I have a question which might quite trivial but I would appreciate any assistance.
Why does it follow that for Sobolev spaces, say $W^{1,p}(\Omega)$, where $\Omega \subset \mathbb{R}^n$, it follows that $$u_n \rightharpoonup^{*} u \in W^{1,p}(\Omega)~~~\Leftrightarrow~~~ u_n \rightharpoonup^{*} u \in L^p(\Omega) \text{ and } \nabla u_n \rightharpoonup^{*} \in L^p(U;\mathbb{R}^n).$$
Thanks for any help.
First of all, we only use weak star convergence in the case of $p=\infty$, i.e., weak star convergence stands for convergence test against the pre-dual, not dual which is the case of weak convergence used for $p<\infty$.
Next, for your question, you can think $W^{1,p}$ space as $N+1$ fold of $L^p$ space, i.e. you have $(u,\partial_1u,\partial_2u,\ldots,\partial_Nu)\in L^p(\Omega)\times L^p(\Omega)\times\cdots\times L^p(\Omega)$ for $N+1$ times. Hence, weak compactness in $W^{1,p}(\Omega)$ is equivalent to weak compactness in $N+1$ times $L^p$ space and hence it gives you what you want.
For the reason why we are using weak star convergence in the case of $L^\infty$ is that $L^\infty$ is not a reflexive space, the dual of $L^\infty$ is not so clear, we only know it is strictly greater then $L^1$. However, the pre-dual of $L^\infty$ is $L^1$ and hence we adapted to use weak star convergence as $p=\infty$. In another word, we say $u_n\to u$ weak star in $L^\infty$ by means of $$ \int u_nv\to\int uv $$ for all $v\in L^1$.