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Unformatted text preview: n Ý. Note that the sequence sup f n is decreasing.
To obtain a contradiction, suppose that the sequence sup f n fails to converge to 0 and, using this
assumption, choose a number 0 such that the inequality
sup f n
holds for every n. For each n, choose a number x n S such that f n x n and, using the fact that x n is
a sequence in the closed bounded set S, choose a partial limit x of the sequence x n .
Using the fact that f n x
0 as n Ý, choose an integer N such that f N x . Now, using the fact that
the function f N is continuous at the number x, choose a number 0 such that the inequality f N t
holds for every number t S
x , x .
Using the fact that x is a partial limit of the sequence x n we now choose an integer n N such that
xn
x , x . Then we have
fn xn
fN xn
contradicting the way in which the number x n was chosen. Some Exercises on the Tests for Uniform Convergence
364 1. Prove that the series
1
converges uniformly in x on the interval n1 sin 1
n x
n 1, 1 and converges pointwise on the entire line R. If x 0 then, starting at a sufficiently large value of n, the sequence of numbers sin 1
decreasing (with limit sin 1) and so the sequence
x
sin 1 n
n x
n is is a decreasing sequence of positive numbers with limit 0 and it follows from Dirichlet’s test that the
series
x
1 n 1 sin 1 n
n
converges. If p is any given positive number and if we define
x
sin 1 n
fn x
n
then starting at a sufficiently large integer N the sequence of functions decreases uniformly to 0 on
the interval 0, p as n Ý. By Dirichlet’s test for uniform convergence the series
x
1 n 1 sin 1 n
n
converges uniformly in x on the interval 0, p .
Now suppose that x 0. It is clear that
sin 1
lim
nÝ
n x
n 0 but, before we can use Dirichlet’s test, we need to show that this sequence is decreasing. We write
sin 1 x
t
ft
t
for t 0 and we observe that for each t we have
t
x cos 1 x
sin 1 x
t
t
2
ft
t 5/2
and it is clear that f t 0 if t is sufficiently large. A similar argument shows that if p 0 then it is
possible to find an integer N such that starting at n N, the sequence
x
sin 1 n
n
decreases to zero for every number x
p, 0 and it is clear that this convergence is uniform. It
therefore follows from Dirichlet’s test for uniform convergence that
x
1 n 1 sin 1 n
n
converges uniformly on the interval p, 0 .
We conclude that the series converges uniformly on every bounded set of real numbers, which is
more than the exercise requested.
2. Prove that if we define
Ý fx
n1 1 n1 sin 1
n x
n for every real number x then the function f is differentiable at every real number and for every number x we
have 365 Ý 1 fx n1 n1 cos 1
n 3/2 x
n The solution of this exercise follows at once from the theorem on term by term differentiation of
series, from Exercise 1 and from the uniform convergence in x of the series
x
1 n 1 cos 1 n
.
3/2
n Exercises on Continuity of a Limit Function
1. Suppose that a and b are real numb...
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 Fall '08
 STAFF
 Math, Calculus

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