Let $H$ be a Hilbert space and $V\subset H$ being a topological dense subspace of it carrying a finer topology such that $V$ is continuously embedded into $H$. Identifying $H$ with its dual $H'$, the inclusions \begin{equation} V\subset H\subset V' \end{equation} hold. The triple $(V,H,V')$ is then referred to as a Gelfand triple, and the duality pairing between $V$ and $V'$ is compatible with the inner product in $H$, in the following sense: \begin{equation} (u,v)_{V,V'}=\langle{u,v\rangle}\quad\text{for all }u\in V,\; v\in H\subset V' \end{equation} with $\langle\cdot,\cdot\rangle$ being the inner product in $H$.
Now, in the case of Sobolev spaces (for simplicity in $\mathbb{R}^n$), the triple $(H^s(\mathbb{R}^n),L^2(\mathbb{R}^n),H^{-s}(\mathbb{R}^n))$ for $s>0$ is known to be a Gelfand triple. In particular, there exist two isometries \begin{equation} \Lambda_\pm:H^{\pm s}(\mathbb{R}^n)\rightarrow L^2(\mathbb{R}^n) \end{equation} such that the duality between $H^{\pm s}(\mathbb{R}^n)$ can be realized as such: \begin{equation} (u,v)_{s,-s}=\langle\Lambda_+u,\Lambda_-v\rangle. \end{equation} My question is: does every Gelfand triple, without further assumptions, carry such a couple of operators $\Lambda_\pm$? If so, are they unique?
This is probably a fairly elementary question, but I am not familiar with the subject. Thanks in advance.
$\Lambda_+:V\to H$ is the continuous embedding. $\Lambda_-$ is an unbounded operator with $\Lambda_- : dom(\Lambda_-)=H \subset V'\to H$, $\Lambda_-u=u$ for $u\in H$.
Usually $\Lambda_+$ is not written explicitly (as all continuous embeddings are), while nobody uses $\Lambda_-$.