Conductivity L2 recovery from fixed-parameter Beltrami solution stability

Fix $\varphi$ from [F3], put $F=\varphi (f_1-f_2)$ and $G=\bar{\mathrm{\partial }}F$. At frequencies $\mathrm{\mid }\xi \mathrm{\mid }<T$, the Wirtinger multiplier has magnitude $\mathrm{\mid }\xi \mathrm{\mid }\mathrm{/}2$, so $\mathrm{\parallel }{\mathbf{1}}_{\{\mathrm{\mid }\xi \mathrm{\mid }<T\}}\widehat{G}{\mathrm{\parallel }}_2\le (T\mathrm{/}2)\mathrm{\parallel }F{\mathrm{\parallel }}_2\le CT\epsilon$. At higher frequencies, apply [F1] to each $f_j$ and use $\theta \le 1$ to get $\mathrm{\parallel }{\mathbf{1}}_{\{\mathrm{\mid }\xi \mathrm{\mid }\ge T\}}\widehat{G}{\mathrm{\parallel }}_2\le CU(A+1)T^{-\theta }$ for $T\ge 6$. Therefore $\mathrm{\parallel }G{\mathrm{\parallel }}_2\le C[T\epsilon +T^{-\theta }]$. For $0<\epsilon \le 1$, choose $T=\max \{6,{\epsilon }^{-1\mathrm{/}(1+\theta )}\}$ to obtain $\mathrm{\parallel }G{\mathrm{\parallel }}_2\le C{\epsilon }^{\theta \mathrm{/}(1+\theta )}$; the range where $T=6$ is absorbed into $C$. For $\epsilon =0$, let $T\to \mathrm{\infty }$ to obtain $G=0$. Since $\varphi =1$ on $D_1$, [F2] implies $\mathrm{\parallel }\mathrm{\partial }(f_1-f_2){\mathrm{\parallel }}_{L^2(D_1)}+\mathrm{\parallel }\bar{\mathrm{\partial }}(f_1-f_2){\mathrm{\parallel }}_{L^2(D_1)}\le 2\mathrm{\parallel }G{\mathrm{\parallel }}_2$.

The negative-power hypothesis makes $\mathrm{\partial }{f}_2\mathrm{\ne }0$ almost everywhere. Subtract the two Beltrami equations in the form $\bar{\mathrm{\partial }}(f_1-f_2)={\mu }_1\overline{\mathrm{\partial }(f_1-f_2)}+({\mu }_1-{\mu }_2)\overline{\mathrm{\partial }{f}_2}$. Thus $\mathrm{\mid }{\mu }_1-{\mu }_2\mathrm{\mid }\le (\mathrm{\mid }\bar{\mathrm{\partial }}(f_1-f_2)\mathrm{\mid }+\kappa \mathrm{\mid }\mathrm{\partial }(f_1-f_2)\mathrm{\mid })\mathrm{\mid }\mathrm{\partial }{f}_2{\mathrm{\mid }}^{-1}$ almost everywhere. Put $s=2\sigma \mathrm{/}(2+\sigma )$, so $0<s<2$ and $1\mathrm{/}s=1\mathrm{/}2+1\mathrm{/}\sigma$. Apply Hölder to the $s$-th powers with conjugate exponents $2\mathrm{/}s$ and $\sigma \mathrm{/}s$, both greater than one. This gives $({\int }_{D_1}\mathrm{\mid }{\mu }_1-{\mu }_2{\mathrm{\mid }}^s{)}^{1\mathrm{/}s}\le CV\mathrm{\parallel }G{\mathrm{\parallel }}_2$, without needing a triangle inequality when $s<1$. Step 1.1 consequently bounds this quantity by $C{\epsilon }^{\theta \mathrm{/}(1+\theta )}$, including $\epsilon =0$.

If $\kappa =0$, both coefficients are zero and the conclusion is immediate. Otherwise $\mathrm{\mid }{\mu }_1-{\mu }_2\mathrm{\mid }\le 2\kappa$, and hence $\int \mathrm{\mid }{\mu }_1-{\mu }_2{\mathrm{\mid }}^2\le (2\kappa {)}^{2-s}\int \mathrm{\mid }{\mu }_1-{\mu }_2{\mathrm{\mid }}^s$. Step 2.1 and the square root yield $\mathrm{\parallel }{\mu }_1-{\mu }_2{\mathrm{\parallel }}_2\le C{\epsilon }^{\theta s\mathrm{/}(2+2\theta )}=C{\epsilon }^{\eta }$. The scalar map $x\mapsto (1-x)\mathrm{/}(1+x)$ has derivative bounded by $2\mathrm{/}(1-\kappa {)}^2$ on $[-\kappa ,\kappa ]$. The mean value theorem therefore gives the same estimate for the two conductivities. Substitution of the stated logarithmic bound for $\epsilon$ gives exponent $b\eta$ for sufficiently small data, and $\epsilon =0$ gives equality by step 2.1.