Banach range invariance and EIT rates

Use the angular-frequency convention $\widehat{f}(\xi )=(2\pi {)}^{-3\mathrm{/}2}\int e^{-ix\cdot \xi }f(x)dx$. The $2\pi$-frequency results in [F1] transfer by $\widehat{f}(\xi )=(2\pi {)}^{-3\mathrm{/}2}({\mathcal{F}}_{2\pi }f)(\xi \mathrm{/}(2\pi ))$, with the same change of variables on distributional tests; hence $\widehat{D^{\alpha }f}=(i\xi {)}^{\alpha }\widehat{f}$ and the transform remains unitary. The fixed frequency dilation changes only equivalence constants in the dyadic norms of [F2]. We record explicitly the smooth-multiplication bound for the stated Besov convention. Write $Q_j={\mathbf{1}}_{A_j}(D)$. For $\chi \in C_c^{\mathrm{\infty }}({\mathbb{R}}^3)$, $Q_j\chi Q_k$ has Fourier kernel $(2\pi {)}^{-3\mathrm{/}2}{\mathbf{1}}_{A_j}(\xi )\widehat{\chi }(\xi -\eta ){\mathbf{1}}_{A_k}(\eta )$. If $\mathrm{\mid }j-k\mathrm{\mid }\le 2$, Young's convolution inequality [F3] bounds its $L^2$ operator norm by a constant times $\mathrm{\parallel }\widehat{\chi }{\mathrm{\parallel }}_1$. If $\mathrm{\mid }j-k\mathrm{\mid }>2$, then $\mathrm{\mid }\xi -\eta \mathrm{\mid }\ge c{2}^{\max (j,k)}$ on these supports. Repeated integration by parts in the Fourier integral of $\chi$ gives $\mathrm{\mid }\widehat{\chi }(\zeta )\mathrm{\mid }\le C_N(1+\mathrm{\mid }\zeta \mathrm{\mid }{)}^{-N}$. Cauchy–Schwarz in the two frequency variables therefore bounds the operator norm by $C_N{2}^{-N\max (j,k)}{2}^{3(j+k)\mathrm{/}2}$. For any real $r$, multiply these bounds by $2^{jr-kr}$ and sum over $k$. Choosing $N>2\mathrm{\mid }r\mathrm{\mid }+8$ makes both off-diagonal geometric sums uniformly bounded in $j$, while the near-diagonal sum has at most five terms. Hence $\mathrm{\parallel }\chi f{\mathrm{\parallel }}_{B_{2,\mathrm{\infty }}^r}\le C_{r,\chi }\mathrm{\parallel }f{\mathrm{\parallel }}_{B_{2,\mathrm{\infty }}^r}$. These calculations first apply to finite block sums; the same estimates give their convergence in distributions, since a Schwartz test has faster-than-polynomial Fourier decay.

The sharp annular definition agrees, up to constants, with any usual smooth inhomogeneous dyadic definition at $p=2$. Indeed a sharp annulus meets at most a fixed number of the smooth supports at neighboring scales. On that annulus, the identity $\sum {\phi }_j=1$ and Cauchy–Schwarz give its $L^2$ norm at most the sum of those finitely many smooth-block $L^2$ norms; conversely each smooth symbol has magnitude at most one and support in finitely many sharp annuli. The weights $2^{jr}$ differ by only a constant at neighboring scales. The low blocks also meet only finitely many supports. This proves both Besov norm comparisons. For Hilbert Sobolev norms, [F1] gives directly $\mathrm{\parallel }f{\mathrm{\parallel }}_{H^r}^2\asymp {\sum }_j{2}^{2jr}\mathrm{\parallel }{Q}_{j}f{\mathrm{\parallel }}_2^2$, consistently with [F2].

Choose ${\chi }_0\in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$ equal to one near $\overline{B_{\rho }}$, and choose $\eta \in C_c^{\mathrm{\infty }}({\mathbb{R}}^3)$ equal to one near $\bar{\mathrm{\Omega }}$. Let ${\tilde{\sigma }}^{†}$ be one whole-space $B_{2,\mathrm{\infty }}^t$ extension, whose existence is part of the restriction-space hypothesis. The compactly supported distribution $f^{†}:={\chi }_0({\tilde{\sigma }}^{†}-1)$ is in $B_{2,\mathrm{\infty }}^t({\mathbb{R}}^3)$ by step 1.1 (the constant term multiplied by ${\chi }_0$ is smooth). It equals ${\sigma }^{†}-1$ on $\mathrm{\Omega }$ and vanishes outside $\overline{B_{\rho }}$, since the conductivity equals one in the collar. Thus $v^{†}=\eta +f^{†}$ is a compactly supported $B_{2,\mathrm{\infty }}^t$ extension of ${\sigma }^{†}$. In particular $v^{†}\in H^s({\mathbb{R}}^3)$: the squared block sum is bounded by a constant times ${\sum }_j{2}^{-2j(t-s)}<\mathrm{\infty }$.

For $e=\sigma -{\sigma }^{†}$, let $e_0$ be its zero extension. Its support lies in $\overline{B_{\rho }}$. Zero extension creates no boundary derivatives because $e$ vanishes on an open collar: local distributional tests near $\mathrm{\partial }\mathrm{\Omega }$ see zero, and interior tests give exactly the zero extensions of the weak derivatives. For $s=m+\theta$ noninteger, the whole-space top-order difference integral equals the integral over ${\mathrm{\Omega }}^2$ plus $2{\int }_{\mathrm{\Omega }}\mathrm{\mid }{D}^{\alpha }e(x){\mathrm{\mid }}^2{\int }_{{\mathbb{R}}^3\setminus \mathrm{\Omega }}\mathrm{\mid }x-y{\mathrm{\mid }}^{-3-2\theta }dydx$. Where $D^{\alpha }e$ can be nonzero, the inner integral is bounded by ${\int }_{\mathrm{\mid }z\mathrm{\mid }\ge 1-\rho }\mathrm{\mid }z{\mathrm{\mid }}^{-3-2\theta }dz<\mathrm{\infty }$. The whole-space difference norm is therefore bounded by $C\mathrm{\parallel }e{\mathrm{\parallel }}_{s,\mathrm{\Omega }}$, and the integer case is immediate. The translation identity $\widehat{g(\cdot +h)}(\xi )=e^{i\xi \cdot h}\widehat{g}(\xi )$ follows by substitution for Schwartz functions and then by $L^2$ density and unitarity for every $g\in L^2$. Thus [F1] and Tonelli [F4] give $\int \int \mathrm{\mid }g(x+h)-g(x){\mathrm{\mid }}^2\mathrm{\mid }h{\mathrm{\mid }}^{-3-2\theta }dxdh=\int \mathrm{\mid }\widehat{g}(\xi ){\mathrm{\mid }}^2\int \mathrm{\mid }{e}^{i\xi \cdot h}-1{\mathrm{\mid }}^2\mathrm{\mid }h{\mathrm{\mid }}^{-3-2\theta }dhd\xi$. The inner integral equals $c_{\theta }\mathrm{\mid }\xi {\mathrm{\mid }}^{2\theta }$ by rotation and dilation; $0<c_{\theta }<\mathrm{\infty }$ follows by the bounds $\mathrm{\mid }{e}^{iu}-1\mathrm{\mid }\le \min \{2,\mathrm{\mid }u\mathrm{\mid }\}$. Summing derivative multipliers shows that this whole-space difference norm is equivalent to $\mathrm{\parallel }{e}_0{\mathrm{\parallel }}_{H^s}$. Thus $\mathrm{\parallel }{e}_0{\mathrm{\parallel }}_{H^s}\le C\mathrm{\parallel }e{\mathrm{\parallel }}_{s,\mathrm{\Omega }}$, while exactly $\mathrm{\parallel }{e}_0{\mathrm{\parallel }}_{H^1}=\mathrm{\parallel }e{\mathrm{\parallel }}_{H^1(\mathrm{\Omega })}$.

Let $B_s(v,w)$ be the whole-space intrinsic bilinear form associated with the derivative and difference norm in step 3.2. Its Fourier multiplier $p_s(\xi )$ equals ${\sum }_{\mathrm{\mid }\alpha \mathrm{\mid }\le m}{\xi }^{2\alpha }+c_{\theta }{\sum }_{\mathrm{\mid }\alpha \mathrm{\mid }=m}{\xi }^{2\alpha }\mathrm{\mid }\xi {\mathrm{\mid }}^{2\theta }$ in the noninteger case, and just the first polynomial in the integer case. Thus $\mathrm{\mid }{p}_s(\xi )\mathrm{\mid }\le C⟨\xi {⟩}^{2s}$, and $P_0:=p_s(D)v^{†}$ belongs to $B_{2,\mathrm{\infty }}^{t-2s}$ by the annular multiplier bound and step 3.1. Its distributional pairing with $e_0$ is $B_s(v^{†},e_0)$. The pairing is defined by convergent Fourier blocks: high blocks have the product bound $C{2}^{-j(t-s)}\mathrm{\parallel }{e}_0{\mathrm{\parallel }}_{H^s}$, a summable sequence. This also justifies the bilinear identity by truncation and limits, without assuming classical derivatives of order $2s$.

In the integer case the whole-space and domain pairings are identical, since every derivative of $e_0$ is supported in $\overline{B_{\rho }}$; take $P=P_0$. For noninteger $s=m+\theta$, their difference is $B_s(v^{†},e_0)-({\sigma }^{†},e{)}_{s,\mathrm{\Omega }}=2{\sum }_{\mathrm{\mid }\alpha \mathrm{\mid }=m}{\int }_{\mathrm{\Omega }}{D}^{\alpha }e(x)(A(x)D^{\alpha }{v}^{†}(x)-b_{\alpha }(x))dx,$ where $A(x)={\int }_{{\mathrm{\Omega }}^c}\mathrm{\mid }x-y{\mathrm{\mid }}^{-3-2\theta }dy$ and $b_{\alpha }(x)={\int }_{{\mathrm{\Omega }}^c}{D}^{\alpha }\eta (y)\mathrm{\mid }x-y{\mathrm{\mid }}^{-3-2\theta }dy$. To obtain the formula, split each whole-space double integral into ${\mathrm{\Omega }}^2$, $\mathrm{\Omega }\times {\mathrm{\Omega }}^c$ and its transpose; $D^{\alpha }{e}_0$ is zero outside and $v^{†}=\eta$ there. On any fixed neighborhood of $\overline{B_{\rho }}$ compact in $\mathrm{\Omega }$, every derivative of $A,b_{\alpha }$ is bounded, by differentiating the separated kernel and using its integrable tail. Choose $\chi \in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$ equal to one on that neighborhood. The domain pairing is therefore the pairing of $e_0$ with $P=P_0-2{\sum }_{\mathrm{\mid }\alpha \mathrm{\mid }=m}(-1{)}^m{D}^{\alpha }(\chi A{D}^{\alpha }{v}^{†}-\chi b_{\alpha })\mathrm{.}$ By step 1.1, its correction terms lie in $B_{2,\mathrm{\infty }}^{t-2m}$: differentiation lowers the block regularity by the derivative order, multiplication by the smooth compact coefficient preserves it, and $\chi b_{\alpha }$ is smooth and compactly supported. Since $2m\le 2s$, they also lie in $B_{2,\mathrm{\infty }}^{t-2s}$. Hence $P\in B_{2,\mathrm{\infty }}^{t-2s}$ in every case, and $⟨P,e_0⟩=({\sigma }^{†},e{)}_{s,\mathrm{\Omega }}$. This treats the exact domain penalty pairing rather than replacing its norm by an equivalent one.

Put $d=t-s>0$ and $a=2s-t-1>0$. Let $P_{\le J}={\sum }_{j\le J}{Q}_{j}P$. By step 5.1 and the Hilbert block norm formula, $\mathrm{\parallel }{P}_{\le J}{\mathrm{\parallel }}_{H^{-1}}^2\le C{\sum }_{j\le J}{2}^{2j(2s-t-1)}\le C{2}^{2Ja}$. Similarly $\mathrm{\parallel }P-P_{\le J}{\mathrm{\parallel }}_{H^{-s}}^2\le C{\sum }_{j>J}{2}^{2j(s-t)}\le C{2}^{-2Jd}$. Weighted Fourier Cauchy–Schwarz and step 3.2 now give $\mathrm{\mid }⟨P,e_0⟩\mathrm{\mid }\le C{2}^{Ja}\mathrm{\parallel }e{\mathrm{\parallel }}_{H^1(\mathrm{\Omega })}+C{2}^{-Jd}\mathrm{\parallel }e{\mathrm{\parallel }}_{s,\mathrm{\Omega }}$, the first claimed estimate.

Expansion of the quadratic form gives exactly $D_R(\sigma ,{\sigma }^{†})=\mathrm{\parallel }e{\mathrm{\parallel }}_{s,\mathrm{\Omega }}^2\mathrm{/}2$. Set $W=\omega (\mathrm{\parallel }G(\sigma )-G({\sigma }^{†})\mathrm{\parallel })$. The conditional stability hypothesis and $uv\le 3u^2\mathrm{/}8+2v^2\mathrm{/}3$ applied to the high-frequency term in step 6.1 yield $-({\sigma }^{†},e{)}_{s,\mathrm{\Omega }}\le 3\mathrm{\parallel }e{\mathrm{\parallel }}_{s,\mathrm{\Omega }}^2\mathrm{/}8+C(2^{Ja}W+2^{-2Jd})$. If $W=0$, stability gives $e=0$. For $0<W\le 1$, take the integer $J\ge 0$ with $2^{-J(a+2d)}\ge W>2^{-(J+1)(a+2d)}$. Then $2^{-2Jd}\le 2^{2d}{W}^{2d\mathrm{/}(a+2d)}$ and $2^{Ja}W\le W^{2d\mathrm{/}(a+2d)}$. Since $a+2d=t-1$, this proves the asserted source inequality with coefficient $3\mathrm{/}4$ on the exact Bregman distance. Substitution of the stated logarithmic stability exponent gives $\nu =2(s-3\mathrm{/}2)(t-s)\mathrm{/}(t-1)$ by multiplying exponents.

We give the Sobolev product bounds used below. They hold for periodic functions on a fixed three-dimensional cube, and for the product of a periodic function and a function with one fixed half-integer Fourier shift. The constants can be chosen uniformly under rotations of the cube. Split Fourier frequencies into dyadic shells and write $f_j,g_j$ for the corresponding blocks. A shell contains at most $C{2}^{3j}$ lattice points, so Cauchy–Schwarz gives $\mathrm{\parallel }{f}_j{\mathrm{\parallel }}_{\mathrm{\infty }}\le C{2}^{3j\mathrm{/}2}\mathrm{\parallel }{f}_j{\mathrm{\parallel }}_2$. For $r>3\mathrm{/}2$ and $0\le s\le r$, these estimates prove $\mathrm{\parallel }fg{\mathrm{\parallel }}_{H^s}\le C\mathrm{\parallel }f{\mathrm{\parallel }}_{H^r}\mathrm{\parallel }g{\mathrm{\parallel }}_{H^s}$. For low $f$ against high $g$, use $\mathrm{\parallel }{f}_{<j-2}{\mathrm{\parallel }}_{\mathrm{\infty }}\le C\mathrm{\parallel }f{\mathrm{\parallel }}_{H^r}$ and square-sum $2^{js}\mathrm{\parallel }{g}_j{\mathrm{\parallel }}_2$. For high $f$ against low $g$, when $s<3\mathrm{/}2$ use $\mathrm{\parallel }{g}_{<j-2}{\mathrm{\parallel }}_{\mathrm{\infty }}\le C{2}^{j(3\mathrm{/}2-s)}\mathrm{\parallel }g{\mathrm{\parallel }}_{H^s}$, and when $s>3\mathrm{/}2$ use the uniform bound by $\mathrm{\parallel }g{\mathrm{\parallel }}_{H^s}$; the critical case has an extra factor $(1+j{)}^{1\mathrm{/}2}$, absorbed by $2^{-j(r-3\mathrm{/}2)}$. The resulting weighted terms are bounded by $2^{-j(r-3\mathrm{/}2)}\mathrm{\parallel }f{\mathrm{\parallel }}_{H^r}\mathrm{\parallel }g{\mathrm{\parallel }}_{H^s}$ or by $2^{js}\mathrm{\parallel }{f}_j{\mathrm{\parallel }}_2\mathrm{\parallel }g{\mathrm{\parallel }}_{H^s}$. For comparable high input shells $k$ that produce an output shell $j\le k+C$, use $\mathrm{\parallel }{Q}_j(f_k{g}_k){\mathrm{\parallel }}_2\le C{2}^{3j\mathrm{/}2}\mathrm{\parallel }{f}_k{\mathrm{\parallel }}_2\mathrm{\parallel }{g}_k{\mathrm{\parallel }}_2$, obtained from the $L^1$ bound on the product and the number of output frequencies. Multiplication by $2^{js}$ and summation over $k\ge j-C$ gives a summable geometric kernel, since $r+s>0$ and $r>3\mathrm{/}2$; at $s=r=0$ this case is not being used. Cauchy–Schwarz for the two input shell sequences and the geometric kernel completes the $H^s$ bound.

Rotate so that $\zeta =s(e_1+i{e}_2)$, where $s=\mathrm{\mid }\zeta \mathrm{\mid }\mathrm{/}\sqrt{2}$, and choose a fixed cube $Q=(-\pi L,\pi L{)}^3$ containing $\overline{B_3}$. Put ${\eta }_k=(k+e_2\mathrm{/}2)\mathrm{/}L$ and use the shifted orthonormal basis $w_k=(2\pi L{)}^{-3\mathrm{/}2}{e}^{i{\eta }_k\cdot x}$. For $A_{\zeta }=D^2+2\zeta \cdot D$, $D=-i\mathrm{\nabla }$, its symbol is $p_k=\mathrm{\mid }{\eta }_k{\mathrm{\mid }}^2+2\zeta \cdot {\eta }_k$, with $\mathrm{\mid }\mathrm{Im}{p}_k\mathrm{\mid }\ge s\mathrm{/}L$. For $\mathrm{\mid }{\eta }_k\mathrm{\mid }\le 4s$, this gives $\mathrm{\mid }{p}_k\mathrm{\mid }\ge s\mathrm{/}L$; for $\mathrm{\mid }{\eta }_k\mathrm{\mid }>4s$, its real part has magnitude at least $\mathrm{\mid }{\eta }_k{\mathrm{\mid }}^2\mathrm{/}2$. Hence for every real $\mu$ and $0\le v\le 2$ the coefficient multiplier $G_{\zeta }:p_k^{-1}$ satisfies $\mathrm{\parallel }{G}_{\zeta }f{\mathrm{\parallel }}_{H^{\mu +v}(Q)}\le C_v{s}^{v-1}\mathrm{\parallel }f{\mathrm{\parallel }}_{H^{\mu }(Q)}$ for $s\ge 1$. The Sobolev spaces here are defined by the shifted weighted coefficient norms. Parseval proves these bounds and $A_{\zeta }{G}_{\zeta }f=f$ in distributions inside $Q$, first on finite sums and then by convergence.

A second bound is $\mathrm{\parallel }fg{\mathrm{\parallel }}_{H^{-1\mathrm{/}2}}\le C_{\epsilon }\mathrm{\parallel }f{\mathrm{\parallel }}_{H^{1\mathrm{/}2+\epsilon }}\mathrm{\parallel }g{\mathrm{\parallel }}_{H^{1\mathrm{/}2}}$ for every $0<\epsilon <1$. Here the low-high and high-low estimates from step 1.1, with output weight $2^{-j\mathrm{/}2}$, have the additional decay $2^{-\epsilon j}$. For comparable high shells, the output estimate becomes $2^j{\sum }_{k\ge j-C}\mathrm{\parallel }{f}_k{\mathrm{\parallel }}_2\mathrm{\parallel }{g}_k{\mathrm{\parallel }}_2=2^{-\epsilon j}{\sum }_{k\ge j-C}{2}^{-(k-j)(1+\epsilon )}{F}_k{G}_k$, with $F_k=2^{k(1\mathrm{/}2+\epsilon )}\mathrm{\parallel }{f}_k{\mathrm{\parallel }}_2$ and $G_k=2^{k\mathrm{/}2}\mathrm{\parallel }{g}_k{\mathrm{\parallel }}_2$. The geometric kernel has finite sum, and Cauchy–Schwarz bounds the product sequence in ${\mathrm{\ell }}^1$ by $\mathrm{\parallel }F{\mathrm{\parallel }}_2\mathrm{\parallel }G{\mathrm{\parallel }}_2$. This proves the claimed bound. All product calculations first hold for finite Fourier sums; the bounds give convergence in the target Sobolev space and identify the product as the distributional limit. Whenever one factor belongs to $H^r$, $r>3\mathrm{/}2$, its Fourier coefficients are absolutely summable by weighted Cauchy–Schwarz, so it is a bounded continuous multiplier and this distributional product agrees with the ordinary product.

The periodic and shifted Sobolev norms of the compactly supported potential are bounded by $C\mathrm{\parallel }q{\mathrm{\parallel }}_{H^{1\mathrm{/}2+\tau }({\mathbb{R}}^3)}\le CM$. For completeness, choose a fixed smooth cutoff equal to one on $B_1$ and supported in $B_2$. Since it multiplies $q$ without changing it, each Fourier coefficient of $q$ is the convolution of its whole-space Fourier transform with a translate of the rapidly decreasing transform of this cutoff, sampled at an integer or shifted lattice point. Weighted Cauchy–Schwarz, followed by summation over the lattice, bounds the coefficient norm by the whole-space norm: the sums ${\sup }_{\xi }{\sum }_k(1+\mathrm{\mid }{\eta }_k-\xi \mathrm{\mid }{)}^{-N}$ and the corresponding integrals are finite for $N>3$, and polynomial Sobolev weights are absorbed by increasing $N$. The reverse local bound needed later follows in the same way by multiplying the Fourier series by a smooth cutoff supported inside $Q$ and equal to one on $B_2$, then estimating the resulting convolution sum in whole-space weighted $L^2$. The cutoffs can be chosen radial before rotation, so all constants are independent of the phase direction.

Put $r_0=3\mathrm{/}2+\tau \mathrm{/}2$ and $h_0=r_0-1=1\mathrm{/}2+\tau \mathrm{/}2$. Split the periodic potential into $q=q_{\mathrm{♯}}+q_{\mathrm{♭}}$ by retaining frequencies of size at most $2^J$ in $q_{\mathrm{♯}}$. Then $\mathrm{\parallel }{q}_{\mathrm{♭}}{\mathrm{\parallel }}_{H^{h_0}}\le C{2}^{-J\tau \mathrm{/}2}M$, whereas $q_{\mathrm{♯}}$ has bounded Sobolev norms of every order with bounds depending only on $J,M$. Choose $J$, depending only on $\tau ,M$, so that $G_{\zeta }{q}_{\mathrm{♭}}$ as a multiplication-composition operator has norm at most $1\mathrm{/}4$ both on shifted $H^{r_0}$ and on shifted $H^{1\mathrm{/}2}$. On $H^{r_0}$ this follows from the product bound in step 1.1, with multiplier $r\in H^{r_0}$ and $q_{\mathrm{♭}}\in H^{h_0}$, and the $v=1$ bound of step 1.2. On $H^{1\mathrm{/}2}$ it follows from step 2.1 with $\epsilon =\tau \mathrm{/}2$ and the same $v=1$ bound. Next choose $s$ large enough, depending only on $J,M,\tau$, that $G_{\zeta }{q}_{\mathrm{♯}}$ has norm at most $1\mathrm{/}4$ on both spaces, using the smooth multiplier bound from step 1.1 and the $v=0$ bound $C\mathrm{/}s$ from step 1.2. Thus $I+G_{\zeta }q$ has inverse given by a convergent geometric series, of norm at most two, on both spaces.

Define $r_{\zeta }=-(I+G_{\zeta }q{)}^{-1}{G}_{\zeta }q$ in shifted $H^{r_0}$. The forcing $q$ in the last position is the actual compactly supported function on $Q$, represented in the shifted basis; the multiplication operators use the periodic representation of the same function. These agree pointwise inside $Q$. Step 1.2 with $\mu =1\mathrm{/}2+\tau$ and $v=1-\tau \mathrm{/}2$ gives $\mathrm{\parallel }{r}_{\zeta }{\mathrm{\parallel }}_{H^{r_0}(Q)}\le C{s}^{-\tau \mathrm{/}2}M$. Since $r_0>3\mathrm{/}2$, weighted Cauchy–Schwarz for its coefficients gives $\mathrm{\parallel }{r}_{\zeta }{\mathrm{\parallel }}_{\mathrm{\infty }}\le C\mathrm{\parallel }{r}_{\zeta }{\mathrm{\parallel }}_{H^{r_0}(Q)}$. This solution is also the unique shifted $H^{1\mathrm{/}2}$ solution, by the second invertibility statement in step 3.1; using step 1.2 with $\mu =1\mathrm{/}2,v=0$ gives $\mathrm{\parallel }{r}_{\zeta }{\mathrm{\parallel }}_{H^{1\mathrm{/}2}(Q)}\le C{s}^{-1}M$. Finally, its equation is $r_{\zeta }=-G_{\zeta }(q+q{r}_{\zeta })$; since $q\in L^2$ and $r_{\zeta }\in L^{\mathrm{\infty }}$, the $v=2$ bound from $L^2$ gives $\mathrm{\parallel }{r}_{\zeta }{\mathrm{\parallel }}_{H^2(Q)}\le CsM(1+M)$, enlarging the threshold to keep $s\ge 1$.

Applying $A_{\zeta }$ to the equation in step 4.1 gives $A_{\zeta }{r}_{\zeta }+q(1+r_{\zeta })=0$ in the interior of $Q$. Direct conjugation and $\zeta \cdot \zeta =0$ show this is exactly $(-\mathrm{\Delta }+q)(e^{i\zeta \cdot x}(1+r_{\zeta }))=0$. The local comparison in step 2.2 transfers the fractional bounds to restriction norms on $B_2$, and restriction gives the integer $H^2$ bound there. Pulling back the initial rotation preserves the whole-space potential norm, the support ball, and the Euclidean PDE. Since $s=\mathrm{\mid }\zeta \mathrm{\mid }\mathrm{/}\sqrt{2}$, step 4.1 yields all displayed estimates. The construction was on a fixed larger cube, so it imposes no boundary condition on the unit ball and supplies regularity on its full neighborhood.

Define $J_b(\eta )={\int }_{\mathrm{\mid }\xi \mathrm{\mid }\le b}⟨\xi {⟩}^{-1}⟨\eta -\xi {⟩}^{-1}d\xi$. For $\mathrm{\mid }\eta \mathrm{\mid }\ge 2b$, $\mathrm{\mid }\eta -\xi \mathrm{\mid }\ge \mathrm{\mid }\eta \mathrm{\mid }\mathrm{/}2$ throughout the integration ball, so $⟨\eta -\xi {⟩}^{-1}\le 2⟨\eta {⟩}^{-1}$. Polar integration gives ${\int }_{\mathrm{\mid }\xi \mathrm{\mid }\le b}⟨\xi {⟩}^{-1}d\xi \le C{b}^2$ for $b\ge 1$. Hence $J_b(\eta )\le C{b}^2⟨\eta {⟩}^{-1}$ in this region.

If $\mathrm{\mid }\eta \mathrm{\mid }<2b$, Cauchy–Schwarz gives $J_b(\eta )\le ({\int }_{\mathrm{\mid }\xi \mathrm{\mid }\le b}⟨\xi {⟩}^{-2}d\xi {)}^{1\mathrm{/}2}({\int }_{\mathrm{\mid }\xi \mathrm{\mid }\le b}⟨\eta -\xi {⟩}^{-2}d\xi {)}^{1\mathrm{/}2}$. The first integral is at most $Cb$ by polar integration. In the second, the substitution $z=\eta -\xi$ places its domain inside $B_{3b}$, so it is also at most $Cb$. Thus $J_b(\eta )\le Cb$. Since $⟨\eta ⟩\le \sqrt{1+4b^2}\le \sqrt{5}b$, this is at most $C^{\mathrm{\prime }}{b}^2⟨\eta {⟩}^{-1}$. Together with step 1.1 this proves $J_b(\eta )\le C{b}^2⟨\eta {⟩}^{-1}$ for every $\eta$.

Fourier translation gives $\mathrm{\parallel }{e}^{-i\xi \cdot x}Q{\mathrm{\parallel }}_{H^{-1\mathrm{/}2}}^2=\int ⟨\eta -\xi {⟩}^{-1}\mathrm{\mid }\widehat{Q}(\eta ){\mathrm{\mid }}^{2}d\eta$. This identity holds for Sobolev distributions as well as smooth functions, because the Fourier transform is represented by a locally square integrable function with the indicated weight, and translation preserves weighted square integrability for every fixed $\xi$. Tonelli [F2] and step 2.1 now give ${\int }_{\mathrm{\mid }\xi \mathrm{\mid }\le b}⟨\xi {⟩}^{-1}\mathrm{\parallel }{e}^{-i\xi \cdot x}Q{\mathrm{\parallel }}_{H^{-1\mathrm{/}2}}^{2}d\xi =\int \mathrm{\mid }\widehat{Q}(\eta ){\mathrm{\mid }}^2{J}_b(\eta )d\eta \le C{b}^2\mathrm{\parallel }Q{\mathrm{\parallel }}_{H^{-1\mathrm{/}2}}^2$.

For the bilinear duality pairing, weighted Fourier Cauchy–Schwarz gives $\mathrm{\mid }⟨e^{-i\xi \cdot x}Q,a_{\xi }⟩\mathrm{\mid }\le \mathrm{\parallel }{e}^{-i\xi \cdot x}Q{\mathrm{\parallel }}_{H^{-1\mathrm{/}2}}\mathrm{\parallel }{a}_{\xi }{\mathrm{\parallel }}_{H^{1\mathrm{/}2}}$. The absence of complex conjugation changes the second Fourier factor to its reflection and does not change its Sobolev norm. Squaring, integrating with weight $⟨\xi {⟩}^{-1}$, using the bound $L\mathrm{/}t$, and applying step 3.1 prove the remainder estimate.

If $u-v\in H_0^1(\mathrm{\Omega })$, then [F2] gives $Tu=Tv$, so $\bar{T}[u]:=Tu$ is well-defined on $\mathcal{D}(\mathrm{\Omega })$. By [F1], $\mathrm{\parallel }\bar{T}[u]{\mathrm{\parallel }}_{W^{1\mathrm{/}2,2}(\mathrm{\partial }\mathrm{\Omega })}\le C\mathrm{\parallel }u{\mathrm{\parallel }}_{H^1(\mathrm{\Omega })}$ for every representative $u$; taking the infimum over its coset gives $\mathrm{\parallel }\bar{T}[u]\mathrm{\parallel }\le C\mathrm{\parallel }[u]{\mathrm{\parallel }}_{\mathcal{D}}$. If $\bar{T}[u]=0$, [F2] gives $u\in H_0^1$, hence $[u]=0$; thus the induced map is injective.

By [F1], every $f\in W^{1\mathrm{/}2,2}(\mathrm{\partial }\mathrm{\Omega })$ is $Tu$ for some $u\in H^1(\mathrm{\Omega })$, so $\bar{T}$ is onto. More quantitatively, choose the lift $Rf$ from [F3]. Then $\bar{T}[Rf]=f$ and $\mathrm{\parallel }[Rf]{\mathrm{\parallel }}_{\mathcal{D}}\le \mathrm{\parallel }Rf{\mathrm{\parallel }}_{H^1}\le C\mathrm{\parallel }f{\mathrm{\parallel }}_{W^{1\mathrm{/}2,2}}$, proving that the inverse is bounded.

The trace of a constant function is that same constant, so $\bar{T}$ maps the closed constant subspace $\mathbb{K}\mathbf{1}\subset \mathcal{D}(\mathrm{\Omega })$ onto the constant subspace of $W^{1\mathrm{/}2,2}(\mathrm{\partial }\mathrm{\Omega })$. The bounded isomorphism and its bounded inverse from steps 1.1–1.2 therefore induce a bounded isomorphism on the two quotient spaces, with equivalent quotient norms.

If $\mathrm{\ell }$ is a continuous linear functional on $W^{1\mathrm{/}2,2}(\mathrm{\partial }\mathrm{\Omega })$ that annihilates constants, then $\dot{\mathrm{\ell }}([f]):=\mathrm{\ell }(f)$ is independent of the representative because $\mathrm{\ell }(f+c)=\mathrm{\ell }(f)$ for every constant $c$. By the quotient-norm definition, $\mathrm{\mid }\dot{\mathrm{\ell }}([f])\mathrm{\mid }\le \mathrm{\parallel }\mathrm{\ell }\mathrm{\parallel }{\inf }_c\mathrm{\parallel }f+c{\mathrm{\parallel }}_{W^{1\mathrm{/}2,2}}$, so $\dot{\mathrm{\ell }}$ is continuous with quotient dual norm at most $\mathrm{\parallel }\mathrm{\ell }\mathrm{\parallel }$. Conversely, any functional that factors as $\mathrm{\ell }=\phi \circ q$ vanishes on constants because $q(c)=0$; and $\mathrm{\mid }\phi (qf)\mathrm{\mid }\le \mathrm{\parallel }\phi \mathrm{\parallel }\mathrm{\parallel }qf\mathrm{\parallel }\le \mathrm{\parallel }\phi \mathrm{\parallel }\mathrm{\parallel }f\mathrm{\parallel }$ because the quotient map $q$ has norm at most one. Hence $\mathrm{\parallel }\mathrm{\ell }\mathrm{\parallel }\le \mathrm{\parallel }\phi \mathrm{\parallel }$, while the first inequality gives $\mathrm{\parallel }\phi \mathrm{\parallel }\le \mathrm{\parallel }\mathrm{\ell }\mathrm{\parallel }$ for the induced functional. This proves the equivalence and equality of dual norms.