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By Choudhary P.

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S) implies that satisfies condition (ii) of Grothendieck's Theorem B4 (in Appendix) and, by the latter Theorem, Thus (i) implies (iii). € ~e A is relatively weakly compact. Now suppose that S is locally compact for the remainder of this proof. We assume that (iii) holds and prove (i). Let C be a given compact subset of S and V be a compact neighbourhood of C. fl. Thus~ E Me(S). Finally we show that (ii) implies (i). Let£> 0 be given. We can choose a compact subset F such that II 1~1-1~1IF II < £.

H(x*"*Y-*"> ,. - ,. h(x*"> ,. h ,. - X (x)x (y) Xh ( xy) .. 6, we have 0 ~ h(~) 2 • h(~*~) • f h(x*~)d~(x) - and so the function x ~ h(x*~) is non-zero. We now set Hence xh € s • A Clearly T is a well-defined mapping of M (S)A into s . a To see that T is onto and one-to-one fix x € S and, corresponding to this Xo define h € Ma(S)* by h(v) := x(x)dv(x). Since x is non-zero and continuous, and Ma (S) is solid, we certainly have that h is non-zero. (y) h(p)x(x) f and so X • Xh• and one-to-one.

First we prove items (i) and (ii). 5. Let u1 ,u2 ,v be open subsets of S such that u1 £ u1 • u 2 £ u 2 • 1 £ V and vu1u 2 S U. l(a) we have that and X £ This completes our proof for (i) and (ii). Second we prove item (iii). Suppose t £ x(O n s 1 )-l and choose w £ 0 n s 1 with tw .. x. · By (ii) and the fact that s 1 is dense we can find v w(W n s1) £ -1 n v n s1 • W n s 1 with vy • w. -1 -1 -1 We note that (tv)V S x(O n s1) for if p E (tv)V then So we can find y £ x = tw • tvy and so p t £ £ £ pVy £ pV(W n s 1) £ p(O n S1) x(O n s 1) -1 • Hence, with item (ii) in mind, we have that int((tv)V-1) S int(x(O n s 1>-1> and so (iii) follows.

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