The problem: Assume $F:\mathbb{R} \to \mathbb{R}$ is $\mathcal{C}^1$, with $F'$ is bounded. $U$ is bounded and $u \in W^{1,p}(U)$, for $1 < p < \infty$. Show $v := F(u)$ is still in $W^{1,p}(U)$, and $v_{x_{i}}=F'(u)u_{x_{i}}$.
I try to use approximation to prove it. Let $u_m$ be in $\mathcal{C}^{\infty}(U)$, s.t. $u_m \to u \in W^{1,p}(U)$. Then, I can get
$$-\int_Uv_{x_i}\phi=\int_UF(u)\phi_{x_i}$$
and the R.H.S. can be approximated by $\int_UF(u_m)\phi_{x_i}$ which equals to $-\int_UF'(u_m)u_{m_{x_i}}\phi_{x_i}$.
I get stuck here, wondering how the convergence of $u_m$ in $L^p$ can imply convergence of $F'(u_m)$. Can the Hölder inequality be helpful?
One more question, what if $p=1$?
Any help would be appreciated.
So you are basically asking why
$$\int_U F'(u_m) D_i u_m \phi \to \int_U F'(u) D_i u \phi .$$
Note
$$\begin{split} &\int_U \left| F'(u_m) D_i u_m \phi- F'(u) D_i u \phi \right|\\ & = \int_U \left| F'(u_m) D_i u_m \phi- F'(u_m) D_i u \phi+ F'(u_m) D_i u \phi-F'(u) D_i u \phi \right|\\ &\le\|\phi\|_\infty \int_U \left| F'(u_m) D_i u_m - F'(u_m) D_i u\right| + \|\phi\|_\infty\int_U\left| F'(u_m) D_i u -F'(u) D_i u \right| \\ &\le \|\phi\|_\infty \|F'\|_\infty\int_U \left| D_i u_m - D_i u\right| + \|\phi\|_\infty\int_U\left| F'(u_m) -F'(u) |\ |D_i u \right| \end{split}$$
The first term goes to zero as $m\to \infty$, for the second term, as $u_m \to u$ in $L^1$, by passing to a subsequence one can assume $u_m \to u$ a.e. on $U$. Thus $F'(u_m) \to F'(u)$ a.e. on $U$. Now as
$$| F'(u_m) -F'(u) |\ |D_i u | \le 2\|F'\|_\infty |D_i u| \in L^1,$$
one can use Lebesgue dominated convergence theorem to show that
$$\int_U\left| F'(u_m) -F'(u) |\ |D_i u \right| \to 0.$$
The theorem works also for $p=1$.