2
$\begingroup$

Let $H$ be a separable, infinite dimensional Hilbert space. For an ONB $(e_n)_{n \in \mathbb{N}}$ of $H$ together with a series $(\alpha_n)_{n \in \mathbb{N}} \subset (0,\infty)$ such that $\sum\limits_{n=1}^\infty \alpha_n^2 < \infty$ and an $r>0$ we call the set \begin{align*} & E_{\alpha,e,r} := \left\{ x = \sum\limits_{n=1}^\infty \langle x,e_n \rangle_H \, e_n \ \middle| \ \sum\limits_{n=1}^{\infty} \frac{\langle x,e_n \rangle_H^2}{\alpha_n^2} \leq r \right\} \end{align*} a Schmidt-ellipsoid.

My question is if we can pack any Schmidt ellipsoid into another Schmidt-ellipsoid, which was constructed with regards to another ONB of $H$. In other words let $(e_n)_{n \in \mathbb{N}}$ and $(b_n)_{n \in \mathbb{N}}$ be two ONBs of $H$ and let $E_{\alpha,e,r}$ be as defined above. Does there exists a series $(\beta_n)_{n \in \mathbb{N}} \subset (0,\infty)$ with $\sum\limits_{n=1}^\infty \beta_n^2 < \infty$ and a $R>0$ such that $E_{\alpha,e,r} \subset E_{\beta,b,R}$?

It might be helpful to intepret the Ellipsoids $E_{\alpha,e,r}$ as Images $T(B_r)$ of certain operators $T: H\rightarrow H$, with $T(e_n) = \alpha_n \, e_n$, where $B_r := \{x \in H \mid ||x||_H^2 \leq r\}$.

$\endgroup$
2
  • $\begingroup$ In case someone else is interested, I just got an answer from an acquaintance. It turns out that you can't pack Schmidt ellipsoids to different ONBs into each other. As a counter example let $H=\ell^2(\mathbb{N})$ and let $(b_n)_{n \in \mathbb{N}}$ be the canonical ONB $b_n = (\delta_{k=n})_{k \in \mathbb{N}}$. Now any Schmidt ellipsoid $E_{\beta,b,R}$ will be a subset of $\ell^1(\mathbb{N})$, which means there are elements in $H=\ell^2(\mathbb{N})$, which can never be in such a Schmidt ellipsoid $E_{\beta,b,R}$. ...(see next comment) $\endgroup$ Commented Nov 8, 2021 at 16:28
  • $\begingroup$ For one such element $x \in H$ one can now easily construct an ONB $(e_n)_{n \in \mathbb{N}}$ such that $e_1 = \frac{x}{||x||_H}$. Now is is a simple matter to choose $\alpha$ and $r>0$ such that $x \in E_{\alpha,e,r}$. We have thus found an example, where $E_{\alpha,e,r}$ can never be a subset of $E_{\beta,b,R}$ regardless of the choice of $\beta$ and $R>0$. $\endgroup$ Commented Nov 8, 2021 at 16:31

0

You must log in to answer this question.

Browse other questions tagged .