# Help understanding the proof of Trace Theorem given in Evans

I need help to understand the proof of the Trace Theorem given in Evans L.C. Partial differential equations (AMS, 1997):

Asume $U$ is a bounded open set and that $\partial U$ is $C^1$. Then there exists a bounded linear operator $T:W^{1,p}(U) \to L^p(\partial U)$ such that:

$\quad$(i) $Tu=u|_{\partial U}$ if $u \in W^{1,p}(U) \cap C(\bar{U})$

and

$\quad$(ii) $\|Tu\|_{L^p(\partial U)} \le C \| U \|_{W^{1,p} (U)},$ for each $u \in W^{1,p}(U)$, with the constant $C$ depending only on $p$ and $U$.

The proof provided starts like this: Okay, so we first asume that there is a neighborhood $U_{x^0}$ of $x^0$ such that $U_{x^0} \subset \{x \in \mathbb{R}^n \ | \ x_n=0\}$ and prove (1).

Then the proof continues by saying "If $\partial U$ is not flat near $x^0$", we straighten out the boundary near $x^0$ to obtain the setting above. Applying estimate (1) and changing variables... (The proof continues)

Could someone clarify me a little more precisely what straighten out the boundary near $x^0$ means?

Updated: This is the definition of open set of class $C^1$ which gives us a diffeomorphism to the set $Q_0$ which is "flat".

We define the following sets:

• $R_+ = \{x=(x_1,...,x_n) \in \mathbb{R}^n \ | \ x_n \geq 0\}$
• $Q = \{x=(x_1,...,x_n) \in \mathbb{R}^n \ | \ (\sum_{i=1}^{n-1} x_i^2)^{1/2} < 1 \ y \ |x_n|<1 \}$
• $Q_+=R_+ \cap Q$
• $Q_0=\{(x_1,...,x_{n-1},0) \in \mathbb{R}^n \ | \ (\sum_{i=1}^{n-1} x_i^2)^{1/2} < 1 \}$

An open set $\Omega$ is of class $C^1$ if for every $x \in \partial \Omega$ there exists a neighborhood $U_x$ of $x$ in $\mathbb{R}^n$ and a bijective map $H: Q \to U_x$ such that:

• $H \in C^1(\overline{Q})$
• $H^{-1} \in C^1(\overline{U_x})$
• $H(Q_+)=U_x \cap Q$
• $H(Q_0)= U_x \cap \partial \Omega$
• Find a differentiable map that transforms a piece of $\partial U$ near $x^0$ into a piece of a plane. Then you change variables in the integrals. – DisintegratingByParts May 5 '16 at 17:43
• @TrialAndError Could you please explain it a bit more? I suppose you mean transform a neighborhood of $x^0$ into a piece of a hyperplane (into a piece of a $n-1-$ dimensional surface), Right? Why do you have the guarantee to have that difeomorphism, is it simply because $U$ is open, right? – D1X May 5 '16 at 17:57
• Having the diffeomorphism is essentially assumed by asserting that $\partial U$ is a $C^1$ manifold. – DisintegratingByParts May 5 '16 at 18:08
• @TrialAndError And shouldn't the Jacobian appear due to the change of variables? The book says "Applying estimate (1) and changing variables we obtain the bound: $\int_{\Gamma} |u|^p dS \leq C \int_{U} |u|^p + |Du|^p dx$, where $\Gamma$ is some open subset of $\partial U$ cointaining $x_0$." – D1X May 6 '16 at 16:32
• The Jacobian will be bounded above and below in the local neighborhood that you obtain by move normally off the $C^1$ boundary. – DisintegratingByParts May 6 '16 at 16:37