Let $X = C_{0}(\Omega) := \{ u \in C(\overline{\Omega})\,|\,u|_{\partial\Omega}=0\}$ and define $F : X \to X$ as Lipschitz continuous function and $F(0) = 0$. Let $\Omega\subset \mathbb{R}^{N}$ be a bounded domain with Lipschitz continuous boundary and $u \in X\cap H_{0}^{1}(\Omega)$, show that $F(u) \in H_{0}^{1}(\Omega)$!
This is my attempt so far :
\begin{align*}
||F(u)||_{H_{0}^{1}}^{2} &= ||F(u)||_{L^{2}}^{2} + ||\nabla F(u)||_{L^{2}}^{2}\\
&:= A + B
\end{align*}
First, I show that $A$ is bounded. Observe that \begin{align*} A &= \int_{\Omega}|F(u)|^{2}dx \\ &\leq \int_{\Omega}(\sup\limits_{\Omega}|F(u)|)^{2}dx = \int_{\Omega}||F(u)||_{X}^{2}dx\\ &=|\Omega|\,||F(u)-F(0)||_{X}^{2}\\ &\leq |\Omega|(L||u-0||^{2}_{X})^{2} = L|\Omega|\,||u||_{X}^{2}<\infty \end{align*} Hence, $A$ is bounded.
Now, my problem is how to show that $B$ is bounded as well? Since $F$ is Lipschitz, is there anything I can say for $B := \int_{\Omega}|\nabla F(u)|^{2}dx$?
Any help is pretty much appreciated!
In this question one can find a proof under the assumption that $F$ is continuously differentiable. From here one can proceed by approximation.
Let $(\eta_\epsilon)_{\epsilon>0}$ be a mollifying kernel and $F_\epsilon=F\ast \eta_\epsilon-F\ast\eta_\epsilon(0)$. Then $F_\epsilon\in C^\infty$, $F_\epsilon\to F$ uniformly and $|F_\epsilon'|\leq L$, where $L$ is the Lipschitz constant of $F$. Since the energy functional $u\mapsto \int|\nabla u|^2$ is lower semicontinuous on $L^2$ (this is equivalent to the completeness of $H^1$), one has $$ \int |\nabla (F\circ u)|^2\leq\liminf_{\epsilon\searrow 0}\int|\nabla (F_\epsilon\circ u)|^2=\liminf_{\epsilon\searrow 0}\int|F_\epsilon'\circ u|^2|\nabla u|^2\leq L^2\int|\nabla u|^2. $$