weak derivatives on open subset implies weak derivatives on whole set

Question

Let $V' \subset V\subset U \subset \mathbb{R}^{n}$ be bounded open sets such that $\overline{V'} \subset V$ and $f\in W^{m,p}(V,\mathbb{R})$ such that $f|_{V \backslash V'} \equiv 0$. Do we have then that $f \in W^{m,p}(U,\mathbb{R})$? Do the weak derivatives of $f$ on $U$ even exist? By $W^{m,p}(V,\mathbb{R})$ I denote the Sobolev space of functions with derivative up to $m$ to belong in $L^{p}$.

Answer 1

If $\overline{V^\prime} \subset V$, there is a $C^\infty$ function $\psi$ such that $\text{supp}(\psi)\subset V$ and $\psi = 1$ on an open set containing $\overline{V^\prime}$. Then $f = f \psi$ and I think it becomes easy to prove that $f\in W^{m, p}(U, \mathbb{R})$.

If $T$ is a distribution in $V$ with support in $\overline{V^\prime}$, it can be viewed as a distribution in $U$ by defining for $\phi\in C^\infty_0(U)$ $$\langle T, \phi\rangle = \langle T, \psi\phi\rangle$$ because $\psi\phi\in C^\infty_0(V)$. Then $$\langle \partial^\alpha T, \phi\rangle = \langle \partial^\alpha T, \psi\phi\rangle = \langle T, (-1)^{|\alpha|} \partial^\alpha(\psi \phi)\rangle = \langle T, (-1)^{|\alpha|} \psi\partial^\alpha(\phi)\rangle$$ because the derivatives of $\psi$ vanish in a neighborhood of $\text{supp}(T)$