For $u$ in the Sobolev space $W^{k,p}(U)$, how to know $u$ or $D^\alpha u$ on the boundary of $U$? In Evans PDE, section 5.2, page number 259.
He writes, we denote by 
$W_0^{k,p}(U)$ 
the closure of $C^\infty_c(U)$ in the Sobolev space $W^{k,p}(U)$. 
Then he writes, $W^{k,p}_0(U)$ contains those functions $u$ 
in $W^{k,p}(U)$ such that  $D^\alpha u = 0$ on the boundary of $U$ for all $|\alpha|\le k-1$
My question is, if $U$ is open then how can we discuss the derivative of the functions of $W^{k,p}(U)$ on the boundary of $U$, I mean how can we go beyond the domain $U$? Even he writes $D^\alpha u = 0$ for all $|\alpha|\le k-1$, so the function in $W^{k,p}_0(U)$ is itself $0$ on the boundary, but how to know/define the value of this function on the boundary of $U$ while our domain is only $U$.
 A: The key is the last remark Evans makes: "This will all be made clearer with the discussion of traces in Section 5.5"  The trace is a bounded linear map 
$$
T: W^{1,p}(\Omega) \to L^p(\partial \Omega)
$$
such that $T u = u \vert_{\partial \Omega}$ whenever $u \in W^{1,p} \cap C^0$.  In other words, it is a bounded extension of the usual "restriction to boundary" map for continuous functions.  Moreover, the kernel of this map, i.e. the set of functions in $u \in W^{1,p}$ for which $Tu=0$, is exactly the set $W^{1,p}_0$.
Now, if $u \in W^{k,p}(\Omega)$, then $\partial^\alpha u \in W^{1,p}(\Omega)$ for $|\alpha| \le k-1$, and thus we can apply the trace map to see that $T \partial^\alpha u \in L^p(\partial \Omega)$.  One should think of this as computing $\partial^\alpha u$ inside $\Omega$ and then taking a limit to retrieve the value on the boundary.  
Finally, if $u \in W^{k,p}_0(\Omega)$, then $\partial^\alpha u \in W^{1,p}_0(\Omega)$ for $|\alpha| \le k-1$, and so $\partial^\alpha u =0$ on $\partial \Omega$ in the sense of traces.
