# Gelfand triples and isometries between dual spaces

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 $$$$V\subset H\subset V'$$$$ hold. The triple $$(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$$, in the following sense: $$$$(u,v)_{V,V'}=\langle{u,v\rangle}\quad\text{for all }u\in V,\; v\in H\subset V'$$$$ with $$\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 $$$$\Lambda_\pm:H^{\pm s}(\mathbb{R}^n)\rightarrow L^2(\mathbb{R}^n)$$$$ such that the duality between $$H^{\pm s}(\mathbb{R}^n)$$ can be realized as such: $$$$(u,v)_{s,-s}=\langle\Lambda_+u,\Lambda_-v\rangle.$$$$ My question is: does every Gelfand triple, without further assumptions, carry such a couple of operators $$\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 $$\Lambda_-$$.