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Let$\lbrace f_n \rbrace_{n\in \mathbb{N}}$ be a sequence of biholomorphisms(bijective holomorphic maps with a holomorphic inverse) on a bounded region(open connected set) $\Omega$. If this sequence converges to $f$ uniformly on every compact subset of $\Omega$, does it also follow that $\lbrace f_n^{-1} \rbrace_{n\in \mathbb{N}}$ converges uniformly? (to $f^{-1}$, if defined)

Thank you in advance.

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By "(to $f^{-1}$, if defined)", are you saying that you want to know whether this holds if $f^{-1}$ is defined, or do you want to know whether it holds in general, and in particular if $f^{-1}$ is defined, you want to know whether the limit of the inverses is $f^{-1}$? –  joriki Nov 13 '11 at 17:45
    
@joriki, I want to prove that $f^{-1}$ is defined and also that the sequence of inverses converges to it. –  Chulumba Nov 13 '11 at 19:04
    
So far, the fact that this refers to complex analysis is apparent only from the tag. I think the question text should state what you mean by "automorphisms". –  joriki Nov 13 '11 at 19:10
    
@joriki, edited as per your request. –  Chulumba Nov 13 '11 at 20:22

1 Answer 1

up vote 2 down vote accepted

Basic Idea: You can use the integral representation of $f_n^{-1}$ to decide this question. I.e. use

$$f_n^{-1}(\omega) = \frac1{2\pi i}\int_{|\zeta-f_n^{-1}(\omega)|=r} \frac{\zeta f_n'(\zeta)}{f_n(\zeta) - \omega} \ d\zeta \label{\ast}$$

which holds for appropriate $r>0$.


Derivation of this formula: This follows from the residue theorem: Let $\omega, n$ be fixed and let $z_0 = f_n^{-1}(\omega)$. Then $f_n(z) -\omega = (z-z_0)g(z)$ for some meromorphic function $g:\Omega \to \Omega$, which satisfies $g(z) \ne 0$ on all of $\Omega$. Thus

$$\frac{zf_n'(z)}{f_n(z) - \omega} = z\frac{g(z) + (z-z_0)g'(z)}{(z-z_0)g(z)} = \frac{z}{z-z_0} + z\frac{g'(z)}{g(z)}$$ so we have

$$\mathrm{Res}\left(\frac{zf_n'(z)}{f_n(z) - \omega}; z_0\right) = z_0 = f_n^{-1}(\omega)$$

In particular the above holds as long as $D_r(z_0)$ is contained in $\Omega$ (we can of course take any other path in $\Omega$ going around $z_0$ exactly once!).


Answer to you question: Now let $\omega$ be arbitrary and choose $z_0$ such that $f(z_0) = \omega$. (I'm assuming that $f$ is surjective... I'll think about a proof) Note that for all $\delta > 0$ and for sufficiently large $n$, we have that

$$|f_n(\zeta) - f(\zeta)|< \delta$$

for all $\zeta$ on the circle $|\zeta - z_0| = r$. So we can actually make the path of integration independent of $n$, i.e. we have

$$f_n^{-1}(\omega) = \frac{1}{2\pi i}\int_{|\zeta-z_0|=r} \frac{\zeta f_n'(\zeta)}{f_n(\zeta) - \omega} d\zeta \label{\ast}$$

for all sufficiently large $n$.

Using this, we see that $f_n^{-1}$ converges locally uniformly to some holomorphic function $h:\Omega \to \Omega$ (consider a small enough disc around $\omega$).

We have $z = f_n^{-1}(f_n(z))$ for all $z, \ n$. This together with locally uniform convergence implies that $$f_n^{-1}(f(z)) \to z$$ as $n \to \infty$. But since $$f_n^{-1}(f(z)) \to h(f(z))$$ we must have $h(f(z)) = z$ ! So $h$ is a holomorphic left inverse for $f$.

But now, let's summarize what we have shown above:

If $f_n$ is a sequence of biholomorphic functions converging to $f$ locally uniformly, then $f$ has a left inverse $h$. In particular, $h$ is surjective.

So arguing in the same way, we see that since $f_n^{-1}$ is a sequence of biholomorphic function converging to $h$ locally uniformly, $h$ has a left inverse $g$. In particular $h$ is injective.

But then $h$ is bijective and therefore $f$ must be its inverse. This shows that $f$ is itself biholomorphic.

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I can neither prove, nor find a counterexample to the following: If $f_n$, $\Omega$ are as above, can it be the case that for some $\omega \in \Omega$ we have $f_n^{-1}(\omega) \to \partial \Omega$? The above proof will only work, if $f$ is surjective. Any ideas would be welcome! =) –  Sam Nov 14 '11 at 8:00
    
can you explain $f_n^{-1}(f(z)) \to h(f(z))$ as $n \to \infty$ a bit? –  Chulumba Nov 14 '11 at 13:37
    
Simply because $f_n^{-1} \to h$ pointwise. ($h$ is defined to be the limit of this sequence - which is shown to converge under the assumptions, which includes surjectivity of $f$ in addition to your other assumptions) –  Sam Nov 14 '11 at 15:49

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