245B_W09_hwk2

245B_W09_hwk2 - 245B, Winter 2009, Assignment 2: Notes and...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 245B, Winter 2009, Assignment 2: Notes and selected model answers (Model solutions follow question numbers in bold .) Folland Chapter 5 7. a. We are given T ∈ L ( X ) such that k I- T k < 1 . Let R n := ∑ n n =0 ( I- T ) n for n ≥ 1 . This is a Cauchy sequence in L ( X ) , since for any m > n we have k R n- R m k = m X i = n +1 ( I- T ) i ≤ m X i = n +1 k I- T k i ≤ X i ≥ n +1 k I- T k i = k I- T k n +1 1- k I- T k → as n → ∞ . By the completeness of L ( X ) (proved, for example, in Folland’s Proposition 5.4) we have R n → R in operator norm for some R ∈ L ( X ) . By the continuity of multiplication in the operator norm (an immediate consequence of its submultiplicativity), this implies TR n → TR and R n T → RT . However, k I- TR n k = k I- R n + ( I- T ) R n k = I- ( I + ( I- T ) + ··· + ( I- T ) n ) + ( ( I- T ) + ( I- T ) 2 + ··· + ( I- T ) n +1 ) = k ( I- T ) n +1 k ≤ k I- T k n +1 → , so TR = lim n TR n = I , and exactly similarly RT = I , so R is a (bounded) right- and left-inverse for T , and so is a bounded inverse. b. Now suppose that T is invertible in L ( X ) and k T- S k ≤ k T- 1 k- 1 . Then by the submultiplicativity of the norm we have also k I- ST- 1 k = k ( S- T ) T- 1 k ≤ k S- T kk T- 1 k < 1 , so by part (a) ST- 1 has a bounded inverse, say R 1 . Exactly similarly, T- 1 S has a bounded inverse, say R 2 , and so we have ( ST- 1 ) R 1 = S ( T- 1 R 1 ) = I and 1 R 2 ( T- 1 S ) = ( R 2 T- 1 ) S = I . Therefore we have exhibited both bounded right- and left-inverses for S , so these must agree and S is invertible. This tells us that for any T in the the subset of invertible operators in L ( X ) , its neighbourhood { S : k S- T k < k T- 1 k- 1 } lies in that subset, so the subset is open....
View Full Document

This note was uploaded on 04/06/2010 for the course MATH various taught by Professor Tao/analysis during the Spring '10 term at UCLA.

Page1 / 6

245B_W09_hwk2 - 245B, Winter 2009, Assignment 2: Notes and...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online