What is the Frechet derivative of $(u^+)^q$?

Question

I know that if we define $E[u]=\int_\Omega u^+dx$, where $\Omega$ is compact in $R^n$ and $u\in H_0^1(\Omega)$, $u^+:=\max\{u,0\}$, then $E[u]$ is not Frechet differentiable. However, if now I define $E[u]=\int_\Omega (u^+)^qdx$, for some $q>2$, then is it $E$ is Frechet differentiable?(all other setting are same). If yes, what is it? Thank you!

Answer 1

@Tomas

Sure, let me first show $u^+$ is not Fréchet differentiable. By definition I need to prove that, for some $DE(u)\in H_0^1(\Omega)=X$, we need to have \begin{equation} \frac{|E(u+v)-E(u)-DE(u)v|}{\|v\|_X}\to 0, \end{equation} as $\|v\|\to0$. Then, consider the definition of $E$, we need to prove \begin{equation} \frac{\left|\int (u+v)^+-\int u^+-\int \chi_{u^+}v\right|}{\|v\|_X}\to 0. \end{equation} However, the above limit can not be true for example if we choose $v_n=u^-\cdot \chi_{B(0,1/n)}$.

Why I think $(u^+)^q$ is not Fréchet differentiable is by the same idea... $u^-$ and $u^+$ has totally different support...