I'm having a problem seeing an inequality in this equestion: Interior $H^2$ regularity - inequalities over regions $U$ and $W \subset U$ Many thanks for your help.
How exactly does one get from the first inequality to the second. I'm happy with the fact the integral is taken over $W$ as $\zeta$ vanishes in $\mathbb{R^n}-W$.
However, when i use the product rule on the right hand side of the first inequality, I get a $D\zeta^2$ term that I can't get rid of.
Now Theorem 3(i) in §5.82 implies \begin{align} \int_u |v|^2 \, dx &\le C \int_U |D(\zeta^2 D_k^h u)|^2 \, dx \\ &\le C \int_W |D_k^h U|^2 + \zeta^2 |D_k^h Du|^2 \, dx \\ &\le C \int_U |Du|^2 + \zeta^2 |D_k^h Du|^2 \, dx \end{align}
Theorem 3(i): assume $1\leq p<\infty$ and $u \in W^{1,p}(U)$ then for each $V \subset \subset U$
\begin{equation} \|D^h u\|_{L^p(V)} \leq C\|Du\|_{L^p(U)} \end{equation} for some constant $C$ and all $0<|h|<\text{dist}(V, \partial U)$. Where $D^h$ is a difference quotient. That is \begin{equation} D^h_i u(x)=\frac{u(x+he_i)-u(x)}{h} \end{equation}
Here's a more complete rundown of the steps to get from the first inequality to the second. In my answer (and in the theorem), $C$ stands for an arbitrary constant, and may change from equation to equation or from line to line.
By the Leibnitz rule: $$\int_U |D(\xi^2 D^h_ku)|^2\;dx \le C \int_U |(D\xi^2)D^h_ku|^2 + \xi^2|DD^h_k u|^2\;dx$$ since $(a+b)^2 \le 2(a^2 + b^2)$. Now, $(D\xi^2) = 2\xi D\xi$, and $2D\xi$ is bounded (since $\xi$ is $C^\infty_0$), so we have the estimate $$\int_U |(D\xi^2)D^h_ku|^2\;dx \le \max_{x \in U} |2D\xi(x)|^2 \int_U \xi^2|D^h_ku|^2\;dx \le C \int_U\xi^2 |D^h_ku|^2\;dx$$ where the constant $C$ depends only on $\xi$. (If you are familiar with the theory of $L^p$ spaces, note that this is just Hölder's inequality: $\lVert fg \rVert_{L^2}^2 \le \lVert f \rVert_{L^\infty}^2 \lVert g \rVert_{L^2}^2$.) Since $\xi^2$ has maximum value $1$ and is supported on $W$, we can estimate this last integral by $$\int_U\xi^2 |D^h_ku|^2\;dx \le \int_W |D^h_k u|^2\;dx.$$
Putting together all of the previous inequalities will give you the second line. Then, you use theorem 3(i) to obtain the result on the third line (with $p = 2$).