3
$\begingroup$

Let $\Omega \subset \mathbb{R}^n$ be a $C^{1}$ bounded domain.

It is possible to define the space $L^1(\partial\Omega)$ as the set of functions $u\colon \partial\Omega \to \mathbb{R}$ with the finite norm $$\lVert u \rVert_{L^1(\partial\Omega)} := \sum_i\lVert {\phi_i(u\circ g_i)}\rVert_{L^1(B_i)}$$ where $\phi_i$ partition of unity, $g_i$ is a $C^1$ diffeomorphism and $B_i$ are subsets of $\mathbb{R}^{n}$.

There are lots of definitions like this (eg. see Renardy and Rogers, Krylov, James Robinson's Infinite Dimensional Dynamical System) or with small variations.

There is also the surface integral of a function $v\colon \partial\Omega \to \mathbb{R}$: $$|u| := \int_{\partial \Omega}fdS$$ where $dS$ is the surface density. To compute this quantity, we need to use a parametrisation.

Now my question, is it always the case that $$\lVert u \rVert_{L^1(\partial\Omega)} \qquad\text{and}\qquad |u|$$ are equivalent? Does this hold for all the little variations of the $L^1$ norm in the first equation??

Now when we have a Lipschitz surface, these norms are equivalent (see Necas). BUT their definition of the $L^1$ norm is different so let us not use that result here. Edit: hmm, it seems $C^1$ domains are not a subset of Lipschitz domains.

$\endgroup$
  • $\begingroup$ Dear @matt.x, every bounded $C^1$ domain is Lipschitz. $\endgroup$ – Tomás Jan 2 '14 at 10:13
  • $\begingroup$ Is "$L^2(\partial \Omega)$" a typo? The definitions give equivalent norms, because $g_i$ have derivatives bounded above and below (on the region that contributes to the integral). The constants in the inequality between the norms will depend on the choice of $g_i$. $\endgroup$ – Post No Bulls Jan 3 '14 at 5:44
  • $\begingroup$ @PostNoBulls Thanks, yes, it should be $L^1(\partial\Omega).$ Why would $g_i$ be bounded from below? I guess there is no canonical definition of surface integral $|\cdot|.$ $\endgroup$ – matt.x Jan 4 '14 at 17:25
  • $\begingroup$ The second definition is my preferred one, but it is not the best for every occasion. Definitions are used on the as-needed basis, they are not a canon... Being a diffeomorphism implies having the Jacobian being nonzero, which makes it bounded away from $0$ on compact subsets. $\endgroup$ – Post No Bulls Jan 4 '14 at 17:27
  • 1
    $\begingroup$ Yes @matt.x, that's the idea. I suggest you to try and write this mathematically. $\endgroup$ – Tomás Jan 11 '14 at 23:35
0
$\begingroup$

Not, because your definition depends on the selection of the partition.

The following does not depend of the partition: $$ \|u\|_{L^1(\partial\Omega)}=\Big\|\sum_{i}\int_{B_i}\varphi_i(u\circ g_i)\,dx\,\Big\|. $$ Note that the $B_i$'s should be subsets of $\mathbb R^{n-1}$.

$\endgroup$

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.