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Assume that $D_n\subset D$, where $D$ is the unit disk is an increasing sequence of Jordan domains with smooth boundaries such that $\cup_{n=1}^\infty D_n=D$ and let $f_n: D \to D_n$ be conformal mapping such that $f(0)=0$ and $\arg(f_n'(0))=0$. Is $f_n$ uniformly convergent to $Id$.

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    $\begingroup$ You mean $f_n(0)=0$ and $\mathrm{arg}f_n'(0)=0$ don't you? $\endgroup$ – Pietro Majer Nov 7 '13 at 21:51
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    $\begingroup$ @PietroMajer : Surely this is what is meant. I assumed it in my answer. $\endgroup$ – Malik Younsi Nov 7 '13 at 21:51
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The answer is no. Consider the following sequence $D_n$. Take the disc $|z|<1-1/n=r_n$, and remove from it the arc $\{ z:|z|=r_{n-1},|\arg z|<\pi-1/n\}$ and the interval $(r_{n-1},r_n)$. The result is a simply connected region $D_n$. This $D_n$ contains $|z|<r_{n-1}$, so the union of $D_n$ is the whole unit disc, and they are nested. Small deformation will make these regions Jordan, smooth, etc.

Evidently the convergence of the conformal maps $f:D\to D_n$ is uniform on compact subsets of $D$ (this follows from Caratheodory thm), but not uniform in $D$, because near the point $z=-1$ the function $f_n(z)$ takes values close to every point of the unit circle. (A small neighborhood of $-1$ blows up under $f_n$ to almost the whole ring $r_{n-1}<|z|<r_n$.)

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  • $\begingroup$ Yes you right, I understand the point. Thanks. $\endgroup$ – user36162 Nov 8 '13 at 7:40
  • $\begingroup$ You're right of course, I was confused and my answer was wrong. Thank you for pointing it out. Great answer, +1 ! $\endgroup$ – Malik Younsi Nov 8 '13 at 12:33

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