Does the Sobolev space $W^{1,\infty}(\Omega)$ consist of all bounded Lipschitz functions on $\Omega$ when $\Omega$ is not quasiconvex? The Wikipedia page for Sobolev spaces states that any function in $W^{1,\infty}(\Omega)$ is Lipschitz continuous, after modifying on a set of measure zero. (See the final paragraph of the "Absolutely continuous on lines (ACL) characterization of Sobolev functions" section.)
However, these lecture notes state that "if $\Omega$ is quasiconvex, then $W^{1,\infty}(\Omega)$ consists of all bounded Lipschitz functions on $\Omega$."
What happens when $\Omega$ is not quasiconvex? Is the Wikipedia claim correct?
 A: The problem is that of locality; if $u \in W^{1,\infty}(\Omega)$ then it admits a locally Lipschitz representative, where for each $x \in \Omega$ there is a neighbourhood $U_x \subset \Omega$ of $x$ on which $u$ is Lipschitz continuous. However in general we may not have global Lipschitz continuity, which is illustrated in the classical example where we take:
$$ \Omega = \{ z \in \Bbb C \setminus [0,\infty) : 1/2 < |z|<2 \},$$
and consider the function $u(z) = \log z$ taking the principal branch to $u(re^{i\theta}) = \log r + i\theta$ where $\theta \in (0,2\pi).$ Then $u$ is smooth in $\Omega$ with bounded derivative, so $u \in W^{1,\infty}(\Omega).$ However it fails to be Lipschitz continuous near $(1,2);$ indeed we have
$$ |\log(e^{i/k}) - \log(e^{i(2\pi - 1/k)})| = 2\pi - 2/k $$
for all $k,$ while $|e^{i/k}-e^{i(2\pi-1/k)}| \to 0$ as $k \to \infty.$ This implies that
$$ \sup_{\substack{z_1,z_2 \in \Omega \\ z_1 \neq z_2}} \frac{|u(z_1)-u(z_2)|}{|z_1-z_2|} = + \infty, $$
so $u$ is not Lipschitz continuous on $\Omega.$
