Is every dual space with strong topology locally convex

analysisfunctional-analysisreal-analysis

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!

Best Answer

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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