Restricting a function in the disk algebra

Question

Let $A$ be the disk algebra, i.e. continuous functions on the closed unit disk in $\Bbb{C}$ that are analytic on the interior of the disk. By the maximum-modulus theorem, we have an isometric morphism of algebras:

$$\varphi: A \to C(S^1): f \mapsto f\vert_{S^1}$$

The book I read claims that $\varphi(A)$ is contained in the closed subalgebra $B$ of $C(S^1)$ generated by $1$ and $z$. Why is this the case?

My intuition is that $f \in A$ can be written as $f(z) = \sum_n a_n z^n$ on the interior of the disk? If this also holds for $|z| = 1$ what I want to prove becomes obvious but I'm not sure this holds.

Also, do we have $B = \varphi(A)$? Or only the inclusion $\varphi(A) \subseteq B?$

Thanks for any help!

Answer 1

It's enough to show that every $f\in A$ can be uniformly approximated by polynomials. The power series doesn't converge uniformly on the closure, but: Let $\epsilon>0$. Since $f$ is uniformly continuous there exists $r\in(0,1)$ such that if $g(z)=f(rz)$ then $$|g(z)-f(z)|<\epsilon\quad(|z|\le1).$$And since the power series for $f$ converges uniformly on $\{|z|\le r\}$ there exists a polynomial $p$ with $$|p(z)-g(z)|<\epsilon\quad(|z|\le 1).$$

Answer 2

David's answer is how one shows that $\varphi(A)\subset B$. I want to add that we actually have $\varphi(A)=B$. Indeed, the subalgebra of $C(S^1)$ generated by $1,z$ is actually the set of all polynomials restricted to $S^1$. Therefore, each element of $B$ is a uniform limit of polynomials on $S^1$. Suppose that $f\in B$ and that $(p_n)$ is a sequence of polynomials such that $\sup_{\zeta\in S^1}|p_n(\zeta)-f(\zeta)|\to0$. Then the sequence $(p_n)$ is uniformly Cauchy on $S^1$. By the maximum modulus principle, we have that $(p_n)$ is uniformly Cauchy on $\mathbb{D}$, so $p_n$ converges uniformly to some function $F$ on $\overline{\mathbb{D}}$. As $F$ is the uniform limit of analytic functions on $\mathbb{D}$, it is also analytic on $\mathbb{D}$. Also, we obviously have $F\vert_{S^1}=f$, so $\varphi(F)=f$.

Tomasz demonstrates another very simple way in their comment.