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 $||Ux||^2=<Ux,Ux>=<U*Ux,x>=<x,x>=||x||^2$ 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 it 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 invertable 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?

  • 1
    $\begingroup$ It's true in general that $\Vert Ax \Vert = \Vert A^* x \Vert = \Vert x \Vert$ for $A$ a unitary operator on a given Hilbert space $H$. Even the converse is true! Maybe the formula $(U-\lambda I)^* =(U^* - \lambda ^* I)$ may help, but otherwise it's seems to be fine. $\endgroup$ – noctusraid Jan 26 '16 at 12:25

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\}. $

  • $\begingroup$ Why does $|\frac{1}{\lambda}| > 1$ imply that $(U^{-1}-\frac{1}{\lambda})$ is invertible? $\endgroup$ – Dan Mar 26 at 10:28

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