If I know that \Phi_\varepsilon is bounded in L^{\infty}(\mathbb{R}^{2d}) and that \nabla \Phi_\varepsilon is bounded in L^{\infty}(\mathbb{R}^{2d}), is it true that \nabla \Phi_\varepsilon \to \nabla \Phi, where \Phi is the limit of \Phi_\varepsilon (in the weak sense)?