Hardy class of bounded analytic functions is Banach space

Question

I need help with the following:

Let $H^{\infty}(\mathbb{D})$ denote Hardy class of bounded analytic functions on unit disc $\mathbb{D} = \{z \in \mathbb{C}: |z|<1\}$. Prove that $$||f|| = \underset{|z|<1}{\sup}|f(z)|$$ is norm on $H^{\infty}(\mathbb{D})$ and $H^{\infty}(\mathbb{D})$ is Banach space in that norm.

I think I have proved that this is norm, if $||f||=0$, then $f(z)=0, \forall z\in \mathbb{D}$, so $f$ is zero function. The rest is trivial.

My attempt to prove that this is Banach space is the following: Let $(f_n)\in H^{\infty}(\mathbb{D})$, such that $\sum_{n=1}^{\infty} ||f_n||$ converges. We want to prove that $\sum_{n=1}^{\infty} f_n$ converges in $H^{\infty}(\mathbb{D})$. Let $z \in \mathbb{D}$: $$ |f_n(z)| \leq ||f_n|| \implies \sum_{n=1}^{\infty} |f_n(z)|<\infty$$ in Banach space $\mathbb{C}$, so $\sum_{n=1}^{\infty} f_n(z)<\infty$. Denote $f(z):=\sum_{n=1}^{\infty} f_n(z)$. Then, $$||f-\sum_{k=1}^n f_k|| \leq \sum_{k=n+1}^{\infty}||f_k||\to 0$$ so $\sum_{n=1}^{\infty} f_n$ converges to $f$.

It seems to me that this is not enough, I think I have to prove also that $f \in H^{\infty}(\mathbb{D})$, i.e. that $f$ is bounded analytic function, but I'm stuck there.

Thanks in advance!

Answer 1

You're almost done. To show that $f$ is bounded and analytic, use the following hints:

1. Morera's theorem implies that uniform limit of analytic functions on an open connected set is analytic (see here).
2. The uniform limit of a sequence of bounded functions is bounded.