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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.

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    $\begingroup$ Please add theorem and page number references for what you are referring to in Gach's thesis. It's 155 pages. $\endgroup$
    – Robert Furber
    Commented Apr 28, 2019 at 15:44
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    $\begingroup$ This seems to be Theorem 6.1.16 on page 106 of Gach's diploma thesis (which is easy to find). $\endgroup$
    – Jochen Wengenroth
    Commented Apr 29, 2019 at 6:17
  • $\begingroup$ In case anyone else wants to look there, in Jarchow, the relevant part is Section 13.2 Theorem 4 on page 275. $\endgroup$
    – Robert Furber
    Commented Apr 30, 2019 at 23:44

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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.

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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.

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  • $\begingroup$ 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). $\endgroup$
    – Robert Furber
    Commented Sep 22, 2019 at 23:55

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