I recently have begun reading about DF-spaces and its clear to me that $C(K)$ is a DF-space for any compact subset (non-empty) $K$ of some $\mathbb{R}^D$ for finite D, since $C(K)$ is Banach. However, what if we relace $K$ by all of $\mathbb{R}^D$ itself? Is $C(\mathbb{R}^D)$ a DF-space when it is equipped with the usual local-uniform convergence topology?



Is $C(\mathbb{R}^n)$ is a DF-Space?
No. With the topology of uniform convergence on compact sets, $C(\mathbb R^d)$ is a Fréchet space (for the sequence of semi-norms $p\_n(f)=\sup\{|f(x)|: |x|\le n\}$) and the intersection of the classes of Fréchet and DF-spaces consists precisely of all Banach spaces.


Here is a proof of this well-known fact: By definition, a DF-space has a fundamental sequence of bounded sets $B\_n\subseteq B\_{n+1}$, i.e., every bounded subset is contained in some $B\_n$. On the other hand, if $p\_n$ is an increasing sequence of semi-norms describing the topology of the Fréchet space and $1/c\_n=\sup\{p\_n(b): b\in B\_n\}+1$, then you can check that $B=\bigcup\_{n\in\mathbb N} c\_n B\_n$ is a bounded set such that each bounded set is contained in a multiple of $B$. Then the closed absolutely convex hull of $B$ is a bounded $0$-neighbourhood and its Minkowski functional is a norm giving the topology of the space.


