An operator satisfying in a sequence of equations

Question

Assume that $H$ is a non-separable Hilbert space. Let $\{\eta_n\}$ be an arbitrary sequence in $H$.

Let $\{\zeta_n\}$ be a sequence in $H$ which forms a linearly independent set. Does there exist any bounded operator $x$ in $B(H)$ satisfying in the equations $x\zeta_n=\eta_n$ ($n=1,2,\cdots$)?

Answer 1

Here is an example where $\|\eta_n\|=\|\zeta_n\|=1$ for all $n$, but $T$ is unbounded.

Let $\zeta_n$ be any orthonormal sequence, and define $$ \eta_n=\frac{\zeta_1+\zeta_{n+1}}{\sqrt2}. $$ Then $\|\eta_n\|=\|\zeta_n\|=1$ for all $n$.

Now consider $$ \xi_n=\frac1{\sqrt n}\sum_{k=1}^n \zeta_k. $$ Then $\|\xi_n\|=1$. If $T\zeta_n=\eta_n$ for all $n$ and $T$ is linear, then \begin{align} \|T\xi_n\|^2&=\left\|\frac1{\sqrt n}\sum_{k=1}^n T\zeta_k \right\|^2 =\left\|\frac1{\sqrt{2n}}\sum_{k=1}^n{\zeta_1+\zeta_{k+1}} \right\|^2\\ \ \\ &=\left\|\frac1{\sqrt{2n}}\left({n\zeta_1+\sum_{k=1}^n\zeta_{k+1}} \right)\right\|^2\\ \ \\ &=\frac{n^2+n}{{2n}}\geq\frac{n}2 \end{align}