Lemma 16.14.7A64885BC3406

Proof. First let $A_0$ be a constant real symmetric matrix with $A_0\geq \lambda I$, and put $M=A_0^{-1/2}$, a symmetric matrix of operator norm at most $\lambda ^{-1/2}$. For $v\in C_ c^\infty (\mathbb R^ d)$ let $\tilde v(y)=v(M^{-1}y)$, so that $v(x)=\tilde v(Mx)$. Then $D^2v(x)=M\, D^2\tilde v(Mx)\, M$ and $A_0:D^2v(x)=\operatorname {tr}(MA_0M\, D^2\tilde v(Mx))=\Delta \tilde v(Mx)$. Since $|MNM|\leq \lambda ^{-1}|N|$ for every matrix $N$, the change of variables $y=Mx$ and (2.9.13) give

This estimate extends to every $w\in W^{2,p}(\mathbb R^ d)$ that vanishes almost everywhere outside a compact set $K$. Indeed, for $0<\varepsilon \leq 1$ the mollifications $w*\rho _\varepsilon $ are test functions supported in the compact set $K'=\{ x:\operatorname {dist}(x,K)\leq 1\} $, their second derivatives are $(\partial _\mu \partial _\sigma w)*\rho _\varepsilon $, by proposition 2.5.3(1), and these converge to $\partial _\mu \partial _\sigma w$ in $L^ p(K')$, by lemma 2.5.1(2), hence in $L^ p(\mathbb R^ d)$, since all these functions vanish almost everywhere outside $K'$. Applying the estimate to $v=w*\rho _\varepsilon $ and letting $\varepsilon \to 0$, we obtain it for $w$.

Now let $A$ and $w$ be as in the statement, put $C_1=\lambda ^{-1}C_{d,p}$, $\delta _0=1/(2C_1)$, and $A_0=A(x_ B)$. Since $|\sum _{\mu ,\sigma }m_{\mu \sigma }m'_{\mu \sigma }|\leq |m|\, |m'|$ for matrices $m$ and $m'$, we have $|A_0:D^2w-A:D^2w|\leq \delta _0|D^2w|$ almost everywhere on $B$, and both sides vanish outside $B$. Hence

and since $\| D^2w\| _ p<\infty $ we obtain the lemma with $C=2C_1$. $\square$