Let's consider a bounded domain $\Omega \subset \mathbb{R}^d$, $d =2,3$, and let $\varphi$ be in $H^1(\Omega) \cap W^{1,\infty}(\Omega)$.

Is it there a smooth (at least $W^{2,4}(\Omega)\cap W^{1,\infty}(\Omega)$) function $\varphi^\varepsilon$ such that

$$\Vert \varphi^\varepsilon - \varphi \Vert_{W^{1,\infty}} \leq C(\varphi) \varepsilon\,?$$

If so, what is the explicit dependence of the constant $C(\varphi)$ on $\varphi$ in terms of its $H^1$- and $W^{1,\infty}$-norm?

And finally, is it there an $\alpha\in\mathbb{R}^+$ such that $$\Vert \varphi^\varepsilon \Vert_{W^{2,4}} = \mathcal{O}(\varepsilon^{-\alpha})\,?$$

If so, what would the sharpest estimate be?


This is not possible. Every $v \in W^{2,4}(\Omega)$ satisfies $\nabla v \in C(\bar\Omega)$. Hence, if $\varphi^\varepsilon \in W^{2,4}(\Omega)$ converges in $W^{1,\infty}(\Omega)$ towards $v$, we get $\nabla v \in C(\bar\Omega)$, since the derivatives converge in $L^\infty(\Omega)$.

But $\nabla \varphi \in L^\infty(\Omega)$ can be discontinuous.


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