Counterexample of Sobolev traces

Question

Does anyone know a counterexample to the following fact: let $\Omega$ be a smooth domain in $\mathbf{R}^n$. Then $T(W^{2,2}) \subset C(\partial \Omega)$. I admit that i do not know much about trace spaces, and looking on adams it seems that this is wrong, but I would like to know if there is a direct counterexample

Answer 1

NOTATION. I write $H^s(\Omega)$ to denote $W^{s, 2}(\Omega)$.

The rationale behind this is the following: letting $$Tu:=\left. u \right|_{\partial \Omega}$$ denote the trace operator, we know that $$T\colon H^s(\Omega)\to H^{s-\frac12}(\partial\Omega), \qquad \forall s>\frac12,$$ and $T$ is surjective. Now, the space $H^{s-\frac12}(\partial \Omega)$ embeds in $C(\partial \Omega)$ if and only if $s-\frac12 >\frac{n-1}{2}$, because $\partial \Omega$ is a dimension $n-1$ manifold. On the other hand, the space $H^s(\Omega)$ embeds in $C(\Omega)$ if and only if $s>\frac n2$, which is exactly the same condition as before.

We conclude that the trace space $H^{s-\frac12}(\partial \Omega)$ embeds into $C(\partial \Omega)$ if and only if the original space $H^s(\Omega)$ embeds into $C(\Omega)$.

You are concerned with $s=2$, in which case we have this embedding if and only if $n\le 3$. For $n\ge 4$ you can construct an explicit example of a function in $H^2(\mathbb R^4_+)$ that has a singularity at the origin, so that its trace to $\mathbb R^3$ is not continuous. (Here $\mathbb R^4_+=\{(x_1, x_2, x_3, x_4)\ :\ x_4\ge 0\}$.)