$\newcommand{\R}{\mathbb R}$ Let $F: \R\to \mathbb R$ be a smooth function such that $F(0) = 0$. Now consider $g\in W^{k,2}(\R^n)$ (Sobolev space) with $k>n/2$.
Is it true that $F\circ g\in W^{k,2}(\R^n) $?
My attempt:
$F\circ g \in L^2$ because by Sobolev embedding theorem $g$ is Holder (hence uniformely continuous) and in $L^2$, thus $g(x)\to 0$ as $|x|\to +\infty$ and $F$ satisfies $F(g(t)) \leq 0 + \sup_{|s|<\epsilon}\{\partial F(s)\} \cdot (g(t)) \leq const. g(t)$ for $t$ large enough (so that $|g(t)|<\epsilon$).
Since $F$ is smooth, then $F\circ g \in W^{k,2}_{loc}(\R^n)$ (locally Sobolev) with weak derivatives given by iterative applications of the chain rule, for example $\partial_x (F\circ g)(x) = \partial F(g(x))\partial_x g(x)$ (this should follow by a density argument+ locally Lipshitzianity of $F$).
It remains to verify that the weak derivatives lie in $L^2$. In the case of the first derivative, $\partial F(g(x))$ is bounded in a neighbourhood of $\infty$ and $\partial_x g(x)\in L^2$, therefore their product is in $L^2$. For higher derivatives, it is more complicated but seems to be true looking for example at the case of $\R^1$ where the second derivative is $\partial^2 F (\partial g)^2 + \partial F \partial^2 g $ which is in $W^{k,2}(\R)$ because $\partial^2 F$ and $\partial F$ are bounded in a neighbourhood of $\infty$, and multiplication $W^{1,2}(\R)\times W^{1,2}(\R)\to W^{1,2}(\R)$ is continuous. Is there a trick for the general case maybe?
What about fractional derivatives?