If is a Banach space and is an isometry , then either or
where
= {z in C: |z| = 1}
Case 1: isometry is bijective
When all points of the specrum are on the boundary
Case 2: the isometry is not surjective.
the spectrum is the whole disk.
If is a Banach space and is an isometry , then either or
where
= {z in C: |z| = 1}
Case 1: isometry is bijective
When all points of the specrum are on the boundary
Case 2: the isometry is not surjective.
the spectrum is the whole disk.
LaTeX note: use \mathcal, not \cal, in formulas (\cal works globally, and converts the rest of the formula to gibberish).
Use the theorem which says that the boundary of the spectrum is contained in the approximate point spectrum. Suppose that is an approximate eigenvalue for T, with approximate eigenvector x. Then x is a unit vector, and . But , and it follows that .
Thus . It then follows on purely geometrical grounds that either or . The first alternative occurs when (in other words, when T is unitary), and the second alternative in the case when (in other words, when T is a proper isometry).