Cauchy–Riemann similarity and unique continuation under countable choice

Define $K(z)=1\mathrm{/}(\pi z)$ as a locally integrable kernel. Its distributional derivative is ${\mathrm{\partial }}_{\bar{z}}K={\delta }_0$. Here is a direct sign and normalization check, requiring no stronger-choice Stokes theorem. For a compactly supported smooth test $\psi$, use $z=r{e}^{i\theta }$ and ${\mathrm{\partial }}_{\bar{z}}=e^{i\theta }({\mathrm{\partial }}_r+i{r}^{-1}{\mathrm{\partial }}_{\theta })\mathrm{/}2$. Then $-\int K{\mathrm{\partial }}_{\bar{z}}\psi dA=-\frac{1}{2\pi }{\int }_0^{2\pi }{\int }_0^{\mathrm{\infty }}({\psi }_r+i{r}^{-1}{\psi }_{\theta })drd\theta =\psi (0)\mathrm{.}$ The angular term integrates to zero by periodicity; it is integrable near zero because ${\psi }_{\theta }=O(r)$. For bounded $f$ supported in a disk of radius $r$, set $Tf=K\ast f$. On that disk, $\mathrm{\parallel }Tf{\mathrm{\parallel }}_{\mathrm{\infty }}\le 4r\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$ by enclosing the translated integration region in the radius-$2r$ disk and integrating $1\mathrm{/}(\pi \mathrm{\mid }z\mathrm{\mid })$. This map is continuous even for measurable $f$: cut the kernel inside a small radius, whose integral norm is arbitrarily small, and use uniform continuity of the remaining nonsingular translated kernel on bounded sets. Distributional convolution gives ${\mathrm{\partial }}_{\bar{z}}Tf=f$.

We prove the actual Sobolev regularity of this transform. For smooth compactly supported $f$, $H_j={\mathrm{\partial }}_{j}Tf=K\ast ({\mathrm{\partial }}_{j}f)$ is continuous, bounded on compact sets, and $O(\mathrm{\mid }z{\mathrm{\mid }}^{-2})$ at infinity. The last bound follows by subtracting $K(z)\int {\mathrm{\partial }}_{j}f=0$ and the estimate $\mathrm{\mid }K(z-\zeta )-K(z)\mathrm{\mid }\le C\mathrm{\mid }\zeta \mathrm{\mid }\mathrm{/}\mathrm{\mid }z{\mathrm{\mid }}^2$ outside twice the support radius. Thus $H_j\in L^2$. The distributional identity $D{H}_j=2{\mathrm{\partial }}_{j}f$, Fourier differentiation and [F1] give ${\widehat{H}}_j(\xi )=\frac{2i{\xi }_j}{i{\xi }_s-{\xi }_t}\widehat{f}(\xi )\text{a.e.}$ with Fourier convention $\widehat{{\mathrm{\partial }}_j}=2\pi i{\xi }_j$. This follows first against Schwartz tests by integration by parts and then in $L^2$; the denominator is nonzero except at the measure-zero origin, so there is no unsupported point-mass term in this $L^2$ identity. Each displayed symbol is smooth homogeneous of degree zero with its first two derivatives bounded on the unit circle. Therefore [F1] gives $\mathrm{\parallel }\mathrm{\nabla }Tf{\mathrm{\parallel }}_q\le C_q\mathrm{\parallel }f{\mathrm{\parallel }}_q$ for every $1<q<\mathrm{\infty }$. For bounded compactly supported $f$, use mollified compact approximants from [F2], converging simultaneously in $L^2$ and any fixed $L^q$, $q>2$. Their transforms converge uniformly on compact sets: Hölder bounds the kernel there in $L^{q^{\mathrm{\prime }}}$, because $q^{\mathrm{\prime }}<2$. Their derivatives converge in $L^q$ by the multiplier estimate. Hence $Tf\in W_{\mathrm{l}\mathrm{o}\mathrm{c}}^{1,q}$ with the same weak derivative formulas. This holds for every finite $q>2$ and for matrix entries as well.

Work on a small closed disk about $z_0$ where $\mathrm{\parallel }B{\mathrm{\parallel }}_{\mathrm{\infty }}$ is bounded. Define a measurable complex-linear matrix $C(z)=-\frac{(B(z)v(z))v(z{)}^{\ast }}{2\mathrm{\mid }v(z){\mathrm{\mid }}^2}$ when $v(z)\mathrm{\ne }0$, and zero otherwise. It satisfies $Cv=-Bv\mathrm{/}2$ everywhere almost surely and $\mathrm{\parallel }C\mathrm{\parallel }\le \mathrm{\parallel }B\mathrm{\parallel }\mathrm{/}2$. Thus ${\mathrm{\partial }}_{\bar{z}}v=Cv$. Extend $C$ by zero off this disk. On continuous complex matrix functions on its closure use $\mathcal{T}(P)=I+T(CP)$. step 1.1 makes its contraction constant at most $4r\mathrm{\parallel }C{\mathrm{\parallel }}_{\mathrm{\infty }}$. Shrink $r$ to make this less than $1\mathrm{/}4$. Iterating from $I$ is an explicit uniformly Cauchy sequence: successive differences have geometric bounds. Its continuous limit $P$ satisfies $P=I+T(CP)$, with $\mathrm{\parallel }P-I\mathrm{\parallel }<1\mathrm{/}3$, hence is pointwise invertible by its finite-matrix Neumann series. step 2.1 proves $P\in W_{\mathrm{l}\mathrm{o}\mathrm{c}}^{1,q}$ for every finite $q>2$ and ${\mathrm{\partial }}_{\bar{z}}P=CP$. No measurable choice of matrix extensions has occurred: the formula for $C$ is specified also at the zeros.

The inverse matrix belongs to the same local Sobolev spaces. Indeed $P^{-1}={\sum }_{n\ge 0}(I-P{)}^n$ converges uniformly, and its differentiated terms have $L^q$ norms bounded by $n(1\mathrm{/}3{)}^{n-1}\mathrm{\parallel }dP{\mathrm{\parallel }}_q$ on compact subsets. Thus the series converges in $W^{1,q}$ and proves $d(P^{-1})=-P^{-1}(dP)P^{-1}$. The weak product rule follows by the mollifier approximation and the bounded continuous factors; the same argument or [F2] gives $h=P^{-1}v\in W_{\mathrm{l}\mathrm{o}\mathrm{c}}^{1,p}$. Since $P$ is complex linear, the ordinary matrix product rule for ${\mathrm{\partial }}_{\bar{z}}$ gives ${\mathrm{\partial }}_{\bar{z}}h=-P^{-1}CP{P}^{-1}v+P^{-1}Cv=0$. Thus $\mathrm{\Delta }h=4{\mathrm{\partial }}_z{\mathrm{\partial }}_{\bar{z}}h=0$ distributionally. By [F3] it is smooth, and its now classical Cauchy–Riemann equation makes it holomorphic. Replacing $P$ by $PP(z_0{)}^{-1}$ and $h$ by $P(z_0)h$ gives the stated normalization without changing the factorization.

We detail finite order without a stronger-choice complex-analysis supplier. For a smooth holomorphic scalar component $h$ on a round disk, its circle Fourier coefficients $a_n(r)$ satisfy $a_n^{\mathrm{\prime }}(r)=n{a}_n(r)\mathrm{/}r$, by its polar Cauchy–Riemann equation and integration by parts in the angle. Hence $a_n(r)=c_n{r}^n$. Boundedness near the center forces every negative coefficient to vanish. At a fixed smaller radius $R$, $\mathrm{\mid }{a}_n(R)\mathrm{\mid }\le {\sup }_{\mathrm{\mid }z\mathrm{\mid }=R}\mathrm{\mid }h\mathrm{\mid }$, so ${\sum }_{n\ge 0}{c}_n{z}^n$ converges uniformly on any still smaller disk by a geometric series. Circle Fourier completeness in [F3] identifies it with $h$ there. Apply this componentwise. Either all vector coefficients vanish, or their least nonzero index $k$ exists and $h(z_0+z)=z^k(a+O(z)),a\mathrm{\ne }0.$ The continuous invertible factor from step 4.1 and its locally bounded inverse give the two-sided norm bound stated for $v$. At a zero the index is positive. In particular zeros of a nonzero local solution are isolated, and no such solution can vanish to infinite order. The set of points near which $v$ is identically zero is open and also closed in the connected source: at a limit point of that set the local factorization has accumulating zeros, so its holomorphic factor cannot have a finite-order isolated zero and must be identically zero. This proves the global alternative and continuation assertion.

For the variable principal structure, choose fixed real coordinates making $J(z_0)=i$ and set $\mathrm{\Phi }(z)=(I-J(z)i)\mathrm{/}2$. At $z_0$ it is the identity, hence is smoothly invertible nearby, and $J\mathrm{\Phi }=(J+i)\mathrm{/}2=\mathrm{\Phi }i\mathrm{.}$ Substitution $v=\mathrm{\Phi }w$ reduces $v_s+J{v}_t+Av=0$ to $Dw+{\mathrm{\Phi }}^{-1}({\mathrm{\Phi }}_s+J{\mathrm{\Phi }}_t+A\mathrm{\Phi })w=0.$ The new zeroth-order matrix is locally bounded, so steps3.1–5.1 apply and give a real invertible continuous combined factor for $v$. Fixed complex coordinates exist by choosing a finite real complex basis; no infinite basis or extra axiom is involved.

Finally take two smooth solutions $u_1,u_2$ with equal value at an interior point and use a convex target coordinate ball near that value. Their coordinate difference $w=u_1-u_2$ satisfies $w_s+J(s,t,u_1)w_t+A(s,t)w=0,$ where $A(s,t)a=({\int }_0^1{d}_{x}J(s,t,u_2+\lambda w)[a]d\lambda )(u_2{)}_t\mathrm{.}$ This identity follows by subtracting the two actual equations; its coefficients are smooth and bounded on a smaller source disk. The variable-principal result therefore applies. Equal infinite jets give infinite-order zero of $w$ and force equality locally. If the two maps agree on an open set, the same local argument propagates equality to its closure, and connectedness gives global equality. In the infinite-jet case first obtain that nonempty equality neighborhood and propagate it. All constructions used explicit kernels, Fourier bounds with the stated CC suppliers, geometric iteration and finite-dimensional formulas. This completes the similarity and unique-continuation assertions.