How to show that the disc algebra $A(D)$ is a Banach space?

Question

I just know that $A(D)$ is a space of functions analytic on the open unit disc and it is a subspace of $H^\infty$ and $H^\infty$ is an hardy space while $H^\infty$ is defined as the vector space of bounded holomorphic functions on the disk. How can I show the disc algebra is a Banach space?

I would be so appreciated if you help me.

Answer 1

The magic phrase is Weierstraß-convergence-theorem (I am not 100 % sure if this is its name in the english literature). The point is that $A(D)$ is obviously a subalgebra of $C(D)$ (set of continuous functions on $D$) which is a Banach space (even Banach algebra) w.r.t the maximumnorm/supremumnorm (again I'm not sure about the english word, anyway I think you know this statement). An important and well known theorem of Weierstraß states that a sequence of holomorhpic locally uniformly convergent functions converge to a holomorhic function. That is to say that $A(D)$ is closed inside $C(D)$.

Answer 2

$\scriptstyle \text{Let me add some details :}$

If a sequence of bounded continuous functions converges in the $\sup$ norm then the limit is continuous and bounded.

If a sequence $f_n(z)$ of holomorphic functions converges uniformly to $f(z)$ then $\int_\gamma f(z) dz= \lim_{n \to \infty} \int_\gamma f_n(z)dz = 0$ for any closed contour $\gamma$, so that $F(z) = \int_a^z f(s)ds$ is well-defined (it doesn't depend on the path $a \to z$) and holomorphic, which means by the 'holomorphic $\implies$ analytic' theorem that $f(z) = F'(z)$ is analytic too.

Altogether, a Cauchy sequence in the disk algebra converges to an analytic function continuous on the boundary.

Answer 3

$ \def \d{\mathrm{d}}$ Suppose $\{f_n\}$ is a Cauchy sequence in the disc algebra, than it converges to a continuous function $f$ on the disc. For every closed triangle $\Delta\subset D$, we have \begin{align*} &\quad\int_{\partial\Delta}f(z)\d z \\&=\lim_{n\to\infty}\int_{\partial\Delta}f_n(z)\d z ({\text{the Domainated Convergence Theorem}})\\ &=0({\text{Cauchy's Theorem in a Convex Set}}). \end{align*} Hence $f$ is holomorphic by Morera's Theorem.

All of these can be found in chapter 10 of Real and Complex Analysis by Rudin