= 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> $\phi$, the classical <product rule> and the <weak derivative> identity for $u$, tested with $\psi\phi$, give
$$
\begin{aligned}
\int_{B_1}\psi u\,\partial_i\phi
&=\int_{B_1}u\,\partial_i(\psi\phi)-\int_{B_1}u(\partial_i\psi)\phi\\
&=-\int_{B_1}\bigl(\psi\partial_i u+u\partial_i\psi\bigr)\phi.
\end{aligned}
$$
Therefore
$$
\boxed{\psi u\in W^{1,p}(B_1),\qquad
\partial_i(\psi u)=\psi\partial_i u+u\partial_i\psi.}
$$
In 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, take
$$
u\equiv1,\qquad\psi(x)=\frac1{1-|x|^2}.
$$
Here $u\in W^{1,p}(B_1)$ and $\psi$ is a <smooth function> on the open ball, while <polar coordinates> show
$$
\int_{B_1}|\psi|^p\,dx
\geq c_n\int_{1/2}^1(1-r^2)^{-p}\,dr=\infty\qquad(p\geq1).
$$
Thus $\psi u\notin L^p(B_1)$, and the global assertion is false. Boundedness of $\psi$ and its first <derivatives> would suffice without <compact support>.
Back to article page
