{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

4-Hilbert

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

This preview shows pages 1–7. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 δ 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 ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 22

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

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online