K\"othe-Bochner spaces and some geometric properties related to rotundity and smoothness

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 preprint the author introduced some further related notions and investigated the behaviour of these geometric properties under the formation of absolute sums. This paper is in a sense a continuation of the previous work. Here we will study the behaviour of said properties under the formation of K\"othe-Bochner spaces, thereby generalising some results of Sirotkin on the acs, luacs and uacs properties of $L^p$-Bochner spaces.


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 1.1. A Banach space X is called (i) rotund (R in short) if for any two elements x, y ∈ S X the equality x + y = 2 implies x = y, (ii) locally uniformly rotund (LUR in short) if for every x ∈ S X the implication x n + x → 2 ⇒ x n − x → 0 holds for every sequence (x n ) n∈N in S X , (iii) weakly locally uniformly rotund (WLUR in short) if for every x ∈ S X and every sequence (x n ) n∈N in S X we have 1. Introduction (iv) uniformly rotund (UR in short) 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 holds, (v) weakly uniformly rotund (WUR in short) 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 → 0 weakly.
The chart below shows the obvious implications between these notions. No other implications are valid in general (see the examples in [19]). Note, however, that all these notions coincide in finite-dimensional spaces, by the compactness of B X .
For the local version one defines δ X (x, ε) = inf{1 − 1/2 x + y : y ∈ B X and x − y ≥ ε} for every x ∈ S X and each ε ∈]0, 2]. Then X is LUR iff 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 in short) if its norm is Gâteaux-differentiable at every non-zero point (equivalently at every point of S X ), which is the case iff for every x ∈ S X there is a unique functional x * ∈ S X * with x * (x) = 1 (cf. [9,Lemma 8.4 (ii)]). X is called Fréchet-smooth (FS in short) if the norm is Frécht-differentiable at every non-zero point. The norm of the space X is said to be uniformly Gâteaux-differentiable (UG in short) if for each y ∈ S X the limit lim τ →0 ( x + τ y − 1)/τ exists uniformly in x ∈ S X . Finally, X is called uniformly smooth (US in short) if lim τ →0 ρ X (τ )/τ = 0, where ρ X denotes the modulus of smoothness of X defined by ρ X (τ ) = sup{1/2( x + τ y + x − τ y − 2) : x, y ∈ S X } for every τ > 0.
In [12] the following notions were introduced (in connection with the so called Anti-Daugavet property).

Introduction
Definition 1.2. A Banach space X is called (i) alternatively convex or smooth (acs in short) if for every x, y ∈ S X with x + y = 2 and every x * ∈ S X * with x * (x) = 1 we have x * (y) = 1 as well, (ii) locally uniformly alternatively convex or smooth (luacs in short) if for every x ∈ S X , every sequence (x n ) n∈N in S X and every functional x * ∈ S X * we have x n + x → 2 and x * (x n ) → 1 ⇒ x * (x) = 1, (iii) uniformly alternatively convex or smooth (uacs in short) if for all sequences (x n ) n∈N , (y n ) n∈N in S X and (x * n ) n∈N in S X * we have x n + y n → 2 and x * n (x n ) → 1 ⇒ x * n (y n ) → 1.
The author introduced the following related notions in [11]. Definition 1.3. A Banach space X is called (i) strongly locally uniformly alternatively convex or smooth (sluacs in short) if for every x ∈ S X and all sequences (x n ) n∈N in S X and (x * n ) n∈N in S X * we have x n + x → 2 and x * n (x n ) → 1 ⇒ x * n (x) → 1, (ii) weakly uniformly alternatively convex or smooth (wuacs in short) if for any two sequences (x n ) n∈N , (y n ) n∈N in S X and every functional x * ∈ S X * we have x n + y n → 2 and x * (x n ) → 1 ⇒ x * (y n ) → 1.
The obvious implication between the acs properties and the rotundity properties are indicated in the following chart. No other implications are generally valid (see the examples in [11]), but note again that the properties acs, luacs, sluacs, wuacs and uacs coincide in finite-dimensional spaces, by compactness.

Introduction
The connection between some of the acs properties to smoothness properties is illustrated in the diagram below.

US
UG S uacs sluacs acs Fig. 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 resp. sluacs we still obtain the same classes of spaces. For uacs spaces this was first proved by G. Sirotkin in [18] 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 [11,Proposition 2.1]).
Then X is uacs iff δ X uacs (ε) > 0 for every ε ∈]0, 2] and we clearly have 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 [14] and our modulus δ X uacs is the same as the modulus of u-convexity from [10]. Also, the notion of u-spaces which was introduced in [7] coincides with the notion of acs spaces.
Recall that a Banach space X is said to be uniformly non-square 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 non-square and hence by a well-known theorem of James (cf. [3, p.261]) they are superreflexive, as was observed in [12,Lemma 4.4]. For a proof of the superreflexivity of uacs spaces that does not rely on James' result on uniformly non-square spaces, see [11,Proposition 2.8].
Let us also restate here the following auxiliary result [11, Lemma 2.30] (it is the generalisation of [1, Lemma 2.1] to sequences, with a completely analogous proof). Lemma 1.5. 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 non-negative real numbers we also have α n x n + β n y n − α n x n − β n y n → 0.
Finally, we will need two more definitions from [11].
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.

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 realvalued measurable 1 functions on S modulo equality µ-almost everywhere 2 such that (iii) if g is measurable and f ∈ E such that |g(t)| ≤ |f (t)| µ-a. e. then g ∈ E and g ≤ f .
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 iff 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 [15, Proposition 3.1.5] E is order continuous iff E is σ-order continuous iff E is order complete and order continuous. Also, reflexivity of E implies order continuity, for any σ-order complete Banach lattice which is not 1 i. e. A-Borel-measurable 2 We will henceforth abbreviate this by µ-a. e. or simply a. e. if µ is tacitly understood. Let us also mention the following well-known fact that will be needed later.
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 Then (E ′ , · E ′ ) is again a Köthe function space, the so called Köthe dual of E. The operator T : E ′ → E * defined by is well-defined, linear and isometric. Moreover, T is onto iff E is order continuous (cf. [15, p.149]), thus for order continuous E we have E * = E ′ .
We refer the reader to [16] or [15] 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 measurable sets A 1 , . . . , A n ∈ A such that ∞ i=1 A i = S and f is constant on each A 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) = 0 µ-a. e. and weakly measurable if x * • f is measurable for every functional x * ∈ X * . According to Pettis' measurability theorem (cf. [15,Theorem 3.2.2]) f is Bochner-measurable iff f is weakly measurable and almost everywhere separably valued (i. e. there is a separable subspace Y ⊆ X such that f (t) ∈ Y µ-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 (·) E E(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 iff F (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 Then the following deep theorem holds.
Theorem 2.2 (cf. [4]). 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 : is an isometric isomorphism and moreover every equivalence class L in Sirotkin proved in [18] that for 1 < p < ∞ the Lebesgue-Bochner space L p (X) is acs resp. luacs resp. uacs whenever X has the respective property. In the next section we will study the more general case of Köthe-Bochner spaces.

Results and proofs
We begin with the acs spaces, for which we have the following result.
Proposition 3.1. 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 [11,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 2.2 l can be represented via and

Results and proofs
We also have Since E is acs it follows from (3.1) and (3. In a similar way as we have obtained (3.3) we can also show f (·) + g(·) + f (·) + g(·) E = 4.
Now we will show that F (t)(g(t)) = F (t) g(t) a. e. (3.8) To this end, let us denote by N 1 resp. N 2 the null sets on which the equality from (3.2) resp. (3.7) does not hold. Let N = N 1 ∪ N 2 . Put B = {t ∈ S \ N : F (t) = 0 and g(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 (3.1) since E is acs. Taking into account (3.4) we arrive at 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 as well as .
Since X is acs it follows that F (t) g(t) = F (t)(g(t)).
) for every t ∈ S \ M and (3.8) is proved. Now combining (3.4) and (3.8) we obtain which finishes the proof.
Before we turn to the case of luacs spaces, let us recall Egorov's theorem (cf. [20, Satz IV.6.7]), 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 3.2. 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 el- As before, we can represent l by an element [F ] ∈ E ′ (X * , w * ). We then have By passing to a subsequence we may also assume that We further have and thus lim An analogous argument also shows Moreover, the inequality Analogously one can see that (3.14) Finally, we have Since E is in particular luacs we get from (3.9) and (3.11) that Because E is in any case luacs + it follows from (3.13), (3.15) and (3.16) that So by passing to a further subsequence we may assume Next we will show that Denote by N 1 resp. N 2 the null sets on which the convergence statement from (3.10) resp. (3.17) does not hold and let N = 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 and 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=1 A 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 and moreover this sequence is decreasing, so the order continuity of E ′ implies lim 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 Now if (a) holds, i. e. if E is WLUR then by (3.11) the sequence ( f k (·) ) k∈N must be weakly convergent to f (·) in E and hence lim k→∞ C∩(Am\Bn,m) for all n, m ∈ N. Since ( f (·) χ C∩(Am\Bn,m) ) n∈N dereases 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 Combinig this with (3.9) leaves us with Since E is luacs and because of (3.11) it follows that Taking into account (3.16) we get By passing to a subsequence we may assume that ( f n (t) ) n∈N is bounded away from zero. Then it follows from Lemma 1.5 that Since X is luacs we can conclude that and we are done.

Recall that a subset
It is well-known that for a finite measure µ a bounded subset A ⊆ L 1 (µ) is relatively weakly compact in L 1 (µ) if and only if A is equi-integrable (see for instance [21, Satz VIII.6.9]). One ingredient for the usual proof of this fact is the following lemma (see [21, Lemma VIII.6.7]), which we will also need in the sequel.
We will also need Vitali's Lemma, which reads as follows (see for example [15,Lemma 3.1.13]) for an even more general version).
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 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. [16, p.28]). With this in mind we can prove the following result concerning luacs + spaces. Theorem 3.5. 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 1.6 (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 and Also, just as we have done in the previous proof, we find that Hence by passing to a subsequence we may assume that (cf. Lemma 2.1) Since X is luacs + it follows from (3.20), (3.28) and (3.29) that Thus by Lemma 3.3 the sequence ( F (·) f n (·) χ B ) n∈N and hence also the 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 m 0 ∈ N such that F (·) χ S\Am 0 E ′ ≤ ε/3.
It follows that for every n ≥ N and because of (3.30) it follows as before that finishing the proof. Now we turn to the sluacs spaces. An easy normalisation argument shows that a Banach space X is sluacs iff 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 iff for every x ∈ S X and every 0 < ε ≤ 2 the number Next we will prove an easy Lemma on the continuity of β X .
Lemma 3.6. For all 0 < ε,ε, ≤ 2 and all x,x ∈ S X we have i. e. β X is 1-Lipschitz continuous with respect to the norm of X ⊕ 1 R.
Proof. First we fix 0 < ε ≤ 2 and x,x ∈ S X . Put δ = x −x and take Thus we get and since 0 < τ < 1 was arbitrary it follows that Again, since (y, x * ) ∈ V x,ε was arbitray we can conclude that and by symmetry it folows that Analogously one can prove that for all x ∈ S X and all 0 < ε,ε, ≤ 2. An application of the triangle inequality then yields the result.
In the paper [13] A. Kamińska and B. Turett proved various theorems concerning different rotundity properties of Köthe-Bochner spaces. For example, by [13,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 [13,Theorem 5] to show the following result.
Theorem 3.7. If E is LUR and X is sluacs 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 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=1 A 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 n 0 ∈ N with 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 3.1 we can conclude and g(t)) a. e. (3.34) Next we define C := {t ∈ S : F (t) = 0} and Then B is measurable and Let us fix 0 < η < min{ε/16, 1/2n 0 } such that Now consider the sets and again by definition of B 1 we obtain In the case 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 (3.34) does not hold and suppose that t ∈ B 2 ∩ A n 0 ∩ (S \ N ). Then in particular t ∈ B and f (t) ≥ g(t) and hence Moreover, by the definitions of B 2 and A n 0 and the choice of η we have Since t ∈ (S \ N ) we also have So by definition of β X we must have Once more by the definition of B 1 this implies Since F (t) g(t) = F (t)(g(t)) the definition of β X implies that 1 2 where the latter inequality holds because of t ∈ A n 0 . It follows that where α 2 := 1/n 0 −η(2−2η) −1 which by (3.36) is greater than zero. Becuase of f (t) ≤ g(t) it follwos that ).
So if we put α = min{α 1 , α 2 } and P Now we will show that if i = 2 resp. i = 3 then Let us first assume i = 2, i. e.
Since f (t) ≥ g(t) for t ∈ B 2 it follows that Because N is a null set we have where the second last inequality holds because of (3.32). Now assume that i = 3, i. e.
It follows that and hence as before we get We further have Altogether we have shown that for we have for every g ∈ S E(X) and every l ∈ S E(X) * with l(g) = 1 and By the aforementioneed characterisation of sluacs spaces ([11, Proposition 2.1]) this implies that E(X) is sluacs.
Next we will have a look at the case of wuacs spaces.
Theorem 3.8. 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 3.2 we find and lim n→∞ f n (·) + g n (·) + f n (·) + g n (·) E = 4. and 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) ) = 0 a. e. (3.44) 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 h 1 ∈ B E resp. h 2 ∈ B E . In view of (3.38) and (3.42) it follows that hence h 1 E = h 2 E = 1 and moreover Because of (3.45) and since E is in particular acs we can show just as in the proof of Proposition 3.1 that C is a null set. The fact that X is wuacs together with Lemma 1.5 easily implies that By the weak convergence of ( g n (·) ) n∈N to h 2 we have Since E is reflexive E ′ is order continuous and thus we can deduce as in the proof of Theorem 3.5, with the aid of Vitali's Lemma, (3.48), (3.47) 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.
If we combine the techniques of the proofs of Theorem 3.8 and Theorem 3.5 we can also obtain another result concerning luacs + spaces (we omit the details). Theorem 3.9. 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 3.10. 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 (3.49) and by passing to a subsequence also lim n→∞ ( F n (t) f n (t) − F n (t)(f n (t))) = 0 a. e. (3.50) as well as So we can pass to another subsequence such that 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 (3.56) that S g(t) f (t) dµ(t) = 1 (3.59) and hence g ∈ S E * . Because of (3.59), (3.51) and the fact that E is sluacs + we get that whence h ∈ S E . Since both E and E * have the Kadets-Klee property it follows that Thus we can pass once more to a subsequence such that Let N be a null set such that the convergence statements of (3.50), (3.58) and (3.62) hold for every t ∈ S\N and put B = {t ∈ S \ N : g(t) = 0 andf (t) = 0} as well as C = {t ∈ B : h(t) = 0}. Similar to the arguments in the proof of Theorem 3.8 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))) = 0 a. e.
By our usual method based on Vitali's Lemma we can conclude that for every A ∈ A with µ(A) < ∞ we have Now we fix an increasing sequence (A m ) m∈N in A such that µ(A m ) < ∞ for all m ∈ N and ∞ m=1 A 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 3.5 this together with (3.64) leads to Taking into account (3.56) we arrive at lim n→∞ l n (f ) = lim n→∞ S F n (t)(f (t)) dµ(t) = 1 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 3.11. 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 the Theorems 3.7 and 3.10 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 and by passing to a subsequence also lim n→∞ ( F n (t) f (t) − F n (t)(f (t))) = 0 a. e. so that by passing to another subsequence we can assume 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 (3.65) that and hence g E ′ = 1. Since E * has the Kadets-Klee* property we get that and thus we can, by passing to yet another subsequence, assume that lim n→∞ F n (t) = g(t) a. e. (3.77) Next we claim that there is an h ∈ S E such that S g(t)h(t) dµ(t) = 1 (3.78) and, after passing to a subsequence once more, For in the case (b) E is LUR and thus by (3.67) and (3.75) we can take h = f (·) . In the case (a) E is reflexive and hence we can assume that ( f n (·) ) n∈N is weakly convergent to some h ∈ B E . Then (3.78) follows from (3.76) and (3.72). This also implies h E = 1 and by the Kadets-Klee property of E we have (3.79). By (3.79) we may assume that ( F n (t) f n (t) − F n (t)(f n (t))) dµ(t) = 0 (3.82) 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 3.10.
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 m 0 ∈ N with f (·) χ S\Am 0 E ≤ δ/2. Because of (3.67), (3.65) and (3.82) there is an N ∈ N such that for all n ≥ N the inequalities It follows that for every n ≥ N we have and hence by the choice of δ Consequently, for every n ≥ N we have Thus we have shown Together with (3.72) it follows l n (f n ) → 1, as desired. Now let us treat the case of uacs spaces. In analogy to [11,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 This property certainly implies that E is uacs. Every UR space has property (u + ). The following theorem holds. Its proof is completely analogous to the one of [11,Theorem 3.16] (which is a modification of the proof of [5, Theorem 3]) but we will explicitly give it here, for the readers convenience.
Theorem 3.12. 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.
Since E is uacs we may find τ > 0 such that for all a, b ∈ B E and every l ∈ B E * we have Next we fix 0 < ρ < min{ε/2, 2τ, ω} and find a numberτ 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 Then z is Bochner-measurable and z(t) = f (t) for all t ∈ S (hence z ∈ E(X)). Furthermore, (3.93) As before we have so the choice ofτ together with (3.93) implies Next we observe that and (because of (3.94) and (3.95)) Using (3.95) and (3.96) we can conclude By the choice of η this implies L(z) ≥ 1 − ε/2. But by (3.95) we also have The above theorem admits the following corollary.
Corollary 3.13. If E is a US Köthe function space and X is a uacs Banach space then E(X) is also uacs.
Proof. Since uacs is a self-dual property (cf. [11,Corollary 2.13]) X * is also uacs and since E is US we have that E * = E ′ is UR (cf. [9, 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 [4] 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 in short) if for any two sequences (x n ) n∈N and (y n ) n∈N in S X and every x ∈ S X we have x − x n + y n 2 → 0 ⇒ x n − y n → 0.
X is called weakly midpoint locally uniformly rotund (WMLUR in short) if it satisfies the above condition with x n − y n → 0 replaced by x n − y n σ − → 0, where the symbol σ − → denotes the convergence in the weak topology of X. The notion of MLUR spaces was originally introduced by Anderson in [2].
In [11] the author introduced the following analogous definitions.
Definition 3.14. Let X be a Banach space.
(i) The space X is said to be midpoint locally uniformly alternatively convex or smooth (mluacs in short) 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 * we have that x − x n + y n 2 → 0 and x * (x n ) → 1 ⇒ x * (y n ) → 1.
(ii) The space X is called midpoint strongly locally uniformly alternatively convex or smooth (msluacs in short) 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 * we have that x − x n + y n 2 → 0 and x * n (x n ) → 1 ⇒ x * n (y n ) → 1.
The chart below summarises the obvious implications. No other implications are true in general (see the examples in [11] Concerning the properties msluacs and mluacs for Köthe-Bochner spaces we have the following result. Theorem 3.15. 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. [15,Theorem 2.1.5]) and so by [15, 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 [11,Proposition 4.7], which in turn uses techniques from the proof of [8,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 funcionals 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 (3.97) and, after passing to an appropriate subsequence, lim n→∞ ( F n (t) f n (t) − F n (t)(f n (t))) = 0 a. e. (3.98) Using this together with (3.99), f n (·) + g n (·) + 2a n = 4 f (·) and the fact that E is MLUR we get that Again, since E is MLUR this implies lim n→∞ f n (·) − g n (·) E = 0. Since B E * is weak*-sequentially compact we may also asssume that ( F n (·) ) n∈N weak*-converges to some g ∈ B E ′ . 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 ( F n (t) f (t) − F n (t)(f (t))) = 0 a. e., since X is msluacs. Using our usual argument via equi-integrability and Vitali's Lemma this leads to lim By the order continuity of E we can derive from this also in the σ-finite case (cf. the proof of Theorem 3.10). Combining (3.107) and (3.105) gives us l n (f ) → 1 and we are done. The statement about mluacs spaces can be proved similarly.
In the last section we will establish some further connections between the various properties that we considered in this paper.

Miscellaneous
In [17] A. 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 and x * (x) = 1 ⇒ x * (x n ) → 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. 3 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 4.1. 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 [11,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 is 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. Becuase of y * (x) = 1 our assumption imlies x * (x) = 1 and we are done.
The following assertion is also easy to prove (we omit the details).
Proposition 4.2. If X is a Banach space which 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. (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 4.1 and [11,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. [11,Proposition 2.16]) and thus (because of x * n (x n ) → 1) we can conclude y * (x n ) → 1, whence y * (y) = 1 = y * (x), which implies x + y = 2, which by the rotundity of X implies x = y. (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.
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.