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Let $U$ be a unitary operator on a Hilbert space $\mathcal{H}$. By the spectral theorem, it is known that $\sigma(U)\subseteq \{z\in \mathbb{C}:|z|=1\}$. How can the explicit inverse of $\lambda-U$ be calculated for $|\lambda|<1$ (or $|\lambda|>1$)?

Next let $R$ be the right shift operator on $l^{2}(\mathbb{N})$. Consider $[R]\in \frac{B(l^{2}(\mathbb{N}))}{K(l^{2}(\mathbb{N}))}$, where $K(l^{2}(\mathbb{N})$) is the set of compact operators on $l^{2}(\mathbb{N})$, and so the algebra considered is the Calkin algebra. $R$ is then unitary mod a compact operator, i.e., $[R]$ is unitary in the Calkin algebra. How can the explicit inverse of $\lambda-[R]$ be found for $|\lambda|<1$ (or $|\lambda>1$)?

Arundhathi
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    I do not think that you can have an explicit formula for the resolvent of $R$ in the Calkin algebra. The problem is that you cannot write explicitely the Hilbert space, where the Calkin algebra acts. On the other hand, you can consider the image of $R$ in the Toeplitz algebra modulo compact operators. It can be realized explicitely as algebra of multiplication operators on the unit circle. Thereby $[R]$ is multiplication by $z$. – Yurii Savchuk Apr 23 '15 at 18:43
  • What have you tried? For the explicit formula, have a look at this question and answer or this answer. – ViktorStein Mar 07 '20 at 00:24

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