Let $U \subseteq \mathbb{R}^n$ be open and denote by $\mathcal{D}(U)$ the space of all compactly supported smooth functions $U \to \mathbb{R}$. Let $\mathcal{D}^\prime(U)$ be the space of all distributions $\mathcal{D}(U) \to \mathbb{R}$ with the standard topology.
Given a distribution $T$, I would like to prove that there exists a sequence $(\psi_n)$ in $\mathcal{D}(U)$ such that
\begin{equation}\label{eq:1}\tag{$\ast$}
\lim_{n \to \infty} \left\langle \psi_n, \varphi \right\rangle = \left\langle T , \varphi\right\rangle
\end{equation}
for all $\varphi \in \mathcal{D}(U)$. I became interested in this question while the following paragraph from this Wikipedia article:
The test functions are themselves locally integrable, and so define distributions. As such they are dense in $\mathcal{D}^\prime(U)$ with respect to the topology on $\mathcal{D}^\prime(U)$ in the sense that for any distribution $T \in \mathcal{D}^\prime(U)$, there is a sequence $\psi_n \in \mathcal{D}(U)$ such that
$$
\left\langle \psi_n, \varphi \right\rangle \to \left\langle T, \varphi \right\rangle
$$
for all $\varphi \in \mathcal{D}(U)$. This fact follows from the Hahn-Banach theorem, since the dual of $\mathcal{D}^\prime(U)$ with its weak*-topology is the space $\mathcal{D}(U)$.
My question is as follows: how does this follow from the Hahn-Banach theorem? I understand why $(\mathcal{D}^\prime(U))^\ast \cong \mathcal{D}(U)$ when the former is given the weak*-topology, but I fail to see how \eqref{eq:1} follows from the Hahn-Banach theorem.
Best Answer
Your scepticism ist justified. A consequence of the Hahn-Banach theorem is that a continuous linear map $f:X\to Y$ between two locally convex Hausdorff spaces has dense range if and only if the transposed $f^t: Y'\to X'$ is injective. Applying this to the inclusion $f:\mathscr D \hookrightarrow \mathscr D'$ one gets that $\mathscr D$ is weak$^*$-dense in $\mathscr D'$ -- but in general this does not imply sequential density.