TITLE: A perturbation of an unconditionally convergent series in $\ell_2$
QUESTION [5 upvotes]: For two functions $x,y:\omega\to\mathbb R$ let $xy:\omega\to\mathbb R$, $xy:n\mapsto x(n)y(n)$, be their pointwise product.
It is clear that for any elements $x,y\in\ell_2$ their pointwise product $xy$ is an element of $\ell_2$.

Problem. Let $\sum_i x_i$ be an unconditionally convergent series in $\ell_2$. Is it true that for any sequence $\{y_i\}_{i\in\omega}\subseteq\{y\in\ell_2:\|y\|\le 1\}$, the series $\sum_i x_iy_i$ converges in $\ell_2$?

REPLY [3 votes]: Volodymyr Kadets kindly informed me that the answer to this problem is affirmative. His argument easily generalizes to prove the following

Theorem. For any $p\in[1,\infty)$, any unconditionally convergent series $\sum_{n\in\omega}x_n$ in the Banach space $\ell_p$ and any sequence $\{y_n\}_{n\in\omega}\subseteq\{y\in\ell_2:\|y\|\le 1\}$, the series $\sum_{n\in\omega}x_ny_n$ converges unconditionally in $\ell_p$.

Proof. Let $q\in(1,\infty]$ be such that $\frac1q+\frac1p=1$. By Bessaga-Pelczynski Theorem (6.4.3 in this book), to show that the series $\sum_{n\in\omega}x_ny_n$ is unconditionally convergent in $\ell_p$, it suffices to check that it is weakly absolutely convergent, i.e., for every $v\in\ell_q=\ell_p^*$ we have $\sum_{n\in\omega}|\langle v,x_ny_n\rangle|<\infty$.
Holder's inequality ensures that for every $x\in\ell_p$ the product $vx$ belongs to $\ell_1$ and the operator $T_v:\ell_p\to\ell_1$, $T_v:x\mapsto vx$, is continuous. Then the series $\sum_{n\in\omega}vx_n$ is unconditionally convergent in $\ell_1$.
The absolute summability of the identity inclusion $\ell_1\to\ell_2$ (which follows from Grothendieck's Theorem 4.3.2 in this book) implies that $\sum_{n\in\omega}\|vx_n\|_{\ell_2}<\infty$.
Now observe that Cauchy-Bunyakovsky-Schwartz inequality ensures that
$$\sum_{n\in\omega}|\langle v,x_ny_n\rangle|=\sum_{n\in\omega}|\langle vx_n,y_n\rangle|\le\sum_{n\in\omega}\|vx_n\|_{\ell_2}\cdot\|y_n\|_{\ell_2}\le\sum_{n\in\omega}\|vx_n\|_{\ell_2}<\infty.$$