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 .
How to show set of all bounded, analytic function forms a Banach space?
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0Which norm are you using? – 2012-05-14
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0I 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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1You 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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0@ 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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1In 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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0Thanks. 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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0What are you calling the normal one? – 2012-05-14
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0oh sorry i meant "natural". – 2012-05-14
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0In 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
1 Answers
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'|
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0Tnx ! well explained – 2012-05-14
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0Hi Giraudo, why is it possible that the limit to infinity of the integrals is the integral of limit? – 2017-04-20