Continuity of functionals on Sobolev space Let $U$ be a bounded set in $R^n$ and $W^{1,p}(U)$ denote a Sobolev space.
Suppose $\{w_n\}\subset W^{1,p}(U)$ converges to $w \in W^{1,p}(U)$.
Let $I[w]=\int_U F(Dw,w,x)dx$ for $w\in W^{1,p}(U)$, where $F$ is smooth.
Then, does $I[w_n]$ converge to $I[w]$? 
If not, can I force $I[w_n]$ converge to $I[w]$ by making more assumptions on F?
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I forgot to mention the assumption that $1<p<\infty$
 A: In general, no.  Take $U = (0,1) \subset \mathbb{R}^1$, $p=2$ and $F(z,w,x) = z^4$.  Then take 
$$w_n(x) = \begin{cases} n^{1/4} x, & 0 < x < n^{-1} \\ n^{-3/4}, & x \ge n^{-1} \end{cases}$$
so that $w_n' = n^{1/4} 1_{(0, n^{-1})}$.   It's easy then to see that $w_n' \to 0$ and $w_n \to 0$ in $L^2$, so $w_n \to 0$ in $W^{1,2}(U)$.  Taking $w = 0$, we have $I[w] = \int_0^1 0^4\,dx = 0$.  But $I[w_n] = \int_0^1 (w_n'(x))^4\,dx = \int_0^{1/n} n\,dx = 1$ for every $n$.
If $F$ is bounded, yes.  Use the double subsequence trick and dominated convergence.  If $I[w_n] \not\to I[w]$ we may pass to a subsequence to find some $c \in [-\infty, \infty]$ such that $I[w_n] \to c \ne I[w]$.  Now since $w_n \to w$ and $Dw_n \to Dw$ in $L^p$, we can pass to a further subsequence so that they converge almost everywhere on $U$.  On this subsequence we still have $I[w_n] \to c \ne I[w]$.  By the continuity of $F$, we then have $F(Dw_n(x), w_n(x), x) \to F(w_n(x), w(x), x)$ for almost every $x \in U$.  Then using dominated convergence with $\|F\|_\infty$ as our dominating function, we get that $I[w_n] \to I[w]$ which is a contradiction.
