Real parts of normal extensions of subnormal operators (Q1097493)

A bounded linear operator S on a separable Hilbert space H is said to be subnormal if S has a normal extension N, to a Hilbert space \(K\supset H\). As H is invariant under N, then \(H^{\perp}=KH\) is invariant under N *. The operator \(T=N\) \(*/H^{\perp}\) is called the dual of \(S=N/H\). In case S has no normal part then S is said to be a pure subnormal operator. N is called the minimal normal extension if the only reducing space of N which contains H is K. The following theorem is proved and two examples are discussed. Theorem 1. Let S be a pure subnormal operator on H with the minimal normal extension N on \(K\supset H\), and let T be the dual of S. Suppose that (*) \(D^{1/2}\) is of trace class, where S *S-SS \(*=D(\geq 0)\). Then Re(N), on K, has an absolutely continuous part, which, on the corresponding absolutely continuous subspace of K, is unitarily equivalent to Re(S)\(\oplus Re(T)\) on \(K=H\oplus H^{\perp}\). More generally, if a and b are real and a \(2+b\) \(2>0\) then a Re(N)\(+b Im(N)\) has an absolutely continuous part which is unitarily equivalent to [a Re(S)\(\oplus b Im(S)]\oplus [a Re(T)+b Im(T)].\) In the first example, it is shown that there exists a pure subnormal operator S, in fact, an analytic Toeplitz operator, having a self commutator D satisfying (*) and a minimal normal extension N for which Re(N) has both an absolutely continuous part and a purely singular continuous part. In the second example, a pure subnormal analytic Toeplitz operator S is given for which S *S-SS * is of trace class and for which Im(N) is purely singular, where as before, N is the minimal normal extension of S.