1873_solutions

Take x r and 2 h x 1 x x 1 k x 1 x x 1 and 11

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Unformatted text preview: r x p we have log x   . 2 Now choose a positive integer n such that p log p  n. Since x log x  n for x sufficiently large x, we can use the Bolzano intermediate value theorem to choose a number a  p such that a log a  n. Now since a   log a    a log a     log a    a log a     n   2 2 we can use the Bolzano intermediate value theorem again to choose a number b a, a   such that . b log b  n  2 We now observe that 231 |f a sin n   2 f b |  sin n 1 and so the proof is complete. 5. a. A function f from a metric space X to a metric space Y is said to be Lipschitzian on a set S if there exists a number k such that the inequality d f t ,f x kd t, x holds for all points t and x in S. Prove that every lipschitzian function is uniformly continuous. Suppose that f is a function from a metric space X to a metric space Y, that k is a positive number and that the inequality d f t ,f x kd t, x holds for all points t and x in the space X. Suppose that  0. We define   /k and observe that, whenever t and x belong to X and d t, x   we have kd t, x  k d f t ,f x b. Given that f x  x for all x k . 0, 1 prove that f is uniformly continuous but not lipschitzian on 0, 1 . Solution: The fact that f is uniformly continuous on the closed bounded set 0, 1 follows at once from the fact that f is continuous there. Now, to prove that f fails to be Lipschitzian, suppose that k is any positive number. Given x 0, 1 we see that |f x f 0 | 1 |x 0 | x and this exceeds k whenever x  1/k 2 . 6. a. Prove that if f is a uniformly continuous function from a totally bounded metric space X onto a metric space Y then Y is also totally bounded. We suppose that f is a uniformly continuous function from a totally bounded metric space X onto a metric space Y. To show that Y is totally bounded, suppose that  0. Choose   0 such that whenever t and x belong to X and d x, t   we have d f x , f t  . Using the fact that X is totally bounded, choose finitely many points x 1 , x 2 , , x n in the space X such that n X  B xj,  . j1 We shall now show that n Y  B f xj ,. j1 Suppose that y Y. Using the fact that the function f is onto the space Y, we choose a member x of X such that y  f x . Choose j such that x B x j ,  . We see that y B f x j , . b. Give an example of a uniformly continuous function from a bounded metric space onto an unbounded metric space. If X is the discrete space in which d x, y  0 if x  y 1 if x y then every function from X to another metric space is continuous. c. Prove that if a function f is uniformly continuous on a bounded subset S of R k into a metric space Y then the range of f is a bounded subset of Y. We know that whenever S is a bounded subset of R k , the subspace S is totally bounded. d. Give a quick proof that if f x  1/x for all x interval 0, 1 . 232 0, 1 then f fails to be uniformly continuous on the Had this function been uniformly continuous then its range would have had to be bounded. But its range i...
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