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Let $\Omega\subset \mathbb{R}^n$ be open and bounded. For $0<K\leq \infty$, $\text{Lip}_K(\Omega)$ denotes the set of lipschitz-continuous functions on $\Omega$ with lipschitzconstant less than or equal to $K$. Suppose that $f\in \text{Lip}_K(\Omega)$ is a function with $f\geq 0$ and $f=0$ on $\partial \Omega$. Is it then possible to approximate $f$ by a sequence of functions $f_n\in \text{Lip}_K(\Omega), f_n \geq 0,$ which have support in $\Omega$ and satisfy either

i) $\lvert\lvert Df_n - Df\rvert\rvert_{L^{\infty}(\Omega)}\to 0$ or

ii) $\lvert\lvert Df_n-Df\rvert\rvert_{L^{1}(\Omega)}$ and $\lvert\lvert Df_n\rvert\rvert_{L^{\infty}(\Omega)}\leq const$?

Since $f=0$ on $\partial \Omega$ and $f\in \text{Lip}(\Omega)$, we have $f\in \mathring{W}^{1,\infty}(\Omega)$, so we can extend $f$ to a function $f\in W^{1,\infty}(\mathbb{R}^n)$ with $f=0$ on $\mathbb{R}^n\setminus \Omega$. I was then trying to use mollifiers but the function which is obtained in this way doesn't have support in $\Omega$. The condition $f_n \in \text{Lip}_K(\Omega)$ also gives me some trouble.

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1 Answer 1

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Condition (i) is not possible: Consider $\Omega = (-1,1)$ and $f(x) = 1 - |x|$. Then, $f_n$ should be supported on some $[-1+\varepsilon, 1-\varepsilon]$. Hence, $\|f_n - f\|_{L^\infty}$ will always be greater than $1$.

Condition (ii) is easy to achieve by using the shrinkage $$ f_n = \max( |f| - 1/n, 0) \operatorname{sign}(f).$$ Then, $f_n = 0$ on a strip of width $1/(L n)$ around $\partial\Omega$. Moreover, $$ \nabla f_n = \nabla f\qquad \text{ a.e. where } |f| > 1/n $$ and $\nabla f_n = 0$ elsewhere. Thus, $\| \nabla f_n \|_{L^\infty}$ is bounded by $\|\nabla f\|_{L^\infty}$. Moreover, we get $\nabla f_n \to \nabla f$ a.e. on $\{|f| > 0\}$ and on $\{f = 0\}$ we have $\nabla f_n = \nabla f = 0$ a.e. Thus, $\nabla f_n \to \nabla f$ a.e. on $\Omega$. Lebesgue's dominated convergence theorem now yields $$ \| \nabla f_n - \nabla f\|_{L^p} \to 0 $$ for all $p \in [1,\infty)$.

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  • $\begingroup$ Thank you! Just to be sure: In my example we have $f\geq 0$ so your function becomes $$f_n = \text{max}(f - 1/n,0)$$ and since $0,1/n, f \in \text{Lip}_K(\Omega)$, it follows that $f_n \in \text{Lip}_K(\Omega)$, right? $\endgroup$
    – Mandelbrot
    Commented Aug 27, 2021 at 13:38
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    $\begingroup$ Actually, $1/n \in Lip_K$ is not enough, since this would result in $f - 1/n \in Lip_{2K}$. Luckily, we have $1/n \in Lip_0$ ;) $\endgroup$
    – gerw
    Commented Aug 27, 2021 at 14:20
  • $\begingroup$ I somehow thought that the Lipschitz constant of the sum of lipschitzcontinuous functions is bounded by the maximum of their respective Lipschitz constants, which obviously isn't true (in general). Thanks again! $\endgroup$
    – Mandelbrot
    Commented Aug 28, 2021 at 8:02

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