Why is $C^\infty$ a subspace of its dual?

Question

Actually, let me start with understanding the argument for $S(\mathbb{R}^n) \subset S'(\mathbb{R}^n)$, where $S$ is Schwartz and $S'$ is the space of tempered distributions. That is, we can identify any $\Psi \in S(\mathbb{R}^n)$, with a $u \in S'(\mathbb{R}^n)$ defined to be $u\phi = \int \Psi \phi$, $\phi \in S(\mathbb{R}^n)$. All I know is from the density of
$S(\mathbb{R}^n)$ in $L^2(\mathbb{R}^n)$, there exists a sequence $\phi_n \in S(\mathbb{R}^n)$ such that $\int \phi_n \Psi \to \int |\Psi|^2$. How do I go from here?

I imagine the answer to my original question will also be along the same lines? thank you very much for your time!

Answer 1

The space $C^\infty(\mathbb R^n)$ does not naturally imbed into its dual which is the space of distributions with compact support -- the reason is that for $f\in C^\infty(\mathbb R^n)$ the attempted map $\psi_f(g)=\int f(x)g(x)dx$ is not well defined (e.g., for $f=g=1$). It is well-defined for $f\in\mathscr D(\mathbb R^n)$, the space of smooth functions with compact support, and a precise meaning of natural could be that $\mathscr D(\mathbb R)\to C^\infty(\mathbb R^n)'$, $f\mapsto \psi_f$ does not have a continuous extension to a map $C^\infty(\mathbb R^n)\to C^\infty(\mathbb R^n)'$.

Answer 2

I just wanted to remark that, in the complex setting at least, there is actually an interesting embedding in the opposite direction, $\mathscr F \colon(C^\infty (\mathbb R^n))’\to C^\infty (\mathbb R^n) $ given by the Fourier transform. In fact, every compactly supported distribution is a tempered distribution (i.e., $(C^\infty(\mathbb R^n))’\hookrightarrow \mathscr S’(\mathbb R^n)$) and, by the Paley Wiener theorem, its Fourier transform is smooth. This also means that, given two compactly distributions $u,v$, it is well-defined the pairing $\langle{u,\mathscr Fv}\rangle$.