I just looked at the fact (at least according to Definition 2.8.1. in Distribution Theory by Friedlander et al.) that for $K_0\subseteq{\bf R}^{n(0)}$ compact, $\Omega_1\subseteq{\bf R}^{n(1)}$ open and $f:{\cal D}(K_0)\rightarrow{\cal D}(\Omega_1)$ linear continuous the image of $f$ is contained in ${\cal D}(K_1)$ for some compact $K_1\subseteq\Omega$.
Is there a more general rule behind this? It seems like one of those cases where images of somewhat small sets remain small, like images of finite sets are finite, continuous images of compact sets are compact, linear continuous images of bounded sets are bounded, etc.
Is there something similar going on here?
The only idea that came to my mind is the following:
If a space $X$ has the property that any countably dimensional subspace is also boundedly generated and $f:X\rightarrow Y$ is continuous linear then also $f(X)$ has this property. Also any metrizable topological vector space has this property, so does ${\cal D}(K_0)$.
This is because for a countably dimensional subspace $Y'\subseteq F(X)$ with basis $\{y_i\}_{i\in{\bf N}}$ we can choose $x_i\in f^{-1}(\{y_i\})$ and the space $X'$ generated by the $x_i$ is also countably dimensional. Choose a bounded set $B\subseteq X$ which generates $X'$, then $f(B)$ is bounded and generates $Y'$.
If $X$ is metrizable then it has a countable local base $\{U_i\}_{i\in{\bf N}}$. For $X'$ generated by $\{x_i\}_{i\in{\bf N}}$ we can choose $\lambda_i>0$ such that $\lambda_i x_i\in\cap_{j\le i}U_i$, then $\{\lambda_i x_i\}_{i\in{\bf N}}$ is a bounded set which generates $X'$.
Now, any subspace of ${\cal D}(\Omega_1)$ which has this property must be contained in some ${\cal D}(K_1)$ because (i) any bounded subset of ${\cal D}(\Omega_1)$ is contained in some ${\cal D}(K_1)$ and (ii) any subset that is not contained in any ${\cal D}(K_1)$ has a countable subset that is not contained in any ${\cal D}(K_1)$.
Seems rather lengthy and unelegant though...