Infinite products in normed algebras

a) If $(\mathbf{x}_n)$ is multipliable and has for product an invertible element of $A$, for every $\varepsilon > 0$, there exists a finite subset $J_0$ of $\mathbf{N}$ such that, for every finite subset $L$ of $\mathbf{N}$ not meeting $J_0$, one has $\| \mathbf{e} - \mathbf{p}_L \| \leq \varepsilon$.

b) Conversely, if the sequence $(\mathbf{x}_n)$ satisfies this condition, it is multipliable. Moreover, if each of the $\mathbf{x}_n$ is invertible, $\prod_{n \in \mathbf{N}} \mathbf{x}_n$ is invertible.

a) Let $\mathbf{p}$ be the product of the multipliable sequence $(\mathbf{x}_n)$, and suppose that $\mathbf{p}$ is invertible in $A$; then (IX, p. 40, prop. 14), there exist $\alpha > 0$ and $a > 0$ such that, for every $y \in A$ such that $\| y - \mathbf{p} \| \leq \alpha$, $y$ is invertible, and $\| y^{-1} \| \leq a$. By assumption, for every $\varepsilon$ such that $0 < \varepsilon < \alpha$, there exists a finite subset $H_0$ of $\mathbf{N}$ such that, for every finite subset $H$ of $\mathbf{N}$ containing $H_0$, one has $\| \mathbf{p}_H - \mathbf{p} \| \leq \varepsilon$. Let $J_0 = \{0, m\}$ be an interval of $\mathbf{N}$ containing $H_0$; for every finite subset $L$ of $\mathbf{N}$ not meeting $J_0$ the integers belonging to $L$ are all greater than those belonging to $H_0$; therefore, if one sets $H = H_0 \cup L$, one has $\mathbf{p}_H = \mathbf{p}_{H_0} \mathbf{p}_L$. Now, since $\| \mathbf{p}_{H_0} - \mathbf{p} \| \leq \varepsilon \leq \alpha$, $\mathbf{p}_{H_0}$ is invertible, and one has $\| \mathbf{e} - \mathbf{p}_{H_0}^{-1} \mathbf{p} \| \leq \varepsilon \| \mathbf{p}_{H_0}^{-1} \| \leq a \varepsilon$; from the relation $\| \mathbf{p}_{H_0} \mathbf{p}_L - \mathbf{p} \| \leq \varepsilon$, one deduces $\| \mathbf{p}_L - \mathbf{p}_{H_0}^{-1} \mathbf{p} \| \leq \varepsilon \| \mathbf{p}_{H_0}^{-1} \| \leq a \varepsilon$ and finally $\| \mathbf{e} - \mathbf{p}_L \| \leq 2a \varepsilon$.

b) Suppose that, for every $\varepsilon > 0$, there exists a finite subset $J_0$ of $\mathbf{N}$ such that, for every finite subset $L$ of $\mathbf{N}$ not meeting $J_0$, one has $\| \mathbf{e} - \mathbf{p}_L \| \leq \varepsilon$. Let $H_0 = \{0, p\}$ be an interval of $\mathbf{N}$ containing $J_0$; every finite subset $H$ of $\mathbf{N}$ containing $H_0$ can be written $H_0 \cup L$, where the integers belonging to $L$ are greater than those belonging to $H_0$; we therefore have $\mathbf{p}_H = \mathbf{p}_{H_0} \mathbf{p}_L$, and since $L$ does not meet $J_0$, $\| \mathbf{p}_H - \mathbf{p}_{H_0} \| \leq \varepsilon \| \mathbf{p}_{H_0} \|$, and consequently $\| \mathbf{p}_H \| \leq (1 + \varepsilon) \| \mathbf{p}_{H_0} \|$. If $\mathbf{p}_{H_0} = 0$, the sequence $(\mathbf{x}_n)$ is obviously multipliable and has product 0; excluding this trivial case, there exists an interval $H_1 = \{0, q\}$, containing $H_0$ and such that, for every finite subset $L$ of $\mathbf{N}$ not meeting $H_1$, one has $\| \mathbf{e} - \mathbf{p}_L \| \leq \varepsilon (\| \mathbf{p}_{H_0} \|)^{-1}$. We deduce as above that, for every finite subset $H \supset H_1$

The Cauchy criterion therefore shows that $J \mapsto \mathbf{p}_J$ has a limit in $A$ following the filtered set $\mathscr{F}(\mathbf{N})$.

If all the $x_n$ are invertible, the same is true of all the finite partial products $p_J$; for every finite subset $H$ containing $H_0$, one can therefore write

this shows that, in the multiplicative group $G$ of the invertible elements of $A$, the image by the mapping $J \mapsto p_J$ of the filter of sections of $\mathfrak{F}(N)$ is a Cauchy filter basis for the left uniform structure of the group $G$; but since $G$ is complete ($IX$, p. 40, prop. 14), the limit of the mapping $J \mapsto p_J$ belongs to $G$.

This is what the criterion of th. 1 ($IX$, p. 79) immediately shows.

Apply the criterion of $IX$, p. 79, th. 1; for every finite subset $L$ of $N$, we have $p_L = e - \prod_{n \in L} (e + u_n) - e = \sum_M \left( \prod_{n \in M} u_n \right)$, $M$ ranging over the set of the non-empty subsets of $L$ (totally ordered by the induced order). Since $\left| \prod_{n \in M} u_n \right| \leq \prod_{n \in M} \| u_n \|$, one can write

Now, since the series of general term $\| u_n \|$ is convergent by assumption, the sequence $(1 + \| u_n \|)$ is multipliable in $\mathbf{R}_+^*$ ($IV$, p. 35, th. 4); for every $\varepsilon > 0$, there therefore exists a finite subset $J_0$ of $N$ such that, for every finite subset $L$ of $N$ not meeting $J_0$, we have $\left| \prod_{n \in L} (1 + \| u_n \|) - 1 \right| \leq \varepsilon$; whence the theorem.

Indeed, there are only a finite number of integers $n$ such that $\|u_n\| > 1$. Let $J = \{0, m\}$ be an interval of $\mathbf{N}$ containing all these integers. The product of the sequence $(\mathbf{e} + u_n)$ is the product of $p_J$ and of the element $\prod_{n > m} (\mathbf{e} + u_n)$, all of whose factors are invertible (IX, p. 39, corollary), and which is consequently itself invertible; since $p_J$ is the product of a finite number of elements which are not divisors of 0, it is not a divisor of 0, and the same is true of $\prod_{n \in \mathbf{N}} (\mathbf{e} + u_n)$.

The sufficient condition for multipliability given in th. 2 is not necessary in general (cf. IX, p. 126, exerc. 6). It is, however, necessary in the important case where $A$ is an algebra of finite rank over the field $\mathbf{R}$ (in particular when $A$ is the field of quaternions $\mathbf{K}$, or a matrix algebra $\mathbf{M}_n(\mathbf{R})$):

It is known (VII, p. 16, prop. 2) that there exists a number $c > 0$ such that, for every finite family $(\mathbf{x}_i)_{i \in I}$ of points of $A$, one has

(1)

Let $(a_n)_{n \in \mathbf{N}}$ be any sequence of elements of $A$. For every finite subset $I$ of $\mathbf{N}$, put

Indeed, since $\sigma_J \leq \sigma_I$ for every subset $J$ of $I$, one has, after (2),

since, by virtue of (1), one has $\sigma_I \leq c \cdot \sup_{J \subset I} \| s_J \|$, it follows that $\sigma_I \leq c \varphi(I)(1 + \sigma_I)$, whence the lemma.

That being so, let $(e + u_n)$ be a multipliable sequence in $A$, whose product is invertible; according to th. 1 (IX, p. 79), there exists a finite subset $J_0$ of $\mathbf{N}$ such that, for every finite subset $H$ of $\mathbf{N}$ not meeting $J_0$, one has

a) If the infinite product of general factor $x_n$ is convergent and if $\prod_{n=0}^\infty x_n$ is invertible, for every $\varepsilon > 0$, there exists $n_0$ such that, for $n_0 \leq m \leq n$, one has $\left\| \prod_{k=m}^n x_k - e \right\| \leq \varepsilon$.

b) Conversely, if the sequence $(x_n)$ satisfies this condition, the infinite product of general factor $x_n$ is convergent; moreover, if each of the $x_n$ is invertible, $\prod_{n=0}^\infty x_n$ is invertible.

We leave to the reader the development of the proof of this proposition, which is copied step by step from that of th. 1 of IX, p. 79 (the finite parts $L$ of $\mathbf{N}$ which occur in the latter having simply 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 remainder of index h of the infinite product of general factor $x_n$.

Assuming still that $\prod_{n=0}^{\infty} x_n$ is invertible, we deduce also from prop. 2 (IX, p. 82) that, if $(z_n)$ is a sequence such that $z_n = x_n$ except for a finite number of indices, the product of general factor $z_n$ is convergent.

Indeed, the sequence of partial products of the sequence $(u_n)$ is extracted from the sequence of partial products of the sequence $(x_n)$.

Finally, by the same reasoning as for commutative groups (III, p. 44), one sees that if, in a normed algebra $A$, a sequence $(x_n)$ is multipliable, the product of general factor $x_n$ is convergent, and one has $\prod_{n=0}^{\infty} x_n = \prod_{n \in \mathbf{N}} x_n$ (which is also written $\prod_{n=0}^{\infty} x_n$); the converse is of course false (cf. IX, p. 126, exerc. 7).

Exercises