Composition operator on Sobolev space I am trying to consider hypothesis on $g$ such that the operator
$$ H_0^1 (\Omega) \to H^{-1}(\Omega), \qquad v \mapsto g(v) $$
is $\mathcal C^1$. Of course, $g(0) = 0$.
I believe that $g \in W^{2,\infty} (\mathbb R)$ is sufficient, but I am quite stuck on the proof.
Thanks in advance,
D
Edit: It is, of course, enough that it be differentiable $H^1 (\Omega) \to L^2 (\Omega)$.
 A: If $G$ is bounded, then for $w\in H^1(\Omega)$,
$$
|\langle g(v),w\rangle| \le  \int_\Omega |g(v)|\cdot|w| dx \le M \|w\|_{L^1(\Omega)},
$$
and you are done. As you can see, there is a lot of room to relax the assumptions on $g$. In fact, one can allow a certain growth of $g(x)$ for $|x|\to\infty$.
Then $g$ is well-defined.
Since $g \in W^{2,\infty}$ it is of class $C^1$. Thus for $x,y\in\mathbb R$,
$$
g(y)-g(x)-g'(x)(x-y)=\frac12 g''(\xi)\cdot|x-y|^2
$$ 
with $\xi$ between $x,y$. Since $g''$ is essentially bounded, we obtain
$$
|g(y)-g(x)-g'(x)(x-y)|\le M|x-y|^2\quad \forall x,y\in\mathbb R.
$$
Now take $v,w,u\in H^1(\Omega)$, then
$$
\int_\Omega |(g(v)-g(u)-g'(u)(v-u))w|dx  \le
M \int_\Omega |v-u|^2 w \le M \|v-u\|_{L^3(\Omega)} \|w\|_{L^3(\Omega)}.
$$
Since $H^1(\Omega)$ is continuously embedded into $L^3(\Omega)$ for spatial dimension $\le 6$, we found that the Frechet derivative $g'$ of $g$ at $v$ is given by
$$
\langle g'(v)z,w\rangle = \int_\Omega g'(v)zw \ dx.
$$
Again, there is some room for improvement: if spatial dimension is strictly less than $6$, you can even allow for a certain growth rate of $g''(x)$ as $x\to\infty$.
Continuity of $v\mapsto g'(v)$ can be proven similarly, by using
$$
|g'(y)-g'(x)| = |g''(\xi)(y-x)| \le M |y-x|,
$$
the growth requirements for $g''$ are the same as in the proof of Frechet differentiability.
