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I am trying to understand more about the Gelfand-Pettis integral. From wikipedia:

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What does it mean that $V$ "admits a dual space?" When $V$ is a Banach space, $V^{\ast}$ is taken to be the space of bounded linear functionals on $V$. When $V$ is a smooth (locally constant) representation of a totally disconnected group, $V^{\ast}$ is usually taken to mean the space of smooth linear functionals on $V$.

Is there a general notion of a topological vector space admitting a dual space? Or is this just context dependent?

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    $\begingroup$ The key phrase seems to be "that separates points". $\endgroup$
    – j.c.
    Commented Oct 14, 2018 at 16:05
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    $\begingroup$ Every topological space admits a dual space (this is just the space of continuous, linear functionals). The full statement is "assume [...] admits a dual space which separates points." Not all topological vector spaces admit a dual space which separates points. For example, $L_\frac{1}{2}[0,1]$ has only two open, convex subsets (the empty set and the whole space). Thus there is a surprising lack of continuous, linear functionals on this space. Not enough to separate points. $\endgroup$
    – user114263
    Commented Oct 14, 2018 at 16:11
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    $\begingroup$ In addition, "separates points" means that the original space embeds canonically into its double dual. $\endgroup$
    – Todd Trimble
    Commented Oct 14, 2018 at 16:25
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    $\begingroup$ Yes, the dual of a topological vector space is (by convention) the set of all continuous linear functionals. That is, when writing "Hom", you mean Hom in the category of topological vector spaces. $\endgroup$
    – Gerald Edgar
    Commented Oct 14, 2018 at 17:54
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    $\begingroup$ math.stackexchange.com/questions/1922567/… $\endgroup$
    – Andrew
    Commented Oct 21, 2018 at 16:57

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