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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 $L,C,d$ and $n$, 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 $L$?

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    $\begingroup$ Errrr, what's the role of $k$, then? I guess there's a typo and $k=L$? $\endgroup$
    – leo monsaingeon
    Commented Apr 7, 2022 at 14:39
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    $\begingroup$ Known estimate... by what? It can't be by any norm of $f$: the quantity you wrote is not scaling homogeneous. (It may be if instead of $L$ you used $L \|f\|_{L^1}$ or something like that.) // By looking at functions like $\sin(nx)$ as $n\to \infty$, it seems that regardless of what you set as $L$, you have that $$\sup_{f\in L^1} \frac{\| f - \mathrm{Lip}_L(\mathbb{R}^d, \mathbb{R}^n) \|_{L^1}}{\|f\|_{L^1}} \geq \frac12$$. $\endgroup$
    – Willie Wong
    Commented Apr 7, 2022 at 18:47
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    $\begingroup$ The space of Lipschitz functions allows functions with arbitrarily large Lipschitz constants. Each one of $\sin(nx)$ is perfectly approximated by a Lipschitz function with Lipschitz constant $n$. // You know, the usual thing about order of quantifiers. $\endgroup$
    – Willie Wong
    Commented Apr 7, 2022 at 18:59
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    $\begingroup$ $L^1$ is too flabby; you need some way to control the size of the set on which the "derivative" is large for something uniform or quantitative. For example, you can say something if instead of $f$ being in the unit ball in $L^1$, you take $f$ in the unit ball of the Sobolev space $W^{1,1}$. (But still measuring the distance between $f$ and the approximant in $L^1$ only.) $\endgroup$
    – Willie Wong
    Commented Apr 7, 2022 at 19:12
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    $\begingroup$ I would maybe approach this by first considering that $f$ is really smooth, say infinitely differentiable with compact support and then use a density argument to yield the bound. I am not sure this is really easier. $\endgroup$
    – Theleb
    Commented Apr 7, 2022 at 20:50

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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 $k=1$).

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  • $\begingroup$ No, I do mean $L^\infty$, since we want to obtain an $L^\infty$-bound on the gradient of the mollified function (which is a bound on the Lipschitz constant). $\endgroup$
    – Mizar
    Commented Apr 8, 2022 at 19:41
  • $\begingroup$ Ah, nevermind. I confused myself for some reason. $\endgroup$
    – Willie Wong
    Commented Apr 8, 2022 at 19:45
  • $\begingroup$ what is the interpretation argument (i know little about interpolation spaces). $\endgroup$
    – ABIM
    Commented Apr 8, 2022 at 23:07
  • $\begingroup$ Here "interpolation" is just fancy language: simply note that $\|\rho_\epsilon\|_{L^p}^p=\int\rho_\epsilon^p\le\max|\rho_\epsilon|^{p-1}\int\rho_\epsilon=\|\rho_\epsilon\|_{L^\infty}^{p-1}$ for all $1\le p<\infty$, which gives that estimate (taking $p:=k'$) for all $1<k\le\infty$ (it is trivial for $k=1$). $\endgroup$
    – Mizar
    Commented Apr 11, 2022 at 22:00

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