Besov smoothness gives the exact collared Sobolev penalty a cutoff source bound

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.