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I am trying to prove that set of bounded, analytic functions $A(\mho)$, $u:\mho\to\mathbb{C}$ forms a Banach space. It seems quite clear using Morera's theorem that if we have a cauchy sequence of holomorph functions converge uniformly to holomorph function. Now i am a bit confused what norm would be suitable in order to make it complete .

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    Which norm are you using?2012-05-14
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    I was just thinking whether any of the L-norms would work. But i am puzzled how to deduce the conclusion . suggestions and explanations are welcome :)2012-05-14
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    You can define a norm of the form $\sum_{n\geq 1}2^{-n}\sup_{z\in K_n}|f(z)|$, where $K_n$ are compacts such that $\Omega=\bigcup_{n\geq 1}K_n$.2012-05-14
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    @ David Thanks, can you explain a bit more , may be as an answer. I would like to appreciate the norm that you have defined . Why did you choose $K_n$ to be compact?2012-05-14
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    In fact it's not necessary at all. A natural norm for $A(\Omega)$ is $\lVert f\rVert=\sup_{z\in\Omega}|f(z)|$; since $f$ is bounded it's well-defined. If $\{f_k\}$ is a Cauchy sequence, then $\{f_k(z)\}$ is Cauchy for each $z$ hence you can define $f(z)$ as $\lim_{n\to \infty}f_n(z)$. Such a function is bounded, and we have to show that $f$ is analytic. Morera is a good idea, otherwise you can use Cauchy's integral formula, and show that a function which satisfies this identity is necessarily analytic.2012-05-14
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    Thanks. I am little curious about the way you defined rather than the natural one, can you explain it:)2012-05-14
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    What are you calling the normal one?2012-05-14
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    oh sorry i meant "natural".2012-05-14
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    In fact it was a confusion with the general case, where $f$ is not assumed to be bounded (but in fact, in order to make this well-defined, we have to use $\min(1,\sup_{K_n}|f|)$, which only gives a semi-norm).2012-05-14

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We endow $A(\Omega)$ with the norm $\lVert f\rVert:=\sup_{z\in\Omega}|f(z)|$, which is well-defined since $f$ is bounded. Let $\{f_n\}\subset A(\Omega)$ a Cauchy sequence, then for a fixed $z$, $\{f_n(z)\}\subset \Bbb C$ is a Cauchy sequence, and has a limit, denoted $f(z)$.

$f$ is bounded, since we can find $N$ such that if $m,n\geq N$ then for all $z\in\Omega$: $|f_n(z)-f_m(z)|\leq 1$ so $|f_n(z)-f(z)|\leq 1$ and $|f(z)|\leq 1+\sup_{\Omega}|f_n|$. Fix $\varepsilon>0$, and take $N\in\Bbb N$ such that if $n,m\geq N$ then for each $z\in\Omega$: $|f_n(z)-f_m(z)|\leq \varepsilon$. We have, letting $m\to +\infty$, that $|f_n(z)-f(z)|\leq \varepsilon$ for $n\geq N$ hence $\lVert f-f_n\rVert\to 0$.

Now we show that $f$ is analytic. We have for each $z$ and each $n$ that $$f_n(z)=\frac 1{2\pi i}\int_{C(z,r)}\frac{f_n(\xi)}{\xi-z}d\xi,$$ where $r$ is such that $\{z'\mid |z-z'|$$f(z)=\frac 1{2\pi i}\int_{C(z,r)}\frac{f(\xi)}{\xi-z}d\xi.$$ For $h$ small enough we have $$f(z+h)=\frac 1{2\pi i}\int_{C(z,r)}\frac{f(\xi)}{\xi-(z+h)}d\xi$$ hence $$\frac{f(z+h)-f(z)}h=\frac 1{2\pi ih}\int_{C(z,r)}\frac{f(\xi)}{(\xi-z)(\xi-(z+h))}(-h)d\xi,$$ which have a limit when $h\to 0$ (namely $-\frac 1{2\pi i}\int_{C(z,r)}\frac{f(\xi)}{(\xi-z)^2}d\xi$).

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    Tnx ! well explained2012-05-14
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    Hi Giraudo, why is it possible that the limit to infinity of the integrals is the integral of limit?2017-04-20