Proposition 2

Let $u\in \mathscr{L}(E)$ be such that $\|1_E-\pi (u)\|<1$ in the algebra $\mathscr{C}$alk(E). Then $u$ is a Riesz endomorphism of E.

Let $r\geqslant 1$ be an integer such that $\|1_E-\pi (u)\|^r<\frac{1}{2}$. Let $P\in \mathbf{C}[X]$ be the polynomial $\frac{1-(1-X)^r}{X}$. Let $v= 1_E-(1_E-u)^r$. We thus have $v=uP(u)$ and $\|1_E-\pi (v)\|<\frac{1}{2}$. Since the endomorphisms $u$ and $P(u)$ of E commute, it suffices to prove that $v$ is a Riesz endomorphism of E (III, p. 49, prop. 9).

Since $\|1_E-\pi (v)\|<1/2$, there exists, by definition of the quotient norm in the space $\mathscr{C}$alk(E), a compact endomorphism $h$ of E and an endomorphism $w$ of E such that $v= 1_E+h+w$ and $\|w\|<\frac{1}{2}$. According to Corollary 1 of I, p. 22, the element $1_E+w$ is an automorphism of E. Since $h$ is compact, $v= (1_E+w) +h$ is a Fredholm endomorphism of E of index 0 (Cor. 2 of III, p. 74). For every integer $n\geqslant 0$, denote by $N_n$ the kernel of $v^n$. To prove that $v$ is a Riesz endomorphism of E, it suffices to prove that there exists an integer $n\geqslant 0$ such that $N_n= N_{n+1}($III, p. 46, def. 2 and III, p. 46, remark).

Reason by contradiction by supposing the sequence $(N_n)$ strictly increasing. For every $n\in \mathbf{N}$, let $p_n$ be the canonical mapping of E onto the normed space $E/N_n$. Let $c$ be a real number such that $2\|w\|< c <1$. Let $n\in \mathbf{N}$. Since $N_{n+1}$ is different from $N_n$, there exists an element $x_n\in N_{n+1}$ such that $\|p_n(x_n)\|=c$ and such that $\|x_n\|<1$ (indeed, there exists $y\in N_{n+1}/N_n$ of norm $c$ so, for every $\varepsilon >0$, there exists $x_n\in N_{n+1}$ such that $p_n(x_n) =y$ and $\|x_n\|\leqslant c+\varepsilon$ ).

Let $m,n$ be two integers such that $m > n\geqslant 0$. We have

since $\|x_m\|\leqslant 1$ and $\|x_n\|\leqslant 1$. Moreover, note that

But since $n < m$ and $v(N_{m+1})\subset N_m$, we have $v(x_m-x_n) +x_n\in N_m$, whence

The inequality (1) therefore gives the lower bound

The sequence $(h(x_m))_{m\in\mathbf{N}}$ therefore has no value of adherence in E, which contradicts the fact that the image by $h$ of the unit ball of E is relatively compact, since $h$ is compact (remark 1 of III, p. 2).