I have to prove the following statements.

Let $U$ be an open subset of $\mathbb{R}^n$ and $u:U\rightarrow\mathbb{R}$ a locally Lipschitz function. Then if $U$ is convex, $$\|Du\|_{L^{\infty}(\Omega)}=L(u, U). \quad(\star)$$ ($L(u, U)$ is the Lipschitz constant of $u$ in $U$). If $U$ is non-convex, then $$\|Du\|_{L^{\infty}(U)}<L(u, U).$$

My attempt. If $U$ is non-convex (and so the general case), the result follows from the Rademacher's Theorem that states that if $u$ is locally Lipschitz continuous in $U$ then $u$ is differentiable a.e.. Is it right? If $U$ is convex, then, given $x, y\in\overline{U}$ (and hence the segment joining $x$ and $y$ is all contained in $\overline{U}$), by approximating $u$ by a sequence of mollifiers, we get $$|u(x)-u(y)|\leq\|Du\|_{L^{\infty}(U)}|x-y|$$ Can we conclude that the equality $(\star)$ holds?

Thank You

In general, we have $\|Du(x)\| \le L(u,U)$ for any $x \in U$ at which the derivative is defined. To see this, let $h={Du(x)^T \over \|Du(x)\|}$ and note that $L(u,U) \ge \lim_{t \to 0} {u(x+th)-u(x) \over t} = Du(x)h = \|Du(x)\|$. Hence $L(u,U) \ge \operatorname{ess sup}_U \|Du\|$
By taking $U=(0,1)\cup(2,3)$ and defining $u=1_{(2,3)}$ we see that $\operatorname{ess sup}_U \|Du\| = 0 < L(u,U) = 1$, so the strictness is possible for non convex $U$.
For convex $U$, let $x,y \in U$ and then $u(x)-u(y) = \int_0^1 Du(y+t(x-y)) dt (x-y)$ (the derivative is defined ae.) from which we get $\|u(x)-u(y)\| \le \operatorname{ess sup}_U \|Du\| \|x-y\|$ hence $L(u,U) \le \operatorname{ess sup}_U \|Du\||$, and so we have equality.