Integral convergence and weak convergence Given that $\Omega \subset \mathbb{R}^{n}$ is a connected bounded Lipshitz domain and $u_{k} \rightharpoonup u$ in $W^{1,p}(\Omega)$. We denote $\Gamma$ as the boundary of the domain. We have the following additional results:
$a(u_{k},\nabla  v) \rightarrow a(u, \nabla v)$ in  $L^{p'}(\Omega;\mathbb{R}^{n})$, $b(u_{k}) \rightarrow b(u)$ in $L^{q'}(\Gamma)$, $\nabla (u_{k}-u) \rightharpoonup 0$ in $L^{p}(\Omega;\mathbb{R}^{n})$ and $(u_{k}-u)|_{\Gamma} \rightharpoonup 0$ in $L^{q}(\Gamma)$.
How does it follow then that:
$\int_{\Omega}a(u_{k},\nabla v)\cdot\nabla(u_{k}-u)dx + \int_{\Gamma}b(u_{k})(u_{k}-u)dS \rightarrow 0$ 
My idea is that it seems that Holder's Inequality should be used so for the first integral term:
$\int_{\Omega}a(u_{k},\nabla v)\cdot\nabla(u_{k}-u)dx \leq (\int_{\Omega}|a(u_{k},\nabla v)|^{p'})^{\frac{1}{p'}}(\int_{\Omega}|\nabla(u_{k}-u)|^{p})^{\frac{1}{p}}$ and since $a(u_{k},\nabla v)$ is convergent it is bounded, I am just not sure we can state that $\int_{\Omega}|\nabla(u_{k}-u)^{p}|^{\frac{1}{p}} \rightarrow 0$ simply from the assumption $\nabla (u_{k}-u) \rightharpoonup 0$ in $L^{p}(\Omega;\mathbb{R}^{n})$? 
Any ideas of how this would be resolved? I can add information if some important info is missing. Thanks.  
 A: You are almost there:
For the first term it is convenient to subtract and add $a(u,\nabla v)\nabla(u_k-u)$, so you get to terms:
$$
A = \int_\Omega (a(u_k,\nabla v) - a(u,\nabla v))\nabla(u_k-u)
$$
and
$$
B = \int_\Omega a(u, \nabla v) \nabla(u_k-u).
$$
$B$ clearly tends to $0$ be the hypothesis of weak convergence. For the $A$ term note that weakly convergent sequences are uniformly bounded in $L^p$ (this is a consequence of the uniform boundedness principle and the duality of $L^p$). So as you said, by Holder's inequality you can estimate $|A|$ by
$$
\|a(u_k,\nabla v)-a(u,\nabla v)\|_{L^{p'}} \|\nabla(u_k-u)\|_{L^{p}},
$$
the first factor tends to $0$ by the strong convergence hypothesis, while the second factor remains uniformly bounded by the above remark.
The second integral is estimated similarly.
The basic argument using the principle of uniform boundedness above goes like this: if $f_n \rightharpoonup 0$ in $L^p$ then you can define the functionals $\Lambda_n(\psi) = \int f_n \psi$. By weak convergence, for all $\psi \in L^{p'}$ we have $\Lambda_n(\psi) \to 0$, in particular $\sup_n |\Lambda_n(\psi)| < \infty$, so by the uniform boundedness principle we must have $\sup_n \|\Lambda_n\|_{L^{p'} \to \mathbb{C}} < \infty$. Now by duality, $\|\Lambda_n\|_{L^{p'} \to \mathbb{C}} = \|f_n\|_{L^p}$, and you are done.
