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?


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).

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.