1873_solutions

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

Ask a homework question - tutors are online