4-Hilbert - MATHEMATICS 116, FALL 2007 CONVEXITY AND...

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Unformatted text preview: MATHEMATICS 116, FALL 2007 CONVEXITY AND OPTIMIZATION WITH APPLICATIONS Outline #4 (Hilbert Space) Last modified: October 22, 2007 Reading. Luenberger, Chapter 3, omitting any consideration of complex vector spaces. Lecture topics. 1. Axioms for an inner product in a pre-Hilbert space List four axioms for an inner product on a real vector space. These all generalize properties of the dot product on R n . Given vectors x and y in a pre-Hilbert space, we can define an l 2 norm from the inner product and convert both vectors to unit vectors. By considering the norm of the difference of these unit vectors, prove the Cauchy-Schwarz inequality. 1 2. Properties of the inner product Prove the triangle inequality for an arbitrary pre-Hilbert space, without using the Minkowski inequality. Prove the parallelogram law in R 2 using only the law of cosines. Prove the parallelogram law for an arbitrary pre-Hilbert space. Continuity: prove that if x n x and y n y , then ( x n | y n ) ( x | y ). 2 3. Projection theorem version 1 This version requires only a pre-Hilbert space (not necessarily complete) but it does not prove existence. x is a vector in a pre-Hilbert space X , and M is a subspace of X . Vector m is the closest vector in M to x . To show that x- m is orthogonal to M , assume the existence of a unit vector m M that is not orthogonal to x- m , and show that you could construct a vector m 1 that is closer to x than m is. A diagram in the plane will suggest exactly how to to the proof. Show the the vector m for which x- m is orthogonal to M is unique. Just consider x- m and show that || x- m || 2 > || x- m || 2 . 3 4. Projection theorem, version 2 Same hypotheses before, except that now we assume that X is complete: not merely a pre-Hilbert space, but a Banach space also, and hence a Hilbert space. Using the completeness of X , prove the existence of m . The trick is to consider a sequence of vectors for which || x- m i || converges to the minimum distance between x and closed subspace M , then show that { m i } is a Cauchy sequence and take its limit. 4 5. Orthogonal complements At first the definition in Luenberger looks wrong. Let S be a subset (not subspace) of Hilbert space X and define its orthogonal complement S to be the set of all vectors orthogonal to S . If S consists of the single vector (1,1,1) in R 3 what is S ? S ? S ? Prove the following rather obvious properties. S is a closed subspace of X . S S . S = S . If S T then T S . 5 6. Orthogonal complement of a closed subspace First a definition: Vector space X is a direct sum, X = M N , if every x X can be written uniquely in the form x = m + n with m M , n N ....
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4-Hilbert - MATHEMATICS 116, FALL 2007 CONVEXITY AND...

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