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Unformatted text preview: Lecture 3 Recall that if A = {k T ( x ) k : x X and k x k 6 1 } and B = { M : x X = k T ( x ) k 6 M k x k} , we claim that sup A = inf B. We have already shown that sup A 6 inf B , so all that remains is to prove that sup A > inf B . Proof. Suppose sup A 6 B . Then there is an x X such that k T ( x ) k > (sup A ) k x k ; so x 6 = 0 and T x k x k > sup A. But k T ( x/ k x k ) k A , a contradiction; hence sup A B and thus sup A > inf B . Definition 1. The common value sup A = inf B in the preceding theorem is denoted k T k (the operator norm of T ). Notice that x X = k T ( x ) k 6 k T kk x k (because sup A B ). Theorem 1. L ( X,Y ) is a vector space upon which the operator norm is a norm. 1 Proof. Note that addition and scalar multiplication are pointwise: ( A )( x ) = A ( x ) ( B + C )( x ) = B ( x ) + C ( x ) . So a sample of the standard types of arguments needed is as follows: if B,C L ( X,Y ) and x X , then k ( B + C )( x ) k = k Bx + Cx k 6 k Bx k + k Cx k ( 4 inequality in Y ) 6 k B kk x k + k C kk x k = ( k B k + k C k ) k x k , therefore B + C L ( X,Y ) and k B + C k 6 k B k + k C k ....
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This note was uploaded on 07/08/2011 for the course MAT 6932 taught by Professor Staff during the Summer '10 term at University of Florida.
 Summer '10
 Staff
 Algebra

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