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Math Help - can such a map be surjective?

  1. #1
    Member Mauritzvdworm's Avatar
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    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.
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  2. #2
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    Quote Originally Posted by Mauritzvdworm View Post

    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}.
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