# Approximate $f\in \text{Lip}_K(\Omega)$ with $f\geq 0,f=0$ on $\partial \Omega$ by $f_n\in C_c(\Omega)\cap \text{Lip}_K(\Omega)$ with $f_n\geq 0$.

Let $$\Omega\subset \mathbb{R}^n$$ be open and bounded. For $$0, $$\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.

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 $$p \in [1,\infty)$$.
• 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? Commented Aug 27, 2021 at 13:38
• Actually, $1/n \in Lip_K$ is not enough, since this would result in $f - 1/n \in Lip_{2K}$. Luckily, we have $1/n \in Lip_0$ ;)