Infinite products in normed algebras

a) Let p be the product of the multipliable sequence (x_n), and suppose that p is a unit in A; then (Chapter IX, § 3, no. 7, Proposition 13) there exist $\alpha > 0$ and $a > 0$ such that, for all $y \in A$ for which we have

y is a unit and $\|y^{-1}\| \leq a$. By hypothesis, for every $\varepsilon$ such that $0 < \varepsilon < \alpha$, there is a finite subset H_0 of \mathbf{N} such that, for every finite subset H of \mathbf{N} containing H_0, we have $\|p_H - p\| \leq \varepsilon$. Let $J_0 = [0, m]$ be an interval of \mathbf{N} which contains H_0; for each finite subset L of \mathbf{N} which does not meet J_0, the integers belonging to L are all greater than those belonging to H_0; hence, if $H = H_0 \cup L$, we have $p_H = p_{H_0} p_L$. Now, since $\|p_{H_0} - p\| \leq \varepsilon \leq \alpha$, $p_{H_0}$ is a unit, and since $\|p_{H_0} p_L - p\| \leq \varepsilon$, we deduce and finally $\|e - p_L\| \leq 2a \varepsilon$.

b) Suppose that, for each $\varepsilon > 0$, there exists a finite subset J_0 of \mathbf{N} such that, for every finite subset L of \mathbf{N} which does not meet J_0, we have $\|e - p_L\| \leq \varepsilon$. Let $H_0 = [0, p]$ be an interval of \mathbf{N} which contains J_0; then every finite subset H of \mathbf{N} which contains H_0 can be written in the form $H_0 \cup L$, where the integers in L are all greater than those in H_0; hence we have $p_H = p_{H_0} p_L$, and since L does not meet $J_0$, $\|p_H - p_{H_0}\| \leq \varepsilon \|p_{H_0}\|$, and consequently $\|p_H\| = (1 + \varepsilon)\|p_{H_0}\|$. If $p_{H_0} = 0$, the sequence $(x_n)$ is evidently multipliable and its product is 0; excluding this trivial case, there is an interval $H_1 = [0, q]$ containing $H_0$ and such that, for every finite subset $L$ of $\mathbf{N}$ which does not meet $H_1$, we have $\|e - p_L\| \leq \varepsilon (\|p_{H_0}\|)^{-1}$. As above, it follows that, for each finite subset $H \supset H_1$,

Cauchy’s criterion therefore shows that $J \to p_J$ has a limit in $A$ with respect to the directed set $\mathfrak{F}(\mathbf{N})$.

If all the $x_n$ are units, then so are all the finite partial products $p_J$; hence for each finite subset $H$ containing $H_0$ we have

and this shows that, in the multiplicative group $G$ of units of $A$, the image under the mapping $J \to p_J$ of the section filter of $\mathfrak{F}(\mathbf{N})$ is a Cauchy filter base with respect to the left uniformity of $G$; but since $G$ is complete (Chapter IX, § 3, no. 7, Proposition 13), the limit of the mapping $J \to p_J$ belongs to $G$.

This follows immediately from the criterion of Theorem 1.

Let us apply the criterion of Theorem 1. For every finite subset $L$ of $\mathbf{N}$, we have $p_L - e = \prod_{n \in L} (e + u_n) - e = \sum_M \left( \prod_{n \in M} u_n \right)$, where $M$ runs through the set of all non-empty subsets of $L$ (linearly ordered by the induced ordering). Since $\prod_{n \in M} u_n \leq \prod_{n \in M} |u_n|$, we may write

Now since the series whose general term is $|u_n|$ is convergent by hypothesis, the sequence $1 - |u_n|$ is multipliable in $\mathbf{R}_+^*$ (Chapter IV, § 7, no. 4, Theorem 4). Hence for each $\varepsilon > 0$ there exists a finite subset $J_0$ of $\mathbf{N}$ such that, for every finite subset $L$ of $\mathbf{N}$ which does not meet $J_0$, we have $\left| \prod_{n \in L} (1 + ||u_n||) - 1 \right| \leq \varepsilon$; hence the result.

There is only a finite number of integers $n$ such that $|u_n| > 1$. Let $J = [a, m]$ be an interval of $\mathbf{N}$ containing all these integers. The product of the sequence $(e + u_n)$ is the product of $p_J$ and the element $\prod_{n \notin J} (e - u_n)$, all of whose factors are units (Chapter IX, § 3, no. 7, Corollary to Proposition 12), and is therefore itself a unit; since $p_J$ is the product of a finite number of non-zero divisors, it is not a zero divisor, and hence $\prod_{n \in \mathbf{N}} (e + u_n)$ is not a zero divisor.

The sufficient condition for multipliability given by Theorem 2 is not in general necessary (cf. Exercise 6). However, it is necessary in the important case where $A$ is an algebra of finite rank over the field $\mathbf{R}$ (i.e., $A$ is finite-dimensional as a vector space over $\mathbf{R}$); in particular this is the case if $A$ is the division ring of quaternions $\mathbf{H}$, or a matrix algebra $\mathbf{M}_n(\mathbf{R})$:

From Chapter VII, § 3, no. 1, Proposition 2, there exists a number $c > 0$ such that, for every finite family $(x_i)_{i \in I}$ of points of $A$, we have

Let $(a_i)_{i \in \mathbf{N}}$ be an arbitrary sequence of elements of $A$. For each finite subset $I$ of $\mathbf{N}$, put

The lemma is obvious if $J$ is empty; we shall prove it by induction on the number of elements in $J$. Let $J = K \cup \{j\}$, where $\{j\}$ is strictly larger than every $i \in K$. Then $p_J = p_K(e + a_j)$ and

so that

and by the inductive hypothesis and the definition of $\varphi(I)$, we have

which proves the lemma.

For since $\sigma_J \leq \sigma_I$ for every subset $J$ of $I$, we have from (2)

and since also $\sigma_I \leq c \cdot \sup_{J \subset I} ||s_J||$, from (1), it follows that

which leads to the result.

Now let $(e + u_n)$ be a multipliable sequence in $A$, whose product is a unit; by Theorem 1 there exists a finite subset $J_0$ of $\mathbf{N}$ such that, for each finite subset $H$ of $\mathbf{N}$ which does not meet $J_0$, we have

By Lemma 2, it follows that $\sum_{i \in H} ||u_i|| - 1$ for every finite subset $H$ of $\mathbf{N}$ which does not meet $J_0$, and hence (Chapter IV, § 7, no. 1, Theorem 1) the family $(||u_n||)$ is summable in $\mathbf{R}$.

a) If the infinite product whose general factor is $x_n$ is convergent and if $\prod_{n=0}^\infty x_n$ is a unit in $A$, then for each $\varepsilon > 0$ there exists an integer $n_0$ such that

whenever $n_0 \leq m \leq n$.

b) Conversely, if the sequence $(x_n)$ satisfies this condition, the infinite product with general factor $x_n$ is convergent; and if each of the $x_n$ is a unit in $A$, then $\prod_{n=0}^\infty x_n$ is a unit.

The proof of this proposition follows step by step the proof of Theorem 1, and is left to the reader (the finite subsets $L$ of $N$ in the proof of Theorem 1 are to be replaced by intervals).

The product of the sequence $(y_n)$ is denoted by $\prod_{n=h}^\infty x_n$, and is also called the residue of index $h$ of the infinite product with general factor $x_n$.

Still under the assumption that $\prod_{n=h}^\infty x_n$ is a unit, it follows from Proposition 2 that if $(z_n)$ is a sequence such that $z_n = x_n$ for all but a finite number of indices, then the product with general factor $z_n$ is convergent.

then the infinite product whose general factor is $u_n$ is convergent and we have

For the sequence of partial products of the sequence $(u_n)$ is a subsequence of the sequence of partial products of the sequence $(x_n)$.

Finally, the same argument as was used for abelian groups (Chapter III, § 5, no. 7) shows that if a sequence $(x_n)$ in a normed algebra $\mathbf{A}$ is multiplicable, then the product whose general factor is $x_n$ is convergent, and

(which is also written as $\prod_{n=0}^{\infty} x_n$); the converse is of course not true (cf. Exercise 7).