$W^{1,p}$ compact in $L^\infty$? Is $W^{1,p}(0,1)$ compactly contained in $L^\infty(0,1)$?  Can I use this to show that I can select a sequence $(u_{n_k})$ from every bounded sequence $(u_n)$ in $W^{1,p}(0,1)$ such that $\lVert u_{n_k} - u \rVert_\infty \to 0$?  
Also, how do I show $u_{n_k}' \rightharpoonup u'$ weakly in $L^p(I)$ if $p < \infty$ and $u_{n_k}' \overset{*}{\rightharpoonup} u'$ in the weak* topology on $L^\infty(0,1)$ if $p = \infty$?

That $W^{1,p}(0,1)$ is contained in $L^{\infty}(0,1)$ follows by one of the Sobolev Embedding Theorems, since $W^{1,p}(0,1)$ is a subset of $C^{0}([0,1])$ (at least for $p>1$) (if I am not wrong :-)). The compactness may also follow from one of these Theorems for some $p$. Maybe the book, not to say the bible :), "Adams, Robert A. (1975), Sobolev spaces" can give further information regarding the question. If it is compact, You can use this, since "compactness makes weak convergence strong" (somehow :)).
