When these conditions hold, the series $u = \sum_{\lambda \in L} u_\lambda$ is equal to $\sum \alpha_\nu X^\nu$ with $a_\nu = \sum_{\lambda \in L} \alpha_{\lambda,\nu}$ for each $\nu \in \mathbf{N}^{(I)}$.

The equivalence of (i) and (ii) follows from Cor. 2 of Gen. Top., III, p. 263. The equivalence of (ii) and (iii) follows from the properties of limits in a product space (Gen. Top., I, p. 55, Cor. 1). The last assertion follows from Prop. 4 of Gen. Top., III, p. 266.

Let us give some examples of summable families. a) Let $u \in A[[I]]$ and let $a_\nu$ be the coefficient of $X^\nu$ in $u$. The family $(\alpha_\nu X^\nu)_{\nu \in \mathbf{N}^{(I)}}$ is then summable, with sum $u$ (which justifies writing $u = \sum \alpha_\nu X^\nu$).

b) Let $u \in A[[I]]$; for every integer $p \geq 0$ let $u_p$ be the homogeneous component of degree $p$ of $u$. Then the family $(u_p)_{p \geq 0}$ is summable and we have $u = \sum_{p \geq 0} u_p$.

c) Let $(u_\lambda)_{\lambda \in L}$ be a family of elements of $A[[I]]$ and suppose that for every integer $n \geq 0$ the set of $\lambda \in L$ such that $\omega(u_\lambda) < n$ is finite. Then the family $(u_\lambda)_{\lambda \in L}$ is summable.

Let $(\alpha_{\lambda, \nu})_{\nu \in \mathbf{N}^{(I)}}$ (resp. $(\beta_{\mu, \nu})_{\nu \in \mathbf{N}^{(I)}}$) be the family of coefficients of $u_\lambda$ (resp. $v_\nu$). For each $\nu \in \mathbf{N}^{(I)}$ there exists only a finite number of pairs $(\nu_1, \nu_2) \in \mathbf{N}^{(I)} \times \mathbf{N}^{(I)}$ such that $\nu_1 + \nu_2 = \nu$, hence only a finite number of pairs $(\lambda, \mu) \in L \times M$ such that the coefficient of $X^\nu$ in $u_\lambda v_\mu$ is $\neq 0$. Hence the family $(u_\lambda v_\mu)_{(\lambda, \mu) \in L \times \mathscr{J}}$ is summable. Now the formula (2) follows from the associativity of the sum (Gen. Top., III, p. 265, formula (2)).

In $\mathbf{A}[[\mathbf{I}]]$ the product is an associative and commutative composition law. We may therefore speak of a multipliable family of elements of $\mathbf{A}[[\mathbf{I}]]$ and of the product of a multipliable family (Gen. Top., III, p. 262, remark 3).

Let us define the ideals $\mathfrak{a}_\beta$ as at the beginning of this No., and let $\beta \in \mathbf{N}^{(I)}$. There exists a finite subset $L_0$ of $L$ such that $u_\lambda \in \mathfrak{a}_\beta$ for $\lambda \notin L_0$. Then for every $M \in \mathscr{S}$ such that $M \not\subset L_0$ we have $u_M \in \mathfrak{a}_\beta$. It follows that the family $(u_M)_{M \in \mathfrak{T}}$ is summable. On the other hand, for any finite subset $M_0$ of $L$ we have

Taken along the filtered ordered set $\mathfrak{T}$, the left-hand side has as limit $\sum_{M \in \mathfrak{T}} u_M$. Hence the right-hand side has as limit $\sum_{M \in \mathfrak{T}} u_M$, which proves (i) and (ii) at the same time.

Let $\nu \in \mathbf{N}^{(I)}$ and $n \in \mathbf{N}$. We have

If $n \geq |\nu| + m$ and $\nu(1) + \cdots + \nu(n) = v$, we have $|\nu(1)| + \cdots + |\nu(n)| \leq n - m$. We thus have $v(r) = 0$ and so $\alpha_{v(r)} = a_r$, for at least $m$ distinct values of $r$; it follows that $\alpha_{\nu(1)} \cdots \alpha_{\nu(n)} = 0$, whence the result.

Let $a_0$ be the constant term of $u$. The constant term of $u^n$ is $\alpha_0^n$, hence the stated condition is necessary; it is sufficient by Prop. 3.

(i) Let $\varphi$ be a continuous homomorphism of $A[[I]]$ into $E$ and $x_i = \varphi(X_i)$. Then: (a) for all $i \in I$, $x_i^n$ tends to 0 as $n$ tends to $+ \infty$; (b) if $I$ is infinite, $x_i$ tends to 0 along the filter of complements of finite subsets of $I$.

(ii) Let $x = (x_i)_i$, $_I$ be a family of elements of $E$ satisfying a) and b) of (i). Then there exists one and only one unital continuous homomorphism $\varphi$ of $A[[I]]$ into $E$ such that $\varphi(X_i) = x_i$ for all $i \in I$.

For all $i \in I$, $X_i^n$ clearly tends to 0 in $A[[I]]$ as $n$ tends to $+ \infty$; on the other hand when $I$ is infinite, $X_i$ tends to 0 along the filter of complements of finite subsets of $I$. This proves (i).

Let $(x_i)_{i \in I}$ be a family of elements of $E$ satisfying conditions a) and b) of (i), let $\psi$ be the homomorphism $u \mapsto u((x_i)_{i \in I})$ of $A[(X_i)_{i \in I}]$ into $E$, and let $V$ be a neighbourhood of 0 of $E$ which is an ideal of $E$. By b) there exists a finite subset $J$ of $I$ such that $x_i \in V$ for all $i \in I - J$. Next there exists by a) an integer $n \geq 0$ such that $x_i^n \in V$ for all $i \in J$. Let $\beta$ be the element of $\mathbf{N}^{(I)}$ such that $\beta_i = n - 1$ for $i \in J$ and $\beta_i = 0$ for $i \in I - J$. If we define the ideal $a_\beta$ of $A[[I]]$ as at the beginning of No. 2 (IV, p. 26), then

This shows that $\psi$ is continuous if we equip $A[(X_i)_{i \in I}]$ with the topology induced by that of $A[[I]]$. Since $E$ is separated and complete, $\psi$ extends to a continuous unital homomorphism $\varphi$ of $A[[I]]$ into $E$. We have $\varphi(X_i) = \psi(X_i) = x_i$ for all $i \in I$. Finally let $\varphi'$ be a continuous unital homomorphism of $A[[I]]$ into $E$ such that $\varphi'(X_i) = x_i$. We have $\varphi'(u) = \varphi(u)$ for all $u \in A[(X_i)_{i \in I}]$, hence $\varphi' = \varphi$ because $A[(X_i)_{i \in I}]$ is dense in $A[[I]]$.

Let us keep the previous notation. If $u \in A[[I]]$, the image of $u$ by $\varphi$ is denoted by $u(x)$ or $u((x_i)_{i \in I})$ (or also $u(x_1, ..., x_n)$ if $I = (1, 2, ..., n)$) and is called the element of $E$ obtained by substitution of $x_i$ for $X_i$ in $u$, or the value of $u$ for the values $x_i$ of the $X_i$ or also the value of $u$ for $X_i = x_i$. In particular we have $u = u((X_i)_{i \in I})$.

Let $E'$ be an associative commutative and unital linearly topologized separated and complete $A$-algebra. Let $\lambda$ be a continuous unital homomorphism of $E$ into $E'$, and $(x_i)_{i \in I}$ a family of elements of $E$ satisfying conditions a) and b) of Prop. 4 (IV, p. 28). The family $(\lambda(x_i))_{i \in I}$ satisfies the same conditions a) and b). For every $u \in A[[I]]$ we have

for the mapping $u \mapsto A(u((x_i)_{i \in I}))$ is a continuous unital homomorphism of $A[[I]]$ into $E'$ which transforms $X_i$ into $\lambda(x_i)$ for all $i \in I$.

If $J$ and $K$ are two sets, we denote by $A_{J, K}$ the set of all families $(g_j)_{j \in J}$ satisfying the following conditions: (i) for all $j \in J$, $g_j$ is an element of $A[[K]]$ whose constant term is nilpotent; (ii) if $J$ is infinite, $g_j$ tends to 0 along the filter of complements of finite subsets of $J$.

We note that if $J$ is finite, every family of formal power series $(g_j)_{j \in J}$ without constant term in $A[[K]]$ belongs to $A_{J, K}$.

Let $(g_j)_{j \in J}$ be in $A_{J, K}$. By the Cor. of Prop. 3 (IV, p. 28) we have $\lim_{n \to \infty} g_j^n = 0$ for all $j \in J$. Let $f \in A[[J]]$; we can substitute $g_j$ for the variable of index $j$ in $f$ and obtain a formal power series $f((g_j)_{j \in J})$ belonging to $A[[K]]$. Moreover, the mapping $f \mapsto f((g_j)_{j \in J})$ is a continuous homomorphism of $A$-algebras $A[[J]]$ into $A[[K]]$.

In particular if $J = (1, ..., p)$ and $f \in A[[X_1, ..., X_p]]$, we can substitute for each $X_j$ a formal power series $g_j \in A[[K]]$ without constant term; the result of this substitution is written $f(g_1, ..., g_p)$.

Let $x = (x_k)_{k \in K}$ be a family of elements of $E$ satisfying conditions a) and b) of Prop. 4 (IV, p. 28). Let us apply (3), taking for $A$ the homomorphism $u \mapsto u(x)$ of $A[[K]]$ into $E$; we obtain

Let $f = (f_i)_{i \in I} \in (A[[J]])^I$ and $g = (g_j)_{j \in J} \in A_{J, K}$. We denote by $f(g)$ or $f \circ g$ the element $(f_i((g_j),_{\epsilon_J}))_{i \in I}$ of $(A[[K]])^I$. If $f \in A_{I, J}$, we have $f \circ g \in A_{,,K}$ because the mapping $f \mapsto f((g_j)_{i, J})$ of $A[[I]]$ into $A[[K]]$ is continuous.

Let $f \in (A[[J]])^I$, $g \in A_{J, K}$, $h \in A_{K, L}$. Then $g \circ h \in A_{J, L}$ and by (4), we have

For

Suppose that there exists $v \in A[[I]]$ such that $uv = 1$. Let $a, \beta$ be the constant terms of $u$ and $v$, then $\alpha \beta = 1$, so $a$ is invertible.

Conversely, suppose that the constant term $a$ of $u$ is invertible. Then there exists a formal power series $t \in A[[I]]$ such that $u = \alpha (1 - t)$ and $\omega(t) > 0$. Now there is a ring homomorphism $\varphi : A[[T]] \to A[[I]]$ such that $\varphi(T) = t$, and $1 - T$ is invertible in $A[[T]]$ (Prop. 5); consequently $1 - t$ is invertible in $A[[I]]$, and hence so is $u$.

Since 0 admits in E a fundamental system of neighbourhoods consisting of ideals, the family $((D_i u)(x) \cdot y_i)_{i \in I}$ is summable in E for all $u \in A[[I]]$ (Gen. Top., III, p. 263, Cor. 2). Formula (16) thus defines an A-linear mapping $D : A[[I]] \to E$. We leave the reader to verify that D is a continuous derivation.

Let $D_1$ be a continuous A-derivation of $A[[I]]$ into E, such that $D_1(X_i) = y_i$ for all $i \in I$. The kernel of the continuous derivation $D - D_1$ is a closed subalgebra B of $\mathbf{A}[[\mathbf{I}]]$ containing 1 and the indeterminates $X_i$. Since the polynomial algebra $\mathbf{A}[(X_i)_{i \in I}]$ is dense in $\mathbf{A}[[\mathbf{I}]]$, we have $B = \mathbf{A}[[\mathbf{I}]]$ and so $D_1 = D$.

This follows from Prop. 7 because the mapping $u \mapsto \Delta(u(x))$ is a continuous derivation of $\mathbf{A}[[\mathbf{I}]]$ into the $\mathbf{A}[[\mathbf{I}]]$-module E.