Chapter 3: Sequence of Functions
14
Section B: Uniform Convergence of functions
By the end of this section you will be able to



understand what is meant by uniform convergence
test a sequence for uniform convergence
compare and contrast between uniform and pointwise convergence
B1 Uniform Convergence of Functions
In the last section we covered pointwise convergence. Remember pointwise convergence depends
on the value of x in the domain of the function. However in uniform convergence the convergence is
independent of x.
The sequence of functions given in Example 6 on page 9 demonstrates uniform convergence:
fn  x  
sin  nx 
n
Some of these functions are illustrated in Figure 11 below. The function f n  x  shown below is for
n  N 0 where the natural number N 0 depends only on  not x. Note that the sequence f n  x  for
n  N 0 is within  of the function f  x  whatever the value of x. Compare this with Fig 9 on page
6 which shows pointwise convergence.
f n  x  for n  N 0
f  x  
f  x  
Fig 11
Example 9
Let f n :  0,     be a sequence of functions given by:
fn  x  
sin  nx 
n
(i) Plot f1  x  , f 2  x  , f 3  x  and f10  x  on the same axes.
Chapter 3: Sequence of Functions
(ii) What is the value of f n  x  
15
sin  nx 
as n   ?
n
Solution
(i) Substituting n  1, 2, 3 and 10 into f n  x  
f1  x  
sin  nx 
gives
n
sin  x 
sin  2 x 
sin  3 x 
sin 10 x 
 sin  x  , f 2  x  
, f3  x  
and f10  x  
1
2
3
10
Plotting these gives
Fig 12
(ii) In Example 6 on page 7 we have already shown that this sequence of functions converges to the
zero function, f  x   0 .
By Example 6 on page 9 the size of the natural number n for f n  x  such that f n  x  is within  of
zero (because these functions f n  x  
sin  nx 
1
converge to zero) is n  N 0  .
n

We examine some numerical values for the above example. For   0.0001 what natural number
N 0 ensures that f n  x  
sin  nx 
is within   0.0001 of zero?
n
We need to find a N 0 such that for all n  N 0 we have
0.0001  f n  x  
sin  nx 
 0.0001
n
By the answer given in the above example we select our N 0 
N 0 is greater than 10 000, N 0  10 001 say, then
1


1
 10 000 . Hence if our
0.0001
Chapter 3: Sequence of Functions
16
0.0001  f10 001  x  
sin 10001x 
 0.0001
10001
Remember the sine function lies between 1 and 1 therefore
0.0001 
1
 0.0001
10001
Putting this on a graph we have:
0
0
0
Fig 13
What value of N 0 do we need for the given sequence of functions to be within   0.000001 ?
N0 
1


1
 1 000 000
0.000001
For N 0  1 000 001 and n  N 0  1 000 001 we have
0.000001  f1 000 001  x  
sin 1000001x 
 0.000001
1000001
Similarly the graph of this is given by:
0
0
0
Fig 14
We need to write the formal definition of uniform convergence.
Chapter 3: Sequence of Functions
17
Definition (3.3). The sequence of functions
 f  x   converges uniformly to a function f  x  in the
n
domain D  for every x in D and for every   0 there exists a natural number N 0 (depending
only on  ) such that
f n  x   f  x    provided n  N 0
Example 10
Let f n :  2, 2   be a sequence of functions given by:
fn  x  
x
n
Show that f n  x  converges uniformly on  2, 2 to the zero function f  x   0 by using the
above Definition (3.3).
Solution
Our domain is  2, 2 which means that:
-2  x  2
Fig 15
-3
-2
-1
0
1
2
Let   0 be given. Then there is a natural number N 0 such that for all n  N 0 we have
fn  x   f  x  
To show that f n  x  
x
x
2
0  
because 2  x  2
n
n N0
x
2
converges to zero we need to find a N 0 which gives
 .
n
N0
Transposing this gives the inequality
2

 N 0 or writing this the other way N 0 
2

. Hence for all
n  N 0 we have
fn  x   0 
2
2
  provided N 0 

N0
By the above Definition (3.3) we conclude that f n  x  
zero function f  x   0 .
x
converges uniformly on  2, 2 to the
n
3
Chapter 3: Sequence of Functions
18
Note that in the above example the natural number N 0 
2

depends only on  and not on x.
Example 11
Let f n : 1,
   be a sequence of functions given by:
f n  x   e  nx
Show that f n  x  converges uniformly on the interval x  1,
 to the zero function f  x   0
by using the above Definition (3.3).
Solution
Let   0 be given. Then there exists a natural number N 0 such that for all n  N 0 we have
f n  x   0  e  nx
 †
e  nx


Because e nx  0
The function e  nx is a decreasing function which means that if b  a  0 then e  nb  e  na . How do
we know that e  nx is a decreasing function on the domain 1,
 ?
We can show the derivative of g  x   e  nx is negative. Hence g '  x    ne  nx  0 because n is a
natural number and e  nx  0 .
Therefore with x  1,
 which means that 1  x   we have
A decreasing function.
Fig 16
Since e  nx is a decreasing function we have
e  nx  e
 n 1
 e  n for x  1
(*)
From the definition we have n  N 0 and because e  nx is a decreasing function therefore
e  n  e  N0
Note that e  N0 
1
. By  †  and (*) we have
e N0
Chapter 3: Sequence of Functions
19
f n  x   0  e  nx  e  n 
1
e N0
How do we show that f n  x   e  nx is uniformly convergent on 1,
Required to prove that f n  x   0  e  nx 
 ?
1
  for any   0 .
e N0
If   1 then by the properties of the exponential function f n  x   0  e  nx  1   for
x  1,  .
If   1 then transposing the above inequality
1

e N0
1


 e N0 
1
 :
e N0
1
1
ln    N 0 or the other way N 0  ln  
 
 
1
1
 . Therefore with n  N 0  ln   we have
 
 
We choose N 0  ln 
f n  x   0  e  nx 
1

e N0
1
 nx
 depends only on   0 which means that f n  x   e converges
 
Note that N 0  ln 
uniformly in the interval 1,
 .
B2 Non Uniform Convergence
Is there a relationship between pointwise and uniform convergence?
If a sequence of functions
numbers then
 f  x   is uniformly convergent to f  x  on an interval of the real
n
 f  x   is pointwise convergent to f  x  on the same interval. You are asked to
n
show this result in the next set of exercise.
It is not true the other way round, that is if
 f  x   converges pointwise to f  x  then  f  x  
n
n
converges to f  x  uniformly is false. How do we prove a sequence of functions is not uniformly
convergent?
To prove that a sequence of functions is not uniformly convergent we negate the Definition (3.3) of
uniform convergence given on page 3:
Chapter 3: Sequence of Functions
The sequence of functions
20
 f  x   converges uniformly to a function f  x  in the domain D 
n
for every x in D and for every   0 there exists a natural number N 0 (depending only on  ) such
that f n  x   f  x    provided n  N 0 .
Negating this statement we have:
Definition (3.4). The sequence of functions
 f  x   does not converges uniformly to a function
n
f  x  in the domain D  there exists a x in D and an   0 such that for every natural number n
we have f n  x   f  x    .
We can apply this definition to show a given sequence of functions is not uniformly convergent. We
need to find a value of x and a value for  such that f n  x   f  x    to show the sequence of
functions
 f  x 
n
does not converge uniformly to f  x  .
Example 12
Let f n :    be a sequence of functions given by:
fn  x  
x
n
Show that f n  x  converges pointwise for all x   to the zero function f  x   0 but not
uniformly.
Solution
We have already shown in Example 4 on page 7 that f n  x  
x
converges pointwise to the zero
n
function.
To show that the sequence of functions
 f  x   does not converge uniformly to zero we need to
n
find an   0 and x such that f n  x   0   .
Consider an 0    2 and x  2n then for all natural numbers n we have
f n  2n   0 
2n
0  2
n
Hence by the above definition (3.4) we conclude that f n  x  
zero for x   .
x
does not converge uniformly to
n
Chapter 3: Sequence of Functions
21
Example 13
Let f n : 0,
   be a sequence of functions given by:
fn  x  
1
nx
Show that f n  x  converges pointwise on the interval 0,
 to the zero function f  x   0 but
not uniformly.
Solution
Applying the limit theorems we have
 1  1   1 
lim  f n  x    lim     lim     0
n 
n  nx
  x  n  n  


1
 0


 Because lim
n  n
 


Hence f n  x  converges pointwise to 0.
In order to show that the sequence of functions
 f  x   does not converge uniformly to zero we
n
need to find a  and a value for x such that f n  x   0   as described in the above definition
(3.4).
Let 0    1 and take x 
1
then for every natural number n we have
n
1
1
1
fn    f   
 0  1 0  1  
n
 n  n 1/ n 
Since f n  x   0   we conclude that
 f  x   does not converge uniformly to zero.
n
Example 14
Let f n :    be a sequence of functions given by:
fn  x  
x 2  nx
n
Show that f n  x  converges pointwise for all x   to the f  x   x but not uniformly.
Solution
 x 2  nx 
  x . How?
 n 
We need to first show that lim  f n  x    lim 
n 
n 
Chapter 3: Sequence of Functions
22
By applying the limit theorems. We have
 x 2  nx 
 x2 
 nx
 
lim 

lim
 lim  



n 
 n  n  n  n  n 
  1 
 x 2  lim     lim  x   x 2  0   x  x
 n  n   n
Hence the given sequence of functions
 f  x   converges pointwise to x.
n
What else do we need to show?
Required to prove that the convergence is not uniform. This means that we need to find a x in 
and an   0 such that f n  x   x 
x 2  nx
 x   . Consider this distance function:
n
x 2  nx
x 2  nx  nx
x2 x2
fn  n   x 
x 


n
n
n
n
Let x  n and 0    n then we have
x2 n2
fn  x   x 

n
n
n
The given sequence of functions
 f  x   does not converge uniformly to x.
n
SUMMARY
The sequence of functions
 f  x   converges uniformly to f  x  means that for all n  N
n
0
fn  x   f  x   
where the N 0 depends on   0 and not on x.
If a sequence of functions converges uniformly to f  x  then it converges pointwise. However if a
sequence of functions converges pointwise then it may not converge uniformly. We have
fn  x  

Uniformly
f  x  implies that f n  x  
 f  x
Pointwise
fn  x  
 f  x  does not imply that f n  x  

Pointwise
Uniformly
f  x