Skip to main content
added 40 characters in body
Source Link
Tanmoy Paul
  • 521
  • 3
  • 13

Let $D$$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 $D$?

A measure $\mu$ on a compact convex set $D$ is said to be a boundary measure if $|\mu|$ is maximal; two measure $\mu, \nu$ on a compact convex set $D$ are such that $\mu\prec\nu$ if $\mu(f)\leq\nu(f)$ for all continuous convex functions on $D$. It can be proved every positive measure is dominated by a positive maximal measure.

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

Let $D$ be a closed convex subset of a locally convex topological vector space $E$. 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 $D$?

A measure $\mu$ on a compact convex set $D$ is said to be a boundary measure if $|\mu|$ is maximal; two measure $\mu, \nu$ on a compact convex set $D$ are such that $\mu\prec\nu$ if $\mu(f)\leq\nu(f)$ for all continuous convex functions on $D$. It can be proved every positive measure is dominated by a positive maximal measure.

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

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?

Source Link
Tanmoy Paul
  • 521
  • 3
  • 13

Dual of the space of affine functions

Let $D$ be a closed convex subset of a locally convex topological vector space $E$. 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 $D$?

A measure $\mu$ on a compact convex set $D$ is said to be a boundary measure if $|\mu|$ is maximal; two measure $\mu, \nu$ on a compact convex set $D$ are such that $\mu\prec\nu$ if $\mu(f)\leq\nu(f)$ for all continuous convex functions on $D$. It can be proved every positive measure is dominated by a positive maximal measure.

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