500

J Steffens

for some weak unit и (hence is \\-\\^,-continuous on ^J Conversely, any positive linear functional on Sf which is sequentially continuous from below and finite on some weak unit is given by L{m,') for some unique meExc

Proof Immediate by (2.2) Note that the fimteness assumption on the functional implies the (т-fîniteness of the associated excessive measure П

Let the order isomorphism referred to in (2 4) be denoted by ip, precisely let

( 2 6) cp E\c^-^p%, m\-^L{m,')

For the rest of this section we work with the base norm space ^ with dual norm „II • II and positive cone induced by p^ In addition, there is the balayage order < on ^' as defined m (2 3), p(^\ <) consists of those iie^' which are positive on ^ Consider m the following the dual pair (^, ^ ) Let т (resp x') denote the associated weak (resp weak*) topology on ^ (resp ^) Then It is well known that (^,т) =^' and (^,т') =^ In particular, the space of т'-continuous linear functional on ^ coincides via the evaluation map with the space ^ We prove m the theorem below that the cone ^ is in 1 — 1 spondence with the cone of those positive т'-continuous hnear functionals on ^ which are sequentially continuous from below on p(^, -<) Let for simphcity ^ denote the space {^\т) Note that lepM iff X is positive on pQ)\ and that Я IS sequentially continuous from below on p{3\<) iff Я(/£„) increases to X(p) whenever ^u„(s) increases to ^(5) for any se5^

The subsequent theorem extends results obtained by Mokobodzki [30] and Boboc et al [7] under the hypothesis of absolute continuity of (V^

( 2 . 7 ) Theorem. There is a 1 — 1 correspondence between ^ and the set of al in pJi' which are sequentially continuous from below on p{^, <) The spondence is given by s\-^s for any se^, where ^ denotes the evaluation map on Q) Moreover, on ç (ExcJ {see 2 6), s = L( •, s) о ф ~ ^ holds

Proof Let l^pM be sequentially continuous from below on p(ß\ <) According to the above remarks there exists a/iep^ s th Я^Я, le there exist /, ge^ sth /^g and X=J—g Let s £-^R+ be the excessive regularization of the supermedian function So =(/-g) l[„<oo] + QO l[„-^j,ie s^jlimaF^So To see

that So IS supermedian, note that for xe\u<oo\ e^ and a8^ V^ are m ^, hence o^^xya{so) = ocSxK{(f-g) hu<oo])=J^^xVÀf)-ocSxVÀg) = oc8^V^(h) incrcascs to £x{h) = So{x) as a->oo since Я = Я is sequentially continuous from below on p{^\<), and for xg[w = oo], ae^ F^(so)g go =£^(so)These computations show m particular that So = s on [u<oo] Therefore we have /=g + s on [м<оо], hence everywhere since both sides represent excessive functions This proves 5=/—g and thus À = s. The element se6^ is uniquely determined since s is uniquely defined off a polar set Let finally inecpiExcJ, le pL = (p{m) for some meExc« Then X{ii) = s{^) = ii{s) = L{m,s) = L{-,s)oq)~^{li) This establishes the first part of the proof.

As to the second part, if se5^ then evaluation at s is strongly, thus m particular, т'-continuous, besides it is positive and sequentially continuous from below on p(^, -<) To see the latter let /i„ increase to /л m {^, <), then m particular s(/z„) = /i„(s) mcreases to /i(s) = s(^) This finishes the proof Q