Approximation of $u \in \mathrm{H}^1_0(\Omega)$ by a sequence whose gradient is smaller than that of $u$ I'm looking for some help about the following question
Let $u \in \mathrm{H}^1_0(\Omega)$ with $\nabla u \in \mathrm{L}^{\infty}(\Omega)$ where $\Omega$ stands for a regular bounded subset of $\mathbb{R}^d$ (actually, $d=3$).
I want to show that there exists a sequence of test functions $(u_n)$ with $u_n \in  C_c^{\infty}(\Omega)$ such that :

*

*$u_n \rightarrow u \ \ $   in   $\ \ \mathrm{L}^1(\Omega)$

*$\Vert \nabla u_n \Vert_{\mathrm{L}^{\infty}(\Omega)} \lesssim \Vert \nabla u \Vert_{\mathrm{L}^{\infty}(\Omega)}$ for all $n$, with $\lesssim$ independant of $n$
I tried different regularization/truncature procedures but without success...
Any help is welcomed !
 A: Let $u_\epsilon = \phi_\epsilon * u$ the the mollification of $u$ with the standard mollifier $\{\phi_\epsilon\}$ on $\mathbb{R}^d$. It is standard to show that $u_\epsilon \to u$ in $H^1(\mathbb{R}^d)$ and so in particular
$$
\|u_\epsilon - u\|_{L^1} \leq |\Omega|^{1/2}\|u_\epsilon - u\|_{L^1} \to 0.
$$
As for the $L^\infty$ bound, we have
$$
|\partial_{x_i}u_\epsilon(x)| \leq \int_\Omega |\partial_{x_i}u(y)|\phi_\epsilon(x - y)\, dy \leq \|\partial_{x_i}u(y)\|_{L^\infty}\|\phi\|_{L^1}  \leq \|\nabla u\|_{L^\infty}.
$$
This requires some extra explaining, actually, because this is only valid for $x \in \Omega_\epsilon = \{x \in \Omega | \text{dist}(x, \partial \Omega) < \epsilon\}$. However, since the result holds for $\epsilon > 0$ arbitrary, it thus holds for any $x \in \bigcup_{\epsilon > 0}\Omega_\epsilon = \Omega$. Another way of justifying it might be using the extension theorem to extend $u$ to a function $\bar{u}$ on all of $\mathbb{R}^d$ with $\|\bar{u}\|_{H^1} \leq C\|u\|_{H^1(\Omega)}$, although I'm not sure if you'd get that $\nabla \bar{u}$ is in $L^\infty$ like you would need.
A: First of all, the function $u$ is Lipschitz on $\Omega$. For $\varepsilon > 0$, we consider the shrinkage
$$
u_\varepsilon(x) = \operatorname{sgn}(u(x)) \, \max\{0, |u(x)| - \varepsilon\}.$$
This function has the following properties:

*

*$u_\varepsilon \in H_0^1(\Omega)$ and $|\nabla u_\varepsilon(x)| \le |\nabla u(x)|$

*$u_\varepsilon \to u$ in $H_0^1(\Omega)$

*$u_\varepsilon$ is compactly supported in $\Omega$ (since $u$ is Lipschitz).

[Feel free to comment if you need help w.r.t. any of these statements]
Now, (as in the answer by Chris) we can mollify $u_{\delta,\varepsilon} = \phi_\delta * u_\varepsilon$. For a suitable coupling of $\delta$ and $\varepsilon$, this sequence has the desired properties.
