compactness in the space of analytic functions I am always getting confused by the idea of compactness so I would like some help to see whether a set is compact. (I need this to prove existence of a solution of a map)
So let $D\in\mathbb{C}$ be open and bounded and let $C^\omega(D)$ be the space of functions analytic on $D$ equipped with the supremum norm. Let also $F\in C^\infty(\mathbb{R}_+;\mathbb{R}_+)$, with $\sup_{z\in D} F(|z|)=M>0$.
Now I define $A=\{f\in C^\omega(D):|f(z)|\le F(|z|) \}$. Is this set compact in $C^\omega(D)$?
I know that for it to be compact I need to prove that any open set in $A$ admits a finite cover. Yet I cannot see how to prove this. Any ideas?
 A: The "supremum norm" is NOT a norm on your space because analytic functions needn't be bounded. If you consider instead the space of bounded analytic functions you get a Banach space and for the constant function $F(x)=1$ your set is the unit ball which is not compact (because this is so in every infinite dimensional Banach space). 
On the other hand, the space of all analytic functions is usually endowed with the topology of uniform convergence on all compact sets (instead of a single norm you have a sequence of norms $\|f\|_n=\sup\lbrace |f(z)|: z\in K_n\rbrace$ where $K_n$ form a compact exhaustion). By a theorem of Montel, all bounded subsets are relatively compact (therefore, such topological vector spaces are called Montel spaces). The main ingredient in the proof of Montel's theorem is the Arzela-Ascoli theorem. 
By definition of your set $A$ it is quite easy to see that $A$ is bounded, that is $\sup\lbrace \|f\|_n: f\in A\rbrace <\infty$ for all $n\in\mathbb N$.

Edit. A nice exposition of the functional analytic view to complex analysis is the book of Luecking and Rubel (Springer 1984). If you fix $D$ compact the space of functions analytic in a neighborhood is a so-called LB-space (inductive limit of Banach spaces) and it is also Montel (see also the book of Luecking and Rubel or the Introduction to Functional Analysis by Meise and Vogt).
