If $U$ is unitary operator then spectrum $\sigma(U)$ is contained inside the unit circle

In a Hilbert space, let $U$ be a continuous operator which it unitary. Prove $\sigma (U)\subseteq \Bbb{S}^1$.

It is important for me to know how I am doing, and I didn't come by a clear explanation so it would be appreciated to have your correcting and guiding.

Attempt: $U^{*}U=UU^{*}=I$ and therefore $U^{-1}=U$. I have $$ \| U x \|^2 = \langle U x, U x \rangle = \langle U^* U x, x \rangle = \langle x, x \rangle = \|x\|^2 $$ and the same goes for $U^*$ and therefore $\| U \| = \| U^{-1} \| = 1$.

Looking at $U-\lambda$ I can take $$ -\lambda U(U^{-1}-{1\over \lambda}I) =U-\lambda I. $$ Clearly, for $\lambda \ne 0$, $-\lambda U$ is defined and is a linear operator. $(U^{-1}-{1\over \lambda}I)$ is invertible if $|\lambda|\ge 1$ and therefore $\sigma (A^{-1})=\sigma (A)\subseteq \{|z|\le1\}$. But from the RHS, $U-\lambda I$ is invertible if $|\lambda|\le 1$ and therefore $\sigma(A)\subseteq \Bbb{S}^1$. I have noticed my claims tend to have holes in them many times which I cannot spot. If it is substantially true, how can I make sure it is completely formal?

Solution 1:

In your proof slight modification is needed.

We have, $UU^* =U^*U =I$. So $U$ is invertible with $U^{-1}= U^*$.

Also as you have already shown that $||U || = 1 = ||U^{-1}||$, we must have $$\sigma (U) \subset \{\lambda \in \mathbb{C} : |\lambda| \leq 1\}\;\; \text{and}\;\; \sigma (U^{-1}) \subset \{\lambda \in \mathbb{C} : |\lambda| \leq 1\}.$$

Now to show $\sigma (U) \subset \{\lambda \in \mathbb{C} : |\lambda| = 1\}, $ it is sufficient show that for $(U-\lambda I)$ is invertible for every $\lambda$ satisfying $|\lambda| <1$. As $U$ is already invertible, sufficient to show that $(U-\lambda I)$ is invertible for every $\lambda$ satisfying $0 <|\lambda| <1$.

Now from the relation $-\lambda U (U^{-1} - \frac{1}{\lambda}I) = (U- \lambda I)$, It is clear that, for $\lambda \neq 0$,

if $(U^{-1} - \frac{1}{\lambda}I)$ is invertible then $(U - \lambda I)$ becomes invertible.

Now note that If $0 <|\lambda| < 1,$ then $|\frac{1}{\lambda}| > 1$, then in that case $(U^{-1} - \frac{1}{\lambda}I)$ invertible and hence $(U- \lambda I)$ invertble. This gives us $\sigma (U) \subset \{\lambda \in \mathbb{C} : |\lambda| = 1\}. $