# Complete dual of bornological space

A bornologigal topological vector space is such that any bounded linear function on it is continuous. It is a standard result [Jarchow, Locally convex spaces, 1981] that if the dual $$E'$$ of a Mackey space $$E$$ is complete for the topology $$\mathcal{T}_{\mathcal{B}_0}$$ of uniform convergence on bipolars of null sequences, then the Mackey space is bornological.

In a report thesis [Gach, Topological versus Bornological Concepts in Infinite Dimensions, 2004, Thm 6.1.16], this result even in stated for the strong topology on $$E'$$. Does this result appears anywhere else ? It seems quite strong to me, and I am not sure I understand the proof.

• Please add theorem and page number references for what you are referring to in Gach's thesis. It's 155 pages. – Robert Furber Apr 28 at 15:44
• This seems to be Theorem 6.1.16 on page 106 of Gach's diploma thesis (which is easy to find). – Jochen Wengenroth Apr 29 at 6:17
• In case anyone else wants to look there, in Jarchow, the relevant part is Section 13.2 Theorem 4 on page 275. – Robert Furber Apr 30 at 23:44

If I understand properly, I doubt very much that this is true.

I the article On different types of non-distinguished Frechet spaes (Note di Mat. 10 (1990), 149-165), Bonet, Dierolf, and Fernandez write that for a Frechet space $$X$$ with dual $$(X',\beta(X',X))$$ the Mackey topology $$\mu(X',X'')$$ is bornological if and only if every linear form on $$X'$$ which is bounded on bounded sets is already continuous. Moreover, they show examples that this not always the case. It thus seems to me that $$(E,\tau)=(X',\mu(X',X''))$$ is a counterexample.

Jochen is quite right. I have another example, just using any irreflexive Banach space $$A$$. The space $$E = (A^*,\mu(A^*,A))$$ is Mackey, by definition. The bounded sets in $$E$$ are the same as the norm-bounded sets, because $$A$$ is Banach, and therefore barrelled, so all dual topologies on $$A^*$$ have the same bounded sets (because $$\sigma(A^*,A)$$-bounded $$\Leftrightarrow$$ equicontinuous). So the canonical embedding $$i(A)$$ of $$A$$ in the strong dual of $$E$$ is an isomorphism, and so $$E^* \cong A$$ is complete.

But $$E$$ is not bornological, because by the same characterization of the bounded sets of $$E$$, the bounded linear functionals on $$E$$ are exactly $$A^{**}$$, which contains $$E^* = i(A)$$ as a proper subspace by the assumption that $$A$$ be irreflexive.

I think this also shows that Gach's proof of (4) $$\Rightarrow$$ (1) is at fault when he says "apply (4.1.5)", because if we apply his proof to $$E$$, the $$F$$ obtained in the proof will be $$A^*$$ with its norm topology, and $$A^*$$ and $$E$$ don't have the same set of continuous linear functionals.

• It occurs to me that I should emphasize that I use $A^*$ to mean the continuous dual space, notated $A'$ in the question and in Jochen Wengenroth's answer. I do this because of my operator algebra background, where $A'$ means the commutant. Usually when $A'$ is the notation for the continuous dual, $A^*$ means the algebraic dual, but this is not the case in my answer (I never find that I need the algebraic dual, except for when I want to point out that the weak-* topology is not complete). – Robert Furber Sep 22 at 23:55