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If $\Omega\subset\mathbb{R}^n$ ($n\geq 1$), then $L^p(\Omega)$ is generally (in the context of PDEs) defined with respect the Lebesgue measure in $\mathbb{R}^n$.

For $n>1$, we can see $\partial \Omega$ as a subset of $\mathbb{R}^{n-1}$ and thus define $L^p(\partial\Omega)$ with respect the Lebesgue measure in $\mathbb{R}^{n-1}$.

For an interval $\Omega\subset\mathbb{R}^{1}$, the boundary $\partial\Omega$ will be a set with two points, say $\{a,b\}$. What is $L^p(\{a,b\})$?

Thanks.

Pedro
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2 Answers2

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The bug is when you say that you can see $\partial \Omega $ as a subset of $\mathbb R^{n-1}$. Your example show exactly that.

And in all generality, I don't know if you can give to $\partial \Omega$ a "natural measure".

D.L.
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  • I think that there is some common definition or convention, but I couldn't find it. Here and here $L^p(\partial\Omega)$ is considered with $\Omega$ being an interval. In the comment was said that, in this case, "integrating over the boundary is the same as evaluating pointwise there". Do you understand what is the measure that is being considered there? – Pedro Sep 07 '15 at 20:18
  • In a general case, you could try to puy on $\partial \omega$ a structure of sub variety of $R^n$ of dimension $n-1$. Then, it is possible to use the differential structure to build a theory of the integration larger than the one of Lebesgue, to give a general meaning to $\int_{\partial \Omega}$ – D.L. Sep 09 '15 at 07:19
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If $\Omega = (a,b)$ is an interval, in most cases the "right" way to interpret $\partial \Omega$ is as the two-point set $\{a,b\}$, equipped with counting measure. So $L^p(\partial \Omega)$ is the two-dimensional vector space of functions $f : \{a,b\} \to \mathbb{R}$ (or $\mathbb{C}$) with the norm $\|f\|_{L^p(\partial \Omega)} = (|f(a)|^p + |f(b)|^p)^{1/p}$.

Of course, in any given context, it is worth checking to see if this interpretation makes sense, or if something else is intended.

Nate Eldredge
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