Using standard notation, we refer to $H^s(\mathbb R) = W^{s,2}(\mathbb R)$ to be the Sobolev Hilbert spaces. As is often the case, it's natural to then consider properties of functions in $H^s(\mathbb R)$ by looking at how it behaves in the Fourier domain, more specifically, we have that for any $s\in \mathbb{R}$,
$
\begin{align*}
H^s(\mathbb{R}) = \{ g \in L_2(\mathbb{R}) : \int (1+t^2)^{s}|\mathcal{F} g(t)|^2dt <\infty\},
\end{align*}
$
where $\mathcal{F}$ is the Fourier transform. In particular to note is that this works for $s<0$ as well, and so negative order Sobolev spaces are easily defined. In particular, one can define the norm of $H^s(\mathbb{R})$ to be defined by the relation $\|g\|_s = \int (1+t^2)^{s}|\mathcal{F}g(t)|^2dt$.
$\textbf{My first question:}$ I have seen, in particular the book Sobolev Spaces by Adams, Fournier (page 64), that another way to define the norm of the dual space of $H^s$, which appears to be $H^-s$ is given by the following:
$$
\begin{align*}
\|g\|^*_{-s} = \sup_{h\in H^{s}}\frac{\langle g,h\rangle}{\|h\|_s},
\end{align*}
$$
where $\langle g,h\rangle$ is the standard $L_2$ inner product. I am having difficulty figuring out whether the norm defined by Fourier transformations or the one defined by using the dual are equivalent or not. The furthest I have been able to show is that $\|g\|^*_{-s} \leq (2\pi)^{-1}\|g\|_{-s}$ by Parceval's relation.
$\textbf{My second question:}$ If there appears to be a relationship, then I would like to restrict attention to spaces of the form $H^{s}(A)$ where $A\subset \mathbb{R}$ with defined norm $\|g\|_{s,A} = \inf\{ \|g^{'}\|_s : g^{'}_{|A} = g\}$ for all $g\in H^s(A)$. Then is there a relationship between $\|g\|_{s,A}$ and
$$
\begin{align*}
\|g\|_{-s,A}^* = \sup_{h\in H^s(A)}\frac{\langle g,h\rangle_{A}}{\|h\|_{s,A}},
\end{align*}
$$
where $\langle g,h\rangle_{A}$ is the standard $L_2$ inner product restricted to $A$? I am having difficulty trying to relate the two, because there seems to be a natural relationship globally, which I would hope to think there is a local relationship as well.
Best Answer
Q1 has been addressed by Mark in the comments (absolute values are missing in your formula), but let me quickly look at this again: Just take Fourier transforms to see that $$ \|g\|_{-s}^*=\sup_{\|h\|_s=1} |\langle g,h\rangle | = \sup \left\{ |\langle \widehat{g}, \widehat{h}\rangle | : \|(1+t^2)^{s/2}\widehat{h}\|_2 = 1 \right\} = \sup \left\{ |\langle (1+t^2)^{-s/2}\widehat{g}, (1+t^2)^{s/2}\widehat{h} \rangle | : \ldots\right\}=\|g\|_{-s} . $$ Notice that in the third $\sup$, we can simply interpret $\langle. , .\rangle$ as the standard scalar product of $L^2(\mathbb R)$; this identifies this $\sup$ as the $L^2$ norm of the first argument of the scalar product, which is $\|g\|_{-s}$.
As for Q2, this formula will fail badly for restrictions because (unlike the case $U=\mathbb R$) the action of $H^{-s}(U)$ distributions on test functions does not extend to $H^s(U)$ functions. First of all, we probably want to insist on open sets $A=U$, so that we can meaningfully restrict distributions. (We restrict a distribution to an open set by only applying it to test functions with support in that set.) Then consider, for example, $g(x)=1/x$ on $U=(0,1)$. Then $g\in H^{-1/2-\epsilon}(U)$ according to your definition because $g$ is the restriction of $PV-1/x$ to $U$. However, if we test against an $h\in H^{1/2+\epsilon}(0,1)$ with $h=1$ near zero, then $\langle g, h\rangle $ isn't even defined.