If the algebra A is unital, the mappings $x\mapsto \|\boldsymbol{\gamma }_x\|$ and $x\mapsto \|\boldsymbol{\delta }_x\|$ are norms on A, defining the topology of A. They satisfy the inequality (1).

If the unital normed algebra A is non-zero, that is, if $e\not= 0$, we also have $\|\boldsymbol{\gamma }_e\|=\|\boldsymbol{\delta }_e\|= 1$.

It is immediate that $\|\boldsymbol{\gamma }_x\|\leqslant \|x\|$ and that $\|\boldsymbol{\delta }_x\|\leqslant \|x\|$.

If A has a unit element $e$, then we have $x=\boldsymbol{\gamma }_xe=\boldsymbol{\delta }_xe$, hence: (2) $\|x\|\leqslant \|\boldsymbol{\gamma }_x\| \cdot \|e\|\|x\|\leqslant \|\boldsymbol{\delta }_x\| \cdot \|e\|$.

In this case, $x\mapsto \|\boldsymbol{\gamma }_x\|$ and $x\mapsto \|\boldsymbol{\delta }_x\|$ are therefore norms equivalent to the norm of A. They satisfy (1) since $\boldsymbol{\gamma }$ and $\boldsymbol{\delta }$ are representations.

The last assertion follows from the fact that $\boldsymbol{\gamma }_e=\boldsymbol{\delta }_e=$ Id$_A$.

lim$_{n\rightarrow+\infty}\frac{a_n}{n}=$ inf$_{n\geqslant 1}\frac{a_n}{n}$.

Put $a_0= 0$ ; the inequality $a_{n+m}\leqslant a_n+a_m$ remains valid for all $n\geqslant$ 0 and all $m\geqslant 0$. Fix an integer $m\geqslant$ 1. For every integer $n\geqslant 1$, let $q(n)$ and $r(n)$ be the integers such that $n=q(n)m+r(n)$ and $0\leqslant r(n)< m$ (E, III, p. 39, th. 1). The hypothesis then implies

Letting $n$ tend to $+\infty$, we deduce that lim sup$_n(a_n/n)\leqslant a_m/m$ since $q(n)/n\rightarrow 1/m$. Since this holds for every $m\geqslant 1$, we therefore have

lim sup$_{n\rightarrow+\infty}\frac{a_n}{n}\leqslant$ inf$_{m\geqslant 1}\frac{a_m}{m}\leqslant$ lim inf$_{n\rightarrow+\infty}\frac{a_n}{n}$.

These inequalities prove the convergence of the sequence $(a_n/n)_{n\geqslant 1}$ as well as the formula lim $a_n/n=$ inf$_{n\geqslant 1}a_n/n$.

$\varrho (x) =$ lim$_{n\rightarrow+\infty}\|x^n\|^{1/n}_1=$ inf$_{n\geqslant 1}\|x^n\|^{1/n}_1$.

If $x$ is nilpotent, one has $\|x^n\|^{1/n}_1= 0$ for every sufficiently large integer $n$ and every norm $x\mapsto \|x\|_1$ defining the topology of A.

Suppose now that $x$ is not nilpotent, and put $\alpha_n=$ $\|x^n\|$. One has $\alpha_n>0$ for every integer $n\geqslant 1$, and $\alpha_{n+m}\leqslant \alpha_n\alpha_m$ for all $n, m\in \mathbf{N}$ according to (1). Lemma 2, applied to the sequence $a_n=$ log($\alpha_n$), shows the existence of the limit $\varrho (x)$ and the formula $\varrho (x) =$ inf$_{n>0}\alpha_n^{1/n}$.

Let $x\mapsto \|x\|_1$ be a norm defining the topology of A. There exist real numbers $a >0$ and $b >0$ such that

for all $x\in A$ (EVT, II, p. 7, cor. 2). Consequently,

for all $n\geqslant 1$, whence, by passing to the limit, or by taking the greatest lower bound, the equality

$\varrho (x) =$ lim$_{n\rightarrow+\infty}\|x^n\|^{1/n}_1=$ inf$_{n>0}\|x^n\|^{1/n}_1$.

The series $\sum^{\infty}_{n=1}\lambda^nx^n$ has radius of convergence

(lim sup$_{n\rightarrow+\infty}\|x^n\|^{1/n}$)$^{-1}=\varrho (x)^{-1}$

(cf. VAR, R1, p. 23, 3.1.4). Suppose that A admits a unit element. If $\varrho (x)<1$, the series $\sum_{n=0}^{\infty}x^n$ is therefore absolutely convergent. Since

for every integer $k\geqslant 0$, the element $1-x$ is invertible and its inverse is equal to $\sum_{n=0}^{\infty}x^n$.

This is immediate since $\|x\|<1$ implies $\varrho (x)<1$.

Let $(\widetilde{A}, e)$ be the unital Banach algebra deduced from A by adjunction of a unit element. There exists a maximal left ideal (resp. right) J of $\widetilde{A}$ such that $J\cap A = I$ (A, VIII, p. 428, prop. 4). Then J is disjoint from the open ball of center $e$ and of radius 1 (cor. 1), and therefore $\overline{J}\not=\widetilde{A}$. Since J is a maximal ideal, this implies that $\overline{J}= J$, and consequently that $I = J\cap A =\overline{J}\cap A$ is closed in A.

Indeed the radical is the intersection of the regular maximal left ideals (A, VIII, p. 430, déf. 3).

a) If $x\in A$ admits a left inverse (resp. right) $y$, every element $x'\in A$ such that $\|x'-x\|<\|y\|^{-1}$ admits a left inverse (resp. right).

b) The set of elements of A which are invertible (resp. left, resp. right) is open in A.

c) Let $(x_n)$ be a sequence of elements of A admitting left inverses (resp. right)$y_n$, and converging to an element $x\in A$. If the sequence $(y_n)$ is bounded, then $x$ is invertible on the left (resp. right).

It suffices to treat the case of left inverses; that of right inverses follows by considering the opposite algebra.

Let $x, y, x'\in A$ be such that $yx= 1$ and $\|x'-x\|<\|y\|^{-1}$. We have

therefore $yx'$ is invertible: there exists $z\in$ A such that $z(yx') = 1$. Thus the element $x'$ is left invertible with inverse $zy$. This proves assertion a) and assertion b) follows from it immediately.

Let $(x_n)$ be a sequence of elements of A admitting left inverses $y_n$, which converges to an element $x\in$ A, and such that the sequence $(y_n)$ is bounded. If $M\geqslant 1$ is a real number such that $\|y_n\|\leqslant M$ for all $n\geqslant 1$, then we have $\|x_n-x\|<M^{-1}\leqslant \|y_n\|^{-1}$ for $n$ sufficiently large, and therefore $x$ admits a left inverse according to a).

Let $y_n$ be a left inverse of $x_n$. By Prop. 3 (ii), $\|y_n\|$ tends to $+\infty$. Let $z_n=\|y_n\|^{-1}y_n$. We have $\|z_n\|= 1$, and $z_nx_n=\|y_n\|^{-1}$ tends to 0, hence $z_nx=z_nx_n+z_n(x-x_n)$ tends to 0. One then concludes with the aid of the remark following Definition 4.

In fact, let $x$ be an element of $\overline{B}$ invertible in A, and $(x_n)$ a sequence of elements of B tending to $x$. Then, for $n$ sufficiently large, $x_n$ is invertible in A and $x^{-1}_n$ tends to $x^{-1}$. Since the subalgebra B is full, we have $x^{-1}_n\in B$, whence $x^{-1}\in \overline{B}$.

Let A be a unital Banach algebra and let $(y_i)_{i\in I}$ be a family of elements of A. Let B be the full subalgebra of A generated by the elements $y_i$. Then $\overline{B}$ is the smallest closed full subalgebra of A containing the $y_i$. It is called the closed full subalgebra generated by the elements $y_i$.

a) The set Sp$_A(x)$ is a compact subset of $\mathbf{C}$;

b) The spectral radius $\varrho (x)$ is the radius of the smallest closed disk with center 0 in $\mathbf{C}$ which contains Sp$_A(x)$ ;

c) The resolvent $\lambda \mapsto R(x, \lambda ) = (\lambda -x)^{-1}$ of $x$ is holomorphic in $\mathbf{C}-$ Sp$_A(x)$ and zero at infinity. Moreover, for every integer $k\geqslant 0$, we have the formula

$\frac{\partial^k}{\partial \lambda^k}R(x, \lambda ) = (-1)^kk$! $R(x, \lambda )^{k+1}$;

d) For every complex number $\lambda$ such that $|\lambda |>1/\varrho (x)$, we have

The complement of the spectrum Sp$_A(x)$ is the inverse image of the group G of invertible elements of A by the continuous mapping $\lambda \mapsto$ $x-\lambda$ of $\mathbf{C}$ into A; by Proposition 3, b) of I, p. 23, the spectrum Sp$_A(x)$ is a closed subset of $\mathbf{C}$. On the other hand, let $\lambda \in \mathbf{C}$ be such that $|\lambda |> \varrho (x)$. We have $\lambda -x=\lambda (1-\lambda^{-1}x)$. Since $\varrho (\lambda^{-1}x) =|\lambda |^{-1}\varrho (x)<1$, the element $1-\lambda^{-1}x$, hence also the element $\lambda -x$, is invertible and

(I, p. 22, Proposition 2). In particular, $\lambda \notin$ Sp$_A(x)$. This proves that Sp$_A(x)$ is contained in the disc with center 0 and radius $\varrho (x)$. Hence Sp$_A(x)$ is compact. This formula (5) also proves that the resolvent of $x$ is defined and holomorphic in the complement of the closed disc $\Delta_{\varrho(x)}$ with center 0 and radius $\varrho (x)$, and tends to 0 at infinity.

Let $\lambda_0\in \mathbf{C}-$ Sp$_A(x)$. Put $y=\lambda_0-x$. Let $\mu\in \mathbf{C}$ be such that $|\lambda_0-\mu|<\|y^{-1}\|^{-1}$. We have

therefore $\mu-x$ is invertible and has inverse

by Proposition 2 of I, p. 22. Therefore the resolvent of $x$ is defined and holomorphic in the open disc with center $\lambda_0$ and radius $\|y^{-1}\|^{-1}$. Hence the resolvent of $x$ is a holomorphic mapping of $\mathbf{C}-$ Sp$_A(x)$ into A.

Formula (1) of I, p. 4 implies $\frac{\partial}{\partial \lambda}R(x, \lambda ) =-R(x, \lambda )^2$, whence, by induction on $k$,

$\frac{\partial^k}{\partial \lambda^k}R(x, \lambda ) = (-1)^kk$! $R(x, \lambda )^{k+1}$.

Let $a >0$ be a real number such that Sp$_A(x)$ is contained in the closed disc $\Delta_a$ with center 0 and radius $a$. The function $\lambda \mapsto (\lambda^{-1}-x)^{-1}$ is then defined and holomorphic for $0<|\lambda |< a^{-1}$ and tends to 0 when $\lambda$ tends to 0. The unique continuous function on the open disc with center 0 and radius $a^{-1}$ which extends this holomorphic function is then holomorphic (VAR, R1, 3.3.9), hence the radius of convergence of the series (5) which defines it is $\geqslant a^{-1}$ (VAR, R1, 3.2.9). By Proposition 2 of I, p. 22, we therefore have $a\geqslant \varrho (x)$.

Suppose first that A is complete. If one had Sp($x$) $=\emptyset$, the resolvent of $x$ would be holomorphic in $\mathbf{C}$ and zero at infinity, hence identically zero (VAR, R., 3.3.6, p. 29). Since $R(x, \lambda ) = (\lambda -x)^{-1}$ is invertible, it would follow that 1 = 0 and therefore that $A =\{0\}$.

In the general case, let $\widehat{A}$ be the completed algebra of A; the relation Sp$_A(x) =\emptyset$ would entail Sp$_{\widehat{A}}(x) =\emptyset$, whence $\widehat{A} =\{0\}$ and $A =\{0\}$.

If $x\in A$, there exists $\lambda \in \mathbf{C}$ such that $x-\lambda$ is not invertible (cor. 1), whence $x-\lambda = 0$ and $x\in \mathbf{C}\cdot 1$.

Let Δ be the disc with center 0 and radius 1 in $\mathbf{C}$. By theorem 1 b) and the fact that $\varrho (x)\leqslant \|x\|$, one has Sp($x$)$\subset \Delta$. Analogously, Sp($x$)$^{-1}$ = Sp($x^{-1}$)$\subset \Delta$, whence the corollary (cf. I, p. 2, remark 4).

a) Let $u$ be an endomorphism of E, not necessarily continuous, which commutes with A. Then $u$ is a homothety;

b) Let $u$ be an endomorphism of E, not necessarily continuous. For every integer $n\geqslant 1$ and every $(\xi_1, . . . , \xi_n)\in E^n$, there exists $v\in A$ such that

Let us prove a). Let $\widetilde{A}$ be the algebra obtained from A by adjunction of a unit element. Since E is a simple A-pseudomodule, it is a simple $\widetilde{A}$-module.

Let B be the commutant of $\widetilde{A}$ in the ring of endomorphisms of the $\mathbf{C}$-vector space E. The algebra B contains 1 and is the endomorphism algebra of the $\widetilde{A}$-module E. Since E is a simple $\widetilde{A}$-module, Schur’s Lemma (A, VIII, p. 43, corollary) shows that B is a field.

Let $\xi_0\in E$ be such that $A\xi_0\not=\{0\}$. One has therefore $A\xi_0= E$. For every $u\in B$, let $A_u$ be the set of $v\in A$ such that $v(\xi_0) =u(\xi_0)$. This set is nonempty, since $A\xi_0= E$. One then puts

$\|u\|_B=$ inf$_{v\in A_u}\|v\|$.

The mapping $u\mapsto \|u\|_B$ is a seminorm on B.

Let us prove that this mapping is a norm. Let $u$ be a nonzero element of B. For every $v\in A_u$, one has $\|v\|\geqslant \|v(\xi_0)\|/\|\xi_0\|=\|u(\xi_0)\|/\|\xi_0\|$, so that $\|u\|_B\geqslant \|u(\xi_0)\|/\|\xi_0\|$. It is therefore enough to prove that $u(\xi_0)\not= 0$. Let $\xi_1\in E$ be such that $u(\xi_1)\not= 0$. Since $A\xi_0= E$, there exists $w\in A$ such that $\xi_1=w(\xi_0)$. Then $wu(\xi_0) =uw(\xi_0) =u(\xi_1)\not= 0$, hence $u(\xi_0)\not= 0$.

On the other hand, let $u$ and $u'$ be elements of B. For every $\varepsilon >0$, there exist $v, v'\in A$ such that $v(\xi_0) =u(\xi_0),v'(\xi_0) =u'(\xi_0)$ and $\|v\|\leqslant \|u\|_B+\varepsilon$, $\|v'\|\leqslant \|u'\|_B+\varepsilon$. Then one has $vv'(\xi_0) =vu'(\xi_0) =u'v(\xi_0) =u'u(\xi_0)$, whence

and finally $\|u'u\|_B\leqslant \|u\|_B\|u'\|_B$. This shows that B, endowed with the norm $u\mapsto \|u\|_B$, is a normed algebra. Since it is a field, corollary 2 implies that $B =\mathbf{C}\cdot 1$, which is the desired conclusion.

Let us prove b). According to a), the commutant of A in End$_{\mathbf{C}}(E)$ is reduced to the homotheties of E. Its bicommutant is therefore End$_{\mathbf{C}}(E)$. Assertion b) therefore results from Jacobson's density theorem (theorem 1 of A, VIII, p. 434).

a) Sp$'(x)$ is a compact subset of $\mathbf{C}$;

b) The spectral radius $\varrho (x)$ is the radius of the smallest closed disk with center 0 in $\mathbf{C}$ that contains Sp$'(x)$;

c) In order that $x$ be quasi-nilpotent, it is necessary and sufficient that one have Sp$'(x) =\{0\}$.

Assertions a) and b) result from th. 1 by considering the Banach algebra deduced from A by adjunction of a unit element. Assertion c) results from b).