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I am getting a bit confused. In the definition of $C^1$ manifold in Renardy and Rogers, they say that $\partial\Omega$ is of class $C^1$ if every point on $\partial\Omega$ has a neighbourhood within which $\partial\Omega$ can be represented as a graph of a $C^1$ function.

There is no transformation of the coordinate system in this definiton. BUT when they define Lipschitz domain, there is an affine transformation involved of the form $\tilde X = AX + C$ where $A$ is a matrix and $C$ is a vector.

The transformation messes things up for me (because it is present in the surface integral of functions on $\partial\Omega$) and I'd prefer not to use it.

Am I right that the transformation is not needed when we have $C^1$ domain, and is only needed when all we know of the domain is that it is Lipschitz?

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In a word: Yes. The transformation adds nothing in the $C^1$ case, but is essential in the Lipschitz case. If you take the graph of a Lipschitz function with sufficiently large Lipschitz constant and rotate it 45 degrees, the result is not necessarily the graph of any function. Do the same to a $C^1$ function, and it's still a graph (locally – though you may have to permute coordinates).

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  • $\begingroup$ Thanks. So if I say my domain is $C^1$ then I don't bother using transformations in norms. $\endgroup$ – matt.x Dec 31 '13 at 15:44
  • $\begingroup$ @HaraldHanche-Olsen, any chance you can expand on your second sentence? $\endgroup$ – soup Jan 10 '14 at 17:12
  • $\begingroup$ @soup I am not sure what you are referring to. But if it's about rotatin a Lipshitz function, just take a sawtoothy function with sufficiently steep sides of the teeth. Rotate it 45 degrees, and you no longer have a graph of a function. You can ensure it has enough small teeth everywhere so that the rotated graph is not a graph in any open set, if you so desire. (But that takes a bit more work than I am ready to invest at this moment.) $\endgroup$ – Harald Hanche-Olsen Jan 10 '14 at 19:29
  • $\begingroup$ Thanks @HaraldHanche-Olsen for explaining. One last question: the reason that the transformation of coordinates is not needed for $C^1$ domains is because by definition, there is a $C^1$ diffeomorphism between the boundary (locally) and open balls. The diffeomorphism "takes care" of it, so to speak. Is that right? $\endgroup$ – soup Jan 11 '14 at 16:23
  • $\begingroup$ @soup More than that, even: Say the boundary $B$ is a $C^1$ hypersurface in $\matbb{R}^n$. Then at any point, the projection on $\mathbb{R}^{n-1}$ obtained by dropping one of the coordinates of $\mathbb{R}^n$ and keeping the rest, is a local diffeomorphism of $B$. (Which coordinate you can drop, will in general vary, of course.) $\endgroup$ – Harald Hanche-Olsen Jan 11 '14 at 16:46

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