Spectrum bilateral shift Let $U \in \mathbb{B}(\ell^2(\mathbb{Z}))$ be the bilateral shift. I want to show that $\sigma(U)=\mathbb{T}$. Using functional Calculus I have shown that $\sigma(U)\subseteq\mathbb{T}$. In order to prove $\mathbb{T} \subseteq \sigma(U)$ I have googled a bit and found this proof, but I do not understand the argument the author is making. By showing that the norm converges to zero, is he not showing that $\lambda$ is an eigenvalue, which it is not.
Thanks

What is being shown is that $U-\lambda I$ cannot have a bounded inverse for $\lambda\in\mathbb{T}$, even though $\mathcal{N}(U-\lambda I)=\{0\}$. If $U-\lambda I$ were to have a bounded inverse, then there would exist $m > 0$ such that $\|(U-\lambda I)x\| \ge m\|x\|$ for all $x$. By showing that there is a sequence of unit vectors $\{\varphi_{n}\}$ such that $\lim_{n}\|(U-\lambda I)\varphi_{n}\|=0$, the existence of $m > 0$ is denied.
