# How to show set of all bounded, analytic function forms a Banach space?

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? – Davide Giraudo May 14 '12 at 19:30
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 :) – Theorem May 14 '12 at 19:32
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 $\Omega=\bigcup_{n\geq 1}K_n$. – Davide Giraudo May 14 '12 at 19:36
@ 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? – Theorem May 14 '12 at 19:41
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. – Davide Giraudo May 14 '12 at 19:50

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 $\{z'\mid |z-z'|<r\}\subset\Omega$. Using the uniform convergence, we get that $$f(z)=\frac 1{2\pi i}\int_{C(z,r)}\frac{f(\xi)}{\xi-z}d\xi.$$ 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 $-\frac 1{2\pi i}\int_{C(z,r)}\frac{f(\xi)}{(\xi-z)^2}d\xi$).