Let $\Omega\_1 \subset \mathbb R^2$ be a bounded simply-connected Lipschitz domain, and $f: \bar \Omega\_1 \rightarrow \bar \Omega\_2$ be a homeomorphism, which is a diffeomorphism on $\Omega\_1$ such that $f$ and its inverse have first derivatives in $L^p$ with $p>2$.


The Morrey inequality implies that $f$ is Hölder. What can be said about its inverse provided that $p$ is large enough? The Morrey inequality cannot be directly applied to the inverse since the boundary of $\Omega\_2$ might be not Lipschitz, but could it be salvaged with the additional information we have?



Hölderness of the inverse to a $W^{1,p}$-homeomorphism (with additional conditions) of a Lipschitz domain
What you want is not possible. The basic idea is that it is possible for a diffeo $\Omega\_1\to \Omega\_2$ that extends to a homeo $\bar{\Omega}\_1 \to \bar{\Omega}\_2$ to "almost fold up" $\partial\Omega\_1$.


To give an explicit example: let $\Omega\_1$ be the upper half-unit-disk, in polar coordinates the set $\{0< r < 1, 0 < \theta < \pi\}$.


Let $f(r,\theta) = (r,(2 - r^q)\theta)$ for some $q \gg 1$. Then $\Omega\_2 = \{0 < r < 1, 0 < \theta < (2-r^q) \pi \}$. Let $g$ be the inverse mapping, we find $g(s,\phi) = (s, (2 - s^q)^{-1} \phi)$. Both $f$ and $g$ are clearly smooth, and extend continuously to the boundary of the domains $\Omega\_1$ and $\Omega\_2$ respectively.


Furthermore,
$$ |\nabla f|^2 = 1 + q^2 r^{2q} \theta^2 + (2-r^q)^2 $$
and
$$ |\nabla g|^2 = 1 + q^2 s^{2q} \phi^2 (2 - s^q)^{-4} + (2 - s^q)^{-2} $$
are uniformly bounded on their domains. So we have $W^{1,p}$ automatically.


Now consider the two points $(r,0)$ and $(r,\pi)\in \bar{\Omega}\_1$, these correspond to $(r,0)$ and $(r, 2\pi - r^q\pi)\in \bar{\Omega}\_2$. In the domain their distance is $2r$. In $\bar{\Omega}\_2$ their distance is on the order of $r^{1+q}$. This shows that $g$ is not in $C^{0,\alpha}$ for any $\alpha > \frac{1}{1+q}$.


Replacing $r^q$ by a function vanishing to infinite order at $0$ will get you then an example that is not in $C^{0,\alpha}$ for any $\alpha > 0$.


(Note that this is firmly an issue related to the fact that $\Omega\_2$ is not Lipschitz, and then fact that we are measuring Hoelder-ness using the ambient metric on $\mathbb{R}^2$.)


