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Here is a question about topological vector spaces. Consider a t.v.s. $X$. Than on the continuous dual you can define the strong topology, given by the uniform convergence on the bounded setes. Let $B = \{A \mid A \, \mathrm{bounded} \}$. Than the topology on the dual is induced by the family of seminorms $\big\{\Vert.\Vert_A\big\}_{A \in B}$ $$\Vert f\Vert_A = \sup_{x \in A} \mid f(x)\mid $$

Now, we know that a topological vector space is locally convex if and only if its topology is induced by a family of seminorms.

Does this mean that every dual space with this topology is locally convex? Or I'm missing something?

Thanks in advance for your answers!

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Does this mean that every dual space with this topology is locally convex? YES! Or I'm missing something? NOT! Your thought is correct.

Another way to think. The strong topology is a polar topology (see the book Topological Vector Space, Lawrence Narici and Edward Beckenstein, Example 8.5.5) and every polar topology is locally convex (see the same book 8.5 Polar Topology, or Example 11.2.5).

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  • $\begingroup$ Great! So, also the weak topology and the weak* are locally convex, I guess? (but this doesn't require duality theory, the general form of a neighborhood should take care of it, right?). I don't know much of duality theory, but I need these results for my thesis, so I was looking for some easy explanation that doesn't require too much theory of general topological vector spaces. $\endgroup$ Aug 26, 2019 at 8:02
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    $\begingroup$ Exactly. The weak and weak* topologies are locally convex. Caution!: You defined the Strong Topology in $X'$ with the collection B={$A\subset X$; A bounded}, but the definition must be with the collection C={$A\subset X$; A is $\sigma(X,X')$-bounded} (this ensures the good definition of the seminorms). In special cases B coincides with C (in Locally Convex Hausdorff Spaces for example, see Narici and Beckenstein Thm 8.8.7.), but it is not true in general. $\endgroup$
    – L26
    Aug 27, 2019 at 1:12

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