I'm working through Evans PDE $2$nd edition Chapter $5$, question $17$:
Assume $F:\mathbb{R}\rightarrow\mathbb{R}$ is $C^1$ with $F'$ bounded. Suppose $U\in \mathbb{R}^n$ is bounded and $u\in W^{1,p}(U)$ for some $1\leq p \leq \infty$. Show $$v:=F(u) \in W^{1,p}(U) \quad \text{and}\quad v_{x_i}=F'(u)u_{x_i} \quad (i \in [1,n])$$
I have been able to show that this is true for $1\leq p < \infty$ where we know that smooth functions $\{u_m\} \in C^{\infty}(U) \cap W^{1,p}(U)$ can approximate $u$ (and therefore can show that the well-defined, unique weak derivatives $F'(u_m)Du_m \rightarrow F'(u)Du$).
My issue is when $p=\infty$ and we do not have this approximation. Can someone help show me how this is true for $W^{1,\infty}(U)=Lip(U)$? Cheers.
2 Answers
For all $x,y\in U$ you have \begin{equation} |v(x)-v(y)|=|F(u(x))-F(u(y))|\leq Lip_{F}|u(x)-u(y)|\leq Lip_{F}Lip_{u}|x-y|, \end{equation} since $u\in Lip(U)=W^{1,\infty}(U)$. Then you know $v\in Lip(U)=W^{1,\infty}(U)$.
I'm answering my own question, starting from where @Cuteboy ended. A fellow student helped walk me through the remaining steps:
We can separately consider $p=\infty$, such that $W^{1,p}(U) = Lip(U)$. Because $F \in C^1$, we immediately also have that it is Lipschitz continuous. So, $$|F(u(x))-F(u(y))| \leq Lip_{F} |u(x)-u(y)| \leq Lip_{F}Lip_{u}|x-y|$$ and $F(u) \in Lip(U)$ as well.
Furthermore, Lipschitz continuous functions are absolutely continuous, and therefore have a derivative almost everywhere. This derivative follows the product rule, so $$D(F(u(x))) = F'(u(x)) Du(x)\text{ almost everywhere in }U$$ Lastly, because the weak derivative of $F(u)$ is uniquely defined, and we have: $$\int_U F'(u) u_{x_i} \phi dx = -\int_U F(u) \phi' dx$$ for any smooth test function $\phi \in \mathcal{D}(U)$, we have that the weak derivatives satisfy: $$\quad v_{x_i} = F'(u) u_{x_i}\quad i \in [1, n]$$ for $p=\infty$ as well.
