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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 leftinverse 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 leftinverses 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....
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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.
 Spring '10
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