Proof that estimate holds Theorem: Let $U$ be a bounded , open subset of $\mathbb{R}^n$ , and suppose $\partial{U}$ is $C^1$. Assume $1 \leq p<n$, and $u \in W^{1,p}(U)$. Then $u \in L^{p^{\ast}}(U)$ , with the estimate $||u||_{L^{p^{\ast}}}(U) \leq C ||u||_{W^{1,p}(U)}$, the constant $C$ depending only on $p,n$, and $U$.
We will use the following three theorems for the proof:
Theorem 1: Assume $U$ is bounded and $\partial{U}$ is $C^1$. Select a bounded open set $V$ such that $U \subset \subset V$. Then there exists a bounded linear operator $E: W^{1,p}(U) \to W^{1,p}(\mathbb{R}^n)$ such that for each $u \in W^{1,p}(U)$: 


*

*$Eu=u$ a.e. in $U$

*$Eu$ has support within $V$, and

*$||Eu||_{W^{1,p}(\mathbb{R}^n)} \leq C ||u||_{W^{1,p}(U)}$


Theorem 2: Assume $u \in W^{k,p}(U)$ for some $1 \leq p<\infty$, and set $u^{\epsilon}=\psi_{\epsilon} \ast u$ in $U_{\epsilon}$.  Then $u^{\epsilon} \in C^{\infty}(U_{\epsilon})$ for each $\epsilon>0$ and $u^{\epsilon}\to u$ in $W_{\text{loc}}^{k,p}(U)$ , as $\epsilon \to 0$.
Theorem 3: Assume $1 \leq p<n$. There exists a constant $C$ , depending only on $p$ and $n$ , such that $||u||_{L^{p^{\ast}}(\mathbb{R}^n)} \leq C ||Du||_{L^p(\mathbb{R}^n)}$, for all $u \in C_C^1(\mathbb{R}^n)$. 
Since $\partial{U}$ is $C^1$ , there exists according to Theorem 1 an extension $Eu=\overline{u} \in W^{1,p}(\mathbb{R}^n)$ , such that
$(\star) \left\{\begin{matrix}
\overline{u}=u \text{ in U}, \overline{u} \text{ has compact support , and}\\ 
||\overline{u}||_{W^{1,p}(\mathbb{R}^n)} \leq ||u||_{W^{1,p}( U )} 
\end{matrix}\right.$
Because $\overline{u}$ has compact support, we know from Theorem 2 that there exist functions $u_m \in C_C^{\infty}(\mathbb{R}^n)$ such that $u_m \to \overline{u}$ in $W^{1,p}(\mathbb{R}^n) (1)$.
Now according to Theorem 3, $||u_m-u_l||_{L^{p^{\ast}}(\mathbb{R}^n)} \leq C ||Du_m-Du_l||_{L^p(\mathbb{R}^n)}$ for all $m,l \geq 1$.
Thus $u_m \to \overline{u}$ in $L^{p^{\ast}}(\mathbb{R}^n) (2)$ as well.
Since Theorem 3 also implies $||u_m||_{L^{p^{\ast}}(\mathbb{R}^n)} \leq C ||D u_m||_{L^p(\mathbb{R}^n)}$, assertions (1) and (2) yield the bound $||\overline{u}||_{L^{p^{\ast}}(\mathbb{R}^n)} \leq C ||D \overline{u}||_{L^p(\mathbb{R}^n)}$.
This inequality and $(\star)$ yield the proof.
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First of all, why does the inequality $||u_m-u_l||_{L^{p^{\ast}}(\mathbb{R}^n)} \leq C ||Du_m-Du_l||_{L^p(\mathbb{R}^n)}$ for all $m,l \geq 1$ imply that $u_m \to \overline{u}$ in $L^{p^{\ast}}(\mathbb{R}^n) $ ?
$$$$
How do we get from $||\overline{u}||_{L^{p^{\ast}}(\mathbb{R}^n)} \leq C ||D \overline{u}||_{L^p(\mathbb{R}^n)}$ and $(\ast)$ that $||u||_{L^{p^{\ast}}(U)} \leq C ||u||_{W^{1,p}(U)}$ ?
 A: The first question follows from the fact that $u_m \rightarrow \overline{u}$ in $W^{1,p}(\mathbb{R}^n)$ implies
$||u_m-u_l||_{L^{p^{\ast}}(\mathbb{R}^n)} \leq C ||Du_m-Du_l||_{L^p(\mathbb{R}^n)} \leq C[||Du_m-D\overline{u}||_{L^p(\mathbb{R}^n)} + ||D\overline{u}-Du_l||_{L^p(\mathbb{R}^n)}] \rightarrow 0$
as $m,j \rightarrow \infty$, and $\lbrace u_m \rbrace \subset L^{p^*}(\mathbb{R}^n)$ is a Cauchy sequence, and the only possibility is that $u_m \rightarrow \overline{u}$ in $L^{p^*}(\mathbb{R}^n)$. The second question follow form the fact that we have (i) $||\overline{u}||_{L^{p^*}(\mathbb{R}^n)} \leq C ||D \overline{u} ||_{L^{p}(\mathbb{R}^n)}$, and from point $(\star)$ $\overline{u}=u=0$ in $\mathbb{R}^n \setminus U$ since $\mathrm{supp}(\overline{u})=\mathrm{supp}(u) \subset U$ is a compact set, again $\mathrm{supp}(D\overline{u})\subseteq \mathrm{supp}(\overline{u})$, and $u=\overline{u}$ on $U$, consider the restrictions on $U$ (i) becomes
$||u||_{L^{p*}(U)} \leq C ||Du||_{L^p(U)} \leq C ||u||_{W^{1,p}(U)}$
with the latest estimate follows by definition of $W^{1,p}$-norm.
