Page 188 - DMTH401_REAL ANALYSIS

Real Analysis

Notes
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Task Show that the sequence (f ) where f (x) = , x R, is pointwise convergent.
n n n
Also find the pointwise limit.

If the sequence of functions (f ) converges pointwise to a function f on a subset A of R, then the
n
following question arises: “If each member of (f ) is continuous, differentiable or integrable, is the
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limit function f also continuous; differentiable or integrable?”. The answer is no if the convergence
is only pointwise. For instance each of the functions f is continuous (in fact uniformly continuous)
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but the sequence of these functions converges to a limit function f(x)
<

ì 0 for 0 x 1
f(x) = í
î 1, for x = 1
which is not continuous. Thus, the pointwise convergence does not preserve the property of
continuity. To ensure the passage of the properties of continuity, differentiability or integrability
to the limit function, we need the notion of uniform convergence which we introduce in the next
section.

14.2 Uniform Convergence

From the definition of the convergence of the sequence or real numbers, it follows that
the sequences (f ) of functions converges pointwise to the function f on A if and only if for each
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x A and for every number  > 0, there exists a positive integer m such that
|f (x) – f(x)| <  whenever n  m.
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Clearly for a given sequence (f,,) of functions, this m will, in general, depend on the given  and
the point x under consideration. Therefore it is, sometimes, written as m (, x). The following
example illustrates this point.

x
Example: Define f (x) = for < x < ¥.
n
n
For each fixed x the sequence (f (x)) clearly converges to zero. For a given  > 0, we must show
n
the existence of an m, such that for all n  m,
x x
|f (x) – f(x)| = - 0 = < 
n
n n
é |x| ù é |x| ù |x|
This can be achieved by choosing m = ê ú + 1 where ê ú denotes the integral part of (i.e.
ë  û ë  û 
the integer m is next to |x| in the real line). Clearly this choice of m depends both on  and x.

1 1 |x|
For example, let = 3 . If x = 3 then = 1 and, so, m can be chosen to be 2. If x = 1, then
10 10 
6
3
|x| = 10 and, so, m should be larger than 10 . Note that it is impossible to find an m that serve
|x|
for all x. For, if it were, then > E, for all x,
m
Consequently |x| is smaller than m, which is not possible. Geometrically, the f ’s can be
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described as shown in the Figure.

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