Surjectivity of the trace operator in Sobolev spaces

Question

Suppose $U$ is an open bounded set with $C^1$ boundary. It is a well-known result in the theory of Sobolev spaces $W^{1,p}$ that there is a continuous linear operator $T:W^{1,p}(U)\rightarrow L^p(\partial U )$ that equals ordinary restriction on continuous functions. Wikipedia tells me that this operator is in general not surjective. I know that one way to prove this is to show that functions in the image are actually more regular than your general $L^p$-function, and I have somewhat understood the proof for that. However, I feel that there should be some elementary counterexample. In fact, any function that is a trace of some $u\in W^{1,p}(U)$ is a limit of a Cauchy sequence of functions in $C^{\infty}(\bar{U})$ (and vice versa), and that could be exploited to exhibit some example where $T$ is not surjective. Does anyone have a good example?

Answer 1

At least in the best circumstances, it is easy to prove that the trace/restriction map loses ${\ell\over p}+\epsilon$ in "Sobolev units" (for arbitrarily small $\epsilon>0$) where codimension is $\ell$. And an easy extension of Sobolev imbedding theorems shows that (for example, for $L^2$ Sobolev spaces so I don't mess up the indexing shift...) $H^{k+{n\over 2}+\epsilon}\subset C^{k,\epsilon}$, the latter being $C^k$ functions with an $\epsilon$ Lipschitz condition.

Thus, with $p=2$ for example, $H^1(\Omega)$ maps to $C^{0,{1\over 2}-\epsilon}(\partial \Omega)$ for every $0<\epsilon<{1\over 2}$.