TITLE: Trace regularity result $\lVert n \times u\rVert_{H^{-1/2}}$
QUESTION [6 upvotes]: There is a result in a paper I am reading : 
Let $\Omega$ be a bounded domain.  For any $\epsilon > 0$, there is a constant $C(\epsilon)$ such that 
$$\lVert n \times u\rVert_{H^{-1/2}(\partial \Omega)} \leq \epsilon\lVert \nabla \times u\rVert_{L^2(\Omega)} + C(\epsilon)\lVert u\rVert_{L^2(\Omega)} $$ where $n$ is the normal.

Here is the proof provided : For any $\phi \in H^{1/2}(\partial \Omega) $, consider the problem 
$$\nabla \times (\nabla \times w) + \frac{1}{\epsilon^2}w = 0 \text{ in } \Omega $$
$$ -n \times (n \times w) = \phi \text{ on } \partial \Omega$$
Then, the result follows immediately from estimating $|(n\times \phi)|$.

I am somewhat new to these types of arguments.  Can I get a first step here?  Thanks in advance.

REPLY [2 votes]: The author meant to resolve one problem: How to define the tangential trace of a vector field as a bounded linear functional on surfaces.
I do not think the result is as obvious as the author claimed. 
Bold faced letters are used for vector field and its function space here just to avoid confusion. $\newcommand{\v}[1]{\boldsymbol{#1}}$
Suppose $\v{w}$ satisfies:
$$
\begin{aligned}
\nabla \times (\nabla \times \v{w}) + \frac{1}{\epsilon^2}\v{w }&= 0 \quad\text{ in } \Omega
\\
-\v{n} \times (\v{n} \times \v{w}) &= \v{\phi} \quad\text{ on } \partial \Omega
\end{aligned}\tag{1}
$$
and $\v{\phi}\in \v{H}^{1/2}(\partial \Omega)$. The corresponding variational problem is to find: $\v{w} \in \v{H}^1(\Omega)$ satisfying the boundary condition and:
$$
\int_{\Omega} \epsilon(\nabla  \times \v{w})\cdot (\nabla  \times \v{v}) +\int_{\Omega} \frac{1}{\epsilon}\v{w}\cdot \v{v} = 0
$$
for any test function $\v{v}\in \v{H}^1(\Omega)\cap\{\v{v}=0 \text{ on } \partial \Omega\}$. This problem is well-posed in a subspace of $\v{H}^1$ vector field: namely divergence free vector field with a zero normal component on boundary. Then the well-posedness can be established by Lax-Milgram lemma. And for divergence free vector field, the $\v{H}^1$-norm is equivalent to $\|\cdot\|_{\epsilon}:=\left(\frac{1}{\epsilon}\| \cdot\|_{\v{L}^2}^2 + \epsilon\| \nabla \times \cdot\|_{\v{L}^2}^2\right)^{1/2}$ for fixed $\epsilon$. Therefore there exists a solution for (1) weakly. 
So for any $\v{\phi}\in \v{H}^{1/2}(\partial \Omega)$ that is defined on boundary, we could associate with it a divergence free $\v{H}^1$-vector field satisfying (1) inside the domain, and the Dirichlet type tangential projection boundary condition. And following natural estimate "solution bounded by data" for $\|\cdot\|_{\epsilon}$ holds:
$$
\|\v{w}\|_{\v{H}^1(\Omega)} \leq C(\epsilon)\|\v{w}\|_{\epsilon}\leq C(\epsilon)\|\v{\phi}\|_{\v{H}^{1/2}(\partial \Omega)}\tag{2}
$$
Now for any $\v{f}\in \v{H}^{1/2}(\partial \Omega)$, with its corresponding problem (1)'s solution $\v{v}\in \v{H}^1(\Omega)$ with boundary data $\v{n}\times(\v{f}\times\v{n})$. We have
$$\int_{\partial \Omega} \v{\phi} \times \v{n} \cdot \v{f} = \int_{\partial \Omega} \v{\phi} \times \v{n} \cdot (\v{n}\times(\v{f}\times\v{n}))  = \int_{\Omega} ( \nabla\times \v{w} \cdot \v{v} - \nabla\times \v{v}\cdot\v{w}) $$
Hence the dual norm by definition can be estimated by:
$$
\begin{aligned}
&\|\v{\phi} \times\v{n} \|_{\v{H}^{-1/2}} = \sup_{\v{f}} \frac{\left|\displaystyle\int_{\partial \Omega} \v{\phi} \times \v{n} \cdot \v{f}\,\right|}{\|\v{f}\|_{\v{H}^{1/2}(\partial \Omega)}}
\\
&\leq \sup_{\v{v}}  \frac{
\sqrt{\epsilon}\|\nabla\times \v{w}\| \frac{1}{\sqrt{\epsilon}}\|\v{v}\| + \sqrt{\epsilon}\| \nabla\times \v{v}\| \frac{1}{\sqrt{\epsilon}}\|\v{w}\| }{\|\v{v}\|_{\v{H}^{1}( \Omega)}}
\\
&\leq \sup_{\v{v}}  \frac{(\epsilon\|\nabla\times \v{w}\|^2 + \frac{1}{\epsilon}\|\v{w}\|^2 )^{1/2} (\epsilon\|\nabla\times \v{v}\|^2 + \frac{1}{\epsilon}\|\v{v}\|^2 )^{1/2}}{\|\v{v}\|_{\v{H}^{1}( \Omega)}}
\\
&\leq c(\epsilon)(\epsilon\|\nabla\times \v{w}\|^2 + \frac{1}{\epsilon}\|\v{w}\|^2 )^{1/2}
\end{aligned}\tag{3}
$$
We also need to balance the constant related to $\epsilon$ from (2) in the first inequality in (3). The last step is to notice that 
$$
\v{\phi} \times \v{n}= -\v{n} \times (\v{n} \times \v{w}) \times\v{n} = \v{w} \times \v{n}
$$
Hence for $\v{w}\in \v{H}^1$:
$$
\|\v{w} \times\v{n} \|_{\v{H}^{-1/2}}^2\leq C( \epsilon\|\nabla\times \v{w}\|^2_{\v{L}^2(\Omega)} + \frac{1}{\epsilon}\|\v{w}\|^2_{\v{L}^2(\Omega)})
$$
As Glitch pointed, the constant $C$ depends on $\epsilon$, so the result is rather pointless.

Remark: The key step is essentially the same with Glitch. The differences are:

I didn't introduce the extra $\v{H}(\mathrm{curl})$ space like he did. We could just restrict ourselves for $\v{H}^1$-vector field why this estimate is correct. Therefore the boundary condition given by the paper is correct. Otherwise, we have to consider a problem like Glitch proposed, using Neumann boundary condition: $\v{n}\times(\nabla \times \v{w}) = \v{\phi}$ on $\partial \Omega$.
I didn't use the extension operator. When passing the first inequality in (3), Glitch used extension operator's property. I used the estimate from the boundary value problem.
The proof here I cooked up is solely for catering the boundary value problem that paper claimed to use. I still prefer Glitch's more "canonical" proof which is the same with what I learned from my electromagnetics class.
(EDIT)As Glitch pointed out, this is only true for some $\v{H}^1$-vector field. I am conjecting might be true for any $\v{H}^1$-vector field, there is a boundary Helmholtz decomposition such that the $\v{L}^2$-trace space can be decomposed into the trace spaces of divergence free $\v{H}^1$ and curl free $\v{H}^1$, such that the tangential trace is bounded by this $\v{H}^1$-vector field's $\v{H}(\mathrm{curl})$-norm inside, and the normal trace is bounded by its $\v{H}(\mathrm{div})$-norm inside. Apparently this generates a new question for myself...