= Solution
Write $Lu=L_0u+(a^{ij}-A^{ij})D_{ij}u$, with repeated indices summed. The <triangle inequality> and the <Cauchy-Schwarz inequality> over the $n^2$ coefficient pairs give
$$
\begin{aligned}
\|Lu\|_2
&\geq\|L_0u\|_2-\left\|\sum_{i,j}(a^{ij}-A^{ij})D_{ij}u\right\|_2\\
&\geq\theta\|D^2u\|_2-n\varepsilon\|D^2u\|_2.
\end{aligned}
$$
Choose
$$
\boxed{\varepsilon=\frac{\theta}{2n}.}
$$
Then the strict coefficient bound in the question implies
$$
\boxed{\frac\theta2\|D^2u\|_{L^2(B_r(x_0))}\leq\|Lu\|_{L^2(B_r(x_0))}.}
$$
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