HW103 - k n kk x k so that (as k n k is nonzero) n 6 k x k...

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Analysis 1 Homework 03 1. Let X be the vector space of all scalar sequences (with termwise operations) and for n N let φ n : X F be the linear functional defined by φ n ( x ) = x n whenever x X . Prove that there is no norm on X relative to which φ n is continuous for each n N ; in fact, prove that there is no norm on X relative to which φ n is continuous for infinitely many n N . Note : The vector space X does have norms; how? Solution 1. Let F N . Suppose that k • k is a norm on X relative to which φ n is continuous for each n F . Define x X by requiring that x n = n k φ n k when n F and (say - the choice is arbitrary) x n = 0 when n / F ; note that here k φ n k denotes the operator norm of φ n relative to the supposed norm on X . Now, if n F then n k φ n k = φ n ( x ) 6
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Unformatted text preview: k n kk x k so that (as k n k is nonzero) n 6 k x k and k x k is an upper bound for F . Recalling (Archimedes) that the only subsets of N that are bounded above are nite, it follows that F is nite. Note : As noted in class, the existence of norms on X (indeed, on any real/complex vector space) follows easily from Zorns Lemma. Remark : As remarked in class, if F N is nite then X does indeed carry a norm kk F relative to which n is continuous whenever n F : simply let k k be any norm on X and for x X dene k x k F = k x k + X n F | x n | . 1...
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This note was uploaded on 07/08/2011 for the course MHF 3202 taught by Professor Larson during the Spring '09 term at University of Florida.

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