The Beltrami Equation and Measurable Riemann Mapping

Fix $U_0⋐U$. Its closure is compact, so continuity gives a compact set $K:=u(\overline{U_0})\subset V$. By [F2] choose $\chi \in C_c^{\mathrm{\infty }}(V)$ equal to $1$ on a neighborhood of $K$. Define $\tilde{G}=\chi G$ on $V$ and extend it by $0$ to $\mathbb{C}\setminus V$. Since $\chi$ has compact support in $V$, $\tilde{G}$ is globally $C^1$, bounded, and has bounded derivative, and it agrees with $G$ on a neighborhood of $K$.

The restriction $u{\mathrm{\mid }}_{U_0}$ belongs to $W^{1,2}(U_0;\mathbb{C})$. By [F1] choose $u_j\in C^{\mathrm{\infty }}(U_0;\mathbb{C})$ converging to it in $W^{1,2}(U_0)$. Each $\tilde{G}\circ u_j$ is $C^1$ and its classical derivative is $D\tilde{G}(u_j)D{u}_j$ by [F7]; by [F5] this is also its weak derivative.

Since $u_j\to u$ in $L^2(U_0)$, [F3] gives a subsequence, still denoted $u_j$, converging to $u$ almost everywhere. Thus $\tilde{G}(u_j)\to \tilde{G}(u)$ almost everywhere. The functions are uniformly bounded by $\mathrm{\parallel }\tilde{G}{\mathrm{\parallel }}_{\mathrm{\infty }}$, and $U_0$ has finite measure by [F8]; dominated convergence gives $\tilde{G}(u_j)\to \tilde{G}(u)$ in $L^2(U_0)$.

Continuity of $D\tilde{G}$ gives $D\tilde{G}(u_j)\to D\tilde{G}(u)$ almost everywhere along the subsequence of step 2.1, and these matrices are bounded by $M:=\mathrm{\parallel }D\tilde{G}{\mathrm{\parallel }}_{\mathrm{\infty }}$. Write $D\tilde{G}(u_j)D{u}_j-D\tilde{G}(u)Du=D\tilde{G}(u_j)(D{u}_j-Du)+(D\tilde{G}(u_j)-D\tilde{G}(u))Du\mathrm{.}$ The first term tends to $0$ in $L^2$ because $D{u}_j\to Du$ in $L^2$ and the matrices have norm at most $M$. The second tends to $0$ in $L^2$ by [F4], since it converges almost everywhere and its squared norm is bounded by $4M^2\mathrm{\mid }Du{\mathrm{\mid }}^2\in L^1(U_0)$. Therefore the weak derivatives of $\tilde{G}\circ u_j$ converge in $L^2(U_0)$ to $D\tilde{G}(u)Du$.

Apply [F6] to the function convergence in step 2.1 and the derivative convergence in step 3.1. It gives $\tilde{G}\circ u\in W^{1,2}(U_0)$ with weak derivative $D\tilde{G}(u)Du$. Since $u(U_0)\subseteq K$ and $\tilde{G}=G$ on a neighborhood of $K$, this is $G\circ u$ with derivative $DG(u)Du$. As $U_0⋐U$ was arbitrary, the asserted local Sobolev membership and chain rule hold on $U$. The empty-domain case is vacuous.

On each relatively compact $K⋐\mathrm{\Omega }$, $\mathrm{\parallel }\mu f_z{\mathrm{\parallel }}_{L^2(K)}\le \mathrm{\parallel }\mu {\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }{f}_z{\mathrm{\parallel }}_{L^2(K)}$, so $h:=f_{\bar{z}}-\mu f_z\in L^2(K)\subseteq L^1(K)$ by [F5]. If $h=0$ almost everywhere, its regular distribution is zero; conversely, if its regular distribution is zero, [F4] gives $h=0$ almost everywhere. Thus the almost-everywhere and distributional equations in (b) are equivalent. Applying the signed weak-derivative identity to the real partials and combining them as in [F8] gives $⟨f_{\bar{z}},\eta ⟩=-\int f{\eta }_{\bar{z}}dA$, which yields the test-function form in (b).

Replacing $f$, $\mu$, or either weak derivative by an almost-everywhere equal representative changes the equation only on the finite union of the corresponding null sets. The weak derivative classes are representative-independent by [F2], and the coefficient class is representative-independent by [F1]. Therefore the plane weak-solution condition is well-defined on these classes.

Let $\psi :{\mathrm{\Omega }}^{\mathrm{\prime }}\to \mathrm{\Omega }$ be biholomorphic. For each relatively compact $U_0⋐{\mathrm{\Omega }}^{\mathrm{\prime }}$, choose $V_0⋐\mathrm{\Omega }$ containing $\psi ({\bar{U}}_0)$; the derivatives of $\psi$ and ${\psi }^{-1}$ are bounded on these compact patches. Applying [F7] with $k=1,p=2$ and using the real chain rule [F10], then rewriting the real derivative matrix by [F8], gives the displayed weak chain-rule formulas on $U_0$. Since ${\psi }^{-1}$ maps null sets to null sets by [F6], the almost-everywhere equation for $f$ remains valid after composition.

Substitute $f_{\bar{z}}=\mu f_z$ into the second identity of step 1.3 and use the pullback formula from [F1]: $(f\circ \psi {)}_{\bar{\zeta }}=(\mu \circ \psi )(f_z\circ \psi )\overline{{\psi }^{\mathrm{\prime }}}=({\psi }^{\ast }\mu )(f_z\circ \psi ){\psi }^{\mathrm{\prime }}=({\psi }^{\ast }\mu )(f\circ \psi {)}_{\zeta }$ almost everywhere on $U_0$. The patches cover ${\mathrm{\Omega }}^{\mathrm{\prime }}$, so $f\circ \psi$ is a weak solution for ${\psi }^{\ast }\mu$. Applying the same argument to ${\psi }^{-1}$ proves the converse.

For the sphere clause, continuity of $f$ ensures that near any source point its image lies in a target chart, so the local coordinate maps in (d) are defined on open plane domains. The chart transitions are biholomorphic by [F9] and the sphere atlas is given by [F11]. On overlaps, source-chart changes preserve the equation by steps 1.3 and 2.1. A target-chart change is a local biholomorphism $\tau$; after shrinking the source neighborhood so its compact image lies in the overlap, [F12] gives $(\tau \circ F{)}_z=({\tau }^{\mathrm{\prime }}\circ F)F_z$ and $(\tau \circ F{)}_{\bar{z}}=({\tau }^{\mathrm{\prime }}\circ F)F_{\bar{z}}$. Multiplication by ${\tau }^{\mathrm{\prime }}\circ F$ proves preservation without division. Thus the local definition is independent of both chart choices and agrees with (a) in the finite chart.

For $z\mathrm{\ne }0$, direct differentiation of $\mathrm{\Gamma }$ gives $a(z):={\mathrm{\partial }}_z\mathrm{\Gamma }(z)=-1\mathrm{/}(4\pi z)$; polar coordinates give ${\int }_{D_r}\mathrm{\mid }a\mathrm{\mid }dA=r\mathrm{/}2$. Integrating by parts outside $D_{\epsilon }$ against a compactly supported smooth test function leaves an inner boundary term bounded by $C\epsilon \mathrm{\mid }\log \epsilon \mathrm{\mid }$, which tends to zero, so the distributional derivative of $\mathrm{\Gamma }$ is the regular distribution of $a$.

Since $Nq\in C^2$ and $-\mathrm{\Delta }Nq=q$, the identity $\mathrm{\Delta }=4{\mathrm{\partial }}_z{\mathrm{\partial }}_{\bar{z}}$ gives ${\mathrm{\partial }}_{\bar{z}}Tq=-4{\mathrm{\partial }}_{\bar{z}}{\mathrm{\partial }}_{z}Nq=q$ pointwise. On every compact subset of $\mathbb{C}\setminus {\bar{D}}_2$, the Newtonian kernel $\mathrm{\Gamma }(z-\zeta )$ and all its $z$-derivatives are bounded uniformly for $\zeta \in {\bar{D}}_2$; differentiation under the integral sign in $Nq$ therefore makes $Nq$, and hence $Tq$, smooth there.

By Fubini and integration by parts in the compactly supported $q$ variable, for every multi-index $\beta$ with $\mathrm{\mid }\beta \mathrm{\mid }\le k$ the distributional identity $D^{\beta }Nq=N(D^{\beta }q)$ holds. The right side is $C_{\mathrm{l}\mathrm{o}\mathrm{c}}^{2,\alpha }$ by [F4]. Starting with $Nq\in C^2$, induction on $\mathrm{\mid }\beta \mathrm{\mid }$ identifies each already-classical derivative $D^{\beta }Nq$ with this continuous representative: distributional injectivity gives equality almost everywhere, and [F8] rules out a nonzero continuous difference on any ball. Each such derivative is then $C^2$. The local estimate in [F4], applied to each $D^{\beta }q$ with support in ${\bar{D}}_2$, yields $Nq\in C_{\mathrm{l}\mathrm{o}\mathrm{c}}^{k+2,\alpha }$ with its norm on ${\bar{D}}_R$ bounded by $C_{k,\alpha ,R}\mathrm{\parallel }q{\mathrm{\parallel }}_{C^{k,\alpha }}$. Since $Tq=-4{\mathrm{\partial }}_{z}Nq$, this gives the asserted $C^{k+1,\alpha }(D_R)$ bound.

The cancellation formula [F5], combined as $Sq=-({\mathrm{\partial }}_{xx}-2i{\mathrm{\partial }}_{xy}-{\mathrm{\partial }}_{yy})Nq$, cancels the two diagonal correction terms. Away from zero the resulting kernel is $-4{\mathrm{\partial }}_z^2\mathrm{\Gamma }(w)=-1\mathrm{/}(\pi w^2)$, so $Sq(z)=-(1\mathrm{/}\pi )\mathrm{p.v.}\int q(\zeta )\mathrm{/}(z-\zeta {)}^{2}dA(\zeta )$. The integral of $(z-\zeta {)}^{-2}$ over every centered annulus is zero because its angular factor is $e^{-2i\theta }$; hence subtracting $q(z)$ only on $\mathrm{\mid }z-\zeta \mathrm{\mid }<1$ gives the displayed subtracted formula. Its near integral is bounded absolutely by $C[q{]}_{0,\alpha }{\int }_0^1{r}^{\alpha -1}dr$, and the far integral is absolutely finite because it avoids the singularity and $q$ has compact support.

For every compactly supported smooth test function $\phi$, Fubini and the distributional derivative identity in step 1.1 give $⟨{\mathrm{\partial }}_{z}Nq,\phi ⟩=⟨a\ast q,\phi ⟩$, with $(a\ast q)(z)={\int }_{\mathbb{C}}a(z-\zeta )q(\zeta )dA(\zeta )\mathrm{.}$ This integral is absolutely finite for each $z$, since $q$ is bounded, supported in ${\bar{D}}_2$, and $a$ is locally integrable. It is continuous: on a compact set of $z$-values let $\delta =\mathrm{\mid }z-z^{\mathrm{\prime }}\mathrm{\mid }<1$; the two disks of radius $2\delta$ around $z,z^{\mathrm{\prime }}$ contribute at most $C\mathrm{\parallel }q{\mathrm{\parallel }}_{\mathrm{\infty }}\delta$, while on their complement $\mathrm{\mid }\mathrm{\nabla }a(w)\mathrm{\mid }\le C\mathrm{\mid }w{\mathrm{\mid }}^{-2}$ and the mean value theorem bounds the difference by $C\mathrm{\parallel }q{\mathrm{\parallel }}_{\mathrm{\infty }}\delta \log (C_0\mathrm{/}\delta )$ for a fixed $C_0$. Since $Nq\in C^2$, both sides are continuous; [F6] makes them equal almost everywhere, and [F8] then makes them equal everywhere. Therefore $Tq=-4{\mathrm{\partial }}_{z}Nq=(1\mathrm{/}\pi )\int q(\zeta )\mathrm{/}(z-\zeta )dA(\zeta )$.

For the global Hölder seminorm when $k=0$, put $K(w)=-1\mathrm{/}(\pi w^2)$ and $k_{ij}(w)={\mathrm{\partial }}_i{\mathrm{\partial }}_j\mathrm{\Gamma }(w)$. Let $\mathrm{\Omega }$ be a disk with $\mathrm{supp}q⋐\mathrm{\Omega }$ and $x\in \mathrm{\Omega }$, and choose a larger disk $D_s(x)\supset \bar{\mathrm{\Omega }}$. On the outer circle the explicit derivative ${\mathrm{\partial }}_i\mathrm{\Gamma }(w)=-w_i\mathrm{/}(2\pi \mathrm{\mid }w{\mathrm{\mid }}^2)$ and polar symmetry give ${\int }_{\mathrm{\partial }{D}_s(x)}{\mathrm{\partial }}_i\mathrm{\Gamma }(x-y){\nu }_j(y)dS(y)={\delta }_{ij}\mathrm{/}2$. Also $k_{ij}(x-y)=-{\mathrm{\partial }}_{y_j}{\mathrm{\partial }}_i\mathrm{\Gamma }(x-y)$, so the divergence theorem gives ${\int }_{D_s(x)\setminus \mathrm{\Omega }}{k}_{ij}(x-y)dA(y)=-{\delta }_{ij}\mathrm{/}2+g_{ij,\mathrm{\Omega }}(x)$, where $g_{ij,\mathrm{\Omega }}(x):={\int }_{\mathrm{\partial }\mathrm{\Omega }}{\mathrm{\partial }}_i\mathrm{\Gamma }(x-y){\nu }_j(y)dS(y)\mathrm{.}$ Split the centered-disk cancellation formula [F5] into $\mathrm{\Omega }$ and $D_s(x)\setminus \mathrm{\Omega }$; on the latter $q(y)=0$, so the ${\delta }_{ij}\mathrm{/}2$ terms cancel and ${\mathrm{\partial }}_i{\mathrm{\partial }}_{j}Nq(x)={\int }_{\mathrm{\Omega }}{k}_{ij}(x-y)(q(y)-q(x))dA(y)-q(x)g_{ij,\mathrm{\Omega }}(x)\mathrm{.}$ Taking the linear combination $-{\mathrm{\partial }}_{xx}+2i{\mathrm{\partial }}_{xy}+{\mathrm{\partial }}_{yy}$ gives the corresponding formula for $Sq$ with kernel $K$ and boundary factor $G_{\mathrm{\Omega }}=-g_{xx,\mathrm{\Omega }}+2i{g}_{xy,\mathrm{\Omega }}+g_{yy,\mathrm{\Omega }}$.

Fix distinct $x,x^{\mathrm{\prime }}$, put $\delta =\mathrm{\mid }x-x^{\mathrm{\prime }}\mathrm{\mid }$ and $m=(x+x^{\mathrm{\prime }})\mathrm{/}2$, and take $\mathrm{\Omega }=D_R(m)$ with $\mathrm{supp}q⋐\mathrm{\Omega }$ and $R\ge 2\delta$. Reflection through $m$ sends $x$ to $x^{\mathrm{\prime }}$, reverses both ${\mathrm{\partial }}_i\mathrm{\Gamma }(x-y)$ and the normal ${\nu }_j(y)$, and preserves arc length, so $G_{\mathrm{\Omega }}(x)=G_{\mathrm{\Omega }}(x^{\mathrm{\prime }})$. Also $\mathrm{\mid }{G}_{\mathrm{\Omega }}(x)\mathrm{\mid }\le C$, since $\mathrm{\mid }x-y\mathrm{\mid }\ge 3R\mathrm{/}4$ on $\mathrm{\partial }\mathrm{\Omega }$, $\mathrm{\mid }\mathrm{\nabla }\mathrm{\Gamma }(w)\mathrm{\mid }\le C\mathrm{/}\mathrm{\mid }w\mathrm{\mid }$, and $\mathrm{\partial }\mathrm{\Omega }$ has length $2\pi R$. Thus the boundary-term difference is at most $C[q{]}_{0,\alpha }{\delta }^{\alpha }$.

Split the integral difference over $D_{\delta }(m)$ and $\mathrm{\Omega }\setminus {\bar{D}}_{\delta }(m)$. On the inner disk, $\mathrm{\mid }K(x-y)(q(y)-q(x))\mathrm{\mid }\le C[q{]}_{0,\alpha }\mathrm{\mid }x-y{\mathrm{\mid }}^{\alpha -2}$ and likewise for $x^{\mathrm{\prime }}$, so polar integration bounds both contributions by $C_{\alpha }[q{]}_{0,\alpha }{\delta }^{\alpha }$. On the outer region, write the difference integrand as $[K(x-y)-K(x^{\mathrm{\prime }}-y)](q(y)-q(x))-K(x^{\mathrm{\prime }}-y)(q(x)-q(x^{\mathrm{\prime }}))\mathrm{.}$ With $\rho =\mathrm{\mid }y-m\mathrm{\mid }\ge \delta$, the segment between $x-y$ and $x^{\mathrm{\prime }}-y$ stays at distance at least $\rho \mathrm{/}2$ from zero; $\mathrm{\mid }\mathrm{\nabla }K(w)\mathrm{\mid }\le C\mathrm{\mid }w{\mathrm{\mid }}^{-3}$ and the mean value theorem bound the first term by $C[q{]}_{0,\alpha }\delta {\rho }^{\alpha -3}$. Its area integral is at most $C[q{]}_{0,\alpha }\delta {\int }_{\delta }^R{\rho }^{\alpha -2}d\rho \le C_{\alpha }[q{]}_{0,\alpha }{\delta }^{\alpha }$, using $\alpha <1$. For the second term, $\mathrm{\mid }q(x)-q(x^{\mathrm{\prime }})\mathrm{\mid }\le [q{]}_{0,\alpha }{\delta }^{\alpha }$ and each real component ${\int }_{\mathrm{\Omega }\setminus {\bar{D}}_{\delta }(m)}{k}_{ij}(x^{\mathrm{\prime }}-y)dA(y)$ is bounded by the divergence theorem: its boundary fluxes are ${\mathrm{\partial }}_i\mathrm{\Gamma }$ on the outer circle and inner circle, each bounded by $C$ using $\mathrm{\mid }{x}^{\mathrm{\prime }}-y\mathrm{\mid }\ge 3R\mathrm{/}4$ on the outer circle and $\mathrm{\mid }{x}^{\mathrm{\prime }}-y\mathrm{\mid }\ge \delta \mathrm{/}2$ on the inner one. This proves $[Sq{]}_{0,\alpha ;{\mathbb{R}}^2}\le C_{\alpha }[q{]}_{0,\alpha ;{\mathbb{R}}^2}$.

On ${\bar{D}}_4$, the local Hessian estimate [F4] bounds $\mathrm{\mid }Sq\mathrm{\mid }$ by $C_{\alpha }\mathrm{\parallel }q{\mathrm{\parallel }}_{C^{0,\alpha }}$. For $\mathrm{\mid }z\mathrm{\mid }\ge 3$, the integral formula gives $\mathrm{\mid }Sq(z)\mathrm{\mid }\le \frac{\mathrm{\parallel }q{\mathrm{\parallel }}_{\mathrm{\infty }}}{\pi }{\int }_{{\bar{D}}_2}\mathrm{\mid }z-\zeta {\mathrm{\mid }}^{-2}dA(\zeta )\le 4\mathrm{\parallel }q{\mathrm{\parallel }}_{\mathrm{\infty }},$ since $\mathrm{\mid }z-\zeta \mathrm{\mid }\ge 1$. Hence $\mathrm{\parallel }Sq{\mathrm{\parallel }}_{C^{0,\alpha }({\mathbb{R}}^2)}\le C_{\alpha }\mathrm{\parallel }q{\mathrm{\parallel }}_{C^{0,\alpha }({\mathbb{R}}^2)}$.

For $k\ge 1$, the regularity in step 1.3 makes $D^{\beta }Sq=S(D^{\beta }q)$ classically for every $\mathrm{\mid }\beta \mathrm{\mid }\le k$: expand $S$ as a linear combination of second derivatives of $Nq$ and commute continuous mixed derivatives using [F6]. Each $D^{\beta }q$ is supported in ${\bar{D}}_2$ and has $C^{0,\alpha }$ norm at most $C_k\mathrm{\parallel }q{\mathrm{\parallel }}_{C^{k,\alpha }}$: at top order this is part of the norm; below top order, the mean-value bound controls pairs at distance at most1 by the next derivatives, while twice the supremum controls pairs farther apart. Applying the seminorm and supremum bounds of steps 3.2 and 4.1 to these finitely many derivatives proves $Sq\in C^{k,\alpha }({\mathbb{R}}^2)$ and the stated constant $M_{k,\alpha }$. The zero datum is included, and all estimates use the strict range $0<\alpha <1$; no endpoint $\alpha =1$ or global $L^p$ bound is claimed.