Let $f:\mathbb{R} \to \mathbb{R}$ be differentiable with continuous and bounded derivative. We have a bounded smooth domain $\Omega$ and a superposition map $F:H^1(\Omega) \to H^1(\Omega)$ defined by $$F(u) = f(u(x)).$$ In fact $F(u) \in L^2(\Omega)$ if $u \in L^2(\Omega)$.
We know $f:\mathbb{R} \to \mathbb{R}$ is Frechet differentiable. Does it follow that $F:H^1(\Omega) \to (H^1(\Omega))^*$ is Frechet differentiable, and if not are there additional conditions on $f$ that assure that it is?
I see that by existing literature about superposition operators on $L^p$-type spaces, it's hard to show that $F:L^2 \to L^2$ is Frechet differentiable since the exponents of the $L^p$ spaces are the same.
By Sobolev embeddings, the space $H^1(\Omega)$ is continuously embedded in $L^p(\Omega)$, for some $p>2$. This implies that $L^{p'}(\Omega)$ is continuously embedded in $H^1(\Omega)^*$ for $1/p + 1/p'=1$. Thus it suffices to study the superposition operator $F:L^p(\Omega)\to L^{p'}(\Omega)$.
It is sufficient for Frechet differentiability of $F$ that the superposition operator induced by $f'$ maps $L^p$ to $L^s$ with $1/p + 1/s =1/p'$ (or $s=\frac p{p-2}$), which for $p<\infty$ is true under a suitable growth condition, i.e., $$ |f'(x)| \le a + b |x|^{p/s} = a + b |x|^{p-2} . $$ I suspect that such a growth condition is also necessary, if we set $p$ to the maximal possible value $p^*$ and $p^*<\infty$. In the 2d case, it is known that for $f(x)=e^x$ the corresponding $F$ is still Frechet from $H^1(\Omega)$ to every $L^p(\Omega)$, which violates any polynomial growth condition. In the 1d case, we have $p=\infty$, and the growth condition above changes to the requirement that $f'$ maps bounded sets to bounded sets.