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I have similar question as posted here: Weak differentiability and diffeomorphisms.

I want to prove, that for $\Omega, \tilde{\Omega} \subset \mathbb{R}^d $ open, $\tau: \Omega \to \tilde{\Omega}$ a $C^1$-diffeomorphism with $D_\tau, D_{\tau^{-1}}$ bounded it holds, that for $1\le p \le \infty$:

$\forall f\in W^{1,p}(\Omega): f \circ \tau \in W^{1,p}(\tilde{\Omega}) \;with\; \partial_i (f\circ \tau)= \sum_{j=1}^d (\partial_j f)\circ \tau \partial_i \tau_j.$

Well, for $p<\infty$ one can approximate $f$ by $f_k \in W^{1,p}(\Omega)\cap C^\infty (\Omega)$. Then we use the transformation formula and Lebesque dominated convergence to conclude the claim. I have problems with the $p=\infty$ case. Of course, it is not so hard to show that $f\circ \tau \in L^\infty$. Furthermore for any bounded $U\subset \Omega$ we have $f\in W^{1,p'}(U)$ if $p' <\infty$. By the first part $f\circ \tau \in W^{1,p'}(\tau(U))$ for any $p' <\infty$.

Does anyone know how could I proceed?

I was thinking about proving the statement locally for $p'<\infty$ for any bounded neighbourhood in $\tilde{\Omega}$ and concluding it for the whole $\tilde{\Omega}$ since there exists a unique distribution which coincides with the collection of ditributions on the particular neighbourhoods. But still, I don't know how to get the statement for $p=\infty$.

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    Maybe that's a good idea: locally and for $p<\infty$ we get by the first part $ \partial_i (f\circ \tau)= \sum_{j=1}^d (\partial_j f)\circ \tau \partial_i \tau_j.$ But the RHS is actually in $L^\infty$ and it is well-defined on the whole set $\Omega$ since we get this representation for any bounded neighbourhood of an $x\in\Omega$. We still have to check if it works as a distribution on the whole set, but it does, since a test function is compactly supported (and compact implies bounded so we can use our formula). Does it sound reasonable to you? – 2017-01-03
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    Yes, a good idea. To check that something is in $W^{1,p}$ you need to check that (1) it's locally in $W^{1,1}$, and (2) the partial derivatives are in $L^p$. The second part follows directly from the chain formula, which itself is proved locally, along with part (1). – 2017-01-03

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