Higher-dimensional C1alpha logarithmic Calderón stability

Extend both coefficients by $1$ using [F1], and fix a ball $B$ containing $[0,1{]}^d$ and a neighborhood of it. Put $g_j=\sqrt{{\gamma }_j}$ and $a_j=\mathrm{\nabla }\log {\gamma }_j$. These are globally $C^1$ and $C^{0,\alpha }$, respectively, and $a_j$ has support in $\overline{{\mathrm{\Omega }}^{\mathrm{\prime }}}$. Their supremum and Hölder bounds are uniform: the positive lower bound $m_{\ast }:=\min \{m,1\}$ gives $\mathrm{\mid }{a}_j\mathrm{\mid }\le C_{d}R\mathrm{/}{m}_{\ast }$; on $\mathrm{\mid }x-y\mathrm{\mid }\le 1$, expand $\mathrm{\nabla }{\gamma }_j(x)\mathrm{/}{\gamma }_j(x)-\mathrm{\nabla }{\gamma }_j(y)\mathrm{/}{\gamma }_j(y)$, using the derivative Hölder bound and $\mathrm{\mid }{\gamma }_j(x)-{\gamma }_j(y)\mathrm{\mid }\le \sqrt{d}R\mathrm{\mid }x-y\mathrm{\mid }$; on larger distances use twice the supremum. Thus [F4] applies uniformly. Let $q_j=\frac{1}{2}\mathrm{div}{a}_j+\frac{1}{4}\mathrm{\mid }{a}_j{\mathrm{\mid }}^2$ and $Q=q_1-q_2$, with bilinear distribution pairings as in [F4].

The Liouville form identity is $\int \mathrm{\nabla }v\cdot \mathrm{\nabla }w+⟨q_j,vw⟩=\int {\gamma }_j\mathrm{\nabla }(v\mathrm{/}{g}_j)\cdot \mathrm{\nabla }(w\mathrm{/}{g}_j)$ for $v,w\in H^1$ on a bounded set. Indeed $q_j=g_j^{-1}\mathrm{\Delta }{g}_j$ means $⟨q_j,vw⟩=-\int \mathrm{\nabla }{g}_j\cdot \mathrm{\nabla }(g_j^{-1}vw)$; expanding this derivative gives the displayed identity term by term. All products are integrable since $g_j,g_j^{-1},\mathrm{\nabla }{g}_j$ are bounded and the factors are in $H^1$. The identity extends from smooth factors by local $H^1$ approximation; multiplication by a bounded $C^1$ function with bounded derivative is continuous on $H^1$, as follows from the product rule and the same approximation. Since $g_j=1$ near $\mathrm{\partial }\mathrm{\Omega }$, the trace of $U_j$ equals that of $u_j:=U_j\mathrm{/}{g}_j$. Testing the form identity with $w=g_{j}z$, first for compactly supported tests and then in $H_0^1$ by density, shows that each Schrödinger solution $U_j$ yields a weak conductivity solution $u_j$. Define the symmetric bilinear forms $B_j(V,W):={\int }_{\mathrm{\Omega }}\mathrm{\nabla }V\cdot \mathrm{\nabla }W+⟨q_j,VW⟩$ and $b_j(v,w):={\int }_{\mathrm{\Omega }}{\gamma }_j\mathrm{\nabla }v\cdot \mathrm{\nabla }w$. Let $v_1$ be the ${\gamma }_1$-harmonic representative with trace $f_2$. Then $B_1(U_1,U_2)=b_1(u_1,U_2\mathrm{/}{g}_1)=b_1(u_1,v_1)$, because $U_2\mathrm{/}{g}_1$ has trace $f_2$ and its difference from $v_1$ is in $H_0^1$, which is annihilated by the weak equation for $u_1$. This equals $(({\mathrm{\Lambda }}_{{\gamma }_1}{f}_1)(\overline{f_2}))$ by the weak DN definition [F2]. By symmetry, $B_2(U_1,U_2)=B_2(U_2,U_1)=b_2(u_2,U_1\mathrm{/}{g}_2)=b_2(u_2,v_2)$, where $v_2$ is the ${\gamma }_2$-harmonic representative with trace $f_1$; Since $b_2$ is symmetric for real ${\gamma }_2$, this equals $b_2(v_2,u_2)=(({\mathrm{\Lambda }}_{{\gamma }_2}{f}_1)(\overline{f_2}))$. Hence subtraction gives $⟨Q,U_1{U}_2⟩=(({\mathrm{\Lambda }}_{{\gamma }_1}-{\mathrm{\Lambda }}_{{\gamma }_2})f_1)(\overline{f_2})$. The compact support of $Q$ makes its whole-space pairing identical to its pairing on $\mathrm{\Omega }$. Consequently $\mathrm{\mid }⟨Q,U_1{U}_2⟩\mathrm{\mid }\le \delta \mathrm{\parallel }{u}_1{\mathrm{\parallel }}_{H^1(\mathrm{\Omega })}\mathrm{\parallel }{u}_2{\mathrm{\parallel }}_{H^1(\mathrm{\Omega })}$.

Use the paired phases ${\zeta }_{\pm }=s(\pm \eta )+i(k\mathrm{/}2\pm \sqrt{s^2-\mathrm{\mid }k{\mathrm{\mid }}^2\mathrm{/}4}{\eta }^{\perp })$ of [F4]. For fixed $s$ their remainders satisfy $\mathrm{\parallel }{r}_j{\mathrm{\parallel }}_{{\dot{X}}_{{\zeta }_j}^{1\mathrm{/}2}}\le C\sqrt{s}$. To verify this last uniform bound, the symbol inequality $\mathrm{\mid }\xi {\mathrm{\mid }}^2\mathrm{/}(s+\mathrm{\mid }{p}_{{\zeta }_j}(\xi )\mathrm{\mid })\le Cs$ follows by splitting $\mathrm{\mid }\xi \mathrm{\mid }\le 8s$ and its complement. Apply the negative cutoff bound in [F3] to $\mathrm{div}{a}_j$ and $\mathrm{\mid }{a}_j{\mathrm{\mid }}^2$ as in [F4] to obtain $\mathrm{\parallel }{q}_j{\mathrm{\parallel }}_{{\dot{X}}_{{\zeta }_j}^{-1\mathrm{/}2}}\le C\sqrt{s}$. The two positive cutoff bounds then give $\mathrm{\parallel }{r}_j{\mathrm{\parallel }}_{L^2(B)}\le C$ and $\mathrm{\parallel }\mathrm{\nabla }{r}_j{\mathrm{\parallel }}_{L^2(B)}\le Cs$. Multiplication by $g_j^{-1}{e}^{x\cdot {\zeta }_j}$ yields $\mathrm{\parallel }{u}_j{\mathrm{\parallel }}_{H^1(\mathrm{\Omega })}\le C{e}^{Cs}s$. Thus step 2.1 bounds the exact interior product by $C\delta e^{Cs}{s}^2$.

Fix a compactly supported smooth cutoff $\varphi =1$ near $\overline{{\mathrm{\Omega }}^{\mathrm{\prime }}}$. We first prove the modulated cutoff bound used for each linear remainder term. Let $u$ be Schwartz and write the Fourier convolution for $\varphi e^{ik\cdot x}u$ with $h=\xi -\eta -k$. The symbol comparison and characteristic-sphere estimate proved in [F3] give, uniformly in the center $y$, $\int ⟨\xi -y{⟩}^{-L}(1+D_{\zeta }(\xi {)}^{-1})d\xi \le C_L$ for sufficiently large $L$, and $s+\mathrm{\mid }{p}_{\zeta }(\eta )\mathrm{\mid }\le C(s+\mathrm{\mid }{p}_{\zeta }(\xi )\mathrm{\mid })⟨\xi -\eta {⟩}^2$, where $D_{\zeta }$ is distance to the characteristic sphere of $p_{\zeta }$. Since $s\mathrm{/}\mathrm{\mid }{p}_{\zeta }(\xi )\mathrm{\mid }\le C(1+D_{\zeta }(\xi {)}^{-1})$, these estimates imply ${\sup }_{\eta }\int \mathrm{\mid }\widehat{\varphi }(\xi -\eta -k)\mathrm{\mid }\frac{s+\mathrm{\mid }{p}_{\zeta }(\eta )\mathrm{\mid }}{\mathrm{\mid }{p}_{\zeta }(\xi )\mathrm{\mid }}d\xi \le C⟨k{⟩}^2$. Indeed, after substituting $h=\xi -\eta -k$, the integrand is bounded by $C\mathrm{\mid }\widehat{\varphi }(h)\mathrm{\mid }⟨k+h{⟩}^2(1+D_{\zeta }(\eta +k+h{)}^{-1})$; use $⟨k+h{⟩}^2\le C⟨k{⟩}^2⟨h{⟩}^2$, the rapid decay of $\widehat{\varphi }$, and the uniform-center estimate. The same weighted Schur argument as in [F3] gives $\mathrm{\parallel }\varphi e^{ik\cdot x}u{\mathrm{\parallel }}_{{\dot{X}}_{\zeta }^{-1\mathrm{/}2}}\le C⟨k{⟩}^2\mathrm{\parallel }u{\mathrm{\parallel }}_{X_{\zeta }^{-1\mathrm{/}2}}$. Applying the dual argument of [F3] to $\bar{\varphi }{e}^{-ik\cdot x}$ yields $\mathrm{\parallel }\varphi e^{ik\cdot x}v{\mathrm{\parallel }}_{X_{\zeta }^{1\mathrm{/}2}}\le C⟨k{⟩}^2\mathrm{\parallel }v{\mathrm{\parallel }}_{{\dot{X}}_{\zeta }^{1\mathrm{/}2}}$, and hence also the same bound with the homogeneous positive-half norm on the left. Density in the weighted Fourier completions follows by truncating away from the characteristic sphere, as in [F3]. Since $2\le N=d+10$, this proves the modulation estimate used below with the stated exponent. Because $q_j$ is real, ${\widehat{q}}_j(-\xi )=\overline{{\widehat{q}}_j(\xi )}$, and $p_{-\zeta }(\xi )=p_{\zeta }(-\xi )$, so $\mathrm{\parallel }{q}_j{\mathrm{\parallel }}_{{\dot{X}}_{-\zeta }^{-1\mathrm{/}2}}=\mathrm{\parallel }{q}_j{\mathrm{\parallel }}_{{\dot{X}}_{\zeta }^{-1\mathrm{/}2}}$. Weighted Fourier Cauchy–Schwarz in the bilinear pairing therefore gives $\mathrm{\mid }⟨q_j,e^{ik\cdot x}{r}_l⟩\mathrm{\mid }\le C⟨k{⟩}^N\mathrm{\parallel }{q}_j{\mathrm{\parallel }}_{{\dot{X}}_{{\zeta }_l}^{-1\mathrm{/}2}}\mathrm{\parallel }{r}_l{\mathrm{\parallel }}_{{\dot{X}}_{{\zeta }_l}^{1\mathrm{/}2}}$. For the quadratic term, expand the definition of $q_j$ on $e^{ik\cdot x}{r}_1{r}_2$ and use [F3] on each factor with its own phase. The product and differentiated product have $L^1$ norms at most $C{s}^{-1}$ and $C$, respectively, times the two remainder norms. The derivative of the exponential adds $C\mathrm{\mid }k\mathrm{\mid }{s}^{-1}$, giving $\mathrm{\mid }⟨q_j,e^{ik\cdot x}{r}_1{r}_2⟩\mathrm{\mid }\le C(1+\mathrm{\mid }k\mathrm{\mid })\mathrm{\parallel }{r}_1{\mathrm{\parallel }}_{{\dot{X}}_{{\zeta }_1}^{1\mathrm{/}2}}\mathrm{\parallel }{r}_2{\mathrm{\parallel }}_{{\dot{X}}_{{\zeta }_2}^{1\mathrm{/}2}}$.

Since ${\zeta }_1+{\zeta }_2=ik$, expand $U_1{U}_2=e^{ik\cdot x}(1+r_1+r_2+r_1{r}_2)$ in step 3.1. Move the three remainder terms to the other side and apply step 4.1. Average over $\lambda \le s\le 2\lambda$ and the unit circle in the fixed plane $P\perp k$, dividing by its total measure $2\pi \lambda$. The left side $\mathrm{\mid }⟨Q,e^{ik\cdot x}⟩\mathrm{\mid }$ is independent of these parameters. For each linear or quadratic product, integral Cauchy–Schwarz and [F4] bound its average by $C⟨k{⟩}^N(1+⟨k{⟩}^2\mathrm{/}\lambda ){\lambda }^{-\beta }$, where $\beta =\alpha \mathrm{/}(1+\alpha )$. The same bound applies when the potential $q_j$ and remainder $r_l$ have different indices, because [F4] holds for either coefficient and either phase. Absorbing the fixed Fourier normalization gives $\mathrm{\mid }\widehat{Q}(k)\mathrm{\mid }\le C\delta e^{C\lambda }{\lambda }^2+C⟨k{⟩}^N(1+⟨k{⟩}^2\mathrm{/}\lambda ){\lambda }^{-\beta }$ whenever $\lambda \ge \max \{s_0,100⟨k⟩\}$. Replacing $k$ by $-k$ if needed accounts for the negative-sign Fourier convention.

Define $\mathcal{Q}:=\mathrm{\parallel }(1+\mathrm{\mid }\xi {\mathrm{\mid }}^2{)}^{-1\mathrm{/}2}\widehat{Q}(\xi ){\mathrm{\parallel }}_2$. This is finite since $Q=\frac{1}{2}\mathrm{div}(a_1-a_2)+\frac{1}{4}(\mathrm{\mid }{a}_1{\mathrm{\mid }}^2-\mathrm{\mid }{a}_2{\mathrm{\mid }}^2)$ with both coefficient fields in $L^2$. For $t\ge 1$, its low-frequency contribution is at most $C{t}^{d\mathrm{/}2}{\sup }_{\mathrm{\mid }k\mathrm{\mid }\le t}\mathrm{\mid }\widehat{Q}(k)\mathrm{\mid }$. The high-frequency contribution of the zero-order field is at most $C{t}^{-1}$. For the divergence field $f=a_1-a_2$, split $f={\rho }_h\ast f+(f-{\rho }_h\ast f)$ with a fixed nonnegative unit-mass mollifier. Plancherel bounds its high-frequency contribution by $C{t}^{-1}\mathrm{\parallel }\mathrm{div}({\rho }_h\ast f){\mathrm{\parallel }}_2+\mathrm{\parallel }f-{\rho }_h\ast f{\mathrm{\parallel }}_2\le C(t^{-1}{h}^{-1}+h^{\alpha })$. The last two bounds follow by differentiating the convolution kernel and by the uniform Hölder estimate with fixed compact support, exactly as in [F4]. Taking $h=t^{-1\mathrm{/}(1+\alpha )}$ gives $\mathcal{Q}\le C[t^{d\mathrm{/}2}{\sup }_{\mathrm{\mid }k\mathrm{\mid }\le t}\mathrm{\mid }\widehat{Q}(k)\mathrm{\mid }+t^{-\beta }]$.

Choose $A=2+(N+d+1)\mathrm{/}\beta$ and a fixed $L\ge \max \{s_0,200\}$. Set $\lambda =L{t}^A$. Then $\lambda \ge 100\sqrt{1+t^2}$, $1+⟨k{⟩}^2\mathrm{/}\lambda \le C$ for $\mathrm{\mid }k\mathrm{\mid }\le t$, and $t^{N+d\mathrm{/}2}{\lambda }^{-\beta }\le C{t}^{-\beta }$. Substituting step 5.1 in step 6.1 therefore gives $\mathcal{Q}\le C[\delta e^{D{t}^A}+t^{-\beta }]$ for a fixed $D>0$, after absorbing the polynomial factor $t^{d\mathrm{/}2}{\lambda }^2$ into the exponential. For sufficiently small $0<\delta <t_0$, take $t=(\log (1\mathrm{/}\delta )\mathrm{/}(2D){)}^{1\mathrm{/}A}\ge 1$. The first term becomes $\sqrt{\delta }$, and the second is $C[\log (1\mathrm{/}\delta ){]}^{-\beta \mathrm{/}A}$. The former is bounded by a constant times the latter, since $e^{-x\mathrm{/}2}{x}^{\beta \mathrm{/}A}$ is bounded for $x\ge 1$. Hence $\mathcal{Q}\le C[\log (1\mathrm{/}\delta ){]}^{-\rho }$ with $\rho =\beta \mathrm{/}A>0$. If $\delta =0$, let $t\to \mathrm{\infty }$ in the estimate before this choice to obtain $\mathcal{Q}=0$.

Put $w=\log g_1-\log g_2$ and $H=g_1{g}_2$. Both $w$ and $\mathrm{\nabla }w$ have compact support in $\overline{{\mathrm{\Omega }}^{\mathrm{\prime }}}$, and $w\in H_0^1(B)$. Expanding the formula for $Q$ gives the distribution identity $HQ=\mathrm{div}(H\mathrm{\nabla }w)$: the divergence part is $\mathrm{\Delta }w$, while the zero-order part is $\mathrm{\nabla }\log H\cdot \mathrm{\nabla }w$. Test this identity with $w$ to obtain $\int H\mathrm{\mid }\mathrm{\nabla }w{\mathrm{\mid }}^2=-⟨Q,Hw⟩$. The test is legitimate by $H^1$ approximation, since $Q$ is an $L^2$ field plus a divergence of an $L^2$ field and $Hw\in H_0^1(B)$. Weighted Fourier Cauchy–Schwarz gives $\mathrm{\mid }⟨Q,Hw⟩\mathrm{\mid }\le \mathcal{Q}\mathrm{\parallel }Hw{\mathrm{\parallel }}_{H^1({\mathbb{R}}^d)}\le C\mathcal{Q}\mathrm{\parallel }w{\mathrm{\parallel }}_{H^1(B)}$. Since $H\ge m_{\ast }>0$, Poincaré in [F5] implies $\mathrm{\parallel }w{\mathrm{\parallel }}_{H^1(B)}\le C\mathcal{Q}$; if this norm is zero the estimate is immediate, otherwise divide the resulting quadratic inequality by it. Finally the mean value theorem on the bounded interval of possible $\log g_j$ gives $\mathrm{\mid }{\gamma }_1-{\gamma }_2\mathrm{\mid }=\mathrm{\mid }{e}^{2\log g_1}-e^{2\log g_2}\mathrm{\mid }\le 2\max \{M,1\}\mathrm{\mid }w\mathrm{\mid }$. Thus $\mathrm{\parallel }{\gamma }_1-{\gamma }_2{\mathrm{\parallel }}_{L^2(\mathrm{\Omega })}\le C\mathcal{Q}$. In particular $\delta =0$ forces equality of the coefficients.

Combine steps 7.1–8.1 for $0\le \delta <t_0$, reducing $t_0$ below $1\mathrm{/}2$ if necessary. For $\delta \ge t_0$, use $\mathrm{\parallel }{\gamma }_1-{\gamma }_2{\mathrm{\parallel }}_2\le 2M\mathrm{\mid }\mathrm{\Omega }{\mathrm{\mid }}^{1\mathrm{/}2}$ and enlarge $C$ to at least $2M\mathrm{\mid }\mathrm{\Omega }{\mathrm{\mid }}^{1\mathrm{/}2}[\log (1\mathrm{/}{t}_0){]}^{\rho }$. This is precisely the saturated modulus in the statement. Every bound used only the fixed class parameters, the geometry and $d,\alpha$, and the zero-data case was proved in step 8.1.