The disk algebra and continuous functions The disk algebra is the set of continuous functions $f: D \to \mathbb C$ where $D$ is the closed unit disc in $\mathbb C$ and $f$ is analytic on the interior of $D$. It is endowed with the $\sup$-norm. 
Why this extra restriction to only include $f$ that are analytic on the interior of $D$? Wouldn't $C(D)$, the set of all continuous $f:D\to \mathbb C$ make a fine complete normed algebra, too?
 A: $C(D)$ is a perfectly fine Banach algebra. (But $C(\text{open unit disc})$ is not, at least under the sup-norm.)
A: The algebras $C(K)$ with $K$ compact set are widely studied algebras, and in lots of ways they are very well understood. 
I have always thought that we study the disk algebra, where functions are analytic in the interior of $D$, to be able to translate the "magic" that arises when analytic functions appear to the context of Banach Algebras. Observe that the other natural complex functions spaces $H^p(D), A^p(D)$... are not Banach Algebras.
On the other hand, in general the Weierstrass theorem of density of polynomials is not valid in general when you work in compact sets of the complex plane. Observe that the sup-norm closure of the polynomials only generates analytic functions. If you want to get every continuous function you must consider usual polynomials in $z$ and polynomials in $\overline{z}$. My point is that you can consider that the disk algebra $A(D)$ arises in a natural way when you consider it as the closure of polynomials with the sup-norm defined in the unit disk $D$.
