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How can I prove that the following Functional is Frechet Differentiable and that the Frechet derivative is continuous?
$ I(u)=\int_\Omega |u|^{p+1} dx , \quad 1

where $\Omega$ is a bounded open subset of $\mathbb{R}^n$ and $I$ is a functional on $H^1_0(\Omega).$

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    @DavideGiraudo $ I^\prime(u)(\psi)=(p+1)\int_\Omega |u|^{p-1}u\psi dx $2012-08-14

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As was given in the comments, the Gâteaux derivative is $ I'(u)\psi = (p+1) \int_\Omega |u|^{p-1}u\psi. $ It is clearly linear, and bounded on $L^{p+1}$ since $ |I'(u)\psi| \leq (p+1) \|u\|_{p+1}^p\|\psi\|_{p+1}, $ by the Hölder inequality with the exponents $\frac{p+1}p$ and $p+1$. Here $\|\cdot\|_{q}$ denotes the $L^q$-norm. Then the boundedness on $H^1_0(\Omega)$ follows from the continuity of the embedding $H^1_0\subset L^{p+1}$.

Now we will show the continuity of $I':L^{p+1}\to (L^{p+1})'$, with the latter space taken with its norm topology. First, some elementary calculus. For a constant $a>0$, the function $f(x)=|x|^a x$ is continuously differentiable with $ f'(x) = (a+1)|x|^a, $ implying that $ \left||x|^ax-|y|^ay\right| \leq (a+1)\left|\int_{x}^{y} |t|^a\mathrm{d}t \right| \leq (a+1)\max\{|x|^a,|y|^a\}|x-y|. $ Using this, we have $ |I(u)\psi-I(v)\psi|\leq (p+1)\int_\Omega \left||u|^{p-1}u-|v|^{p-1}v\right|\cdot|\psi| \leq p(p+1) \int_\Omega \left(|u|^{p-1}+|v|^{p-1}\right)|u-v|\cdot|\psi|. $ Finally, it follows from the Hölder inequality with the exponents $\frac{p+1}{p-1}$, $p+1$, and $p+1$ that $ |I(u)\psi-I(v)\psi| \leq p(p+1) \left(\|u\|_{p+1}^{p-1}+\|v\|_{p+1}^{p-1}\right)\|u-v\|_{p+1}\cdot\|\psi\|_{p+1}, $ which establishes the claim. Finally, the continuity of $I':H^1_0\to (H^1_0)'$ follows from the continuity of the embedding $H^1_0\subset L^{p+1}$.