Two basic questions on PDEs (trace operator and Sobolev space) 
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*I am a bit unsure about the role of the trace operator. I understand that if you have a PDE that is solved by a function $u$ in some Sobolev space, then it's not necessarily defined on the boundary since $u$ is in $L^p$ and so can be "redefined" on the boundary as it's a null set. But sometimes we have boundary conditions that $u$ needs to satisfy so that uniquely fixes $u$. If we don't have such boundary conditions, why does it matter what $u$ is on the boundary? What role does the trace operator play?

*If $u$ is in $H^{r+2}(\Omega)$ for all $r$ where $\Omega$ is compact, then $u$ is also in $C^k(\overline{\Omega})$ for all $k$. This is a fact but I don't know why? How does existence and integrability of $u$ and its derivatives imply continuity of $u$ and its derivatives?
 A: As Willie explained the intuition of the definition for the trace operator, maybe I can say some words about the boundary condition in Differential Equations, and hope that it can help. Take the elliptic equation as example, if we want to consider the shape of a membrane under outer force, we are lead to equation like $-\Delta u=f$, assume the boundary of the membrane is good enough and is fixed (actually now the boundary function is the trace of the shape function), then this is the Cauchy problem for the equation. 
Moreover, because in elliptic equation we have good enough a priori estimate(use Newton potential we could see what the solution looks like), we can solve this Cauchy problem for general boundary data(which might not be continuous, for example $L^p$), in this case the solution is not continuous up to the boundary, but converge in $L^p$ sense, which is similar to the definition of trace operator.
So for me, "trace" (generalization of the boundary value problem)just makes the problem well-posed. 
