The ACL and Sobolev analytic definition of quasiconformality

• Mikhail Lyubich, Conformal Geometry and Dynamics of Quadratic Polynomials, vol. I, Ch. 2 §§11.3–11.5. Section 11.3 defines distributional partial derivatives and gives the ACL criterion for homeomorphisms of planar domains; §11.4 defines quasiconformality by local integrability of these derivatives and bounded dilatation; Proposition 2.11 proves local square-integrability for quasiconformal maps.
• Christopher J. Bishop, Quasiconformal Mappings, Ch. 2 §1 and Ch. 3 §4, equation (4.1). The example there uses the Cantor singular function to show that the differential inequality alone does not imply ACL. Its printed derivative sentence has a typo: for $f(x+iy)=x+g(x)+iy$, one has ${\mathrm{\partial }}_{z}f=1$ and ${\mathrm{\partial }}_{\bar{z}}f=0$ almost everywhere; these are the values used below.

Assume the Axiom of Choice (The Axiom of Choice). Let $\mathrm{\Omega },{\mathrm{\Omega }}^{\mathrm{\prime }}\subseteq \mathbb{C}$ be complex domains (A complex domain is a nonempty connected open subset of $\mathbb{C}$) and $f:\mathrm{\Omega }\to {\mathrm{\Omega }}^{\mathrm{\prime }}$ a homeomorphism (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological). Read $f$ as a map into ${\mathbb{R}}^2$; continuity makes its components locally square integrable. Recall that $W_{\mathrm{l}\mathrm{o}\mathrm{c}}^{1,2}(\mathrm{\Omega })$ (Integer-order Sobolev spaces and their norms) means that the components of $f$ admit weak first partial derivatives (Weak derivative of a locally integrable function) in $L_{\mathrm{l}\mathrm{o}\mathrm{c}}^2$; equivalently, their almost-everywhere classes have ACL representatives in both coordinate directions whose measurable classical coordinate derivatives belong to $L_{\mathrm{l}\mathrm{o}\mathrm{c}}^2$ (Absolute continuity on almost every coordinate line, The ACL characterisation of $W^{1,p}$). ACL alone does not assert this derivative integrability. Countable Choice (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)) is included in AC for the completed-product Fubini interfaces. On almost every coordinate line the continuous map $f$ agrees almost everywhere with its ACL representative, hence everywhere by continuity of both restrictions; its classical line derivatives then represent its weak derivatives.

Let $K\ge 1$ and put $k=(K-1)\mathrm{/}(K+1)\in [0,1)$. The homeomorphism $f$ is $K$-quasiconformal in the analytic sense when

(A1) $f\in W_{\mathrm{l}\mathrm{o}\mathrm{c}}^{1,2}(\mathrm{\Omega })$; and

(A2) its weak Wirtinger derivatives ${\mathrm{\partial }}_{z}f,{\mathrm{\partial }}_{\bar{z}}f$ (The Wirtinger derivatives ${\mathrm{\partial }}_{z}f$ and ${\mathrm{\partial }}_{\bar{z}}f$, and antiholomorphic functions) lie in $L_{\mathrm{l}\mathrm{o}\mathrm{c}}^2(\mathrm{\Omega })$ and satisfy $\mathrm{\mid }{\mathrm{\partial }}_{\bar{z}}f(z)\mathrm{\mid }\le k\mathrm{\mid }{\mathrm{\partial }}_{z}f(z)\mathrm{\mid }\text{for almost every }z\in \mathrm{\Omega }\mathrm{.}$

The inequality is a statement about $L_{\mathrm{l}\mathrm{o}\mathrm{c}}^2$ classes: it is independent of the choice of representative of $f$ and of the Borel representatives of its weak Wirtinger derivatives, since each such representative agrees almost everywhere with its class (Weak differentiation ignores null-set changes). The map is analytically quasiconformal if it is $K$-analytically quasiconformal for some finite $K$. Its minimal constant is recovered from the Beltrami coefficient defined later on this page.

Orientation and the regularity requirement. Inequality (A2) gives $J_f=\mathrm{\mid }{\mathrm{\partial }}_{z}f{\mathrm{\mid }}^2-\mathrm{\mid }{\mathrm{\partial }}_{\bar{z}}f{\mathrm{\mid }}^2\ge (1-k^2)\mathrm{\mid }{\mathrm{\partial }}_{z}f{\mathrm{\mid }}^2\ge 0$ almost everywhere, with respect to planar Lebesgue measure (The nonnegative Lebesgue integral). A classical derivative inequality alone does not replace (A1): on $(0,1{)}^2$, the map $f(x+iy)=x+g(x)+iy$, where $g$ is the Cantor function, is a homeomorphism onto $(0,2)\times (0,1)$ (The map $x\mapsto x+c(x)$ is a homeomorphism from $[0,1]$ onto $[0,2]$). The Cantor function is continuous, nonconstant, and locally constant off the null Cantor set, so $g^{\mathrm{\prime }}=0$ almost everywhere (The Cantor function is well defined, satisfies $c(x)\le c(y)$ whenever $x\le y$, is surjective onto $[0,1]$, and is constant on every interval removed from the Cantor set, The Cantor set is an uncountable subset of $\mathbb{R}$ of Lebesgue measure zero). Thus the classical Wirtinger derivatives are ${\mathrm{\partial }}_{z}f=1$ and ${\mathrm{\partial }}_{\bar{z}}f=0$ almost everywhere. But $g$ cannot be absolutely continuous on every compact subinterval of $(0,1)$: the fundamental theorem would make it constant there, and then continuity would contradict $g(0)=0$, $g(1)=1$ (Fundamental theorem of calculus for absolutely continuous functions). Subtracting the identity shows that $f$ fails ACL on every horizontal line and therefore fails (A1). Its classical inequality is not the weak-derivative assertion (A2). Orientation preservation follows from (A1) together with (A2), as proved by the equivalence theorem on this page.