I tried to open every step of trace theorem from the book of Evans, but I still wondering if is everything right?
Thanks in advance.
Trace Theorem: Let $\Omega\subset\mathbb{R}^{n}$ be open, bounded and $\partial \Omega$ of class $C^1$ and $1\leq p< \infty$. Then there exists a bounded linear operator $$T:W^{1, p}(\Omega)\rightarrow L^p(\partial \Omega)$$ such that
- If $u\in W^{1,p}(\Omega)\cap C(\overline{\Omega})$ then $Tu=u_{|{\partial \Omega}}$.
- There exists $C=C(p, \Omega)>0$ such that $$\|Tu\|_{L^p(\partial \Omega)}\leq C \|u\|_{W^{1,p}(\Omega)},\qquad \forall u\in { W^{1,p}(\Omega)}.$$
MY ATTEMPT:
Step 1
Assume first $u\in C^{1}(\overline{\Omega})$. We will proceed as in the first part of the proof of the extension theorem. Thus, initially suppose that $x_0\in \partial \Omega$ and that $\partial \Omega$ is flat near $x_0$, lying in the plane $\{x_n=0\}$. Let us choose an open ball $B_r(x_0)$ such that $B_{\frac{r}{2}(x_0)} \subset B_r(x_0)$. \begin{align} B_{r}^{+}(x_0)&:= B\cap \mathbb{R}_{+}^{n}\subset\overline{\Omega}\\ B_{\frac{r}{2}}^{+}(x_0)&:= B_{\frac{r}{2}}(x_0) \cap \mathbb{R}_{+}^{n} \subset B_{r}^{+}(x_0) \end{align} Let us denote by $\Gamma :=\partial B_{\frac{r}{2}}^{+}(x_0) \cap \partial \mathbb{R}_{+}^{n}$.
Of course \begin{align}\tag{1} \partial \mathbb{R}_{+}^{n}\supset \Gamma \end{align} Choosing that \begin{align}\tag{2} \varphi \in C_{0}^{\infty}(\Omega) \implies \varphi\equiv 0, \forall x \notin \mathscr{supp} \Omega \end{align} satisfying \begin{align} 0\leq \varphi &\leq 1 \in B_{r}^{+}(x_0)\tag{3}\\ \varphi&=1 \in B_{\frac{r}{2}}^{+}(x_0)\tag{4} \end{align}
\begin{align} \tag{5} &\text{And since}\; \Omega = B_{r}^{+}(x_0)\; \text{then}\; \partial \Omega \; \text{ is a hypersurfaceof class}\; C^1 \end{align}
We can apply the divergence theorem to obtain
\begin{align}\tag{6} \int_{\Gamma} |u|^p d\sigma_x & \overset{4}{=} \int_{\Gamma} \varphi |u|^p d\sigma_x\\ &\overset{1}{\leq}\int_{\partial \mathbb{R}_{+}^{n}} \varphi |u|^p dx\\ &\overset{5}{=}-\int_{\mathbb{R}_{n}^{+}} \left(\varphi |u|^p \right) _{x_n} dx \end{align}
Where in equality in ${{5}}$ we use the divergence theorem taking $\eta =(0, 0, ..., -1)$ and $F=(0,0, ..., 0, \varphi |u|^p)$, and we obtained that \begin{align} \int_{\partial \mathbb{R}_{+}^{n}} F\eta d\sigma_{x}&=\int_{\partial \mathbb{R}_{+}^{n}} ( 0,0, ..., 0, \varphi |u|^p)(0, 0, ..., -1) d\sigma_{x}\\ &=- \int_{\mathbb{R}_{+}^{n}} \left(\varphi |u|^p\right)_{x_n} dx \end{align} Now differentiating the last expression in ${{6}}$, we have
\begin{align} \tag{7} &-\int_{\mathbb{R}_{n}^{+}} \varphi_{x_n} |u|^p dx - \int_{\mathbb{R}_{n}^{+}} {\operatorname{sgn}}(u) u_{x_n} p |u|^{p-1}\varphi dx\\ &\leq \int_{\mathbb{R}_{n}^{+}} |\varphi_{x_n}| |u|^p dx +\int_{\mathbb{R}_{n}^{+}} |p{\operatorname{sgn}}(u) \varphi u_{x_n}| |u|^{p-1} dx\\ &=\int_{\mathbb{R}_{n}^{+}} |\varphi_{x_n}| |u|^p dx +p\int_{\mathbb{R}_{n}^{+}} |\varphi u_{x_n}| |u|^{p-1} dx \end{align}
Let's solve separately the integrals that appear in the last equality above. Hence for the first statement we have: \begin{align}\tag{8} \int_{\mathbb{R}_{n}^{+}} |\varphi_{x_n}| |u|^p dx &\leq \int_{B_{r}^{+}(x_0)} |\varphi_{x_n}| |u|^p dx\\ &\leq \|\nabla \varphi\|_{\infty} \int_{B_{r}^{+}(x_0)} |u|^p dx\\ &\leq C_1 \int_{B_{r}^{+}(x_0)} |u|^p dx \end{align}
Now note that in the last statement of ${{7}}$, applying Young's inequality to $|u_{x_n}|$ and $|u|^{p-1}$, considering $\frac{1}{ p}+ \frac{1}{p'}=1$ where $p'= \frac{p}{p-1}$, we have
\begin{align}\tag{9} p\int_{\mathbb{R}_{n}^{+}} |\varphi u_{x_n}| |u|^{p-1} dx &\leq p\int_{B_{r}^{+}(x_0)} |u_{x_n}| |u|^{p-1} dx\\ &\leq p\int_{B_{r}^{+}(x_0)}\left( \frac{1}{p}|u_{x_n}|^p +\frac{1}{p'} |u|^{(p-1)p'} \right)dx\\ &\leq \int_{B_{r}^{+}(x_0)}|\nabla u|^p dx+\frac{p}{p'} \int_{B_{r}^{+}(x_0)} |u|^p dx \end{align} Hence, replacing the equations ${{8}}$ and ${{9}}$ into the equation $(7)$ it follows that
\begin{align}\tag{10} &\int_{\mathbb{R}_{n}^{+}} |\varphi_{x_n}| |u|^p dx +p\int_{\mathbb{R}_{n}^{+}} |\varphi u_{x_n}| |u|^{p-1} dx\\ &\leq C_1 \int_{B_{r}^{+}(x_0)} |u|^p dx + \int_{B_{r}^{+}(x_0)}|\nabla u|^p dx+\frac{p}{p'} \int_{B_{r}^{+}(x_0)} |u|^p dx\\ &\leq C_2\left( \int_{B_{r}^{+}(x_0)} |u|^p dx + \int_{B_{r}^{+}(x_0)}|\nabla u |^p dx+ \int_{B_{r}^{+}(x_0)} |u|^p dx\right)\\ &\leq C_3\left( \int_{B_{r}^{+}(x_0)}|\nabla u|^p dx+ \int_{B_{r}^{+}(x_0)} | u|^p dx\right) \end{align}
Where $C_2:=\max\{1, C_1, \frac{p}{p'}\}$ and $C_3:=\max\{2, C_2\}$. \ Therefore, it follows from ${{10}}$ and the above that \begin{align} \tag{11} \int_{\Gamma} |u|^p d\sigma_x \leq \tilde{C}\left( \int_{B_{r}^{+}(x_0)}|\nabla u|^p dx+ \int_{B_{r}^{+}(x_0)} |u|^p dx \right) \end{align} That is, \begin{align} \tag{12} \|u\|_{L^p(\Gamma)}^{p} \leq \tilde{C}\|u\|_{W{1,p}(B_{r}^{+}(x_0 ))}^{p} \end{align}
Step 2
If $x_0\in \partial \Omega$, and $\partial \Omega$ is not flat near $x_0$, we can straighten out the boundary near $x_0$ in order to obtain the same estimate as in step 1.
As $\partial \Omega$ is compact there exist finitely many points $x^{0}_{j}\in \partial \Omega$ (we are going to denote only by $X_0$) and open subsets $\Omega_j\subset \mathbb{R}^n$ such that $\partial \Omega\subset \displaystyle\mathop{\cup}_{ j=1}^{k}\Omega_j$. With this, define
\begin{align}\tag{13} \psi_j:\Omega_j\rightarrow \overline{B_{r}^{+}(x_0)} \end{align}
More specifically, given that $\Omega$ is $C^1$, it follows from the definition of class $C^1$ boundary that for every point $x^{0}_{j}\in \partial \Omega$ there exist $r>0$ and a function $\gamma:\mathbb{R}^{n-1}\rightarrow \mathbb{R}$ also of class $C^1$, such that
\begin{align} \tag{14} \Omega \cap \Omega_j=&\{x\in {U\subset \mathbb{R}^n: x_n > \gamma(x_1, ... , x_{n-1}) }\}\\ \partial\Omega \cap \Omega_j=&\{x\in \overline{U}\subset\mathbb{R}^n: x_n = \gamma(x_1, . .., x_{n-1})\} \end{align}
Hence $13$ must satisfy
\begin{align}\tag{15} \psi_j(\Omega\cap \Omega_j)&= B_{r}^{+}(x_0)\\ \psi_j(\partial \Omega \cap \Omega_j) &= \partial B_{r}^{+}(x_0)\cap \partial \mathbb{R}_{+}^{n} \end{align}
Define \begin{align}\tag{16} \Gamma_j:= \psi_{j}^{-1}(\Gamma). \end{align}
Therefore, there exist finitely many points $x^{0}_{j}\in \partial \Omega$ and closed subsets $\Gamma_j\subset\mathbb{R}^{n}$, such that
\begin{align}\tag{17} \partial \Omega= \mathop{\bigcup}_{j=1}^{k}\Gamma_j \end{align}
Under these conditions, it also follows from this definition that there exist a bijective function of class $C^1$, say $\psi_j$, and its inverse such that straightens out $\partial \Omega$ near $x_0$.
Explicitly $\psi_j$ is given by
\begin{align}\tag{18} \psi_j\colon \Omega_j\subseteq \mathbb{R}^n & \rightarrow \overline{B^{+}_{r}(x_0)}\\ (x_1, ..., x_n)&\mapsto {(x_1, ..., x_{n-1}, x_n -\gamma(x_1, ..., x_{n-1}))} \end{align}
Thus, we have to show that $\psi_j$ is a bijection of class $C^1$, that is, a diffeomorphism that satisfies the conditions from ${{13}}$ to ${{17}}$. To see this, just consider the change of variables given by
\begin{align}\tag{19} \psi_j(x):=y \end{align}
where $x=(x_1, ..., x_n)$ and ${y=(x_1, ..., x_{n-1}, x_n -\gamma(x_1, ..., x_{n-1}) )}$.
We claim that $\psi_j$ is a diffeomorphism.
Indeed, note that considering the change of variables in ${{19}}$ the determinant of Jacobian is $$ \begin{align*} |J| &= \det \begin{pmatrix}\tag{20} 1 & 0 & \dots & -\frac{\partial \gamma}{\partial x_1} \\ \vdots & \ddots & \ddots & \vdots \\ \vdots & \dots & \ddots & - \frac{\partial \gamma}{\partial x_{n-1}} \\ 0 & \dots & 0 & 1- \frac{\partial \gamma}{\partial x_{n}} \end{pmatrix} \\ &= \left( -\frac{\partial \gamma}{\partial x_1} - \frac{\partial \gamma}{\partial x_2} - \dots - \frac{\partial \gamma}{\partial x_{n-1}} \right) \det \begin{pmatrix} 1 & 0 & \dots & 0 \\ \vdots & \ddots & \ddots & \vdots \\ \vdots & \dots & \ddots & 0 \\ 0 & \dots & 0 & 1 \end{pmatrix}\\ &= -\frac{\partial \gamma}{\partial x_1} - \frac{\partial \gamma}{\partial x_2} - \dots - \frac{\partial \gamma}{\partial x_{n-1}}\neq 0 \end{align*} $$ hence, $\psi_j$ is a diffeomorphism, and the Jacobian $J$ is continuous in $\Omega$ which is bounded, then $|J|_{\psi_j}$ is bounded. Similarly, for $\psi_j^{-1}$, the determinant of Jacobian $|J|_{\psi_j^{-1}}$ is bounded.
From what was exposed in step 1, and the considerations above, we have
\begin{align}\tag{21} \int_{\Gamma_j} |u|^p d\sigma_x & \stackrel{{{20}}}{=} \int_{\Gamma_j} |u|^p |J|_{\psi_j} d\sigma_y\\ &\stackrel{{{11}}}{\leq} C_j \left(\int_{B_{r}^{+}(x_0)}|\nabla u|^p + |u|^p \right ) |J|_{\psi_j} dy\\ &\stackrel{{20}}{=} C_j \left(\int_{\Omega_j}|\nabla u|^p + |u|^p \right) |J|_{\psi_j^{-1}}dx\\ &\leq C \left(\int_{\Omega}|\nabla u|^p + |u|^p \right) dx \end{align}
where in the last inequality we take $C:=\underset{1\leq j\leq n}{\max}\{C_j\}$, where the constant $C_j$ does not depend on $u$. That is, by estimating step 1,
\begin{align}\tag{22} \|u\|_{L^p(\partial \Omega)}^{p} \stackrel{{19}}{=}&\|u\|_{L^p\left(\mathop{ \bigcup}_{j=1}^{k}\Gamma_j\right)}^{p}\\ \leq &\sum_{j=1}^{k}\|u\|_{L^p(\Gamma_j)}^{p}\\ \stackrel{{12}}{\leq} & \tilde{C} \sum_{j=1}^{k}\|u\|_{W^{1, p}(\Omega)}^ {P}\\ = & \tilde{C} k \|u\|_{W^{1,p}(\Omega)}^{p} \end{align}
Therefore, considering only the first and last terms in the inequality above, raising both sides to $\frac{1}{p}$ and noting that the constants in steps 1 and 2 depend on $p$ and $k$. Taking $C:=C(k,p)$ we obtain that
\begin{align}\tag{23} \|u\|_{L^p(\partial \Omega)}\leq C \|u\|_{W^{1,p}(\Omega)} \end{align}
Step 3
Now defining \begin{align} T\colon C^1(\overline{\Omega}) & \rightarrow L^p(\partial\Omega)\\ u &\mapsto Tu\\ \end{align}
so that
\begin{align} \tag{24} Tu:=u_{|{\partial \Omega}} \end{align} So by the inequality in ${{23}}$ we have \begin{align}\tag{25} \|Tu\|_{L^p(\partial \Omega)}\leq C \|u\|_{W^{1,p}(\Omega)} \end{align} Therefore, the theorem is proven for all $u\in C^1(\overline{\Omega})$.
Step 4
We will use the fact that $W^{1,p}(\Omega)$ is dense in $C^1{(\overline{\Omega})}$ to construct a unique extension to a continuous linear operator defined in $W ^{1,p}(\Omega)$.
Indeed, let us suppose that $u\in W^{1,p}(\Omega)$, then there exists a sequence of functions $(u_m)\in C^{1}(\overline{\Omega})$ such that $$u_m \rightarrow u \in W^{1,p}(\Omega).$$ It follows from the inequality in ${{25}}$ that
\begin{align} \tag{26} \|T (u_m - u_l)\|_{L^{p}(\partial \Omega)}&\leq C \|u_m-u_l\|_{W^{1,p}(\Omega)} \end{align}
thus $\{Tu_m\}_{m\in \mathbb{N}}$ is a Cauchy sequence in $L^{p}(\partial \Omega)$ which is complete space, therefore it is a convergent sequence with $\lim_{m\to \infty} Tu_m \in L^{p}(\partial \Omega)$.
We define it like this
\begin{align}\tag{27} Tu := \underset{m\to \infty}{\lim} Tu_m \end{align}
as being the limit reached in $L^p(\partial \Omega)$. Note that $Tu$ does not depend on the choice of sequences $(u_m)$. To see this, consider that there exists other $v_m\in C^1(\overline{\Omega})$ such that $v_m\rightarrow u$ and thus by ${{25}}$
\begin{align} \|Tv_m - Tu_m\|_{L^p(\partial \Omega)}\leq &C \|v_m-u_m\|_{W^{1,p}(\Omega)}\\ \leq & C {\|v_m -u\|_{W^{1,p}(\Omega)}+\|u_m -u\|_{W^{1,p}(\Omega)}} \end{align}
taking the limit $m\rightarrow \infty$ then we obtain that
\begin{align} Tv_m\equiv Tu_m \end{align}
So ${{26}}$ it is well defined, linear and limited.
Finally, if $u\in W^{1,p}(\Omega)\cap C(\overline{\Omega})$, it follows that $(u_m)$ converges uniformly to $u$ on compact subsets of $\overline{\Omega}$, in particular over $\partial \Omega$, this is
\begin{align} u_{|{\partial \Omega}} =Tu =T\left(\lim_{m\to \infty} u_m \right) = \lim_{m\to \infty} Tu_m \end{align}
Therefore, $(Tu_m)$ converges uniformly to $u_{|{\partial \Omega}}$ over $\partial \Omega$, this convergence uniform comes from results of real analysis in several variables.