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Let's assume that $f$ is a quasiconformal homeomorphism of $\mathbb{C}$ with Beltrami coefficient $\mu = \frac{\bar{\partial} f}{\partial f}$. Notice that by definition $\Vert \mu \Vert _{L^{\infty}} < 1$. It is well known that if $\mu_n \rightarrow \mu$ in $L^{\infty}$ then $f_n \rightarrow f$ (the f's are all normalised to fix three points) in the compact open topology (hence pointwise at least). I believe (but am not sure) that the rate of convergence is also known.

My question is - In the above setting if $\mu_n \rightarrow \mu$ in $L^p$ for some $\infty >p>1$ then can we say (preferably with effective convergence estimates) that $f_n$ still goes to $f$ in the compact-open topology (or atleast pointwise)? It is also assumed that $\Vert \mu _n \Vert _{L^{\infty}} \rightarrow \Vert \mu \Vert _{L^{\infty}}$.

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EDIT: The old answer to the originally asked question (without the additional assumption of convergence of the $L^\infty$ norms) is below.

I am not sure about effective convergence estimates, but the answer to the question is yes, even under slightly weaker assumptions. It directly follows from the Bers-Bojarski convergence theorem, which can be found in the book of Lehto and Virtanen. The statement of the theorem is that if $(f_n)$ is a sequence of $K$-qc mappings converging locally uniformly to a $K$-qc mapping $f$, and if their complex dilatations $\mu_n \to \mu$ a.e., then $\mu$ is a.e. the complex dilatation of $f$.

In the situation of your question, if $(f_n)$ is a sequence of normalized $K$-qc maps with dilatations $\mu_n\to\mu$ in $L^p$, any subsequence has a subsequence which converges locally uniformly to a $K$-qc limit (by compactness of the family of normalized $K$-qc maps.) Now you can pass to yet another subsequence where $\mu_n \to \mu$ a.e. Applying the Bers-Bojarski theorem gives that any such subsequential $K$-qc limit $f$ has complex dilatation $\mu$, which implies that the original sequence converges locally uniformly to $f$, where $f$ is the normalized qc map with complex dilatation $\mu$.

Note that you really only need $\mu_n \to \mu$ locally in measure, and uniform quasiconformality of the sequence $(f_n)$.


Old Answer

(Without the assumption that $\| \mu_n \|_\infty \to \| \mu \|_\infty$.)

I assume you want to have $f$ to be normalized in some way, otherwise convergence of the maps will not hold even in the case of $L^\infty$ convergence of the complex dilatations. However, even for normalized qc maps, without additional assumptions $L^p$ convergence of the complex dilatations does not imply convergence of the maps.

As an example, consider the family $f_k$ of quasiconformal mappings defined in polar coordinates as $(r,\theta) \mapsto (g_k(r),\theta)$, where $g_k$ is the identity for $r > 1$, $g_k(r) = r^{k^2}$ for $1-\frac1k \le r \le 1$, and $g_k$ is linear on $[0,1-1/k]$. Then $f_k$ is quasiconformal in the plane, $\mu_k$ is supported on the annulus $1-\frac1k \le r \le 1$, so $\mu_k \to 0$ in $L^p$ for any $p<\infty$, and $f_k$ converges to the identity outside the unit disk, and to the constant $0$ inside the disk, so the limit is not even continuous.

Maybe you want to require that the $f_k$ are uniformly quasiconformal, i.e., that there exists $\kappa<1$ such that $\| \mu_k \|_\infty \le \kappa$ for all $k$?

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  • $\begingroup$ Fair enough. Indeed I want them to be uniformly quasiconformal. In fact, stronger than that. $\endgroup$
    – Vamsi
    Nov 21 '14 at 3:38
  • $\begingroup$ I added the answer to the updated question. $\endgroup$ Nov 21 '14 at 20:47

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