Let $\mathcal{C}_{0}^{\infty}$ denote the space of $\mathcal{C}^{\infty}$ functions with compact support and Let $\operatorname{Lip}_{1}(\mathbb{R}^n)$ be the space of Lipschitz continuous functions having Lipschitz constant equal to 1.
Is it true that the space $\{ u \in \mathcal{C}_{0}^{\infty}(\mathbb{R}^n) \: : \: | \nabla u | \le 1 \}$ is dense in $\operatorname{Lip}_{1}(\mathbb{R}^n)$? And if the answer is yes, how can I prove it?
Any help is appreciated.
The Lipschitz functions need to have compact support. Otherwise, the function $f(x):=\sin x$ is a counterexample as it cannot be approximated uniformly with functions with compact support.
If $f$ is a Lipschitz function with compact support, i.e., $f\in C^{0,1}_c(\mathbb R^n)$, then the convolution with the standard mollification kernel $\rho_\epsilon * f$ is a smooth function with compact support. Since $D (\rho_\epsilon * f) = \rho_\epsilon * (Df)$, the resulting function is 1-Lipschitz as well. In addition, $$\begin{split} |f(x) - \int_{\mathbb R^n} \rho_\epsilon(x-y)f(y)dy| &= |\int_{\mathbb R^n} \rho_\epsilon(x-y)(f(x)-f(y))dy|\\ &\le 1\cdot \int_{\mathbb R^n}\rho_\epsilon(x-y)|x-y|dy \\ &\le \epsilon \int_{\mathbb R^n}\rho_\epsilon(x-y)dy=\epsilon, \end{split}$$ since the support of $\rho_\epsilon$ is contained in a ball of radius $\epsilon$ and $\int_{\mathbb R^n}\rho_\epsilon(y)dy=1$. Hence $\rho_\epsilon * f \to f$ uniformly.