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