Proving $u\mapsto |u|^2u$ is Lipschitz on bounded subsets of $H^2(\Omega)\cap H_0^1(\Omega).$

Question

I'm [reading a paper][1]¹ and am stumped verifying two details.

Let $\Omega$ be a bounded region in $\mathbb{R}^2$ with smooth boundary. I'd like to show that the map $u\mapsto |u|^2u$ is

1. a map from $H^2(\Omega)\cap H_0^1(\Omega)$ to itself.
2. Lipschitz on every bounded subset of $H^2(\Omega)\cap H_0^1(\Omega)$.

The paper establishes as lemmas that

1. If $u\in H^2(\Omega)$ and $||u||_{H^1}\le 1$, then $||u||_{\infty}\lesssim 1+\sqrt{\log(1+||u||_{H^2})}$
2. $|||u|^2u||_{H^2}\lesssim ||u||_{\infty}^2||u||_{H^2}$ for $u\in H^2(\Omega)$.

Combining the two lemmas makes it easy to see that if $u\in H^2(\Omega)$ then $|u|^2u\in H^2(\Omega)$. If $u\in H^2(\Omega)\cap H_0^1(\Omega)$ and $\phi_n\in C_0^{\infty}(\overline{\Omega})$ are such that $||\phi_n-u||_{H^1}\to 0$, it would suffice to show that $|||\phi_n|^2\phi_n-|u|^2u||_{H^1}\to 0$, but I don't know how to go about showing this.

Thanks for your help!

¹Brézis, Haïm; Gallouet, T., Nonlinear Schrödinger evolution equations, Nonlinear Anal., Theory Methods Appl. 4, 677-681 (1980). ZBL0451.35023. [1]: https://www.i2m.univ-amu.fr/perso/thierry.gallouet/art.d/brezisgallouet.pdf

Answer 1

All spaces are assumed complex, so $|u|^2u$ cannot be replaced by $u^3$. Your difficulties with convergence come from inappropriate definition of sequence $\{\phi_n\}$ — it must be a sequence convergent in $H^2(\Omega)\cap H^1_0(\Omega)$. To show that $|u|^2u\in H^1_0(\Omega)$, there is more elegant way. Take an alternative equivalent definition $$ H^1_0(\Omega)=\{w\in H^1(\Omega)\,\colon\; w|_{\partial\Omega}=0\}= \overline{C_0^{\infty}(\Omega)}^{H^1(\Omega)} $$ and consider the trace of $\,|u|^2u\,$ on $\,\partial\Omega\,$ with $\,|u|^2u\in H^2(\Omega)\subset C(\overline{\Omega})$. Since $\, u|_{\partial\Omega}=0\,$ implies $\, (|u|^2u)|_{\partial\Omega}=0$, one concludes that $\,|u|^2u\in H^2(\Omega)\cap H_0^1(\Omega)$. To show that the mapping $$ \,L\colon H^2(\Omega)\cap H_0^1(\Omega)\to H^2(\Omega)\cap H_0^1(\Omega)\,\tag{$\ast$} $$ where $\,L(u)\overset{\rm def}{=}|u|^2u$, is Lipschitz on every bounded set in $H^2(\Omega)\cap H_0^1(\Omega)$, notice that \begin{align*} L(u)-L(v)=|u|^2u-|v|^2v= (u-v)\!\cdot\!|v|^2+(|u|^2-|v|^2)\!\cdot\!v\\ =(u-v)\!\cdot\!|v|^2+(u_1^2+u_2^2-v_1^2-v_2^2)\!\cdot\!v =(u-v)\!\cdot\!|v|^2+(u_1-v_1)\!\cdot\!(u_1+v_1)\!\cdot\!v\\ +(u_2-v_2)\!\cdot\!(u_2+v_2)\!\cdot\!v \end{align*} where $\,u=u_1+iu_2\,$ and $\,v=v_1+iv_2\,$ with $$ u,v\in B_M\overset{\rm def}{=}\{w\in H^2(\Omega)\cap H_0^1(\Omega)\,\colon\; \|w\|_{H^2(\Omega)}\leqslant M\},\quad M>0. $$ Applying to the second order derivatives of products the Leibnitz rule along with the appropriate estimates for embeddings of $\,H^2(\Omega)\,$ into $\,W^{1,p}(\Omega)\,$ and $\,C(\overline{\Omega})$, i.e., following basically the way the two lemmas are being proved, one readily finds that $$ \|L(u)-L(v)\|_{H^2(\Omega)}\leqslant CM^2\|u-v\|_{H^2(\Omega)}\quad\forall\, u,v\in B_M\tag{$\ast\ast$} $$ with some constant $C>0$ depending only on domain $\Omega$.  Inequality $(\ast\ast)$ implies that the nonlinear mapping $(\ast)$ is Lipschitz on every bounded set in $H^2(\Omega)\cap H_0^1(\Omega)$, Q.E.D.