for a compact manifold $M$, is the dual space of $H^1(M)$ equal to $H^{-1}(M)$?

Question

Let $M$ be a compact Riemannian manifold. Is it true that $$(H^1(M))^* = H^{-1}(M)?$$

is there some intuitive explanation why? Or some reference? Thanks

Here $H^1$ is the usual Sobolev space of $u \in L^2(M)$ with $\nabla u \in L^2(M)$.

Answer 1

No, we only define the dual of $(H_0^1(M))$ is $H^{-1}(M)$, but not $H^1(M)$. Although the dual of $H^1(M)$ does exist, and it is smaller then $H^{-1}(M)$.

If you want an intuitive explanation, I recommend you to read the definition by Fourior transform then you will see the dual of $H^s$ is $H^{-s}$, not by definition but by computation.