The Dunford-pettis Property on Tensor Products

We show that, in some cases, the projective and the injective tensor products of two Banach spaces do not have the Dunford-Pettis property (DPP). As a consequence, we obtain that (c0⊗̂πc0)∗∗ fails the DPP. Since (c0⊗̂πc0)∗ does enjoy it, this provides a new space with the DPP whose dual fails to have it. We also prove that, if E and F are L1-spaces, then E⊗̂ǫF has the DPP if and only if both E and F have the Schur property. Other results and examples are given. A Banach space E has the Dunford-Pettis property (DPP, for short) if every weakly compact operator on E is completely continuous, i.e., takes weak Cauchy sequences into norm Cauchy sequences [14]. Equivalently, E has the DPP if, for all weakly null sequences (xn) ⊂ E and (φn) ⊂ E (the dual of E), we have limφn(xn) = 0. If E has the DPP, then so does E, but the converse is not true [25]. A Banach space with the Schur property has the DPP. The DPP is inherited by complemented subspaces. For more on the DPP, the reader is referred to [8, 6]. It was unknown for many years if the Dunford-Pettis property of E and F implies that of their projective tensor product E⊗̂πF and of their injective tensor product E⊗̂ǫF [8]. Talagrand [26] found a Banach space E so that E has the Schur property but C([0, 1], E) = C[0, 1]⊗̂ǫE and L([0, 1], E) = L[0, 1]⊗̂πE∗ fail the DPP. It is proved in [24] that, if E and F have the DPP and contain no copy of l1, then E⊗̂πF has both properties. It is shown in [20] that, if E and F have the Schur property, then E⊗̂ǫF has the Schur property. Answering a question of [6], it is proved in [1] that, for compact spacesK1, . . . , Kn, the space C(K1)⊗̂π · · · ⊗̂πC(Kn) has the DPP if and only if K1, . . . , Kn are scattered. Taking advantage of an idea of [1] we prove, among others, the following results, where L(E,F ) denotes the space of all (linear bounded) operators from E into F while Lcc(E,F ) is the space of all completely continuous operators from E into F : (a) Suppose E is not Schur, F contains a copy of l1, and L(E,F ) = Lcc(E,F ). Then E⊗̂πF does not have the DPP. (b) Suppose E does not have the Schur property, F ∗ contains a copy of l1, and L(E, F ) = Lcc(E∗, F ). Then (E⊗̂ǫF ) does not have the DPP. As a consequence, (c0⊗̂πc0) ∗∗ fails the DPP, which answers a question of [6]. 1991 Mathematics Subject Classification. Primary: 46B20; Secondary: 46B28.