Quasiconformal homeomorphism with dilatation supported in a strip

Question

Consider a $K$-quasiconformal homeomorphism $\varphi\colon \mathbb{C} \to \mathbb{C}$ such that $\varphi(0) = 0$, $\varphi(1) = 1$, and the corresponding Beltrami coefficient $\mu_{\varphi}$ is supported in the strip $A_1 = \{x + i y: x \in \mathbb{R}, y \in [-1, 1]\}$. Can we prove that the image of $A_1$ is contained in some other strip $A_R$ (where $R$ depends on $K$)? Or which assumptions one can impose on $\varphi$ or $\mu_{\varphi}$ so that this would be true?

Answer 1

Here is an attempt of the proof for the case when $\varphi$ preserves the real axis (in fact, similar proof should work if we assume that $\varphi(\mathbb{R}) \subset A_{r'}$ for some $r'$). In the general situation, theoretically, the image $\varphi(A_1)$ can be a topological strip that does not get "fatter", but slowly oscillates from $\mathbb{R}$, i.e., can not be put inside any $A_R$.

All the maps considered in the proof are $K$-q.c., they fix $0$ and $\infty$, and their dilatations are supported in $A_1$. In particular, by the first part of Teichmuller-Wittich-Belinski-Lehto theorem (Theorem 6.1 from O. Lehto, K.I. Virtanen, Quasiconformal Mappings in the Plane) implies that all such maps are complex-differentiable at infinity, i.e., $\partial_z \varphi(\infty) := \partial_z (1/\varphi(1/z))(0) = \varphi'(\infty) = \lim_{z \to \infty} z/\varphi(z)$ is well defined and $\partial_{\overline{z}} \varphi(\infty) := \partial_{\overline{z}}(1/\varphi(1/z))(0) = 0$.

Also we introduce the following two notations: $\mathcal{I}_s := \{s + ti: t \in [-1, 1]\}$ and $l_{\varphi} := \max_{z \in \varphi(\mathcal{I}_0)} |z|$.

First of all, let us prove the following lemma (in fact, in it we don't use the fact that $\varphi(\mathbb{R}) = \mathbb{R}$):

Lemma. Let $\varphi$ be a $K$-q.c. homeomorphism such that $\varphi(0) = 0$, $\varphi(\infty) = \infty$, $\partial_z \varphi(\infty) = \alpha \in \mathbb{C}^*$ and $\mathrm{supp} \mu_{\varphi} \subset A_1$. Then there is a constant $L = L(K, \alpha)$ so that $l_{\varphi} \leq L$.

Proof. Consider the map $\psi(z) := \varphi(z) / \varphi(1)$. It is easy to see that $\psi$ fixes 0, 1 and $\infty$, $\partial_z \psi(\infty) = \alpha \varphi(1)$ and $\mu_{\psi} = \mu_{\varphi}$ pointwise, moreover, $l_{\psi} = l_{\varphi} / |\varphi(1)|$.

Now let us apply the last part Teichmuller-Wittich-Belinski-Lehto theorem. It requires certain preparation. Let $$ I_{\psi} := \frac{1}{2\pi}\iint_{|z| > 1} \frac{|\mu_{\psi}(z)|}{|z|^2}\mathrm{d}x\mathrm{d}y \leq I(K), $$ since $|\mu_{\psi}|$ is bounded from above and $\iint_{A_1 \setminus \mathbb{D}} \frac{\mathrm{d}x\mathrm{d}y}{|z|^2}$ is finite. Then by the theorem we have that $$ \frac{e^{-I_{\psi}}}{\max_{|z| = 1} |\psi(z)|} \leq |\partial_z \psi(\infty)| \leq \frac{e^{I_{\psi}}}{\min_{|z| = 1} |\psi(z)|}. $$

Note that both max and min above are uniformly bounded among all $K$-q.c. $\psi$ fixing 0, 1 and $\infty$ (since the space of all such maps is compact), and $I_{\psi}$ is bounded by $I(K)$ as it was observed before. In other words, there are $0 < m < M$ such that $m \leq |\partial_z \psi(\infty)| = |\alpha\varphi(1)| \leq M$. Finally, we $l_{\varphi}$ is bounded by some $L = L(K, \alpha)$ for all $\varphi$ satisfying $m/\alpha \leq |\varphi(1)| \leq M/\alpha$ and fixing 0 and $\infty$, since they form a compact set.

Now let us return to the map $\varphi$ from the original question. Consider family of maps $(\varphi_t)_{t \in \mathbb{R}}$ where $\varphi_t(z) := \varphi(z + t) - \varphi(t)$. It is easy to see that $\varphi_t$ fixes 0 and $\infty$, moreover, $\partial_z \varphi_t(\infty) = \partial_z \varphi(\infty)$. Thus, we can apply Lemma to show that $l_{\varphi_t}$ is bounded by some uniform constant $L(K, \partial_z \varphi(\infty)) \leq L(K, z_0)$ (again, last equality can be proved by Teichmuller-Wittich-Belinski-Lehto theorem and compactness argument). Now note that $\varphi_t(\mathcal{I}_0)$ equals to real-translated copy (here we use the assumption that $\varphi(t) \in \mathbb{R}$) of $\varphi(\mathcal{I}_t)$ and, therefore, its vertical oscillation is bounded. Hence, there exists $R = R(K, z_0)$ such that $\varphi(A_1) \subset A_r$.