I don't understand the last line of a proof (which is supposed to be obvious...), could you help me?
The context is the following. We have a bounded open set $U$ of $\mathbb{R}^m$, and a $C^\infty$-mapping $F : \mathbb{R}^m \times \mathbb{R} \times U \to \mathbb{R}^m$ We also have a sequence $(u_n)_n$ of $W^{1,p}(U)$ ($p > 1$) and $u \in W^{1,p}(U)$ such that $u_n$ converges uniformly to $u$ and $\nabla u_n$ converges weakly in $L^p(U,\mathbb{R}^m)$ to $\nabla u$. Last, we know that $|u| + |\nabla u| < M$ for some $M < +\infty$.
The first thing said is that $F(\nabla u, u_n, .)$ converges uniformly to $F(\nabla u, u, .)$ on $U$. I have the intuition that it's true but I don't know how to prove it formally. Is it just obvious from some sort of theorem about function composition? I suppose I have to use the condition $|u| + |\nabla u| < M$ somewhere...
Secondly, $\displaystyle\lim\limits_{n\to\infty}\int_U F(\nabla u(x), u_n(x),x)\cdot (\nabla u_n(x) - \nabla u(x)) dx = 0$(We know that the definite integral exists and is finite for each $n$). They said that "weak convergence is, by its very definition, compatible with linear expressions, and so the limit holds", but I don't understand this statement...
Thank you in advance if you can explain me one of the two affirmations (and sorry for my not perfect english) :). I perhaps have forgotten some hypothesis (even if I hope not), so don't hesitate if you think there is something missing or unclear.