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Aug 25, 2022 at 12:36 comment added Jack L. .... or if $X$ is a uniformly smooth space (say $\ell^2$), then $X^*$ has UKK. But $\ell^\infty$, as the dual of $\ell^1$ does not have UKK. The Hardy space, $H^1$ and several classical non-reflexive spaces have the w*-UKK.
Aug 25, 2022 at 12:35 comment added Jack L. .... Indeed, (w*-)UKK is a variation of a topological theme: that of the coincidence of the sequential weak(-star) topology and norm topology in the unit sphere (also known as Kadec-Klee property). Thus, it will appear infeasible to obtain a geometrical characterization as such, beyond that given by the definition .$$~$$ As requested, concerning examples, because Schur spaces (i.e. spaces where the sequential weak and norm topologies agree in the unit ball) and uniformly convex spaces have the UKK, it follows that if $X$ is a predual of $\ell^1$ (say $c_0$) or
Aug 25, 2022 at 12:26 comment added Jack L. The definition, as you have given above, is the uniform Kadec Klee (simply, UKK) property for the dual space, rather than the weak* uniform Kadec Klee (simply, w*-UKK), which requires weak-star convergence. That said, both properties are equivalent in Grothendieck spaces (i.e. spaces where weak and weak-star convergence coincide)..$$~$$That said, it is properly safe to say that (w*-)UKK is a topological, rather than geometrical, property of the unit sphere.......(cont’d below).
S Aug 24, 2022 at 6:55 history suggested Angel Peñaflor CC BY-SA 4.0
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Aug 24, 2022 at 1:59 history edited Tomás Pérez Fernández CC BY-SA 4.0
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S Aug 24, 2022 at 1:48 review First questions
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S Aug 24, 2022 at 1:48 history asked Tomás Pérez Fernández CC BY-SA 4.0