TITLE: Best approximation of L1 function by Lipschitz function
QUESTION [7 upvotes]: Fix constant $L,C>0$ and $k\geq 1$ and let $f\in W^{1,k}(\mathbb{R}^d,\mathbb{R}^n)$ with $\|f\|_{W^{1,k}}\leq C$.
Is there a known estimate on the distance
$$
\|f - \operatorname{Lip}_L(\mathbb{R}^d,\mathbb{R}^n)\|_{L^1(\mathbb{R}^d,\mathbb{R}^n)},
$$
depending on the constants $L,C,$ and $k$, where $\operatorname{Lip}_L(\mathbb{R}^d,\mathbb{R}^n)$ is the set of Lipschitz functions from $\mathbb{R}^d$ to $\mathbb{R}^n$ with Lipschitz constant at-most $L$?

REPLY [2 votes]: Let $(\rho_\epsilon)_{\epsilon>0}$ be a standard family of mollifiers, with $\rho_\epsilon$ supported in the ball $B_\epsilon(0)$. Since $\|\rho_\epsilon\|_{L^1}=1$ and $\|\rho_\epsilon\|_{L^\infty}=c_d\epsilon^{-d}$, by interpolation we get $\|\rho_\epsilon\|_{L^{k'}}=c_d^{1/k}\epsilon^{-d/k}$ (for the dual exponent $k'=\frac{k}{k-1}$). Hence,
$$\|\nabla(\rho_\epsilon*f)\|_{L^\infty}=\|\rho_\epsilon*\nabla f\|_{L^\infty}\le c_d^{1/k}\epsilon^{-d/k}C$$
for your function $f$ (by Holder). In order to get an $L$-Lipschitz function you can take
$$\epsilon:=c_d^{1/d}(C/L)^{k/d}=c_d'(C/L)^{k/d}$$
for another constant $c_d'$ depending only on $d$.
Then you can bound the distance of $\rho_\epsilon*f$ from $f$ as follows:
$$\|\rho_\epsilon*f-f\|_{L^{k}}\le\epsilon\|\nabla f\|_{L^{k}}\le c_d'(C/L)^{k/d}.$$
Note: the question of measuring the distance in $L^1$ seems ill-posed, since $W^{1,k}$ does not embed into $L^1$ (unless $k=1$).