# Relation between $L^1(\partial\Omega)$ and the surface integral on $C^1$ domains

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.

• Dear @matt.x, every bounded $C^1$ domain is Lipschitz. – Tomás Jan 2 '14 at 10:13
• 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 $g_i$. – Post No Bulls Jan 3 '14 at 5:44
• @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 $|\cdot|.$ – matt.x Jan 4 '14 at 17:25
• 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. – Post No Bulls Jan 4 '14 at 17:27
• Yes @matt.x, that's the idea. I suggest you to try and write this mathematically. – Tomás Jan 11 '14 at 23:35

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 $\mathbb R^{n-1}$.