Could we find an inverse inequality from the trace theorem?

Trace theorem: Let $\Omega$ is a bounded domain, with its boudary $\partial\Omega$ is piecewise smooth, then there exist a positive constant C such that $$ \|u\|_{L^2(\partial\Omega)} \le C\|u\|_{H^1(\Omega)} \qquad \forall~ u \in H^1(\Omega). $$ Could we find another positive constant $C_2$ such that $$ C_2\|u\|_{L^2(\partial\Omega)} \ge \|u\|_{H^1(\Omega)} \qquad \forall~ u \in H^1(\Omega)? $$

Solution 1:

Could we find another positive constant $C_2$ such that $$ C_2\|u\|_{L^2(\partial\Omega)} \ge \|u\|_{H^1(\Omega)} \qquad \forall~ u \in H^1(\Omega)? $$

The answer is surely NO for the same $u$ that appears in the first inequality. The counterexample is already given by Davide Giraudo in the comments.

However, if we are given a function $u$ that lives on the boundary of $\Omega$ at the first place, we can extend $u$ into the interior of $\Omega$ by a linear extension operator $E$: $$ E: H^{s-1/2}(\partial \Omega) \to H^s(\Omega), \; v\mapsto E(v) , $$ such that the trace of $E(v)$ is $v$ and the extension operator is bounded in that $$ \|E(v)\|_{H^s(\Omega)} \leq c\|v\|_{H^{s-1/2}(\partial\Omega)}. $$ With slight abuse of notation, you can sure denote this $E(v)$ as $v$, but keep in mind that this extension is not unique.

Remarks: As you said in your profile that you are interested in electromagnetics, I recalled reading Hiptmair's recent paper: Extension by zero in discrete trace spaces Inverse estimates. It discussed the bounds for extension operator of the Raviart-Thomas/Nedelec element spaces measure in $H^{-1/2}(\mathrm{div};\partial \Omega)$ sense.

The references for general extension for the trace space of $H(\mathbf{curl})$ can be found in two widely cited papers in the electromagnetics community:

• A. Alonso, A. Valli: Some remarks on the characterization of the space of tangential traces of $H(\mathbf{rot};\Omega)$ and the construction of an extension operator.

• A. Buffa, M. Costabel, D. Sheen: On traces for $H(\mathbf{curl};\Omega)$ in Lipschitz domains.