Let $X$ be an infinite-dimensional 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? Is it known if this property is inherited by duals or pre-duals? (biduals obviously inherit it). Is it known if such spaces can be reflexive?
It is known, e.g., that spaces with the Daugavet property cannot be reflexive, and neither can those with $n(X) = 1$, at least in the real case; see also
Kadets, Vladimir; Martín, Miguel; Payá, Rafael. Recent progress and open questions on the numerical index of Banach spaces. RACSAM. Rev. R. Acad. Cienc. Exactas Fís. Nat. Ser. A Mat. 100 (2006), no. 1-2, 155--182. MR2267407
Consequently, neither can lush or (almost) CL spaces.
Edit: It is shown in
Kryczka, Andrzej; Prus, Stanisław. Separated sequences in nonreflexive Banach spaces. Proc. Amer. Math. Soc. 129 (2001), no. 1, 155--163. MR1695123
that $\beta(B_X) \ge \sqrt[5]4 \approx 1.3195$ holds for any non-reflexive space $X$. They also construct a non-reflexive space $X$ with $\beta(B_X) \approx 1.7086$.
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 $m \ne n$.
$X = C[0,1]$: Replacing the constant 1-intervals with hat-functions in the previous example produces a sequence of continuous functions that show $\beta(B_X) = 2$. This sequence works in $L^\infty[0,1]$, too, naturally.
$X = \ell^1$: 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 $m \ne n$.
