Functional Analysis: Proceedings of the Seminar at the by Haskell Rosenthal (auth.), Edward W. Odell Jr., Haskell P.

By Haskell Rosenthal (auth.), Edward W. Odell Jr., Haskell P. Rosenthal (eds.)

The articles during this quantity are in accordance with talks given in a seminar at Austin in the course of 1986-87. they vary from these facing clean examine and discoveries to exposition and new proofs of older effects. the most subject matters and subject matters contain geometric and analytic homes of infinite-dimensional Banach areas and their convex subsets in addition to a few points of Banach areas linked to harmonic research and Banach algebras.

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Additional info for Functional Analysis: Proceedings of the Seminar at the University of Texas at Austin, 1986–87

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Krivine. Application des ultraproduits Ii l'etude des espaces et algebres de Banach, Studia Math, 41 (1972), 315-334. 7. W. Feller. II\ Wiley, 1966. 8. S. Guerre. Types et suites symetriques dans L p, 1 p < 00, Israel J. Math, 53 (2) (1986). 9. R. Garling. Stable Banach spaces, random measures and Orlicz function spaces, in "Probability Measures on Groups", Lect. 928, Springer, 1982. 10. S. Guerre, Y. Raynaud. On sequences with no almost symmetric subsequence, University of Texas Functional Analysis Seminar, 1985--86, 83-93.

D . d. Xill n~ liZ + Gil i and as clearly Ilzn + Gil liZ + Gil --+ n~oo we obtain a contradiction. If (zn)n is not supposed to be equiintegrable, we use the splitting lemma and the argument of n03 to reach the contradiction (ifthe disjoint part of (zn) defines a type 8 E Td(L(O x [0, 1])) we obtain limn Ilzn +:Li a:r XiII = 8(Z + G) = limn Ilzn + GIl)· • 6. Relaxing the type 2 condition. In the proof of theorems 1, 1 bis, the type 2 condition is used to ensure that the norm of L([O, 1]) dominates the norm of L2([0, 1]) (up to a constant), and to prove Lemma 2.

Instead we reason as follows. Define T = inf{ p{g) -1: 9 E X with 9 ~ f T by and 9 ~ 0 } . We then claim: (63) T >0. Suppose this were false. Then we could choose a sequence (gn) in C with p(gn) -+ 1 and gn for all n. Then setting h n = gn - f and en = p(h n ) for all n, we have that each n, choose k n in Kn with h n = enk... Since K is bounded, h n -+ -+ f O. Now for O. Hence gn -+ f 2: 0, K is closed and bounded, it is easily seen that C is closed; hence en ~ f. Since a contradiction. Thus (63) holds.

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