# 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 $L^p$?

-
When we work with a PDE, we may ask a condition at the boundary of an open set. Unfortunately, it have a measure $0$ and the function of $W^{m,p}(\Omega)$ are defined up sets of measure $0$. The trace operator avoid these complications. –  Davide Giraudo Jun 3 '12 at 10:30
@DavideGiraudo : Can you please help me to understand this with an example :) –  Theorem Jun 3 '12 at 10:37
For example $\Omega$ is the unit disk in $\Bbb R^2$. Then we can have a PDE of the form $-\operatorname{div}(u)=f$ on $\Omega$ and $u=g$ on the boundary. In fact, we have $T(u)=g$ and it's well-defined (but it's not necessarily the restriction of $u$ at the boundary, except when $u$ is regular). –  Davide Giraudo Jun 3 '12 at 10:40

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.