Let $1\le p <\infty$, $\psi\in C^1(\mathbb{R})$ such that $\psi (0)=0$ and $|\psi '(y)|\le C|y|^{p-1}$ for all $y\in\mathbb{R}$ and for a constant $C>0$.
Prove that $F:L^p(I)\to \mathbb{R}$ defined by $F(u)=\int_I \psi udx$ is welldefined and it's Gateaux derivative is $DF(u)=\psi(u)$.
Well defined: $|\int_I \psi udx|\le \int_I| \psi u|dx$. And now, how to continue? $\psi$ don't has to be bounded or in $L^p(I)$. I stuck with estimate $\psi$.
For the second part let $u,v\in L^p(I)$, then $\frac{F(u +hv)-F(v)}{h}=\int_I \frac{\psi (u+hv)}{h}dx$. I can't derive $u$.. How to continue?
As $\phi \in C^1$, you have that $\phi(y)-\phi(0) = (y-0) \phi'(c)$ with $c \in [0,y]$
This imply that
$$| \phi(y) | = |y| |\phi'(c)| \leq |y|C |c|^{p-1} \leq C|y|^p$$
The majoration follow directly from this.
For the Gateau derivative, $$ \frac{\phi(u+hv) - \phi(u)}{h} = \phi'(u).v + \epsilon(h)$$
So
$$\frac{F(u+hv)-F(u)}{h} = \int_I \phi'(u(x))v(x) + \epsilon(h)(x) dx$$
So this will give us that the Gâteau derivative is the functionnal
$$Df(u)(v) = \langle \phi'(u), v \rangle = \int_I \phi'(u(x)) v(x) dx $$