Is $W^{1,\infty}(\Omega)$ dense on $C^0(\Omega)$? Prove:
(1) $W^{1,\infty}(\Omega)$ is the set of all lipschitz functions on $\Omega$.
(2)$W^{1,\infty}(\Omega)$ is dense on $C^0(\Omega)$
$\Omega\in\mathbb R^k$ is an open, convex set.
(Not homeworks)

I believe that the state (1) is not correct. We can only prove that a function $f\in W^{1,\infty}(\mathbb R)$ coincide almost everywhere with a Lipschitz continuous function. But how to show this?
For (2) I believe it is right? On wikipedia it says that "all L-continuous" function on a compact metric space is a subalgebra of Banach space of continuous functions, and thus dense in it. But how to show this for an open, convex set $\Omega$?
 A: You may object to (1) on the ground that $W^{1,\infty}$ is not a set of functions, but a set of equivalence classes of functions. But I think this would be arguing about a technicality. Each equivalence class in $W^{1,\infty}$ contains a Lipschitz continuous representative, and it is natural to identify the class with this representative. 
Details: let $f\in W^{1,\infty}(\Omega)$. On almost every line segment parallel to a coordinate axis, $f$ is absolutely continuous with derivative in $L^\infty$; but this means it is Lipschitz continuous. Let $A$ be the union of all such line segments contained in $\Omega$. Then $f$ is Lipschitz continuous on $A$, and $\Omega\setminus A$ has measure zero. So if we  redefine $f$ on $\Omega\setminus A$, extending $f$ by continuity, the result is a Lipschitz continuous function on $\Omega$. 

(2) requires one to specify a topology on $C^0(\Omega)$; I suppose it's the topology of uniform convergence on compact subsets. If we tried uniform norm, then it would be necessary to restrict attention to bounded functions; however, density would fail anyway.
First, approximate $f\in C^0(\Omega)$ by compactly supported functions $f_n\in C^0_c(\Omega)$; for this it suffices to take $f_n= f\chi_n$ where $\chi_n$ is a cut-off function that is equal to $1$ on most of $\Omega$ and has compact support. 
Second, approximate $f_n$ by 
$$g_L(x) = \sup_{y\in \Omega} (f_n(y) - L|x-y|) \tag1$$
Note that $g_L$ is $L$-Lipschitz and $g_L\ge f_n$. Also, if $L$ is large enough then $g_L\le f_n+\epsilon$ by virtue of the uniform continuity of $f_n$. Specifically, the supremum in (1) is effectively taken only over $|x-y|<\delta$, as larger values of $|x-y|$ incur too much   penalty from $- L|x-y|$. 
