= Solution
Fix $B_R(x_0)\subset B_1$, put $A_0=A(x_0)$, and let $v$ have the same boundary values as $u$ while solving $\operatorname{div}(A_0\nabla v)=0$. Then $w=u-v\in W_0^{1,2}(B_R)$ satisfies
$$
\operatorname{div}(A_0\nabla w)
=\operatorname{div}((A_0-A(x))\nabla u).
$$
Since $u\in C^1$ gives $|Du|\leq L$ locally, the supplied energy estimate and <Hölder space>[Hölder continuity] of $A$ yield
$$
\int_{B_R}|\nabla w|^2\leq C L^2[A]_{C^\alpha}^2R^{n+2\alpha}.
$$
The constant-coefficient decay estimate gives
$$
\int_{B_\rho}|\nabla v-(\nabla v)_{B_\rho}|^2
\leq C(\rho/R)^{n+2}
\int_{B_R}|\nabla v-(\nabla v)_{B_R}|^2.
$$
Thus, for $\Phi(r)=\int_{B_r}|\nabla u-(\nabla u)_{B_r}|^2$,
$$
\Phi(\rho)\leq C(\rho/R)^{n+2}\Phi(R)+CR^{n+2\alpha}.
$$
Because $n+2>n+2\alpha$, the <Campanato iteration lemma> gives $\Phi(\rho)\leq C\rho^{n+2\alpha}$ on balls in $B_{1/2}$. Hence $\nabla u\in\mathcal L^{2,n+2\alpha}=C^{0,\alpha}$ and $u\in C^{1,\alpha}(B_{1/2})$.
Back to article page
