Spectrum bilateral shift

Question

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

Answer 1

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.

Answer 2

You can get a trivial proof if you understand that the bilateral shift can be seen as the operator of multiplication by the identity on $L^2(\mathbb T)$: $$ M_zf(z)=zf(z). $$ It is an easy exercise that the spectrum of a multiplication operator by a function $g$ is the closure of the range of $g$, which is $\mathbb T$ in this case.

As for the proof linked in the question, the argument shows that $\lambda I-U$ is not bounded below, which prevents it from being invertible. As $\lambda$ in the argument is any element of the unit circle, the argument shows that $\mathbb T\subset \sigma(U)$.

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As a comment, there's no need for functional calculus to establish that $\sigma(V)\subset\mathbb T$ for any unitary $V$. Indeed, if $|\lambda|<1$, then $$ \|I-(I-\lambda V^*)\|=\|\lambda V^*\|≤|\lambda|<1, $$ so $I-\lambda V^*$ is invertible. Then $V-\lambda I=V(I-\lambda V^*)$ is invertible.

And when $|\lambda|>1$ we write $$V-\lambda I=-\lambda\Bigl(I-\frac1\lambda\,V\Bigr)$$ and the above applies to show that $V-\lambda I$ is invertible.