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For a function $x:\omega\to\mathbb R$ let $|x|$ denote the function $|x|:\omega\to[0,\infty)$, $|x|:n\mapsto|x(n)|$.

It is well-know that a series $\sum_{n\in\omega}r_n$ of real numbers converges unconditionally if and only if $\sum_{n\in\omega}|r_n|<\infty$.

On the other hand, there exists an unconditionally convergent series $\sum_{n\in\omega}x_n$ in $\ell_2$ such that the series $\sum_{n\in\omega}|x_n|$ is divergent.

I am interested in finding conditions on an unconditionally convergent series $\sum_{n\in\omega}x_n$ in $\ell_2$ guaranteeing that the series $\sum_{n\in\omega}|x_n|$ converges.

Question. Let $I:\ell_1\to\ell_2$ be the identity operator and $(x_n)_{n\in\omega}$ be a sequence of elements of $\ell_1$ such that $\sum_{n\in\omega}\|x_n\|^2<\infty$ and the series $\sum_{n\in\omega}I(x_n)$ unconditionally converges in $\ell_2$. Is the series $\sum_{n\in\omega}|I(x_n)|$ convergent in $\ell_2$?

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    $\begingroup$ I think your notation is a little confusing, because at first the $x_n$ are the terms of a sequence $x$, but later the $x_n$ are sequences themselves. $\endgroup$
    – LSpice
    Commented Oct 29, 2021 at 17:16
  • $\begingroup$ @LSpice Thank you for your remark. I tried to fix this small problem. Now is it more understandable? $\endgroup$
    – Taras Banakh
    Commented Oct 29, 2021 at 18:18
  • $\begingroup$ Yup, thanks!... $\endgroup$
    – LSpice
    Commented Oct 29, 2021 at 20:25

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If I understand correctly what you are asking, the answer is "certainly not".

Consider $k$ orthogonal vectors $v_j$ with $k$ coordinates $\pm 1$ (the Hadamard matrix). Now multiply them by $t$ and take $n$ such identical bunches. Then the sum of $\ell^1$ norms squared is $nt^2k^3$, the squared $\ell^2$ norm of the sum of absolute value vectors is $n^2t^2k^3$ and the maximum of the squared $\ell^2$ norms of subset sums is $n^2t^2k^2$. Then playing this game on disjoint bunches of coordinates, we see that the question is whether the conditions $\sum_m n_mt_m^2k_m^3<+\infty$ and $\sum_m n_m^2t_m^2k_m^2<+\infty$ imply $\sum_m n_m^2t_m^2k_m^3<+\infty$. Now just take $n_m=k_m=2^m$ and $t_m=2^{-2m}/m$, say.

But perhaps you meant something else? :-)

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  • $\begingroup$ So putting everything together, the counter example has the form $x_j:=c_j v_j$ for $j\in\mathbb N$,with $\sum_{j=1}^\infty c_j^2<\infty$, and $\sum_{j=1}^\infty x_j$ is unconditionally convergent because it is the expansion of $c:=(c_1,c_2,\dots)\in\ell_2$ in the orthogonal basis $(v_j)_j$ (Moreover by construction $\sum_{j=1}^\infty \|x_j\|_1^2<\infty$ but $\sum_{j=1}^\infty |x_j|\notin \ell_2$). Correct? $\endgroup$
    – Pietro Majer
    Commented Oct 31, 2021 at 21:26
  • $\begingroup$ @PietroMajer Not sure if I understand what you wrote. Formally what you are saying is slightly different from what I'm saying but, perhaps, the difference is non-essential. My series is certainly not orthogonal, so the combination of "$x_j=c_jv_j$" and "orthogonal basis $v_j$" was definitely not there. The "moreover" part is correct though. $\endgroup$
    – fedja
    Commented Oct 31, 2021 at 21:44
  • $\begingroup$ @fedja Thank you very much for your answer, which is very helpful. $\endgroup$
    – Taras Banakh
    Commented Nov 1, 2021 at 5:13
  • $\begingroup$ @fedja In fact, this question was motivated by another one: mathoverflow.net/q/407517/61536 , which was my genuine motivation. Maybe you know what is the answer to that motivation question: can the unconditional convergence be destroyed by multiplications of all terms of the series by some vectors of norm $\le 1$? $\endgroup$
    – Taras Banakh
    Commented Nov 1, 2021 at 12:42
  • $\begingroup$ @TarasBanakh That one seems harder because in my construction that multiplication would mean multiplying each coordinate by $1/\sqrt k$, which will exactly erase the size gain from switching to the absolute value, but I'll think of it (if I have free time, which is problematic and any brains, which is even more problematic after the age of 50...) $\endgroup$
    – fedja
    Commented Nov 1, 2021 at 12:57

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