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

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?

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 $$H^{s}$$