Unformatted text preview: e 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 oneone
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 onedimensional 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. 200 2. Given that
x2 1 if 0 fx 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.
1 0.5 0 2 4 x 6 8 10 0.5 1 We define
xn 2n
2 and
t n 2n
for every positive integer n. Since f x n 1 and f t n 0 for every n we know that f x n
not approach 0 as n Ý. Now
x n t n 2n
2n
2
2n
2 2n
2n 2n
2
2 2n 2n
2
2 f t n does 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
a number p such that whenever x p we have 201 b  and f a fb  1. We begin by choosing log x .
2
Now choose...
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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.
 Fall '08
 STAFF
 Math, Calculus

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