Disk algebra

Remarks

$A(\mathbb{D})$ is a closed subalgebra of $C(\overline{\mathbb{D}})$, hence a commutative Banach algebra with unit. By the maximum modulus principle, the norm equals $\sup_{\abs{z} = 1} \abs{f(z)}$, so $A(\mathbb{D})$ may also be viewed as an algebra of functions on the circle. The polynomials are dense: $f(rz) \to f(z)$ uniformly as $r \to 1$, and each $f(rz)$ is the uniform limit of its Taylor polynomials on $\overline{\mathbb{D}}$. The maximal ideals of $A(\mathbb{D})$ are the sets $\set{f : f(a) = 0}$ for $a \in \overline{\mathbb{D}}$.

Example: $f(z) = \sum_{n \ge 1} z^n / n^2$. The terms are bounded by $1/n^2$ on $\overline{\mathbb{D}}$, so the series converges uniformly there; hence $f$ is continuous on $\overline{\mathbb{D}}$ and holomorphic in $\mathbb{D}$.

Non-example: $S(z) = \exp\left(-\frac{1 + z}{1 - z}\right)$ is bounded and holomorphic in $\mathbb{D}$, but it has no continuous extension to $\overline{\mathbb{D}}$. Its radial limit at $1$ is $0$, while its radial limits at the points $e^{it} \ne 1$ have modulus $1$, so no continuous extension to $\overline{\mathbb{D}}$ exists.