hw3sol - Math 125A, Fall 2009 Solutions for Homework 3...

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Unformatted text preview: Math 125A, Fall 2009 Solutions for Homework 3 (Prepared by Matt Low) 1. Let > 0 be given. By the definition of uniform continuity, there exists a number > 0 with the property that if | a- b | < then | f ( a )- f ( b ) | < . Because | x n- y n | 0, we may choose N so that | x n- y n | < for n N . Thus, if n N , then | x n- y n | < , and so | f ( x n )- f ( y n ) | < . 2. (a) Note that f consists of sums and compositions of continuous functions. Per Theorems 17.4 and 17.5, this means that f itself is continuous. It follows from Theorem 19.2 that f is uniformly continuous. (b) Define two sequences ( x n ) , ( y n ) by x n = 1 n and y n = 1 n 2 . It is clear that | x n- y n | 0, but: lim n | f ( x n )- f ( y n ) | = lim n ln n = Because | f ( x n )- f ( y n ) | 6 0, this contradicts the property derived in Problem 1, and so we conclude that f is not uniformly continuous. Or: Since lim x 0+ ln( x ) =- , which is not a finite number, one cannot extend ln( x ) to be a continuous function on [0 , 1]. By the continuous extension theorem (theorem 19.5), ln1]....
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hw3sol - Math 125A, Fall 2009 Solutions for Homework 3...

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