Let $(M,g)$ be the compact manifold with boundary. Let $\mathcal{E}^{\prime}(M)$ denotes the set of distributions (continuous linear functionals $)$ on $C^{\infty}(M)$, we equip this space with the weak* topology. I need to show that Dual space of $\mathcal{E}^{\prime}(M)$ is $C^{\infty}(M)$?
It is enough to show that these spaces are reflexive, so the dual of $\mathcal{E}^{\prime}( M)$ is $C^{\infty}( M)$.
Can any one please give me hint to show this spaces are reflexive? Any hint or help will be greatly appreciated.
Thank you in advanced
Let $\mathcal{E}'(M)^*$ denote the dual of $\mathcal{E}'(M)$.
Let $F \in \mathcal{E}'(M)^*$. Since $F$ is continuous, $O = \{\eta \in \mathcal{E}'(M): |F(\eta)|< 1\}$ is an open neighbourhood of $0$. Then by definition of the weak* topology, we can find $\phi_1, \dots, \phi_n \in C^\infty(M)$ and $\varepsilon > 0$ such that $\{\eta \in \mathcal{E}'(M): |\eta(\phi_i)| < \varepsilon\}$ is contained in $O$.
Hence it follows that $\bigcap_{i=1}^n \ker J \phi_i \subseteq \ker F$ where $J: C^\infty(M) \to \mathcal{E}'(M)^*$ is the obvious embedding. Hence (this is a usual exercise in linear algebra), $F = \sum_{i=1}^n \lambda_i J \phi_i = J \left ( \sum_{i=1}^n \lambda_i \phi_i \right )$ which is the desired result.
Notice that the above only uses that $C^\infty(M)$ is a topological vector space and that we consider its dual as being equipped with the weak* topology.