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Let $M^+(D)$ be the space of all positive measures on a closed convex subset $D$ of a locally convex topological vector space $E$. Two measure $\mu, \nu\in M^+(D)$ one can define a partial ordering $\mu\prec\nu$ if $\mu(f)\leq\nu(f)$ for all continuous convex functions $f$ on $D$. It can be proved every positive measure is dominated by a positive maximal measure. A signed measure $\tau$ on $D$ is said to be a boundary measure if $|\tau|$ is maximal.

Suppose $A(D)$ be the space of all Real-valued affine functions on $D$. With respect to the sup norm, it is a Banach space. Is it true that the dual space of $A(D)$ none other than the space of all Boundary measures on $D$?

Is there any finer result available if $D$ is a Choquet simplex?

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  • $\begingroup$ Thanks for pointing out. Is it okay now? $\endgroup$
    – Tanmoy Paul
    Commented Nov 1, 2018 at 18:52
  • $\begingroup$ I assume that you want $D$ to be compact. If $L$ is in the dual of $A(D)$, extend it to $L'$ in the dual of $C(D)$, represent the latter by a signed measure, and decompose this into positive measures, $\mu=\mu^+-\mu^-$. By Choquet theory, each probabilty measure on $D$ has a barycentre in $D$, which can be represented by a maximal measure. Thus, $\mu^{\pm}\in C(D)^*$, restricted to $A(D)$, can be represented by positive boundary measures -- this is the Choquet resp. Bishop-de Leeuw theorem. In the case of a simplex, the representing boundary measure is unique (Choquet-Meyer's theorem). $\endgroup$
    – Dirk Werner
    Commented Nov 1, 2018 at 20:04
  • $\begingroup$ Yes, of course, $D$ is compact. $\endgroup$
    – Tanmoy Paul
    Commented Nov 2, 2018 at 0:38
  • $\begingroup$ Prof. Werner, thank you for the explanation. One small doubt; If $D$ is not metrizable then what is the significance of Choquet-Bishop-De Leeuw Theorem? It only guarantees that $|\mu|(C)=0$ for any boundary measure $\mu$ and for any Baire set $C\subset D\setminus ext(D)$. Although it is not possible to conclude that $Supp(\mu)\subseteq ext(D)$ or even $\int_Df(t)d\mu(t)=\int_{ext(D)}f(t)d\mu(t)$, for ant $f\in C(D)$. $\endgroup$
    – Tanmoy Paul
    Commented Nov 18, 2018 at 11:00
  • $\begingroup$ The notion of maximal measure seems to be the right substitute for a measure supported on the extreme boundary in the narrow sense, which is not always available as shown by the "porcupine topology'' in the Bishop/DeLeeuw paper. That this is a meaningful generalisation is indicated e.g. by the fact that it allows to prove Rainwater's theorem in the nonseparable case; see R. Phelps's ``Lectures on Choquet's Theorem''. It is also important for uniform algebras. $\endgroup$
    – Dirk Werner
    Commented Nov 18, 2018 at 17:39

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