Let $E$ be a real Banach space. Let $\mathcal L(E)$ be the space of bounded linear operators on $E$. For $T \in \mathcal L(E)$,
- we denote by $N(T)$ its kernel and by $R(T)$ its range.
- we denote by $\rho(T)$ its resolvent set, by $\sigma(T)$ its spectrum, and by $EV(T)$ its set of eigenvalues. Then $EV(T) \subset \sigma(T) = \mathbb R \setminus \rho(T)$.
I'm trying to solve an exercise in Brezis' Functional Analysis, i.e.,
Let $E := \ell^p$ with $1 \leq p \leq \infty$ and let $\left(\lambda_n\right)$ be a bounded sequence in $\mathbb{R}$. Consider the multiplication operator $M \in \mathcal{L}(E)$ defined by $$ M x = (\lambda_1 x_1, \lambda_2 x_2, \ldots, \lambda_n x_n, \ldots) \quad \text {where} \quad x= (x_1, x_2, \ldots, x_n, \ldots) . $$ Determine $E V(M)$ and $\sigma(M)$.
There are possibly subtle mistakes that I could not recognize in below attempt. Could you please have a check on it? I'm also happy to see other approaches.
First, we determine $EV(M)$. Let $\eta \in \mathbb R$. Then $\eta \in EV(M)$ iff $M-\eta I$ is not injective. We have $x \in N(M-\eta I)$ iff $Mx = \eta x$ iff $$ (\lambda_1 x_1, \lambda_2 x_2, \ldots, \lambda_n x_n, \ldots) = (\eta x_1, \eta x_2, \ldots, \eta x_n, \ldots), $$ iff $(\lambda_n -\eta) x_n=0$ for all $n \ge 1$. Then $N(M-\eta I) \neq \{0\}$ iff there is $n$ such that $x_n \neq 0$ iff there is $n$ such that $\eta = \lambda_n$. Hence $EV(M) = \{\lambda_1, \lambda_2, \ldots, \lambda_n, \ldots\}$.
Second, we determine $\sigma(M)$. It remains to find those $\eta \in \mathbb R$ such that $M-\eta I$ is not surjective. For $x \in E$, we have $$ (M-\eta I)x = \big ( (\lambda_1-\eta)x_1, (\lambda_2-\eta)x_2, \ldots, (\lambda_n-\eta)x_n, \ldots \big ). $$
Assume that $\lambda \notin \{\lambda_1, \lambda_2, \ldots, \lambda_n, \ldots\}$ and $y \in E$. Let $$ x := \left ( \frac{y_1}{\lambda_1-\eta}, \frac{y_2}{\lambda_2-\eta}, \ldots, \frac{y_n}{\lambda_n-\eta}, \ldots \right ). $$
Then $(M-\eta I) x=y$ and thus $M-\eta I$ is surjective. As such, $\sigma(M) = EV(M)$.
