Projection Theorem 1


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1 Projection Theorem 1
2 CauchySchwarz Inequality Lemma. (CauchySchwarz Inequality) For all x, y in an inner product space, [ xy, ] x y. Equality holds if and only if x y or y θ. Proof. If y θ, the inequality holds trivially. Therefore, assume y θ. For all scalars, we have [ xy, xy] [ x, x] [ y, x] [ x, y] [ y, y] In particular, for [ xy, ]/[ yy, ], we have [, ] [ xx, ] xy, [ yy, ] or [ xy, ] [ xx, ][ yy, ] x y.
3 CauchySchwarz Inequality Proposition. On a prehilbert space X the function x [ x, x] is a norm. Proof. The only requirement for a norm which has not already been established is the triangle inequality. For any x, y X, we have xy [ xy, xy] [ x, x] [ x, y] [ y, x] [ y, y] x [ x, y] y. By the CauchySchwarz inequality, this becomes xy x x y y x y. The square root of the above inequality is the desired result. 3
4 Parallelogram Law Lemma. (The Parallelogram Law) In a prehilbert space xy xy x y. Proof. The proof is made by direct expansion of the norms in terms of the inner product. 4
5 Proof. Continuity of the Inner Product Lemma. (Continuity of the Inner Product) Suppose that xn x and yn y in a prehilbert space. Then x, y x, y. Since the sequence is convergent, it is bounded; say M. Now x n Applying the CauchySchwarz inequality, we obtain Since is bounded, x n x y x y x y x y x yx y x y y x x y,,,,,,,,. n n n n n n n n n x n x, y x, y x y y x x y. n n n n n x, y x, y M y y x x y. n n n n n n 5
6 Projection Theorem The concept of orthogonality has many of the consequences in prehilbert spaces that it has in plane geometry. For example, the Pythagorean theorem is true in prehilbert spaces. Lemma. If x y, then xy x y. Proof. xy x+y, x+y x x, y y, x y x y. 6
7 Projection Theorem The optimization problem considered is this: Given a vector x in a pre Hilbert space X and a subspace M in X, find the vector mm closest to x in the sense that it minimizes xm. Of course, if x itself lies in M, the solution is trivial. In general, however, three important questions must be answered for a complete solution to the problem. First, is there a vector mm which minimizes xm, or is there no m that is at least as good as all others? Second, is the solution unique? And third, what is the solution or how is it characterized? We answer these questions now. 7
8 Projection Theorem Theorem 1. Let X be a prehilbert space, M a subspace of X, and x an arbitrary vector in X. If there is a vector m M such that xm xm for all mm, then m is unique. A necessary and sufficient condition that m M be a unique minimizing vector in M is that the error vector xm be orthogonal to M. Proof. We show first that if is a minimizing vector, then xm is m orthogonal to M. Suppose to the contrary that there is an mm which is not orthogonal to xm. Without loss of generality, we may assume that m 1 and that x m,m. Define the vector m1 in M as m 1 =m δm. 8
9 Then Projection Theorem xm1 xmδm xm xm, δm δm, xm δ. xm δ xm Thus, if x m is not orthogonal to M, m is not a minimizing vector. We show now that if x m is orthogonal to M, then m is a unique minimizing vector. For any mm, the Pythagorean theorem gives. xm xm m m xm m m Thus, xm xm for m m. 9
10 Classical Projection Theorem Theorem. (The Classical Projection Theorem) Let H be a Hilbert space and M a closed subspace of H. Corresponding to any vector xh, there is a unique vector m M such that x m xm for all mm. Furthermore, a necessary and sufficient condition that m M be the unique minimizing vector is that x m be orthogonal to M. Proof. The uniqueness and orthogonality have been established in Theorem 1. It is only required to establish the existence of the minimizing vector. 1
11 Classical Projection Theorem If xm, then m x and everything is settled. Let us assume xm and define δ inf xm. We wish to produce an m M with x m δ. mm For this purpose, let { m } be a sequence of vectors in M such that xm δ. Now, by the parallelogram law, i i ( m x) ( xm ) ( m x) ( xm ) m x xm j i j i j i. Rearranging, we obtain m m m 4 i j j mi mj x x mi x. 11
12 Classical Projection Theorem mi mj For all i, j the vector is in M since is a linear subspace. M mi mj Therefore, by definition of δ, x δ and we obtain j i j i 4 m m m x x m δ. Since m x δ as i, we conclude that i m m as i, j. j i Therefore, { m } is a Cauchy sequence, and since M is a closed subspace i of a complete space, the sequence { } has a limit m in M. By m i continuity of the norm, it follows that x m δ. 1
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