Peak sets and Choquet boundary of a function algebra I have two problems to ask. 


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*Let $A$ be a function algebra of $C(K)$. $t\in K$ is said to be a peak point of $A$ if $\exists~f\in A$ s.t. $|f(t)|=\|f\|$ and $|f(s)|<|f(t)|$ for any $s\neq t$. Suppose $A=\{f\in A(\mathbb{D}):f(0)=f(1)\}$, $A(\mathbb{D})$ represents the Disc algebra in $\mathbb{C}$. Can someone prove that the set of all peak points, say $\mathbb{P}$, of $A$ is the set $\mathbb{T}\setminus \{1\}$. Here $\mathbb{T}$ represents the unit circle in the plane $\mathbb{C}$. In this case the Choquet Boundary of $A$ coincides with $\mathbb{P}$.


It is not very difficult to prove that $1\notin \mathbb{P}$, in fact for any function in $A$ which attains its norm at $1$ will also attain at $0$. Hence from the analyticity of $f$ it follows that $f$ is constant in $\overline{\mathbb{D}}$. 


*Let us take two disjoint circles in the plane $\mathbb{C}$ and take the convex hull of these two circles. Then what is the Choquet boundary and the set of peak points of the set of all complex affine functions on the above convex set? 


This problem is in the Phelps book on Choquet Theory in Pg. 43, the author states an easy way to visualize is to observe $Ch(A)=Ch(Re A)$, where $A$ is the above subspace and $Re A$ represents the Real part of $f$, $f\in A$. $Ch(A)$ represents the Choquet boundary of $A$.
 A: Solution (1).
Let $A=\{f\in A(\mathbb{D}):f(0)=f(1)\}$. It follows from the Maximum Modulus Principle that the Choquet boundary $Ch(A)$ does not contain $1$, as $B(A)$ and the set of peak points are same for this case. We claim that $Ch(A)=\mathbb{T}\setminus\{1\}$. It remains to prove that for any $z_0\in\mathbb{T}\setminus\{1\}$, $z_0$ is a peak point for $A$. Let us recall the Condition for the local Peak point. 
A point $y\in K$ is said to be a local peak point of an Uniform algebra $A$ over $C(K)$ if for any open neighbourhood $U$ of $y$ and $\epsilon>0$, there exists $f \in A$ such that $\|f\| \leq 1$, $|f(y)|>1-\epsilon$ and 
 $|f|\leq \epsilon$ in $K \backslash U$.
It is known that if $K$ is metrizable then local peak point and peak point are the same.
Now choose $\epsilon>0$ and a neighborhood $U$ of $z_0$ in $\mathbb{T}$. 
Case 1:When $|z_0-1|\geq 1$.
Get $f\in A(\mathbb{D})$ such that $|f(z_0)|>1-\frac{\epsilon}{2}$ and $|f|\lvert_{U^c}<\frac{\epsilon}{2}$.
Case 2:When $|z_0-1|<1$.
Get $f\in A(\mathbb{D})$ such that $|f(z_0)|>1-\frac{\epsilon}{2}|z_0-1|$ and $|f||_{U^c}<\frac{\epsilon}{2}|z_0-1|$.
Now define $g(z)=\frac{z(z-1)}{|z_0-1|}.f(z)$. Clearly $g\in A$ and $|g(z_0)|>1-\epsilon$, $|g||_{U^c}<\epsilon$.
This shows $z_0$ is a peak point for $A$ and hence $Ch(A)=\mathbb{T}\setminus\{1\}$.
Solution (2)

Let $K$ be the subset of the plane consisting of the convex hull of two disjoint circles and let $M$ be the complex-valued 
affine functions on $K$. Let the four tangent points be $p,q,r,s$, denote the open arcs $pr$ and $qs$ as $S_{1}$ and $S_{2}$ and denote the smaller circle as 
$D_{1}$ and the bigger circle as $D_{2}$.
Let $x_{0}$ be any point on the open line segment joining $p$ and $q$. Then there exists distinct points $y$ and $z$ on the line segment such that $x_{0} = \frac{y+z}{2}$. For any $f \in M$, 
$f(x_{0}) =\frac{f(y)+f(z)}{2}$; this implies $f(x_{0}) = \int_{K} fd(\frac{\delta_{y}+\delta_{z}}{2})$, but $\delta_{x_{0}} \neq \frac{\delta_{y}+\delta_{z}}{2}$. Hence, $x_{0}\notin Ch(M)$ and hence is not a peak point for $M$. 
Consider any $z_{0} \in S_{1}\cup S_{2}$ and the tangent line say $ax+by=c$ for some $a,b,c \in \mathbb{R}$ to the point $z_{0}$ as shown in the above 
figure. Then $K \subset \{x+iy \in \mathbb{C}: ax+by \leq c\}$ and $g : \mathbb{C} \rightarrow \mathbb{R}$ defined as $g(x+iy)=ax+by$ is a real linear functional. Without loss of generality assume that $c>0$. Hence, 
$\tilde{g}(z)=g(z)-ig(iz)$, is a complex linear functional. Now $\sup_{z\in K}|\tilde{g}(z)|=\sup_{z\in K}|g(z)|$. Hence the maximum of $\tilde{g}$ is same as that of $g$. Now $\tilde{g}(z_0)=|\tilde{g}(z_0)|e^{i\theta}$, ie. $e^{-i\theta}\tilde{g}(z_0)=|\tilde{g}(z_0)|\geq|g(z_0)|=\sup_K g(z)=c$. Consider the linear functional $e^{-i\theta}\tilde{g}(z)$ which is affine on $K$, $|e^{-i\theta}\tilde{g}(z)|=|\tilde{g}(z)|\leq c$.
Hence, $z_{0}$ is a peak point for $M$. 
Now suppose $p$ is not in the Choquet boundary for $M$. This implies 
$\delta_{p} \not\in ext(K(M))$. Therefore there exists $z_{1},z_{2} \in S_{1} \cup S_{2}$ such that $\delta_{p} = \frac{\delta_{z_{1}}+\delta_{z_{2}}}{2}$. Thiis implies 
$p=\frac{z_{1}+z_{2}}{2}$ which is a contradiction since $p$ lies on the boundary of the circle. This implies $p$ is in $Ch(M)$. Similarly, the remaining tangent 
points also lie in the Choquet boundary for $M$.
Suppose $p$ is a peak point for $M$. Then we will get $f \in M$ such that $|f(x)|<|f(p)|$, for $x \in K\setminus\{p\}$. Now $f(p)=|f(p)|e^{i\theta}$, define $h(z)=Re(e^{-i\theta}f(z))$, then $|h(z)|\leq |f(z)|$, for all $z\in\mathbb{C}$ and $h(p)=|f(p)|=sup_K|f(z)|$. That is $h$ is a real linear functional on $\mathbb{C}$ which attains its supremum over a smooth surface $K$ at $p$. It shows that $h$ must be the tangent to the surface. Hence $h$ must coincides with the line segment $pq$. That is $|f(p)|=h(p)=h(q)\leq|f(q)|<|f(p)|$, a contradiction. Hence such an $f$ does not exist.
This implies $D_{1} \subset \{x \in \mathbb{R}^{2}: f(x)\leq \|f\|\}$ and 
$f(x) = \|f\|$ is the tangent line  to $D_{1}$ at point $p$. This implies $f(x) = \|f\|$, for all $x$ on the line segment joining $p$ and $q$, which is a contradiction. 
Hence, $p$ is not a peak point for $M$. 
Therefore, we get that $Ch(M)$ is precisely $S_{1} \cup S_{2}\cup \{p,q,r,s\}$ and $Ch(M)\setminus \{p,q,r,s\}$ is the set of peak points for $M$. 
