ps4solutions

# ps4solutions - MAT 337 Problem Set 4 Sample Solutions Ian...

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MAT 337 Problem Set 4 Sample Solutions Ian Zwiers - March 25, 2009 This document is for demonstration purposes; it is not a marking key. 5.5(I) Let f be a continuous function on (0 , 1] . Show that f is uniformly continuous if and only if lim x 0 + f ( x ) exists. Note: We assume that f : (0 , 1] 7→ R M , as that matches Deﬁnition 5.5.1. It is essential that the range lives inside a complete space. ( ) Let f be uniformly continuous. Suppose lim x 0 + f ( x ) does not exist. Then sequence a n > 0 with lim n →∞ a n = 0 such that lim n →∞ f ( a n ) does not exist. Let ± > 0 be arbitrary. By uniform continuity, δ > 0 such that | f ( x ) - f ( y ) | < ± whenever | x - y | < δ (for x,y (0 , 1]). Since lim n →∞ a n exists, it is Cauchy, so there exists N such that | a n - a m | < δ for all n,m > N . But then | f ( a n ) - f ( a m ) | < ± for all n,m > N , which proves that ( f ( a n )) n =1 is Cauchy. By the completeness of R M , lim n →∞ f ( a n ) exists. By contradiction, our supposition is false and lim x 0 + f ( x ) exists. ( ) Let lim x 0 + f ( x ) exist. Then we may deﬁne g : [0 , 1] 7→ R M as an extension of f : g ( y ) = ( lim x 0 + f ( x ) if y = 0 f ( y ) if y (0 , 1] For y = 0, g is continuous by deﬁnition. For y (0 , 1], g is continuous since f is continuous. That is, g is continuous on [0 , 1], which is a compact set. By Theorem 5.5.9, g is uniformly continuous. Given ± > 0, there exists δ > 0 such that | g ( x ) - g ( y ) | < ± whenever x,y [0 , 1] with | x - y | < δ . For x,y (0 , 1] with | x - y | < δ , | f ( x ) - f ( y ) | = | g ( x ) - g ( y ) | < ± , which proves that f is uniformly continuous. 5.6(H) (a) Show that a continuous function on ( -∞ , + ) cannot take every real value exactly twice. Suppose that

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ps4solutions - MAT 337 Problem Set 4 Sample Solutions Ian...

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