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By the Teichmuller uniqueness theorem, given a homeomorphism $f:X \rightarrow X$ where $X$ is the $n$-punctured sphere, there is a unique quasiconformal homeomorphism $g$ fixing $0$, $1$, and $\infty$ (in the same homotopy class as $f$) whose Beltrami coefficient $\mu$ has the smallest $L^{\infty}$ norm $x$.

My question is : If $g_n$ are q.c homeomorphisms fixing $0$, $1$, and $\infty$ in the same homotopy class as $f$ such that the $L^{\infty}$ norms $x_n$ of their Beltrami coefficients $\mu_n$ decrease to a limit $x$ (but we don't know whether the $\mu_n$ converge to anything), then do the $g_n$ converge to $g$ in some space (ex: Sobolev space)?

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    $\begingroup$ Do you assume in the 2nd paragraph that $x$ is the norm of the Teichmuller map $g$ (equal to $f$ from the 1st paragraph)? Otherwise, the answer is clearly negative. If $g$ is indeed extremal, then the answer is obviously positive because of the convergence property for qc maps; convergence is uniform on the 2-sphere. $\endgroup$
    – Misha
    Commented Nov 19, 2013 at 5:52
  • $\begingroup$ Yes x is the norm of the Teichmuller map. Sorry, I am a novice in this field. So you are saying that if the $L^{\infty}$ norms of some Beltramis get close to the extremal $L^{\infty}$, then the corresponding q.c maps get close in the uniform topology (also I don't want just a subsequence but the entire sequence to converge)? If so, can you cite a reference. Thanks a million! $\endgroup$
    – Vamsi
    Commented Nov 19, 2013 at 15:10

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Here are some details. First, in your setting, all the maps $g_n$ are K-quasiconfirmal for a certain K. Thus, by the convergence property for qc maps, the sequence $(g_n)$ has a convergent subsequence and the limit is qc. Clearly, the limit is homotopic to the maps $g_n$. It suffices to show then that if $(g_n)$ itself converges to some $h$ then this limit is the extremal map. By the same convergence property, since norms of Beltrami differentials are assumed to converge to the norm of the external map $g$, we get $$ K(h)=K(g). $$ Hence, by uniqueness of the extremal map, $h=g$. qed

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    $\begingroup$ @Vamsi: This is just a general nonsense from the point-set topology: If you have a sequence $z_n$ in a compact Hausdorff space $Z$ and every convergent subsequence in $(z_n)$ has the same limit $z$, then $\lim_n z_n=z$. In our setting the compact space is the space of K-qc maps normalized at 3 points. $\endgroup$
    – Misha
    Commented Nov 19, 2013 at 21:16
  • $\begingroup$ Sorry. I was being silly. $\endgroup$
    – Vamsi
    Commented Nov 19, 2013 at 21:21

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