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Unformatted text preview: 201A, Fall 10, Thomases Homework 2 1. Show that if f is a function, and S is a bounded subset of R , the the following statements are equivalent: (i) The function f is uniformly continuous on S . (ii) If { x n } is a Cauchy sequence in S then { f ( x n ) } is a Cauchy se quence. Proof. (i) (ii) Let { x n } n be a Cauchy sequence in S , and suppose > 0, then a positive > 0 such that if x 1 and x 2 are in S ,  x 1 x 2  < implies  f ( x 1 ) f ( x 2 )  < . Since { x n } n is a Cauchy sequence, we can choose N so that m, n > N implies  x m x n  < . Then writing what the uniform continuity means for x m and x n  f ( x m ) f ( x n )  < , when m, n > N. So { f ( x n ) } n is a Cauchy sequence. (i) (ii) Suppose f is not uniform continuous on the bounded set S . Then there exists a positive number > 0 for each n N there are elements in S , x n and y n satisfying:  x n y n  < 1 n and  f ( x n ) f ( y n )  . Since S is bounded, the Bolzano Weierstrauss theorem guarantees that the sequence { x n } n has a convergent subsequence; say lim n x n k = L . Since { x n y n } n is a null sequence and since y n k = x n k + ( y n k x n k ) it follows that lim n x n k = L (pass to limit in the above equality and you will see that y n k is forced to go to L ). Note that L might not be in S , but this does not matter. Now define the sequence { u n } n by u n = n x n k 1 , y n k 1 , x n k 2 , y n k 2 o . The way that we constructed { u n } n implies that lim n u n = L . In particular, { u n } n is a Cauchy sequence that lies in S . But for each n ,  f ( u 2 n 1 ) f ( u 2 n )  =  f ( x n k ) f ( y n k )  . Hence we got that { f ( u n ) } n is not a Cauchy sequence as we assumed at the begining. So we got a contradiction. Therefore what we have assumed was not true. Done. 2. If X is a normed linear space with norm k k . define = k x k 1 + k x k ....
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 Spring '11
 Wiley
 Differential Equations, Equations

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