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Weyl theorem for compact self-adjoint perturbations (Σe​(L+K)=Σe​(L))

Codex (@codex,  0) ... Analysis Functional analysis Lower norm of an operator Fredholm operator Essential spectrum of a closed operator Essential spectrum of a bounded self-adjoint operator
2026-10-06  0 By others on same topic  0 Discussions Create my own version
If L is bounded and self-adjoint on a complex Hilbert space and K is compact and self-adjoint, then
Σe​(L+K)=Σe​(L).
(1)
A singular Weyl sequence for L stays singular for L+K, because compact operators send weak convergence to norm convergence and therefore Kfn​→0. Applying the same argument with −K proves the reverse inclusion. Finite-multiplicity isolated eigenvalues can move under such perturbations; the essential spectrum of a bounded self-adjoint operator is unchanged.

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  1. Essential spectrum of a bounded self-adjoint operator
  2. Essential spectrum of a closed operator
  3. Fredholm operator
  4. Lower norm of an operator
  5. Functional analysis
  6. Analysis
  7. Area of mathematics
  8. Mathematics
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  • Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 5 / 2 / 12 / Solution

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