I'm wondering in which situations the composition $f \circ v$ of a Lipschitz function $f$ and a $L^p$ function $v$ is still in $L^p$.
So one thing I came up with, is:
Claim: Let $\Omega \subset \mathbb{R}^m$ be bounded, $g : \mathbb{R}^m \rightarrow \mathbb{R}^m$ Lipschitz continuous with a root $v_0$ (i.e. $g(v_0)=0$). Then it holds $$ v \in L^p(\Omega, \mathbb{R}^m) \Rightarrow g \circ v \in L^p(\Omega, \mathbb{R}^m)$$ Proof: As $\Omega$ is bounded the constant function $v_0$ is in $L^p(\Omega, \mathbb{R}^m)$. Let $v \in L^p(\Omega, \mathbb{R}^m)$, then the Lipschitz property yields $$\Vert g(v(x)) \Vert_p^p = \Vert g(v(x)) - g(v_0) \Vert_p^p \leq C \Vert v(x) - v_0 \Vert_p^p$$ and therefore $$ \Vert g \circ v \Vert_{L^p} \leq C \Vert v - v_0 \Vert_{L^p} < \infty$$ which proves the claim.
So my question is, do you know other similar results? And is it even possible to loose the assumptions in my claim? For instance, does the claim still hold, if the root assumption is removed?
Thanks in advance!