What is the minimal radius $r(d)$ such that for every finite set of diameter $d$ there is a closed ball of radius $r(d)$ that contains it? This in the $L^1$ space with $n$ dimensions.
In formulas, given
$$ \|x - y\|_1 = \sum_{i \in \mathbb{Z}_n} |x_i - y_i| $$ $$ d(S) = \sup_{x,y \in S} \| x - y \|_1 $$
what is the minimal $r(d)$ such that:
$$ \forall S \subset \Re^n, |S| \in \mathbb{N}^+ \implies \exists z \in \Re^n : \forall s \in S, \|s - z\|_1 \leq r(d(S)) $$
It's easy to prove that in every normed space $\frac{d}{2} \leq r(d) \leq d$, the first part is from the triangle inequality and the second because with a radius of $d$ all points in S satisfy the requirement.
While searching for an answer, I found https://en.wikipedia.org/wiki/Injective_metric_space which considers a more generic property that allows every point of $S$ to have different radii and different distances between each other. For that reason, it can be used to prove that $n = 2 \implies r(d) = \frac{d}{2}$ but not that $n > 2 \implies r(d) > \frac{d}{2}$.
The ideal would be to know the $r(d)$ for a generic number of dimensions $n$, but any improvement on the $\frac{d}{2} \leq r(d) \leq d$ is welcome, and at worst I'd like to know what happens for $n \to \inf$, as I have an $n$ that is exponential in the size of my problem's input.