TITLE: Approximation of $u \in \mathrm{H}^1_0(\Omega)$ by a sequence whose gradient is smaller than that of $u$
QUESTION [5 upvotes]: 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 !

REPLY [3 votes]: 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.