Thread: can such a map be surjective?

consider the function

where is an analytic function on the space (open unit disk in )

can we show that is in fact surjective?

In the bigger picture I would like to thow that , the Banach algebra whose elements are analytic functions on with pointwise defined operations is contained in the space of all bounded analytic function on

The plan is as follows:
Since any is analytic it will also be continuous. is compact, so if I can show that theses functions are in fact surjective then I can use preservation of compactness which implies the image will be compact, and hence closed and bounded.

Originally Posted by Mauritzvdworm

In the bigger picture I would like to thow that , the Banach algebra whose elements are analytic functions on with pointwise defined operations is contained in the space of all bounded analytic function on

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 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 .

Also 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 and note that is a pole.

Edit2: Ha, even in the general case it's an overkill. Just take a point in the boundary and define .