Composition of an $L^{\infty}$ function with an homeomorphism

Question

Let $\Omega$ be a bounded open subset of $\mathbb{R}^2$ supposed convex with smooth boundary. Let $f\in L^{\infty}(\Omega)$ and let $\varphi\in Id+W_{0}^{1,\infty}(\Omega,\mathbb{R}^2)$ be an homeomorphism from $\bar{\Omega}$ to $\bar{\Omega}$ such that $\forall q>2$, $\varphi^{-1}$ is in $W^{1,q}(\Omega,\mathbb{R}^2)$. Is $f \circ \varphi^{-1}\in L^{\infty}(\Omega)$? If yes, is it true that $\|f \circ \varphi^{-1}\|_{L^{\infty}(\Omega)}\leq \|f\|_{L^{\infty}(\Omega)}$? I do thank you for your attention. Best wishes.

Answer 1

Consider $N\subset \Omega$ such that $|f(x)|\leq M:= \|f\|_{L^\infty}$ for every $x\in \Omega\setminus N$ and $|N|=0$. Let $N':=\varphi(N)$, then we have

1. $|N'|=0$.
2. $|f\circ\varphi^{-1}(y)|\leq M$ for every $y\notin N'$.

1 follows from the fact that Lipschitz maps decrease Lebesgue measure (or more generally any Hausdorff measure), i.e. $|\varphi(A)|\leq \| \varphi\|_{Lip}|A|$ for every set $A$. On the other hand, if $y\notin N'$, then $y=\varphi(x)$ for some $x\notin N$ and so $f\circ \varphi^{-1}(y)= f(x)$ and the claim follows from the definition of $N$.

It seems to me that the extra assumptions on $\varphi^{-1}$ are not needed, but maybe I'm misunderstanding something?