2
$\begingroup$

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.

$\endgroup$
0

1 Answer 1

3
$\begingroup$

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 $$\sum_n\frac{1}{\alpha_n^2}=\sum_n c_n^2\leq 1. $$

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.