Definitizing polynomials of unitary and Hermitian operators in Pontrjagin spaces
Let B be a densely defined Hermitian or a unitary operator in a \(\pi_ k\)-space \(\Pi_ k\). A complex-valued function f, defined on some subset of the complex plane is called definitizing for B if f(B) can be defined in a suitable way on a dense domain D(f(B)) and the relation \([f(B)x,x]\geq 0\) holds for all \(x\in D(f(B)).\) In the present paper, if A is a Hermitian operator, we are only interested in definitizing functions which are polynomials of the form \(f=\bar pp\) with a polynomial p, \(\bar p\) being the polynomial \(p(z)=\overline{p(\bar z)}\). Similarly, if U is a unitary operator in \(\Pi_ k\), the definitizing functions of U we are interested in are of the form \(f=p^*p\) for some polynomial p, where \(p^*(z)=\overline{p(1/\bar z)}.\) In Section 2 we give a new proof for the (known) fact that every unitary operator in \(\Pi_ k\) has a unique (up to scalar multiples) definitizing polynomial of minimal degree. In Section 3 it is proved that also for an essentially selfadjoint operator A, satisfying some additional conditions, the definitizing polynomial \(f_ A\) of minimal degree is uniquely determined. However, in Section 5 we show that A can have irreducible definizing polynomials different from \(f_ A\). Here a definitizing polynomial \(f=\bar pp\) of A is called irreducible if p cannot be written as \(p=p_ 1p_ 2\), such that \(p_ 1\), \(p_ 2\) are nonconstant polynomials and \(f_ 1=\bar p_ 1p_ 1\) is definitizing for A. A Hermitian operator A, which is not essentially selfadjoint, has in general infinitely many irreducible definitizing polynomials. These definitizing polynomials are considered in Section 4. In Section 6 we apply the results of Sections 3-5 to the moment problem for sequences with k negative squares.
scientific article; zbMATH DE number 4136812