Page 190 - DMTH401_REAL ANALYSIS

Real Analysis

Notes

From, the definition of uniform convergence, it follows that uniform convergence of a sequence
of functions implies its pointwise convergence and uniform limit is equal to the pointwise limit.
We will show below by suitable examples that the converse is not true.

x
Example: Show that the sequence (f ) where f (x) = , x  R is pointwise but not uniformly
n n
n
convergent in R.
Solution: You have seen that (f ) is pointwise convergent to f where f(x) = 0 " x  R. In the same
n
example, at the end, it is remarked that given  > 0, it is not possible to find a positive integer m
|x|
such that < for n  m and " x R i.e., |f (x) – f(x)| <  for n  m and " x R. Consequently
n n
(f ) is not uniformly convergent in R.
n

n
Example: Show that the sequence (f ) where f (x) = x is convergent pointwise but not
n n
uniformly on [0, 1].
Solution: You have been shown that (f ) is pointwise convergent to, f on [0, l] where
n
f(x) = 0 " x [0, l [ and f(l) = l
Let  > 0 be any number. For x = 0 or x = 1, |f (x) – f(x)| <  for n  1.
n
log 
n
For 0 < x < l, |f (x) – f(x)| <  if x <  i.e. n log x < log  i.e. n > .
n
log x
é log ù

since log x is negative for 0 < x < 1. If we choose m = ê ú + 1 , then |f (x) – f(x)| <  for n  m.
n
ë log x û
Clearly m depends upon  and x.
We will now prove that the convergence is not uniform by showing that it is not possible to find
an m independent of x.
Let us suppose that 0 <  < l. If there exists m independent of x in [0, 1] so that

<|f (x) – f(x)| <  for all n  m,
n
then x’’ < for all n  m, whatever may be x in 0 < x < 1.

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