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

2026-07-072026-07-07http://salesiana.dossiersoluciones.com/handle/123456789/104926We 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.5 pagesSpectral TheoryMathematical Physics34B30; 47B40; 47A10; 34L05; 34L40When is a non-self-adjoint Hill operator a spectral operator of scalar type?text