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## Homework Statement

Let [itex]F[/itex] be the set of all analytic functions [itex]f[/itex] that map the open unit disc [itex]D(0,1)[/itex] into the set [itex]U = \left\{w=u+iv : -2 < u < 2 \right\}[/itex] such that [itex]f(0)=0[/itex]. Determine whether or not [itex]F[/itex] is a normal family.

## Homework Equations

[itex]DEF'N:[/itex] A normal family on a domain (i.e. open and connected) [itex]U\subset ℂ [/itex] is a family of analytic functions such that every sequence in the family has a subsequence that converges uniformly on compact subsets of [itex]U[/itex].

[itex]THEOREM[/itex](Montel's):

Version 1: A family of analytic functions that is bounded on compact subsets of a domain contains a sequence that converges uniformly on compact subsets (bounded on compact subsets means that for each compact subset there is some positive bound that holds for every function in the family. The constant can depend on the subset but NOT on the function or the point z).

Version 2: If every function in the family is bounded by the same positive constant on the domain then there exists a sequence in the family that converges uniformly on compact subsets.

## The Attempt at a Solution

My first thought was to simplify by considering only functions from [itex]D \rightarrow D [/itex]. Schwarz lemma says that since [itex]f(0)=0[/itex] and [itex]|f(z)|\leq 1 [/itex] then [itex]|f(z)| \leq |z| [/itex]. Hence we can apply the above theorem since we have a bound over the domain and conclude that this "subfamily" is normal. Too bad the real problem isn't this easy! I'm having a tough time figuring this out for the family [itex]F[/itex]. First of all, it only states that the functions are into [itex]U[/itex]. So the mapping can be ANY one-to-one map into ANY subset of [itex]U[/itex]. I've also thought about using maximum modulus principle, but this only gives me bounds on each particular function, rather than the whole family.

Any clues would be greatly appreciated!! (EDIT: not even sure if my above statements are entirely correct since I'm pretty confused at this exact moment)

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