When is a non-self-adjoint Hill operator a spectral operator of scalar type?

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dc.creatorGesztesy, Fritz
dc.creatorTkachenko, Vadim
dc.date2005-11-15
dc.date.accessioned2026-07-07T06:51:15Z
dc.date.available2026-07-07T06:51:15Z
dc.descriptionWe derive necessary and sufficient conditions for a one-dimensional periodic Schrödinger (i.e., Hill) operator H=-d^2/dx^2+V in L^2(R) to be a spectral operator of scalar type. The conditions demonstrate the remarkable fact that the property of a Hill operator being a spectral operator is independent of smoothness (or even analyticity) properties of the potential V.
dc.description5 pages
dc.identifierhttps://arxiv.org/abs/math/0511370
dc.identifierhttp://arxiv.org/abs/math/0511370
dc.identifier.urihttp://salesiana.dossiersoluciones.com/handle/123456789/104926
dc.subjectSpectral Theory
dc.subjectMathematical Physics
dc.subject34B30; 47B40; 47A10; 34L05; 34L40
dc.titleWhen is a non-self-adjoint Hill operator a spectral operator of scalar type?
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