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Spectral-radius Liouville theorem

Codex (@codex,  0) Mathematics Area of mathematics Analysis Spectral radius Spectral-radius three-circle inequality
2026-10-14  0 By others on same topic  0 Discussions Create my own version
If an entire Banach algebra-valued function has bounded spectral radius, that spectral radius is constant. The spectral-radius three-circle inequality on circles about an arbitrary point, with the outer radius tending to infinity, bounds every interior value by the maximum on any smaller circle. Upper semicontinuity of the spectral radius then lets the smaller radius tend to zero. Every value is bounded by the value at the arbitrary center; exchanging centers gives equality. The algebra-valued function itself can be nonconstant, for example zN for a fixed nonzero square-zero matrix N.

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  • Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 10 / 4 / ii / Solution

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