If $\varphi$ is a testfunction with compact support in $U$ then for for example $f \in W^{1,p}(U)$, where $U$ is open.
We have $\int\limits_U \varphi^\prime f= - \int_U\limits f^\prime \varphi $
Sometimes I would like to have
$\int\limits_U \varphi^\prime f= - \int_U\limits f^\prime \varphi $
where this time $\varphi \in W^{1,p}_0(U)$.
Does that actually hold and why?
I think we do not have in generall that $\varphi\vert_{\partial U}=0$ for $\varphi \in W^{1,p}_0(U)$.
I would like to have that at least in the case where $U$ is bounded with boundary in $C^1$.
Let us prove for the case where $p\geq 2$ and $U$ is an open and bounded set.
Let $\phi\in W_0^{1,p}$. By definition, there exist a sequence of test functions $\phi_n$ such that $\phi_n\to\phi$ on $W^{1,p}$. This means that both $\phi_n^\prime\to\phi^\prime$ ,$\phi_n\to\phi$ on $L^p$. By the Inverse Dominated Convergence Theorem There is a subsequence $(k_i)_{i\in\mathbb{N}}$ with $\phi_{k_i}(x)\to\phi(x)$ and $\phi^\prime_{k_i}(x)\to\phi^\prime(x)$ almost everywhere and functions $g_1$, $g_2$ in $L^p$ such that $|\phi_{k_i}(x)|\leq g_1(x)$ and $|\phi^\prime_{k_i}(x)|\leq g_2(x)$ a.e. on $U$. Therefore $|f^\prime(x)\phi_{k_i}(x)|\leq |f^\prime(x)|g_1(x)$ and $|f(x)\phi^\prime_{k_i}(x)|\leq |f(x)|g_2(x)$. Since $p\geq 2$ $fg_2,f^\prime g _1\in L^{p/2}$ and since $U$ is bounded, $|f|g_2,|f^\prime| g_1\in L ^1$. So applying the Dominated Convergence Theorem we get the desired result.