Properties of the $L^2$ dual realizing test functions as distributions Let $U \subset \mathbb{R}^n$ be open; let $C^\infty_c(U)$ denote the smooth real/complex compactly-supported functions on $U$, with the "Canonical LF" topology, i.e. the limit topology from all the restriction maps $C_c^\infty(U) \rightarrow C^\infty(K)$ for compact $K\subset U$. (I assume $C^\infty(K)$ has the the uniform topology.) Let $\langle f,g\rangle = ||\bar{f}g||_{L^1(U)} = \int_{U}|dx|\,\overline{f(x)}\,g(x)$ be the standard inner product. Let $W = C^\infty_c(U)^*$ be the strong dual of $C^\infty_c(U)$.
I have the following questions:

*

*Is the linear injective map $\phi : C^\infty_c(U) \rightarrow W : f \mapsto (g \mapsto \langle f,g\rangle)$ also continuous? If so, is it an embedding of TVSs?


*Is the image in $W$ of $\phi$ dense, for the weak-$\ast$ topology (for $\langle\cdot,\cdot\rangle$) on $W$? What about for the strong dual topology on $W$?
I am trying to examine these but unraveling the limit topology on $C^\infty_c(U)$ and the strong dual topology on $W$ is somewhat complicated. Does anyone know any techniques that can be applied here?
 A: Ad 1:
\begin{align*}
\phi\text{ c'ts.}&\Leftrightarrow(\forall f_n\to0\in C_c^{\infty}(U))(\phi(f_n)\to0\text{ in the strong dual topology})\\
&\Leftrightarrow(\forall f_n\to0)(\forall\text{ bounded }B\subseteq C_c^{\infty}(U))\left(\lim_n{\sup_{g\in B}{\phi(f_n)(g)}}=0\right)
\end{align*}
$C_c^{\infty}(U)$ has no norm, so "bounded" here needs a little definition.  Wikipedia jumps straight to duality, but the simplest definition is von Neumann boundedness: for all neighborhoods $V$ of the origin, there exists $r$ such that $rB\subseteq V$.  To match the Wikipedia definition, $B$ should be von-Neumann bounded in the weak topology on $C_c^{\infty}(U)$.
Since the weak topology is coarser than the standard topology, we may pass to the latter a fortiori.  In fact, we can always expand $B$ to a set of the form $$B=\{f:(\forall x)(|f(x)|\leq R)\}\subseteq C_c^{\infty}(U)$$
But then $$\sup_{g\in B}{\phi(f_n)(g)}=\sup_{g\in B}{\langle f_n,g\rangle}\leq R\|f_n\|_{L^1}$$  Since convergence in $C_c^{\infty}$ implies convergence in $L^1$, the latter tends to $0$.  Thus $\phi$ is indeed continuous.
As to whether $\phi$ is an embedding, it depends on what you mean by embedding.

*

*$\phi$ is injective: if $\phi(f)=0$, then for all $g\in C_c^{\infty}(U)$, we have $0=\phi(f)(g)=\langle f,g\rangle$; taking $g=f$ shows that $f=0$.

*But $\phi$ is not bicontinuous.  For each $n$, let $f_n:C_c^{\infty}(\mathbb{R})$; $f_n(x)=\frac{1}{1+(x-n)^2}$.  By an argument given below, any bounded $B\subseteq C_c^{\infty}(\mathbb{R})$ is contained in $C_c^{\infty}(K)$ for some compact $K$, so that $\phi(f_n)\to0$ on $B$.  But $f_n\not\to0$ as elements of $C_c^{\infty}(U)$, since a sequence converging to $0$ in $C_c^{\infty}(U)$ must have uniformly compact support.

Ad 2:
The image is dense in both topologies; the strong dual topology is finer than the weak-$*$, so it suffices to prove density there.
First, I claim distributions of compact support are dense.  To be precise, fix an enumeration $\{K_n\}_{n=0}^{\infty}$ of increasing compact sets exhausting $U$ and find a "partition of unity": for each $n$, let $\psi_n\in C_c^{\infty}(U)$ be smooth, $1$-bounded, and such that $\psi_n|_{K_n}=1$ but $\psi_n|_{U\setminus K_{n+1}}=0$.  If $v\in W$, then I claim that the functionals $v_n:W$; $v_n(f)=v(f\cdot\psi_n)$ converge to $v$.
To see this, pick bounded $B\subseteq C_c^{\infty}(U)$.  Suppose for contradiction that there exists $\{x_n\}_{n=0}^{\infty}\in U^{\omega}$ such that, for all $n$, there exists $f_n\in B$ with $f(x_n)\neq0$, but $x_n\notin K_n$.  Then $w:W$; $$w(g)=\sum_n{f_n(x)g(x)}$$ is well-defined and continuous (since elements of $C_c^{\infty}(U)$ have compact support) but is unbounded on $B$.  Thus $B\subseteq C_c^{\infty}(K_n)$ for some $n$ (by abuse of notation), so that, for $m\geq n+1$, $v_m$ and $v$ coincide on $B$.
Second, recall that distributions of compact support are the derivatives of functions; that is, if $v\in W$ has compact support, then there exists $g\in C(U)$ and a multiindex $\alpha$ such that, for all $f$, $$v(f)=\langle g,f^{(\alpha)}\rangle$$  This is theorem 6.26 in Rudin, Functional Analysis (1991), and takes some machinery to prove.
Lastly, pick a sequence of smooth functions converging to $g$ and integrate by parts; this gives a sequence from $\operatorname{im}{\phi}$ converging to $v$.
