# When is a Nemytskii map between Sobolev spaces compact?

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?

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.
• For the non-compactness of the set $E$: start with a sequence $u_j\in H^1$ with $\|\nabla u_j\|_2=1$ and $u_j \to 0$ a.e. If $f(t):=\mathrm{dist}(t,\mathbb{Z})$, then $v_j:=f(u_j)$ has still $\|\nabla v_j\|_2=1$, $v_j \to 0$ a.e. and also $\|v_j\|_{\infty}\le1$. May 3 '17 at 11:33
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
• (some argument is still needed to pass from the setting of measurable functions to $H^1$) May 3 '17 at 11:36