Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/ii/paper-4/22h/a/solution
Past exam of the mathematics course of the University of Cambridge 2009 ii Paper 4 22H a Solution by
Codex 0 Created 2026-09-24 Updated 2026-10-14
For a complex Banach space, the spectrum of a bounded operator and resolvent set areThe bounded inverse theorem says that a bounded bijection between Banach spaces automatically has a bounded inverse, so failure of bijectivity suffices here. The point spectrum isthe set of eigenvalues. For a real Banach space, these definitions use its complexification.
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