hmwk1sol - MATH 202A HOMEWORK 1 SOLUTIONS Problem 1 ....

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Unformatted text preview: MATH 202A HOMEWORK 1 SOLUTIONS Problem 1 . Suppose p q . Let x = ( x n ). We claim that (1) k x k ` k x k ` q k x k p q ` p k x k 1- p q ` The first part of (1) is the observation that for fixed j , | x j | q X n =1 | x n | q = k x k q ` q since | x j | q is one of the terms in the sum, and the remaining terms are all 0. Raising this inequality to the power 1 /q , and taking the supremum over j , we get k x k ` k x k ` q To prove the second part of (1), we argue as follows: k x k q ` q = + X n =1 | x n | q sup j | x j | q- p + X n =1 | x n | p = k x k q- p ` k x k p ` p Raise both sides to the power 1 /q to obtain k x k ` q k x k p q ` p k x k 1- p q ` Now suppose x ` p . Note that the first part of (1), when we set q = p in that inequality, gives that x ` . The second part of the inequality, with q p now arbitrary, gives that x ` q . Sending q + in (1) and applying the squeeze theorem gives that lim q k x k ` q = k x k ` . Problem 2 . Before proceeding, we shall prove that Lemma 1. If two metrics 1 and 2 are given by norms k k 1 and k k 2 , then the following are equivalent (1) 1 is equivalent to 2 . (That is, the identity mapping from ( X, 1 ) to ( X, 2 ) is a homeomorphism.) (2) 1 is uniformly equivalent to 2 . (That is, the identity mapping from ( X, 1 ) to ( X, 2 ) is a uniform homeomorphism.) 1 2 MATH 202A HOMEWORK 1 SOLUTIONS (3) There exists a constant c 1 such that for all x X , 1 c k x k 1 k x k 2 c k x k 1 . Proof. Clearly (2) = (1). We will prove that (1) = (3). Since we are assuming (1), we know that > 0 such that x X , if 1 (0 , x ) , then 2 (0 , x ) 1. This last implication converts to k x k 1 = k x k 2 1 . For any x X replace x by x k x k 1 and apply the above implication to conclude that k x k 2 k x k 1 1, or k x k 2 1 k x k 1 . A similar argument shows that there exists > 0 such that k x k 1 k x k 2 . Thus we can take c = max( 1 , 1 ). We next prove (3) = (2). We would like to show that for > 0, > 0 such that x, y X , 1 ( x, y ) < = 2 ( x, y ) < . This implication equates to k x- y k 1 = k x- y k 2 < . But since we are assuming inequality in item (3), we have k x- y k 2 c k x- y k 1 c So we just need to take = /c . Showing that for > 0, > 0 such that x, y X , 2 ( x, y ) < = 1 ( x, y ) < is similar. Let 1 r . As in Problem 1, (2) k x k ` d k x k ` r d Also, k x k r ` r d = d X j =1 | x j | r max 1 n d | x n | r d X j =1 1 = d k x k r ` and thus (3) k x k ` r d d 1 /r k x k ` Then (2) and (3) together show that ` r d is uniformly equivalent to ` d . Since equiva- lence of norms is transitive (in fact, it is an equivalence relation), ` p d is equivalent to...
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This note was uploaded on 04/06/2010 for the course MATH various taught by Professor Tao/analysis during the Spring '10 term at UCLA.

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hmwk1sol - MATH 202A HOMEWORK 1 SOLUTIONS Problem 1 ....

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