Proposition 1

Let E be a complex Banach space and $u$ an endomorphism of E. Let $\lambda \in \mathbf{C}$ and $f\in \mathscr{O}$(Sp($u$)). We have

Ker($u-\lambda 1_E$)$\subset$ Ker($f(u)-f(\lambda )1_E$).

Let $x\in E$ be nonzero. The set A of the $v\in \mathscr{L}(E)$ such that $x$ is an eigenvector of $v$ is a unital subalgebra of $\mathscr{L}(E)$.

The algebra A is full: if $v\in A$ is invertible in $\mathscr{L}(E)$ and if $x$ is an eigenvector of $v$ for the eigenvalue $\lambda$, then $\lambda \not= 0$ and $v^{-1}(x) =\lambda^{-1}x$, which proves that $v^{-1}\in A$.

The algebra A is closed in $\mathscr{L}(E)$. In fact, if $(v_n)_{n\in\mathbf{N}}$ is a sequence in A such that $v_n$ converges to $v\in \mathscr{L}(E)$, the sequence $(\lambda_n)_{n\in\mathbf{N}}$ such that $v_n(x) =\lambda_nx$ is bounded, and therefore admits a subsequence converging to a complex number $\mu$, so that $v(x) =\mu x$.

Suppose that $x$ is an eigenvector of $u$ and that $u(x) =\lambda x$. The algebra A contains $u$, and therefore contains the closed full unital subalgebra B generated by $u$, which is commutative. The mapping $\chi : B\rightarrow \mathbf{C}$ which associates with $v$ the eigenvalue of $v$ relative to $x$ is a character of B such that $\chi (u) =\lambda$. For every $f\in \mathscr{O}$(Sp($u$)), we have $f(u)\in B$ and $\chi (f(u)) =f(\chi (u)) =f(\lambda )$ by Proposition 7 of I, p. 75, whence the proposition.