Measurable-coefficient Beltrami CGO solutions and principal phases

We first establish the factorization needed below. If a continuous $G\in W_{\mathrm{l}\mathrm{o}\mathrm{c}}^{1,2}(\mathbb{C})$ satisfies $\bar{\mathrm{\partial }}G=\nu \mathrm{\partial }G$ with $\nu$ compactly supported and $\mathrm{\mid }\nu \mathrm{\mid }\le \kappa$, let $\mathrm{\Psi }$ be the principal homeomorphic solution for $\nu$ from [F2]. Then $H=G\circ {\mathrm{\Psi }}^{-1}$ is locally $H^1$ and holomorphic. Here are the Sobolev details: on compact sets, mollifications $G_n$ converge uniformly to $G$ and in $H^1$. The classical composition chain rule with the Sobolev inverse, first by smooth approximation of that inverse, gives $D(G_n\circ {\mathrm{\Psi }}^{-1})=(D{G}_n\circ {\mathrm{\Psi }}^{-1})D{\mathrm{\Psi }}^{-1}$. The area formula and inverse distortion bound give $\int \mathrm{\mid }D((G_n-G_m)\circ {\mathrm{\Psi }}^{-1}){\mathrm{\mid }}^2\le K\int \mathrm{\mid }D(G_n-G_m){\mathrm{\mid }}^2$ over corresponding compact neighborhoods. Thus these compositions converge locally in $H^1$, with the chain rule in the limit. At common differentiability points, the two Beltrami equations and the inverse derivative identity cancel the $\bar{\mathrm{\partial }}$ derivative of $H$. Null-set preservation makes these identities valid on a full-measure target set. Hence $\bar{\mathrm{\partial }}H=0$ distributionally; apply $\mathrm{\partial }$ and [F4] to make $H$ smooth harmonic, and then its Cauchy–Riemann equation gives holomorphy.

Suppose a bounded continuous $F\in W_{\mathrm{l}\mathrm{o}\mathrm{c}}^{1,r}$ satisfies $\mathrm{\mid }\bar{\mathrm{\partial }}F\mathrm{\mid }\le \kappa {\mathbf{1}}_{D_1}\mathrm{\mid }\mathrm{\partial }F\mathrm{\mid }+\gamma \mathrm{\mid }F\mathrm{\mid }$, with $\gamma \in L^r$ supported in $D_1$. It can be written $\bar{\mathrm{\partial }}F=\nu \mathrm{\partial }F+AF$ with $\nu ,A$ supported in $D_1$, $\mathrm{\mid }\nu \mathrm{\mid }\le \kappa$, and $\mathrm{\mid }A\mathrm{\mid }\le \gamma$. To construct them measurably, if $\mathrm{\mid }\bar{\mathrm{\partial }}F\mathrm{\mid }\le \kappa \mathrm{\mid }\mathrm{\partial }F\mathrm{\mid }$, take $\nu =\bar{\mathrm{\partial }}F\mathrm{/}\mathrm{\partial }F$ and $A=0$, using zero at zero denominators. Otherwise choose $\nu \mathrm{\partial }F$ of length $\kappa \mathrm{\mid }\mathrm{\partial }F\mathrm{\mid }$ in the direction of $\bar{\mathrm{\partial }}F$, and put $A=(\bar{\mathrm{\partial }}F-\nu \mathrm{\partial }F)\mathrm{/}F$; in this branch $F\mathrm{\ne }0$ wherever the numerator is nonzero, by the assumed inequality. Set both coefficients to zero outside $D_1$. Let $h=(I-\nu \mathcal{B}{)}^{-1}A$ and $\eta =\mathcal{C}h$. The equation for $h$ implies its support lies in ${\bar{D}}_1$. By [F1], $\bar{\mathrm{\partial }}\eta -\nu \mathrm{\partial }\eta =A$ and $\mathrm{\parallel }\eta {\mathrm{\parallel }}_{\mathrm{\infty }}+\mathrm{\parallel }\eta {\mathrm{\parallel }}_{W^{1,r}}\le C_{\kappa ,r}\mathrm{\parallel }\gamma {\mathrm{\parallel }}_r$, with $\eta =O(1\mathrm{/}z)$. The product rule gives $\bar{\mathrm{\partial }}(e^{-\eta }F)=\nu \mathrm{\partial }(e^{-\eta }F)$. This bounded continuous function factors as an entire bounded holomorphic function after a principal change of variables by step 1.1, so it is constant. Thus $F=c{e}^{\eta }$; if $F\to c_{\mathrm{\infty }}$ at infinity, then $c=c_{\mathrm{\infty }}$. In particular a solution of this inequality tending to zero is identically zero.

Put $e_{-k}(z)=e^{-i(kz+\bar{k}\bar{z})}$, $\nu =e_{-k}\mu$ and $\alpha =-i\bar{k}{e}_{-k}\mu$. The ansatz $f=e^{ikz}(1+\eta )$ gives exactly $\bar{\mathrm{\partial }}\eta -\nu \overline{\mathrm{\partial }\eta }=\alpha \bar{\eta }+\alpha$. Define the real-linear operator $Kg=\mathcal{C}(I-\nu \bar{\mathcal{B}}{)}^{-1}(\alpha \bar{g})$ on $L^r(\mathbb{C};\mathbb{C})$, regarded as a real Banach space. The data to its inverse have support in ${\bar{D}}_1$, with norm at most $C_{\kappa ,r}\mathrm{\mid }k\mathrm{\mid }\mathrm{\parallel }g{\mathrm{\parallel }}_r$. Its inverse output has the same support, by its defining equation. The Cauchy map on such data is compact into $L^r$: on the unit ball of data, [F1] gives uniform boundedness and a common Hölder modulus, and gives a uniform $C\mathrm{/}\mathrm{\mid }z\mathrm{\mid }$ bound outside $D_2$. To obtain a finite $ϵ$-net, first make the $L^r$ norm of this exterior tail small by taking a large disk, then partition that disk into sufficiently small squares and quantize the uniformly bounded complex values at their centers to a finite mesh; the Hölder modulus bounds the $L^r$ approximation error inside. Completeness and [F3] turn total boundedness into relative compactness. Thus $K$ is compact.

If $g=Kg$, then $g\in W^{1,r}$ is continuous and bounded by [F1], is $O(1\mathrm{/}z)$ at infinity, and satisfies $\bar{\mathrm{\partial }}g-\nu \overline{\mathrm{\partial }g}=\alpha \bar{g}$. Therefore $\mathrm{\mid }\bar{\mathrm{\partial }}g\mathrm{\mid }\le \kappa \mathrm{\mid }\mathrm{\partial }g\mathrm{\mid }+\mathrm{\mid }\alpha \mathrm{\mid }\mathrm{\mid }g\mathrm{\mid }$, so step 2.1 implies $g=0$. By [F3], $I-K$ is invertible. The equation $\eta -K\eta =\mathcal{C}(I-\nu \bar{\mathcal{B}}{)}^{-1}\alpha =K{\mathbf{1}}_{D_1}$ has a unique $L^r$ solution. The equation itself and [F1] put $\eta \in W^{1,r}$, continuous with $O(1\mathrm{/}z)$ decay. Reversing the ansatz calculation proves existence of $f_{\mu }$ and the asserted normalization. For any other solution, density of compact smooth functions in global $W^{1,r}$ gives $\mathrm{\partial }\eta =\mathcal{B}\bar{\mathrm{\partial }}\eta$; applying the inverse in [F1] and then $\mathcal{C}$ recovers the same integral equation, since its holomorphic residual vanishes at infinity. It therefore gives the same $\eta$. When $k=0$, $\alpha =0$, so $\eta =0$ and $f_{\mu }=1$.

Write $M=1+\eta$ from step 4.1. Its equation is $\bar{\mathrm{\partial }}M=\nu \overline{\mathrm{\partial }M}+\alpha \bar{M}$, so $\mathrm{\mid }\bar{\mathrm{\partial }}M\mathrm{\mid }\le \kappa \mathrm{\mid }\mathrm{\partial }M\mathrm{\mid }+\mathrm{\mid }\alpha \mathrm{\mid }\mathrm{\mid }M\mathrm{\mid }$. Step 2.1 gives $M=e^a$, where $a=\mathcal{C}h$ for data $h$ supported in ${\bar{D}}_1$ with $\mathrm{\parallel }h{\mathrm{\parallel }}_r\le C_{\kappa ,r}\mathrm{\parallel }\alpha {\mathrm{\parallel }}_r\le C_{\kappa ,r}\mathrm{\mid }k\mathrm{\mid }$. Thus $M$ never vanishes and $\mathrm{\parallel }a{\mathrm{\parallel }}_{\mathrm{\infty }}+\mathrm{\parallel }a{\mathrm{\parallel }}_{W^{1,r}}\le C_{\kappa ,r}\mathrm{\mid }k\mathrm{\mid }$. The bounds $\mathrm{\mid }{e}^a-1\mathrm{\mid }\le e^{\mathrm{\parallel }a{\mathrm{\parallel }}_{\mathrm{\infty }}}\mathrm{\mid }a\mathrm{\mid }$ and $\mathrm{\mid }\mathrm{\nabla }{e}^a\mathrm{\mid }\le e^{\mathrm{\parallel }a{\mathrm{\parallel }}_{\mathrm{\infty }}}\mathrm{\mid }\mathrm{\nabla }a\mathrm{\mid }$ prove the quantitative bound on $M-1$. The same Cauchy integral gives $\mathrm{\mid }a(z)\mathrm{\mid }\le C_{\kappa ,r}\mathrm{\mid }k\mathrm{\mid }\mathrm{/}\mathrm{\mid }z\mathrm{\mid }$ for $\mathrm{\mid }z\mathrm{\mid }>2$.

If $k\mathrm{\ne }0$, set $\mathrm{\Phi }=z+a\mathrm{/}(ik)$. Then $f=e^{ik\mathrm{\Phi }}$, $\mathrm{\Phi }=z+O(1\mathrm{/}z)$ and the asserted uniform deformation estimates follow from step 5.1. Since the exponential has no zeros, the chain rule in the Beltrami equation for $f$ gives $\mathrm{\mid }\bar{\mathrm{\partial }}\mathrm{\Phi }\mathrm{\mid }\le \kappa \mathrm{\mid }\mathrm{\partial }\mathrm{\Phi }\mathrm{\mid }$, with $\bar{\mathrm{\partial }}\mathrm{\Phi }=0$ outside ${\bar{D}}_1$. Define its complex coefficient by $\bar{\mathrm{\partial }}\mathrm{\Phi }\mathrm{/}\mathrm{\partial }\mathrm{\Phi }$, using zero where both vanish, and take the principal homeomorphic solution $\mathrm{\Psi }$ for that coefficient from [F2]. Step 1.1 gives $\mathrm{\Phi }=H\circ \mathrm{\Psi }$ with $H$ entire holomorphic. Both maps have the principal expansion, so $H(w)-w\to 0$ at infinity. Liouville in [F4] gives $H(w)=w$, hence $\mathrm{\Phi }=\mathrm{\Psi }$ is a principal quasiconformal homeomorphism. Multiplication and composition with the exponential on $D_S$ now give the stated solution $W^{1,r}$ bound, while for $\mathrm{\mid }k\mathrm{\mid }=1$ the last clause of [F2] gives the inverse-gradient estimate. This completes all claims.