Page 192 - DMTH401_REAL ANALYSIS

Real Analysis

Notes 
|f (x) – f(x)| < for n > k  m.
n
2
Fix k and let n ® ¥. Then f (x) ® f(x) and we get
n

|f(x) – f (x)| < i.e., |f (x) – f(x)| < .
k k
2
This is true for k  m and for all x in A. This shows that (f) is uniformly convergent to f on A,
which proves the sufficient part.
As remarked in the introduction, uniform convergence is the form of convergence of the sequence
of function (f ) which preserves the continuity, differentiability and integrability of each term f
n n
of the sequence when passing to the limit function f. In other words if each member of the
sequence of functions (f ) defined on a set A is continuous on A, then the limit function f is also
n
continuous provided the convergence is uniform. The result may not be true if the convergence
is only pointwise. Similar results hold for the differentiability and integrability of the limit
function f. Before giving the theorems in which these results are proved, we discuss some
examples to illustrate the results.

Example: Discuss for continuity the convergence of a sequence of functions (f), where
n
f (x) = 1 – |1 – x | , x {x||1 – x |  1} = [ –2, 2].
2
2
2
<
ì ï 1, when |1 x | 1
-
Solution: Here limf(x) = í
2
n®¥ 0, when 1 x | 1 i.e. x = ± 2
0
=
-
ï î
Therefore the sequence (f ) is pointwise convergent to f where
n
2
ì ï 1, when | x | 1
-
<
f(x) = í
2
-
=
ï î 0, when |1 x | 1
Now each member of the sequence (f,) is continuous at 0 but f is discontinuous at 0. Here (f ) is
n
not uniformly convergent in [–2, 2] as shown below.
Suppose (f ) is uniformly convergent in [–2, 2], so that f is its uniform limit.
n
1
Taking  = , there exists an integer m such that
2
1
f(x) = < for n  m and " x [2, 2].
2
1
in particular |f (x) – f(x)| < for x [2, 2]
m 2
2
2
ì ï |1 x |"' when |1 x | 1
-
<
-
Now |f (x) – f(x)| = í
m 2
-
=
ï î 0 when |1 x | 1
2
Since lim |1 – x | = 1,  a + v no.  such that
m
x® 0
2
m
|1 – x | – 1| < 1/4 for 0 < |x | < 
m
2
i.e. 3/4 <|1 – x | < 5/4 for |x| < 
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