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 one of the sequence spaces $c_0$, $\ell^1$, and $\ell^\infty$ or one of the function spaces $L^1[0,1]$ or $L^\infty[0,1]$, 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: The following computations show $\beta(B_X) = 2$ for the spaces that were given as examples earlier.

• $X = c_0$ or $X = \ell^\infty$: Consider the sequence $x_1 = (1, 0, \dotsc)$, $x_2 = (-1, 1, 0, \dotsc)$, $x_3 = (-1, -1, 1, \dotsc)$; then $x_m$ and $x_n$ differ by 2 in the coordinate $\min(m,n)$ for $m \ne n$.

• $X = L^\infty[0,1]$: Analogously, consider the sequence $(f_n)$ 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 $m \ne n$.

• $X = \ell^1$: Here, it suffices to consider the canonical sequence of unit vectors $x_1 = (1, 0, \dotsc)$, $x_2 = (0, 1, 0, \dotsc)$, $x_3 = (0, 0, 1, 0, \dotsc)$ and so forth (so that $x_n$ and $x_m$ differ by 1 in the $k$-th coordinate for $k \in \{ m, n \}$ whenever $n \ne m$).

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