Sobolev Preamble
For $N\geqslant 2$, let $\Omega \subset \mathbb{R}^N$ be a connected, bounded, open set with Lipschitz continuous boundary $\partial \Omega$.
Let the space $H^{1/2}(\partial \Omega)$ denote the Sobolev–Slobodeckij space (or fractional Sobolev space) defined by the seminorm
\begin{equation}
\|f\|_{H^{1/2}(\partial \Omega)}^2:= \|f\|^2_{L^2(\partial \Omega)}+ \oint\limits_{\partial \Omega}\oint\limits_{\partial \Omega} \dfrac{|f(\mathbf{x})-f(\mathbf{y})|}{|\mathbf{x}-\mathbf{y}|^{N}}dS_{\mathbf{x}}dS_{\mathbf{y}}.
\end{equation}
Let $H^{-1/2}(\partial \Omega)$ denote the dual of the space $H^{1/2}(\partial \Omega)$, with associated norm
\begin{equation}
\|g\|_{H^{-1/2}(\partial \Omega)}:= \sup\limits_{v \in H^{1/2}(\partial \Omega)\setminus \{0\}}\dfrac{\left<g, v \right>_{L^2(\partial \Omega)}}{\|v\|_{H^{1/2}(\partial \Omega)}}
\end{equation}
EDIT: To clarify $\left<g, v \right>_{L^2(\partial \Omega)}$ is understood in the sense,
\begin{equation}
\left<g, v \right>_{L^2(\partial \Omega)} := \oint\limits_{\partial \Omega} gv \ dS_{\mathbf{x}},
\end{equation}
the $g\in H^{-1/2}(\partial \Omega)$ is the functional that acts upon $v\in H^{1/2}(\partial \Omega)$.
Weak equation
For a given $g \in H^{-1/2}(\partial \Omega)$, I wish to understand the solutions $u \in H^{1}(\Omega)$ to the following equation
\begin{equation}
\left< g, \phi \right>_{H^{-1/2}(\partial \Omega)} = \left< Tu, \phi \right>_{L^2(\partial \Omega)}, \quad \forall \phi \in H^{-1/2}(\partial \Omega),
\end{equation}
where the bounded surjective map $T:H^1(\Omega)\rightarrow H^{1/2}(\partial \Omega)$ is the trace operator.
Problem
I wish to understand if I could express the trace of $u$ explicitly in terms of $g$. For example, if I set $\phi=g$ then I deduce that
\begin{equation}
\|g\|^2_{H^{-1/2}(\partial \Omega)}= \sup\limits_{\|v\|_{H^{1/2}(\partial \Omega)}=1}\left<g, v \right>_{L^2(\partial \Omega)}^2 = \left< Tu, g \right>_{L^2(\partial \Omega)}
\end{equation}
Thus, we have that
\begin{equation}
\sup\limits_{\|v\|_{H^{1/2}(\partial \Omega)}=1}\left<g, v\left<g, v \right>_{L^2(\partial \Omega)}-Tu \right>_{L^2(\partial \Omega)}=0
\end{equation}
In a certain sense, one could view the trace of $u$ as equal to the projection of $g$ onto $H^{1/2}(\partial\Omega)$. However, I am unsure how to generalise this to $\phi\neq g$. Do I require a more restrictive test function space for $\phi$?
Best Answer
The inner product in $H^{1/2}$ is given by $$(f,g)_{H^{1/2}}=\int_{\partial \Omega}fg\,dS+\int_{\partial\Omega}\int_{\partial\Omega}\frac{f(x)-f(y))(g(x)-g(y))}{|x-y|^N}dS_xdS_y$$ By Riesz representation theorem, given a linear continuous function $L:H^{1/2}\to \mathbb{R}$, there is a unique function $f\in H^{1/2}$ such that $$L(\phi)=(f,\phi)_{H^{1/2}}$$ for all $\phi\in H^{1/2}$. Let’s write $L=L_f$ for later.
So in the definition of norm in $H^{-1/2}$ you cannot take the $L^2$ norm. The norm of $L$ is $$\sup_{v\in H^{1/2}\setminus\{0\}}\frac{L(v)}{\|v\|_{H^{1/2}}}= \sup_{v\in H^{1/2}\setminus\{0\}}\frac{(f,v)_{H^{1/2}}}{\|v\|_{H^{1/2}}}$$
The EDIt is not correct.
Now, it is not clear what you are asking. You fix $L\in H^{-1/2}$ and you want to solve $$(L,M)_{H^{-1/2}}=L(Tu)$$ for every $M$ in the dual? If $L,M\in H^{-1/2}$, we have that $L=L_f$ and $M=M_h$ for some $f$ and $h$ in $H^{1/2}$ and their inner product $$(L,M)_{H^{-1/2}}=(f,h)_{H^{1/2}}$$ The dual $H'$ of a Hilbert Space $H$ is a Hilbert Space So what yo are asking is $$(f,h)_{H^{1/2}}=(h,Tu)_{H^{1/2}}$$ for all $h\in H^{1/2}$? Then the only solution is $f=Tu$, so $ u$ is any function whose trace is $f$.