Example: a locally convex TVS which is not compactly generated

Is there an example of a locally convex topological vector space which is not compactly generated?

(any such example must be non-Fréchet, since all Fréchet spaces are compactly generated)

(note: I am using "compactly generated" in the topologists' sense, not the functional analysts' sense)

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I do not have a proof nor a counterexample. But I'd like to note that bornological LCTVS are compactly generated so that the classical examples for LCTVS test functions and distributions will not give us counterexamples. Furthermore the full subcategory of compactly generated LCTVS is closed under colimit and more generally under taking "final objects" (in the sense of "final topology" form point set topology, not in the sense of "terminal object" from category theory. Is there a standard name for analogues of initial and final topologies?) –  Johannes Hahn Apr 28 at 21:54
This question has been answered but the following remark might be of interest since it gives a unified method which allows one to answer many related ones. Suppose that one has a topological property which is stable under the formation of closed subsets as is that of being a $k$-space (the standard name for the property in question). Then if you can find a completely regular space which fails the property, you can find a lcs which fails. This follows immediately from the fact every such space embeds in a canonical manner as a closed subspace of its so-called free lcs. –  janacek Apr 29 at 7:24

If $A$ is uncountable then Problem J(b), p. 240 of Kelley says that $\mathbf R^A$ (or $\mathbf C^A$) with the product topology (a.k.a. uniform convergence on finite subsets) is not compactly generated. This topology is locally convex, being generated by the seminorms $\|z\|_F=\sup_{a\in F}|z_a|$ for all finite $F\subset A$.
"uniform convergence on finite subsets" is also known as "pointwise convergence". $\hspace{1.37 in}$ –  Ricky Demer Apr 28 at 23:57
@NateEldredge Well, only if $|A| \le \mathfrak{c}$, so in a sense, most spaces $\mathbb{R}^A$ are non-separable... –  Henno Brandsma May 5 at 13:31