This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Lecture 12 3. Banach spaces Definition. Let K be one of the fields R or C . A Banach space over K is a normed Kvector space ( X , k . k ), which is complete with respect to the metric d ( x,y ) = k x y k , x,y ∈ X . Example 3.1. The field K , equipped with the absolute value norm, is a Banach space. More generally, the vector space K n , equipped with any of the norms k ( λ 1 ,...,λ n ) k ∞ = max { λ 1  ,...,  λ n } , k ( λ 1 ,...,λ n ) k p =  λ 1  p + ··· +  λ n  p 1 /p , p ≥ 1 , is a Banach space. Remark 3.1. Using the facts from the general theory of metric spaces, we know that for a normed vector space ( X , k . k ), the following are equivalent: (i) X is a Banach space; (ii) given any sequence ( x n ) n ≥ 1 ⊂ X with ∑ ∞ n =1 k x n k < ∞ , the sequence ( y n ) n ≥ 1 of partial sums, defined by y n = ∑ n k =1 x k , is convergent; (iii) every Cauchy sequence in X has a convergent subsequence. This is pretty obvious, since the sequence of partial sums has the property that d ( y n +1 ,y n ) = k y n +1 y n k = k x n +1 k , ∀ n ≥ 1 . Exercise 1* . Let X be a finite dimensional normed vector space. Prove that X is a Banach space. Hints: Use inductionn on dim X . The case dim X = 1 is trivial. Assume the statement is true for all normed vector spaces of dimension d , and let us prove it for a normed vector space of dimension d +1. Fix such an X , and a linear basis { e 1 ,e 2 ,...,e n ,e d +1 } for X . Start with a Cauchy sequence ( x n ) n ≥ 1 ⊂ X . Write each term as x n = d +1 X k =1 α n ( k ) e k . Prove first that ( α n ( d + 1) ) n ≥ 1 ⊂ K is bounded . Then extract a subsequence ( x n p ) p ≥ 1 such that ( α n p ( d + 1) ) p ≥ 1 is convergent. If we take α ( d + 1) = lim p →∞ α n p ( d + 1), then prove that the sequence ( x n p α n p ( d +1) e d +1 ) p ≥ 1 is Cauchy in the space Span { e 1 ,...,e d } . Using the inductive hypothesis, conclude that ( x n p ) p ≥ 1 is convergent in X . Thus, every Cauchy sequence in X has a convergent subsequence , hence X is Banach. Exercise 2* . Let n ≥ 1 be an integer, and let k · k be a norm on K n . Prove that there exist constants C,D > 0, such that C k x k ∞ ≤ k x k ≤ D k x k ∞ , ∀ x ∈ K n . 79 80 LECTURE 12 Hint: Let e 1 ,...,e n be the standard basis vectors for K n , so that α 1 e 1 + ··· + α n e n = ( α 1 ,...,α n ) , ∀ ( α 1 ,...,α n ) ∈ K n . Define D = k e 1 k + ··· + k e n k . The existence of C is equivalent to the existence of some C > such that k x k ∞ ≤ C k x k , ∀ x ∈ K n . (If such a C exists, then we take C = 1 /C .) To prove the existence of C as above, we consider the set T = { x ∈ K n : k x k ≤ 1 } , and we need to prove that sup x ∈ T k x k ∞ < ∞ ....
View
Full Document
 Three '10
 Smith
 Vector Space, Metric space, Banach space, Banach

Click to edit the document details