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

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 of 𝐿^𝑝-Bochner spaces.

1. Introduction

We begin with some notation and definitions. Throughout this paper, 𝑋 denotes a real Banach space, 𝑋^∗ its dual, 𝐵_𝑋 its unit ball and 𝑆_𝑋 its unit sphere.

In the next definition, we summarise the most important rotundity properties.

Definition 1. A Banach space 𝑋 is called(i)rotund (𝑅) if for any two elements 𝑥,𝑦∈𝑆_𝑋 the equality ‖𝑥+𝑦‖=2 implies 𝑥=𝑦, (ii)locally uniformly rotund (LUR) if for every 𝑥∈𝑆_𝑋, the implication 𝑥_𝑛+𝑥⟶2⟹𝑥_𝑛−𝑥⟶0(1) holds for every sequence (𝑥_𝑛)_𝑛∈ℕ in 𝑆_𝑋, (iii)weakly locally uniformly rotund (WLUR) if for every 𝑥∈𝑆_𝑋 and every sequence (𝑥_𝑛)_𝑛∈ℕ in 𝑆_𝑋 one has 𝑥_𝑛+𝑥⟶2⟹𝑥_𝑛⟶𝑥weakly,(2)(iv)uniformly rotund (UR) if for any two sequences (𝑥_𝑛)_𝑛∈ℕ and (𝑦_𝑛)_𝑛∈ℕ in 𝑆_𝑋 the implication 𝑥_𝑛+𝑦_𝑛⟶2⟹𝑥_𝑛−𝑦_𝑛⟶0(3) holds, (v)weakly uniformly rotund (WUR) if for any two sequences (𝑥_𝑛)_𝑛∈ℕ and (𝑦_𝑛)_𝑛∈ℕ the following implication holds: 𝑥_𝑛+𝑦_𝑛⟶2⟹𝑥_𝑛−𝑦_𝑛⟶0weakly.(4)

Figure 1 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 𝐵_𝑋.

Figure 1 

The modulus of convexity of the space 𝑋 is defined by 𝛿_𝑋(𝜀)1=inf1−2𝑥+𝑦∶𝑥,𝑦∈𝐵_𝑋,𝑥−𝑦≥𝜀(5) for every 𝜀 in the interval ]0,2]. Then 𝑋 is UR if and only if 𝛿_𝑋(𝜀)>0 for all 0<𝜀≤2.

For the local version one defines 𝛿_𝑋()1𝑥,𝜀=inf1−2𝑥+𝑦∶𝑦∈𝐵_𝑋,𝑥−𝑦≥𝜀(6) for every 𝑥∈𝑆_𝑋 and each 𝜀∈]0,2]. Then 𝑋 is LUR if and only if 𝛿_𝑋(𝑥,𝜀)>0 for all 𝑥∈𝑆_𝑋 and all 0<𝜀≤2.

Let us also recall some notions of smoothness. The space 𝑋 is called smooth (𝑆) if its norm is Gâteaux-differentiable at every nonzero point (equivalently at every point of 𝑆_𝑋), which is the case if and only if for every 𝑥∈𝑆_𝑋 there is a unique functional 𝑥^∗∈𝑆_𝑋^∗ 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∈𝑆_𝑋 the limit lim_𝜏→0(‖𝑥+𝜏𝑦‖−1)/𝜏 exists uniformly in 𝑥∈𝑆_𝑋. Finally, 𝑋 is called uniformly smooth (US) if lim_𝜏→0𝜌_𝑋(𝜏)/𝜏=0, where 𝜌_𝑋 denotes the modulus of smoothness of 𝑋 defined by 𝜌_𝑋(𝜏)=sup{1/2(‖𝑥+𝜏𝑦‖+‖𝑥−𝜏𝑦‖−2)∶𝑥,𝑦∈𝑆_𝑋} for every 𝜏>0.

In [3] the following notions were introduced (in connection with the so called Anti-Daugavet property).

Definition 2. A Banach space 𝑋 is called (i)alternatively convex or smooth (acs) if for every 𝑥,𝑦∈𝑆_𝑋 with ‖𝑥+𝑦‖=2 and every 𝑥^∗∈𝑆_𝑋^∗ with 𝑥^∗(𝑥)=1 one has 𝑥^∗(𝑦)=1 as well, (ii)locally uniformly alternatively convex or smooth (luacs) if for every 𝑥∈𝑆_𝑋, every sequence (𝑥_𝑛)_𝑛∈ℕ in 𝑆_𝑋 and every functional 𝑥^∗∈𝑆_𝑋^∗ one has 𝑥_𝑛+𝑥⟶2,𝑥^∗𝑥_𝑛⟶1⟹𝑥^∗(𝑥)=1,(7)(iii)uniformly alternatively convex or smooth (uacs) if for all sequences (𝑥_𝑛)_𝑛∈ℕ, (𝑦_𝑛)_𝑛∈ℕ in 𝑆_𝑋 and (𝑥^∗_𝑛)_𝑛∈ℕ in 𝑆_𝑋^∗, one has 𝑥_𝑛+𝑦_𝑛⟶2,𝑥^∗_𝑛𝑥_𝑛⟶1⟹𝑥^∗_𝑛𝑦_𝑛⟶1.(8) The author introduced the following related notions in [4].

Definition 3. A Banach space 𝑋 is called(i)strongly locally uniformly alternatively convex or smooth (sluacs) if for every 𝑥∈𝑆_𝑋 and all sequences (𝑥_𝑛)_𝑛∈ℕ in 𝑆_𝑋 and (𝑥^∗_𝑛)_𝑛∈ℕ in 𝑆_𝑋^∗, one has 𝑥_𝑛+𝑥⟶2,𝑥^∗_𝑛𝑥_𝑛⟶1⟹𝑥^∗_𝑛(𝑥)⟶1,(9)(ii)weakly uniformly alternatively convex or smooth (wuacs) if for any two sequences (𝑥_𝑛)_𝑛∈ℕ, (𝑦_𝑛)_𝑛∈ℕ in 𝑆_𝑋 and every functional 𝑥^∗∈𝑆_𝑋^∗, one has 𝑥_𝑛+𝑦_𝑛⟶2,𝑥^∗𝑥_𝑛⟶1⟹𝑥^∗𝑦_𝑛⟶1.(10)

The obvious implications between the acs properties and the rotundity properties are indicated in Figure 2. 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 2 

The connection between some of the acs properties to smoothness properties is illustrated in Figure 3.

Figure 3 

Let us mention that if we replace the condition 𝑥^∗_𝑛(𝑥_𝑛)→1 by 𝑥^∗_𝑛(𝑥_𝑛)=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 space 𝑋 one defines 𝐷_𝑋(𝜀)=𝑥,𝑦∈𝑆_𝑋×𝑆_𝑋∶∃𝑥^∗∈𝑆_𝑋^∗𝑥^∗(𝑥)𝑥=1,_∗𝑦,𝛿≤1−𝜀_𝑋uacs(𝜀)=inf1−𝑥+𝑦2∶𝑥,𝑦∈𝐷_𝑋(𝜀)∀𝜀∈]0,2].(11) Then 𝑋 is uacs if and only if 𝛿^𝑋_uacs(𝜀)>0 for every 𝜀∈]0,2] and one clearly has 𝛿_𝑋(𝜀)≤𝛿^𝑋_uacs(𝜀) for each 𝜀∈]0,2].

The above characterisation shows that the class of uacs spaces coincides with the class of  𝑢-spaces introduced by Lau in [6] and our modulus 𝛿^𝑋_uacs 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 space 𝑋 is said to be uniformly nonsquare if there is some 𝛿>0 such that for all 𝑥,𝑦∈𝐵_𝑋 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 (𝑥_𝑛)_𝑛∈ℕ and (𝑦_𝑛)_𝑛∈ℕ be sequences in the (real or complex) normed space 𝑋 such that ‖𝑥_𝑛+𝑦_𝑛‖−‖𝑥_𝑛‖−‖𝑦_𝑛‖→0.
Then for any two bounded sequences (𝛼_𝑛)_𝑛∈ℕ, (𝛽_𝑛)_𝑛∈ℕ of nonnegative real numbers one also has ‖𝛼_𝑛𝑥_𝑛+𝛽_𝑛𝑦_𝑛‖−𝛼_𝑛‖𝑥_𝑛‖−𝛽_𝑛‖𝑦_𝑛‖→0.

Finally, we will need two more definitions from [4].

Definition 6. A Banach space 𝑋 is called (i)a luacs⁺ space if for every 𝑥∈𝑆_𝑋, every sequence (𝑥_𝑛)_𝑛∈ℕ in 𝑆_𝑋 with ‖𝑥_𝑛+𝑥‖→2, and all 𝑥^∗∈𝑆_𝑋^∗ one has 𝑥^∗𝑥_𝑛⟶1⟺𝑥^∗(𝑥)=1,(12)(ii)a sluacs⁺ space if for every 𝑥∈𝑆_𝑋, every sequence (𝑥_𝑛)_𝑛∈ℕ in 𝑆_𝑋 with ‖𝑥_𝑛+𝑥‖→2, and all sequences (𝑥^∗_𝑛)_𝑛∈ℕ in 𝑆_𝑋^∗, one has 𝑥^∗_𝑛𝑥_𝑛⟶1⟺𝑥^∗_𝑛(𝑥)⟶1.(13) 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 𝜒_𝐴 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)𝜒_𝐴∈𝐸 for every 𝐴∈𝒜 with 𝜇(𝐴)<∞, (ii)for every 𝑓∈𝐸 and every set 𝐴∈𝒜 with 𝜇(𝐴)<∞𝑓 is 𝜇-integrable over 𝐴, (iii)if 𝑔 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 spaces 𝐿^𝑝(𝜇) for 1≤𝑝≤∞.

Every Köthe function space 𝐸 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 lattice 𝐸 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 space 𝐸 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. If 𝐸 is a Köthe function space, (𝑓_𝑛)_𝑛∈ℕ a sequence in 𝐸, and 𝑓∈𝐸 such that ‖𝑓_𝑛−𝑓‖_𝐸→0, then there is a subsequence of (𝑓_𝑛)_𝑛∈ℕ which converges pointwise almost everywhere to 𝑓.

For a Köthe function space 𝐸 we denote by 𝐸^′ the space of all measurable functions 𝑔∶𝑆→ℝ (modulo equality 𝜇-a.e.) such that 𝑔_𝐸′∶=sup_𝑆𝑓𝑔d𝜇∶𝑓∈𝐵_𝐸<∞.(14) Then (𝐸^′,‖⋅‖_𝐸′) is again a Köthe function space, the so called Köthe dual of 𝐸. The operator 𝑇∶𝐸^′→𝐸^∗ defined by 𝑇𝑔𝑓=_𝑆𝑓𝑔d𝜇∀𝑓∈𝐸,∀𝑔∈𝐸^′(15) 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 if 𝑋 is a Banach space, a function 𝑓∶𝑆→𝑋 is called simple if there are finitely many disjoint measurable sets 𝐴₁,…,𝐴_𝑛∈𝒜 such that 𝜇(𝐴_𝑖)<∞ for all 𝑖=1,…,𝑛, 𝑓 is constant on each 𝐴_𝑖 and 𝑓(𝑡)=0 for every 𝑡∈𝑆⧵⋃^𝑛_𝑖=1𝐴_𝑖. The function 𝑓 is said to be Bochner-measurable if there exists a sequence (𝑓_𝑛)_𝑛∈ℕ of simple functions such that lim_𝑛→∞‖𝑓_𝑛(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 space 𝐸 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 𝐿^𝑝(𝑋) for 1≤𝑝≤∞.

Next we recall how the dual of 𝐸(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 on 𝐸^′(𝑋^∗,𝑤^∗) can be defined by ‖[𝐹]‖_{𝐸′(𝑋∗,𝑤∗)}∶=inf𝑔_𝐸′∶𝑔∈𝐸^′,𝐹(𝑡)≤𝑔(𝑡)a.e..(16) Then the following deep theorem holds.

Theorem 8 (cf. [13]). Let 𝐸 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 𝑉([𝐹])𝑓∶=_𝑆𝐹(𝑡)𝑓(𝑡)d𝜇(𝑡),∀𝑓∈𝐸(𝑋),∀[𝐹]∈𝐸^′𝑋^∗,𝑤^∗(17) 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 𝐿^𝑝(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. If 𝐸 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∈𝑆_𝐸(𝑋) such that ‖𝑓+𝑔‖_𝐸(𝑋)=2 and a functional 𝑙∈𝑆_𝐸(𝑋)^∗ with 𝑙(𝑓)=1.
Since 𝐸 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 1=𝑙𝑓=_𝑆𝐹(𝑡)𝑓(𝑡)(𝑡)≤d𝜇_𝑆‖𝐹(𝑡)‖𝑓(𝑡)d𝜇(𝑡)≤‖‖𝐹(⋅)‖‖_𝐸′𝑓(⋅)_𝐸=‖𝑙‖𝑓_𝐸(𝑋)=1(18) and hence _𝑆‖𝐹(𝑡𝑓(𝑡(𝑡)𝐹(𝑡)𝑓(𝑡)=‖𝐹(𝑡𝑓(𝑡)‖)d𝜇=1,(19))‖)a.e.(20) We also have 2=𝑓+𝑔_𝐸(𝑋)=𝑓(⋅)+𝑔(⋅)_𝐸≤𝑓(⋅)+𝑔(⋅)_𝐸≤𝑓_𝐸(𝑋)+𝑔_𝐸(𝑋)=2(21) and thus 𝑓(⋅)+𝑔(⋅)_𝐸=2.(22) Since 𝐸 is acs it follows from (19) and (22) that _𝑆‖𝐹(𝑡)‖𝑔(𝑡)d𝜇(𝑡)=1.(23) In a similar way as we have obtained (22) we can also show 𝑓(⋅)+𝑔(⋅)+𝑓(⋅)+𝑔(⋅)_𝐸=4.(24) Because 𝐸 is acs this together with (19), (22) and (23) implies _𝑆‖𝐹(𝑡)‖𝑓(𝑡)+𝑔(𝑡)d𝜇(𝑡)=2.(25) From (19), (23) and (25) we get ‖𝐹(𝑡)‖𝑓(𝑡)+𝑔(𝑡)−𝑓(𝑡)+𝑔(𝑡)=0a.e.(26) Now we will show that 𝐹(𝑡)𝑔(𝑡)=‖𝐹(𝑡)‖𝑔(𝑡)a.e.(27) 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.
Put 𝐵={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, define ℎ∶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) _𝑆ℎ(𝑡)𝑓(𝑡)d𝜇(𝑡)=1(28) which also implies ‖ℎ‖_𝐸′=1. Together with (22) we now get _𝑆ℎ(𝑡)𝑔(𝑡)d𝜇(𝑡)=1,(29) since 𝐸 is acs. Taking into account (23) we arrive at (‖𝐹_𝑆(𝑡)‖−ℎ(𝑡))𝑔(𝑡)d𝜇(𝑡)=0.(30) Hence (‖𝐹(𝑡)‖−ℎ(𝑡))‖𝑔(𝑡)‖=0 a.e. and thus 𝐶 must be a null set.
Now if 𝑡∈(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 𝑓(𝑡)+𝑔(𝑡)=𝑓(𝑡)+𝑔(𝑡).(31) By [10, Lemma 2.1] this implies 𝑓(𝑡)𝑓(𝑡+𝑔(𝑡))𝑔(𝑡)=2.(32) Since 𝑋 is acs it follows that ‖𝐹(𝑡)‖‖𝑔(𝑡)‖=𝐹(𝑡)(𝑔(𝑡)).
So 𝑀∶=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 𝑙𝑔=_𝑆𝐹(𝑡)𝑔(𝑡)d𝜇(𝑡)=1,(33) 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 (𝑓_𝑛)_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 (𝑓_𝑛)_𝑛∈ℕ is uniformly convergent to zero on 𝐴.

Now we are ready to prove the following theorem.

Theorem 10. Let 𝐸 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 (𝑓_𝑛)_𝑛∈ℕ in 𝑆_𝐸(𝑋) and an element 𝑓∈𝑆_𝐸(𝑋) such that ‖𝑓_𝑛+𝑓‖_𝐸(𝑋)→2 as well as a functional 𝑙∈𝑆_𝐸(𝑋)^∗ such that 𝑙(𝑓_𝑛)→1. As before, we can represent 𝑙 by an element [𝐹]∈𝐸^′(𝑋^∗,𝑤^∗). We then have 𝑙𝑓_𝑛=_𝑆𝐹(𝑡)𝑓_𝑛(𝑡)d𝜇(𝑡)≤_𝑆‖𝐹(𝑡)‖𝑓_𝑛(𝑡)d𝜇(𝑡)≤1(34) and hence lim_{𝑛→∞𝑆}‖𝐹(𝑡)‖𝑓_𝑛(𝑡)d𝜇(𝑡)=1.(35) By passing to a subsequence we may also assume that lim_𝑛→∞‖𝐹(𝑡)‖𝑓_𝑛(𝑡)−𝐹(𝑡)𝑓_𝑛(𝑡)=0a.e.(36) We further have 𝑓_𝑛+𝑓_𝐸(𝑋)=𝑓_𝑛(⋅)+𝑓(⋅)_𝐸≤𝑓_𝑛(⋅)+𝑓(⋅)_𝐸≤2(37) and thus lim_𝑛→∞𝑓_𝑛(⋅)+𝑓(⋅)_𝐸=2.(38) An analogous argument also shows lim_𝑛→∞𝑓_𝑛(⋅)+𝑓(⋅)+𝑓_𝑛(⋅)+𝑓(⋅)_𝐸=4.(39) Moreover, the inequality 𝑓_𝑛+𝑓_𝐸(𝑋)≥𝑓+1_𝑛(⋅)+𝑓(⋅)+𝑓_𝑛(⋅)_𝐸≥𝑓_𝑛(⋅)+𝑓(⋅)+𝑓_𝑛(⋅)+𝑓(⋅)_𝐸−1(40) holds for every 𝑛∈ℕ. It follows that lim_𝑛→∞𝑓_𝑛(⋅)+𝑓(⋅)+𝑓_𝑛(⋅)_𝐸=3.(41) Analogously one can see that lim_𝑛→∞𝑓_𝑛(⋅)+𝑓(⋅)+𝑓(⋅)_𝐸=3.(42) Finally, we have 𝑓_𝑛(⋅)+𝑓(⋅)+𝑓_𝑛(⋅)_𝐸≥𝑓+3_𝑛(⋅)+𝑓(⋅)+𝑓_𝑛(⋅)+3𝑓(⋅)_𝐸≥2𝑓_𝑛(⋅)+𝑓(⋅)+𝑓(⋅)_𝐸.(43) Consequently lim_𝑛→∞𝑓_𝑛(⋅)+𝑓(⋅)+𝑓_𝑛(⋅)+3𝑓(⋅)_𝐸=6.(44) Since 𝐸 is in particular luacs we get from (35) and (38) that _𝑆‖𝐹(𝑡)‖𝑓(𝑡)d𝜇(𝑡)=1.(45) Because 𝐸 is in any case luacs⁺, it follows from (41), (44), and (45) that lim_{𝑛→∞𝑆}‖𝐹(𝑡)‖𝑓_𝑛(𝑡)+𝑓_𝑛(𝑡)+𝑓(𝑡)d𝜇(𝑡)=3(46) and thus lim_{𝑛→∞𝑆}‖𝐹(𝑡)‖𝑓_𝑛(𝑡+𝑓(𝑡−𝑓))_𝑛(𝑡)+𝑓(𝑡)d𝜇(𝑡)=0.(47) So by passing to a further subsequence we may assume lim_𝑛→∞‖𝐹(𝑡)‖𝑓_𝑛(𝑡)+𝑓(𝑡)−𝑓_𝑛(𝑡)+𝑓(𝑡)=0a.e.(48) Next we will show that 𝐹(𝑡)𝑓(𝑡)=‖𝐹(𝑡)‖𝑓(𝑡)a.e.(49) Since (𝑆,𝒜,𝜇) is 𝜎-finite there is an increasing sequence (𝐴_𝑚)_𝑚∈ℕ in 𝒜 such that 𝜇(𝐴_𝑚)<∞ for every 𝑚∈ℕ and ⋃^∞_𝑚=1𝐴_𝑚=𝑆.
Denote by 𝑁₁ 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∶𝑓_𝑛(𝑡)→0}. We will see that 𝐶 is a null set.
First we define for every 𝑚∈ℕ a function 𝑎_𝑚∶𝑆→ℝ by setting 𝑎_𝑚(𝑡)=‖𝐹(𝑡)‖ for 𝑡∈𝑆⧵(𝐶∩𝐴_𝑚) and 𝑎_𝑚(𝑡)=0 for 𝑡∈𝐶∩𝐴_𝑚. Note that each 𝑎_𝑚 is measurable and since |𝑎_𝑚(𝑡)|≤‖𝐹(𝑡)‖ for every 𝑡∈𝑆 we have 𝑎_𝑚∈𝐵_𝐸′.
We have lim_𝑘→∞‖𝐹(𝑡)‖‖𝑓_𝑘(𝑡)‖𝜒_{𝐶∩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 (𝐵_𝑛,𝑚)_𝑛∈ℕ in 𝒜|_𝐴𝑚 with 𝜇(𝐴_𝑚⧵𝐵_𝑛,𝑚)≤1/𝑛 such that (‖𝐹(⋅)‖‖𝑓_𝑘(⋅)‖𝜒_{𝐶∩𝐴𝑚})_𝑘∈ℕ converges uniformly to zero on each 𝐵_𝑛,𝑚.
It follows that 𝑀_𝑚∶=⋂^∞_𝑛=1𝐴_𝑚⧵𝐵_𝑛,𝑚 is a null set for every 𝑚∈ℕ.
Let us now first suppose that (b) holds, so 𝐸^′ is order continuous. We have lim_𝑛→∞‖𝐹(𝑡)‖𝜒_{𝐶∩(𝐴𝑚⧵𝐵𝑛,𝑚)}(𝑡)=0∀𝑡∈𝑆⧵𝑀_𝑚(50) and moreover this sequence is decreasing, so the order continuity of 𝐸^′ implies lim_𝑛→∞‖𝐹(⋅)‖𝜒_{𝐶∩(𝐴𝑚⧵𝐵𝑛,𝑚)𝐸′}=0.(51) So if 𝑚∈ℕ and 𝜀>0 are given, we can find an index 𝑛∈ℕ such that ‖‖𝐹(⋅)‖𝜒_{𝐶∩(𝐴𝑚⧵𝐵𝑛,𝑚)}‖_𝐸′≤𝜀, and then, by uniform convergence, an index 𝑘₀∈ℕ such that ‖𝐹(𝑡)‖‖𝑓_𝑘(𝑡)‖𝜒_{𝐶∩𝐵𝑛,𝑚}(𝑡)≤𝜀𝜇(𝐴_𝑚)⁻¹ for every 𝑡∈𝑆 and every 𝑘≥𝑘₀.
Then we have _{𝐶∩𝐴𝑚}‖𝐹(𝑡)‖𝑓_𝑘(𝑡(𝑡)=)d𝜇_{𝐶∩𝐵𝑛,𝑚}‖𝐹(𝑡)‖𝑓_𝑘(𝑡(𝑡)+)d𝜇_{𝐶∩(𝐴𝑚⧵𝐵𝑛,𝑚)}‖𝐹(𝑡)‖𝑓_𝑘(𝑡)d𝜇(𝑡)≤_{𝐶∩𝐵𝑛,𝑚}𝜀𝜇𝐴_𝑚d𝜇(𝑡)+‖𝐹(⋅)‖𝜒_{𝐶∩(𝐴𝑚⧵𝐵𝑛,𝑚)𝐸′}≤2𝜀(52) for each 𝑘≥𝑘₀.
In conclusion we have lim_{𝑘→∞𝐶∩𝐴𝑚}‖𝐹(𝑡)‖𝑓_𝑘(𝑡)d𝜇(𝑡)=0∀𝑚∈ℕ.(+) Now if (a) holds, that is, if 𝐸 is WLUR, then by (38) the sequence (‖𝑓_𝑘(⋅)‖)_𝑘∈ℕ must be weakly convergent to ‖𝑓(⋅)‖ in 𝐸, and hence lim_{𝑘→∞𝐶∩(𝐴𝑚⧵𝐵𝑛,𝑚)}‖𝐹(𝑡)‖𝑓_𝑘(𝑡(𝑡)=)d𝜇_{𝐶∩(𝐴𝑚⧵𝐵𝑛,𝑚)}‖𝐹(𝑡)‖𝑓(𝑡)d𝜇(𝑡)(53) for all 𝑛,𝑚∈ℕ. Since (‖𝑓(⋅)‖𝜒_{𝐶∩(𝐴𝑚⧵𝐵𝑛,𝑚)})_𝑛∈ℕ decreases to zero a.e., the order continuity of 𝐸 gives us lim_𝑛→∞‖‖𝑓(⋅)‖𝜒_{𝐶∩(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 lim_{𝑛→∞𝑆}‖𝐹(𝑡)‖−𝑎_𝑚(𝑡)𝑓_𝑛(𝑡)d𝜇(𝑡)=0∀𝑚∈ℕ.(54) Combining this with (35) leaves us with lim_{𝑛→∞𝑆}𝑎_𝑚(𝑡)𝑓_𝑛(𝑡)d𝜇(𝑡)=1∀𝑚∈ℕ.(55) Since 𝐸 is luacs and because of (38), it follows that _𝑆𝑎_𝑚(𝑡)𝑓(𝑡)d𝜇(𝑡)=1∀𝑚∈ℕ.(56) Taking into account (45) we get _𝑆‖𝐹(𝑡)‖−𝑎_𝑚(𝑡)𝑓(𝑡)d𝜇(𝑡)=0∀𝑚∈ℕ,(57) and hence for every 𝑚∈ℕ we have (‖𝐹(𝑡)‖−𝑎_𝑚(𝑡))‖𝑓(𝑡)‖=0 a.e. Consequently, 𝐶∩𝐴_𝑚 is a null set for every 𝑚, and thus 𝐶=⋃^∞_𝑚=1𝐶∩𝐴_𝑚 is also a null set.
Now suppose that 𝑡∈(𝑆⧵𝐶)∩𝐵. Then we have 𝐹(𝑡)≠0, 𝑓(𝑡)≠0 and ‖𝑓_𝑛(𝑡)‖↛0, as well as ‖𝐹(𝑡)‖‖𝑓_𝑛(𝑡)‖−𝐹(𝑡)(𝑓_𝑛(𝑡))→0 and lim_𝑛→∞𝑓_𝑛(𝑡)+𝑓(𝑡)−𝑓_𝑛(𝑡)+𝑓(𝑡)=0.(58) By passing to a subsequence we may assume that (‖𝑓_𝑛(𝑡)‖)_𝑛∈ℕ is bounded away from zero. Then it follows from Lemma 5 that lim_𝑛→∞𝑓_𝑛(𝑡)𝑓_𝑛(𝑡+𝑓(𝑡))𝑓(𝑡)=2.(59) Also, we have lim_𝑛→∞𝐹(𝑡)‖𝐹(𝑡𝑓)‖^𝑛(𝑡)𝑓_𝑛(𝑡)=1.(60) Since 𝑋 is luacs we can conclude that 𝐹(𝑡)(𝑓(𝑡))=‖𝐹(𝑡)‖‖𝑓(𝑡)‖.
So 𝑀∶=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 𝑙𝑓=_𝑆𝐹(𝑡)𝑓(𝑡)d𝜇(𝑡)=1(61) and we are done.

Recall that a subset 𝐴⊆𝐿¹(𝜇) is said to be equi-integrable if for every 𝜀>0 there is some 𝛿>0 such that 𝐵∈𝒜with𝜇(𝐵)≤𝛿⟹_𝐵𝑓d𝜇≤𝜀∀𝑓∈𝐴.(62) 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 (𝑓_𝑛)_𝑛∈ℕ in 𝐿¹(𝜇) is equi-integrable whenever the sequence (∫_𝐵𝑓_𝑛𝑑𝜇)_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 (𝑓_𝑛)_𝑛∈ℕ be a sequence in 𝐿¹(𝜇) such that {|𝑓_𝑛|∶𝑛∈ℕ} is equi-integrable. Let 𝑓 be a measurable function on 𝑆 such that 𝑓_𝑛(𝑡)→𝑓(𝑡)𝜇-a.e. Then 𝑓∈𝐿¹(𝜇) and ‖𝑓_𝑛−𝑓‖₁→0.

Finally, let us recall that a Banach space 𝑋 is said to have the Kadets-Klee property (also known as property (𝐻)) if for every sequence (𝑥_𝑛)_𝑛∈ℕ in 𝑋 and each 𝑥∈𝑋 the implication 𝑥⟶_𝑛^𝜎𝑥,𝑥_𝑛⟶‖𝑥‖⟹𝑥_𝑛−𝑥⟶0(63) 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 measure 𝜇 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 (𝑓_𝑛)_𝑛∈ℕ be a sequence in 𝑆_𝐸(𝑋) and 𝑓∈𝑆_𝐸(𝑋) such that ‖𝑓_𝑛+𝑓‖_𝐸(𝑋)→2, and let 𝑙∈𝑆_𝐸(𝑋)^∗ such that 𝑙(𝑓)=1. It will be enough to show that a subsequence of (𝑙(𝑓_𝑛))_𝑛∈ℕ converges to one.
Since 𝐸 is order continuous, we can as before represent 𝑙 by some [𝐹]∈𝐸^′(𝑋^∗,𝑤^∗) and conclude _𝑆‖𝐹(𝑡𝑓(𝑡(𝑡)‖𝐹(𝑡𝑓(𝑡(𝑡)𝑓(𝑡))‖)d𝜇=1,(64))‖)=𝐹a.e.(65) Also, just as we have done in the previous proof, we find that lim_𝑛→∞𝑓_𝑛(⋅)+𝑓(⋅)_𝐸=2,lim_𝑛→∞𝑓_𝑛(⋅)+𝑓(⋅)+𝑓_𝑛(⋅)+𝑓(⋅)_𝐸=4,lim_𝑛→∞𝑓_𝑛(⋅)+𝑓(⋅)+𝑓_𝑛(⋅)_𝐸=3,lim_𝑛→∞𝑓_𝑛(⋅)+𝑓(⋅)+𝑓(⋅)_𝐸=3,lim_𝑛→∞𝑓_𝑛(⋅)+𝑓(⋅)+𝑓_𝑛(⋅)+3𝑓(⋅)_𝐸=6.(66) Since 𝐸 is LUR, it follows that lim_𝑛→∞𝑓_𝑛(⋅)−𝑓(⋅)_𝐸=0,(67)lim_𝑛→∞𝑓_𝑛(⋅)+𝑓(⋅)+𝑓_𝑛(⋅)−3𝑓(⋅)_𝐸=0.(68) Hence, by passing to a subsequence, we may assume that (cf. Lemma 7) lim_𝑛→∞𝑓_𝑛(𝑡)=𝑓(𝑡)a.e.,lim_𝑛→∞𝑓_𝑛(𝑡)+𝑓(𝑡)=2𝑓(𝑡)a.e.(69) By (67) and (64) we also have lim_{𝑛→∞𝑆}‖𝐹(𝑡)‖𝑓_𝑛(𝑡)d𝜇(𝑡)=1.(70) Since 𝑋 is luacs⁺, it follows from (65), (69) that lim_𝑛→∞‖𝐹(𝑡)‖𝑓_𝑛(𝑡)−𝐹(𝑡)𝑓_𝑛(𝑡)=0a.e.(71) From (67) we also get lim_{𝑛→∞𝐴}‖𝐹(𝑡)‖𝑓_𝑛(𝑡(𝑡)=)d𝜇_𝐴‖𝐹(𝑡)‖𝑓(𝑡)d𝜇(𝑡)∀𝐴∈𝒜.(72) Thus by Lemma 11 the sequence (‖𝐹(⋅)‖‖𝑓_𝑛(⋅)‖𝜒_𝐵)_𝑛∈ℕ and hence also the sequence (‖(𝐹(⋅)‖‖𝑓_𝑛(⋅)‖−𝐹(⋅)(𝑓_𝑛(⋅)))𝜒_𝐵)_𝑛∈ℕ are equi-integrable with respect to (𝐵,𝒜|_𝐵,𝜇_𝒜|𝐵) for every 𝐵∈𝒜 with 𝜇(𝐵)<∞. This combined with Vitali’s lemma and (71) implies Lim_{𝑛→∞𝐵}‖𝐹(𝑡)‖𝑓_𝑛(𝑡)−𝐹(𝑡)𝑓_𝑛(𝑡)(𝑡)(𝐵)d𝜇=0∀𝐵∈𝒜with𝜇<∞.(73) So if 𝜇(𝑆)<∞, we immediately get 𝑙𝑓_𝑛=_𝑆𝐹(𝑡)𝑓_𝑛(𝑡)d𝜇(𝑡)⟶1,(74) because of (70).
If 𝜇 is merely 𝜎-finite but 𝐸^′ is order continuous, we can fix an increasing sequence (𝐴_𝑚)_𝑚∈ℕ in 𝒜 such that ⋃^∞_𝑚=1𝐴_𝑚=𝑆 and 𝜇(𝐴_𝑚)<∞ for every 𝑚∈ℕ. Then the sequence (‖𝐹(⋅)‖𝜒_{𝑆⧵𝐴𝑚})_𝑚∈ℕ decreases pointwise to zero, and, by the order continuity of 𝐸^′, we can conclude that ‖‖𝐹(⋅)‖𝜒_{𝑆⧵𝐴𝑚}‖_𝐸′→0.
Thus given any 𝜀>0, we find an 𝑚₀∈ℕ such that ‖‖𝐹(⋅)‖𝜒_{𝑆⧵𝐴𝑚0}‖_𝐸′≤𝜀/3. Since 𝜇(𝐴_𝑚0)<∞, there exists 𝑁∈ℕ such that _𝐴𝑚0‖𝐹(𝑡)‖𝑓_𝑛(𝑡)−𝐹(𝑡)𝑓_𝑛(𝑡)(𝑡)≤𝜀d𝜇3∀𝑛≥𝑁.(75) It follows that for every 𝑛≥𝑁_𝑆‖𝐹(𝑡)‖𝑓_𝑛(𝑡)−𝐹(𝑡)𝑓_𝑛(𝑡)(𝑡)≤𝜀d𝜇3+2_{𝑆⧵𝐴𝑚0}‖𝐹(𝑡)‖𝑓_𝑛(𝑡(𝑡)≤𝜀)d𝜇3+2‖𝐹(⋅)‖𝜒_{𝑆⧵𝐴𝑚0𝐸′}≤𝜀.(76) So we have lim_{𝑛→∞𝑆}‖𝐹(𝑡)‖𝑓_𝑛(𝑡)−𝐹(𝑡)𝑓_𝑛(𝑡)d𝜇(𝑡)=0,(77) and because of (70) it follows as before that 𝑙𝑓_𝑛=_𝑆𝐹(𝑡)𝑓_𝑛(𝑡)d𝜇(𝑡)⟶1,(78) finishing the proof.

Now we turn to the sluacs spaces. An easy normalisation argument shows that a Banach space 𝑋 is sluacs if and only if for every 𝑥∈𝑆_𝑋, every sequence (𝑥^∗_𝑛)_𝑛∈ℕ in 𝑆_𝑋^∗, and all sequences (𝑥_𝑛)_𝑛∈ℕ in 𝑋 with ‖𝑥_𝑛+𝑥‖→2, ‖𝑥_𝑛‖→1 and 𝑥^∗_𝑛(𝑥_𝑛)→1, we have 𝑥^∗_𝑛(𝑥)→1. In view of this characterisation, 𝑋 is sluacs if and only if for every 𝑥∈𝑆_𝑋 and every 0<𝜀≤2 the number 𝛽_𝑋(𝑥,𝜀)∶=infmax1−𝑥+𝑦2,𝑦,𝑥−1_∗𝑦−1∶𝑦,𝑥^∗∈𝑉_𝑥,𝜀(79) is strictly positive, where 𝑉_𝑥,𝜀∶=𝑦,𝑥^∗∈𝑋×𝑆_𝑋^∗∶𝑥^∗𝑦−𝑥≥𝜀.(80) Next we will prove an easy lemma on the continuity of 𝛽_𝑋.

Lemma 14. For all 0<𝜀, 𝜀,̃≤2 and all 𝑥,𝑥∈𝑆_𝑋, one has 𝛽_𝑋(𝑥,𝜀)−𝛽_𝑋(𝑥,𝜀)̃≤‖𝑥−𝑥‖+|𝜀−𝜀|̃;(81) that is 𝛽_𝑋 is 1-Lipschitz continuous with respect to the norm of 𝑋⊕₁ℝ.

Proof. First we fix 0<𝜀≤2 and 𝑥,𝑥∈𝑆_𝑋. Put 𝛿=‖𝑥−𝑥‖ and take 𝑦∈𝑋, 𝑥^∗∈𝑆_𝑋^∗ such that 𝑥^∗(𝑦−𝑥)≥𝜀. It follows that 𝑥^∗(𝑦−𝑥)≥𝜀−𝛿.
Now let 0<𝜏<1 be arbitrary. We can find 𝑧∈𝑆_𝑋 with 𝑥^∗(𝑧)≥1−𝜏. Define 𝑦=𝑦+𝛿(1−𝜏)⁻¹𝑧. Then 𝑥^∗𝑥=𝛿𝑦−𝑥1−𝜏^∗(𝑧)+𝑥^∗𝑦−𝑥≥𝛿+𝑥^∗𝑦−𝑥=𝜀(82) and hence 𝑦max1−𝑥+2,𝑦−1,𝑥^∗𝑦−1≥𝛽_𝑋(𝑥,𝜀).(83) 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 𝑥+𝑦2−𝑥+𝑦2≤12𝑥‖+𝑦=1‖𝑥−𝑦−2𝛿𝛿+≤𝛿1−𝜏.1−𝜏(84) Thus we get max1−𝑥+𝑦2,𝑦−1,𝑥^∗𝑦−1≥𝛽_𝑋()−𝛿𝑥,𝜀,1−𝜏(85) and since 0<𝜏<1 was arbitrary, it follows that max1−𝑥+𝑦2,𝑦−1,𝑥^∗𝑦−1≥𝛽_𝑋(𝑥,𝜀)−𝛿.(86) Again, since (𝑦,𝑥^∗)∈𝑉_𝑥,𝜀 was arbitrary we can conclude that 𝛽_𝑋(𝑥,𝜀)−𝛽_𝑋(𝑥,𝜀)≤𝛿=‖𝑥−𝑥‖(87) and by symmetry it follows that 𝛽_𝑋(𝑥,𝜀)−𝛽_𝑋(𝑥,𝜀)≤‖𝑥−𝑥‖.(88) Analogously one can prove that 𝛽_𝑋(𝑥,𝜀)̃−𝛽_𝑋(𝑥,𝜀)≤|𝜀−𝜀|̃(89) for all 𝑥∈𝑆_𝑋 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. If 𝐸 is LUR and 𝑋 is sluacs smooth, then 𝐸(𝑋) is also sluacs.

Proof. Since 𝐸 is LUR it is order continuous.
Let 0<𝜀≤2 and 𝑓∈𝑆_𝐸(𝑋) be arbitrary and let 𝐴_𝑛∶=𝑡∈𝑆∶𝑓(𝑡)≠0,𝛽_𝑋𝑓(𝑡)𝑓(𝑡,𝜀)8≥1𝑛(90) for every 𝑛∈ℕ. Since by the previous lemma 𝛽_𝑋(⋅,𝜀/8) is continuous, it follows that the sets 𝐴_𝑛 are measurable. Also, the sequence (𝐴_𝑛)_𝑛∈ℕ is increasing and because 𝑋 is sluacs, we have ⋃^∞_𝑛=1𝐴_𝑛={𝑡∈𝑆∶𝑓(𝑡)≠0}; hence (‖𝑓(⋅)‖𝜒_{𝑆⧵𝐴𝑛})_𝑛∈ℕ decreases pointwise to zero. The order continuity of 𝐸 implies ‖‖𝑓(⋅)‖𝜒_{𝑆⧵𝐴𝑛}‖_𝐸→0, and thus we can find 𝑛₀∈ℕ with 𝑓(⋅)𝜒_{𝑆⧵𝐴𝑛0𝐸}≤𝜀.64(91) Now let us take 𝑔∈𝑆_𝐸(𝑋) and 𝑙∈𝑆_𝐸(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 _𝑆‖𝐹(𝑡𝑔(𝑡(𝑡)‖𝐹(𝑡𝑔(𝑡(𝑡)𝑔(𝑡))‖)d𝜇=1,(92))‖)=𝐹a.e.(93) Next we define {(𝑡)},(𝑡)𝑔(𝑡)(𝑡)≥𝜀𝐶∶=𝑡∈𝑆∶𝐹≠0𝐵∶=𝑡∈𝐶∶𝐹−𝑓4‖𝐹(𝑡)‖max𝑓(𝑡),𝑔(𝑡).(94) Then 𝐵 is measurable, and _𝑆⧵𝐵𝐹(𝑡)𝑔(𝑡)(𝑡)(𝑡)≤𝜀−𝑓d𝜇4_𝑆⧵𝐵‖𝐹(𝑡𝑓(𝑡,𝑔(𝑡(𝑡)≤𝜀)‖max))d𝜇4_𝑆⧵𝐵‖𝐹(𝑡𝑓(𝑡+𝑔(𝑡(𝑡)≤𝜀)‖))d𝜇4𝜀2=2.(95) Since 𝑙(𝑔−𝑓)≥𝜀, it follows that _𝐵𝐹(𝑡)𝑔(𝑡)(𝑡)(𝑡)≥𝜀−𝑓d𝜇2.(96) Let us fix 0<𝜂<min{𝜀/16,1/2𝑛₀} such that 𝜂<21−𝜂𝑛₀.(97) Now consider the following sets: 𝐵₁𝑔(𝑡<𝑓(𝑡,𝐵∶=𝑡∈𝐵∶)1−𝜂)₂𝑓(𝑡≤𝑔(𝑡≤𝑓(𝑡,𝐵∶=𝑡∈𝐵∶1−𝜂)))₃𝑔(𝑡≤𝑓(𝑡<𝑔(𝑡,𝐵∶=𝑡∈𝐵∶1−𝜂)))₄∶=𝑡∈𝐵∶1−𝜂𝑔(𝑡)>𝑓(𝑡).(98) Then 𝐵₁,…,𝐵₄ are measurable, pairwise disjoint, and ⋃⁴_𝑖=1𝐵_𝑖=𝐵. Thus by (96) there exists some 𝑖∈{1,…,4} such that _𝐵𝑖𝐹(𝑡)𝑔(𝑡)(𝑡)(𝑡)≥𝜀−𝑓d𝜇8.(99) If 𝑖=1, then, since ‖𝑔(𝑡)‖≤‖𝑓(𝑡)‖ for 𝑡∈𝐵₁, it follows that _𝐵1‖𝐹(𝑡𝑓(𝑡(𝑡)≥𝜀)‖)d𝜇,16(100) and again by definition of 𝐵₁ we obtain 𝑔(⋅)−𝑓(⋅)_𝐸=𝑔(⋅)−𝑓(⋅)_𝐸≥_𝐵1‖𝐹(𝑡)‖𝑓(𝑡)−𝑔(𝑡)d𝜇(𝑡)≥𝜂_𝐵1‖𝐹(𝑡𝑓(𝑡(𝑡)𝜀)‖)d𝜇≥𝜂16(101) and hence 𝑓+𝑔2_𝐸(𝑋)≤𝑓(⋅)+𝑔(⋅)2_𝐸≤1−𝛿_𝐸𝑓(⋅𝜀),𝜂.16(102) 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 by 𝑁 the null set on which the equality from (93) does not hold and suppose that 𝑡∈𝐵₂∩𝐴_𝑛0∩(𝑆⧵𝑁). Then in particular 𝑡∈𝐵 and ‖𝑓(𝑡)‖≥‖𝑔(𝑡)‖, and hence 𝐹(𝑡)‖𝐹(𝑡𝑔(𝑡))‖𝑓(𝑡−𝑓(𝑡))𝑓(𝑡≥𝜀)4.(103) Moreover, by the definitions of 𝐵₂ and 𝐴_𝑛0 and the choice of 𝜂 we have 𝑔(𝑡)𝑓(𝑡=𝑔(𝑡)−1)𝑓(𝑡1)−1≤𝜂<𝑛₀≤𝛽_𝑋𝑓(𝑡)𝑓(𝑡,𝜀)8≤𝛽_𝑋𝑓(𝑡)𝑓(𝑡,𝜀)4.(104) Since 𝑡∈(𝑆⧵𝑁), we also have 𝐹(𝑡)‖𝐹(𝑡𝑔(𝑡))‖𝑓(𝑡=𝑔(𝑡)−1)𝑓(𝑡)−1<𝛽_𝑋𝑓(𝑡)𝑓(𝑡,𝜀)4.(105) So by definition of 𝛽_𝑋 we must have 12𝑓(𝑡)𝑓(𝑡+𝑔(𝑡))𝑓(𝑡)≤1−𝛽_𝑋𝑓(𝑡)𝑓(𝑡,𝜀)41≤1−𝑛₀.(106) Once more by the definition of 𝐵₁ this implies 𝑓(𝑡)+𝑔(𝑡)2≤11−𝑛₀𝑓(𝑡≤)1−1/𝑛₀2𝑓(𝑡+𝑔(𝑡=11−𝜂))21−𝛼₁𝑓(𝑡)+𝑔(𝑡),(107) where 𝛼₁∶=(1/𝑛₀−𝜂)(1−𝜂)⁻¹>0.
Now suppose that 𝑡∈𝐵₃∩𝐴_𝑛0∩(𝑆⧵𝑁). Then 𝐹(𝑡)‖𝐹(𝑡𝑔(𝑡))‖𝑔(𝑡−𝑓(𝑡))𝑔(𝑡≥𝜀)4.(108) Consequently 𝐹(𝑡)‖𝐹(𝑡𝑔(𝑡))‖𝑔(𝑡−𝑓(𝑡))𝑓(𝑡≥𝜀)4+𝐹(𝑡)‖𝐹(𝑡𝑓(𝑡))‖𝑔(𝑡−𝑓(𝑡))𝑓(𝑡≥𝜀)4−𝑓(𝑡)𝑔(𝑡−𝑓(𝑡))𝑓(𝑡=𝜀)4−𝑓(𝑡)𝑔(𝑡≥𝜀)−14𝜀−𝜂≥8.(109) Since ‖𝐹(𝑡)‖‖𝑔(𝑡)‖=𝐹(𝑡)(𝑔(𝑡)), the definition of 𝛽_𝑋 implies that 12𝑓(𝑡)𝑓(𝑡+𝑔(𝑡))𝑔(𝑡)≤1−𝛽_𝑋𝑓(𝑡)𝑓(𝑡,𝜀)81≤1−𝑛₀,(110) where the latter inequality holds because of 𝑡∈𝐴_𝑛0. It follows that 12𝑓(𝑡)𝑓(𝑡+𝑔(𝑡))𝑓(𝑡1)≤1−𝑛₀+12𝑔(𝑡)𝑓(𝑡−𝑔(𝑡))𝑔(𝑡1)=1−𝑛₀+12𝑔(𝑡)𝑓(𝑡1)−1≤1−𝑛₀+1211−𝜂−1=1−𝛼₂,(111) where 𝛼₂∶=1/𝑛₀−𝜂(2−2𝜂)⁻¹ which by (97) is greater than zero. Because of ‖𝑓(𝑡)‖≤‖𝑔(𝑡)‖, it follows that 𝑓(𝑡)+𝑔(𝑡)≤1−𝛼₂𝑓(𝑡)+𝑔(𝑡).(112) So if we put 𝛼=min{𝛼₁,𝛼₂} and 𝑃=𝐵₂∩𝐴_𝑛0∩(𝑆⧵𝑁), 𝑄=𝐵₃∩𝐴_𝑛0∩(𝑆⧵𝑁) then 𝑓(𝑡)+𝑔(𝑡)≤(1−𝛼)𝑓(𝑡)+𝑔(𝑡),∀𝑡∈𝑃∪𝑄.(113) Now we will show that if 𝑖=2 and 𝑖=3, respectively then _𝑃‖𝐹(𝑡𝑓(𝑡(𝑡)≥𝜀)‖)d𝜇64resp._𝑄‖𝐹(𝑡𝑓(𝑡(𝑡)≥𝜀)‖)d𝜇.64(114) Let us first assume that 𝑖=2, that is, _𝐵2𝐹(𝑡)𝑔(𝑡)(𝑡)(𝑡)≥𝜀−𝑓d𝜇8.(115) Since ‖𝑓(𝑡)‖≥‖𝑔(𝑡)‖ for 𝑡∈𝐵₂ it follows that _𝐵2‖𝐹(𝑡𝑓(𝑡(𝑡)≥𝜀)‖)d𝜇.16(116) Because 𝑁 is a null set, we have _𝑃‖𝐹(𝑡𝑓(𝑡(𝑡)=)‖)d𝜇_{𝐵2∩𝐴𝑛0}‖𝐹(𝑡𝑓(𝑡(𝑡)=)‖)d𝜇_𝐵2‖𝐹(𝑡𝑓(𝑡(𝑡)−)‖)d𝜇_{𝐵2⧵𝐴𝑛0}‖𝐹(𝑡𝑓(𝑡(𝑡)≥𝜀)‖)d𝜇−16_{𝑆⧵𝐴𝑛0}‖𝐹(𝑡𝑓(𝑡(𝑡)≥𝜀)‖)d𝜇−𝜒16𝑓(⋅)_{𝑆⧵𝐴𝑛0𝐸}≥𝜀−𝜀16≥𝜀64,64(117) where the second last inequality holds because of (91).
Now assume that 𝑖=3; that is, _𝐵3𝐹(𝑡)𝑔(𝑡)(𝑡)(𝑡)≥𝜀−𝑓d𝜇8.(118) It follows that 𝜀8≤_𝐵3‖𝐹(𝑡𝑔(𝑡+𝑓(𝑡(𝑡)≤)‖))d𝜇_𝐵3‖𝐹(𝑡1)‖1+𝑓(𝑡(𝑡)1−𝜂)d𝜇≤4_𝐵3‖𝐹(𝑡)‖𝑓(𝑡)d𝜇(𝑡),(119) and hence as before we get _𝑄‖𝐹(𝑡𝑓(𝑡(𝑡)≥𝜀)‖)d𝜇−𝜀32=𝜀64.64(120) So if 𝑖=2, or 𝑖=3 then there is 𝑅∈{𝑃,𝑄} such that 𝑓(⋅)𝜒_𝑅𝐸≥_𝑅‖𝐹(𝑡𝑓(𝑡(𝑡)≥𝜀)‖)d𝜇.64(121) Put ℎ=‖𝑓(⋅)‖(1−2𝛼𝜒_𝑅). Then ℎ∈𝐵_𝐸, and moreover ‖‖𝑓(⋅)‖−ℎ‖_𝐸=2𝛼‖‖𝑓(⋅)‖𝜒_𝑅‖_𝐸≥𝛼𝜀/32; hence 𝑓(⋅)1−𝛼𝜒_𝑅𝐸=12𝑓(⋅)+ℎ_𝐸≤1−𝛿_𝐸𝑓(⋅),𝜀𝛼.32(122) We further have 𝑓+𝑔2_𝐸(𝑋)≤12𝑓(⋅)+𝑔(⋅)𝜒_𝑆⧵𝑅+𝑓(⋅)+𝑔(⋅)_{𝑅𝐸()113}𝜒≤12𝑓(⋅)+𝑔(⋅)𝜒_𝑆⧵𝑅+(1−𝛼)𝑓(⋅)+𝑔(⋅)𝜒_𝑅𝐸≤12𝑔(⋅)+𝑓(⋅)−𝛼𝑓(⋅)𝜒_𝑅𝐸≤12+12𝑓(⋅)1−𝛼𝜒_𝑅𝐸≤12+121−𝛿_𝐸𝑓(⋅),𝜀𝛼132=1−2𝛿_𝐸𝑓(⋅),𝜀𝛼.32(123) Altogether we have shown that, for 1𝛿∶=min2𝛿_𝐸𝑓(⋅),𝜀𝛼32,𝛿_𝐸𝑓(⋅),𝜀𝜂16>0,(124) we have for every 𝑔∈𝑆_𝐸(𝑋) and every 𝑙∈𝑆_𝐸(𝑋)^∗ with 𝑙(𝑔)=1 and 𝑙(𝑓)≤1−𝜀𝑓+𝑔2_𝐸(𝑋)≤1−𝛿.(125) 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. If 𝜇 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 since 𝐸 is reflexive (or since it has the Kadets-Klee property), it is order continuous.
Let us take two sequences (𝑓_𝑛)_𝑛∈ℕ and (𝑔_𝑛)_𝑛∈ℕ in the unit sphere of 𝐸(𝑋) such that ‖𝑓_𝑛+𝑔_𝑛‖_𝐸(𝑋)→2 and a functional 𝑙∈𝑆_𝐸(𝑋)^∗, as usual represented by [𝐹]∈𝐸^′(𝑋^∗,𝑤^∗), with 𝑙(𝑓_𝑛)→1.
As in the proof of Theorem 10 we find lim_{𝑛→∞𝑆}‖𝐹(𝑡)‖𝑓_𝑛(𝑡)d𝜇(𝑡)=1,(126) and by passing to a subsequence also lim_𝑛→∞‖𝐹(𝑡)‖𝑓_𝑛(𝑡)−𝐹(𝑡)𝑓_𝑛(𝑡)=0a.e.(127) It is also easy to see that lim_𝑛→∞𝑓_𝑛(⋅)+𝑔_𝑛(⋅)_𝐸=2,(128)lim_𝑛→∞𝑓_𝑛(⋅)+𝑔_𝑛(⋅)+𝑓_𝑛(⋅)+𝑔_𝑛(⋅)_𝐸=4.(129) Since 𝐸 is wuacs, it follows from (126) and (128) that lim_{𝑛→∞𝑆}‖𝐹(𝑡)‖𝑔_𝑛(𝑡)d𝜇(𝑡)=1.(130) Again since 𝐸 is wuacs and because of (126), (128), (129), and (130) we can deduce that lim_{𝑛→∞𝑆}‖𝐹(𝑡)‖𝑓_𝑛(𝑡)+𝑔_𝑛(𝑡)−𝑓_𝑛(𝑡)+𝑔_𝑛(𝑡)d𝜇(𝑡)=0.(131) Hence we can pass to a further subsequence such that lim_𝑛→∞‖𝐹(𝑡)‖𝑓_𝑛(𝑡)+𝑔_𝑛(𝑡)−𝑓_𝑛(𝑡)+𝑔_𝑛(𝑡)=0a.e.(132) By the reflexivity of 𝐸 we can pass once more to a subsequence such that (‖𝑓_𝑛(⋅)‖)_𝑛∈ℕ and (‖𝑔_𝑛(⋅)‖)_𝑛∈ℕ are weakly convergent to ℎ₁∈𝐵_𝐸 and ℎ₂∈𝐵_𝐸, respectively. In view of (126) and (130) it follows that _𝑆‖𝐹(𝑡)‖ℎ_𝑖(𝑡)d𝜇(𝑡)=1∀𝑖∈{1,2},(133) hence ‖ℎ₁‖_𝐸=‖ℎ₂‖_𝐸=1, and moreover ℎ₁+ℎ_2𝐸=2.(134) The fact that 𝐸 has the Kadets-Klee property implies that 𝑓_𝑛(⋅)−ℎ_1𝐸⟶0,𝑔_𝑛(⋅)−ℎ_2𝐸⟶0,(135) and thus by Lemma 7 we can, for the last time, pass to a subsequence such that lim_𝑛→∞𝑓_𝑛(𝑡)=ℎ₁(𝑡),lim_𝑛→∞𝑔_𝑛(𝑡)=ℎ₂(𝑡)a.e.(136) 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 that 𝑋 is wuacs together with Lemma 5 easily implies that lim_𝑛→∞‖𝐹(𝑡)‖𝑔_𝑛(𝑡)−𝐹(𝑡)𝑔_𝑛(𝑡)().=0,∀𝑡∈𝑆⧵𝑁∪𝐶(137) By the weak convergence of (‖𝑔_𝑛(⋅)‖)_𝑛∈ℕ to ℎ₂ we have lim_{𝑛→∞𝐴}‖𝐹(𝑡)‖𝑔_𝑛(𝑡(𝑡)=)d𝜇_𝐴‖𝐹(𝑡)‖ℎ₂(𝑡)d𝜇(𝑡)∀𝐴∈𝒜.(138) 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 lim_{𝑛→∞𝑆}‖𝐹(𝑡)‖𝑔_𝑛(𝑡)−𝐹(𝑡)𝑔_𝑛(𝑡)d𝜇(𝑡)=0.(139) Because of (130), it follows that lim_𝑛→∞𝑙𝑔_𝑛=lim_{𝑛→∞𝑆}𝐹(𝑡)𝑔_𝑛(𝑡)d𝜇(𝑡)=1,(140) and we are done.

If we combine the techniques of the proofs of Theorems 16 and 13, we can also obtain another result concerning luacs⁺ spaces (we omit the details).

Theorem 17. If 𝜇 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 for 𝐸(𝑋) to be sluacs.

Theorem 18. If 𝜇 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 (𝑓_𝑛)_𝑛∈ℕ be a sequence in 𝑆_𝐸(𝑋) and 𝑓∈𝑆_𝐸(𝑋) such that we have ‖𝑓_𝑛+𝑓‖_𝐸(𝑋)→2. Also, let (𝑙_𝑛)_𝑛∈ℕ be a sequence in 𝑆_𝐸(𝑋)^∗ such that 𝑙_𝑛(𝑓_𝑛)→1. If we represent each 𝑙_𝑛 by [𝐹_𝑛]∈𝐸^′(𝑋^∗,𝑤^∗), we can obtain as usual lim_{𝑛→∞𝑆}𝐹_𝑛(𝑡)𝑓_𝑛(𝑡)d𝜇(𝑡)=1,(141) and by passing to a subsequence also lim_𝑛→∞𝐹_𝑛(𝑡)𝑓_𝑛(𝑡)−𝐹_𝑛(𝑡)𝑓_𝑛(𝑡)=0a.e.(142) as well as lim_𝑛→∞𝑓_𝑛(⋅)+𝑓(⋅)_𝐸=2,(143)lim_𝑛→∞𝑓_𝑛(⋅)+𝑓(⋅)+𝑓_𝑛(⋅)+𝑓(⋅)_𝐸=4,(144)lim_𝑛→∞𝑓_𝑛(⋅)+𝑓(⋅)+𝑓_𝑛(⋅)_𝐸=3,(145)lim_𝑛→∞𝑓_𝑛(⋅)+𝑓(⋅)+𝑓(⋅)_𝐸=3,(146)lim_𝑛→∞𝑓_𝑛(⋅)+𝑓(⋅)+𝑓_𝑛(⋅)+3𝑓(⋅)_𝐸=6.(147) Using the fact that 𝐸 is sluacs⁺ we can conclude that lim_{𝑛→∞𝑆}𝐹_𝑛(𝑡)𝑓(𝑡)d𝜇(𝑡)=1,(148)lim_{𝑛→∞𝑆}𝐹_𝑛(𝑡)𝑓_𝑛(𝑡)+𝑓(𝑡)−𝑓_𝑛(𝑡)+𝑓(𝑡)d𝜇(𝑡)=0.(149) So we can pass to another subsequence such that lim_𝑛→∞𝐹_𝑛(𝑡)𝑓_𝑛(𝑡)+𝑓(𝑡)−𝑓_𝑛(𝑡)+𝑓(𝑡)=0a.e.(150) Since 𝐸 (and hence also 𝐸^∗) is reflexive, we may assume without loss of generality that (‖𝑓_𝑛(⋅)‖)_𝑛∈ℕ is weakly convergent to some ℎ∈𝐵_𝐸 and that (‖𝐹_𝑛(⋅)‖)_𝑛∈ℕ is weakly convergent to some 𝑔∈𝐵_𝐸^∗=𝐵_𝐸′.
It follows from (148) that _𝑆𝑔(𝑡)𝑓(𝑡)d𝜇(𝑡)=1(151) and hence 𝑔∈𝑆_𝐸^∗. Because of (151), (143), and the fact that 𝐸 is sluacs⁺, we get that lim_{𝑛→∞𝑆}𝑔(𝑡)𝑓_𝑛(𝑡)d𝜇(𝑡)=1,(152) and consequently _𝑆𝑔(𝑡)ℎ(𝑡)d𝜇(𝑡)=1,(153) whence ℎ∈𝑆_𝐸. Since both 𝐸 and 𝐸^∗ have the Kadets-Klee property, it follows that 𝑓_𝑛(⋅)−ℎ_𝐸⟶0,𝐹_𝑛(⋅)−𝑔_𝐸′⟶0.(154) Thus we can pass once more to a subsequence such that lim_𝑛→∞𝑓_𝑛(𝑡)=ℎ(𝑡),lim_𝑛→∞𝐹_𝑛(𝑡)=𝑔(𝑡)a.e.(155) Combining (153) and (151) we also obtain ℎ+𝑓(⋅)_𝐸=2.(156) 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 Theorem 16 one can see that 𝐶 is a null set and then, using the fact that 𝑋 is sluacs, deduce that lim_𝑛→∞𝐹_𝑛(𝑡)𝑓(𝑡)−𝐹_𝑛(𝑡)𝑓(𝑡)=0a.e.(157) By our usual method based on Vitali’s lemma we can conclude that for every 𝐴∈𝒜 with 𝜇(𝐴)<∞ we have lim_{𝑛→∞𝐴}𝐹_𝑛(𝑡)𝑓(𝑡)−𝐹_𝑛(𝑡)𝑓(𝑡)d𝜇(𝑡)=0.(158) Now we fix an increasing sequence (𝐴_𝑚)_𝑚∈ℕ in 𝒜 such that 𝜇(𝐴_𝑚)<∞ for all 𝑚∈ℕ and ⋃^∞_𝑚=1𝐴_𝑚=𝑆. The order continuity of 𝐸 implies ‖‖𝑓(⋅)‖𝜒_{𝑆⧵𝐴𝑚}‖_𝐸→0. Analogous to the argument at the end of the proof of Theorem 13, this together with (158) leads to lim_{𝑛→∞𝑆}𝐹_𝑛(𝑡)𝑓(𝑡)−𝐹_𝑛(𝑡)𝑓(𝑡)d𝜇(𝑡)=0.(159) Taking into account (148) we arrive at lim_𝑛→∞𝑙_𝑛𝑓=lim_{𝑛→∞𝑆}𝐹_𝑛(𝑡)𝑓(𝑡)d𝜇(𝑡)=1,(160) and the proof is finished.

Next we will consider sufficient conditions for a Köthe-Bochner space to be sluacs⁺ (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. Let 𝐸 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 𝐵_𝐸^∗ is weak*-sequentially compact,then 𝐸(𝑋) is 𝑠𝑙𝑢𝑎𝑐𝑠⁺.

Proof. By Theorems 15 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 (𝑓_𝑛)_𝑛∈ℕ in 𝑆_𝐸(𝑋) and 𝑓∈𝑆_𝐸(𝑋) such that ‖𝑓_𝑛+𝑓‖_𝐸(𝑋)→2, and let (𝑙_𝑛)_𝑛∈ℕ be a sequence in 𝑆_𝐸(𝑋)^∗ such that 𝑙_𝑛(𝑓)→1. If we represent each 𝑙_𝑛 by [𝐹_𝑛]∈𝐸^′(𝑋^∗,𝑤^∗) we can obtain as usual lim_{𝑛→∞𝑆}𝐹_𝑛(𝑡)𝑓(𝑡)d𝜇(𝑡)=1(161) and by passing to a subsequence also lim_𝑛→∞𝐹_𝑛(𝑡)𝑓(𝑡)−𝐹_𝑛(𝑡)𝑓(𝑡)=0a.e.(162) as well as lim_𝑛→∞𝑓_𝑛(⋅)+𝑓(⋅)_𝐸=2,(163)lim_𝑛→∞𝑓_𝑛(⋅)+𝑓(⋅)+𝑓_𝑛(⋅)+𝑓(⋅)_𝐸=4,(164)lim_𝑛→∞𝑓_𝑛(⋅)+𝑓(⋅)+𝑓_𝑛(⋅)_𝐸=3,(165)lim_𝑛→∞𝑓_𝑛(⋅)+𝑓(⋅)+𝑓(⋅)_𝐸=3,(166)lim_𝑛→∞𝑓_𝑛(⋅)+𝑓(⋅)+𝑓_𝑛(⋅)+3𝑓(⋅)_𝐸=6.(167) Since 𝐸 is sluacs⁺, it follows that lim_{𝑛→∞𝑆}𝐹_𝑛(𝑡)𝑓_𝑛(𝑡)d𝜇(𝑡)=1,(168)lim_{𝑛→∞𝑆}𝐹_𝑛(𝑡)𝑓_𝑛(𝑡+𝑓(𝑡−𝑓))_𝑛(𝑡)+𝑓(𝑡)d𝜇(𝑡)=0,(169) so that by passing to another subsequence we can assume lim_𝑛→∞𝐹_𝑛(𝑡)𝑓_𝑛(𝑡)+𝑓(𝑡)−𝑓_𝑛(𝑡)+𝑓(𝑡)=0a.e.(170) In both cases (a) and (b) the dual unit ball 𝐵_𝐸^∗ is weak*-sequentially compact, so that we can also assume the weak*-convergence of (‖𝐹_𝑛(⋅)‖)_𝑛∈ℕ to some 𝑔∈𝐵_𝐸^∗. It follows from (161) that _𝑆𝑔(𝑡)𝑓(𝑡)d𝜇(𝑡)=1,(171) and hence ‖𝑔‖_𝐸′=1. Since 𝐸^∗ has the Kadets-Klee* property, we get 𝐹_𝑛(⋅)−𝑔_𝐸′⟶0,(172) and thus we can, by passing to yet another subsequence, assume that lim_𝑛→∞𝐹_𝑛(𝑡)=𝑔(𝑡)a.e.(173) Next we claim that there is an ℎ∈𝑆_𝐸 such that _𝑆𝑔(𝑡)ℎ(𝑡)d𝜇(𝑡)=1,(174) and, after passing to a subsequence once more, 𝑓_𝑛(⋅)−ℎ_𝐸⟶0.(175) 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 (‖𝑓_𝑛(⋅)‖)_𝑛∈ℕ is weakly convergent to some ℎ∈𝐵_𝐸. 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 lim_{ℎ𝑛→∞}𝑓_𝑛(𝑡)=ℎ(𝑡)a.e.(176) 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 lim_𝑛→∞𝐹_𝑛(𝑡)𝑓_𝑛(𝑡)−𝐹_𝑛(𝑡)𝑓_𝑛(𝑡)=0a.e.(177) and then lim_{𝑛→∞𝐴}𝐹_𝑛(𝑡)𝑓_𝑛(𝑡)−𝐹_𝑛(𝑡)𝑓_𝑛(𝑡)d𝜇(𝑡)=0,(178) for every 𝐴∈𝒜 with 𝜇(𝐴)<∞.
Let us now fix a sequence (𝐴_𝑚)_𝑚∈ℕ in 𝒜 as in the proof of Theorem 18. The order continuity of 𝐸 implies ‖‖𝑓(⋅)‖𝜒_{𝑆⧵𝐴𝑚}‖_𝐸→0.
Let 𝜀>0 be arbitrary. Since 𝐸 is sluacs⁺ there exists a 𝛿>0 such that for all 𝑏∈𝑆_𝐸 and all 𝑙∈𝐵_𝐸^∗ with ‖𝑏+‖𝑓(⋅)‖‖_𝐸≥2(1−𝛿) and 𝑙(‖𝑓(⋅)‖)≥1−𝛿 one has 𝑙(𝑏)≥1−𝜀.
Fix 𝑚₀∈ℕ with ‖‖𝑓(⋅)‖𝜒_{𝑆⧵𝐴𝑚0}‖_𝐸≤𝛿/2. Because of (163), (161) and (178) there is an 𝑁∈ℕ such that for all 𝑛≥𝑁 the inequalities 𝑓_𝑛(⋅)+𝑓(⋅)_𝐸≥2(1−𝛿_𝑆),𝐹_𝑛(𝑡𝑓(𝑡(𝑡)𝛿))d𝜇≥1−2_𝐴𝑚0,𝐹_𝑛(𝑡)𝑓_𝑛(𝑡)−𝐹_𝑛(𝑡)𝑓_𝑛(𝑡)d𝜇(𝑡)≤𝜀(179) hold.
It follows that for every 𝑛≥𝑁 we have _𝐴𝑚0𝐹_𝑛(𝑡𝑓(𝑡(𝑡)≤𝑓(⋅𝜒))d𝜇−1)_{𝑆⧵𝐴𝑚0𝐸}+_𝑆𝐹_𝑛(𝑡)𝑓(𝑡)d𝜇(𝑡)−1≤𝛿,(180) and hence by the choice of 𝛿, _𝐴𝑚0𝐹_𝑛(𝑡)𝑓_𝑛(𝑡)d𝜇(𝑡)≥1−𝜀.(181) Consequently, for every 𝑛≥𝑁 we have _𝑆𝐹_𝑛(𝑡)𝑓_𝑛(𝑡)−𝐹_𝑛(𝑡)𝑓_𝑛(𝑡)d𝜇(𝑡)≤𝜀+_{𝑆⧵𝐴𝑚0}𝐹_𝑛(𝑡)𝑓_𝑛(𝑡)−𝐹_𝑛(𝑡)𝑓_𝑛(𝑡)d𝜇(𝑡)≤𝜀+2_{𝑆⧵𝐴𝑚0}𝐹_𝑛(𝑡)𝑓_𝑛(𝑡(𝑡)(()d𝜇≤𝜀+21−1−𝜀))=3𝜀.(182) Thus we have shown lim_{𝑛→∞𝑆}𝐹_𝑛(𝑡)𝑓_𝑛(𝑡)−𝐹_𝑛(𝑡)𝑓_𝑛(𝑡)d𝜇(𝑡)=0.(183) Together with (168) it follows 𝑙_𝑛(𝑓_𝑛)→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∈𝑆_𝐸 and every ℎ∈𝑆_𝐸′ we have 𝑓+𝑔_𝐸≥2(1−𝛿_𝑆),𝑓ℎd𝜇=1⟹_𝑆|ℎ|𝑓−𝑔d𝜇≤𝜀.(184) 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. If 𝐸 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. Let 0<𝜀≤2 be arbitrary. Since 𝐸 is in particular uacs there is a number 𝜂>0 such that for all functions 𝑎,𝑏∈𝐵_𝐸 and every functional 𝑙∈𝐵_𝐸^∗ with 𝑙(𝑎)=1 one has 𝑙(𝑏)𝜀<1−4𝛿^𝑋_uacs𝜀2⟹‖𝑎+𝑏‖_𝐸≤21−𝜂.(185) Now let 𝑓,𝑔∈𝑆_𝐸(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).
Let 𝐿 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 0≤𝛾(𝑡)≤2𝜈(𝑡)𝛽(𝑡)∀𝑡∈𝑆.(186) As before we can deduce from 𝐿(𝑓)=1 that _𝑆‖𝐹(𝑡)‖𝑓(𝑡)d𝜇(𝑡)=1(187) and 𝐹(𝑡)(𝑓(𝑡))=‖𝐹(𝑡)‖‖𝑓(𝑡)‖ a.e., hence 𝐹(𝑡)𝑓(𝑡)=𝜈(𝑡)𝛽(𝑡)a.e.(188) Next we define 𝛼(𝑡)=12𝛿^𝑋_uacs𝛾(𝑡)𝜈(𝑡)𝛽(𝑡)(𝑡)(𝑡)𝛽(𝑡),(𝑡)1if0<𝛾<𝜈0if𝛾=0,2𝛿^𝑋_uacs(1)otherwise.(189) Note that since 𝛿^𝑋_uacs 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 𝑓(𝑡)+𝑔(𝑡)≤2(1−𝛼(𝑡))𝛽(𝑡)a.e.(190) By (186) and (187) we have ∫_𝑆𝛾(𝑡)d𝜇(𝑡)≤2. Furthermore, we also have 𝜀<1−𝐿𝑔=𝐿𝑓−𝑔=_𝑆𝐹(𝑡)𝑓(𝑡)(𝑡)(𝑡)≤−𝑔d𝜇_𝑆𝛾(𝑡)d𝜇(𝑡);(191) thus 𝜀<_𝑆𝛾(𝑡)d𝜇(𝑡)≤2.(192) Now put 𝐴={𝑡∈𝑆∶2𝛾(𝑡)>𝜀𝜈(𝑡)𝛽(𝑡)} and 𝐵=𝑆⧵𝐴. We then have (because of (187)) _𝐵𝛾(𝑡)(𝑡)≤𝜀d𝜇2_𝐵𝜈(𝑡)𝛽(𝑡)(𝑡)≤𝜀d𝜇2_𝑆𝜈(𝑡)𝛽(𝑡)(𝑡)=𝜀d𝜇2.(193) Together with (192) it follows that _𝐴𝛾(𝑡)(𝑡)𝜀d𝜇>𝜀−2=𝜀2.(194) Taking into account (186) we get _𝐴𝜈(𝑡)𝛽(𝑡)(𝑡)>𝜀d𝜇4.(195) Next we define ℎ=𝛽𝜒_𝐵 and ℎ^′=𝛽𝜒_𝐴 as well as ℎ^′′=(1−𝛿^𝑋_uacs(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), 𝑙ℎ+ℎ^′′=1−𝛿^𝑋_uacs𝜀2𝑙ℎ^′=1−_𝐴𝜈(𝑡)𝛽(𝑡)(𝑡)𝜀d𝜇<1−4𝛿^𝑋_uacs𝜀2.(196) So by our choice of 𝜂 we get ‖2ℎ+ℎ^′+ℎ^′′‖_𝐸≤2(1−𝜂); that is, 1ℎ+1−2𝛿^𝑋_uacs𝜀2ℎ^′_𝐸≤1−𝜂.(197) By monotonicity of 𝛿^𝑋_uacs we have 𝛼(𝑡)≥12𝛿^𝑋_uacs𝜀2∀𝑡∈𝐴.(198) Using (190), (198), and (197) we obtain 𝑓+𝑔_𝐸(𝑋)=𝑓(⋅)+𝑔(⋅)_𝐸≤2(1−𝛼)𝛽_𝐸≤21−2⁻¹𝛿^𝑋_uacs𝜀2ℎ^′+ℎ_𝐸≤21−𝜂.(199) 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.
Since 𝐸 is uacs, we may find 𝜏>0 such that for all 𝑎,𝑏∈𝐵_𝐸 and every 𝑙∈𝐵_𝐸^∗ we have 𝑙(𝑎)≥1−𝜏,‖𝑎+𝑏‖_𝐸≥2(1−𝜏)⟹𝑙(𝑏)≥1−𝜔.(200) 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}.
Let 𝑓,𝑔∈𝑆_𝐸(𝑋) be arbitrary and 𝐿∈𝑆_𝐸(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 define 𝑧∶𝑆→𝑋 by 𝑧(𝑡)=𝑓(𝑡)𝑔(𝑡𝑔(𝑡)(𝑡)𝑓(𝑡)(𝑡))if𝑔≠0,if𝑔=0.(201) Then 𝑧 is Bochner-measurable, and ‖𝑧(𝑡)‖=‖𝑓(𝑡)‖ for all 𝑡∈𝑆 (hence 𝑧∈𝐸(𝑋)). Furthermore, 𝑧(𝑡)−𝑔(𝑡)=𝑓(𝑡)−𝑔(𝑡)∀𝑡∈𝑆.(202) As before we have _𝑆‖𝐹(𝑡)‖𝑓(𝑡)d𝜇(𝑡)=1.(203) Also, 2(1−𝜏)≤2(1−𝜉)≤𝑓+𝑔_𝐸(𝑋)≤𝑓(⋅)+𝑔(⋅)_𝐸,(204) so the choice of 𝜏 together with (202) implies _𝑆‖𝐹(𝑡)‖𝑧(𝑡)−𝑔(𝑡)d𝜇(𝑡)≤𝜌.(205) Next we observe that 𝑓(⋅)+𝑔(⋅)+𝑓(⋅)+𝑔(⋅)_𝐸≥2𝑓+𝑔_𝐸(𝑋)≥4(1−𝜉)≥4(1−𝜏),(206) and (because of (203) and (205)) _𝑆‖𝐹(𝑡)‖𝑓(𝑡)+𝑔(𝑡)d𝜇(𝑡)=1+_𝑆‖𝐹(𝑡)‖𝑔(𝑡)d𝜇(𝑡)≥1+_𝑆‖𝐹(𝑡𝑓(𝑡(𝑡)−)‖)d𝜇_𝑆‖𝐹(𝑡)‖𝑓(𝑡)−𝑔(𝑡)d𝜇(𝑡)=2−_𝑆‖𝐹(𝑡𝑓(𝑡−𝑔(𝑡(𝑡)().)‖))d𝜇≥2−𝜌≥21−𝜏(207) So (200) implies _𝑆‖𝐹(𝑡)‖𝑓(𝑡)+𝑔(𝑡)d𝜇(𝑡)≥2(1−𝜔).(208) Using (205) and (208) we can conclude 𝑓+𝑧_𝐸(𝑋)≥_𝑆‖𝐹(𝑡𝑓(𝑡)(𝑡(𝑡)≥)‖+𝑧)d𝜇_𝑆‖𝐹(𝑡𝑓(𝑡)(𝑡(𝑡)−)‖+𝑔)d𝜇_𝑆‖𝐹(𝑡𝑔(𝑡)(𝑡(𝑡)().)‖−𝑧)d𝜇≥21−𝜔−𝜌>21−𝜂(209) 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. If 𝐸 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 space 𝑋 is said to be midpoint locally uniformly rotund (MLUR) if for any two sequences (𝑥_𝑛)_𝑛∈ℕ and (𝑦_𝑛)_𝑛∈ℕ in 𝑆_𝑋 and every 𝑥∈𝑆_𝑋 we have 𝑥𝑥−^𝑛+𝑦_𝑛2⟶0⟹𝑥_𝑛−𝑦_𝑛⟶0.(210)𝑋 is called weakly midpoint locally uniformly rotund (WMLUR) if it satisfies the above condition with ‖𝑥_𝑛−𝑦_𝑛‖→0 replaced by 𝑥_𝑛−𝑦→−_𝑛^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. Let 𝑋 be a Banach space.(i)The space 𝑋 is said to be midpoint locally uniformly alternatively convex or smooth (mluacs) if for any two sequences (𝑥_𝑛)_𝑛∈ℕ and (𝑦_𝑛)_𝑛∈ℕ in 𝑆_𝑋, every 𝑥∈𝑆_𝑋, and every 𝑥^∗∈𝑆_𝑋^∗, one has 𝑥𝑥−^𝑛+𝑦_𝑛2⟶0,𝑥^∗𝑥_𝑛⟶1⟹𝑥^∗𝑦_𝑛⟶1.(211)(ii)The space 𝑋 is called midpoint strongly locally uniformly alternatively convex or smooth (msluacs) if for any two sequences (𝑥_𝑛)_𝑛∈ℕ and (𝑦_𝑛)_𝑛∈ℕ in 𝑆_𝑋, every 𝑥∈𝑆_𝑋 and every sequence (𝑥^∗_𝑛)_𝑛∈ℕ in 𝑆_𝑋^∗, one has 𝑥𝑥−^𝑛+𝑦_𝑛2⟶0,𝑥^∗_𝑛𝑥_𝑛⟶1⟹𝑥^∗_𝑛𝑦_𝑛⟶1.(212)
Figure 4 summarises the obvious implications. No other implications are true in general (see the examples in [4]).

Figure 4 

Concerning the properties msluacs and mluacs for Köthe-Bochner spaces we have the following result.

Theorem 23. Let 𝐸 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 𝐵_𝐸^∗ is weak*-sequentially compact, then 𝐸(𝑋) is also msluacs.

Proof. Let us first recall that ℓ^∞ 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 that 𝑋 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 (𝑓_𝑛)_𝑛∈ℕ, (𝑔_𝑛)_𝑛∈ℕ in 𝑆_𝐸(𝑋) and 𝑓∈𝑆_𝐸(𝑋) such that ‖𝑓_𝑛+𝑔_𝑛−2𝑓‖_𝐸(𝑋)→0. Also, take a sequence (𝑙_𝑛)_𝑛∈ℕ of norm-one functionals on 𝐸(𝑋) such that 𝑙_𝑛(𝑓_𝑛)→1. As usual, 𝑙_𝑛 will be represented by [𝐹_𝑛]∈𝐸^′(𝑋^∗,𝑤^∗), and we conclude lim_{𝑛→∞𝑆}𝐹_𝑛(𝑡)𝑓_𝑛(𝑡)d𝜇(𝑡)=1,(213) and, after passing to an appropriate subsequence, lim_𝑛→∞𝐹_𝑛(𝑡)𝑓_𝑛(𝑡)−𝐹_𝑛(𝑡)𝑓_𝑛(𝑡)=0a.e.(214) We also have 2𝑓(⋅)−𝑓_𝑛(⋅)+𝑔_𝑛(⋅)_𝐸=2𝑓(⋅)−𝑓_𝑛(⋅)+𝑔_𝑛(⋅)_𝐸≤2𝑓(⋅)−𝑓_𝑛(⋅)−𝑔_𝑛(⋅)_𝐸=2𝑓−𝑓_𝑛−𝑔_{𝑛𝐸(𝑋)};(215) hence 2𝑓(⋅)−𝑓_𝑛(⋅)+𝑔_𝑛(⋅)_𝐸⟶0.(216) As before we can also show 𝑓_𝑛(⋅)+𝑔_𝑛(⋅)_𝐸⟶2.(217) Also, because of ‖𝑓_𝑛+𝑔_𝑛−2𝑓‖_𝐸(𝑋)→0, we may pass to a further subsequence such that lim_𝑛→∞𝑓_𝑛(𝑡)+𝑔_𝑛(𝑡)−2𝑓(𝑡)=0a.e.(218) Let us define for every 𝑛∈ℕ𝑎_𝑛(𝑡)𝑓(𝑡−1∶=2)2𝑓_𝑛(𝑡)+𝑔_𝑛(𝑡,𝑏_𝑛)(𝑡)𝑓(𝑡−1∶=)2𝑓_𝑛(𝑡)+𝑔_𝑛(𝑡).(219) Note that 𝑓(𝑡)≤𝑏_𝑛(𝑡)+12𝑓_𝑛(𝑡)+𝑔_𝑛(𝑡).(220) So if 𝑎_𝑛(𝑡)≥0, then 𝑎_𝑛(𝑡𝑓(𝑡−1)=2)2𝑓_𝑛(𝑡)+𝑔_𝑛(𝑡𝑏)≤2_𝑛(𝑡+1)2𝑓_𝑛(𝑡)+𝑔_𝑛(𝑡).(221) If 𝑎_𝑛(𝑡)<0, then 𝑎_𝑛(𝑡=1)2𝑓_𝑛(𝑡)+𝑔_𝑛(𝑡𝑓(𝑡𝑏)−2)≤2_𝑛(𝑡+1)2𝑓_𝑛(𝑡)+𝑔_𝑛(𝑡).(222) So we always have 𝑎_𝑛(𝑡)≤2𝑏_𝑛(𝑡+1)2𝑓_𝑛(𝑡)+𝑔_𝑛(𝑡).(223) It follows that 12𝑓_𝑛(⋅)+𝑔_𝑛(⋅)_𝐸+2𝑏_𝑛𝐸≥2𝑏_𝑛+12𝑓_𝑛(⋅)+𝑔_𝑛(⋅)_𝐸≥𝑎_𝑛𝐸1≥2−2𝑓_𝑛(⋅)+𝑔_𝑛(⋅)_𝐸,(224) and we can conclude with (216) and (217) that ‖𝑎_𝑛‖_𝐸→1.
Using this together with (216), ‖𝑓_𝑛(⋅)‖+‖𝑔_𝑛(⋅)‖+2𝑎_𝑛=4‖𝑓(⋅)‖, and the fact that 𝐸 is MLUR we get lim_𝑛→∞2𝑓(⋅)−𝑓_𝑛(⋅)−𝑔_𝑛(⋅)_𝐸=0.(225) Again, since 𝐸 is MLUR, this implies lim_𝑛→∞𝑓_𝑛(⋅)−𝑔_𝑛(⋅)_𝐸=0.(226) Because of (225) and (226), we can pass to a further subsequence such that lim_𝑛→∞𝑓_𝑛(𝑡)=𝑓(𝑡)=lim_𝑛→∞𝑔_𝑛(𝑡)a.e.(227) Since 𝐵_𝐸^∗ is weak*-sequentially compact we may also asssume that (‖𝐹_𝑛(⋅)‖)_𝑛∈ℕ weak*-converges to some 𝑔∈𝐵_𝐸′.
Equations (225) and (226) imply ‖‖𝑓_𝑛(⋅)‖−‖𝑓(⋅)‖‖_𝐸→0. Together with (213) this gives us lim_{𝑛→∞𝑆}𝐹_𝑛(𝑡)𝑓(𝑡)d𝜇(𝑡)=1;(228) hence we also have _𝑆𝑔(𝑡)𝑓(𝑡)d𝜇(𝑡)=1;(229) thus ‖𝑔‖_𝐸′=1. Since 𝐸^∗ has the Kadets-Klee* property, it follows that ‖‖𝐹_𝑛(⋅)‖−𝑔‖_𝐸′→0, so if we pass again to a subsequence, we may assume lim_𝑛→∞𝐹_𝑛(𝑡)=𝑔(𝑡)a.e.(230) Now if we combine (214), (216), (227), and (230) we obtain lim_𝑛→∞𝐹_𝑛(𝑡)𝑓(𝑡)−𝐹_𝑛(𝑡)𝑓(𝑡)=0a.e.,(231) since 𝑋 is msluacs.
Using our usual argument via equi-integrability and Vitali’s lemma this leads to lim_{𝑛→∞𝐴}𝐹_𝑛(𝑡)𝑓(𝑡)−𝐹_𝑛(𝑡)𝑓(𝑡)d𝜇(𝑡)=0(232) for every 𝐴∈𝒜 with 𝜇(𝐴)<∞.
By the order continuity of 𝐸 we can derive from this lim_{𝑛→∞𝑆}𝐹_𝑛(𝑡)𝑓(𝑡)−𝐹_𝑛(𝑡)𝑓(𝑡)d𝜇(𝑡)=0(233) also in the 𝜎-finite case (cf. the proof of Theorem 18).
Combining (233) and (228) gives us 𝑙_𝑛(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 to 𝐿^𝑝 spaces for 1<𝑝<∞ (as we said before, for the properties acs/luacs/uacs this was already proved by Sirotkin in [5]).

Corollary 24. If 𝑋 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 𝐿^𝑝(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 (𝑥_𝑛)_𝑛∈ℕ in 𝑆_𝑋, every 𝑥∈𝑆_𝑋 and each 𝑥^∗∈𝑆_𝑋^∗, the implication 𝑥_𝑛+𝑥⟶2,𝑥^∗(𝑥)=1⟹𝑥^∗𝑥_𝑛⟶1(234) 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 space 𝑋 is 𝑙𝑢𝑎𝑐𝑠⁺ if and only if 𝑋 is WLUR in the sense of Lovaglia and for all 𝑥^∗,𝑦^∗∈𝑆_𝑋^∗ with ‖𝑥^∗+𝑦^∗‖=2 and every 𝑥∈𝑆_𝑋 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 (𝑥_𝑛)_𝑛∈ℕ in 𝑆_𝑋 and 𝑥∈𝑆_𝑋 such that ‖𝑥_𝑛+𝑥‖→2 as well as 𝑥^∗∈𝑆_𝑋^∗ with 𝑥^∗(𝑥_𝑛)→1. Since 𝐵_𝑋^∗∗ is weak*-compact, we can find 𝑥^∗∗∈𝐵_𝑋^∗∗ and a subnet (𝑥_𝜑(𝑖))_𝑖∈𝐼 which is weak*-convergent to 𝑥^∗∗. It follows that 𝑥^∗∗(𝑥^∗)=1=‖𝑥^∗∗‖.
Now fix a sequence (𝑦^∗_𝑛)_𝑛∈ℕ in 𝑆_𝑋^∗ such that 𝑦^∗_𝑛(𝑥_𝑛+𝑥)→2. Then 𝑦^∗_𝑛(𝑥_𝑛)→1 and 𝑦^∗_𝑛(𝑥)→1. There are 𝑦^∗∈𝐵_𝑋^∗ and a subnet (𝑦^∗_𝜓(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^∗(𝑥_𝑛)→1. It follows that 𝑥^∗∗(𝑦^∗)=1=𝑥^∗∗(𝑥^∗); hence ‖𝑥^∗+𝑦^∗‖=2.
Because of 𝑦^∗(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. If 𝑋 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 space 𝑋 it is possible to prove some more results.

Proposition 27. Let 𝑋 be a reflexive Banach space. (i)If 𝑋 is WLUR in the sense of Lovaglia, then 𝑋 is 𝑙𝑢𝑎𝑐𝑠⁺.(ii)If 𝑋 is sluacs and 𝑙𝑢𝑎𝑐𝑠⁺, then 𝑋 is wuacs.(iii)If 𝑋 is wuacs and 𝑅, then 𝑋 is WLUR.

Proof. (i) follows directly from the Proposition 25 and [4, Proposition 2.15]. Of the remaining assertions we will only prove (iii) explicitly.
Let (𝑥_𝑛)_𝑛∈ℕ be a sequence in 𝑆_𝑋 and 𝑥∈𝑆_𝑋 such that ‖𝑥_𝑛+𝑥‖→2. We can find a sequence (𝑥^∗_𝑛)_𝑛∈ℕ in 𝑆_𝑋^∗ such that 𝑥^∗_𝑛(𝑥_𝑛+𝑥)→2, and hence 𝑥^∗_𝑛(𝑥_𝑛)→1 and 𝑥^∗_𝑛(𝑥)→1.
Since 𝑋 is reflexive, we may assume that (𝑥^∗_𝑛)_𝑛∈ℕ is weak*-convergent to some 𝑦^∗∈𝐵_𝑋^∗ and (𝑥_𝑛)_𝑛∈ℕ is weakly convergent to some 𝑦∈𝐵_𝑋. It follows that 𝑦^∗(𝑥)=1 and hence ‖𝑥^∗_𝑛+𝑦^∗‖→2.
Since 𝑋 is wuacs, the dual space 𝑋^∗ is sluacs (cf. [4, Proposition 2.16]), and thus (because of 𝑥^∗_𝑛(𝑥_𝑛)→1) we can conclude that 𝑦^∗(𝑥_𝑛)→1, whence 𝑦^∗(𝑦)=1=𝑦^∗(𝑥), which implies ‖𝑥+𝑦‖=2, which by the rotundity of 𝑋 implies 𝑥=𝑦.

Proposition 28. Let 𝑋 be a reflexive Banach space with the Kadets-Klee property. (i)If 𝑋 is acs then 𝑋, is luacs.(ii)If 𝑋 is WLUR in the sense of Lovaglia, then 𝑋 is wuacs and 𝑠𝑙𝑢𝑎𝑐𝑠⁺.(iii)If 𝑋 is WLUR in the sense of Lovaglia and 𝑅, then 𝑋 is wuacs and LUR.

Proof. (i) Let (𝑥_𝑛)_𝑛∈ℕ, 𝑥, and 𝑦 be as in the proof of (iii) of the previous proposition and let 𝑥^∗∈𝑆_𝑋^∗ with 𝑥^∗(𝑥_𝑛)→1. Then 𝑥^∗(𝑦)=1 and hence ‖𝑦‖=1. Since 𝑋 has the Kadets-Klee property, it follows that ‖𝑥_𝑛−𝑦‖→0 and thus ‖𝑥+𝑦‖=2. Because 𝑋 is acs, we obtain 𝑥^∗(𝑥)=1, as desired.
(ii) We first show that 𝑋 is wuacs. Take two sequences (𝑥_𝑛)_𝑛∈ℕ and (𝑦_𝑛)_𝑛∈ℕ in 𝑆_𝑋 such that ‖𝑥_𝑛+𝑦_𝑛‖→2 and a functional 𝑥^∗∈𝑆_𝑋^∗ with 𝑥^∗(𝑥_𝑛)→1. By the reflexivity of 𝑋 we may assume that (𝑥_𝑛)_𝑛∈ℕ is weakly convergent to some 𝑥∈𝐵_𝑋. Then 𝑥^∗(𝑥)=1; hence ‖𝑥‖=1.
But 𝑋 has the Kadets-Klee property, so this implies ‖𝑥_𝑛−𝑥‖→0.
Now fix a sequence (𝑦^∗_𝑛)_𝑛∈ℕ in 𝑆_𝑋^∗ such that 𝑦^∗_𝑛(𝑥_𝑛)→1, and 𝑦^∗_𝑛(𝑦_𝑛)→1. It follows that 𝑦^∗_𝑛(𝑥)→1 and consequently ‖𝑦_𝑛+𝑥‖→2.
Since 𝑥^∗(𝑥)=1 and 𝑋 is WLUR in the sense of Lovaglia, we get 𝑥^∗(𝑦_𝑛)→1, proving that 𝑋 is wuacs.
Now we will show that 𝑋 is sluacs. Take (𝑥_𝑛)_𝑛∈ℕ and 𝑥 in 𝑆_𝑋 with ‖𝑥_𝑛+𝑥‖→2 and a sequence (𝑥^∗_𝑛)_𝑛∈ℕ in 𝑆_𝑋^∗ such that 𝑥^∗_𝑛(𝑥_𝑛)→1. Also, fix a sequence (𝑦^∗_𝑛)_𝑛∈ℕ in 𝑆_𝑋^∗ with 𝑦^∗_𝑛(𝑥_𝑛)→1 and 𝑦^∗_𝑛(𝑥)→1.
We may assume that (𝑥_𝑛)_𝑛∈ℕ is weakly convergent to some 𝑦∈𝐵_𝑋 and (𝑦^∗_𝑛)_𝑛∈ℕ is weak*-convergent to some 𝑦^∗∈𝐵_𝑋^∗. It follows that 𝑦^∗(𝑥)=1, and hence ‖𝑦^∗+𝑦^∗_𝑛‖→2.
Since 𝑋 is wuacs, 𝑋^∗ is sluacs, and thus we get 𝑦^∗(𝑥_𝑛)→1. It follows that 𝑦^∗(𝑦)=1; hence ‖𝑦‖=1 and ‖𝑥+𝑦‖=2. The Kadets-Klee property of 𝑋 gives us ‖𝑥_𝑛−𝑦‖→0.
Because of 𝑥^∗_𝑛(𝑥_𝑛)→1, we can now infer 𝑥^∗_𝑛(𝑦)→1. Since 𝑋 is in particular acs, this implies 𝑥^∗_𝑛(𝑥)→1 (cf. [4, Proposition 2.19]).
We will skip the last part of the proof, the reverse implication in the definition of sluacs⁺.
(iii) By (ii) 𝑋 is wuacs and sluacs⁺. Let us take a sequence (𝑥_𝑛)_𝑛∈ℕ in 𝑆_𝑋 and an element 𝑥∈𝑆_𝑋 such that ‖𝑥_𝑛+𝑥‖→2. Fix a sequence (𝑥^∗_𝑛)_𝑛∈ℕ in 𝑆_𝑋^∗ such that 𝑥^∗_𝑛(𝑥_𝑛)=1, for every 𝑛∈ℕ. Since 𝑋 is sluacs, it follows that 𝑥^∗_𝑛(𝑥)→1.
Assume that (𝑥_𝑛)_𝑛∈ℕ is weakly convergent to 𝑦∈𝐵_𝑋 and that (𝑥^∗_𝑛)_𝑛∈ℕ is weak*-convergent to 𝑥^∗∈𝐵_𝑋^∗. It follows that 𝑥^∗(𝑥)=1 and hence 𝑥^∗∈𝑆_𝑋^∗. Moreover, since 𝑋 is WLUR in the sense of Lovaglia, we get that 𝑥^∗(𝑥_𝑛)→1.
Since (𝑥_𝑛)_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 ‖𝑥_𝑛−𝑦‖→0.
Because of 𝑥^∗(𝑥)=𝑥^∗(𝑦)=1, we must have ‖𝑥+𝑦‖=2 and thus the rotundity of 𝑋 implies 𝑥=𝑦.

Proposition 29. Let 𝑋 be a Banach space such that 𝑋^∗ has the Kadets-Klee* property and 𝐵_𝑋^∗ is weak*-sequentially compact. (i)If 𝑋 is 𝑆, then it is also WLUR in the sense of Lovaglia.(ii)If 𝑋^∗ is acs, then 𝑋 is 𝑙𝑢𝑎𝑐𝑠⁺, and for all sequences (𝑥_𝑛)_𝑛∈ℕ  in 𝑆_𝑋, (𝑥^∗_𝑛)_𝑛∈ℕin 𝑆_𝑋^∗ and every 𝑥∈𝑆_𝑋 with ‖𝑥_𝑛+𝑥‖→2 and 𝑥^∗_𝑛(𝑥)→1 one has 𝑥^∗_𝑛(𝑥_𝑛)→1. (iii) If 𝑋^∗ is WLUR* in the sense of Lovaglia, then 𝑋 is sluacs.

Proof. We will only prove (iii), so let (𝑥_𝑛)_𝑛∈ℕ and 𝑥 be in 𝑆_𝑋 with ‖𝑥_𝑛+𝑥‖→2 and (𝑥^∗_𝑛)_𝑛∈ℕ a sequence in 𝑆_𝑋^∗ such that 𝑥^∗_𝑛(𝑥_𝑛)→1. Let (𝑦^∗_𝑛)_𝑛∈ℕ be a sequence in 𝑆_𝑋^∗ with 𝑦^∗_𝑛(𝑥_𝑛)→1 and 𝑦^∗_𝑛(𝑥)→1.
By assumption, we may suppose that (𝑦^∗_𝑛)_𝑛∈ℕ is weak*-convergent to some 𝑦^∗∈𝐵_𝑋^∗. Then 𝑦^∗(𝑥)=1; hence 𝑦^∗∈𝑆_𝑋^∗. By the Kadets-Klee* property of 𝑋^∗ we must have ‖𝑦^∗_𝑛−𝑦^∗‖→0.
It follows that 𝑦^∗(𝑥_𝑛)→1; hence ‖𝑥^∗_𝑛+𝑦^∗‖→2. Since 𝑋^∗ is WLUR* in the sense of Lovaglia, we obtain 𝑥^∗_𝑛(𝑥)→1.