# What is the spectrum of the sequence operator $B: (x_1,x_2,\ldots) \rightarrow (0,x_1,\frac{1}{2}x_2,\ldots,\frac{1}{n}x_n,\ldots)$?

The question is stated in the title, and the operator is defined on $\ell^2$. I have determined that $||B|| = 1$, and therefore $\sigma(B) \subset \{\lambda \in \mathbb{C} : \, |\lambda| \le 1 \}$. Furthermore, I know that the point spectrum of $B$ is empty. Also, the adjoint is defined by $B^*:(x_1,x_2,\ldots) \rightarrow (x_2,\frac{1}{2}x_3,\ldots,\frac{1}{n}x_{n+1},\ldots)$, and since $0$ is an (the only, I think) eigenvalue of $B^*$, $0 \in \sigma(B)$ as well. How do I determine whether $\lambda \in \sigma(B)$ for any $\lambda$ such that $0 < |\lambda| \le 1$?

Note that $\|B^n\| \le 1/n!$. From this you can show that the spectral radius of $B$ is $0$.

• You're right, I hadn't thought about that, thanks! For anyone else reading this, the theorem being referred to is that $r(B) = \{\text{the spectral radius of B} \} = \lim_{n \rightarrow \infty} ||B^n||^{1/n}$. – user218389 Jul 9 '15 at 22:28