Dual of Sobolev space $H^{s}(\mathbb{R}^n)$ Taylor Michael.

Question

Why the dual of $H^{s}(\mathbb{R}^n)$ is $H^{-s}(\mathbb{R})$? I know that dual of $H^{s}(\mathbb{R}^n)$ is $\left\{T:H^{s}(\mathbb{R}^n)\to \mathbb{C}:T \text{ bounded and linear functional} \right\}$

Is it because $\Lambda^{-s}\Lambda^{s}u=u$ and $\Lambda^{s}\Lambda^{-s}u=u$? Is that argument enough?

Answer 1

You have already solved it with $(u,v)=\int \hat{u}(\xi)\tilde{v}(\xi)d\xi=\int\hat{u}(\xi)<\xi>^s\tilde{v}(\xi)<\xi>^{-s}d\xi$ is a isomorphism of $H^{-s}$ and the dual of $H^{s}$