When is a Nemytskii map between Sobolev spaces compact?

Question

Let $f:\mathbb{R} \to \mathbb{R}$ be a smooth function with bounded derivative. Define the Nemytskii map $F:H^1(\Omega) \to H^1(\Omega)$ by $F(u)(x) := f(u(x))$. Here $\Omega$ is a bounded smooth domain.

There exists work where we can deduce continuity and differentiability of $F$ under some assumptions on $f$.

What conditions on $f$ will ensure that $F$ satisfies the following compactness criterion: if $u_n \rightharpoonup u$ in $H^1(\Omega)$ (weakly), then there is a subsequence of the $u_n$ such that $F(u_{n_j}) \to F(u)$ strongly in $H^1(\Omega)$?

If convergence is $H^1$ is too difficult, $L^2$ (in both domain and range of $F$) can be ok. Does anyone come across this before?

Answer 1

I think in your setting $F$ is not compact, unless $f$ is a constant map. Indeed, if $f$ is not a constant map, it coincides on some non-trivial interval $[a,b]$ with a smooth diffeomorphism $g:\mathbb{R}\to\mathbb{R}$ with derivatives $g'$ and $(g^{-1})'$ bounded on $\mathbb{R}$. Thus the corresponding Nemytskii map $G:u\mapsto g\circ u$ is a homeomorphism of $H^1(\Omega)$ into itself.

The set $E$ of all $u$ in the unit ball of $H^1(\Omega)$ with $a\le u \le b$ is of course bounded, but never relatively compact in $H^1$, so neither is $G(E)$. But the latter set coincides with $F(E)$, so $F$ is not a compact map in the said sense.

Answer 2

There is a general theorem about superposition (i.e., Nemytskii) operators between spaces of measurable functions: If the measure is non-atomic, the superposition operator defined by $f$ is compact if and only if $f$ is constant, see Theorem 1.8 in

Appell, Jürgen; Zabrejko, Petr P., Nonlinear superposition operators, Cambridge Tracts in Mathematics 95. Cambridge: Cambridge University Press (ISBN 978-0-521-09093-3/pbk). 311~p. (2008). ZBL1156.47052.