# Weak convergence in $L^p$ and uniform convergence

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.

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To conclude with $F(\nabla u(x),u_n(x),x)\to F(\nabla u(x),u(x),x)$ uniformly, we want $F$ to be uniformly continuous on the set $\{p\in \mathbb R^n: |p|\le M\}\times [-M,M]\times U$. Unfortunately, this set is not compact because of the factor $U$. We can get uniform convergence on any compact subset of $U$, but to get it on all of $U$ an additional hypothesis is needed, such as $F$ being continuous on $\mathbb R^m\times\mathbb R\times \overline{U}$.
The second question is easier to answer. Here is a more general version: if $f_n\to f$ strongly in $L^q(U)$ and $g_n\rightharpoonup g$ weakly in $L^p(U)$ (here $p^{-1}+q^{-1}=1$ as usual), then $\int_U f_ng_n\to \int_U fg$. Proof:
• $\int_U f(g_n-g)\to 0$ by the definition of weak convergence.
• $\int_U (f_n-f)g_n \le \|f_n-f\|_q \|g_n\|_p\to 0$, using Hölder's inequality and the fact that weakly convergent sequences are bounded
• Adding the two, we get $\int_U(f_ng_n-fg)\to 0$. $\quad\Box$
Of course, uniform convergence on a bounded set implies strong convergence in every $L^q$.
Oh thank you very very much! I wasn't expecting a so quick and precise answer. You are totally right for the first point, I was wrong with the hypothesis on $F$, it is actually $C^\infty$ on $\mathbb{R}^m \times \mathbb{R} \times \overline{U}$ which gives $F$ uniformly continuous on the set you mention. And your proof is clear for the second question, a reasoning in functional analysis is rarely so clear to me :) – Nicolas Dec 21 '12 at 8:26