Trace regularity result $\lVert n \times u\rVert_{H^{-1/2}}$ 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.
 A: I think that there is a typo in the equation for $w$.  Here's how I would approach this issue.
First note that we can integrate by parts with sufficiently smooth functions to deduce the equality
\begin{equation}
\int_\Omega \text{curl}X \cdot Y = \int_\Omega X \cdot \text{curl} Y + \int_{\partial \Omega} n \times X \cdot Y.
\end{equation}
Now, if we view the term on $\partial \Omega$ as the dual pairing between $H^{1/2}(\partial\Omega)$ and $H^{-1/2}(\partial \Omega) = (H^{1/2}(\partial\Omega))^*$, then we have
\begin{equation}
<n \times X, Y>= \int_\Omega \text{curl}X \cdot Y - \int_\Omega X \cdot \text{curl} Y.
\end{equation}
We derived this assuming that $Y$ was already defined in $\Omega$, but we can extend to general $Y \in H^{1/2}(\partial \Omega)$ by using an extension operator: for $Y \in H^{1/2}(\partial \Omega)$ we let $EY \in H^1(\Omega)$ be such that $EY = Y$ on $\partial \Omega$ and 
\begin{equation}
\lVert EY \rVert_{H^1(\Omega)} \lesssim \lVert Y\rVert_{H^{1/2}(\partial \Omega)}.
\end{equation}
Then 
\begin{multline}
\lvert <n \times X, Y> \rvert= \left\lvert \int_\Omega \text{curl}X \cdot EY - \int_\Omega X \cdot \text{curl} EY \right\rvert \\
\lesssim \lVert EY \rVert_{H^1} \lVert X\rVert_{H_{curl}} 
\lesssim \lVert Y \rVert_{H^{1/2}} \lVert X\rVert_{H_{curl}},
\end{multline}
where we have defined the norm
\begin{equation}
\lVert X\rVert_{H_{curl}}^2 = \int_\Omega \lvert \text{curl}X\rvert^2 + \lvert X\rvert^2
\end{equation}
with corresponding Hilbert space 
\begin{equation}
H_{curl}(\Omega) = \{ u \in L^2 \vert \text{curl}u \in L^2 \}.
\end{equation}
Since this works for any $Y \in H^{1/2}(\partial \Omega)$ we deduce, upon taking the supremum over all $Y$ with $\lVert Y \rVert_{H^{1/2}}\le 1$, that 
\begin{equation}
\lVert n \times X \rVert_{H^{-1/2}(\partial \Omega)} \lesssim \lVert X \rVert_{H_{curl}}.
\end{equation}
This works for $X$ smooth, but by a density argument (here some work is needed, but it's in the literature -- e.g in one of the books by J.-L. Lions) the same inequality holds for all $X \in H_{curl}$.  This means that having $L^2$ control on $X$ and $\text{curl}X$ is enough to get control on $n \times X$ on the boundary.  
Now fix $\varepsilon >0$ and define an equivalent inner-product on $H_{curl}$ given by 
\begin{equation}
(X,Y)_\varepsilon = \int_\Omega \varepsilon \text{curl}X \cdot \text{curl}Y + \frac{1}{\varepsilon} X \cdot Y.
\end{equation}
Write $\lVert X\rVert_{\varepsilon} = \sqrt{(X,X)_\varepsilon}$. For any $\Phi \in H^{1/2}(\partial \Omega)$ we know from the above discussion that the mapping 
\begin{equation}
H_{curl} \ni X \mapsto <n \times X, \Phi>
\end{equation}
is a bounded linear functional.  By the Riesz representation theorem there exists a unique $W_\Phi \in H_{curl}$ so that 
\begin{equation} 
<n \times X, \Phi> = (X,W_\Phi)_\varepsilon \;\forall X \in H_{curl}
\end{equation}
and 
\begin{equation}
\lVert W_\Phi \rVert_{\varepsilon} = \sup_{\lVert X \rVert_{\varepsilon} \le 1} <n \times X, \Phi>  \lesssim \lVert \Phi \rVert_{H^{1/2}(\partial \Omega)}.
\end{equation}
Then for any $\Phi \in H^{1/2}(\partial \Omega)$ and any $X \in H_{curl}$ we have
\begin{multline}
\lVert n\times X\rVert_{H^{-1/2}}  = \sup_{\lVert \Phi \rVert_{H^{1/2}} \le 1}
\lvert <n \times X, \Phi> \rvert = \sup_{\lVert \Phi \rVert_{H^{1/2}} \le 1} \lvert (X,W_\Phi)_\varepsilon \rvert \\
\le \sup_{\lVert \Phi \rVert_{H^{1/2}} \le 1} \lVert X \rVert_{\varepsilon} \lVert W_\Phi\rVert_\varepsilon  \lesssim \lVert X \rVert_{\varepsilon}   \sup_{\lVert \Phi \rVert_{H^{1/2}} \le 1} \lVert \Phi \rVert_{H^{1/2}(\partial \Omega)} = \lVert X \rVert_{\varepsilon}.
\end{multline}
This produces the estimate
\begin{equation}
\lVert n\times X\rVert_{H^{-1/2}} \le 
\sqrt{\int_\Omega C\varepsilon \lvert \text{curl}X \rvert^2 + \frac{C}{\varepsilon} \lvert X \rvert^2}   \le 
\sqrt{C \varepsilon} \lVert \text{curl}X \rVert_{L^2} + \sqrt{\frac{C}{\varepsilon}} \lVert X \rVert_{L^2},
\end{equation}
which immediately yields the desired estimate upon setting $\epsilon = \sqrt{C \varepsilon}$.
Now, to wrap up, why did I say there was a typo on the $w$ equation?  Well, the function $W_\Phi$ satisfies
\begin{equation}
<n \times X, \Phi> = 
\int_\Omega \varepsilon \text{curl}X \cdot \text{curl}W_\Phi + \frac{1}{\varepsilon} X \cdot W_\Phi
\end{equation}
for all $X$.  This is a weak formulation of the PDE
\begin{equation}
\begin{cases}
\varepsilon \text{curl}^2 W_\Phi + \varepsilon^{-1}W_\Phi =0 &\text{in }\Omega\\
\varepsilon \text{curl}W_\Phi\times n = \Phi \times n &\text{on }\partial \Omega,
\end{cases}
\end{equation}
which is pretty close to what was written in the paper.
Incidentally, a very similar argument produces the estimate 
\begin{equation}
\lVert n\cdot X\rVert_{H^{-1/2}} \le 
\epsilon \lVert \text{div}X \rVert_{L^2} + \frac{C}{\epsilon} \lVert X \rVert_{L^2},
\end{equation}
for all 
\begin{equation}
X \in H_{div}(\Omega) = \{ u \in L^2 \vert \text{div} u \in L^2\}
\end{equation}
with the obvious inner-product.

A remark about the extension operator and Shuhao Cao's answer:
An extension operator is definitely in use in this method.  In order to solve the PDE for $w$ using Lax-Milgram, one must first transform the problem with inhomogeneous boundary conditions to one with homogeneous boundary conditions (L-M works on linear spaces rather than affine ones).  To do so one must first take $\phi \in H^{1/2}(\partial \Omega)$ and extend it to $E\phi \in X$, where $X$ is the Hilbert space in which one wants to apply L-M, and $-n \times (n \times E\phi) = \phi$ when traced onto $\partial \Omega$. Then one uses L-M to find a $\tilde{w}$ (with homogeneous BCs $\tilde{w}=0$ on $\partial \Omega$) satisfying 
$$
\int_{\Omega} \epsilon(\nabla  \times \tilde{w})\cdot (\nabla  \times v) + \frac{1}{\epsilon}\tilde{w} \cdot v = -\int_{\Omega} \epsilon(\nabla  \times E\phi)\cdot (\nabla  \times v) + \frac{1}{\epsilon} E\phi \cdot v
$$
for all $v$.  Then $w = \tilde{w} + E\phi$ is the desired solution.  Note that it's not enough to have any extension $E\phi$, but rather one must have a bounded extension in order to estimate $\tilde{w}$ in terms of $\phi$ rather than $E\phi$.  In the space $X$ proposed to work in, the extension is actually more complicated than the usual $E: H^{1/2}(\partial \Omega) \to H^1(\Omega)$ since it also requires that $\text{div}E\phi =0$.
A: 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...
