# Is the space of bounded continuous functions on the reals with the topology of uniform convergence on compact sets fully barrelled?

Let us consider the space $$X=C_b(\mathbb{R},\mathbb{R})$$ endowed with the following topology: $$f_n\rightarrow f$$ iff $$f_n\rightarrow f$$ uniformly on compact subsets of $$\mathbb{R}$$. This space is a locally convex topological vector space which is metrizable. Unless I'm really mistaken it is not a Frechet space, since a Cauchy sequence in $$X$$ could potentially converge to an unbounded (but continuous) function.

A barrelled space is a space for which each barrel (i.e. each convex, balanced, absorbing and closed subset) is in fact a neighborhood of zero. A fully barrelled space is a barrelled space for which each closed subspace is also barrelled. Frechet spaces (and therefore also Banach spaces) are examples of fully barrelled spaces. However, barrelled spaces need not be complete, but I have not found anywhere if fully barrelled spaces are complete.

My question is whether the space $$X$$ with this topology is a fully barrelled space.

• Do you know whether this space is barrelled? I do not know but if I should make a guess I would say no. – Jochen Oct 18 '18 at 7:17
• No, I didn't know that either, but the answer below shows that it isn't. – WillemMSchouten Oct 18 '18 at 11:25

The space is not even barrelled: $$T=\{f\in X: |f(x)|\le 1$$ for all $$x\in\mathbb R\}$$ is absolutely convex, closed, and absorbing in $$X$$, hence a barrel. But $$T$$ is not a $$0$$-neighbourhood because every $$0$$-neighbourhood $$U$$ in $$X$$ contains a set of the form $$U_n=\{f\in X: |f(x)|\le 1/n$$ for all $$|x|\le n\}$$ for some $$n\in\mathbb N$$ and no such $$U_n$$ is contained in $$T$$.

• Well, not entirely sure that every 0-neighbourhood should contain such a set since I think the set $\{f\in X:|f(x)|\leq 1/2n$ for all $|x|\leq n\}$ for some $n\in\mathbb{N}$ is also a 0-neighborhood which does not contain any of the $U_n$. Your idea seems correct though, since every 0-neighbourhood should contain a set of the form $\{f\in X:|f(x)|\leq \epsilon$ for all $x\in K\}$ for some $\epsilon>0$ and compact $K$ and none of these sets are contained in $T$. So thanks for the answer. – WillemMSchouten Oct 18 '18 at 11:24
• Your set contains my $U_{2n}$ and I stand by my claim that $U_n$ form a $0$-neighbourhood basis-- but I admit that it is more natural to consider two parameters: one for the compact set and one for the size of the functions on that set. – Jochen Oct 18 '18 at 13:23
• Ow right my bad. – WillemMSchouten Oct 20 '18 at 9:11