Area and $L^2$ derivative bounds for quasiconformal homeomorphisms

The ACL interface [F3] gives finite classical coordinate partials almost everywhere. The general core differentiability interface [F4] therefore gives total differentiability almost everywhere. Intersecting this full-measure set with the ACL set in [F3] and the weak inequality set in [F1] gives a full-measure subset $G\subseteq \mathrm{\Omega }$ where $f$ is totally differentiable and its classical Wirtinger derivatives agree with the weak Wirtinger classes. At every $x\in G$ the pointwise inequality gives $J_f(x)=\mathrm{\mid }{f}_z(x){\mathrm{\mid }}^2-\mathrm{\mid }{f}_{\bar{z}}(x){\mathrm{\mid }}^2\ge (1-k^2)\mathrm{\mid }{f}_z(x){\mathrm{\mid }}^2\ge 0.$ Also $J_f\in L_{loc}^1$ because $\mathrm{\mid }{J}_f\mathrm{\mid }\le \mathrm{\mid }{f}_z{\mathrm{\mid }}^2+\mathrm{\mid }{f}_{\bar{z}}{\mathrm{\mid }}^2$ and both weak derivatives are locally square-integrable.

Fix a rational box $Q⋐\mathrm{\Omega }$ and $x\in G\cap Q$ with $J_f(x)>0$. Set $A:=Df(x)$ and $m:=\mathrm{\parallel }{A}^{-1}{\mathrm{\parallel }}_{op}^{-1}>0$. Given $0<\delta <1$, differentiability gives, for all sufficiently small $r>0$, $\mathrm{\mid }f(x+h)-f(x)-Ah\mathrm{\mid }<\frac{1}{2}m\delta r(\mathrm{\mid }h\mathrm{\mid }\le r)\mathrm{.}$ For $\mathrm{\mid }h\mathrm{\mid }=r$ and $\mathrm{\mid }v\mathrm{\mid }\le (1-\delta )r$, the inequality $\mathrm{\mid }A(h-v)\mathrm{\mid }\ge m\mathrm{\mid }h-v\mathrm{\mid }\ge m\delta r$ shows that $f(\mathrm{\partial }B(x,r))$ misses the open ellipsoid $f(x)+A(B(0,(1-\delta )r))$. Since $f$ is a homeomorphism, $f(\mathrm{\partial }B(x,r))=\mathrm{\partial }f(B(x,r))$; the ellipsoid is connected, contains $f(x)\in f(B(x,r))$, and avoids that boundary, so it lies in $f(B(x,r))$. By [F8], $\mathrm{\mid }f(B(x,r))\mathrm{\mid }\ge (1-\delta {)}^2{J}_f(x)\mathrm{\mid }B(x,r)\mathrm{\mid }\mathrm{.}$

Define the finite Borel measure ${\nu }_Q(S):=\mathrm{\mid }f(S\cap Q)\mathrm{\mid }$ for Borel $S\subseteq {\mathbb{R}}^2$. Countable additivity follows from injectivity of $f$, and finiteness follows from [F6]. Let $g_Q$ be the Radon–Nikodym density of the absolutely continuous part of ${\nu }_Q$. For almost every $x\in Q$, [F7] gives ${\lim }_{r\to 0^{+}}\frac{\mathrm{\mid }f(B(x,r))\mathrm{\mid }}{\mathrm{\mid }B(x,r)\mathrm{\mid }}=g_Q(x),$ where $r$ is small enough that $B(x,r)\subset Q$. At points in $G$ with $J_f(x)>0$, step 2.1 and then $\delta \downarrow 0$ show that this limit is at least $J_f(x)$. At points with $J_f(x)=0$ the same inequality follows from $g_Q\ge 0$. Thus $J_f\le g_Q$ almost everywhere on $Q$. Therefore, for every Borel $E\subseteq Q$, ${\int }_E{J}_{f}dA\le {\int }_E{g}_{Q}dA={\nu }_{Q,a}(E)\le {\nu }_Q(E)=\mathrm{\mid }f(E)\mathrm{\mid }\mathrm{.}$

Enumerate the countable rational boxes $Q_j⋐\mathrm{\Omega }$ covering $\mathrm{\Omega }$, and for a Borel $E\subseteq \mathrm{\Omega }$ set $E_j:=E\cap (Q_j\setminus {\bigcup }_{i<j}{Q}_i)\mathrm{.}$ The Borel sets $E_j$ are disjoint and each lies in $Q_j$. Step 3.1 gives ${\int }_{E_j}{J}_f\le \mathrm{\mid }f(E_j)\mathrm{\mid }$; the sets $f(E_j)$ are pairwise disjoint Borel sets because $f$ is injective. Countable additivity yields ${\int }_E{J}_{f}dA={\sum }_j{\int }_{E_j}{J}_{f}dA\le {\sum }_j\mathrm{\mid }f(E_j)\mathrm{\mid }=\mathrm{\mid }f(E)\mathrm{\mid }\mathrm{.}$

Let $E\subseteq \mathrm{\Omega }$ be Lebesgue measurable. By [F10], write $E=F\cup N$ where $F\subseteq E$ is Borel and $N\subseteq Z$ for a Borel null set $Z$. Since $J_f\in L_{loc}^1$ and $E\setminus F$ is null, ${\int }_E{J}_f={\int }_F{J}_f$ as extended nonnegative integrals. Step 4.1 and [F11] now give ${\int }_E{J}_{f}dA\le \mathrm{\mid }f(F)\mathrm{\mid }\le {\lambda }_2^{\ast }(f(E))\mathrm{.}$ In particular the image-area expression is ordinary Lebesgue measure whenever $f(E)$ is measurable, and finite outer image area implies ${\int }_E{J}_f<\mathrm{\infty }$.

By step 1.1, $(1-k^2)\mathrm{\mid }{f}_z{\mathrm{\mid }}^2\le J_f$ almost everywhere, so integration and step 5.1 give the $f_z$ bound. If $k>0$, the inequality $\mathrm{\mid }{f}_{\bar{z}}{\mathrm{\mid }}^2\le k^2\mathrm{\mid }{f}_z{\mathrm{\mid }}^2\le \frac{k^2}{1-k^2}{J}_f$ gives the other bound by integration. If $k=0$, [F1] gives $f_{\bar{z}}=0$ almost everywhere and hence its squared integral is zero for every $E$, without multiplying infinite image area by zero.

The identity in [F2] and the pointwise estimates of step 6.1 give $\mathrm{\mid }Df{\mathrm{\mid }}_{HS}^2=2(\mathrm{\mid }{f}_z{\mathrm{\mid }}^2+\mathrm{\mid }{f}_{\bar{z}}{\mathrm{\mid }}^2)\le \frac{2(1+k^2)}{1-k^2}{J}_f\mathrm{.}$ For the normalized sphere map and bounded $B$, [F6] makes $f(B)$ measurable and finite-area; integrating this inequality and using step 5.1 proves the normalized-family bound with the displayed factor. For a Lebesgue-null set, choose a Borel null superset and apply [F5] to that superset; its image is Borel and null, so every subset is Lebesgue-null by completeness [F10]. This proves the retained Lusin-N assertion separately. The area and derivative estimates above prove all remaining claims of the Statement.