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

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MATH 202A HOMEWORK 1 SOLUTIONS Problem 1 . Suppose p q . Let x = ( x n ). We claim that (1) x x q x p q p x 1 - p q The first part of (1) is the observation that for fixed j , | x j | q n =1 | x n | q = x 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 x x q To prove the second part of (1), we argue as follows: x q q = + n =1 | x n | q sup j | x j | q - p + n =1 | x n | p = x q - p x p p Raise both sides to the power 1 /q to obtain x q x p q p x 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 →∞ x q = x . Problem 2 . Before proceeding, we shall prove that Lemma 1. If two metrics ρ 1 and ρ 2 are given by norms · 1 and · 2 , then the following are equivalent (1) ρ 1 is equivalent to ρ 2 . (That is, the identity mapping from ( X, ρ 1 ) to ( X, ρ 2 ) is a homeomorphism.) (2) ρ 1 is uniformly equivalent to ρ 2 . (That is, the identity mapping from ( X, ρ 1 ) to ( X, ρ 2 ) is a uniform homeomorphism.) 1

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2 MATH 202A HOMEWORK 1 SOLUTIONS (3) There exists a constant c 1 such that for all x X , 1 c x 1 x 2 c x 1 . Proof. Clearly (2) = (1). We will prove that (1) = (3). Since we are assuming (1), we know that δ > 0 such that x X , if ρ 1 (0 , x ) δ , then ρ 2 (0 , x ) 1. This last implication converts to x 1 δ = x 2 1 . For any x X replace x by δx x 1 and apply the above implication to conclude that δ x 2 x 1 1, or x 2 1 δ x 1 . A similar argument shows that there exists δ > 0 such that δ x 1 x 2 . Thus we can take c = max( 1 δ , 1 δ ). We next prove (3) = (2). We would like to show that for > 0, δ > 0 such that x, y X , ρ 1 ( x, y ) < δ = ρ 2 ( x, y ) < . This implication equates to x - y 1 δ = x - y 2 < . But since we are assuming inequality in item (3), we have x - y 2 c x - y 1 So we just need to take δ = /c . Showing that for > 0, δ > 0 such that x, y X , ρ 2 ( x, y ) < δ = ρ 1 ( x, y ) < is similar. Let 1 r ≤ ∞ . As in Problem 1, (2) x d x r d Also, x r r d = d j =1 | x j | r max 1 n d | x n | r d j =1 1 = d x r and thus (3) x r d d 1 /r x 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 q d for any 1 p = q ≤ ∞ . Problem 3 . By Lemma 1, it suffices to show that there does not exist a constant c such that for all x X , (4) x 1 c x 2
MATH 202A HOMEWORK 1 SOLUTIONS 3 Suppose there were such a constant c . We know that A n =1 1 n 2 < .

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

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