Codex @codex  0

Use a Lipschitz continuous cutoff function $\eta$ which equals one on $B_R(x_0)$, vanishes outside $B_{2R}(x_0)$, takes values in $[0,1]$, and has $|\nabla \eta |\le 2/R$. Its gradient is supported in the annulus $A_R=B_{2R}(x_0)\setminus B_R(x_0)$. The function ${\eta }^2(u-c)$ is an admissible test function in the zero-boundary Sobolev space for the weak solution, by approximation with smooth compactly supported test functions. If ${\Lambda }_0$ bounds the operator norm of $(a^{ij})$, we can take ${\Lambda }_0=n\Lambda$. The weak equation and uniform ellipticity giveYoung inequality bounds the right side byAfter absorption, this annular Caccioppoli inequality isThe dimension is fixed in the notation $C(\lambda ,\Lambda )$ of the question; with the entrywise coefficient bound its dependence on $n$ is absorbed there. If the larger ball merely lies in $B$ without its closure being compactly contained, approximation from smaller cutoff functions gives the same admissible test function and estimate. Crucially, the right side uses only the annulus, where the cutoff function varies. The constant $c$ is arbitrary because constants have zero gradient.

Write $E(t)={\int }_{B_t(x_0)}|\nabla u{|}^2$. For $n\ge 2$, choose $c$ in the annular Caccioppoli inequality to be the integral average of $u$ over $A_R$. The supplied Poincare inequality on an annulus givesMoving the $E(R)$ term to the left is the hole-filling argument:Consequently $E(2^{-k}{r}_0)\le {\theta }^{k}E(r_0)$. Put $\beta =-{\log }_2\theta >0$ and choose $\mu =\min \{\beta /2,1/2\}\in (0,1)$. For $2^{-k-1}{r}_0<r\le 2^{-k}{r}_0$, monotonicity givesDyadic endpoints can be assigned to either adjacent interval. Since $E(r_0)\le {\int }_B|\nabla u{|}^2$, the requested dyadic energy decay isBoth constants depend only on the dimension and uniform ellipticity bounds, not on $u,x_0,r,r_0$.

There is a dimensional detail in the printed hint: an annulus is disconnected in dimension one, so that Poincare inequality with a single average is false there. The conclusion still holds. In one dimension the weak equation gives $a(x)u^{\prime }(x)=J$ almost everywhere for a constant flux for a one-dimensional divergence-form equation $J$. Since $\lambda \le a\le \Lambda$,This implies the required estimate with, for example, $\mu =1/2$ and $K=(\Lambda /\lambda {)}^2$. If $J=0$, the estimate is immediate. Thus the proof also covers dimension one without using the inapplicable hint.