Extend $u$ by zero to $\mathbb R^n$; its <compact support> inside the ball makes the extension <smooth>. The <Fourier transform of a derivative> gives
$$
\widehat{L_0u}(\xi)=-(A\xi\mathbin\cdot\xi)\widehat u(\xi).
$$
The assumption says that $L_0$ is a <uniformly elliptic operator>, so
$$
|A\xi\mathbin\cdot\xi|^2\geq\theta^2|\xi|^4.
$$
Moreover,
$$
\sum_{i,j=1}^n|\xi_i\xi_j|^2
=\left(\sum_{i=1}^n\xi_i^2\right)^2
=|\xi|^4.
$$
The <Plancherel theorem> therefore yields
$$
\|L_0u\|_2^2
=\int_{\mathbb R^n}|A\xi\mathbin\cdot\xi|^2|\widehat u|^2\,d\xi
\geq\theta^2\int_{\mathbb R^n}|\xi|^4|\widehat u|^2\,d\xi
=\theta^2\|D^2u\|_2^2.
$$
All integrands vanish outside the original support where appropriate, so
$$
\boxed{\theta\|D^2u\|_{L^2(B_r(x_0))}\leq\|L_0u\|_{L^2(B_r(x_0))}.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 105 / 4 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-105/4/a/solution