Proposition 1

Let $\lambda = (\lambda_i)_{i\in I}$ be a bounded family of elements of $\mathbf{C}$. Let $p$ be an element of $[1,+\infty ]$. Let E be the Banach space $\ell^p_{\mathbf{C}}(I)$.

a) For every $x\in E$, one has $\lambda x\in E$ and the mapping $u:x\mapsto \lambda x$ is a continuous endomorphism of E, whose spectrum in the Banach algebra $\mathscr{L}(E)$ is equal to the closure in $\mathbf{C}$ of the set of the $\lambda_i$, and whose norm is equal to sup$_i|\lambda_i|$;

b) The endomorphism $u$ is compact if and only if for every real number $\varepsilon >0$, the set of elements $i\in I$ such that $|\lambda_i|\geqslant \varepsilon$ is finite.

With the notation of the proposition, the endomorphism $x\mapsto \lambda x$ is called a diagonal endomorphism of $\ell^p_{\mathbf{C}}(I)$.

Let us prove the proposition. Let C = sup$_i|\lambda_i|$. Let $x\in E$. One has the inequalities

$\sum_{i\in I}|\lambda_ix_i|^p\leqslant C^p\|x\|^p$, if $p\not = +\infty$

sup$_{i\in I}|\lambda_ix_i|\leqslant C\|x\|$, if $p= +\infty$.

This proves that $\lambda x\in E$. The mapping $x\mapsto \lambda x$ is therefore an endomorphism of E, and these inequalities prove that it is of norm $\leqslant C$.

For $j\in I$, denote by $e_j$ the element $(x_i)_{i\in I}$ of $\ell^p_{\mathbf{C}}(I)$ such that $x_j= 1$ and $x_i= 0$ if $i\not =j$. We then have $u(e_j) =\lambda_je_j$ for all $j\in I$, which shows that $\lambda_j$ belongs to the spectrum of $u$. Since the spectrum of $u$ is closed, the closure in $\mathbf{C}$ of the set of the $\lambda_i$ is contained in the spectrum of $u$. Since the spectrum of $u$ is contained in the disk centered at 0 and of radius $\|u\|($I, p. 24, Theorem 1 and formula (3) of I, p. 21), we deduce the inequality $C\leqslant \|u\|$, whence $\|u\|= C$.

Conversely, if $\lambda \in \mathbf{C}$ is not adherent to the set of the $\lambda_i$, then the family $((\lambda_i-\lambda )^{-1})_{i\in I}$ is bounded. What precedes therefore shows that the linear mapping $x\mapsto ((\lambda -\lambda_i)^{-1}x_i)_{i\in I}$ is an endomorphism of E. It is inverse to $u-\lambda 1_E$, and hence $\lambda$ does not belong to the spectrum of $u$. This proves a).

Let us now prove b). Suppose that the endomorphism $u$ is compact. Let $\varepsilon >0$. Denote by J the set of the $i\in I$ such that $|\lambda_i|\geqslant \varepsilon$. The set of elements $e_i$ for $i\in J$ is bounded in E. Its image by $u$, formed by the elements $\lambda_ie_i$ for $i\in J$, is therefore relatively compact in E (Remark 1 of III, p. 2). Since $\|\lambda_ie_i-\lambda_je_j\|\geqslant \varepsilon$ for every pair of elements $i\not =j$ in J, this is possible only if the set J is finite.

Let us prove the converse assertion. Let J be a finite subset of I and denote by $z_J= (z_{J,i})_{i\in I}$, where $z_{J,i}=\lambda_i$ for $i\in J$ and $z_{J,i}= 0$ for $i\in I$- J ; the family $z_J$ defines an endomorphism of finite rank $u_J:x\mapsto z_Jx$ of E. According to what precedes, the norm of the linear mapping $u-u_J$ is bounded above by the upper bound in $\mathbf{R}_+$ of the family $(|\lambda_i|)_{i\in I-J}$. The assumption implies that for every $\varepsilon >0$, there exists a finite subset $J\subset I$ such that this upper bound is $\leqslant \varepsilon$. It follows that the mapping $u$ is adherent to $\mathscr{L}^f$(E), and hence that it is compact (III, p. 4, cor. of Prop. 2).