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**Nusc** If $\displaystyle \cal{X} $ is a Banach space and $\displaystyle T: \cal{X} \rightarrow \cal{X} $ is an isometry , then either $\displaystyle \sigma(T) \subseteq \partial \cal{D}$ or $\displaystyle \sigma(T) = cl \cal{D} $

where

$\displaystyle \partial\cal{D} $ = {z in C: |z| = 1}

Case 1: isometry is bijective

When $\displaystyle \sigma(T) \subseteq \partial \cal{D} $ all points of the specrum are on the boundary

Case 2: the isometry is not surjective.

$\displaystyle \sigma(T) = cl \cal{D} $

the spectrum is the whole disk.