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In the theory of stereotype spaces, it is known that for a locally convex space $X$,

  1. If $X$ is pseudocomplete, then $X^{\star}$ is pseudosaturated, and
  2. If $X$ is pseudosaturated, then $X^{\star}$ is pseudocomplete.

This result is presented in (2003) Pontryagin Duality in the Theory of Topological Vector Spaces and in Topological Algebra, S. S. Akbarov. (Thrm 2.16)

I'm curious about a similar duality between quasi-completeness and some form of barrelledness (or quasi-barrelledness). Of course, in this case the topology on the dual space is the topology of uniform convergence on von Neumann bounded sets. Is there any such a result?

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    $\begingroup$ IV. 6.1 in Schaefer's Topological Vector Spaces states that if $E$ is barrelled, $E'$ is quasicomplete for any $\mathfrak{S}$ topology defined by bounded sets, so in particular, the strong topology (as you define it, though not necessarily by Schaefer's definition of strong topology). So one direction is right. $\endgroup$ Commented Apr 5, 2017 at 5:50
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    $\begingroup$ @RobertFurber Thank you. I'll look at the reference you give. $\endgroup$ Commented Apr 7, 2017 at 2:28

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There is rather deep result of Laurent Schwartz for the class of Schwartz locally convex spaces $X$ (that is, for every continuous seminorm $p$ there is another one $q\ge p$ such that the unit ball of $q$ is precompact with respect to $p$, the name was coined by Grothendieck and because of the Arzela-Ascoli theorem many non-Banach spaces appearing in analysis are Schwartz): The strong dual of a complete Schwartz space is bornological (that is, a linear map is already continuous if it is bounded on bounded sets). A proof is, e.g., in the book Introduction to Functional Analysis by Meise and Vogt.

A rather elementary converse is that the strong dual of every bornological locally convex space is complete.

You can find more on the relation between barrelledness conditions and completeness properties of the dual in book Barrelled Locally Convex Spaces of Bonet and Perez Carreras.

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