Regarding the endpoints of the line segments if the Dirac measure is not an extreme point of the state space then there exists a convex combination of the extreme points of the state space which also represents that point. Since affine functions on this compact convex separate points so the point is the same as that convex combination. Because of the strict convexity of the circle, the two points must be outside of the circle, which is also absurd.

Can someone tell me if this argument is valid or not? If we choose any point on the open line segment which is a part of the tangent to both the circles then that point cannot be a point in the Choquet boundary of the subspace. Because if we choose any such point than not the Dirac measure of that point only probability measure which represents that point. We can write this point as a midpoint of two distinct points lying on the tangent line. Hence the corresponding convex combination of the Dirac measures of those points is also representing measure for that point.

Can we say anything about the space of type $C(K)$? If we follow the above argument, if $f\in C(K)$ is a smooth point then is it necessary that $f$ remains smooth in $C(K)^{**}$? As we know bi-dual of $C(K)$ is of the form $C(\Omega)$ for some compact $T_2$ $\Omega$.

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 $f\in C(D)$.

According to my notations is it possible to conclude that, $\int_Sf(t)d\mu (t)=\int_Kf(t)d\mu(t)$ if $S$ is Borel and $f\in C(K)$. Here remember that $\mu(E)=0$ if $E\subseteq K\setminus S$ is a Baire set.