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Unformatted text preview: Lecture 2 Exercise : If dim V > 1 then ‘the’ ‘ P norm arises from an I.P. precisely when p = 2. Proof. ( ⇒ ) Since dim V > 1, there are at least two vectors v 1 , v 2 in the basis used to define the norm. If the ‘ p nrom comes from an I.P. then the parallelogram law must hold: k v 1 + v 2 k 2 p + k v 1 v 2 k 2 p = 2( k v 1 k 2 p + k v 2 k 2 p ) ((1 p + 1 p ) 1 p ) 2 ((1 p + 1 p ) 1 p ) 2 1 2 1 2 So 2 · 2 1 p = 2 · 2 and hence 2 p = 1 or p = 2. ( ⇐ ) This was seen before. Exercise : If V is infinitedimensional, then it is not complete for the ‘ ∞ norm associated to a basis. Proof. Since the dimension of V is infinite, we can find a countable (sub)collection ( v m ) ∞ m =1 of basis vectors used to define the norm. Define a new sequence ( w m ) ∞ m =1 by setting w m = ∑ m k =1 1 k v k . Then for n > m we have: k w n w m k ∞ = k 1 n v n + 1 n 1 v n 1 + ··· + 1 m + 1 v m +1 k ∞ Thus d ∞ ( w n ,w m ) = max { 1 n , 1 n 1 ,..., 1 m +1 } = 1 m +1 ≤ if n > m > 1 and so ( w m ) ∞ m =1 is Cauchy. Informally, if ( w m ) ∞ m =1 converges then w m → ∞ X k =1 1 k v k / ∈ V (since every vector is a finite linear combination of the...
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 Summer '10
 Staff
 Linear Algebra, Algebra, Vectors, Vector Space, 1 M, Cauchy, normed, cm vm

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