Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 .

share|cite|improve this question
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
up vote 2 down vote accepted

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 $-\frac 1{2\pi i}\int_{C(z,r)}\frac{f(\xi)}{(\xi-z)^2}d\xi$).

share|cite|improve this answer
Tnx ! well explained – Theorem May 14 '12 at 20:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.