Logarithmic EIT stability on explicit conductivity classes

Rotate coordinates so the two given vectors are the first coordinate vectors. Write $\xi =(a,s{e}_2^{\mathrm{\prime }}+r\omega )$ with $a\in \mathbb{R}$, $r\ge 0$, and $\omega \in S^{d-2}$ in the remaining coordinates. Then $p_{\zeta }(\xi )=s^2-a^2-r^2+2isa$, and its zero set is $\mathrm{\Sigma }=\{a=0,r=s\}$. The exact distance to this sphere is $D(\xi )=\sqrt{a^2+(r-s{)}^2}$. For each fixed $K$, on $\mathrm{\mid }\xi \mathrm{\mid }\le Ks$ one has $\mathrm{\mid }{p}_{\zeta }(\xi )\mathrm{\mid }{\asymp }_{K}sD(\xi )$. To check the lower bound, if $\mathrm{\mid }a\mathrm{\mid }\ge D\mathrm{/}4$, use the imaginary part. Otherwise $\mathrm{\mid }r-s\mathrm{\mid }>D\mathrm{/}2$ and $\mathrm{\mid }a\mathrm{\mid }\le \mathrm{\mid }r-s\mathrm{\mid }\mathrm{/}2$; since $\mathrm{\mid }r-s\mathrm{\mid }\le r+s$, $\mathrm{\mid }{s}^2-r^2-a^2\mathrm{\mid }\ge (s+r)\mathrm{\mid }s-r\mathrm{\mid }-a^2\ge (3\mathrm{/}4)(s+r)\mathrm{\mid }s-r\mathrm{\mid }\ge (3\mathrm{/}8)sD$. The upper bound follows by expanding $s^2-a^2-r^2$ and using $\mathrm{\mid }a\mathrm{\mid }+r+s\le C_{K}s$. Also, if $\mathrm{\mid }\xi \mathrm{\mid }\ge 8s$, the triangle inequality gives $\mathrm{\mid }{p}_{\zeta }(\xi )\mathrm{\mid }\asymp \mathrm{\mid }\xi {\mathrm{\mid }}^2$. Therefore $s\mathrm{/}\mathrm{\mid }{p}_{\zeta }(\xi )\mathrm{\mid }\le C(1+D(\xi {)}^{-1})$ everywhere away from $\mathrm{\Sigma }$.

For $N>d+3$, the integral $\int ⟨\xi -\eta {⟩}^{-N}D(\xi {)}^{-1}d\xi$ is bounded uniformly in $s\ge 1$ and $\eta \in {\mathbb{R}}^d$, where $⟨x⟩=(1+\mathrm{\mid }x{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}$. Off $D<1\mathrm{/}4$ this follows from $D^{-1}\le 4$ and integrability of the bracket power. On $D<1\mathrm{/}4$, write $\eta =(b,s{e}_2^{\mathrm{\prime }}+t\nu )$ with $t\ge 0$ and any unit $\nu$ if $t=0$. The inequality $\mathrm{\mid }r\omega -t\nu \mathrm{\mid }\ge (r\mathrm{/}2)\mathrm{\mid }\omega -\nu \mathrm{\mid }$ follows from $(r-t{)}^2+rt\mathrm{\mid }\omega -\nu {\mathrm{\mid }}^2$ by minimizing in $t\ge 0$. Hence the angular integral of $⟨\xi -\eta {⟩}^{-N}$ is at most $C{r}^{-(d-2)}$ for $r\ge 3\mathrm{/}4$. Indeed, in polar angle $\theta$ from $\nu$, split at $\theta =1\mathrm{/}r$ and use $\mathrm{\mid }\omega -\nu \mathrm{\mid }=2\sin (\theta \mathrm{/}2)\ge 2\theta \mathrm{/}\pi$ and the spherical measure bound $C{\theta }^{d-3}d\theta$; the two integrals are bounded by $C{r}^{-(d-2)}$. This cancels the radial Jacobian $r^{d-2}$. The remaining integral over $\mathrm{\mid }a\mathrm{\mid },\mathrm{\mid }r-s\mathrm{\mid }<1\mathrm{/}4$ of $(a^2+(r-s{)}^2{)}^{-1\mathrm{/}2}$ is finite with an absolute bound by planar polar integration. This proves the uniform assertion, including all centers $\eta$.

Put $h=\xi -\eta$. The identity $p_{\zeta }(\eta )-p_{\zeta }(\xi )=\mathrm{\mid }\xi {\mathrm{\mid }}^2-\mathrm{\mid }\eta {\mathrm{\mid }}^2-2i\zeta \cdot h$ and the bound $\mathrm{\mid }\xi \mathrm{\mid }\le C(s+\sqrt{\mathrm{\mid }{p}_{\zeta }(\xi )\mathrm{\mid }})$ imply $s+\mathrm{\mid }{p}_{\zeta }(\eta )\mathrm{\mid }\le C(s+\mathrm{\mid }{p}_{\zeta }(\xi )\mathrm{\mid })⟨h{⟩}^2\mathrm{.}$ For the auxiliary bound, use $\mathrm{\mid }\xi {\mathrm{\mid }}^2\le \mathrm{\mid }{p}_{\zeta }(\xi )\mathrm{\mid }+2\sqrt{2}s\mathrm{\mid }\xi \mathrm{\mid }$ and solve this quadratic inequality. In the difference identity, the terms $s\mathrm{\mid }h\mathrm{\mid }$, $\sqrt{\mathrm{\mid }{p}_{\zeta }(\xi )\mathrm{\mid }}\mathrm{\mid }h\mathrm{\mid }$, and $\mathrm{\mid }h{\mathrm{\mid }}^2$ are absorbed by the displayed right side using $s\ge 1$ and $2ab\le a^2+b^2$. Since $\widehat{\varphi }$ is rapidly decreasing, steps 1.1–2.1 now give ${\sup }_{\eta }\int \mathrm{\mid }\widehat{\varphi }(\xi -\eta )\mathrm{\mid }\frac{s+\mathrm{\mid }{p}_{\zeta }(\eta )\mathrm{\mid }}{\mathrm{\mid }{p}_{\zeta }(\xi )\mathrm{\mid }}d\xi \le C_{\varphi }\mathrm{.}$ Indeed the integrand is bounded by $C_{\varphi }⟨\xi -\eta {⟩}^{-N}(1+D(\xi {)}^{-1})$ after choosing sufficiently many Schwartz powers.

For nonnegative weights $v,w$ positive almost everywhere and an integrable kernel $K$, Cauchy–Schwarz and Tonelli yield $\int w(\xi )\mathrm{\mid }(K\ast f)(\xi ){\mathrm{\mid }}^{2}d\xi \le \mathrm{\parallel }K{\mathrm{\parallel }}_1({\sup }_{\eta }\int \mathrm{\mid }K(\xi -\eta )\mathrm{\mid }\frac{w(\xi )}{v(\eta )}d\xi )\int v(\eta )\mathrm{\mid }f(\eta ){\mathrm{\mid }}^{2}d\eta \mathrm{.}$ To see this directly, bound $\mathrm{\mid }K\ast f{\mathrm{\mid }}^2$ by $\mathrm{\parallel }K{\mathrm{\parallel }}_1\int \mathrm{\mid }K(\xi -\eta )\mathrm{\mid }\mathrm{\mid }f(\eta ){\mathrm{\mid }}^{2}d\eta$, multiply by $w$, and interchange nonnegative integrals. Apply this with $K=(2\pi {)}^{-d\mathrm{/}2}\widehat{\varphi }$, $v=(s+\mathrm{\mid }{p}_{\zeta }\mathrm{\mid }{)}^{-1}$ and $w=\mathrm{\mid }{p}_{\zeta }{\mathrm{\mid }}^{-1}$. Step 3.1 and [F2] prove the negative-half cutoff bound. The locally integrable singularity proved in step 2.1 and the quadratic large-frequency growth show that every Schwartz function has finite negative-half norm.

The positive-half estimate is the dual of step 4.1. More explicitly, use the same negative-half bound for the cutoff $\bar{\varphi }$ and the Fourier $L^2$ pairing. Cauchy–Schwarz with reciprocal weights gives $\mathrm{\mid }(\varphi u,v{)}_{L^2}\mathrm{\mid }=\mathrm{\mid }(u,\bar{\varphi }v{)}_{L^2}\mathrm{\mid }\le \mathrm{\parallel }u{\mathrm{\parallel }}_{{\dot{X}}_{\zeta }^{1\mathrm{/}2}}\mathrm{\parallel }\bar{\varphi }v{\mathrm{\parallel }}_{{\dot{X}}_{\zeta }^{-1\mathrm{/}2}}\le C\mathrm{\parallel }u{\mathrm{\parallel }}_{{\dot{X}}_{\zeta }^{1\mathrm{/}2}}\mathrm{\parallel }v{\mathrm{\parallel }}_{X_{\zeta }^{-1\mathrm{/}2}}$. The supremum over Schwartz $v$ of the left side divided by its negative-half norm equals $\mathrm{\parallel }\varphi u{\mathrm{\parallel }}_{X_{\zeta }^{1\mathrm{/}2}}$. For this equality, truncate the weighted dual extremizer to a ball and away from $\mathrm{\Sigma }$, approximate it by smooth compactly supported frequency functions there, and use ordinary $L^2$ density where the weights are bounded above and below; monotone truncation recovers the full weighted integral. Thus the claimed supremum uses no unattained extremizer.

The positive-half weight is at least $s$, so step 5.1 immediately gives $\mathrm{\parallel }\varphi u{\mathrm{\parallel }}_2\le C{s}^{-1\mathrm{/}2}\mathrm{\parallel }u{\mathrm{\parallel }}_{{\dot{X}}_{\zeta }^{1\mathrm{/}2}}$. Split the gradient Fourier integral into $\mathrm{\mid }\xi \mathrm{\mid }\le 8s$ and its complement. The first part has norm at most $8s\mathrm{\parallel }\varphi u{\mathrm{\parallel }}_2$, hence at most $C{s}^{1\mathrm{/}2}\mathrm{\parallel }u{\mathrm{\parallel }}_{{\dot{X}}_{\zeta }^{1\mathrm{/}2}}$. On the complement $\mathrm{\mid }\xi {\mathrm{\mid }}^2\le C\mathrm{\mid }{p}_{\zeta }(\xi )\mathrm{\mid }$, so its norm is at most $C\mathrm{\parallel }\varphi u{\mathrm{\parallel }}_{X_{\zeta }^{1\mathrm{/}2}}\le C\mathrm{\parallel }u{\mathrm{\parallel }}_{{\dot{X}}_{\zeta }^{1\mathrm{/}2}}$. Since $s\ge 1$, their sum gives the gradient bound. A bounded map sends norm-Cauchy sequences to norm-Cauchy sequences, with the same bound; passing to equivalence classes therefore extends each map uniquely to the stated completions.

For $f\in C_c^{\mathrm{\infty }}$, [F1] shows that $Sf:=\mathrm{\partial }\mathcal{C}f$ is bounded locally and decays as $O(\mathrm{\mid }z{\mathrm{\mid }}^{-2})$, hence is in $L^2$. Distributional commutation gives $\bar{\mathrm{\partial }}Sf=\mathrm{\partial }f$. Fourier transformation gives $(i{\xi }_1-{\xi }_2)\widehat{Sf}=(i{\xi }_1+{\xi }_2)\widehat{f}$, so almost everywhere off the origin $\widehat{Sf}=(({\xi }_1-i{\xi }_2)\mathrm{/}({\xi }_1+i{\xi }_2))\widehat{f}$. The multiplier has modulus one. [F2] extends $S$ uniquely to the isometry $\mathcal{B}$ on $L^2$. Multiplication by the translation character commutes with this multiplier, giving translation commutation there.

The kernel $K(z)=-1\mathrm{/}(\pi z^2)$ has annular integral $2\log 2$. Its gradient is bounded by $C\mathrm{\mid }z{\mathrm{\mid }}^{-3}$, so for $\mathrm{\mid }z\mathrm{\mid }\ge 2\mathrm{\mid }h\mathrm{\mid }$, the segment estimate gives $\mathrm{\mid }K(z-h)-K(z)\mathrm{\mid }\le C\mathrm{\mid }h\mathrm{\mid }\mathrm{\mid }z{\mathrm{\mid }}^{-3}$. Polar integration from $2\mathrm{\mid }h\mathrm{\mid }$ to infinity bounds its integral by an absolute constant. The off-support representation for compactly supported $L^2$ inputs follows by approximating them by smooth inputs in one fixed compact neighborhood of their support: on any compact set at positive distance from that neighborhood the kernel integrals converge uniformly by Cauchy–Schwarz, while the operator outputs converge in $L^2$ by step 1.1. Exhausting the complement of the essential support by compact sets and shrinking the neighborhoods gives the representation almost everywhere off the essential support. Thus [F3] applies. Let $A=\max \{1,\mathrm{\parallel }\mathcal{B}{\mathrm{\parallel }}_{4\to 4}\}$. Interpolating the norm-one $L^2$ bound and the $L^4$ bound gives $\mathrm{\parallel }\mathcal{B}{\mathrm{\parallel }}_{p\to p}\le A^{\theta }$ with $1\mathrm{/}p=(1-\theta )\mathrm{/}2+\theta \mathrm{/}4$, that is $\theta =2-4\mathrm{/}p$. Density makes all extensions compatible and carries translation commutation to every finite $p$.

Let $p>2$, let $q=p\mathrm{/}(p-1)<2$, and let $f$ have support in $D_R$. Hölder gives absolute convergence of $\mathcal{C}f$ at every $z$: the singularity $\mathrm{\mid }z-w{\mathrm{\mid }}^{-q}$ is locally integrable, and the support is bounded. For $\delta =\mathrm{\mid }z-z^{\mathrm{\prime }}\mathrm{\mid }>0$, split its kernel difference into the union of the two disks of radius $2\delta$ about $z,z^{\mathrm{\prime }}$ and its complement. On the first region, integrating the two singularities with Hölder bounds their contributions by $C_p\mathrm{\parallel }f{\mathrm{\parallel }}_p{\delta }^{1-2\mathrm{/}p}$. On the complement, $\mathrm{\mid }(z-w{)}^{-1}-(z^{\mathrm{\prime }}-w{)}^{-1}\mathrm{\mid }\le C\delta \mathrm{\mid }z-w{\mathrm{\mid }}^{-2}$; its $L^q$ norm is at most $C_p\delta ({\int }_{2\delta }^{\mathrm{\infty }}{r}^{1-2q}dr{)}^{1\mathrm{/}q}=C_p{\delta }^{1-2\mathrm{/}p}$. This proves the global Hölder bound. On $\mathrm{\mid }z\mathrm{\mid }\le 2R$, the same singular-kernel estimate gives $\mathrm{\mid }\mathcal{C}f(z)\mathrm{\mid }\le C_{p,R}\mathrm{\parallel }f{\mathrm{\parallel }}_p$; on $\mathrm{\mid }z\mathrm{\mid }>2R$, the bound is $C\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{/}\mathrm{\mid }z\mathrm{\mid }\le C_R\mathrm{\parallel }f{\mathrm{\parallel }}_p\mathrm{/}\mathrm{\mid }z\mathrm{\mid }$. The latter is $L^p$ because $p>2$, proving $\mathrm{\parallel }\mathcal{C}f{\mathrm{\parallel }}_p\le C_{p,R}\mathrm{\parallel }f{\mathrm{\parallel }}_p$ and the stated decay. Choose a fixed $\chi \in C_c^{\mathrm{\infty }}(D_{R+1})$ equal to $1$ on ${\bar{D}}_R$. By the global density in [F2], take $g_n\in C_c^{\mathrm{\infty }}(\mathbb{C})$ with $g_n\to f$ in $L^p$, and set $f_n=\chi g_n$. Since $\chi f=f$ almost everywhere, $\mathrm{\parallel }{f}_n-f{\mathrm{\parallel }}_p\le \mathrm{\parallel }\chi {\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }{g}_n-f{\mathrm{\parallel }}_p\to 0$, and each $f_n$ is smooth and supported in $D_{R+1}$. Their Cauchy transforms converge in $L^p$, and their two weak derivatives converge to $f$ and $\mathcal{B}f$ by step 2.1. Passing the derivative identities against tests proves the two weak identities and the $W^{1,p}$ bound.

If $\kappa =0$, the final operator is the identity. Otherwise choose $2<p_{\kappa }\le 4$ so close to $2$ that $\kappa A^{2-4\mathrm{/}{p}_{\kappa }}\le (1+\kappa )\mathrm{/}2<1$; this is possible by continuity and the value $1$ of the bound at $p=2$. For $2\le p\le p_{\kappa }$, the operator $Tf=\mu \mathcal{B}f+\nu \overline{\mathcal{B}f}$ has pointwise modulus at most $(\mathrm{\mid }\mu \mathrm{\mid }+\mathrm{\mid }\nu \mathrm{\mid })\mathrm{\mid }\mathcal{B}f\mathrm{\mid }$, and therefore norm at most $(1+\kappa )\mathrm{/}2$. Regard $L^p(\mathbb{C};\mathbb{C})$ as a real Banach space. The norm-convergent series ${\sum }_{j=0}^{\mathrm{\infty }}{T}^j$ is a two-sided inverse to $I-T$, since the finite geometric-series remainder has norm tending to zero. Its norm is at most $1\mathrm{/}(1-(1+\kappa )\mathrm{/}2)=2\mathrm{/}(1-\kappa )$, proving the assertion.

Direct expansion gives ${\zeta }_{\pm }\cdot {\zeta }_{\pm }=0$, $\mathrm{\mid }{\zeta }_{\pm }{\mathrm{\mid }}^2=2s^2$, and ${\zeta }_{+}+{\zeta }_{-}=ik$. For either sign, writing $a=\xi \cdot \eta$, $b=\xi \cdot {\eta }^{\perp }$, the symbol satisfies $\mathrm{\mid }{p}_{\zeta }(\xi )\mathrm{\mid }\asymp \mid \mathrm{\mid }\xi {\mathrm{\mid }}^2+k\cdot \xi \pm 2r(s)b\mid +2s\mathrm{\mid }a\mathrm{\mid }$, with absolute constants. Put $R=\mathrm{\mid }\xi \mathrm{\mid }$, $A=\mathrm{\mid }{\mathrm{\Pi }}_P\xi \mathrm{\mid }$, and $W(s,\eta ,\xi )=R^2\mathrm{/}(s+\mathrm{\mid }{p}_{\zeta }(\xi )\mathrm{\mid })$. We prove the uniform kernel bound ${\lambda }^{-1}{\int }_{S^1}{\int }_{\lambda }^{2\lambda }Wdsd\eta \le C(1+(1+\mathrm{\mid }k{\mathrm{\mid }}^2)\mathrm{/}\lambda )$. If $R\le 100\sqrt{1+\mathrm{\mid }k{\mathrm{\mid }}^2}$, the pointwise bound $W\le R^2\mathrm{/}\lambda$ proves it. Otherwise $\mathrm{\mid }k\cdot \xi \mathrm{\mid }\le R^2\mathrm{/}100$. If $A\le R^2\mathrm{/}(100\lambda )$, then $2r(s)\mathrm{\mid }b\mathrm{\mid }\le 4\lambda A\le R^2\mathrm{/}25$ and the real part has absolute value at least $R^2\mathrm{/}2$, giving $W\le 2$. Thus only the case $A>R^2\mathrm{/}(100\lambda )$ remains; here $R^2\mathrm{/}A<100\lambda$.

In this last case, on $\mathrm{\mid }a\mathrm{\mid }\ge A\mathrm{/}2$ the imaginary part gives $W\le R^2\mathrm{/}(\lambda A)\le 100$. On the remaining angular set, partition into shells $E_j=\{2^{-j-2}A<\mathrm{\mid }a\mathrm{\mid }\le 2^{-j-1}A\}$, $j\ge 0$, ignoring the measure-zero set $a=0$. Each shell has arclength at most $C{2}^{-j}$, since $a=A\cos \theta$ and $\mathrm{\mid }\sin \theta \mathrm{\mid }\ge \sqrt{3}\mathrm{/}2$ there. On each shell, $\mathrm{\mid }b\mathrm{\mid }\ge \sqrt{3}A\mathrm{/}2$ and the denominator is bounded below, up to an absolute constant, by $\mathrm{\mid }B\pm 2r(s)b\mathrm{\mid }+\lambda (1+2^{-j-1}A)$, where $B=R^2+k\cdot \xi$ is independent of $s,\eta$. Since $dr\mathrm{/}ds=s\mathrm{/}r\ge 1$ and the $r$-interval is contained in $[\lambda \mathrm{/}2,2\lambda ]$, integration in $s$ is bounded by integration in $r$ over this latter interval. For any interval $I$ of length at most $2\lambda$ and $c>0$, splitting at the zero of the affine function gives ${\int }_I(\mathrm{\mid }B+2br\mathrm{\mid }+c{)}^{-1}dr\le C\mathrm{\mid }b{\mathrm{\mid }}^{-1}\log (1+\lambda \mathrm{\mid }b\mathrm{\mid }\mathrm{/}c)$. Therefore ${\int }_{\lambda }^{2\lambda }Wds\le C(R^2\mathrm{/}A)\log (1+CA\mathrm{/}(1+2^{-j-1}A))\le C\lambda (j+1)$. Multiplying by the shell measure and summing ${\sum }_{j\ge 0}(j+1)2^{-j}<\mathrm{\infty }$ proves the kernel bound of step 1.1. This proof treats all $A$, including $A=0$ via its earlier case, and its constants are independent of $k,P$.

Initially let $f$ be Schwartz. By [F1], $\mathrm{\parallel }\varphi \mathrm{div}f{\mathrm{\parallel }}_{{\dot{X}}_{\zeta }^{-1\mathrm{/}2}}^2\le C\int \mathrm{\mid }\xi \cdot \widehat{f}(\xi ){\mathrm{\mid }}^2\mathrm{/}(s+\mathrm{\mid }{p}_{\zeta }(\xi )\mathrm{\mid })d\xi \le C\int W(s,\eta ,\xi )\mathrm{\mid }\widehat{f}(\xi ){\mathrm{\mid }}^{2}d\xi$. Interchange the nonnegative integrals using [F2] and apply step 2.1 and Plancherel. This proves the first estimate. For $f\in H^1$, the simpler denominator bound $s+\mathrm{\mid }{p}_{\zeta }\mathrm{\mid }\ge s\ge \lambda$ gives $\mathrm{\parallel }\varphi \mathrm{div}f{\mathrm{\parallel }}_{{\dot{X}}_{\zeta }^{-1\mathrm{/}2}}^2\le C{\lambda }^{-1}\mathrm{\parallel }\mathrm{\nabla }f{\mathrm{\parallel }}_2^2$; averaging over the parameter interval of length $\lambda$ and circle length $2\pi$ proves the second estimate. The same estimates hold for general $L^2$ and $H^1$ fields by Fourier truncation and smooth approximation. The limits match the distribution $\varphi \mathrm{div}f$: convergence in the weighted negative space implies distributional convergence when tested against Schwartz functions, by reciprocal-weight Cauchy–Schwarz, and $L^2$ convergence of $f$ already implies distributional convergence of its derivative.

Write $f=f_h+(f-f_h)$ with $f_h={\rho }_h\ast f$. Differentiation under the convolution, first for smooth fields and then by $L^2$ approximation, gives $\mathrm{\parallel }\mathrm{\nabla }{f}_h{\mathrm{\parallel }}_2\le \mathrm{\parallel }\mathrm{\nabla }{\rho }_h{\mathrm{\parallel }}_1\mathrm{\parallel }f{\mathrm{\parallel }}_2=C_{\rho }{h}^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}_2$ by [F3]. Also $\mathrm{\parallel }f-f_h{\mathrm{\parallel }}_2\le {\omega }_2(f,h)$: at each $x$ apply the integral Cauchy–Schwarz inequality to the probability density ${\rho }_h(y)dy$, then integrate in $x$ and interchange nonnegative integrals to bound the squared norm by $\int {\rho }_h(y)\mathrm{\parallel }f-f(\cdot -y){\mathrm{\parallel }}_2^{2}dy$. The support of ${\rho }_h$ is inside $\mathrm{\mid }y\mathrm{\mid }\le h$. Since $\mathrm{\parallel }v+w{\mathrm{\parallel }}^2\le 2\mathrm{\parallel }v{\mathrm{\parallel }}^2+2\mathrm{\parallel }w{\mathrm{\parallel }}^2$, apply the $H^1$ estimate of step 3.1 to $f_h$ and its $L^2$ estimate to $f-f_h$. This yields the claimed mollified estimate after enlarging the constant.

Choose $\chi \in C_c^{\mathrm{\infty }}(D_3)$ equal to $1$ on a neighborhood of ${\bar{D}}_2$. The function $\chi f$, extended by zero outside $D_3$, belongs to $W^{1,r}(\mathbb{C})$: for each weak derivative, testing against a global compactly supported smooth function and using the local product rule gives the zero extension of $\chi \mathrm{\nabla }f+f\mathrm{\nabla }\chi$, with no boundary term because $\chi$ vanishes near $\mathrm{\partial }{D}_3$. By [F2], choose global $g_n\in C_c^{\mathrm{\infty }}(\mathbb{C})$ converging to this extension in $W^{1,r}(\mathbb{C})$. On a neighborhood of $\mathrm{supp}\varphi$, $g_n\to f$ in $W^{1,r}$, so the ordinary smooth product rule for $\varphi g_n$ passes to the limit and gives $F=\varphi f\in W^{1,r}(\mathbb{C})$ with $\bar{\mathrm{\partial }}F=\varphi \bar{\mathrm{\partial }}f+f\bar{\mathrm{\partial }}\varphi$ and $\mathrm{\partial }F=\varphi \mathrm{\partial }f+f\mathrm{\partial }\varphi$. Since $r>2$ and $F$ has bounded support, $F\in H^1(\mathbb{C})$. In Fourier variables the two Wirtinger multipliers give $\mathrm{\partial }F=\mathcal{B}\bar{\mathrm{\partial }}F$: this holds first in $L^2$ and then in $L^r$ by the compatible Beurling extension. Setting $G=\bar{\mathrm{\partial }}F$ and $H=f\bar{\mathrm{\partial }}\varphi -\mu f\mathrm{\partial }\varphi -\nu \bar{f}\overline{\mathrm{\partial }\varphi }$, the Beltrami equation becomes $(I-\mu \mathcal{B}-\nu \bar{\mathcal{B}})G=H$. Thus [F1] gives $\mathrm{\parallel }G{\mathrm{\parallel }}_r\le C_{\kappa ,r}\mathrm{\parallel }H{\mathrm{\parallel }}_r\le C_{\kappa ,r}\mathrm{\parallel }f\mathrm{\nabla }\varphi {\mathrm{\parallel }}_r$ and $\mathrm{\parallel }\mathrm{\partial }F{\mathrm{\parallel }}_r\le \mathrm{\parallel }\mathcal{B}{\mathrm{\parallel }}_{r\to r}\mathrm{\parallel }G{\mathrm{\parallel }}_r$. This is the first assertion.

Write ${\tau }_{h}g=g(\cdot -h)$. Translating the equation in step 1.1, subtracting it, and using translation commutation gives $(I-\mu \mathcal{B}-\nu \bar{\mathcal{B}})({\tau }_{h}G-G)=({\tau }_h\mu -\mu )\mathcal{B}{\tau }_{h}G+({\tau }_h\nu -\nu )\overline{\mathcal{B}{\tau }_{h}G}+{\tau }_{h}H-H$. Inverse boundedness on $L^2$ and Hölder therefore yield $\mathrm{\parallel }{\tau }_{h}G-G{\mathrm{\parallel }}_2\le C_{\kappa }[(\mathrm{\parallel }{\tau }_h\mu -\mu {\mathrm{\parallel }}_q+\mathrm{\parallel }{\tau }_h\nu -\nu {\mathrm{\parallel }}_q)\mathrm{\parallel }\mathcal{B}G{\mathrm{\parallel }}_r+\mathrm{\parallel }{\tau }_{h}H-H{\mathrm{\parallel }}_2]$. All terms are in $L^2$: the differentiated cutoff has compact support, $G\in L^r$, and the coefficient translation differences are in $L^q$ because the coefficients are bounded with fixed bounded support.

For any smooth cutoff derivative $\psi$ supported in $D_2$, extend $f\psi$ by zero. It belongs to $H^1(\mathbb{C})$ and $\mathrm{\parallel }f\psi {\mathrm{\parallel }}_{H^1}\le C_{r,\psi }U$ by Hölder on the bounded support. The fundamental theorem along line segments gives $\mathrm{\parallel }{\tau }_h(f\psi )-f\psi {\mathrm{\parallel }}_2\le \mathrm{\mid }h\mathrm{\mid }\mathrm{\parallel }\mathrm{\nabla }(f\psi ){\mathrm{\parallel }}_2$ for smooth inputs; smooth approximation in [F2] gives the same inequality for this Sobolev function. Expand each coefficient product in $H$ by ${\tau }_h(\mu f\psi )-\mu f\psi =({\tau }_h\mu -\mu ){\tau }_h(f\psi )+\mu ({\tau }_h(f\psi )-f\psi )$, and the analogous expression with $\nu \bar{f}$. Hölder gives $\mathrm{\parallel }{\tau }_{h}H-H{\mathrm{\parallel }}_2\le CU(\mathrm{\parallel }{\tau }_h\mu -\mu {\mathrm{\parallel }}_q+\mathrm{\parallel }{\tau }_h\nu -\nu {\mathrm{\parallel }}_q+\mathrm{\mid }h\mathrm{\mid })$. Step 1.1 also gives $\mathrm{\parallel }\mathcal{B}G{\mathrm{\parallel }}_r\le CU$. Substitution in step 2.1, followed by the supremum over $\mathrm{\mid }h\mathrm{\mid }\le t$, proves the desired bound for $G$. Finally $\varphi \bar{\mathrm{\partial }}f=G-f\bar{\mathrm{\partial }}\varphi$, and the just-proved translation bound for $f\psi$ yields the second bound.

Average $\mathrm{\parallel }{\tau }_{h}G-G{\mathrm{\parallel }}_2^2$ over the square $[-2\mathrm{/}T,2\mathrm{/}T{]}^2$ with normalized Lebesgue measure. Plancherel identifies the resulting multiplier as $2[1-\mathrm{sinc}(2{\xi }_1\mathrm{/}T)\mathrm{sinc}(2{\xi }_2\mathrm{/}T)]$, where $\mathrm{sinc}(x)=\sin (x)\mathrm{/}x$ and its value at zero is $1$. If $\mathrm{\mid }\xi \mathrm{\mid }\ge T$, one coordinate has magnitude at least $T\mathrm{/}\sqrt{2}$, so one sinc factor has absolute value at most $1\mathrm{/}\sqrt{2}$, while the other is at most $1$. The multiplier is therefore at least $2(1-1\mathrm{/}\sqrt{2})>0$ on this frequency set. Every shift in the square has norm at most $2\sqrt{2}\mathrm{/}T<3\mathrm{/}T$. Hence $\mathrm{\parallel }{\mathbf{1}}_{\{\mathrm{\mid }\xi \mathrm{\mid }\ge T\}}\widehat{G}{\mathrm{\parallel }}_2\le C{\omega }_2(G,3\mathrm{/}T)$. Since $3\mathrm{/}T\le 1\mathrm{/}2$, step 3.1 applies and gives the claimed tail bound.