1873_solutions

# Integers n for which x n b the sequence f x n has no

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Unformatted text preview: s to be continuous? Hint: Look at the case in which S is the union of two mutually disjoint open intervals. See the remarks about Exercise 3. 5. Is it true that if a set S of real numbers is not an interval and is not bounded then there must exist a one-one continuous function on S whose inverse function fails to be continuous? Again the answer is no. The remarks about Exercise 3 are not confined to bounded intervals. 6. Prove that if f is a continuous function from the interval 0, 1 into 0, 1 then there must be at least one number x 0, 1 such that f x  x. This assertion is the one-dimensional form of the Brouwer fixed point theorem. Solution: For every number x 0, 1 we define g x  f x x. The function g defined in this way is continuous on the interval 0, 1 . We see that g0 f0 0 0 and g1 f1 1 0 and we conclude from the Bolzano intermediate value theorem that there is at least one number x for which g x  0. 0, 1 Exercises on Uniform Continuity 1. Is it true that if S is an unbounded set of real numbers and f x  x 2 for every number x S then the function f fails to be uniformly continuous? The assertion given here is false. Every function defined on the set Z of integers must be uniformly continuous. 2. Given that fx  1 if 0 x2 0 if 2  x 3 prove that f is continuous but not uniformly continuous on the set 0, 2 Þ 2, 3 . Since the number 2 does not belong to the domain of f, the function f is constant in a neighborhood of every number in its domain. Therefore f is continuous on the set 0, 2 Þ 2, 3 . To see why f fails to be uniformly continuous we observe that given any positive number  we can find a number t 0, 2 and a number x 2, 3 such that |t x |  , and for any such choice of numbers x and t we must have |f t f x |  |1 0 |  1. 3. Given that f x  sin x 2 for all real numbers x, prove that f is not uniformly continuous on the set R. 230 1 0.5 0 2 4 x 6 8 10 -0.5 -1 We define xn  2n   2 and for every positive integer n. Since f x n not approach 0 as n Ý. Now xn tn    t n  2n  1 and f t n  0 for every n we know that f x n 2n   2 2n  2n  2 2n 2n  2n  f t n does  2  2  2n 2n   2  2  2n  2n 0 as n Ý and so it follows from the relationship between limits of sequences and uniform continuity that the function f fails to be uniformly continuous. 4. Ask Scientific Notebook to make some 2D plots of the function f defined by the equation f x  sin x log x for x  0. Plot the function on each of the intervals 0, 50 , 50, 100 , 100, 150 and 150, 200 . Revise your plot and increase its sample size if it appears to contain errors. Why do these graphs suggest that f fails to be unformly continuous on the interval 0, Ý ? Prove that this function does, indeed, fail to be uniformly continuous. Solution: To prove that f fails to be uniformly continuous we shall show that for every number   0 there exist two positive numbers a and b such that |a b |   and |f a f b | 1. We begin by choosing a number p such that wheneve...
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## This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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