Weak derivative of restriction and restriction of weak derivative

Question

Suppose $D^kf$ is weak derivative of $f$ of order $k$ on a domain $U \subseteq \mathbb{R}^n$. Let $V$ be any open connected subset of U. Is it true that $D^k(f\chi_V)=(D^kf)\chi_V$?

Answer 1

No. This is not true, even for $k = 1$.

Formally, if you apply Leibniz formula, you get $D^1 (f \chi_V) = (D^1 f) \chi_V + f (D^1 \chi_V)$. So, with respect to your formula, there is a supplementary term. You could argue that, since $\chi_V$ is locally constant, $D^1 \chi_V = 0$. But this is not the case, the (weak) derivative of $\chi_V$ contains a singular term on the boundary of $V$.

Take $n = 1$, $U = \mathbb{R}$ and say $f(x) = 1$ everywhere to make things very simple. Also, take $V = (0,1)$. On the one hand $(D^1 f) \chi_V = 0$ because $D^1 f = 0$. On the other hand, since $f \chi_V = \chi_V$, $D^1(f\chi_V) = D^1(\chi_V) = \delta_{x=0} - \delta_{x=1}$ in $\mathcal{D}'(\mathbb{R})$.