Talk:Characterization of Continuity of Linear Functional in Weak-* Topology

Characterization of Continuous Linear Functionals in Weak-* Topology? Caliburn (talk) 21:19, 16 June 2023 (UTC)
 * Singular noun -- the theorem applies to an arbitrary continuous linear functional. "Characterization of" suits well enough, despite our aim to avoid words that are different either side of the pond. But I thin k I'm okay to be firm on the singularness of the noun.


 * Same applies to the other page. It's a style, it's near enough 100% consistent throughout, nice to sustain that vibe. --prime mover (talk) 23:07, 16 June 2023 (UTC)

And yep, I seem to have over-generalised without thinking enough. $X^\ast$ has not been given a topology here. We can of course define the $w^\ast$ topology still, but I think there is a more standard strong topology called the polar topology. I don't know the details so I will just scale back this page to an NVS. Caliburn (talk) 22:08, 16 June 2023 (UTC)
 * Do you mean the weak topology? So, is the equality $\struct {X^\ast, w^\ast}^\ast = \struct {X^\ast, w}^\ast$ true? --Usagiop (talk) 22:43, 16 June 2023 (UTC)
 * No, explicitly:
 * $\struct {X^\ast, w^\ast}^\ast = \set {x^\wedge : x \in X} = \iota X$
 * as sets, (there is no topology given to the LHS) where $\map {x^\wedge} f = \map f x$ for each $f \in X^\ast$. As to $\iota X$ vs $\iota \sqbrk X$, I'm not sure. I guess it's whatever looks better. I also would not object to going back to writing $J$ or $j$ instead, as is often found in literature.
 * The equality you write is true $X$ is reflexive, since it is precisely the statement that $\iota X = X^{\ast \ast}$. (the weak dual is the same as the norm dual) This is essentially proven in Normed Vector Space is Reflexive iff Weak and Weak-* Topologies on Normed Dual coincide. Caliburn (talk) 22:48, 16 June 2023 (UTC)
 * OK, so you mean:
 * $\struct {X^\ast, w^p}^\ast = \iota X$
 * where $w^p$ is the polar topology, should be true. --Usagiop (talk) 23:11, 16 June 2023 (UTC)
 * I don't think I do? I definitely mean the $w^\ast$ topology on $X^\ast$. Then the only continuous linear functionals $\struct {X^\ast, w^\ast} \to \GF$ are the evaluation maps. I don't know anything about the polar topology yet. This is Theorem 3.16 in Functional Analysis and Infinite-Dimensional Geometry and it appears precisely like this. Caliburn (talk) 23:15, 16 June 2023 (UTC)
 * OK, I just misunderstood. You then only wanted to mention the polar topology of $X^\ast$, nothing more. --Usagiop (talk) 23:24, 16 June 2023 (UTC)
 * I thought the polar topology was something different. Caliburn (talk) 23:25, 16 June 2023 (UTC)
 * So you mean you think:
 * $\struct {X^\ast, w^p}^\ast \ne \iota X$
 * --Usagiop (talk) 23:27, 16 June 2023 (UTC)
 * To be clear, my problem was that I wrote $X^{\ast \ast}$ (meaning continuous linear functionals on $X^\ast$) without having given $X^\ast$ a topology. Caliburn (talk) 22:56, 16 June 2023 (UTC)

Also it says so we get the equality. Caliburn (talk) 22:13, 16 June 2023 (UTC)