Family of sequences in a Hilbert space with certain property

Question

Suppose $\mathcal{F}$ is a family of sequences on the unit sphere of $l_2$ with the following property: For any sequence $\varepsilon_n\downarrow 0$ but which is not eventually identically $0$, there exists $(x_n)\in\mathcal{F}$ and $z\in S_{l_2}$ such that, for any $n$, $\langle z, x_n\rangle\leq \varepsilon_n$. Can we find a sequence $(y_n)\in\mathcal{F}$ and $z\in S_{l_2}$ such that for any $n$ $\langle z, y_n\rangle=0$?

Answer 1

In general, it does not work.

Construct $\mathcal F=\{z_n, \ n\in\mathbb N\}$ as the set of sequences defined in the following way: $z_n=(z_{n,k})$ will be the sequence of elements in $l^2$ with $$ z_{n,k} = \begin{cases}\frac1n e_{1} + \frac{\sqrt{n^2-1}}ne_{2} & \text{ if } k=1\\ \frac1n e_{2} + \frac{\sqrt{n^2-1}}ne_{k+1} & \text{ if } k>1.\\ \end{cases} $$

If $\epsilon_m\to0$ is given, take $n>\epsilon_1^{-1}$, $x=z_n$, $z=e_1$. Then $$ \langle x_k,z\rangle = \langle z_{n,k},e_1\rangle = \begin{cases}\frac1n \le \epsilon_1& \text{ if } k=1\\ 0\le \epsilon_k& \text{ otherwise. } \end{cases} $$

Now let $z\in l^2$ with $\|z\|=1$. Assume $\langle z_{n,k},z\rangle=0$ for all $k$. Let $k\ge 2$. Then $$ 0 = \langle z_{n,k} - z_{n,k+1} ,z\rangle = \frac{\sqrt{n^2-1}}n (z_{k+1}-z_{k+2}). $$ Hence $z_{k+1}=z_{k+2}$ for $k\ge 2$. Since $z\in l^2$ this implies $z_k=0$ for all $k\ge 3$. From $\langle z_{n,2},z\rangle=0$ it follows $z_2=0$, and $\langle z_{n,1},z\rangle=0$ implies $z_1=0$. Hence $z=0$ a contradiction.