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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!

Evan William Chandra
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  • I think you made a mistake in the problem. Take any function $\phi\in X\setminus H^1(\Omega)$ and let $F(f)=f(p)\phi$ for some $p\in \Omega$. This map $F$ satisfies all you assumptions, yet $F(f)\notin H^1_0(\Omega)$ unless $f(p)=0$. Instead I think $f$ should be a map from $\mathbb{R}$ to $\mathbb{R}$ and $F(u)$ denotes the composition of $u$ and $F$. – MaoWao Jan 26 '19 at 22:01
  • Thank you for your correction! So, supposing $F(u) : \mathbb{R} \to \mathbb{R}$, can I do something to prove the boundedness of $B$? – Evan William Chandra Jan 28 '19 at 01:40
  • I think your question is answered here (in more generality): https://math.stackexchange.com/questions/1110231/chain-rule-in-the-sobolev-space-w1-p – MaoWao Jan 28 '19 at 12:49
  • From what I've seen, that question assumes the boundedness of $F$ as a once differentiable function. Can I assume the same thing given I have locally Lipschitz function? I know Lipschitz continuous function implies differentiability a.e. but I am not sure about the boundedness of derivatives – Evan William Chandra Jan 29 '19 at 01:49

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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 $$ \int |\nabla (F\circ u)|^2\leq\liminf_{\epsilon\searrow 0}\int|\nabla (F_\epsilon\circ u)|^2=\liminf_{\epsilon\searrow 0}\int|F_\epsilon'\circ u|^2|\nabla u|^2\leq L^2\int|\nabla u|^2. $$

MaoWao
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