1873_solutions

# Above 6 prove that if h is a set of real numbers and

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Unformatted text preview: 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 1 |f a f b |  sin n sin n   2 and so the proof is complete. 5. a. A function f is said to be Lipschitzian on a set S if there exists a number k such that the inequality |f t f x | k|t x | holds for all numbers t and x in S. Prove that every Lipschitzian function is uniformly continuous. Suppose that f is a function defined on a set S, that k is a positive number and that the inequality |f t f x | k|t x | holds for all numbers t and x in the set S. Suppose that  0. We define   /k and observe that, whenever t and x belong to S and |t x |   we have |f t b. Given that f x  x for all x fx | k|t x|  k 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. Suppose that f is uniformly continuous on a set S, that x n is a sequence in the set S and that x n has a partial limit x R. Prove that it is impossible to have f x n Ý as n Ý. Using the fact that f is uniformly continuous on S we choose a number   0 such that the inequality |f t f s |  1 holds whenever t and s belong to S and |t s |  . Since x is a partial limit of the sequence x n we know that the condition xn x  ,x   2 2 holds for infinitely many positive integers n. Choose a positive integer N such that xN x  ,x   . 2 2 For every one of the infinitely many positive integers n for which the condition x  ,x   xn 2 2 holds, since 202 |x n xN | x |  |x |x n x N |      , 2 2 we have 1  f xN . f xn Therefore the sequence of numbers f x n is frequently in the interval cannot approach Ý. Of course, this sequence cannot approach Ý either. Ý, 1  f x N and so it b. Did you assume that x S in Part a? If you did, go back and do the exercise again. You have no information that x S. If you didn’t assume x S, you can sit this question out. c. Suppose that f is uniformly continuous on a bounded set S and that x n is a sequence in S. Prove that it is impossible to have f x n Ý as n Ý. The assertion follows at once from Part a and the fact that every bounded sequence of numbers has a partial limit in R. d. Prove that if f is uniformly continuous on a bounded set S then the function f is bounded. Solution: To obtain a contradiction, assume that f is unbounded above. Choose a sequence xn in S such that f x n  n for each n. Now use part c...
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