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

Question

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?

Answer 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\}. $

Answer 2

We can use the following fact:

If a map $S:X\rightarrow Y$ between complete metric spaces is a contraction, that is: $$d(S(x_1),S(x_2)) \leq c d(x_1,x_2) \quad \text{for $1>c\geq0$}$$ Then there is a unique fixed point $x_o$ such that $S(x_o)=x_o$.

This is for example, Theorem 3.2 here. You basically take any point and start iteratively computing $S$ in this point and you converge to a certain $x_o$ which is the fixed point.

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Because $\lVert U (x)\rVert=\lVert x\rVert$, then $\lVert U \rVert=1$.

Case I:

If $|\lambda|>1$, then we have that for a fixed $y$, there will be a $x$ that satisfies $Ux-\lambda x=y $ if and only if:

$$x=\frac{Ux}{\lambda}-y$$

However the map $S_y(x)=\frac{Ux}{\lambda}-y$ is a contraction with $c=\frac{1}{|\lambda|}<1$ so there will be a unique fixed point which solves our equation! Hence, $U-\lambda I$ is invertible!

Case II:

If $|\lambda|<1$, then we have that for a fixed $y$, there will be a $x$ that satisfies $Ux-\lambda x=y $ if and only if:

$$x=\lambda U^{-1} x+ U^{-1}y $$

However the map $S_y(x)=\lambda U^{-1} x+ U^{-1}y$ is a contraction with $c=|\lambda|<1$ so there will be a unique fixed point which solves our equation! Hence, $U-\lambda I$ is invertible!

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From cases I and II, we conclude that the only way for our operator not to be invertible is if $|\lambda|=1 $. In other words, we need $\sigma(U)\subseteq \{z\in \mathbb{C}\:| \: |z|=1\}$