Extreme Points of Disc Algebra

Question

I know that the extreme points of the closed unit ball of $\mathcal{H}^\infty(\mathbb{D})$, the space of all bounded holomorphic functions on the unit disc are the functions $f\in\mathcal{H}^\infty(\mathbb{D})$ such that $\int\limits_{-\pi}^\pi\log(1-|f(e^{i\theta})|)d\theta=-\infty$. What are the extreme points of the closed unit ball of the disc algebra, i.e. all functions continuous on $\overline{\mathbb{D}}$ and holomorphic on $\mathbb{D}$?
It is clear that all the extreme points of the closed unit ball of $\mathcal{H}^\infty(\mathbb{D})$, that are in the disc algebra are extreme points of the closed unit ball of the disc algebra. But are these the only extreme points?
Are there any extreme points of the closed unit ball of $\mathcal{H}^\infty(\mathbb{D})$ that is not in the disc algebra?
Is there any extreme point of the closed unit ball of the disc algebra that is not a finite Blaschke product?

Answer 1

$\newcommand\TT{\mathbb T}$ $\newcommand\AA{\mathbb A}$ $\newcommand\abajo{\\[0.3cm]}$ $\newcommand\e{\varepsilon}$ $\newcommand\DD{\mathbb D}$

The same condition characterizes the extreme points. Since $\AA\subset \mathcal H^\infty$, every $f$ satifying the integral condition is an extreme point.

Conversely, suppose that $\log(1-|f(e^{i\theta})|)$ is integrable. The argument is roughly the same as in $H^\infty$, with a twist to make the function $g$ stay in $\AA$. Let $w\in C([0,2\pi])$ with $w(2\pi)=w(0)$, $0\leq w(t)\leq1-|f(e^{it})|$, with $\log w\in L^1([0,2\pi])$, and $e^{iw}$ continuously differentiable on the open arcs in the set $\{|f|\ne1\}\cap\TT$ (details at the end). Define $$ g(z)=\exp\bigg[\frac1{2\pi}\,\int_0^{2\pi}\frac{e^{it}+z}{e^{it}-z}\,\log w(t)\,dt\bigg] $$ By the Schwarz formula, the expression inside the brackets is analytic on $\DD$ and has real part $\log w(t)$ when $z=e^{it}$; then $|g(e^{it})|=w(t)$ is continuous on $\TT$.

Since $|g(e^{it})|\leq 1-|f(e^{it})|$ we have, via the maximum modulus principle, that $\|f\|_\infty+\|g\|_\infty\leq 1$. Then $g\in\AA$, $f\pm g\in \AA$, and $$ f=\tfrac12\,(f+g)+\tfrac12\,(f-g) $$ is not extreme.

---

How to construct the function $w$: (not claiming that this is a smart way to do it)

The set $\{|f|\ne1\}\cap\TT=|f|^{-1}(-\infty,1)\cap\TT$ is open in $\TT$, so it is a countable union of disjoint arcs. Say (the union might be finite, but that only simplifies the argument) $$ \{|f|\ne1\}\cap\TT=\bigcup_n \exp(i\,(a_n,b_n)), $$ with $0<a_n<b_n<a_{n+1}<b_{n+1}<2\pi$, for all $n$. Let $h_n\in C[0,1]$ be continuously differentiable, with $0\leq h_n\leq1$, $h_n(0)=h_n(1)=0$, and $\int_{h_n<1}|\log t|<2^{-n}$ (this condition is achievable due to $\log t$ being integrable in $[0,1]$). Define $$ w(t)=(1-|f(e^{it})|)\,\sum_n h_n\big(\frac{t-a_n}{b_n-a_n}\big)\,1_{(a_n,b_n)}. $$ The sum is trivially defined because only one term is nonzero at a time; for the same reason, $w$ is continuously differentiable on each $(a_n,b_n)$. Now \begin{equation*} \begin{split} \int_0^{2\pi}|\log w(t)|\,dt &=-\sum_n\int_{a_n}^{b_n}\log h_n\big(\frac{t-a_n}{b_n-a_n}\big)\,dt =-\sum_n\int_{0}^{1}\log h_n\big(t\big)\,dt\abajo &=-\sum_n\int_{h_n\ne1}\log h_n\big(t\big)\,dt\abajo &\leq\sum_n2^{-n}<\infty \end{split} \end{equation*}