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

*

*a map from $H^2(\Omega)\cap H_0^1(\Omega)$ to itself.

*Lipschitz on every bounded subset of $H^2(\Omega)\cap H_0^1(\Omega)$.

The paper establishes as lemmas that

*

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

*$|||u|^2u||_{H^2}\lesssim ||u||_{\infty}^2||u||_{H^2}$ 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
 A: 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.
