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 $L^\infty[0,1]$, 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?

**Edit**: To see that $\beta(B_{c_0}) = 2$, consider the sequence $x_1 = (1, 0, ...)$, $x_2 = (-1, 1, 0, ...)$, $x_3 = (-1, -1, 1, ...)$; then $x_m$ and $x_n$ differ by 2 in the coordinate $\min(m,n)$ for $m \ne n$.

Analogously, consider the sequence $(f_n)$ in $L^\infty[0,1]$ given by
$$f_n = \begin{cases}
  -1 & \text{on $[0, 1-2^{1-n})$}\\
   1 & \text{on $[1-2^{1-n},1-2^{-n})$}\\
   0 & \text{on $[1-2^{-n},1]$}
\end{cases}$$
for $n \ge 1$. Then $f_n$ and $f_m$ differ by 2 on the interval $[1-2^{1-k},1-2^{-k}]$ with $k = \min(m,n)$ whenever $m \ne n$