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While writing arXiv:1510.05757v2, I found myself proving some basic facts about products of Banach algebra elements over an infinite totally ordered set. (Statements and proofs are in Appendix C.) The product over a totally ordered set $A$ is defined as you'd expect.

Definition. The product $\prod_{p \in A} x_p$ is the limit of the net that gives the product over each finite subset of $A$.

I prove things like the following.

Equivariance property. If $\prod_{p \in A} x_p$ converges, then $\phi\left(\prod_{p \in A} x_p\right) = \prod_{p \in A} \phi(x_p)$ for any continuous homomorphism $\phi$ from the algebra to itself.

Convergence condition. If the sum $C = \sum_{p \in A} \|x_p - 1\|$ converges, the product $\prod_{p \in A} x_p$ converges as well.

Estimates. When the sum $C$ converges, the norm of $\prod_{p \in A} x_p$ and its distance from $1$ are bounded in terms of $C$. (See the preprint for explicit bounds.)

For products over $\mathbb{N}$, these facts are well-established. The convergence condition, for example, is proven in Welstead's "Infinite products in a Banach algebra". My arguments for products over $A$ are small (though sometimes tricky) modifications of the arguments for products over $\mathbb{N}$.

Can I find these sorts of results in the literature instead of using my own proofs?

I don't need the results to be stated for general Banach algebras; I only use them for matrix algebras in the preprint. I'd be happy even to find a text that defines infinite ordered products like this, or to learn of a standard name for them.

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    $\begingroup$ Several years ago I briefly tried to look this up for a MathReview that I was writing and I don't remember seeing anything other than Welstead's paper and some articles it referred to. My suspicion is that such results have been reinvented independently as and when needed, just like you have found yourself doing. $\endgroup$
    – Yemon Choi
    Commented Mar 17, 2021 at 4:31
  • $\begingroup$ Random question: why do you need $A$ to be totally ordered? Your product definition makes perfect sense over any set $A$. $\endgroup$ Commented Mar 17, 2021 at 16:21
  • $\begingroup$ Finally, I am not 100% sure about the Banach algebra case, but isn't it the case that your infinite product is only well defined if $x_p \neq 1$ for at most countably many $p$s? At least, if you wish to impose the convergence condition throughout, it is clear that $C$ converges only when at most countably many $p$ has $x_p \neq 1$. And since the order structure is unimportant, for the definition, you are automatically back in the case over $\mathbb{N}$ and not an arbitrary set $A$. $\endgroup$ Commented Mar 17, 2021 at 16:25
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    $\begingroup$ @WillieWong—When the Banach algebra is non-commutative, we can't just multiply a finite set of elements; we have to multiply them in some (total) order. My definition of the product over $A$ only makes sense if the product over each finite subset of $A$ makes sense, so each finite subset of $A$ has to be totally ordered. $\endgroup$
    – Vectornaut
    Commented Mar 19, 2021 at 12:56
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    $\begingroup$ @Vectornaut My recollection was that it wasn't a good paper: mathscinet.ams.org/mathscinet-getitem?mr=2969534 $\endgroup$
    – Yemon Choi
    Commented Mar 19, 2021 at 15:00

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