# Are the unconditionally convergent series, with terms in a Banach algebra, closed under the Cauchy product?

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?

• 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? Dec 6, 2019 at 1:51
• 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. Dec 6, 2019 at 2:02
• 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. Dec 6, 2019 at 2:21
• 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 $l^1$. Dec 6, 2019 at 2:22
• ah, forgot about that. it's probably false Dec 6, 2019 at 2:34