Proposition 14

Let E be a complex Banach space. Let $u$ be a closed operator with dense domain on E and U its resolvent set.

a) For every $\lambda \in U$, the open disc in $\mathbf{C}$ with center $\lambda$ and radius $\|R(u, \lambda )\|^{-1}$ is contained in U;

b) The set U is open in $\mathbf{C}$;

c) Suppose that Sp($u$) is not empty. Let $\lambda \in U$ and denote by $\delta$ the distance in $\mathbf{C}$ from $\lambda$ to the spectrum of $u$. One has $\delta >0$ and $\|R(u, \lambda )\|\geqslant 1/\delta$;

d) The mapping $\lambda \mapsto R(u, \lambda )$ is a holomorphic mapping of U into $\mathscr{L}(E)$. For every integer $k\in \mathbf{N}$ and every $\lambda \in U$, one has

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

Let $\lambda \in U$. For every $\mu\in \mathbf{C}$ such that $\|(\mu-\lambda )1_E\|<\|R(u, \lambda )\|^{-1}$, lemma 4 applied to the injective partial operator $\lambda 1_E-u$ and to $v= (\mu-\lambda )1_E$ implies that the partial operator $\mu1_E-u=\lambda 1_E-u+v$ is injective and has a continuous inverse. This implies a). According to loc. cit., one also has the formula

which implies that the resolvent of $u$ is holomorphic in U.

Assertion b) results at once from a). If Sp($u$) is nonempty, the distance from $\lambda$ to Sp($u$) is strictly positive (TG, IX, p. 13, Prop. 2), whence c).

The last part of assertion d) is proved by induction on $k$, the case $k= 1$ being a consequence of formula (8), p. 245.