Let $f : \mathbb{R}\to\mathbb{R}$ be a locally Lipschitz continuous function and $f(0)=0$. I want to show that if $u \in C([0,T];L^{2}(\mathbb{R}^{N}))$, then $f(u(\,\cdot\,))$ is locally Lipschitz in $L^{2}(\mathbb{R}^{N})$. Is there any natural assumption in which the argument can work? I know that replacing $L^{2}(\mathbb{R}^{N})$ with $C_{0}(\Omega)$ for $\Omega$ a bounded subset of $\mathbb{R}^{N}$ works well.
Any help is much appreciated here! Thank you very much!
On
This does not work without any further assumptions on $f$. There is no guarantee that $t\mapsto f(u(t))$ is in $L^2$ for some $t$. For instance, take $v\in L^2\setminus L^4$. Set $u(t):=v$, $f(u):=u^2$, then $f(u(t))\not\in L^2$ for all $t$.
Also local Lipschitz continuity of $f$ does not go well with $L^2$-functions. Such functions are potentially unbounded, so $f$ is potentially applied to values from an unbounded set.
Even in the simple case of $f(x) = x^2$, $u\mapsto f(u)$ is not locally Lipschitz in $L^2(\mathbb{R}^N)$: for $$ \|f(u+v) - f(u)\|_{L^2} = \|(u+v)(u-v)\|_{L^2}. $$ By taking $v=0$, we see that for $f$ to be locally Lipschitz, you need that for all $R>0$, there is a constant $C_R$ so that for $u\in B(0,R) \subset L^2$, you have $$ \|u^2\|_{L^2} \leq C_R\|u\|_{L^2}. $$ However, the change of variable $u_\lambda(x) = \lambda^{N/2}u(\lambda x)$ preserves the $L^2$-norm. Putting $u_\lambda$ into the above inequality, we find that $$ \lambda^{N/2}\|u^2\|_{L^2} = \|u_\lambda^2\|_{L^2} \leq C_R\|u_\lambda\|_{L^2} = C_R\|u\|_{L^2} $$ and taking $\lambda\to\infty$ we see that no such $C_R$ can exist. By a similar argument no function of the form $f(x) = |x|^r$, $r>1$ can give you such a Lipschitz bound.
Typically, to conclude this sort of bound one needs much more information. You need to know the structure of $f$, the function space you are working on, and what specific estimates (e.g. Sobolev embedding and the like) are available in your function spaces.