Consider the bounded mapping $A:W^{1,p}(\Omega) \rightarrow W^{1,p}(\Omega)^{*}$ where $A$ is defined as:
$\langle A(u),v \rangle\text{ } := \int_{\Omega}a(x,u,\nabla u)\cdot \nabla v + c(x,u,\nabla u)v dx + \int_{\Gamma}b(x,u)vdS$
Where $\Omega \subset \mathbb{R}^{n}$ is a bounded Lipschitz domain and $\Gamma$ is the boundary. $a: \Omega \times \mathbb{R} \times \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$, $b: \Gamma \times \mathbb{R} \rightarrow \mathbb{R}$, $c:\Omega \times \mathbb{R} \times \mathbb{R}^{n} \rightarrow \mathbb{R}$ are measurable Caratheodory functions.
To distinguish between the highest and lower order terms, we define $B(w,u) \in W^{1,p}(\Omega)^{*}$ by $\langle B(w,u),v\rangle \text{ }:= \int_{\Omega}a(x,w,\nabla u)\cdot \nabla v + c(x,w,\nabla w)v dx + \int_{\Gamma}b(x,w)vdS$ for $u,v \in W^{1,p}(\Omega)$, so obviously $A(u) = B(u,u)$.
Assumptions: Assume that $u_{k} \rightharpoonup u$ weakly in $W^{1,p}(U)$ and $\text{limsup}_{k \rightarrow \infty}\langle A(u_{k},u_{k}-u)\rangle \text{} \leq 0$.
The following three things have already been proved:
(1) $\epsilon \langle A(u_{k}), u-v\rangle \text{ } \geq -\langle A(u_{k}),u_{k}-u \rangle + \langle B(u_{k},u_{\epsilon}),u_{k}-u\rangle + \epsilon \langle B(u_{k},u_{\epsilon}),u-v\rangle \text{ ,where }$ $u_{\epsilon} := (1-\epsilon)u + \epsilon v \text{ and }\epsilon \in [0,1]$
(2) $\text{lim}_{k\rightarrow \infty}\langle B(u_{k},v),u_{k}-u\rangle = 0$
(3)$\text{w-lim}_{k \rightarrow \infty}B(u_{k},v) = B(u,v)$ (the weak limit in $W^{1,p}(\Omega)^{*}$)
We use these in the following inequality:
$\epsilon \text{liminf}_{k \rightarrow \infty} \langle A(u_{k}),u-v \rangle \text{} \geq -\text{limsup}_{k \rightarrow \infty}\langle A(u_{k}),u_{k}-u\rangle + \text{lim}_{k \rightarrow \infty}\langle B(u_{k},u_{\epsilon}),u_{k}-u\rangle + \epsilon\text{lim}_{k \rightarrow \infty}\langle B(u_{k},u_{\epsilon}),u-v\rangle \text{} \geq \text{}\epsilon\langle B(u,u_{\epsilon}),u-v\rangle \text{} $
Questions:
1.How do we know that $\text{lim}_{k \rightarrow \infty}\langle B(u_{k},u_{\epsilon}),u-v\rangle $ exists?
2.How does $\text{lim}_{k \rightarrow \infty}\langle B(u_{k},u_{\epsilon}),u-v\rangle \text{} =\text{} \langle B(u,u_{\epsilon}),u-v\rangle $? I suspect we use the assumptions $u_{k} \rightharpoonup u$ and $\text{w-lim}_{k \rightarrow \infty}B(u_{k},v) = B(u,v)$ (the weak limit in $W^{1,p}(\Omega)^{*}$), but I don't see how it follows?
Thanks for any assistance. Notify me if regarding mistakes or if something is unclear.
I'm going to assume that in (3) you are using weak-$\star$ convergence, otherwise I am not sure what you mean by w-limit.
So, a sequence $\Lambda_n$ ($B(u_k,u_\epsilon)$ in your notation) in $X^*$ conrveges to $\Lambda \in X^*$ in the weak-$\star$ topology if for every $x \in X$ we have $$ \Lambda_n(x) \to \Lambda(x) $$ as $n \to \infty$.
Now, your $B(u,v)$ acts on elements $\phi$ of $W^{1,p}(\Omega)$ by the pairing $$ B(u,v)(\phi) = \langle B(u,v),\phi \rangle, $$ so number (3) is precisely telling you that, not only does $\lim_{k \to \infty} \langle B(u_k,u_\epsilon), \phi \rangle$ exist, it actually converges to $\langle B(u,u_\epsilon), \phi \rangle$. You are just using this fact with $\phi = u-v$.
You don't actually need $u_k$ to converge weakly to $u$ (although this fact was most probably used when proving (3)).