Unformatted text preview: j! 2 jn1 Using the fact that
k lim
x1 jn1 2j ! j
x
j j! 2
4 k 4j jn1 2j !
j! 2 we choose a number t 1 such that
k
jn1 4j 2j ! j
t 1.
j! 2 We therefore know that
Ý sup
j1 n 2j ! j
x
4 j j! 2 j1 2j ! j
x
4 j j! 2 Ý 0 x1
jn1 2j ! j
t 1.
4 j j! 2 x log x n converges uniformly in x on the interval 0, 1 .
6. Prove that the series
We begin by observing that x log x 0 as x 0 (from the right). Now the expression x log x takes its
minimum value when log x 1 0, in other words, when x 1 . Therefore the maximum value of
e
x log x  is 1 and the fact that
e
1
x log x n
en
for every positive integer n and every number x
0, 1 allows us to use the comparison test to
x log x n converges uniformly in x on the interval 0, 1 .
deduce that
7. Given that f n and g n are sequences of real valued functions defined on a set S, that f and g are functions
defined on S and that f n f and g n g pointwise as n Ý, prove that
a. f n g n f g pointwise as n Ý.
Given any x S, the fact that f n x
fn x gn x
f x g x .
b. f n gn c. f n g n f g pointwise as n fg pointwise as n d. In the event that g x f x and g n x g x as n Ý guarantees that Ý. Ý. 0 for every number x in the set S we have f n /g n f/g pointwise as n Ý. 8. Given that f n and g n are sequences of real valued functions defined on a set S, that f and g are functions
defined on S and that f n f and g n g boundedly as n Ý, prove that
a. f n g n
b. f n
c. f n g n gn f g boundedly as n Ý. f Ý. g boundedly as n fg boundedly as n Ý. d. In the event that there exists a number 0 such that g n x  362 for each n and every number x in the set S we have f n /g n f/g boundedly as n Ý.
These assertions follow at once from Exercise 7 after we have observed that if f n and g n
are bounded sequences of functions then so are f n g n etc.
9. Suppose that f n is a sequence of real valued functions defined on a set S and that f is a given function
defined on S. Prove that the following conditions are equivalent:
a. The sequence f n converges uniformly to the function f on the set S.
b. For every number
holds for all n 0 there exists an integer N such that the inequality
supf n f  N. 0 there exists an integer N such that the inequality
f n x f x 
holds for all n N and all x S.
Conditions a and b are obviously the same and it is clear that these imply condition c. To show
that condition c implies condition b, suppose that 0. Using the fact that 2 is a positive
number, choose an integer N such that the inequality
f n x f x 
2
holds whenever n N and x S. Then for all n N we have
.
supf n f 
2 c. For every number 10. Suppose that f n is a sequence of real valued functions defined on a set S and that f is a given function
defined on S. Examine the following two conditions:
0 there exists an integer N such that the inequality
f n x f x 
N and all x S. For every number
holds for all n For every number
holds for all n 0 and every number x S there exists an integer N such that th...
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 Fall '08
 STAFF
 Math, Calculus

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