The

The Dunford-Pettis property is shown to hold for the uniform algebra A(Cl) and its dual for some standard domains ft, including strongly pseudoconvex bounded domains in C", pseudoconvex bounded domains of finite type in C 2 , and bounded domains in C. Previously the result was known for the unit ball and unit polydisc in C n . Techniques used involve Bourgain algebras, Hankel operators, properties of the Bergman kernel, quasi-metrics on the boundary, and d-theory.


Introduction
A Banach space X over the complex field is said to have the Dunford-Pettis property if for each Banach space Y, every weakly compact linear operator T: X -> Y is completely continuous.There are several conditions equivalent to this which have been discussed in [12].Here is one of them which is often used as a definition of the Dunford-Pettis property.THEOREM 1.1 [15].Let X be a Banach space.Then X has the Dunford-Pettis property if and only if for every weakly null sequence (x n ) in X and weakly null sequence (x*) in X*, we have lim n _ oo <x*,x n > = 0.
No infinite dimensional reflexive Banach space Xhas the Dunford-Pettis property since the identity operator /: X -• X is a weakly compact operator, but the unit ball of X is not norm compact.The basic result proved by Dunford and Pettis in [13] is that L l (ji) has the Dunford-Pettis property.Since the canonical map of X into X** is weakly continuous, Theorem 1.1 implies that X must have the Dunford-Pettis property if X* has it.In particular, the space C 0 (K) of all continuous functions which vanish at infinity on a locally compact Hausdorff space K has the Dunford-Pettis property.
Work has been done by several authors to determine which Banach spaces have the Dunford-Pettis property.The C*-algebras and their duals and preduals which have the Dunford-Pettis property have been completely analysed by Chu and Iochum in [6] (see also [7]).In [3], Bourgain proved that H™ of the unit disk has the Dunford-Pettis property.In [4], Bourgain proved that the Ball algebra A(B n ) and the polydisc algebra A(D n ) (and their dual spaces) have the Dunford-Pettis property, thereby generalizing in two directions the result of J. Chaumat for the disc algebra (see [22,Section 8]).Recall that A(Q) is the subspace of C(Q) of functions which are holomorphic on Q.Now the unit ball and unit polydisc in C n are examples of bounded symmetric domains, and the unit ball is a special case of a strongly pseudoconvex domain.The purpose of the present paper is to prove that both A(Q) and its dual space A(Cl)* have the Dunford-Pettis property for a quite large class of domains in C n including the unit ball, strongly pseudoconvex domains, and finite type domains in C 2 .Our purpose is facilitated by exploiting the connection between the Dunford-Pettis property holding for A(Q) and the boundedness and compactness of certain Hankel operators The paper is organized as follows.In Section 2, we give some relations between A(Q) or A(O)* having the Dunford-Pettis property and the complete continuity of some Hankel operators on A(Q).In Section 3, we consider more general integral operators which are modelled on the Bergman projection, and we prove that they satisfy the conditions used in Section 2. In Section 4, several examples of standard domains which satisfy the conditions, and hence for which A(Q) and A(Q)* have the Dunford-Pettis property, are given.Finally, in Section 5, we show, as an application of ^-theory, that A(Cl) and A(Q)* have the Dunford-Pettis property when Q is a bounded domain in the complex plane with C 1 boundary.
Bourgain's methods in [4] are very useful for finding subspaces of C(K) (where K is compact) having the Dunford-Pettis property.This theory has been developed by Cima and Timoney in [9], where they define the so-called Bourgain algebra of a subspace of C{K).Here we describe briefly their result.Let <f>eC(K) and let Md enote the multiplication operator on C(K).We let X b be the space of all functions <fi e C(K) satisfying: if (x n ) c: X is a weakly null sequence, then lim,,,,^ dist ((/)x n , X) = 0.It has been proved in [9] that X B and X b are algebras, which are the so-called Bourgain algebras of X. From [4, Proposition 2] and the proof of [4, Theorem 1], Cima and Timoney in [9] formulated and proved the following theorem.
THEOREM 1.2 [9].Let Kbe a compact Hausdorff space.Let X be a closed subspace of C{K).Then we have (i) if X B -C{K), then both X and X* have the Dunford-Pettis property ; (ii) if X b = C(K), then X has the Dunford-Pettis property.

Sufficient Condition on the Bergman Kernel
Let Q be a bounded domain in C n with C 1 boundary.We let dV denote Lebesgue volume measure over O.For 0 <p ^ oo, we let L P (Q) be the usual Lebesgue space with respect to the measure dV.We let A P (Q) be the subspace of L P (C1) consisting of holomorphic functions.It is well known and easy to show that A p (£l) is a closed subspace of L V (Q).In particular, A 2 (£l) is a Hilbert space.By subharmonicity of \f\ p for/e A P (Q), it follows easily that the evaluation functional eJJ) = f{z) is bounded on A v (Cl) for every point zeQ.By the Riesz representation theorem, there is K z eA 2 (Q) so that for all fe A\Q).Let K(z, w) = K z (w).It is well known that K(z, w) is a reproducing kernel of A\Q), called the Bergman kernel.Let P:L\£l) -> A\Q) be the orthogonal projection, the so-called Bergman projection.Then we have for all/GL 2 (Q).For <psL 2 (Q), we define the multiplication operator formally by MJJ) = <f>f.The commutator of M+ and P is defined by [M^P] = M^P-PM^, and we define the Hankel operator H^ as follows: H^ = [Af 0 , P]P.From these definitions, one can easily see that H^ = [M^P] on A 2 (Q).In other words, we can also look at [M^ P] as an integral operator with kernel function: The first relation between the Dunford-Pettis property of the Banach space A(Gl) and the boundedness and complete continuity of the Hankel operators H^ on A{Q) is given by the following simple proposition.Proof.It is required to prove that for each weakly null sequence (/")*_!<= A(Q), and all <fi e C(Q), we have O. (2.1) Since C^Q) is dense in C(Q), it suffices to prove (2.1) for all <f>eC\Ci).Let ^eC\Cl).Since H^AiQ)) c C(Q), we have P(0/ n )eC(Q) n ^42 ( for all n.Since //^ is completely continuous from A(Q) to C(Q), we have However, This implies that (2.1) holds for all ^eC^Q), and completes the proof.
From the assumption of Proposition 2.1, one realizes that in order to prove that A(Q) has the Dunford-Pettis property, it is important to understand when Hankel operators carry A(Q) into C(Q) and are completely continuous from A(Q) to C(Q).Conditions for the boundedness and compactness of the Hankel operators on Bergman spaces or Hardy spaces have been studied by many authors (see for example [2,1] and references therein).
In the next two theorems, we first give a sufficient condition for the complete continuity of H^.A^) -> C(Q) and then we show that this condition implies the Dunford-Pettis property for A(Q)* and A(Q).In the following two sections, we shall show that certain standard domains satisfy this condition.THEOREM 2.2.Let Ci.be a bounded domain in C n with C 1 boundary.Let 0e C(Q).Then the following statements hold.
(i) If ||A,(z,-)lli ^ C^for all zeO, then H^:A p (Cl) -L P (Q) is bounded for all 1 < p ^ oo.Here we use the notation A™ = i/ 00 , and C^ is a constant depending only on <f) and Q. (ii (iii) If the set {K^z, -)zeQ,}is relatively compact in L\Q) and ifK(z, w) e C(Q x Cl e ) for all 0 < £ < 1, where Q e = {zeQ: dist (z,dQ) ^ e}, then H^: Proof.Since the Bergman kernel is conjugate symmetric, we have |A^(z, w)\ = \K^(w, z)\ for all z, wefi.Applying Schur's lemma [25] with the assumption in (i), the integral operator with kernel function K^ is bounded on L P (Q) for all 1 ^p ^oo, so Now we prove (ii).By (i), H^.A P (Q) -> L P (Q) is bounded for all 1 ^p <oo.In particular, H^:A(Q) -> L°°(Q) is bounded.Let {/"} c A(Q) be a weakly null sequence.We shall show that {H^(f n )} is strongly null in L°°(Q).Suppose not.Then there is a positive number e 0 and a subsequence z n/c e Q such that I^C/» t )(*n fc )l > 4 for /: = 1,2,.... (2.2) Since {K^z, •): z e Q} is relatively compact in L 1 (Q), there is a further subsequence (we use the same notation) {A^(z n ,•)} converging strongly to some geL^Q).Since {/"} is weakly null, it is bounded in L°°(Q).Thus
(2.3) Since {K^z, •): z e Q} is relatively compact in L X (Q), without loss of generality, we may assume that there are functions g v g 2 eL\Q) such that d(e) = max so that lim^o<5(e) = 0. Since K(z,w) is uniformly continuous onflxQ £ , for each e > 0, there is ^(e) > 0 such that for all Z 1 ,Z 2 GQ and all weQ 6 we have |A,(Z\MO-A,(Z",MOI ^ S(e) if l^-z 2 !^ r}(e).The above inequality contradicts (2.3) as e -• 0. Therefore, (iii) follows.In order to prove Theorem 2.3, we need the following well-known elementary proposition.
By Proposition 2.4 with X = A(Q), for each n there is a net (x a ) in A(Q) so that (a) wioo^rilw; Since {A^(z, •)} is relatively compact in L 1 ^), an argument similar to the proof of (2.4) above shows that zetl zed when a is sufficiently large.We shall show that dist («£***, ,4(Q)**)<£ (or n^N e .
In the first place, with for all 0 < e « l , then { ^( z , ) : z e Q } w relatively compact in //(Q).
Let N = N(z) be the smallest integer such that dQ. a B(n(z), 2 N r(z)).Then where cW(z, S) is shorthand for W(x, cS).
Next we establish (b).Let {K^{z k , •)} be any sequence with z k ed (for k = 1,2,...).We show that it has a convergent subsequence in L^Q).Since Q is compact, there is a subsequence {z k } and an element z o eQ so that lim^^z^.= z 0 .We shall show that K^{z k , ) is a convergent sequence in L X (Q).It suffices to show it is a Cauchy sequence in L^Q).For any 0 < 6 <^ 1, since A^( •, •) is uniformly continuous on Q x Q s , there is a positive integer N s such that if /, m > N d , then for all weCl s .Thus for l,m> N s , we have Proof.We let d be the quasi-metric defined in [20,21].For convenience, we recall the definition.LetpedQ, and let A(p,S) be the quantity defined in [19, p. 116].Fix S > 0. Since A(p, z) is strictly increasing in z and A(p, 0) = 0, there is a unique zr(S,p) such that A(p,z) = S. Let X X ,X 2 be real vector fields such that X X ,X 2 and T span the real tangent space to dQ at each point p.Here X x , X 2 span the complex tangent space over R at each point and Tpoints in the 'complex normal' direction.Then we define the ball B(p, 3) on dQ by B(p,S) = {qedQ.q= exp^X.+ ^X.+ CT), \a,\ ^ z(p,3) forj = 1,2, \{\ ^ 3).

Application of ^-theory
In many cases, the integral representing kernels for solutions of the 5-equations are simpler than the Bergman kernel, for example, for a bounded domain in the complex plane.In this section, we shall apply 5-theory (for which see [17]) and the ideas from Section 2 to prove that A(Q) and A(Q)* have the Dunford-Pettis property for bounded Q with smooth boundary.Let us start with the following more general theorem.For each n, we let u n = T(f n d<f>).Since T: C(Q) -> C(Q), it follows that u n eC(Q) for all n and g n :=f n (j) -u n eA(Q).Since f n ~d<j) -> 0 weakly and If we know the integral kernel of the solutions of the d-equation on Q, then we can check for complete continuity to obtain the Dunford-Pettis property.For example, in the case of the complex plane, we can prove the following theorem.In order to prove that A(Gl)* has the Dunford-Pettis property, we need to repeat some parts of the proof of Theorem 2.3.In fact, for e > 0 arbitrary, if we replace K^{z,w) in (2.Therefore, we have dist(0x**,,4(Q)**) ^0 asn ^oo.
Applying Theorem 1.2, the proof of Theorem 5.2 is complete.
REMARK.The main results of this paper can be proved alternatively by using the technique of 5-theory as represented in the proof of Theorem 5.2.The complete continuity of the relevant ~d solution operators can be found in [23,5,17].
It was pointed out by the referee that Theorem 5.2 can be obtained in another way, using a result by Cima and Timoney in [9].

PROPOSITION 2 . 1 .
Let O be a bounded domain in C n with C 1 boundary.Assume that for all <f>e C^Q), the Hankel operator H^ maps A(Q) into C(Q.) and is completely continuous.Then A(Cl) b = C(Q).Hence, by Theorem 1.2, A(D) has the Dunford-Pettis property.

THEOREM 2 . 3 .
Let Qbe a bounded domain in C n with C l boundary.Suppose that for all 0eC 1 (Q) the set {A^(z, •): z e Q} is relatively compact in L 1 ^), and K(z,w)e C(Q x£l e )for allO<e<\, where Q e = {zeQ:dist(z, dQ) ^ e}.Then A(Q) B = C(Q), and hence by Theorem 1.2, A(Q)* and A(Q) have the Dunford-Pettis property.

THEOREM 5 . 1 .
Let Q be a bounded domain in C n with C 1 boundary.If the d operator T: C(Q) -> C(Q) is completely continuous (here HT(f) = / ) , then A(Q) has the Dunford-Pettis property.Proof.According to Theorem 1.2, it suffices to prove that for each weakly null sequence (/")*_!<= A(Q), for all <j>eC(Q).Since C\Q) is dense in C(Q), it suffices to prove that (5.1) holds for Let ^EC\Q).

THEOREM 5 . 2 .
Let Qbe a bounded domain in the complex plane with C 1 boundary.Then A(Q) and A(Gl)* have the Dunford-Pettis property.Proof.The d-solution operator T: C(Q) -> C(O) is given by where dA is Lebesgue area measure.It is easy to see that {l/(w -z):zeQ} is a relatively compact set in L\Q).Thus T: C(Q) -> C(Q) is completely continuous.By Theorem 5.1, A(Q) has the Dunford-Pettis property.Alternatively, one can see the complete continuity by recalling that T: C(Q) -> Lip a (Q) for any ex < 1 and that the embedding Lip a (O) -• C(Q) is compact.
4) by d(/>(w)/(z-w), then we have sup zed n > N e , where N e is some integer depending only on e.By the related arguments in the proof of Theorem 2.3, if we let x a eA(Q) with HxJ^ ^ 1 be such that \im a (x a ,x*} = <x**,x*> for all x*eA(Q)*, and if we define then we have u a sA(D) and= \T(xJ(j>)(z)\ =x a (w)d<t>(w)