I was wondering if the following fact is true:
I consider a sequence $(u_h)\subset W^{1, \infty}(U, \mathbb{R}^N)$, with $U$ open bounded set in $\mathbb{R}^n$, such that $$u_h \rightharpoonup u$$ $*w-W^{1,\infty}$ (the * weak convergence).
Can I say that this $(u_h)$ is equi-lipschitz, i.e $\exists$ $M$ such that $|u_h(x)-u_h(y)\leq M|x-y|$ for every $x,y\in\Omega$ and for every index $h$?
Can I use perhaps the fact that this sequence is bounded in $W^{1,\infty}$?
I would like to know if this fact is true without any other assumption on $\Omega$.
I think that the answer is no, because if $\Omega $ is general, I cannot say that a function in $W^{1,\infty}$ is also Lipschitz. Am I right?
Thanks a lot for the help!