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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 $0 < |\lambda| \le 1$?

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

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

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