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I am not 100% clear what you are asking, but I will answer according to two interpretations: a) Suppose that $F$, a metrizable TVS, is the strong dual of $E$, a locally convex TVS. Need $F$ be comple …

This is true. To show it, in the following I will use $\langle \mbox{-}, \mbox{-} \rangle$ for the pairing between a space and its dual (with the vectors from the space on the left, the dual on the ri …

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- …

The answer is yes. First, since $\newcommand{E}{\mathcal{E}}\E$ is a Fréchet space, it is barrelled, and so any $\sigma(\E',\E)$-bounded subset of $\E'$ is equicontinuous, and therefore bounded in any …

YCor has given a counterexample for topological vector spaces. The statement is still false for locally convex spaces. Consider the space $X$ defined to be a locally convex coproduct of $\newcommand{\ …

The answer is no. I know that for some people here, saying "It's false for $X = \ell^1$" would be a good enough hint, but I also know that this question originated on Math StackExchange, so I've inclu …

A good resource for these things is Section IV.10 of Schaefer's Topological Vector Spaces, so you should look there for the proofs of the following statements. For $E$ a locally convex space, let $\el …

If you work in terms of topologies on M, then the way to go is to use the theory of dualities from functional analysis. The basic theory of these things can be found in books called "Topological Vecto …

Judging by what you say in the question, I think you are referring to what Świrszcz called "Saks spaces" in this article: Monadic functors and convexity, Bull. Acad. Polon. Sci. Ser. Sci. Math. Astro …

The criterion suggested in the question works fine for $\sigma$-finite spaces, and Michael Greinecker's answer is correct under this assumption. However, the suggested criterion is not (provably) suf …

I agree with Dmitri Pavlov that separability is not so important in the modern theory of von Neumann algebras, and this answers the second question. However, an example answering the first question ha …

For any C$^*$-algebra $A$, we can define its opposite algebra $A^{\mathrm{op}}$, which is the algebra where $ab$ is defined to be $ba$, as calculated in $A$. Let's restrict to unital algebras for simp …

There's a general principle for proving that a topology on a vector space $E$ is not a weak topology (in the general sense, a topology of the form $\sigma(E,F)$ for some $F \subseteq E^*$). For $\sigm …

Yes. In the next paragraph I will show that if $X$ is a Fréchet space (without requiring separability) then $X'_c$ with the compact-open topology is a $k$-space. As you note, this implies sequentialit …

I will pull together the comments into a community wiki answer with some of my own remarks so that the question isn't left on the unanswered questions list. If you're willing to accept that it is cons …