 Solution

For $\psi \in C_c^{\infty }(B_1)$, the functions $\psi$ and ${\partial }_i\psi$ are bounded. Hence $\psi u$ and $\psi {\partial }_{i}u+u{\partial }_i\psi$ belong to $L^p(B_1)$. For a test function $\varphi$, the classical product rule and the weak derivative identity for $u$, tested with $\psi \varphi$, giveThereforeIn fact $\psi u$ has compact support in $B_1$ and belongs to the corresponding zero-boundary Sobolev space, by mollification within that support.

For arbitrary $\psi \in C^{\infty }(B_1)$, the same identity still proves the local product rule for weak derivatives, but membership in the global Sobolev space can fail. To give a counterexample for every dimension and every $p$ in the question, takeHere $u\in W^{1,p}(B_1)$ and $\psi$ is a smooth function on the open ball, while polar coordinates showThus $\psi u\notin L^p(B_1)$, and the global assertion is false. Boundedness of $\psi$ and its first derivatives would suffice without compact support.