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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.

Is

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

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?

mr_e_man
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    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? – mathworker21 Dec 06 '19 at 01:51
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    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. – mr_e_man Dec 06 '19 at 02:02
  • 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. – mr_e_man Dec 06 '19 at 02:21
  • 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 $l^1$. – mathworker21 Dec 06 '19 at 02:22
  • ah, forgot about that. it's probably false – mathworker21 Dec 06 '19 at 02:34
  • Any $\ell^p$ space ($p\geq1$) with pointwise multiplication is a Banach algebra, because $$\lVert ab\rVert^p=\sum_n|a_nb_n|^p=\sum_n|a_n|^p|b_n|^p$$ $$\leq\sum_{n,m}|a_n|^p|b_m|^p=\left(\sum_n|a_n|^p\right)\left(\sum_m|b_m|^p\right)=\lVert a\rVert^p\lVert b\rVert^p$$ – mr_e_man Dec 06 '19 at 06:16
  • Here's an idea. Let's work in $l^2$. Take two "random" orthonormal bases, $(w_n)_n$ and $(u_n)_n$. The point is that, unlike the standard orthonormal basis $(e_n)_n$, these bases have orthogonality due to negativity, in that the inner product $\langle w_1, w_2 \rangle$ is a sum of countably many nonzero reals that turns out to be $0$ (for $\langle e_1, e_2 \rangle$, we just have $1\cdot0+0\cdot1+\sum 0\cdot 0 = 0$). Let $A_n = \frac{1}{n}w_n$ and $B_n = \frac{1}{n}u_n$. As you mentioned, both $\sum A_n, \sum B_n$ converge unconditionally due to orthogonality. – mathworker21 Dec 06 '19 at 12:59
  • Now, if $||w_lu_m||$ is bounded away from $0$, then each $A_lB_m$ has norm around $\frac{1}{lm}$ and $\sum_{l+m = n} A_lB_m$ will be the sum of $n$ things each of norm at least $\frac{1}{n^2}$. However, we won't have orthogonality between the terms, so the norm should be at least $\frac{1}{n}$, nor will we have orthogonality between that sum $\sum_{l+m = n} A_lB_m$ and the sum $\sum_{l'+m'=n'} A_{l'}B_{m'}$. The point is that $\sum_n (\sum_{l+m=n} A_lB_m)$ is the sum of $\frac{1}{n}$-norm things that are not orthogonal to each other, and thus should not converge unconditionally. – mathworker21 Dec 06 '19 at 13:03
  • The three issues with my above comments are (1) idk what I mean by "random" orthonormal basis; (2) I'm not sure whether it's possible to have $\inf_{l,m} ||w_l u_m|| > 0$; (3) The norm of $\sum_{l+m=n} A_lB_m$ might be a bit less than $\frac{1}{n}$ due to 'partial orthogonality' but we could just take $A_n$ to be $\frac{1}{\sqrt{n}\log n}w_n$ at the start instead, or something. – mathworker21 Dec 06 '19 at 13:09
  • I think it might be possible to fix issues (1)-(3) by theoretical arguments. However, you could just try to do the construction I suggested with some explicit $(w_n)_n,(u_n)_n$ (possible $(u_n)_n = (w_n)_n$). The issue is just that I don't know any explicit orthonormal bases of $l^2$ besides the standard one, but maybe you do. – mathworker21 Dec 06 '19 at 13:11
  • No, I haven't found a way to fix those three issues. For an orthonormal basis, we could take an infinite direct sum of finite-dimensional rotations, but that probably wouldn't have the properties you want. – mr_e_man Dec 10 '19 at 01:21
  • @Integrand - I don't understand the reason for your edit. The commas don't make much difference either way. But the original title was intended to mean "Is the set of all unconditionally convergent series [...] closed [...]?" Without the "the", the question doesn't sound right to me. For comparison, consider: "Are the natural numbers closed under subtraction?" versus "Are natural numbers closed under subtraction?" Do you really think the latter is more clear? Some natural numbers are closed under subtraction: $0-0=0$. – mr_e_man Oct 25 '20 at 22:00
  • @mr_e_man apologies if I have substantially altered the content inadvertently. Feel free to re-edit the post or rollback to a previous version. – Integrand Oct 25 '20 at 22:50
  • @mr_e_man I have a similar question and was wondering if you have progress on your question. – raegakj Dec 23 '23 at 05:27
  • @raegakj - No, sorry. – mr_e_man Dec 27 '23 at 16:49

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