Theoretical Computer Science Adjoints, absolute values and polar decompositions.pdf

CDMTCS Research Report Series Adjoints, Absolute Values and Polar Decompostions Douglas Bridges Department of Mathematics University of Waikato Fred Richman Department of Mathematics Florida Atlantic University Peter Schuster Mathematics Institut der Universitat Munchen, Germany CDMTCS-069 November 1997 Centre for Discrete Mathematics and Theoretical Computer Science Adjoints, absolute values and polar decompositions Douglas Bridges, Fred Richman, Peter Schuster November 24, 1997 Abstract. Various questions about adjoints, absolute values and polar decompositions of operators are addressed from a constructive point of view. The focus is on bilinear forms. Conditions are given for the existence of an adjoint, and a general notion of a polar decomposition is developed. The Riesz representation theorem is proved without countable choice. 1. Introduction. Let H be an inner product space over the real or complex numbers. An operator is an everywhere defined linear transformation from H to H. If H is finite-dimensional, then every bounded operator has an adjoint, a theorem that can be proved in general using the Law of Excluded Middle. In the constructive framework of this paper, however, it cannot be shown that every bounded operator on an infinite-dimensional Hilbert space has an adjoint. In order to explain this by means of a Brouwerian example, we need a lemma whose proof is a straightforward application of the Cauchy-Schwarz inequality. Lemma 1. Let (αn), (βn) be sequences of complex numbers such that ∑∞ n=1 |αn|2 converges and ∑∞ n=1 |βn|2 is bounded. Then ∑∞ n=1 |αnβn| converges. 2 A Brouwerian example Q: Let (an) be a binary sequence with a1 = 0 and at most one term equal to 1, and let (en) be an orthonormal basis of an infinite-dimensional Hilbert space. Lemma 1 enables us to define a bounded operator Q such that Qen = ane1 for each n. That is Qx = ( ∞∑ n=1 anxn ) e1. If Q has an adjoint, then either ‖Q?e1‖ 0 or else ‖Q?e1‖ 1. Since 〈Q?e1, en〉 = 〈e1, Qen〉 = a