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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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    @joriki, edited as per your request.2011-11-13

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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

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

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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    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)2011-11-14