Conductivity CGO solutions and continuous logarithmic phases

Put $M_{\pm }=e^{-ikz}{f}_{\pm }$. Their equations are $\bar{\mathrm{\partial }}{M}_{\pm }=\pm \mu e^{-i(kz+\bar{k}\bar{z})}(\overline{\mathrm{\partial }{M}_{\pm }}-i\bar{k}{\bar{M}}_{\pm })$. For every real $t$, the combination $G=M_{+}-it{M}_{-}$ satisfies the same displayed equation with its plus coefficient, because conjugation changes $-it$ to $it$. It therefore obeys the similarity inequality in [F1], with $B=\mathrm{\mid }k\mathrm{\mid }\kappa$, and tends to $1-it$ at infinity. Thus $G=(1-it)e^a$ and has no zeros. The continuous ratio $M_{+}\mathrm{/}{M}_{-}$ tends to $1$ at infinity and never meets the imaginary axis: a value $it$ would give a zero of $G$. If its real part were nonpositive anywhere, continuity along a path to a sufficiently large point would force such an imaginary-axis value. Hence its real part is strictly positive everywhere. Since the exponential prefactors cancel, this is also the ratio $f_{+}\mathrm{/}{f}_{-}$.

Write $f_{+}=a+ib$. The equation $\bar{\mathrm{\partial }}{f}_{+}=\mu \overline{\mathrm{\partial }{f}_{+}}$ with real $\mu$ gives $b_y=\gamma a_x$ and $b_x=-\gamma a_y$. Equality of the mixed weak derivatives of $b$ implies $\mathrm{div}(\gamma \mathrm{\nabla }a)=0$. The same computation for $f_{-}$, with coefficient $-\mu$, shows that its imaginary part solves the $\gamma$ equation. Thus their stated complex combination $u_{\gamma }$ solves that equation. Setting $A=f_{+}+f_{-}$ and $B=f_{+}-f_{-}$ gives $u_{\gamma }=(A+\bar{B})\mathrm{/}2$. Step 1.1 is equivalent to $\mathrm{\mid }B\mathrm{\mid }<\mathrm{\mid }A\mathrm{\mid }$, so $u_{\gamma }\mathrm{\ne }0$. Its normalized remainder is $R_{\gamma }=\frac{1}{2}[M_{+}+M_{-}+e^{-i(kz+\bar{k}\bar{z})}({\bar{M}}_{+}-{\bar{M}}_{-})]-1$. The multiplier bounds in [F1] and differentiation of the unit-modulus exponential give $R_{\gamma }\in W^{1,r}$ and its $O(1\mathrm{/}z)$ normalization; the case $k=0$ follows from $f_{\pm }=1$.

For every constant $\mathrm{\mid }\lambda \mathrm{\mid }=1$, uniqueness of the CGO solution gives $f_{\lambda \mu }=\frac{1}{2}[(1+\lambda )f_{+}+(1-\lambda )f_{-}]=(A+\lambda B)\mathrm{/}2$. To check its equation, the constants $a=(1+\lambda )\mathrm{/}2$, $b=(1-\lambda )\mathrm{/}2$ satisfy $a=\lambda \bar{a}$ and $-b=\lambda \bar{b}$; substitution proves the required real-linear Beltrami equation, and $a+b=1$ gives the normalization. At a point where $B\mathrm{\ne }0$, choose $\lambda (z,k)=\bar{B}\mathrm{/}B$; where $B=0$, choose any unit value. Then $u_{\gamma }(z,k)=f_{\lambda (z,k)\mu }(z,k)$. Each right side is $e^{ik{\mathrm{\Phi }}_{\lambda \mu }(z,k)}$.

For fixed $k\mathrm{\ne }0$, the phases ${\mathrm{\Phi }}_{\lambda \mu }$ depend continuously on $\lambda$ and $z$. Indeed, the coefficients in the Fredholm construction of [F1] depend continuously on $\lambda$ in operator norm; the resolvent identity gives continuity of its solution locally in $\lambda$, hence on the compact parameter circle. Its nonvanishing normalized multiplier admits the uniquely determined logarithm tending to zero at infinity. That logarithm depends continuously on $\lambda$ on compact sets: start at a uniformly large base point where all multipliers are close to $1$, use the principal logarithm there, and extend along a finite covering of any fixed compact path by disks where the multipliers stay away from zero. The resulting phase is the normalized principal phase in [F2]. Therefore $\tilde{\delta }(z)=ik{\mathrm{\Phi }}_{\lambda (z,k)\mu }(z,k)$ is continuous where $B\mathrm{\ne }0$. If $B(z_0,k)=0$, all values $f_{\lambda \mu }(z_0,k)$ coincide. Their phases have equal exponentials and vary continuously with $\lambda$, so their differences belong to the discrete set $2\pi i\mathbb{Z}$ and are constant; they are thus all equal. Compactness of the rotation circle and joint continuity then give continuity of $\tilde{\delta }$ also at $z_0$. Its exponential is $u_{\gamma }$ by step 3.1, and [F2] gives its required normalization and bounds. Uniqueness follows because any two continuous logarithms on the connected plane differ by a constant $2\pi in$, which the normalization forces to be zero.

The phase ${\delta }_{\gamma }=\tilde{\delta }$ differs from $ikz$ by a bounded function. For any target $w$, choose a large circle $\mathrm{\mid }z\mathrm{\mid }=R$ on which $\mathrm{\mid }{\delta }_{\gamma }(z)-ikz\mathrm{\mid }+\mathrm{\mid }w\mathrm{\mid }<\mathrm{\mid }k\mathrm{\mid }R$. The straight homotopy between ${\delta }_{\gamma }(z)-w$ and $ikz$ avoids zero on this circle, so their winding numbers are both one. If ${\delta }_{\gamma }-w$ had no zero in the disk, its boundary loop would extend to a loop through a disk in $\mathbb{C}\setminus \{0\}$ and would have winding zero, a contradiction. Thus the phase is onto. Finally the Hölder and supremum bounds on $M_{\pm }$ in [F1]–[F2], combined with the smooth exponential on a fixed disk, give the asserted Hölder estimate for $f_{\pm }$ and hence for their real and imaginary combination $u_{\gamma }$.