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Lecture03

# Lecture03 - Lecture 3 Recall that if A = cfw T(x x X and x...

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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 k k x k (because sup A B ). Theorem 1. L ( X, Y ) is a vector space upon which the operator norm is a norm. 1

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Proof. Note that addition and scalar multiplication are pointwise: ( αA )( x ) = αA ( 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 k k x k + k C k k 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 . Exercise : View the matrix T = a b c d as a linear map from R 2 to R 2 ; then T x y = ax + by cx + dy . Give R 2 the l 2 norm; then x y = p x 2 + y 2 . Calculate k T k .
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