Centered NtD data control the exact CGO identity with its modulation factor

The energy argument in [F1] applies to every mean-zero $f\in L^2(S^2)$: the functional $v\mapsto \int f\overline{\mathrm{Tr}v}$ is bounded on the boundary-mean-zero $H^1$ subspace by [F4], and its uniformly coercive conductivity form has a unique solution with norm at most $C\mathrm{\parallel }f{\mathrm{\parallel }}_2$. This defines a bounded map ${\mathcal{N}}_j:L_0^2\to L_0^2$. Its centered sequence columns are those in [F1], so their difference has finite squared sum. For a finite current expansion $f=\sum c_{\mathrm{\ell }k}{Y}_{\mathrm{\ell }k}$, Cauchy–Schwarz gives $\mathrm{\parallel }({\mathcal{N}}_1-{\mathcal{N}}_2)f{\mathrm{\parallel }}_2\le \delta (\sum \mathrm{\mid }{c}_{\mathrm{\ell }k}{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}$. Density and boundedness extend this to every $f\in L_0^2$. Finally the bilinear symmetry $\int f{\mathcal{N}}_{j}h=\int h{\mathcal{N}}_{j}f$ follows by testing the two weak equations against one another, using real coefficients and complex bilinear extension.

Put $u_j=U_j\mathrm{/}{g}_j$. The weak Liouville identity gives $\mathrm{div}({\sigma }_j\mathrm{\nabla }{u}_j)=0$: expand ${\sigma }_j\mathrm{\nabla }(U_j\mathrm{/}{g}_j)=g_j\mathrm{\nabla }{U}_j-U_j\mathrm{\nabla }{g}_j$, whose divergence is $g_j\mathrm{\Delta }{U}_j-U_j\mathrm{\Delta }{g}_j=0$. Products are legitimate by [F2]–[F3]; one may also verify the equality first on smooth approximations and pass in distributions. Since $g_j=1$ on a full collar of $S^2$, the boundary normal flux of $u_j$ is exactly $f_j={\mathrm{\partial }}_{\nu }{U}_j$. It belongs to $L^2$ because $U_j\in H^2(B_2)$ and [F4] applies to each first derivative. The divergence theorem [F5] gives the Green identity for smooth approximants; density [F6] and trace continuity [F4] pass it to the $H^2$ CGO solutions on $B_1$. On the collar $U_j$ is harmonic and the coefficient is identically one. Taking the test function one shows $\int f_j=0$. Thus $\mathrm{Tr}{u}_j={\mathcal{N}}_j{f}_j+c_j$ for a constant $c_j$.

Subtract the two bilinear Schrödinger Green identities [F5]–[F6], testing the $j=1$ equation against $U_2$ and the $j=2$ equation against $U_1$. Their gradient terms cancel, giving $\int Q{U}_1{U}_2={\int }_{S^2}(f_1\mathrm{Tr}{U}_2-f_2\mathrm{Tr}{U}_1)$. Since $U_j=u_j$ on the boundary, step 2.1 identifies these traces with ${\mathcal{N}}_j{f}_j+c_j$. The constant terms disappear because the fluxes have mean zero; the symmetry from step 1.1 gives exactly the displayed centered NtD identity. All volume terms are integrable since $Q\in L^2$ and $U_j\in H^2(B_2)\subset L^{\mathrm{\infty }}(B_1)$; boundedness follows, for example, by weighted Fourier Cauchy–Schwarz after multiplication by an interior cutoff.

If $\mathrm{\mid }{\zeta }_j\mathrm{\mid }=\sqrt{2}t$, [F3] and differentiation of the exponential give $\mathrm{\parallel }{U}_j{\mathrm{\parallel }}_{H^2(B_1)}\le C{e}^{Ct}{t}^3$ for $t\ge T$. The trace bound on first derivatives then gives $\mathrm{\parallel }{f}_j{\mathrm{\parallel }}_2\le C{e}^{Ct}$, increasing $C$ to absorb the polynomial. The operator-norm bound in step 1.1 and Cauchy–Schwarz in the bilinear boundary pairing yield $\mathrm{\mid }\int Q{U}_1{U}_2\mathrm{\mid }\le C\delta e^{Ct}$.

For smooth compactly supported $w$ put $[w{]}_{1\mathrm{/}2}^2:=\iint \mathrm{\mid }w(x)-w(y){\mathrm{\mid }}^2\mathrm{\mid }x-y{\mathrm{\mid }}^{-4}dxdy$. Translation, Plancherel and Tonelli [F7] give $[w{]}_{1\mathrm{/}2}^2=\int \mathrm{\mid }\widehat{w}(\xi ){\mathrm{\mid }}^2(\int \frac{\mathrm{\mid }{e}^{i\xi \cdot h}-1{\mathrm{\mid }}^2}{\mathrm{\mid }h{\mathrm{\mid }}^4}dh)d\xi =c_3\int \mathrm{\mid }\xi \mathrm{\mid }\mathrm{\mid }\widehat{w}(\xi ){\mathrm{\mid }}^{2}d\xi ,$ where rotation and dilation give the inner integral, and $0<c_3<\mathrm{\infty }$ follows from $\mathrm{\mid }{e}^{iu}-1\mathrm{\mid }\le \min \{2,\mathrm{\mid }u\mathrm{\mid }\}$. Thus $\mathrm{\parallel }w{\mathrm{\parallel }}_{H^{1\mathrm{/}2}}^2\asymp \mathrm{\parallel }w{\mathrm{\parallel }}_2^2+[w{]}_{1\mathrm{/}2}^2$. Restrict each $H^2(B_2)$ remainder to $B_1$ and use [F6] to approximate it there by ambient smooth functions; multiplication by a cutoff supported in $B_1$ and [F8] extend the products to $H^2({\mathbb{R}}^3)$, so the norm comparison extends to these cutoff remainders. For bounded compactly supported factors the identity $f(x)g(x)-f(y)g(y)=f(x)(g(x)-g(y))+g(y)(f(x)-f(y))$ then yields $\mathrm{\parallel }fg{\mathrm{\parallel }}_{H^{1\mathrm{/}2}}\le C(\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }g{\mathrm{\parallel }}_{H^{1\mathrm{/}2}}+\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }f{\mathrm{\parallel }}_{H^{1\mathrm{/}2}})$. For $\xi \mathrm{\ne }0$, choose real orthonormal $d_1,d_2$ perpendicular to $\xi$ and set ${\zeta }_1=-\xi \mathrm{/}2+\sqrt{t^2-\mathrm{\mid }\xi {\mathrm{\mid }}^2\mathrm{/}4}{d}_2+it{d}_1$, ${\zeta }_2=-\xi \mathrm{/}2-\sqrt{t^2-\mathrm{\mid }\xi {\mathrm{\mid }}^2\mathrm{/}4}{d}_2-it{d}_1$; for $\xi =0$ choose any orthonormal pair. Direct expansion gives ${\zeta }_j\cdot {\zeta }_j=0$, ${\zeta }_1+{\zeta }_2=-\xi$ and $\mathrm{\mid }{\zeta }_j\mathrm{\mid }=\sqrt{2}t$. Hence $U_1{U}_2=e^{-i\xi \cdot x}(1+r_1+r_2+r_1{r}_2)$. Choose cutoffs $\chi ,\eta \in C_c^{\mathrm{\infty }}(B_1)$ with $\chi =1$ on $B_{\rho }$ and $\eta =1$ on $\mathrm{supp}\chi$. Applying the product estimate to $\chi r_j$ and $\eta r_j$, and [F3], gives $\mathrm{\parallel }\chi (r_1+r_2+r_1{r}_2){\mathrm{\parallel }}_{H^{1\mathrm{/}2}({\mathbb{R}}^3)}\le C\mathrm{/}t$. Fourier translation and $⟨\eta +\xi ⟩\le \sqrt{2}⟨\eta ⟩⟨\xi ⟩$ then give $\mathrm{\parallel }{e}^{-i\xi \cdot x}\chi (r_1+r_2+r_1{r}_2){\mathrm{\parallel }}_{H^{1\mathrm{/}2}}\le C⟨\xi {⟩}^{1\mathrm{/}2}\mathrm{/}t$.

Extract the Fourier coefficient of $Q$ from the exact product identity in step 5.1. Its remainder is precisely the signed pairing of $Q$ with $e^{-i\xi \cdot x}\chi (r_1+r_2+r_1{r}_2)$, not with the unmodulated remainder. Weighted Fourier Cauchy–Schwarz bounds its absolute value by $CN⟨\xi {⟩}^{1\mathrm{/}2}\mathrm{/}t$. Combine this with step 4.1 and absorb the fixed unitary Fourier factor $(2\pi {)}^{-3\mathrm{/}2}$ into $C$. This proves the stated estimate with its modulation factor retained.