Let $G:\mathbb{R}\to\mathbb{R}$ be a Lipschitz function and $L$ its Lipschitz constant. Suppose that $G(0)=0$. It is known that $G$ is almost everywhere differentiable and $G'\in L^\infty$ with $\|G'\|_{\infty}\leq L$.
My question is: Can I find a sequence $G_n:\mathbb{R}\to\mathbb{R}$ such that, $G_n(0)=0$, $G_n\in C^1(\mathbb{R})$, each $G_n$ is Lipschitz with constant $L_n$, $G_n(x)\to G(x)$, $G'_n(x)\to G'(x)$ almost everywhere and $L_n\to L$?
Thank you
Yes. The continuous functions are dense in $L^1([a,b])$ on any interval. So for each $n$, choose a continuous $F_n$ such that $\int_{-n}^n |F_n(x)-G'(x)|\,dx \to 0$. Replacing $F_n$ by $F_n \vee -L \wedge L$ only makes the $L^1$ distance smaller. Passing to a subsequence, we can also get $F_n \to G'$ almost everywhere. Set $G_n(x) = \int_0^x F_n(y)\,dy$; then for $x \in [-n,n]$ we have $$|G(x) - G_n(x)| = \left\vert \int_0^x G'(y) - F_n(y)\,dy\right\vert \le \int_0^n |G'(y) - F_n(y)|\,dy \to 0.$$ In particular, we have $G_n \to G$ uniformly on compact sets.
I guess I didn't show that the Lipschitz constants converge; this is left as an exercise because I'm too lazy.