 Solution

For any $R>0$, set $U_R(y)=u(Ry)$ and $a_R^{ij}(y)=a^{ij}(Ry)$. These coefficients retain the same bounds and ellipticity, and a change of variables in the weak formulation shows that $U_R$ is a nonnegative weak subsolution on $B_1$. Part (a) gives the scale-correct estimateThe constant is independent of $R$. Fix a Euclidean ball $B_s$ and let $R\to \infty$ with $R>2s$. The right side tends to zero, so $u=0$ almost everywhere on $B_s$. Since $s$ is arbitrary, the Liouville theorem for integrable nonnegative elliptic subsolutions provesThis is $u\equiv 0$ as a Sobolev space element. Any continuous representative is identically zero pointwise; an arbitrary measurable representative can be modified at isolated points without changing the weak equation. Integrability and the explicit scaling force the conclusion.