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A Physical Interpretation of Tight Frames

# A Physical Interpretation of Tight Frames - A Physical...

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A Physical Interpretation of Tight Frames P. G. Casazza 1 , M. Fickus 2 , J. Kovaˇ cevi´ c 3 , M. T. Leon 4 , and J. C. Tremain 5 1 Mathematics Department, University of Missouri, Columbia, Missouri 65211 USA, [email protected] 2 Department of Mathematics and Statistics, Air Force Institute of Technology, Wright-Patterson AFB, Ohio 45433 USA, [email protected] 3 Department of Biomedical Engineering, Carnegie Mellon University, Pittsburgh, Pennsylvania 15213, USA, [email protected] 4 Mathematics Department, University of Missouri, Columbia, Missouri 65211 USA, [email protected] 5 Mathematics Department, University of Missouri, Columbia, Missouri 65211 USA, [email protected] Summary. We find finite tight frames when the lengths of the frame elements are predetermined. In particular, we derive a “fundamental inequality” which completely characterizes those sequences which arise as the lengths of a tight frame’s elements. Furthermore, using concepts from classical physics, we show that this characteriza- tion has an intuitive physical interpretation. 1 Introduction Let H N be a finite N -dimensional Hilbert space. A finite sequence { f m } M m =1 of vectors is A -tight for H N if there exists A 0 such that, A f 2 = M m =1 | f, f m | 2 , for all f H N . An A -tight frame is an A -tight sequence for which A > 0. By polarization, { f m } M m =1 is A -tight for H N if and only if, Af = M m =1 f, f m f m , for all f H N . Clearly, any orthonormal basis is a 1-tight frame. However, the converse is false. For example, the vertices of a tetrahedron, appropriately centered and scaled, form a 1-tight frame of four elements for R 3 . Moreover, while the elements of an orthonormal basis are of unit length a priori , there are no explicit assumptions made about the lengths of a tight frame’s elements. This raises the question,

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2 P. G. Casazza, M. Fickus, J. Kovaˇ cevi´ c, M. T. Leon, and J. C. Tremain Given positive integers M and N , for what sequences of nonnegative numbers { a m } M m =1 do there exist tight frames { f m } M m =1 for H N , such that f m = a m for all m ? The answer to this question is the subject of this chapter. To be more precise, we first derive a necessary condition upon the lengths. Proposition 1. If { f m } M m =1 is A -tight for H N then, max m =1 ,...,M f m 2 A = 1 N M m =1 f m 2 . (1) Proof. First note that for any m = 1 , . . . , M , f m 4 = | f m , f m | 2 M m =1 | f m , f m | 2 = A f m 2 . Thus, f m 2 A for all m = 1 , . . . , M , yielding the inequality in (1). For the equality, let { e n } N n =1 be an orthonormal basis for H N . By Parseval’s identity, 1 N M m =1 f m 2 = 1 N M m =1 N n =1 | f m , e n | 2 = 1 N N n =1 A e n 2 = A. Thus, the lengths { a m } M m =1 of a tight frame of M elements for an N - dimensional space must satisfy the fundamental inequality , max m =1 ,...,M a 2 m 1 N M m =1 a 2 m . (2) Remarkably, this easily-found necessary condition will prove suﬃcient as well. That is, we shall show that for any sequence { a m } M m =1 which satisfies (2), and for any N -dimensional Hilbert space H N , there exists a tight frame { f m } M m =1 for H N for which f m = a m for all m .
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