On a closed convex set $Z$ in ${ℝ}^N$ with sufficiently smooth (${𝒲}^{2,\infty }$) boundary, the stop operator is locally Lipschitz continuous from ${𝐖}^{1,1}([0,T],{ℝ}^N)\times Z$ into ${𝐖}^{1,1}([0,T],{ℝ}^N)$. The smoothness of the boundary is essential: A counterexample shows that ${𝒞}^1$-smoothness is not sufficient.