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SOME REMARKS ON THE DUNFORD-PETTIS PROPERTY 5 9 9 NARCISSE RANDRIANANTOANINA 1 n a J Abstract. Let A be the disk algebra,Ω be a compact Hausdorff 4 2 space andµ be a Borelmeasure onΩ. It is shownthatthe dual of C(Ω,A)hastheDunford-Pettisproperty. Thisprovedinparticular ] thatthespacesL1(µ,L1/H1)andC(Ω,A)havetheDunford-Pettis A 0 property. F . h t a 1. Introduction m [ Let E be a Banach space, Ω be a compact Hausdorff space and µ 1 be a finite Borel measure on Ω. We denote by C(Ω,E) the space of all v 4 E-valued continuous functions from Ω and for 1 ≤ p < ∞, Lp(µ,E) 1 stands for the space of all (class of) E-valued p-Bochner integrable 2 1 functions with its usual norm. A Banach space E is said to have the 0 5 Dunford-Pettispropertyifevery weakly compact operatorwithdomain 9 / E is completely continuous (i.e., takes weakly compact sets into norm h at compact subsets of the range space). There are several equivalent def- m initions. The basic result proved by Dunford and Pettis in [7] is that : v the space L1(µ) has the Dunford-Pettis property. A. Grothendieck i X [8] initiated the study of Dunford-Pettis property in Banach spaces r a and showed that C(K)-spaces have this property. The Dunford-Pettis property has a rich history; the survey articles by J. Diestel [4] and A. Pe lczyn´ski [10] are excellent sources of information. In [4], it was asked if the Dunford-Pettis property can be lifted from a Banach E to C(Ω,E) or L1(µ,E). M. Talagrand [13] constructed counterexam- ples for these questions so the answer is negative in general. There 1991 Mathematics Subject Classification. 46E40,46E25. Key words and phrases. Bochner spaces, Dunford-Pettis property, weakly com- pact sets. 1 2 NARCISSE RANDRIANANTOANINA are however some positive results. For instance, J. Bourgain showed (among other things) in [2] that C(Ω,L1) and L1(µ,C(Ω)) both have the Dunford-Pettis property; K. Andrews [1] proved that if E∗ has the Schur property then L1(µ,E) has the Dunford-Pettis property. In [12], E. Saab and P. Saab observed that if A is a C∗-algebra with the Dunford-Pettisproperty then C(Ω,A)has theDunford-Pettis property and they asked (see [12] Question 14, p.389) if a similar result holds if one considers the disk algebra A. In this note we provide a posi- tive answer to the above question by showing that the dual of C(Ω,A) has the Dunford-Pettis property. This implies in particular that both L1(µ,L1/H1) and C(Ω,A) have the Dunford-Pettis property. Our ap- 0 proach is to study a “Random version” of the minimum norm lifting from L1/H1 into L1. 0 The notation and terminology used and not defined in this note can be found in [5] and [6]. 2. Minimum norm lifting Let us beging by fixing some notations. Throughout, m denotes the normalized Haar measure on the circle T. The space H1 stands for the 0 spaceofintegrablefunctionsonTsuchthatfˆ(n) = f(θ)e−inθdm(θ) = T R 0 for n ≤ 0. It is a well known fact that A∗ = L1/H1 ⊕ M (T) where M (T) is 0 1 S S thespace ofsingular measures onT(see forinstance [10]). Consider the quotient map q : L1 → L1/H1. This map has the following important 0 property: for each x ∈ L1/H1, there exists a unique f ∈ L1 so that 0 q(f) = x and kfk = kxk. This fact provides a well-defined map called the minimum norm lifting σ : L1/H1 L1 such that q(σ(x)) = x and kσ(x)k = kxk . 0 One of the many important features of σ is that it preserves weakly compact subsets, namely the following was proved in [10]. Proposition 1. If K is a relatively weakly compact subset of L1/H1 0 then σ(K) is relatively weakly compact in L1. SOME REMARKS ON THE DUNFORD-PETTIS PROPERTY 3 Our goal in this section is to extend the minimum norm lifting to certain classes of spaces that contains L1/H1. In particular we will 0 introduce a random-version of the minimum norm lifting. First we will extend the minimum norm lifting to A∗. We define a map γ : L1/H1 ⊕ M (T) L1 ⊕ M (T) as follows: 0 1 s 1 s γ({x,s}) = {σ(x),s}. Clearly γ defines a minimum norm lifting from A∗ into M(T). In order to procede to the next extension, we need the following proposition: Proposition 2. Let σ and γ as above then a) σ : L1/H1 L1 is norm-universally measurable (i.e., the inverse 0 image of every norm Borel subset of L1 is norm universally mea- surable in L1/H1); 0 b) γ : A∗ M(T) is weak*-universally measurable (i.e, the inverse image of every weak*-Borel subset of M(T) is weak*-universally measurable in A∗). Proof. For a), notice that L1/H1 and L1 are Polish spaces (with the 0 norm topologies) and so is the product L1 × L1/H1. Consider the 0 following subset of L1 ×L1/H1: 0 A = {(f,x); q(f) = x, kfk = kxk} . The set A is a Borel subset of L1×L1/H1. In fact, A is the intersection 0 of the graph of q (which is closed) and the subset A = {(f,x),kfk = 1 kxk} which is also closed. Let π be the restriction on A of the second projection of L1 × L1/H1 onto L1/H1. The operator π is of course 0 0 continuous and hence π(A) is analytic. By Theorem 8.5.3 of [3], there exists a universally measurable map φ : π(A) → L1 whose graph be- longs to A. The existence and the uniqueness of the minimum norm lifting imply that π(A) = L1/H1 and φ must be σ. 0 The proof of b) is done with simmilar argument using the fact A∗ and M(T) with the weak* topologies are countable reunion of Polish 4 NARCISSE RANDRIANANTOANINA spaces and their norms are weak*-Borel measurable. The proposition is proved. Let (Ω,Σ,µ) be a probability space. For a measurable function f : Ω → L1/H1, the function ω 7→ σ(f(ω)) (Ω → L1) is µ-measurable by 0 Proposition 2. We define an extension of σ on L1(µ,L1/H1) as follows: 0 σ˜ : L1 µ,L1/H1 L1(µ,L1) with σ˜(f)(ω) = σ(f(ω)) for ω ∈ Ω . 0 (cid:16) (cid:17) Themapσ˜ iswelldefinedandkσ˜(f)k = kfkforeachf ∈ L1(µ,L1/H1). 0 Also if we denote by q˜: L1(µ,L1) → L1(µ,L1/H1), the map q˜(f)(ω) = 0 q(f(ω)), we get that q˜(σ˜(f)) = f. Similarly if f : Ω → A∗ is weak*-scalarly measurable, the function ω 7→ γ(f(ω)) (Ω → M(T)) is weak*-scalarly measurable . As above we define γ˜ as follows: for each measure G ∈ M(Ω,A∗), fix g : Ω → A∗ its weak*-density with respect to its variation |G|. We define γ˜(G)(A) = weak*− γ(g(ω)) d|G|(ω) for all A ∈ Σ. Z A Clearly γ˜(G) is a measure and it is easy to check that ||γ˜(G)|| = ||G|| (in fact |γ˜(G)| = |G|). The rest of this section is devoted to the proof of the following result that extends the property of σ stated in Proposition 1 to σ˜. Theorem 1. LetK be arelativelyweaklycompactsubsetof L1(µ,L1/H1). 0 The set σ˜(K) is relatively weakly compact in L1(µ,L1). We will need few general facts for the proof. In the sequel, we will identify (for a given Banach space F) the dual of L1(µ,F) with the space L∞(µ,F ∗) of all map h from Ω to F∗ that are weak*-scalarly σ measurable and essentially bounded with the uniform norm (see [14]). ∞ Definition 1. Let E be a Banach space. A series x in E is said n nP=1 to be weakly unconditionally Cauchy (WUC) if for every x∗ ∈ E∗, the ∞ series |x∗(x )| is convergent. n nP=1 The following lemma is well known: SOME REMARKS ON THE DUNFORD-PETTIS PROPERTY 5 Lemma 1. If S is a relatively weakly compact subset of a Banach space ∞ E, then for every WUC series x∗ in E∗, lim x∗(x) = 0 uniformly nP=1 n m→∞ n on S. The following proposition which was proved in [11] is the main in- gredient for the proof of Theorem 1. For what follows (e ) denote the n n unit vector basis of c and (Ω,Σ,µ) is a probability space. 0 Proposition 3. [11] Let Z be a separable subspace of a real Banach space E and (f ) be a sequence of maps from Ω to E∗ that are weak*- n n scalarly measurable and sup kf k ≤ 1. Let a < b (real numbers) n n ∞ then: There exist a sequence g ∈ conv{f ,f ,...} measurable subsets n n n+1 C and L of Ω with µ(C ∪L) = 1 such that (i) If ω ∈ C and T ∈ L(c ,Z), kTk ≤ 1; then for each h ∈ 0 n conv{g ,g ,...}, eitherlimsuphh (ω),Te i ≤ b orliminfhh (ω),Te i ≥ n n+1 n n n n n→∞ n→∞ a; (ii) ω ∈ L, there exists k ∈ N so that for each infinite sequence of zeroes and ones Γ, there exists T ∈ L(c ,Z), kTk ≤ 1 such that 0 for n ≥ k, Γ = 1 =⇒ hg (ω),Te i ≥ b n n n Γ = 0 =⇒ hg (ω),Te i ≤ a . n n n We will also make use of the following fact: Lemma 2. ([10], p.45) Let (U ) be a bounded sequence of positive ele- n n ments of L1(T). If (U ) is not uniformly integrable, then there exists a n n ∞ W.U.C. series a inthe discalgebraA suchthat: limsupsup |ha ,U i| > ℓ n ℓ n ℓP=1 ℓ→∞ 0. PROOF OF THEOREM 1. Assume without loos of generality that K is a bounded subset of L∞(µ,L1/H1). The set σ˜(K) is a bounded 0 subset of L∞(µ,L1(T)). Let |σ˜(K)| = {|σ˜(f)|;f ∈ K}. Notice that for each f ∈ L1(µ,L1/H1), there exists h ∈ L∞(µ,H∞) = L1(µ,L1/H1)∗ 0 σ 0 with khk = 1 and |σ˜(f)(ω)| = σ˜(f)(ω).h(ω) (the multiplication of the 6 NARCISSE RANDRIANANTOANINA function σ˜(f)(ω) ∈ L1(T) with the function h(ω) ∈ H∞(T)) for a.e. ω ∈ Ω. Consider ϕ = |σ˜(f )| be a sequence of L1(µ,L1(T)) with (f ) ⊂ K n n n n and choose (h ) ∈ L∞(µ,H∞) so that ϕ (ω) = σ˜(f )(ω).h (ω) ∀ n ∈ n n σ n n n N. Lemma 3. There exists ψ ∈ conv{ϕ ,ϕ ,...} so that for a.e. ω ∈ n n n+1 Ω, limhψ (ω),Te i exists for each T ∈ L(c ,A) . n n 0 n→∞ To prove thelemma, let (a(k),b(k)) beanenumeration ofallpairs k∈N of rationals with a(k) < b(k). We will apply Proposition 3 successively starting from (ϕ ) for E = C(T) and Z = A. Note that Proposition 3 n n is valid only for real Banach spaces so we will separate the real part and the imaginary part. Inductively, weconstructsequences (ϕ(k)) andmeasurablesubsets n n≥1 C , L of Ω satisfying: k k (i) C ⊆ C , L ⊆ L , µ(C ∪L ) = 1 k+1 k k k+1 k k (ii) ∀ ω ∈ C and T ∈ L(c ,A), kTk ≤ 1 and j ≥ k, either k 0 limsupRehϕ(j)(ω),Te i ≤ b(k) or n n n→∞ liminfRehϕ(j)(ω),Te i ≥ a(k) n→∞ n n (iii) ∀ ω ∈ L , there exists ℓ ∈ N so that for each Γ infinite sequences k of zeroes and ones, there exists T ∈ L(c ,A), kTk ≤ 1 such that 0 if n ≥ ℓ, Γ = 1 ⇒ Rehϕ(k)(ω),Te i ≥ b(k) n n n Γ = 0 ⇒ Rehϕ(k)(ω),Te i ≤ a(k) ; n n n (iv) ϕ(k+1) ∈ conv{ϕ(k),ϕ(k) ,...} . n n n+1 Again this is just an application of Proposition 3 starting from the se- quenceΩ → C(T)∗ (ω 7→ Re(ϕ (ω)))wherehRe(ϕ (ω)),fi = Rehϕ (ω),fi n n n ∀ f ∈ C(T). Let C = C and L = L . k k k k T S Claim: µ(L) = 0. SOME REMARKS ON THE DUNFORD-PETTIS PROPERTY 7 To see the claim, assume that µ(L) > 0. Since L = L , there k k exists k ∈ N so that µ(L ) > 0. Consider ϕk ∈ conv{ϕS,ϕ ,...} k n n n+1 and let P = {k ∈ N, b(k) > 0} and N = {k ∈ N, a(k) < 0}. Clearly N = P ∪N. Let us assume first that k ∈ P. Using (iii) with Γ = (1,1,1,...), for eachω ∈ L ,thereexistsT ∈ L(c ,A),kTk ≤ 1sothatRehϕ(k)(ω),Te i ≥ k 0 n n b(k). Using similar argument as in [11] Lemma 4., one can construct a map T : Ω → L(c ,A) with: 0 a) ω 7→ T(ω)e is measurable for every e ∈ c ; 0 b) kT(ω)k ≤ 1 ∀ ω ∈ Ω and T(ω) = 0 for ω ∈ Ω\L . k c) Rehϕ(k)(ω),T(ω)e i ≥ b(k) ∀ ω ∈ L . n n k So we get that liminf Rehϕ(k)(ω),T(ω)e idµ(ω) ≥ b(k)µ(L ) n→∞ ZL n n k k which implies that liminf hϕ(k)(ω),T(ω)e idµ(ω) ≥ b(k)µ(L ) . n→∞ (cid:12)(cid:12)ZLk n n (cid:12)(cid:12) k (cid:12) (cid:12) If k ∈ N, we rep(cid:12)eat the same argument wit(cid:12)h Γ = (0,0,0,...) to get that liminf hϕ(k)(ω),T(ω)e idµ(ω) ≥ |a(k)|µ(L ) . n→∞ (cid:12)(cid:12)ZLk n n (cid:12)(cid:12) k (cid:12) (cid:12) So in both cases(cid:12), if δ = max(b(k)µ(L ),|a(cid:12)(k)|µ(L )), there exists a k k map T : Ω → L(c ,A) (measurable for the strong operator topology) 0 so that liminf hϕk(ω),T(ω)e idµ(ω) ≥ δ . (1) n→∞ (cid:12)(cid:12)ZLk n n (cid:12)(cid:12) (cid:12) (cid:12) To get the contradic(cid:12)tion, let (cid:12) qn qn ϕ(k) = λn|σ˜(f )(ω)| = λnσ˜(f )(ω).h (ω) n i i i i i X X i=pn i=pn qn with λn = 1, p < q < p < q < ··· and h ∈ L∞(µ,H∞). i 1 1 2 2 i σ i=Ppn Condition (1) is equivalent to: qn liminf λn hσ˜(f )(ω).h (ω),T(ω)e i dµ(ω) ≥ δ. n→∞ (cid:12)(cid:12)(cid:12)iX=pn i ZLk i i n (cid:12)(cid:12)(cid:12) (cid:12) (cid:12) 8 NARCISSE RANDRIANANTOANINA Therefore there exists N ∈ N so that for each n ≥ N, qn λn hσ˜(f )(ω).h (ω),T(ω)e i dµ(ω) ≥ δ/2; i(cid:12)Z i i n (cid:12) iX=pn (cid:12)(cid:12) Lk (cid:12)(cid:12) (cid:12) (cid:12) for each n ≥ N, choose i(n) ∈ [p ,q ] so that n n hσ˜(f )(ω).h (ω),T(ω)e i dµ(ω) ≥ δ/2 (cid:12)Z i(n) i(n) n (cid:12) (cid:12) Lk (cid:12) (cid:12) (cid:12) and we obta(cid:12)in that for each n ≥ N, (cid:12) hσ(f (ω)),T(ω)e .h (ω)i dµ(ω) ≥ δ/2. (2) (cid:12)Z i(n) n i(n) (cid:12) (cid:12) Lk (cid:12) (cid:12) (cid:12) Notice that(cid:12)for every ω ∈ Ω, T(ω)e ∈ A and h ((cid:12)ω) ∈ H∞(T) so the n i(n) product T(ω)e .h (ω) ∈ H∞(T) and therefore n i(n) hσ(f (ω)),T(ω)e .h (ω)i = hf (ω),T(ω)e .h (ω)i. i(n) n i(n) i(n) n i(n) For n ≥ N, fix T(ω)e .h (ω) ω ∈ L n i(n) k φ (ω) =  . n 0 ω ∈/ L k  ∞ If we set φ = 0 for n < N then the series φ is a W.U.C. series n i iP=1 ∞ in L∞(µ,H∞): to see this notice that for each ω ∈ Ω, T(ω)e is a σ n nP=1 ∞ W.U.C. series in A (hence in C(T)) so |T(ω)e | is a W.U.C. series n nP=1 in C(T). Now let x ∈ L1(µ,L1/H1) (the predual of L∞(µ,H∞)) and 0 σ fix v ∈ L1(µ,L1) with q˜(v) = x, we have ∞ ∞ |hφ ,xi| = |hφ ,vi| n n X X n=1 n=1 ∞ = |hT(·)e .h (·).χ (·),vi| n i(n) Lk X n=N ∞ ≤ kh kh|T(·)e |,|v|i i(n) n X n=N ∞ ≤ h|T(·)e |,|v|i < ∞ . n X n=1 Now (2) is equivalent to: for each n ≥ N, hφ ,f i ≥ δ/2 (cid:12) n i(n) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) SOME REMARKS ON THE DUNFORD-PETTIS PROPERTY 9 which is a contradiction since {f , i ∈ N} ⊆ K is relatively weakly i ∞ compact and φ is a W.U.C. series. The claim is proved. n nP=1 To complete the proof of the lemma, let us fix a sequence (ξ ) so n n that ξ ∈ conv{ϕ(k),ϕ(k) ,... ,} for every k ∈ N, we get by (ii) that n n n+1 lim Rehξ (ω),Te i exists for every T ∈ L(c ,A); we repeat the same n n 0 n→∞ argument as above for the imaginary part (starting from (ξ ) ) to get a n n sequence(ψ ) withψ ∈ conv{ξ ,ξ ,...}sothat lim Imhψ (ω),Te i n n n n n+1 n n n→∞ exists for every T ∈ L(c ,A). The lemma is proved. 0 To finish the proof of the theorem, we will show that for a.e. ω, the sequence (ψ (ω)) is uniformly integrable. If not, there would n n≥1 be a measurable subset Ω′ of Ω with µ(Ω′) > 0 and (ψ (ω)) is not n n≥1 uniformly integrable for each ω ∈ Ω′. Hence by Lemma 2, for each ω ∈ Ω′, there exists T ∈ L(c ,A) so that: 0 limsup sup|hψ (ω),Te i| > 0. n m m→∞ n So there would be increasing sequences (n ) and (m ) of integers, δ > 0 j j so that |hψ (ω),Te i| > δ ∀ j ∈ N; choose an operator S : c → c nj mj 0 0 so that Se = e ; we have |hψ (ω),TSe i| > δ. But by Lemma 3, nj mj nj nj lim |hψ (ω),TSe i| exists so lim |hψ (ω),TSe i| > δ. We have just n n n n n→∞ n→∞ shown that for each ω ∈ Ω′, there exists an operator T ∈ L(c ,A) so 0 that lim |hψ (ω),Te i| > 0 and same as before, we can choose the n n n→∞ operator T measurably, i.e., there exists T : Ω → L(c ,A), measurable 0 for the strong operator topology so that: a) kT(ω)k ≤ 1 for every ω ∈ Ω; b) lim |hψ (ω),T(ω)e i| = δ(ω) > 0 for ω ∈ Ω′; n→∞ n n c) T(ω) = 0 for ω ∈/ Ω′. These conditions imply that lim |hψ (ω),T(ω)e i|dµ(ω) = δ(ω) = δ > 0 n→∞Z n n ZΩ′ and we can find measurable subsets (B ) so that n n δ liminf hψ (ω),T(ω)e idµ(ω) > n→∞ (cid:12)(cid:12)ZBn n n (cid:12)(cid:12) 4 (cid:12) (cid:12) (cid:12) (cid:12) 10 NARCISSE RANDRIANANTOANINA and one can get a contradiction using similar construction as in the proof of Lemma 3. We have just shown that for each sequence (f ) in K, there exists n n a sequence ψ ∈ conv(|σ˜(f )|, |σ˜(f )|,...) so that for a.e ω ∈ Ω, the n n n+1 set {ψ (ω), n ≥ 1} is relatively weakly compact in L1(T). By Ulger’s n criteria of weak compactness for Bochner space ([15]), the set |σ˜(K)| is relatively weakly compact in L1(µ,L1(T)) = L1(Ω×T,µ⊗m). Hence σ˜(K) is uniformly integrable in L1(Ω× T,µ⊗ m) which is equivalent to σ˜(K) is relatively weakly compact in L1(µ,L1(T)). This completes the proof. Theorem 1. can be extended to the case of spaces of measures. Corollary 1. Let K be a relatively weaklycompactsubset of M(Ω,A∗). The set γ˜(K) is relatively weakly compact in M(Ω,M(T)). The following lemma will be used for the proof. Lemma 4. Let Π : M(T) → L1 be the usual projection. The map Π is weak* to norm universally measurable. Proof. For each n ∈ N and 1 ≤ k < 2n, let D = {eit; (k−1)π ≤ n,k 2n−1 t < kπ }. Define for each measure λ in M(T), R (λ) = g ∈ L1 be 2n−1 n n 2n the function 2nλ(D )χ . It is not difficult to see that the map n,k D n,k kP=1 λ 7→ λ(D ) is weak*-Borel, so the map R is weak* Borel measur- n,k n able as a map from M(T) into L0. But R (λ) converges a.e. to the n derivative of λ with respect to m. If R(λ) is such limit, the map R is weak* Borel measurable and therefore M (T) = R−1({0}) is weak* s Borel measurable. Now fix B a Borel measurable subset of L1. Since L1 is a Polish space and the inclusion map of L1 into M(T) is norm to weak* continious, B is a weak* analytic subset of M(T) which implies that Π−1(B) = B +M (T) is a weak* analytic (and hence weak* uni- s versally measurable) subset of M(T). Thus the proof of the lemma is complete.

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