# Thread: can such a map be surjective?

1. ## can such a map be surjective?

consider the function $f:\mathbb{D}\rightarrow \mathbb{C}$

where $f$ is an analytic function on the space $\mathbb{D}$ (open unit disk in $\mathbb{C}$)

can we show that $f$ is in fact surjective?

In the bigger picture I would like to thow that $\mathcal{A}$, the Banach algebra whose elements are analytic functions on $\mathbb{D}$ with pointwise defined operations is contained in the space $H^{\infty}(\mathbb{D})$ of all bounded analytic function on $\mathbb{D}$

The plan is as follows:
Since any $f\in\mathcal{A}$ is analytic it will also be continuous. $\mathbb{D}$ is compact, so if I can show that theses functions $f$ are in fact surjective then I can use preservation of compactness which implies the image $f(\mathbb{D})$ will be compact, and hence closed and bounded.

2. Originally Posted by Mauritzvdworm

In the bigger picture I would like to thow that $\mathcal{A}$, the Banach algebra whose elements are analytic functions on $\mathbb{D}$ with pointwise defined operations is contained in the space $H^{\infty}(\mathbb{D})$ of all bounded analytic function on $\mathbb{D}$
I don't get what you mean in the first paragraph, but this is (I think) false. It is a standard (although not that well known apparently) result that any open set in $\mathbb{C}$ has a function defined on it that cannot be extended at all (Domains of Holomorphy). The proof for this (see Remmert, Classical topics in Complex function theory) builds a function that has a sequence accumulating everywhere in the boundary with $f(x_n) \rightarrow \infty$.

Also $\mathbb{D}$ is open so it's not compact, which is the problem.

Edit: Mybe this was overkill (although it proves that this inclusion is not valid for any open set). For your case take $f(z)=\frac{1}{1-z}$ and note that $1$ is a pole.

Edit2: Ha, even in the general case it's an overkill. Just take a point in the boundary $p$ and define $f(z)=\frac{1}{p-z}$.