Uniform interior higher integrability for conductivity solutions

On vector fields on ${\mathbb{R}}^d$ put $P=-(R_i{R}_j{)}_{i,j}$, with multiplier $\xi {\xi }^T\mathrm{/}\mathrm{\mid }\xi {\mathrm{\mid }}^2$ away from zero. This is the orthogonal gradient projection, so $\mathrm{\parallel }P{\mathrm{\parallel }}_{L^2({\mathrm{\ell }}^2)\to L^2({\mathrm{\ell }}^2)}=1$ by [F2]. The finitely many compositions in [F1] give a finite $L^4({\mathrm{\ell }}^4)$ bound $C_4\ge 1$. Apply scalar [F2] on the product of Lebesgue measure and counting measure on $\{1,\dots ,d\}$ to this matrix operator, whose $L^2$ norm there is exactly one. If $1\mathrm{/}p=(1-\theta )\mathrm{/}2+\theta \mathrm{/}4$, then $\mathrm{\parallel }P{\mathrm{\parallel }}_{L^p({\mathrm{\ell }}^p)\to L^p({\mathrm{\ell }}^p)}\le C_4^{\theta }$. Since $\mathrm{\parallel }z{\mathrm{\parallel }}_{{\mathrm{\ell }}^p}\le \mathrm{\parallel }z{\mathrm{\parallel }}_{{\mathrm{\ell }}^2}\le d^{1\mathrm{/}2-1\mathrm{/}p}\mathrm{\parallel }z{\mathrm{\parallel }}_{{\mathrm{\ell }}^p}$ for $p>2$, its Euclidean-vector $L^p$ norm is at most $D_p=d^{1\mathrm{/}2-1\mathrm{/}p}{C}_4^{\theta }$, which tends to one as $p\downarrow 2$. These extensions agree with the $L^2$ projection on common inputs by density.

Put $a=(m+M)\mathrm{/}2$ and $k=(M-m)\mathrm{/}(M+m)<1$. Extend $\gamma$ outside $\mathrm{\Omega }$ by $a$ and set $c=(a-\gamma )\mathrm{/}a$, so $\mathrm{\parallel }c{\mathrm{\parallel }}_{\mathrm{\infty }}\le k$. Choose $p\in (2,4)$ sufficiently close to two that $k{D}_p<1$ and, when $d>2$, $p<2d\mathrm{/}(d-2)$. The operators $Pc$ then have norms less than one both on $L^2({\mathrm{\ell }}^2)$ and on $L^p({\mathrm{\ell }}^2)$. For every vector field $F$ in their intersection, the series ${\sum }_{j\ge 0}(Pc{)}^j{a}^{-1}PF$ converges in BOTH spaces. The two limits agree as distributions, since convergence in either space implies convergence against compactly supported smooth tests. In particular, any $v\in H^1({\mathbb{R}}^d)$ solving $\mathrm{div}(\gamma \mathrm{\nabla }v)=\mathrm{div}F$ satisfies $\mathrm{\nabla }v=a^{-1}PF+Pc\mathrm{\nabla }v$ in $L^2$: subtracting the two sides gives a gradient field of zero divergence, whose Fourier transform vanishes away from zero and hence vanishes in $L^2$. Thus $\mathrm{\parallel }\mathrm{\nabla }v{\mathrm{\parallel }}_p\le a^{-1}{D}_p(1-k{D}_p{)}^{-1}\mathrm{\parallel }F{\mathrm{\parallel }}_p$.

Choose $\chi \in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$ equal to one on a neighborhood of $\bar{U}$, set $u_0=u-u_{\mathrm{\Omega }}$ and extend $v=\chi u_0$ by zero. The product rule and the weak equation give $\mathrm{div}(\gamma \mathrm{\nabla }v)=\mathrm{div}(\gamma u_0\mathrm{\nabla }\chi )+g$, where $g=\gamma \mathrm{\nabla }u\cdot \mathrm{\nabla }\chi$ is compactly supported, $\mathrm{\parallel }g{\mathrm{\parallel }}_2\le C\mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_2$, and $\int g=0$ by testing the original weak equation with $\chi$. Put $\mathrm{\Phi }=-{\mathrm{\Phi }}_{-}$ using [F6], so $\mathrm{\Delta }\mathrm{\Phi }={\delta }_0$, and set $W=\mathrm{\nabla }\mathrm{\Phi }\ast g$. The radial formula in [F6] gives ${\mathrm{\partial }}_i\mathrm{\Phi }=x_i\mathrm{/}(\mathrm{\mid }{S}^{d-1}\mathrm{\mid }\mathrm{\mid }x{\mathrm{\mid }}^d)$ and ${\mathrm{\partial }}_{ij}\mathrm{\Phi }={\delta }_{ij}\mathrm{/}(\mathrm{\mid }{S}^{d-1}\mathrm{\mid }\mathrm{\mid }x{\mathrm{\mid }}^d)-d{x}_i{x}_j\mathrm{/}(\mathrm{\mid }{S}^{d-1}\mathrm{\mid }\mathrm{\mid }x{\mathrm{\mid }}^{d+2})$ away from zero, hence $\mathrm{\mid }\mathrm{\nabla }\mathrm{\Phi }(x)\mathrm{\mid }\le C\mathrm{\mid }x{\mathrm{\mid }}^{1-d}$ and $\mathrm{\mid }{D}^2\mathrm{\Phi }(x)\mathrm{\mid }\le C\mathrm{\mid }x{\mathrm{\mid }}^{-d}$. This convolution defines a locally integrable distribution with $\mathrm{div}W=g$: integrate the fundamental-solution identity against compactly supported tests and use local integrability of $\mathrm{\mid }x{\mathrm{\mid }}^{1-d}$. Its low-frequency behavior is controlled by the zero integral of $g$, as quantified next.

Enclose $\mathrm{supp}g$ in a fixed ball of radius $L$. On $\mathrm{\mid }x\mathrm{\mid }\le 2L+1$, $W$ only uses the restriction of $\mathrm{\nabla }\mathrm{\Phi }$ to a fixed larger ball; that kernel is in $L^r$ for $1\le r<d\mathrm{/}(d-1)$. The convolution inequality with $1+1\mathrm{/}p=1\mathrm{/}2+1\mathrm{/}r$ gives $\mathrm{\parallel }W{\mathrm{\parallel }}_{L^p(\mathrm{\mid }x\mathrm{\mid }\le 2L+1)}\le C\mathrm{\parallel }g{\mathrm{\parallel }}_2$, because the chosen $p<2d\mathrm{/}(d-2)$ makes this $r$ admissible (there is no finite upper restriction for $d=2$). The same argument with $r=1$ gives the local $L^2$ bound. Outside that ball, subtract $\mathrm{\nabla }\mathrm{\Phi }(x)\int g=0$ in the integral and use $\mathrm{\mid }{D}^2\mathrm{\Phi }(x-y)\mathrm{\mid }\le C\mathrm{\mid }x{\mathrm{\mid }}^{-d}$ to obtain $\mathrm{\mid }W(x)\mathrm{\mid }\le C\mathrm{\mid }x{\mathrm{\mid }}^{-d}\mathrm{\parallel }g{\mathrm{\parallel }}_1$. This is integrable to both the second and the $p$th power at infinity. Thus $\mathrm{\parallel }W{\mathrm{\parallel }}_2+\mathrm{\parallel }W{\mathrm{\parallel }}_p\le C\mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_2$. The convolution estimates are [F5].

Set $F=\gamma u_0\mathrm{\nabla }\chi +W$. By [F3], the first term belongs to $L^2\cap L^p$ with norm at most $C\mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_2$; step 4.1 gives the same bound for the second. Step 3.1 now reads $\mathrm{div}(\gamma \mathrm{\nabla }v)=\mathrm{div}F$. Apply step 2.1 and use $\mathrm{\nabla }v=\mathrm{\nabla }u$ on $U$ to obtain $\mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_{L^p(U)}\le C\mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_2$, with all constants depending only on the fixed data. For the Neumann conclusion on an arbitrary bounded smooth $\mathrm{\Omega }$, choose a similarity $S(x)=a+\mathrm{\ell }x$, $\mathrm{\ell }>0$, with $\widehat{\mathrm{\Omega }}=S(\mathrm{\Omega })\subset (0,1{)}^d$. Put $\widehat{u}=u\circ S^{-1}$ and $\widehat{\gamma }=\gamma \circ S^{-1}$. Pullback $\widehat{v}\mapsto \widehat{v}\circ S$ is a bounded isomorphism of $H^1$ spaces, carries $H_0^1$ onto $H_0^1$ and preserves constants, hence induces a bounded isomorphism of the trace quotients by constants. Define $\widehat{F}([\mathrm{Tr}\widehat{v}])={\mathrm{\ell }}^{d-2}F([\mathrm{Tr}(\widehat{v}\circ S)])$. The energy change of variables gives ${\int }_{\widehat{\mathrm{\Omega }}}\widehat{\gamma }\mathrm{\nabla }\widehat{u}\cdot \overline{\mathrm{\nabla }\widehat{v}}dy={\mathrm{\ell }}^{d-2}{\int }_{\mathrm{\Omega }}\gamma \mathrm{\nabla }u\cdot \overline{\mathrm{\nabla }(\widehat{v}\circ S)}dx=\widehat{F}([\mathrm{Tr}\widehat{v}])$, and $\mathrm{\parallel }\widehat{F}\mathrm{\parallel }\le C_S\mathrm{\parallel }F\mathrm{\parallel }$. Apply [F4] there and use $\mathrm{\parallel }\mathrm{\nabla }\widehat{u}{\mathrm{\parallel }}_{L^2(\widehat{\mathrm{\Omega }})}={\mathrm{\ell }}^{d\mathrm{/}2-1}\mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_{L^2(\mathrm{\Omega })}$ to get the claimed energy bound on $\mathrm{\Omega }$. Every construction used a fixed finite cutoff and bounded operators, so the two choice assumptions enter only through [A1].