In order to define what it means for a map $f \colon \Omega \subseteq \mathbb R^n \to \mathbb R^n$ to be conformal, it is sufficient to require that $f$ is everywhere differentiable. Does conformality automatically implies that $f$ is $\mathcal C^1$ (hence real-analytic, see below)? 


By complex analysis, we know the answer is positive when $n=2$. 


In higher dimensions, [Liouville's theorem](https://en.wikipedia.org/wiki/Liouville%27s_theorem_(conformal_mappings)) characterizes conformal maps as Möbius transformations, but it is stated for $f \in W^{1,n}$ in Wikipedia. Is it known whether it also holds when $f$ is assumed everywhere differentiable?



Regularity of conformal maps

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> Let $n\geq 3.$ Let $\Omega$ be an open connected subset of $\mathbb R^n,$ and let $f:\Omega\to\mathbb R^n$ be a function having a pointwise derivative $Df(x)$ everywhere satisfying $(Df)^T(Df)=g(x)I$ with $g(x)>0.$ Then $f$ is continuously differentiable.
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By the [inverse function theorem](https://terrytao.wordpress.com/2011/09/12/the-inverse-function-theorem-for-everywhere-differentiable-maps/), $f$ is a local homeomorphism. By shrinking $\Omega$ we can assume $f$ maps $\Omega$ homeomorphically to $f(\Omega),$ and that $\Omega$ and $f(\Omega)$ are bounded.
Note $$\|Df\|^2=c\_ng(x)=c\_n\det(g(x)I)^{1/n}=c\_n|\det Df(x)|^{2/n}$$
where $\|\cdot\|$ is Frobenius norm, and $c\_n$ is a constant depending on $n.$
By the change of variables formula ([proof sketch](https://math.stackexchange.com/questions/59105/is-the-image-of-a-null-set-under-a-differentiable-map-always-null)), $$\int\_\Omega |\det Df(x)|\;dx=\mu(f(\Omega))<\infty.$$
So $f\in W^{1,n}(\Omega,\mathbb R^n)$ and you can use the $W^{1,n}$ result you mentioned.


