Asymptotics of orthonormal basis in $\ell^2$

Question

Is it true that for all orthognal basis $\{e_n\}_{n=1}^\infty$ of $\ell^2(\mathbb N)$, we have $$\sup_{n}n^2 \sum_{k=1}^\infty (e_n(k))^2 k^{-2}<\infty? $$

where $e_n=(e_n(1),e_n(2),\cdots,e_n(k),\cdots)\in \ell^2(\mathbb N)$ and $$ \langle e_n,e_m \rangle_{\ell^2}=\sum_{k=1}^\infty e_n(k)e_m(k)=\delta(m,n).$$

For the special case: the standard orthognal basis $\bar {e_n}(k)=\delta(k,n),$ it's clearly true $$ n^2 \sum_{k=1}^\infty (\bar {e_n}(k))^2 k^{-2}=1,\quad \text{for all $n$.}$$ For general orthognal basis, I think it's not true, but I can't find a counter example.

Update: At first I thought this argument is far too strong, construct some other ONB and reorder them would make a counterexample, but I'm not so sure about it now. Note this truth, define $T\in \mathscr L(\ell (\mathbb N))$: $$T(\bar{e_n})=n^{-1}\bar {e_n},$$ it's straightforward that $T$ is a Hilbert–Schmidt operator: $$\|T\|^2_{HS}=\sum_n \|T({e_n})\|^2=\sum_n \|T(\bar{e_n})\|^2=\sum_n n^{-2}<\infty. $$ Note that the asymptotic above is $$\sum_{k=1}^\infty (e_n(k))^2 k^{-2}=\|T({e_n})\|^2,$$ and the question turns to $$\|T({e_n})\|=O(n^{-1})?$$ That is, is the "tail" of the Hilbert–Schmidt norm of $T$ decrease in the same speed no matter what ONB we choose?

Answer 1

It is false in general. Start by constructing a bijective $\varphi$ from $\mathbb N$ to $\mathbb N$ such that $\varphi\left(n^4\right) = n^2$. This should not be difficult since $\mathbb N\backslash \left\{n^4: n\in \mathbb N\right\}$ and $\mathbb N\backslash \left\{n^2: n\in \mathbb N\right\}$ are both infinite countable sets. Now let $e_n = \overline e_{\varphi(n)}$. For every $n$,

$$n^8 \sum_{k=1}^{\infty} e_{n^4}(k)k^{-2} = n^8 \sum_{k=1}^{\infty} \overline e_{\varphi\left(n^4\right)}(k)k^{-2} = \sum_{k=1}^{\infty} \overline e_{n^2}(k)k^{-2} = n^{8} \left(n^{2}\right)^{-2} = n^4 \to \infty$$

Answer 2

I guess you can do the following, let $$v=\frac{\sqrt6}{\pi}(1,\frac{1}{2},\frac{1}{3},\ldots)$$ then $v$ is a unit vector.

Now take an ONB that contains $v$, such system exists since we can e.g., replace $e_0$ with $v$ in the standard ONB, and apply a Gram-Schmidt procedure.

Next we rearrange this ONB, so that after $n$-th re-ordering $v$ appears as the $n$-th basis vector, then the sup in OP is at least $$n^2\sum_{k=1}^{\infty}v_k^2k^{-2}=n^2\sum_{k=1}^{\infty}k^{-4}=\frac{n^2\pi^4}{90}\to\infty$$