Maximum number of 1-separated points in the unit ball of an $n$-dimensional Banach space.

Question

In an infinite dimensional Banach space $X$, we can find a sequence of unit vectors $x_n$ such that $||x_n-x_m||\geq 1, n \neq m$.

For a finite dimensional Banach space, the number of such 1-separated points has to be finite, since the unit ball is compact.

Let $M_1(X)$ be the maximum number of 1-separated points in the set $\{x\in X: ||x||\leq 1 \}$. How is $M_1(X)$ related to $\dim(X)$?

If $X$ is $n$-dimensional, with a suitable choice of basis, $\vert \vert.\vert \vert_\infty \leq \vert \vert .\vert \vert_X \leq \vert \vert .\vert \vert_1 $. This gives $M_1(\ell_\infty^n ) \leq M_1(X) \leq M_1(\ell_1^n) $

So I have multiple questions:

1. What is the explicit value of $M_1(\ell_1^n)$ and $M_1(\ell_\infty^n)$?
2. Is every value between $M_1(\ell_\infty^n)$ and $M_1(\ell_1^n)$ attained?
3. How is $M_1(\ell_p^n)$ related to $p$ and $n$?

Edit: as pointed out by gerw in the comments, it is not true that $M_1(\ell_\infty^n) \leq M_1(X) $.

So my modified question is: can we estimate $M_1(\ell_p^n)$?

Answer 1

In addition to my comments above, I have some rather trivial lower bounds for $M_1(\ell^n_\infty)$ and $M_1(\ell^n_1)$:

For the case $p = \infty$ it is not hard to see, that the set of points ${-1,0,1}^n$ is $1$-separated. Hence, $M_1(\ell^n_\infty) \ge 3^n$.

For the case $p = 1$ it is not hard to see, that the set of the following points:

• The origin $0$,
• All points $(0,\ldots, 0, \pm 1, 0, \ldots, 0)$, count: $2 \, n$
• All points $(0, \ldots, 0, \pm\frac12,0, \ldots, 0, \pm\frac12,0,\ldots,0)$, count: $4 \, n \, (n-1)/2 = 2 \, n \, (n-1)$. Hence, $M_1(\ell^n_1) \ge 2 \, n^2 + 1$.

From some simple sketching in dimensions $1,2,3$ it seems that these bounds are sharp, but things may change in higher dimensions ;)