Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've been working on the following problem:


Let $D$ and $D'$ be simply connected plane domains, each different from the whole plane. Suppose that $z_1 \in D$ and $z_2 \in D'$. Let $\mathcal{F}$ be the family conisting of all functions $f$ that are defined, analytic, and one-to-one in $D$, and satisfy $f(D) \subset D'$ and $f(z_1) = z_2$. Prove that $\mathcal{F}$ is a normal family.


I know that the solution is trivial if I use Montel's 2nd Theorem (aka the Fundamental Normality Test). I think I'm supposed to use more basic results here though.

My idea was to invoke the Riemann Mapping Theorem to get a conformal map $h$ from $D'$ to the unit disk with $h(z_2) = 0$. The family $$ \{ h \circ f: f \in \mathcal{F} \}$$ is then a normal family by Montel's $1$st theorem, since it is uniformly bounded by $1$.

Is there some way I can get from here to normality of the original family; or some completely different approach I should be trying?

Thanks.

Edit: I've come up with a solution that might work. However, I'm a bit doubtful about it since it doesn't seem to use the one-to-one assumption anywhere. Can anyone point out a hole in my argument, or an implicit use of the one-to-one assumption?

By the Riemann Mapping Theorem, there exists a one-to-one analytic map $g$ from the unit disk $\mathbb{D}$ onto $D$ such that $g(0) = z_1$, and a one-to-one analytic map $h$ from $D'$ onto $\mathbb{D}$ such that $h(z_2) = 0$.

Then for each $f \in \mathcal{F}$, the composition $h \circ f \circ g$ is a one-to-one map from $\mathbb{D}$ to itself which fixes the origin.

Let $K$ be a compact subset of $D$, and let $K' = g^{-1}(K)$. Since $g^{-1}$ is continuous, $K'$ is a compact subset of $\mathbb{D}$. In particular, there exists $0 < r < 1$ such that $K' \subseteq \{ z: |z| \leq r\}$.

By the Schwarz Lemma, we may conclude that $$( h \circ f \circ g )(K') = (h \circ f)(K) \subseteq \{ z : |z| \leq r\}. $$

Now $$f (K) = h^{-1}((h \circ f)(K)) \subset h^{-1}(\{z: |z| \leq r\}).$$ The right-hand side of this relation is a compact set which is not dependent on the choice of $f$. It is therefore bounded, with a uniform bound for all $f \in \mathcal{F}$. By Montel's first theorem, $\mathcal{F}$ is a normal family.

share|improve this question

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.