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What is the geometric meaning or interpretation of spaces that possess the weak* uniform Kadec-Klee property?

I am writing the last part of my undergraduate thesis and I would like to add a comment in which I explain the geometric idea or, in it's absence, the inspiration to define the concept mentioned, as well as some examples of spaces that fulfill the property and examples of spaces that do not comply. I would be very grateful if you could help me with any of those things.

The definition of the topological dual space of a Banach space having the uniform weak* Kadec-Klee property is the following:

For every $\varepsilon > 0$, there exists a $\delta>0$ such that if:

  • $\{f_n^*\}_n$ is a sequence in the unit ball of the dual space converges weakly to $f^*$.
  • The separation constant is greater than $\varepsilon$, where the separations constant is defined as the infimum of the distances between any two different elements of the sequence,

Then $\|f^{*}\|< 1 - \delta$.

I find this definition a little bit complicated to visualize geometrically.

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    $\begingroup$ 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). $\endgroup$
    – Jack L.
    Aug 25, 2022 at 12:26
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    $\begingroup$ .... 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 $\endgroup$
    – Jack L.
    Aug 25, 2022 at 12:35
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    $\begingroup$ .... 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. $\endgroup$
    – Jack L.
    Aug 25, 2022 at 12:36

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