→1, proving that X is wuacs.
Now we will show thatX is sluacs. Take (xn)n∈ℕ and x in SX with ∥xn+x∥→2 and a sequence (xn*)n∈ℕ in SX* such that xn*(xn)→1. Also, fix a sequence (yn*)n∈ℕ in SX* with yn*(xn)→1 and yn*(x)→1.
We may assume that(xn)n∈ℕ is weakly convergent to some y∈BX and (yn*)n∈ℕ is weak*-convergent to some y*∈BX*. It follows that y*(x)=1, and hence ∥y*+yn*∥→2.
SinceX is wuacs, X* is sluacs, and thus we get y*(xn)→1. It follows that y*(y)=1; hence ∥y∥=1 and ∥x+y∥=2. The Kadets-Klee property of X gives us ∥xn-y∥→0.
Because ofxn*(xn)→1, we can now infer xn*(y)→1. Since X is in particular acs, this implies xn*(x)→1 (cf. [4, Proposition 2.19]).
We will skip the last part of the proof, the reverse implication in the definition ofsluacs+.
(iii) By (ii)X is wuacs and sluacs+. Let us take a sequence (xn)n∈ℕ in SX and an element x∈SX such that ∥xn+x∥→2. Fix a sequence (xn*)n∈ℕ in SX* such that xn*(xn)=1, for every n∈ℕ. Since X is sluacs, it follows that xn*(x)→1.
Assume that(xn)n∈ℕ is weakly convergent to y∈BX and that (xn*)n∈ℕ is weak*-convergent to x*∈BX*. It follows that x*(x)=1 and hence x*∈SX*. Moreover, since X is WLUR in the sense of Lovaglia, we get that x*(xn)→1.
Since(xn)n∈ℕ converges weakly to y this implies x*(y)=1 and hence ∥y∥=1. Now the Kadets-Klee property of X allows us to conclude ∥xn-y∥→0.
Because ofx*(x)=x*(y)=1, we must have ∥x+y∥=2 and thus the rotundity of X implies x=y.Proposition 29.
LetX be a Banach space such that X* has the Kadets-Klee* property and BX* is weak*-sequentially compact. (i)
IfXisS,
then it is also WLUR in the sense of Lovaglia.(ii)
IfX*is acs, thenXisluacs+,
and for all sequences(xn)n∈ℕinSX,(xn*)n∈ℕinSX*and everyx∈SXwith∥xn+x∥→2andxn*(x)→1one hasxn*(xn)→1.(iii)
IfX*is WLUR*in the sense of Lovaglia, thenXis sluacs.Proof.
We will only prove (iii), so let(xn)n∈ℕ and x be in SX with ∥xn+x∥→2 and (xn*)n∈ℕ a sequence in SX* such that xn*(xn)→1. Let (yn*)n∈ℕ be a sequence in SX* with yn*(xn)→1 and yn*(x)→1.
By assumption, we may suppose that(yn*)n∈ℕ is weak*-convergent to some y*∈BX*. Then y*(x)=1; hence y*∈SX*. By the Kadets-Klee* property of X* we must have ∥yn*-y*∥→0.
It follows thaty*(xn)→1; hence ∥xn*+y*∥→2. Since X* is WLUR* in the sense of Lovaglia, we obtain xn*(x)→1.



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