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Wiki says " A bounded linear operator $U: H \to H$ on a Hilbert space $H$ is called a unitary operator if it satisfies $U^{*}U=UU^{*}=I$ , where $U^{*}$ is the adjoint of $U$, $I$ is the identity operator "

  1. Isn't any unitary operator $U$ bounded and its norm $||U||=1$ since $||Ux||^{2}=\langle Ux,Ux\rangle=\langle x,U^{*}Ux\rangle=||x||^{2}$ ?

  2. Isn't $U-\lambda I$ invertible since U and $\lambda I$ are invertible for any $\lambda \in \mathbb{C}$?

Can any one give an example of unbounded unitary operator? Thanks!

Davide Giraudo
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Ergin Süer
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    As you note correctly, every unitary operator is isometric, hence bounded with norm $1$. But your second point is false, because the difference of invertible operators is in general not invertible. E.g. $U = \rm{id}$ is unitary, but $U - \lambda I$ is not invertible (in fact $\equiv 0$) for $\lambda = 1$. But if $U - \lambda I$ is not invertible, then $|\lambda| = 1$, i.e. the spectrum lies on the unit circle. – PhoemueX Jun 21 '14 at 15:39
  • While you can define the notion of "adjoint" for unbounded operators, this is far more involved than simply asking for $U$ to be bounded (which you then get for free once you have a notion of adjoint and the relaton$U^U=1=UU^$). – s.harp Sep 19 '20 at 09:15

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  1. Here you are absolutely correct, and in fact that is exactly how you would prove both statements you made. To be verbose, the definition of a bounded operator gives that $\|Ux\| \leq C\|x\|$ for some constant $C$, and the norm of an operator is defined as $\|U\| = \sup_{x \in H}\{\|Ux\|:\|x\|=1\}$ or equivalently $\|U\| = \sup_{x \in H}\left\{\frac{\|Ux\|}{\|x\|}:x\neq 0\right\}$. To reiterate your proof with a slight modification: $$\|Ux\|=\sqrt{\langle Ux,Ux \rangle}=\sqrt{\langle U^*Ux,x \rangle}=\sqrt{\langle Ix,x \rangle}=\sqrt{\langle x,x \rangle}=\|x\|$$ Clearly, $\|Ux\| \leq C\|x\|$ is a true statement for $C=1$ since $\|Ux\|=\|x\|$, and thus $U$ is bounded. Likewise, it should be clear the norm of $U$ is $1$, and we can use the definition to further illustrate this as such: $$\|Ux\|=\|x\| \Rightarrow \frac{\|Ux\|}{\|x\|}=1 \Rightarrow \sup_{x \in H}\left\{\frac{\|Ux\|}{\|x\|}:x \neq 0\right\}=1 \Rightarrow \|U\|=1$$
  2. This point, however, is not necessarily true. The operation of taking an inverse is not distributive across addition or subtraction, so $(U- \lambda I)^{-1} \neq \left(U^{-1}- \frac{1}{\lambda}I\right)$ and as such $U- \lambda I$ is not inherently invertible simply because $U$ and all $\lambda I$ are invertible. If this were the case, the spectrum of $U$ would be the empty set; however, it is a fact that iff a linear operator on a Hilbert space is unitary then all of the eigenvalues that appear in its spectral decomposition are located on the unit circle of a complex plane. A proof for the forward direction of this can be found here. The reverse direction is trivial.
Mohsen Shahriari
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Joseph L.
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