On square roots of normal operators
Calvin R. Putnam · Proceedings of the American Mathematical Society · 1957
It is clear that if N possesses the spectral resolution N=fzdK(z), then any operator of the form A =fzl12dK(z), where, for the value of z1/2, the choice of the branch of the function may depend on z, is a solution of (1). Moreover, all such operators are even normal. Of course, equation (1) may have other, nonnormal, solutions A. The object of this note is to point out a simple condition to be satisfied by a square root A guaranteeing that it be normal. This criterion will involve the (closed, convex) set W= WA consisting of the closure of the set of values (Ax, x) where lxi| = 1. (Cf. also [2] wherein is discussed a connection between commutators and the set W.) The following theorem will be proved: (I) Let N be a fixed normal operator and let A denote an arbitrary solution of (1). Suppose that there exists a line L in the complex plane passing through the origin and lying entirely on one side of (and possible lying all, or partly, in) the set WA. Then A is necessarily normal. It is easy to see that the hypothesis of (I) concerning the line L is surely satisfied if W is a single point or a straight line segment. In this case, A is even the sum of multiples of a self-adjoint operator and the unit operator I. (In fact, there exists some angle C and some complex number z such that the set W belonging to ei6A +zI is a point or a segment of the real axis, and hence eiOA +zI is self-adjoint.) In case the set W is actually two-dimensional, the assumption amounts to supposing that 0 is not in the interior of W, although it is allowed of course that 0 be on the boundary.