## Derivative of functional

Define $J\colon W_0^{1,2}(\Omega)\to \mathbb{R}$ by $J(u)=\int_\Omega u^{p+1}dx$ for $p\in (1,\frac{n+2}{n-2})$.

Is $J'(u)(v)=\int_\Omega (p+1)u^pvdx$?

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 Here, $J'(u)$ is the Frechet derivative of $J$ at $u$, of course. – goooooogler Dec 30 2010 at 10:53 $u^{p+1}:=u|u|^{p}$ and $\Omega$ is bounded right? – Pietro Majer Dec 30 2010 at 12:29 You Right. $\Omega$ is bounded. – goooooogler Dec 30 2010 at 12:40

Start with the case of $J$ on $L^{p+1}(\Omega)$, here with any measure space $\Omega$ and $p>1$. Integrate the first order Taylor expansion for $(u+v)^{p+1}$ at $u$ and get $$\bigg|\ J(u+v)-J(u)-(p+1)\int_\Omega |u|^p v\ dx\ \bigg| \le p(p+1)\int_\Omega \big(|u|+|v|\big)^ {p-1} |v|^2 \ dx.$$ Use the Hoelder inequality to check that the RHS is actually $o(\|v\|_{p+1}).$ You can also use he dominated convergence theorem to show that $J$ is $C^1$. Your Sobolev space case follows immediately by composition with the Gagliardo-Nirenberg-Sobolev embedding (assuming you have $\Omega$ bounded in the case $p+1\neq 2^*$, otherwise $J(u)$ may be not even defined for some $u$).