We have a Banach algebra $\mathbb L$, and two sequences $(A_0,A_1,A_2,\cdots),\;(B_0,B_1,B_2,\cdots)\in\mathbb L^{\mathbb N}$, for which the sums $\sum_{n\in\mathbb N}A_n,\;\sum_{n\in\mathbb N}B_n$ are unconditionally convergent.


$$\sum_{n\in\mathbb N}\left(\sum_{l+m=n}A_lB_m\right)$$

also unconditionally convergent?

If it helps, you may assume commutativity ($A_lB_m=B_mA_l$), or that they're power series with scalar coefficients ($A_n=a_nX^n,\;B_n=b_nX^n,\;X\in\mathbb L$).

The case with absolute convergence is easy. (There, we just need to replace $|a_nb_k|=|a_n||b_k|$ with $|a_nb_k|\leq|a_n||b_k|$.)

Possibly related: Is the sequence space $\ell^p$ closed under the Cauchy product?

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    $\begingroup$ Hi. I only now got the time to think about this. Can you give an example of an unconditionally convergent series that is not absolutely convergent? $\endgroup$ Dec 6, 2019 at 1:51
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    $\begingroup$ It requires infinite dimensions. In the sequence space $\ell^2$ with basis $(e_n\mid n\in\mathbb N)$, $$\sum_n\frac1ne_n$$ converges unconditionally, with the norms of remainders being bounded by $\sum_n\tfrac1{n^2}$ and approaching $0$, but doesn't converge absolutely, because $\sum_n\lVert\tfrac1ne_n\rVert=\sum_n\tfrac1n$ diverges. $\endgroup$
    – mr_e_man
    Dec 6, 2019 at 2:02
  • $\begingroup$ I think the Hilbert-Schmidt operators form a Banach algebra and a Hilbert space, and the Banach space structure (no inner product or multiplication) is isomorphic to $\ell^2$. So the above applies. $\endgroup$
    – mr_e_man
    Dec 6, 2019 at 2:21
  • $\begingroup$ I actually think $l^2$ is a Banach algebra, right? if $(a_n)_n,(b_n)_n \in l^2$, then $(a_nb_n)_n$ is in $l^2$ since it's in $l^1$. $\endgroup$ Dec 6, 2019 at 2:22
  • $\begingroup$ ah, forgot about that. it's probably false $\endgroup$ Dec 6, 2019 at 2:34


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