Let $(\lambda_{n})$ be a bounded sequence of complex numbers, and, $T:\ell^{1}\rightarrow \ell^{1}$ a operator defined by $T(x)=(\lambda_{n}x_{n})$ for $x=(x_{n})\in\ell^{1}$. Find the spactrum of $T$, that is, $\sigma(T)$.
Remark Note that $(T-\lambda I)$ is injective for all $\lambda\neq \lambda_{n}$ and $T(e_{n})=\lambda_{n}e_{n}$ then the point spectrum $\sigma_{p}(T)=\{\lambda_{n}\}$.
For other hand, $T^{*}:\ell^{\infty}\rightarrow \ell^{\infty}$ is given by $T^{*}(y)=(\lambda_{n}y_{n})$ for $y=(y_{n})\in\ell^{\infty}$, then $\sigma_{p}(T^{*})=\{\lambda_{n}\}$. Therefore, we have $$\lambda\in\{\overline{\lambda}_{n}\} \Longleftrightarrow \overline{\lambda}\in\{\lambda_{n}\} \Longleftrightarrow \{0\}\neq N(T^{*}-\overline{\lambda} I)=R(T-\lambda I)^{\bot}.$$ Then, we have $\overline{R(T-\lambda I)}\neq E$. Therefore, the residual spectrum is $\sigma_{r}(T)=\{\overline{\lambda}_{n}\}$. Is this correct?
So, we can conclude $\sigma(T)=\{\lambda_{n}\}\cup\{\overline{\lambda}_{n}\}\cup\sigma_{c}(T).$
The problem: I could not calculate $\sigma_{c}(T)$, I think $\sigma_{c}(T)=\overline{\{\lambda_{n}\}}$ but I do not know how to show it.