Residual spectrum of a multiplication operator on $l_{\infty}$ and $c_0$ Let $a \in l_{\infty}$ and $$T: l_{\infty} → l_{\infty}, x↦ax$$. I was able to find that $\sigma(T) = \overline{\{a_n:n∈\mathbb{N}\}}$ and tried to find residual spectrum of this operator, however I'm unable to find it. Can somebody help me? Thank you!
I'm also wondering, what if we consider the same operator in $c_0$, where $c_0$ is a space of sequences converging to zero?
 A: Clearly $\sigma_r(T) \subseteq \sigma(T) \setminus \{a_n:n \in \mathbb{N}\}$. Vice versa, if $\lambda \in  \sigma(T) \setminus \{a_n:n \in \mathbb{N}\}$, then $T-\lambda I$ is injective and $(1)_{n=1}^\infty \notin \overline{ran(T-\lambda I)}$: Let $(n_j)$ be a sequence in
$\mathbb{N}$ such that $a_{n_j} \to \lambda$ $(j \to \infty)$ and $x \in l_\infty$. Then $a_{n_j}x_{n_j} -  \lambda x_{n_j} \to 0$ $(j \to \infty)$, so $\|(Tx-\lambda x) - (1)_{n=1}^\infty\|_\infty \ge 1$. Thus  $ran(T-\lambda I)$ is not dense. Thus $\sigma_r(T) = \sigma(T) \setminus \{a_n:n \in \mathbb{N}\}$.
If we consider $T:c_0 \to c_0$ then $\sigma(T)$ is unchanged but $\sigma_r(T)$ is empty: Again $a_n \in \sigma_r(T)$ is not possible and if $\lambda \in \sigma(T) \setminus \{a_n:n \in \mathbb{N}\}$ then $ran(T-\lambda I)$ is dense: If $y =(y_n)\in c_0$ consider
$$
x:=(y_1/(a_1-\lambda), \dots, y_m/(a_m- \lambda),0,0,\dots) \in c_0.
$$
Then $Tx-\lambda x - y=(0,\dots,0, -y_{m+1},-y_{m+2}, \dots)$, hence
$\|Tx-\lambda x - y\|_\infty$ is small for $m$ sufficiently big.
Hope all is correct, please check.
A: Let $a$ be an accumulation point of $\{a_n\,:\, n\ge 1\}$ and $a\neq a_n$ for all $n.$ Then $a=\lim_ka_{n_k}$ for a subsequence $n_k.$ The operator $T-aI$ is injective. Its range is contained in
$$\{y\in \ell^\infty\,:\, \lim_k y_{n_k}=0\}$$
Thus the range is not dense. Summarizing the residual spectrum is equal
$$\sigma(T)\setminus \{a_n\,:\, n\ge 1\}$$
Concerning $c_0,$ for $a\in \sigma(T)\setminus \{a_n\,:\, n\ge 1\}$ the range $T-aI$ contains all the basic elements $\delta_n$ in $c_0.$ Therefore it is dense.
