Theorem 1

Let A be a unital Banach algebra and $x\in A$.

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 $a\geqslant \varrho (x)$.