Quantitative CGO construction for C1alpha conductivity potentials

Fix $\varphi \in C_c^{\mathrm{\infty }}$ equal to $1$ on the ball consisting of points of distance at most $2$ from $B$. For Schwartz $u,v$, the integrands defining $m_q$ are supported where $\varphi =1$, so they can be written using $u_B=\varphi u,v_B=\varphi v$. By [F1], $\mathrm{\parallel }{u}_B{v}_B{\mathrm{\parallel }}_1\le C{s}^{-1}\mathrm{\parallel }u{\mathrm{\parallel }}_{{\dot{X}}_{\zeta }^{1\mathrm{/}2}}\mathrm{\parallel }v{\mathrm{\parallel }}_{{\dot{X}}_{\zeta }^{1\mathrm{/}2}}$ and $\mathrm{\parallel }\mathrm{\nabla }(u_B{v}_B){\mathrm{\parallel }}_1\le C\mathrm{\parallel }u{\mathrm{\parallel }}_{{\dot{X}}_{\zeta }^{1\mathrm{/}2}}\mathrm{\parallel }v{\mathrm{\parallel }}_{{\dot{X}}_{\zeta }^{1\mathrm{/}2}}$, the latter from the product rule and the two combinations $s^{1\mathrm{/}2}{s}^{-1\mathrm{/}2}$. Thus a bounded coefficient multiplying $u_B{v}_B$ costs $C{s}^{-1}$, whereas a bounded vector coefficient multiplying its gradient costs $C$. The same estimates hold when $u,v$ have any two null phases of the same length, because [F1] is uniform in the phase directions. These bounds also prove continuity of the bilinear form on the weighted completions and local $H^1$ representatives.

Choose a nonnegative smooth mollifier $\rho$ of integral $1$ supported in the unit ball and put $a_h={\rho }_h\ast a$ for $0<h\le 1$. The bounds $\mathrm{\parallel }\mathrm{div}{a}_h{\mathrm{\parallel }}_{\mathrm{\infty }}\le C_{\rho }{h}^{-1}\mathrm{\parallel }a{\mathrm{\parallel }}_{\mathrm{\infty }}$ and $\mathrm{\parallel }a-a_h{\mathrm{\parallel }}_{\mathrm{\infty }}\le [a{]}_{C^{0,\alpha }}{h}^{\alpha }$ follow, respectively, by differentiation of the kernel and integration of $\mathrm{\mid }a(x)-a(x-y)\mathrm{\mid }\le K\mathrm{\mid }y{\mathrm{\mid }}^{\alpha }$. The support of $a_h$ stays inside the region where $\varphi =1$. Split the derivative term of $m_q$ into $a_h$ and $a-a_h$, and integrate the former by parts. Step 1.1 gives $\mathrm{\mid }⟨m_{q}u,v⟩\mathrm{\mid }\le C[s^{-1}{K}^2+s^{-1}{h}^{-1}K+K{h}^{\alpha }]\mathrm{\parallel }u{\mathrm{\parallel }}_{{\dot{X}}^{1\mathrm{/}2}}\mathrm{\parallel }v{\mathrm{\parallel }}_{{\dot{X}}^{1\mathrm{/}2}}$. Take $h=s^{-1\mathrm{/}(1+\alpha )}$; since $s\ge 1$, this is bounded by $C{s}^{-\alpha \mathrm{/}(1+\alpha )}$ times the product of norms. Weighted Fourier duality identifies the bounded bilinear form with a map to ${\dot{X}}^{-1\mathrm{/}2}$: in Fourier coordinates the dual weight is $\mathrm{\mid }{p}_{\zeta }(-\xi ){\mathrm{\mid }}^{-1}$ for a bilinear pairing, and reflection changes the test phase to $-\zeta$; the cutoff bounds of step 1.1 are identical for $\zeta$ and $-\zeta$. Therefore the same proof with $v$ measured in ${\dot{X}}_{-\zeta }^{1\mathrm{/}2}$ gives exactly the asserted map norm. Enlarge $s_0$ until this norm is at most $1\mathrm{/}2$.

The weighted completions are realized as the Fourier $L^2$ spaces with weights $\mathrm{\mid }{p}_{\zeta }{\mathrm{\mid }}^{\pm 1}$: smooth functions supported away from the characteristic set and inside a bounded frequency ball are dense by truncation and ordinary $L^2$ approximation there. They define tempered distributions because reciprocal-weight Cauchy–Schwarz controls their pairing with a Schwartz test; the only new local singularity is $\mathrm{\mid }{p}_{\zeta }{\mathrm{\mid }}^{-1}$, integrable by the codimension-two computation in [F1]. Multiplication of the Fourier transform by $p_{\zeta }^{-1}$ is consequently an isometry $G_{\zeta }:{\dot{X}}_{\zeta }^{-1\mathrm{/}2}\to {\dot{X}}_{\zeta }^{1\mathrm{/}2}$ and is a distributional right inverse of $\mathrm{\Delta }+2\zeta \cdot \mathrm{\nabla }$. The characteristic set has measure zero, so its values need not be assigned. Also $q\in {\dot{X}}_{\zeta }^{-1\mathrm{/}2}$: its compactly supported divergence part follows from [F1] applied to $\mathrm{div}a\in X_{\zeta }^{-1\mathrm{/}2}$, and its bounded compactly supported zero-order part follows from the same cutoff bound. For the former, $\mathrm{\mid }\xi {\mathrm{\mid }}^2\mathrm{/}(s+\mathrm{\mid }{p}_{\zeta }(\xi )\mathrm{\mid })$ is bounded for fixed $s$, so $a\in L^2$ suffices; for the latter, the inhomogeneous negative norm is at most $s^{-1\mathrm{/}2}$ times its $L^2$ norm.

By steps 2.1–3.1, the series $r_{\zeta }={\sum }_{j=0}^{\mathrm{\infty }}(G_{\zeta }{m}_q{)}^j{G}_{\zeta }q$ converges in ${\dot{X}}_{\zeta }^{1\mathrm{/}2}$, has norm at most $2\mathrm{\parallel }q{\mathrm{\parallel }}_{{\dot{X}}_{\zeta }^{-1\mathrm{/}2}}$, and satisfies $(\mathrm{\Delta }+2\zeta \cdot \mathrm{\nabla })r_{\zeta }=q+m_q{r}_{\zeta }$. For a cutoff equal to $1$ on any fixed compact set, [F1] gives convergence in $H^1$ there, so $r_{\zeta }\in H_{\mathrm{l}\mathrm{o}\mathrm{c}}^1$. The product-form definition agrees with the distributional equation by this local convergence. Conjugating $-\mathrm{\Delta }$ by $e^{x\cdot \zeta }$ gives $-\mathrm{\Delta }-2\zeta \cdot \mathrm{\nabla }-\zeta \cdot \zeta$; the last term vanishes. Hence the displayed remainder equation is exactly $(-\mathrm{\Delta }+q)e^{x\cdot \zeta }(1+r_{\zeta })=0$. This proves existence and the remainder bound.

Since $a$ is supported in $\bar{B}$, for $0<h\le 1$ its translation difference has support in a fixed bounded ball and absolute value at most $K{h}^{\alpha }$. Consequently ${\omega }_2(a,h)\le C_{B}K{h}^{\alpha }$ and $\mathrm{\parallel }a{\mathrm{\parallel }}_2\le C_{B}K$. Apply [F2] to $a$ with a cutoff equal to $1$ near its support. The zero-order part $\mathrm{\mid }a{\mathrm{\mid }}^2$ has squared negative-half norm at most $C{s}^{-1}\mathrm{\parallel }\mathrm{\mid }a{\mathrm{\mid }}^2{\mathrm{\parallel }}_2^2\le C{s}^{-1}{C}_B{K}^4$ by [F1]. The inequality $\mathrm{\parallel }v+w{\mathrm{\parallel }}^2\le 2\mathrm{\parallel }v{\mathrm{\parallel }}^2+2\mathrm{\parallel }w{\mathrm{\parallel }}^2$ therefore bounds the average squared norm of $q$ by $C[{\lambda }^{-1}+{\lambda }^{-1}{h}^{-2}+(1+(1+\mathrm{\mid }k{\mathrm{\mid }}^2)\mathrm{/}\lambda )h^{2\alpha }]$. Set $h={\lambda }^{-1\mathrm{/}(2+2\alpha )}$. Both ${\lambda }^{-1}{h}^{-2}$ and $h^{2\alpha }$ equal ${\lambda }^{-\alpha \mathrm{/}(1+\alpha )}$, and ${\lambda }^{-1}$ is no larger. The pointwise remainder bound from step 4.1 then gives the asserted average for $q$ and $r$ together. The parameterized norms and integrals are measurable: first for Fourier-truncated smooth data this follows directly from the measurable weights; the Neumann partial sums are measurable as iterated multipliers and fixed coefficient forms, and their norm limits retain measurability.