I have the following problem: $\Omega\subset \mathbb{R}^N$ is open and bounded with Lipschitz boundary.I have a sequence of functions $(u_n)\subset W^{1,\infty}(\Omega)$ that converges to $u\in W^{1,\infty}(\Omega)$. In order to finish a proof I'm working on I need to prove that this assumption implies that $\nabla u_n $ converges to $\nabla u $ in $L^\infty$ (or $L^1$, both will work).
I have found the following theorem in the functional analysis book by Brezis (corollary 9.14)
For $1\leq p \leq \infty$ $W^{1,p}(\Omega) \subset L^\infty(\Omega)$ for $p>N$ the injections is continuous and for all $u\in W^{1,p}$ we have $$|u(x)-u(y)|\leq C_{\Omega,N,p} \| u\|_{W^{1,p}} |x-y|^\alpha$$ where $\alpha = 1 - N/p$.
By taking $p=\infty$ I can get $$\frac{|u(x) - u(y)|}{|x-y|} \leq C_{\Omega,N,\infty} \|u\|_{W^{1,\infty}}$$
My question is: Is it possible to conclude from this that $$\|\nabla u\|_{L^1} \leq C \| u\|_{W^{1,\infty}}$$ or $$\|\nabla u\|_{L^\infty} \leq C \| u\|_{W^{1,\infty}}$$ or none of the above.