composition of functions in sobolev space

Question

If we have $u\in H^s(\Omega)$ where $\Omega$ is a domain of $\mathbb R^n,s>0$ and $ g : \mathbb R^n \to \mathbb R^n$, what is the least condition on $g$ to ensure $g(u) \in H^s.$ I think $g\in C^\infty$ is enough, but it seems not optimal. Need some references. Thanks!

Answer 1

If $\Omega$ is unbounded, then $C^{\infty}$ is not sufficient without growth conditions on $g.$ For example taking $\Omega = \mathbb R$ and $g(x) = 1$ (which is smooth) we get $g \circ u \equiv 1$ which is not integrable regardless of what $u$ is.

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For the case $s = 1,$ we have the following result.

Theorem: If $\Omega \subset \mathbb R^n$ is bounded and $g : \mathbb R \rightarrow \mathbb R$ is globally Lipschitz continuous, then $g \circ u \in H^1(\Omega)$ whenever $u \in H^1(\Omega).$ For general $\Omega,$ the same holds if $g(0) = 0.$

The proof is a standard approximation argument, starting with the case where $g$ is $C^1$ with bounded derivative. We take a sequence $u_n \in C^{\infty}(\Omega) \cap H^1(\Omega)$ such that $u_n \rightarrow u$ in $H^1(\Omega)$ and show that $g(u_n)$ converges to $g(u)$ in $H^1(\Omega)$ also. The general case follows by a similar argument, except we approximate $g$ by $C^1$ functions.

The $C^1$ should be covered in any introductory text on Sobolev spaces. The extension to Lipschitz $g$ is generally omitted because there's a few subtle points regarding what exactly the derivative of $g(u)$ is (it's $g'(u)Du,$ for an appropriate representative of $g'$).

By iterative differentiation we get for $u \in H^k$ with $k$ integer, we need $g \in C^{k-1,1},$ so $(k-1)$-times differentiable with Lipschitz $(k-1)$st derivatives, with everything globally bounded.

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For general $s>0$ I do not know what happens.