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Let $\Omega$ be a bounded open subset of $\mathbb R^n$, and $T: \mathbb R \to \mathbb R$ an absolutely continuous function with sublinear growth, in the sense that

$$|T(x)| \leq 1 + |x|, \forall x \in \mathbb R.$$

Question: Let $1 \leq p < \infty$, and suppose further than $T’ \in L^p$. Is it true that for all $u \in W^{1, p}(\Omega)$, we have $T \circ u \in W^{1, p}(\Omega)$?

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Take $\Omega=(0,1)$. As it was pointed out in another post, $T$ absolutely continuous is equivalent to $\partial_x T \in L^1(0,1)$. If you want $T\circ u \in W^{1,p}$, it is a good idea to have $\partial_x T\in L^p$. I don't see why sublinearity would help for the properties of the derivative - I can see why you imposed it, since it is useful to stay in $L^p$ for $T\circ u$. But if you take $u:x\to x$, you only have $T\circ u=T$ with a derivative in $L^1$.

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  • $\begingroup$ Ah, that makes sense thanks. I will impose the additional constraint on the derivative in the original post. $\endgroup$
    – Nate River
    Commented Aug 7, 2021 at 7:18

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