Consider the fractional Sobolev spaces on $\mathbb R^n$

$H^s(\mathbb R^n) := \left\{ u \in \mathcal S'(\mathbb R^n) \; : \; ( 1 + |\xi|^2 )^{s/2} \hat u \in L^2(\mathbb R^n) \right\}$.

Let $\Omega$ be any open subset of $\mathbb R^n$, we then define

$H^s(\Omega) := \left\{ u \in \mathcal D'(\Omega) \; : \; u = R_\Omega U, U \in H^s(\mathbb R^n) \right\}$,

where $R_\Omega U$ denotes the restriction of distributions. Note that $H^s(\Omega)$ is defined equivalently by factoring those elements of $H^s(\mathbb R^n)$ which have no support on $\Omega$. Let $H^s(\mathbb R^n)$ and $H^s(\Omega)$ be equipped with the canonical scalar products, making them Hilbert spaces.

I wonder whether we do have a bounded linear pairing

$\langle u, v \rangle_{H^s(\Omega) \times H^{-s}(\Omega)} := \inf \left\{ \langle u, v \rangle \; : \; u = R_\Omega U, U \in H^s(\mathbb R^n), v = R_\Omega V, V \in H^{-s}(\mathbb R^n) \right\}. $

The pairing of $H^s(\mathbb R^n) \times H^{-s}(\mathbb R^n)$ is in fact a duality pairing, and so it is well-defined and bounded. But $H^s(\Omega) \times H^{-s}(\Omega)$ is generally not a duality pairing to my best knowledge. As it is well-defined (the infimum is a real number), is it a bounded pairing?

Consider the fractional Sobolev spaces on $\mathbb R^n$

$H^s(\mathbb R^n) := \left\{ u \in \mathcal S'(\mathbb R^n) \; : \; ( 1 + |\xi|^2 )^{s/2} \hat u \in L^2(\mathbb R^n) \right\}$.

Let $\Omega$ be any open subset of $\mathbb R^n$, we then define

$H^s(\Omega) := \left\{ u \in \mathcal D'(\Omega) \; : \; u = R_\Omega U, U \in H^s(\mathbb R^n) \right\}$,

where $R_\Omega U$ denotes the restriction of distributions. Note that $H^s(\Omega)$ is defined equivalently by factoring those elements of $H^s(\mathbb R^n)$ which have no support on $\Omega$. Let $H^s(\mathbb R^n)$ and $H^s(\Omega)$ be equipped with the canonical scalar products, making them Hilbert spaces.

I wonder whether we do have a bounded linear pairing $\langle \cdot, \cdot \rangle_{H^s(\Omega) \times H^{-s}(\Omega)}$ that extends the $L^2$-product on, say, smooth test functions.

The pairing of $H^s(\mathbb R^n) \times H^{-s}(\mathbb R^n)$ is in fact a duality pairing, and so it is well-defined and bounded. But $H^s(\Omega)$ and $H^{-s}(\Omega)$ do generally not form a duality pairing to my best knowledge. Can you still define the product between them?