By Isaac Todhunter

ISBN-10: 1402167474

ISBN-13: 9781402167478

This Elibron Classics variation is a facsimile reprint of a 1861 version via Macmillan and Co., Cambridge.

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If X is not reflexive then they can be different as is shown by the following classical example. 3. ) Assume that co is the Banach space of real sequences for which zn = 0, with the norm Let us define f ( t ) = (X(O,lI(t),2 X ( o , ; ] ( t ) ) , . , n X ( o , $ ] ( t ) , * ) * * for t E [O,13. Evidently f(0) = 0 = (0, . . , 0, . . ) and if t E (0,1] then +] and f(t) = there is an n* E N such that t E ( 1 , 2 , . . , n*,0, . . ) for this t E (0,13. The values of f belong t o co. 2) that the Dunford integral is 1' (27) f ( t ) d t = (1,1, .

Hence and this implies lim 11 fr r-+m - fill = 0. 8. e. the set J’ of simple functions is dense in B with respect to the seminorm 11 . [I1. 9. The space B equipped with the seminorm is complete. 111 Proof. 111. 8for every q E N there exists a simple function fq E J’ such that 1194 -fqlh 1 < -. Q Hence and therefore the sequence (fq)is L-Cauchy. 2 the sequence (fq) contains a subsequence (f q s ) which converges almost everywhere in I to a certain function f : I + X and this Bochner Integral 21 subsequence is L-Cauchy.

Then there is a K > 0 and a sequence E, C I , n E N of measurable sets with p ( E n ) -+ 0, n + 00 and r for some XI*, E B(X*). Since the measures of En tend to zero, it is possible to take a subsequence of En assuming that for m < n we have Dunford and Pettis Integrals Take A, = En,\ Urn<,Em;A, m # 1 and 33 c I are measurable, A, n Al = 0, 6 Iz;(f>l Hence there exist B, disjoint) such that > 5' c A,, B, measurable (B, are pairwise and therefore IIT*(XB,>II> f for every n. Therefore the series co T*(xB,)= v(B,) cannot converge and d) is not satisfied.

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