Past exam of the functional analytic methods for partial differential equations course of the University of Oxford / 2023 / 2 / a

A locally integrable function $u$ has weak derivative $g_i$ in direction $i$ when $g_i$ is locally integrable andThe test functions have compact support inside the domain, so no boundary values enter this definition. Such $g_i$ is unique almost everywhere and is denoted ${\partial }_{i}u$.

For $1\le p<\infty$, the first-order Sobolev space consists of the Lp space classes whose first weak derivatives are also in that Lp space:This Sobolev norm is equivalent to $\Vert u{\Vert }_p+\Vert \nabla u{\Vert }_p$; either convention is acceptable. When $p=2$, this is the Hilbert space $H^1(B_1)$.

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.