Hi. I have a space defined as follows:

Let D be a bounded simply connected domain in C and let HC(D) be the space of all analytic functions on D which are continuous on Closure(D). How can I show that HC(D) is Banach W.r.s.p Sup norm on Closure(D)?

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- Aug 5th 2011, 02:17 AMmghasemi62Why this space is a Banach space?
Hi. I have a space defined as follows:

Let D be a bounded simply connected domain in C and let HC(D) be the space of all analytic functions on D which are continuous on Closure(D). How can I show that HC(D) is Banach W.r.s.p Sup norm on Closure(D)? - Aug 5th 2011, 03:17 AMgirdavRe: Why this space is a Banach space?
Let $\displaystyle \{f_n\}\subset HC(D)$ a Cauchy sequence. Since $\displaystyle D$ is simply connected, you can use Cauchy formula to show that the limit $\displaystyle f$ is analytic.

- Aug 5th 2011, 07:18 AMmghasemi62Re: Why this space is a Banach space?
Thank you for the help.

I want to know that, is there any other norm which HC(D) can be Banach with that norm? For example is it Banach with ||f||=Sup|f^(k)(x)| k=0,1,... (x in closure(D)).

If it is not then which norm I can use except norm ||f||=Sup|f(x)|? - Aug 6th 2011, 11:17 AMJose27Re: Why this space is a Banach space?
For the first post, I'd like to note that the simply-connectedness is not necessary (instead of Cauchy/Morera use Cauchy's integral formula).

For the other questions: The norms of the derivatives may not make sense unless you can control the derivatives at the boundary. On the other hand, Cauchy's estimates allow you to control all derivatives locally (away from the boundary), so in this spirit we have the norms

$\displaystyle \| f \| _{k, z_1, \ldots , z_k} = \sup |f| + \sum_{j=1}^k |f^{(j)} (z_j)|$

where $\displaystyle z_j \in \mbox{int} (D)$ are arbitrary (but fixed) points. Then all these are equivalent to the one you're given. At the moment I don't know if you can estimate the derivatives on all of $\displaystyle D$ with only the original function (For what it's woth my guess is no).

Edit: In the definition of the norms, you can substitute the j-th term in the sum by the supremum of the j-th derivative in some (fixed) compact subset of the domain and the argument still works. - Aug 6th 2011, 03:38 PMmghasemi62Re: Why this space is a Banach space?
hmm. Good Idea. But still I have a problem !!?

- Aug 6th 2011, 03:47 PMmghasemi62Re: Why this space is a Banach space?
Sorry i'm a little confused. How can I prove the completeness with respect to your suggested norm.

In the first post when I have define $\displaystyle \sup$ norm on $\displaystyle HC(D)$ I needed the uniform convergence of $\displaystyle f_{n}$ to $\displaystyle f$, in order to use Cauchy/Morera theorem to prove that $\displaystyle HC(D)$ is Banach w.r.p $\displaystyle \sup$ norm. But now I don't have the uniform convergence with this new norm. How can I do it now? (Maybe I did it in a wrong way already) - Aug 6th 2011, 09:51 PMJose27Re: Why this space is a Banach space?
The norms are equivalent so they have the same Cauchy sequences. To prove the first statement the Cauchy estimates are the key: We can find constants $\displaystyle b,c>0 $ (dependent on k, and the $\displaystyle z_j$) such that

$\displaystyle b\sup |f| \leq \| f\| _{k,z_1, \ldots ,z_k} \leq c\sup|f|$