# Dual of the space of affine functions

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?

• Thanks for pointing out. Is it okay now? – Tanmoy Paul Nov 1 '18 at 18:52
• 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). – Dirk Werner Nov 1 '18 at 20:04
• Yes, of course, $D$ is compact. – Tanmoy Paul Nov 2 '18 at 0:38
• 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 $f\in C(D)$. – Tanmoy Paul Nov 18 '18 at 11:00
• 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. – Dirk Werner Nov 18 '18 at 17:39