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Let $\Omega \subseteq \mathbb{R}^n$ be open, bounded with smooth boundary. Does $u \in H^1(\Omega)$ imply that $(1+u^2)^{\frac{1}{2}}\in H^1(\Omega)$?

If yes, I would expect a more general result holds about composing $H^1(\Omega)$ with $C^1$ sufficiently slowly increasing functions with sufficiently slowly increasing derivatives. I am not really sure about how one goes about proving this kind of result.

Nicolò Cavalleri
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  • In general, if $u\in H^1(\Omega)$ and $\phi$ is Lipschitz, then $\phi\circ u\in H^1(\Omega)$. That $\phi\circ u\in L^2(\Omega)$ is not hard to see and $\nabla(\phi\circ u)\in L^2(\Omega)$ follows from the chain rule. Proving the chain rule for Sobolev functions takes some approximation arguments, but should be covered in standard sources on the topic. – MaoWao Dec 03 '20 at 19:12

1 Answers1

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Define $f(x)=(1+x^2)^{1/2}$. Note that $f \in C^1(\mathbb{R})$ and $f'(x)=x/(1+x^2)^{1/2}$ is bounded in the sense $|f'(x)|\leq 1$ for all $x \in \mathbb{R}$. Then by the result in this question we have $f(u) \in H^1(\Omega)$, exactly as you wanted.

Cahn
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