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I am reading Sobolev space in the book Partial Differential Equation by Evan and I do not understand some point in the proof of the trace theorem.
Let $U$ is open, bounded and $\partial U$ is $C^1$. Let $x_0\in\partial U$ and assume that $\partial U$ is flat near $x_0$ lying in the plane $\{x_n=0\}$. Choose a ball $B(x_0,r)$ such that $B^+=B\cap \{x_n\ge 0\}\subset U$ and $B^-=B\cap \{x_n\le 0\}\subset \mathbb{R}^n-U$. Let $\hat{B}=B(x_0,r/2)$. Choose $\zeta\in C_c^{\infty}(B)$, with $\zeta=1$ on $\hat{B}$. Define $\Gamma=\hat{B}\cap\partial U$. Set $x'=(x_1,\ldots,x_{n-1})\in\mathbb{R}^{n-1}=\{x_n=0\}$. The author then claims that
$$\int_{\Gamma}|u|^pdx'\le\int_{x_n=0}\zeta|u|^pdx'=-\int_{B^+}(\zeta|u|^p)_{x_n}dx.$$ What is the meaning of $\int_{\Gamma}|u|^pdx'$ ? How do we get the equality ?

Omega
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  • The equality is just integration by parts with respect to $x_n$ over $B^+$, taking into account that $\zeta$ is compactly supported. I'm not sure I understand your first question (it is a surface integral over $\Gamma$, precisely the sort of object that the trace theorem is about). – Ian Jun 29 '15 at 14:13
  • what about the meaning the first integration? is it a surface integration or just integration on R^(n-1) – Omega Jun 29 '15 at 14:18
  • Technically it's a surface integration, but because $\partial U$ is flat there, it's more or less the same as an integral over $\mathbb{R}^{n-1}$. This is why you use a smooth transformation to force the boundary to be straight in the proof. (I'm not very good with manifolds, so I don't think I can help very much with the technical issues about the surface integral). – Ian Jun 29 '15 at 14:30

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