Is the map $u\mapsto |u|^2 u$ globally or locally Lipschitz continuous in the $H_0^1$ norm? Currently, I want to know if the non-linearity in my differential equation is locally or globally Lipschitz. The linear part is the following map $\mathcal{L}:D_{\mathcal{L}}\subset X \longrightarrow X$, where $D_\mathcal{L}(\mathcal{L}) = H_0^1(\mathbb{R},\mathbb{C}) \cap H^2(\mathbb{R},\mathbb{C})$ and $X=L^2(\mathbb{R},\mathbb{C})$. Therefore, the functions inputted into the nonlinearity belong to this domain as well. 
My nonlinearity is given by $|u|^2u$, and thus I want to show that for two functions $u,v \in D_{\mathcal{L}}$ that 
\begin{equation}
||( |u|^2u - v|v|^2)||_{L^2} \leq K||u-v||_{L^2},
\end{equation} where $K$ is either a constant defined for all $u,v \in D_\mathcal{L}$ (globally Lipschitz) or $K$ is a constant that is dependent on  each $u,v$ chosen (locally Lipschitz). 
I thinking of perhaps using this inequality?
\begin{equation}
|u|u|^2 - v |v|^2| \leq 3(|u|^2 - |v|^2)|u-v|.
\end{equation}
 A: The map $F(u) = |u|^2 u$ is homogeneous of degree 3, meaning $F(\lambda u) = \lambda^3 u$ for all positive scalars $\lambda$. A homogeneous map cannot be globally Lipschitz unless the degree of homogeneity is $1$. Indeed, $\lambda\mapsto \lambda ^d$ is not Lipschitz on $(0, \infty)$ unless $d=1$. 
The best one can hope for is an inequality of the form
$$
\|F(u)-F(v)\| \leq L(\|u\|+\|v\|)^{d-1}\|u-v\|
\tag1$$
where $d$ is the degree of homogeneity. Both sides of (1) are homogeneous of degree $d$; thus we can divide both $u, v$ by the same constant, reducing the problem to the case $\|u\|, \|v\|\le 1$. In other words: if $F$ is degree $d$ homogeneous and Lipschitz on the unit ball, then it satisfies (1). 
Unfortunately, $F(u)=|u|^2u$ does not satisfy any such bound in the $L^2$ norm. Consider the function on the interval $[0, 1]$ given by 
$$
u_\epsilon (t) = (t+\epsilon )^{-1/3}, \quad \epsilon>0
$$
It is smooth, is in $L^2$ and its $L^2$ norm stays bounded as $\epsilon \to0$. Yet, the $L^2$ norm of $|u_\epsilon|^2u_\epsilon$ blows up as $\epsilon\to 0$.
A similar example for $H^1$ norm can be obtained by increasing the exponent by $1$ (this will be taken out by the derivative): 
$$
u_\epsilon (t) = (t+\epsilon )^{2/3}, \quad \epsilon>0
$$
Or, one can put $\epsilon$ in the exponent in such a way that it will produce blow up for $F(u)$ but not for $u$. This kind of example can be turned into an $H^1_0([-1, 1])$ function by multiplying it with a smooth cut-off.
