Let $\Omega \subset \mathbb{R}^3$ be a lipschitz domain. We then have the trace operator $\tau : H^1(\Omega) \to L^2(\partial \Omega)$ and can define the space $H^{1/2}(\partial \Omega) := \tau(H^1(\Omega)),$ which is equipped with the norm
$$\lVert g\rVert_{H^{1/2}(\partial \Omega)}:=\inf_{v\in H^1(\Omega), \tau(v)=g} \lVert v\rVert_{H^1(\Omega)}.$$
Finally, let $H^{-1/2}(\partial \Omega)$ denote the dual space of $H^{1/2}(\partial \Omega)$. We can view $L^2(\partial \Omega)$ as a subspace of $H^{-1/2}(\partial \Omega)$ via the linear map
$$
\begin{split}
L:L^2(\partial \Omega)&\to H^{-1/2}(\partial \Omega),\\
f &\mapsto L_f,
\end{split}
$$
where
$$L_f(\tau(\phi)):= \int_{\partial \Omega} f\cdot \tau(\phi)\,dS,\quad \phi \in H^1(\Omega).$$
$L$ is injective by the fundamental lemma of the calculus of variations and also bounded: The inequality
$$\lvert L_f(\tau(\phi))\rvert \leq \lVert f\rVert_{L^2(\partial \Omega)} \lVert \tau(\phi)\rVert_{L^2(\partial \Omega)}\leq C \lVert f\rVert_{L^2(\partial \Omega)} \lVert\phi\rVert_{H^1(\Omega)}$$
implies that
$$\lvert L_f(\tau(\phi))\vert\leq C\lVert f\rVert_{L^2(\partial \Omega)}\lVert\tau(\phi)\rVert_{H^{1/2}(\partial \Omega)},$$
so
$$\lVert L_f\rVert_{H^{-1/2}(\partial \Omega)}\leq C\lVert f\rVert_{L^2(\partial \Omega)}.$$
Does there exists a constant $C>0$ such that
$$(+)\quad\lvert\lvert f\rvert\rvert_{L^2(\partial \Omega)}\leq C \lVert L_f\rVert_{H^{-1/2}(\partial \Omega)}\quad \forall f\in L^2(\partial \Omega)$$
or a similar inequality holds?
My question is motivated by the following: For $u\in L^2(\Omega)^3$, a function $v\in L^2(\Omega)^3$ is called the weak curl of $u$ if
$$\int_{\Omega} u\cdot \text{curl }\phi\,dx = \int_{\Omega} v\cdot \phi\,dx\quad\forall \phi\in C_0^{\infty}(\Omega)^3.$$
We write $v=\text{curl } u$. We can introduce the hilbert space $H(\text{curl}, \Omega)$ of all functions $u\in L^2(\Omega)^3$ for which a weak curl exists with the norm
$$\lvert\lvert u\rvert\rvert_{H(\text{curl},\Omega)}^2:= \lvert\lvert u\rvert\rvert_{L^2(\Omega)^3}^2 + \lvert\lvert \text{curl } u\rvert\rvert_{L^2(\Omega)^3}^2.$$
One can then show the following theorem (for a reference, see Theorem 3.29 in "Finite Element Methods for Maxwell's Equations" by Monk):
$$ $$
Let $\Omega\subset \mathbb{R}^3$ be a bounded lipschitz domain. Then the trace map $\gamma_t$ which is defined classically by
$$\gamma_t(v) := \nu \times (v\rvert_{\partial \Omega}),\quad v\in C^{\infty}(\overline{\Omega})^3$$
(where $\nu$ is the outward normal vector of $\Omega$) can be extended by continuity to a continuous linear map $\gamma_t:H(\text{curl},\Omega)\to H^{-1/2}(\partial \Omega)^3$. Furthermore, the following Green's theorem holds for any $v\in H(\text{curl},\Omega)$ and $\phi\in H^1(\Omega)^3$:
$$\gamma_t(v) (\phi) = \int_{\Omega} \text{curl } v \cdot \phi – v\cdot \text{curl } \phi\,dx.$$
$$ $$
By the discussion above, we have $L^2(\partial \Omega)^3 \subset H^{-1/2}(\partial \Omega)^3$. In other words, for some $v\in H(\text{curl},\Omega)$, the functional $\gamma_t(v)$ can be expressed by a function $f\in L^2(\partial \Omega)^3$, i.e.
$$\gamma_t(v)(\phi) = \int_{\partial \Omega} f\cdot \tau(\phi)\,dx\quad \forall \phi \in H^1(\Omega)^3.$$
Since $\gamma_t$ is bounded, we have
$$\lvert\lvert \gamma_t(v)\rvert\rvert_{H^{-1/2}(\partial \Omega)^3}\leq C\lvert\lvert v\rvert\rvert_{H(\text{curl},\Omega)}.$$
My question is whether we have a similar inequality for $\lvert\lvert f\rvert\rvert_{L^2(\partial \Omega)^3}$:
$$\lvert\lvert f\rvert\rvert_{L^2(\partial \Omega)^3} \leq C\lvert\lvert v\rvert\rvert_{H(\text{curl},\Omega)}.$$
This would follow immediatly from $(+)$.
Best Answer
I think the last desired inequality cannot be right, but a direct counterexample escaped me, so here is an argument with a bit of a detour.
Let me maybe first frame this with an abstract result.
Lemma. If $A \colon X \supseteq D \to Y$ is a closed operator between Banach spaces with domain $D$, then $$\|Ax\|_Y \lesssim \|x\|_X \quad \text{for all}~x\in D$$ if and only if $D$ is a closed subspace of $X$. In particular, if $A$ is in fact densely defined, then the foregoing inequality is equivalent to $D = X$.
The "if" direction follows with the closed graph theorem. For the "only if" part, observe that if $D \supseteq (x_k) \to x$ in $X$, then by the assumed inequality, $(Ax_k)$ is a Cauchy sequence in $Y$ and thus convergent, hence the closedness of $A$ implies that $x \in D$.
That being said, consider $\gamma_t$ as an unbounded operator $D \subseteq H(\text{curl},\Omega) \to L^2(\partial\Omega)$ with the domain $$D = \Bigl\{ v \in H(\text{curl},\Omega) \colon \gamma_t(v) \in L^2(\partial\Omega)\Bigr\}.$$
Using the Green type formula, you can show that this is in fact a closed operator. Moreover, since, say, continuous differentiable functions on the closure of $\Omega$ are included in $D$, it is also densely defined. Hence, with the Lemma, it follows that the desired inequality $$\|\gamma_t(v)\|_{L^2(\partial\Omega)} \lesssim \|v\|_{H(\text{curl},\Omega)} \qquad \text{for all}~v \in D$$ is equivalent to $D = H(\text{curl},\Omega)$ which is, I suppose, certainly not correct.