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anonymous
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Geometric implications of $\beta(B_X) = 2$

Let $X$ be a Banach space and $\beta$ denote Istrățescu's spreading measure of noncompactness, i.e. $$\beta(M) = \sup \{ \varepsilon > 0 \colon \exists_{(x_n)^{\mathbb N} \in X^{\mathbb N}} \forall_{m \ne n} \colon \|x_n-x_m\| > \varepsilon \}$$ for any $M \subset X$.

If we denote the (closed) unit ball of $X$ by $B_X$, it is clear that we have $1 \le \beta(B_X) \le 2$ (by Riesz's lemma and the triangle inequality). I believe one has $\beta(B_X) = \sqrt 2$ whenever $X$ is a Hilbert space and $\beta(B_X) = 2$ if $X$ is the sequence space $c_0$, $\ell^\infty$ or the function space $L^\infty$, so that intuitively, the less round a space, the larger the value $\beta(B_X)$.

Q: There is a large zoo of notions of non-roundness, like the (alternative) Daugavet property, the notion of (almost) CL-spaces, lushness, $n(X) = 1$ (where $n$ denotes the numerical index). Is it known how the property $\beta(B_X) = 2$ fits in here?

anonymous
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