Suppose that $X$ is a metric space and let $C\_k(X)$ denote the space of real functions on $X$ with the topology of uniform convergence on compact sets. Then $C\_k(X)$ is a topological vector space. Let $C\_k(X)^\*$ denote the strong dual space. For each $x\in X$ the point evaluation functionals $\delta\_x$ belong to $C\_k(X)^\*$. 


Is the set $\{\delta\_x\colon x\in X\}$ bounded in $C\_k(X)^\*$?


When $X$ is compact this is trivial.



Boundedness of Dirac deltas
$S \subseteq C\_k(X)^\*$ is bounded in the strong dual topology if and only if $\sup \_{\omega \in S, f \in B} |\omega (f)| < +\infty$ for every bounded $B \subseteq C\_k(X)$.


$B \subseteq C\_k (X)$ is bounded if and only if $\sup \_{f \in B, x \in K} |f(x)| < +\infty$ for every $K \subseteq X$ compact.


Call your subset $S$. Assume that $X$ is not compact; [for metric spaces, compactness is equivalent to pseudocompactness](https://en.wikipedia.org/wiki/Pseudocompact_space#Properties_related_to_pseudocompactness), therefore there exists an *unbounded* $F \in C\_k(X)$; notice that $\{F\} \subset C\_k(X)$ is obviously bounded. Then


$$\sup \_{\omega \in S, f \in \{F\}} |\omega(f)| = \sup \_{x \in X} |F(x)| = +\infty$$


so $S$ is not bounded. Together with the fact (which you claim to know) that if $X$ is compact then $S$ is bounded, we get that *$S$ is bounded if and only if $X$ is compact*.


