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The Berger-Shaw Theorem for multi-rationally cyclic hyponormal operators gives an estimate on the trace of the self-commutator of the operator. If S is a rationally cyclic subnormal (or hyponormal) operator then the Berger-Shaw theorem says that the $trace[S*,S] \leq 1/pi Area[\sigma(S)]$. That is, the trace is at most (1/pi) times the Area of the spectrum of the operator.

In this paper we give an exact computation of the trace of the self-commutator when S is a cyclic subnormal operator. More precisely, if S is a cyclic subnormal operator then $trace[S*,S] = (1/pi) Area[\sigma(S) - \sigma_e(S)]$. Thus, the trace is equal to (1/pi) times the area of the spectrum minus the essential spectrum of S. The set $G = \sigma(S) - \sigma_e(S)$ is the set of analytic bounded point evaluations for S. If f is a bounded analytic function on G then we also compute the trace of the self-commutator of f(S) as the Dirichlet integral of f over the set G.

We also give some partial results for rationally cyclic subnormal operators. We also give some examples showing that things are more complicated for rationally cyclic subnormal operators. We show that there exists a pure rationally cyclic subnormal operator S such that the trace of the self-commutator is strictly between the two values (1/pi) Area[\sigma(S) - \sigma_e(S)] and (1/pi) Area[\sigma(S)].

Another natural question is answered in the negative. Let R(K) denote the uniform closure of the rational functions with poles off K. It is well known that if R(\sigma(S)) = C(\sigma(S)), then S is a normal operator. Likewise, Conway & Feldman proved that if R(\sigma_e(S)) = C(\sigma_e(S)), then S is an essentially normal operator. It is natural to ask that if R(\sigma_e(S)) = C(\sigma_e(S)) and the index function ind(S - z) is integrable with respect to area meassure on \sigma(S) - \sigma_e(S), then does S have trace class self-commutator? The answer is yes, if Area[\sigma_e(S)] = 0. Our question is a natural improvement over this. In fact, an example is given in this paper showing that the answer to this question is NO!