Let $H$ be a separable Hilbert space of infinite dimension and let $(e\_n)\_{n \in \mathbb{N}}$ be an orthonormal basis of $H$. For a series $(\alpha\_n)\_{n \in \mathbb{N}} \subset \mathbb{R^+}$ we are interested in whether or not the unit ball $$B\_1 := \{h \in H \mid 1 \geq \|h\|\_H\}$$ is a subset of the closed symmetric convex hull $\overline{\operatorname{sco}}(\alpha\_n \, e\_n \mid n \in \mathbb{N})$ of the series $(\alpha\_n \, e\_n)\_{n \in \mathbb{N}} \subset H$. It is clear, that such a series exists, since we can iteratively construct one, such that for every $n \in \mathbb{N}$ the $n$-dimensional sphere $B\_{1+\frac{1}{n}} \cap \langle e\_1,...,e\_n \rangle$ is in the convex hull $\overline{\operatorname{sco}}(\alpha\_1 \, e\_1,...,\alpha\_{n} \, e\_n)$.


My question is if there are any known bounds on the rate at which such a series $(\alpha\_n)\_{n \in \mathbb{N}}$ must tend to infinity, i.e. if one can construct the series in a way to have it grow relatively slowly.


I would also be interested in any useful literature and maybe a lower bound on the rate, other than $\mathcal{O}(1)$.


Any help is be much appreciated.



Geometry in Hilbert spaces / spheres in high dimensions
Set $c\_n:=\frac{1}{\alpha\_n}$. Clearly, for any $N>0$ one must have $\DeclareMathOperator{\spa}{span}$ $\DeclareMathOperator{\conv}{conv}$ $\newcommand{\bsB}{\boldsymbol{B}}$
$$
\bsB\_1\cap\spa\{e\_1,\dotsc, e\_N\} \subset\conv\{ \pm\alpha\_1 e\_1,\dotsc,\pm\alpha\_N e\_n\}.
$$
This happens if and only if
$$
f\_N(\alpha):=\min\{ x\_1^2+\cdots +x\_N^2;\;\;c\_1x\_1+\cdots +c\_Nx\_N=1\}\geq 1.
$$
(The hyperplane $c\_1x\_1+\cdots +c\_Nx\_N=1$ cuts the $k$-th axis at the point $\frac{1}{c\_k}=\alpha\_k$.) Using Lagrange multipliers we get
$$
f\_N(\alpha)=\frac{1}{c\_1^2+\cdots +c\_N^2}\geq 1.
$$
Thus we need
$$\sum\_n\frac{1}{\alpha\_n^2}=\sum\_n c\_n^2\leq 1. $$


