Let $u \in H^{m}(U)$ and $W \subset\subset U$ that is $W \subset \bar{W} \subset U$ with $\bar{W}$ is compact in $U$ and $W$ is compactly contained in $U$. Is this true that
$$||D^{\alpha}u||_{L^{2}(W)} \leq K. ||u||_{L^{2}(U)}$$
holds true for any multiindex $\alpha$ with $|a|\leq m$ and some constant $K$?
Thank you very much!
This is too good to be true.
Assume the inequality holds for all $\alpha$ with $|\alpha |\le m$. Let $g\in L^2(U)$ and $g|_W \notin H^m(W)$, consider a sequence $f_n\in C^\infty_0(U)$ so that $f_n \to g$ in $L^2(U)$. Your inequality implies
$$\| f_n - f_m \|_{H^m (W)} \le K \|f_n - f_m\|_{L^2(U)}$$
and thus $\{f_n\}$ converges in $H^m(W)$ to some function, which must be $g|_W$. Thus $g|_W$ is in $H^m(W)$, which is a contradiction.