CHAPTER 1 Preliminaries We begin with a number of short topics. Since the first two are fairly standard, we will not give all details. The last topic is rather specialized and, in its discussion, we will use results from Chapters 2 and 3. 1.1. Absolute Value and Polar Decomposition Throughout, all our Hilbert spaces will be complex and separable (are there any others?) and our inner product is linear in the second factor and conjugate linear in the first. Recall that a bounded operator, A, is called positive (written A 0) if and only if (0, A(f) 0 for all / it then follows by polarization that A* = A. If A — B 0, we write A B or B A. It can be proven that for any A 0, there is a unique B 0 with B2 = A (see [250, Section VI.4]) we write B = VI. For any bounded operator, A, the operator A*A is positive and one defines We warn the reader that |^ + S | | ^ | + |B| (1.1) is false in general. For example, if - C O -(!-!) then So (0, \A + J3|£) (/,(\A\ + \B\)/) for 0 = (J), showing that (1.1) does not hold. Notice that |||^|0|| 2 = ||A^||2. From this, it is easy (see [250, Section VI.4]) to construct uniquely an operator U so that (i) A = U\A\ (ii) \\Uil\\ = U\\ for V G Ran|A| = (ker A)^ (in) \\U^\\ = 0 for ^ G (Ran^l)- 1 - ker A Notice that \A\ = U*A. The formula, A = U\A\, is called the polar decomposition of A. 1.2. Compact Operators and the Canonical Decomposition We call a bounded operator A on a Hilbert space finite rank if dim(Ran A) oc (the dimension of Ran^l is called the rank of A). A bounded operator, A, is called compact if and only if it is a norm limit of finite rank operators. (For a Hilbert space, but not a general Banach space, this is equivalent to the more usual Riesz definition that A[{(j) \ \\cj)\\ 1}] has compact closure.) The following two results are standard: l http://dx.doi.org/10.1090/surv/120/01

Purchased from American Mathematical Society for the exclusive use of nofirst nolast (email unknown) Copyright 2005 American Mathematical Society. Duplication prohibited. Please report unauthorized use to cust-serv@ams.org. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.