Does the trace operator commute with partial differentiation?

Question

Let $u\in H^2(\mathbb R^n)$. According to the trace theorem, $u$ has a trace $Tu\in H^1(\mathbb R^{n-1}\times\{0\})$ (at least locally). Suppose you know more about that trace, for example that $Tu$ equals $f\in C^\infty(\mathbb R^{n-1})$ a.e. on $\mathbb R^{n-1}\times\{0\}$. The question is now the following: Do taking the trace and taking a derivative commute, i.e. is it true that $T(\partial_i u)=\partial_if$, if $i\ne n$?

Answer 1

The trace operator is the continuous extension of the densely defined operator $T: C^1_0(\mathbb{R}^n) \to H^1(\mathbb{R}^{n-1} \times\{0\})$. This $T$ is in particularly linear, and hence commutes with partial differentiation tangential to $\mathbb{R}^{n-1}\times \{0\}$. Thus, appealing to any of the standard approximation procedures you see that indeed, the extended operator also commutes with partial differentiation.

(For example, we know that mollified $u_\epsilon\in H^2\cap C^\infty$ converges to $u$ as $\epsilon \to 0$ in $H^2$, and that its partials converge in $H^1$. If the traces of $\partial_i u_\epsilon$ converges in $L^2$, say, then by continuity that limit is also the trace of $\partial_i u$. )