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 $\Phi_\varepsilon$ as $\varepsilon \to 0$ (in the weak sense)?