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Unformatted text preview: p IS COMPLETE Let 1 p , and recall the definition of the metric space p : For 1 p < , p = ( sequences a = ( a n ) n =1 in R such that X n =1  a n  p < ) ; whereas consists of all those sequences a = ( a n ) n =1 such that sup n N  a n  < . We defined the pnorm as the function k k p : p [0 , ) , given by k a k p = X n =1  a n  p ! 1 /p , for 1 p < , and k a k = sup n N  a n  . In class, we showed that the function d p : p p [0 , ) given by d p ( a , b ) = k a b k p is actually a metric. We now proceed to show that ( p ,d p ) is a complete metric space for 1 p . For convenience, we will work with the case p < , as the case p = requires slightly different language (although the same ideas apply). Suppose that a 1 , a 2 , a 3 ,... is a Cauchy sequence in p . Note, each term a k in the se quence is a point in p , and so is itself a sequence: a k = ( a k 1 ,a k 2 ,a k 3 ,... ) . Now, to say that ( a k ) k =1 is a Cauchy sequence in p is precisely to say that > K N s.t. k,m K, k a k a m k p < . That is, for given > and sufficiently large k,m , we have X n =1  a k n a m n  p = k a k a m k p p < p . Now, the above series has all nonnegative terms, and hence is an upper bound for any fixed term in the series. That is to say, for fixed n N ,  a k n a m n  X n =1  a k n a m n  p < p , and so we see that the sequence ( a k n ) k =1 is a Cauchy sequence in R . But we know that R is a complete metric space, and thus there is a limit a n R to this sequence. This holds for each n N . The following diagram illustrates whats going on. a 1 = a 1 1 a 1 2 a 1 3 a 1 4 a 2 = a 2 1 a 2 2 a 2 3 a 2 4 a 3 = a 3 1 a 3 2 a 3 3 a 3 4 a 4 = a 4 1 a 4 2 a 4 3 a 4 4 ....
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This note was uploaded on 12/07/2011 for the course MATH 18.100B taught by Professor Prof.katrinwehrheim during the Fall '10 term at MIT.
 Fall '10
 Prof.KatrinWehrheim

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