Q: Why do we talk about Trace Operator? What is the importance of a Trace operator in PDE . Although I have read the Wiki page on this but I am not able to connect it to the aspect of solving PDE's.
Particularly why do we define Trace operator as $T\colon W^{1,p}(u) \to L^p(\partial U)$. 
Also why do we deal with only exponent $1$ . What is great about operator from $W^{1,p}$ to $L^p$? 
Please help me to understand this concept of trace operator. Thank you very much . 

A: We take $U$ a smooth open set, regular enough in order to have density of $C^{\infty}_0(\overline U)$ in $W^{1,p}(U)$. 
We define $T$ on $C^{\infty}_0(\overline U)$ by taking the restriction of these functions to the boundary (it's well-defined, these one are not equivalence classes of functions). We check that this operator is continuous on this space of functions, then we extend it by continuity to $W^{1,p}(U)$. 
We get functions in $L^p(\partial U)$, which is a space of which can be defined using charts. Thanks to that, we can define an element of the Sobolev space on the boundary, even if this one has measure $0$, and it extends the concept of trace for function. 
