# Understanding the Trace Theorem for Sobolev spaces?

I'm reading Evans book on PDEs and I'm not sure of the meaning behind the Trace Theorem for Sobolev spaces and what it aims to accomplish. The theorem is as follows:

Assume a domain $U$ is bounded and that $\partial U$ is $C^1$. Then there exists a bounded linear operator

$$T: W^{1,p}(U) \to L^p(\partial U)$$ such that

i) $Tu = u|_{\partial U}$ if $u \in W^{1,p}(U) \cap C(\overline{U})$

ii) $||Tu||_{L^p(\partial U)} \le C ||u||_{W^{1,p}(U)}$

for each $u \in W^{1,p}(U)$, with $C$ depending only on $p$ and $U$.

I get that if $u$ is continuous we have already have a definition of $u$ on the boundary and this leads to i).

I don't see what is meant by ii) though. I read it as 'for any $u \in W^{1, p}$, the size of $T$ acting on $u$ in $L^p$ over the boundary $\partial U$ will always be less than or equal to a constant times the size of $u$ in $W^{1, p}$ over the entire domain $U$'.

1. How is this useful, we haven't even given an explicit definition of $T$, we have only made a statement about its size? We are relating norms over different domains ($\partial U$ and $U$), this seems like a meaningless thing to do..so why do we do it?
2. Furthermore, what is the relevance of $||Tu||_{L^p(\partial U)}$ being less than or equal to a constant times $||u||_{W^{1,p}(U)}$? The constant could be arbitrarily large so I don't see the point of making this statement?
• It would be helpful to learn some functional analysis before reading the later chapters (5 and onwards) in Evans. The inequality $\|Tu\| \leq C\|u\|$ shows that $T$ is bounded, and bounded linear operators are continuous.
– Jeff
Commented Jan 10, 2017 at 17:12
• @Jeff But what use is it that the operator is $T$ continuous? I really don't see what this Trace theorem is providing us with. All we have is: from (i) if $u$ is continuous on $\overline{U}$ we can use this $u$ 'as is' on the boundary..and from (ii) if $u$ is not continuous, there exists a continuous operator $T$ that acts on $u$..but I don't see how this is providing anything useful as it doesn't alter the value of the function $u$ on the boundary..its just a statement about the existance of a continous operator? Commented Jan 11, 2017 at 15:02
• The point is that $C(\bar{U})\cap W^{1,p}(U)$ is dense in $W^{1,p}(U)$. So you first define $Tu = u\vert_{\partial U}$ for $u \in C(\bar{U})\cap W^{1,p}(U)$, and then by (ii), $T$ is continuous, and can be uniquely extended to all of $W^{1,p}$. The extension goes like this: For $u \in W^{1,p}(U)$ take a sequence $u_k\in C(\bar{U})\cap W^{1,p}(U)$ converging to $u$ in $W^{1,p}(U)$ and define $Tu = \lim_{k\to \infty} Tu_k$, where the limit is taken in $L^2(\partial U)$. The continuity of $T$ guarantees this is well-defined and does not depend on the choice of approximating sequence $u_k$.
– Jeff
Commented Jan 11, 2017 at 17:23
• By the way, this density argument (proving something on a dense subset and then extending to the whole space) is extremely common in PDE. If you are having trouble understanding this, it would be good to go back and review or learn some functional analysis before attempting to read Evans. I suppose you have already encountered weak derivatives. What do you think of these; a similar issue must have occurred to you (what is the weak derivative of a non-differentiable function?).
– Jeff
Commented Jan 11, 2017 at 17:26
• @Jeff Thanks for the clarification. I now understand the implication of precisely why we need condition ii). I think the problem was that I was originally looking at this theorem in isolation, but looking at it in the context of its utility once we construct an approximating sequence of continuous functions it makes complete sense. Commented Jan 12, 2017 at 8:45

The point is, that the property (ii) is equivalent to the linear(!) map $T \colon W^{1,p}(U) \to L^p(\partial U)$ being continuous. This allows us to construct a unique $T$ from (i) by approximating $W^{1,p}(U)$-functions by continuous ones.
• Ok so the operator $T$ is continuous, but how does that help us with defining a function on the boundary when case ii) is for functions that are not continuous in $\overline{U}$? What is our goal with this trace operator $T$..what kind of functions is it specifying on the boundary? Commented Jan 10, 2017 at 14:50