Talk:Normed Vector Space is Reflexive iff Weak and Weak-* Topologies on Normed Dual coincide

in general "normed dual" is just what I'm calling the dual of an NVS. "topological dual" is a more general concept that people reading pages on NVSs may not have met.

in this instance, "reflexive" considers a second dual and for this $X^\ast$ needs a topology, which is given by the norm. Caliburn (talk) 09:37, 25 June 2023 (UTC)

$X$ has only one topology, by which $X^\ast$ is uniquely determined as a set.

Then you consider three topological spaces based on $X^\ast$:

$\struct {X^\ast , \norm \cdot _{X^\ast} }$

$\struct {X^\ast , w }$

$\struct {X^\ast , w^\ast }$

Finally, you consider three kinds of $X^{\ast \ast}$ as sets:

$(1):\quad X^{\ast \ast} := \struct {X^\ast , \norm \cdot_{X^\ast} }^\ast$

$(2):\quad \struct {X^\ast , w }^\ast$

$(3):\quad \struct {X^\ast , w^\ast }^\ast$

In addition, in proof, you also use the non-trivial identity $(1) = (2)$.

Now you mean, calling $X^\ast$ "normed dual" indicates that you mean $X^{\ast \ast}$ being defined as $(1)$.

I don't believe that many readers can follow your logic.

Not harmful itself but not helpful. --Usagiop (talk) 10:09, 25 June 2023 (UTC)

$X^{\ast \ast}$ has always meant the former, this is made explicit if you follow definitions. I don't see where I've used $(1) = (3)$, that equality is true if and only if $X$ is reflexive. I will edit the page for topological dual to cover topology-sensitive contexts. Caliburn (talk) 10:17, 25 June 2023 (UTC)

Yes, $X^\ast$ and $X^{\ast \ast}$ are uniquely defined as sets. In particular, $X^\ast$ is really unique by nature, while $X^{\ast \ast}$ is only specified by you in another page. That is why I am wondering why you write "normed dual" at that place. Sorry I meant $(1) = (2)$ (corrected). --Usagiop (talk) 11:14, 25 June 2023 (UTC)

See new revision of Definition:Topological Dual Space. (edit: this doesn't deal with the ambiguity here lol, oops, gimme a sec) Caliburn (talk) 10:22, 25 June 2023 (UTC)