Proposition 9

Let A be a unital Banach algebra and $a\in A$. Let U be an open neighbourhood of Sp$_A(a)$ and $f\in \mathscr{O}(U)$. Let moreover V be a compact neighbourhood of Sp$_A(a)$ contained in U such that V is a piece of U with oriented boundary $\partial V$.

For every integer $n\geqslant 0$, the mapping $z\mapsto f(z)(z-a)^{-n-1}$ is continuous on $\partial V$, the differential form $z\mapsto f(z)(z-a)^{-n-1}dz$ is integrable on $\partial V$ and one has

(7) $f^{(n)}(a) =2ni\pi$! $\int_{\partial V}f(z)(z-a)^{-n-1}dz$

where $f^{(n)}\in \mathscr{O}(U)$ is the $n$-th derivative of $f$.

We proceed by induction on $n$. When $n= 0$, the result follows from Prop. 1 of I, p. 61. Suppose now that the assertion of the proposition is true for the integer $n\geqslant 0$. Let $g\in \mathscr{O}(\mathbf{C}-$ Sp$_A(a); A)$ be the holomorphic function defined by $g(z) = (z-a)^{-n-1}f(z)$. The differential form $g'(z)dz$ = $dg$ is of class $C^1$; since the piece V is compact, Stokes's formula (VAR, R2, p. 47, 11.2.3) implies

Since $g'(z) = (z-a)^{-n-1}f'(z)-(n+1)(z-a)^{-n-2}f(z)$, we deduce

Applying the induction hypothesis to $f'$, we therefore obtain

$2ni\pi$! $f^{(n+1)}(a) = (n+ 1)\int_{\partial V}f(z)(z-a)^{-n-2}dz$,

which is the assertion of the proposition for the integer $n+ 1$. This completes the proof.