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

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

2005-11-15

arxiv.org/abs/math/0511370v1

Mathematics

Spectral Theory

5 pages

Scientific paper

We derive necessary and sufficient conditions for a one-dimensional periodic
Schr\"odinger (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.

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