Page 170 - DMTH401_REAL ANALYSIS

Real Analysis

Notes

Task Show that the equation 16x + 64x – 32x – 117 = 0 has a real root > 1.
4
2
3
12.2 Pointwise Continuity and Uniform Continuity

Here you will be introduced with the concept of uniform continuity of a function. The concept of
uniform continuity is given in the whole domain of the function whereas the concept of continuity
is pointwise that is it is given at a point of the domain of the function. If a function f is continuous
at a point a in a set A, then corresponding to a number E > 0, there exists a positive number (a)
(we are denoting 6 as (a) to stress that 6 in general depends an the point a chosen) such that
|x – a| < (a) implies that |f(x) – f(a)| < . The number (a) also depends on E. When the point a
varies (a) also varies. We may or may not have a 6 which serves for all points a in A. If we have
such a 6 common to all points a in A, then we say that f is uniformly continuous on A. Thus, we
have the following definition of uniform continuity.
Definition 1: Uniform Continuity of a Function
Let f be a function defined on a subset A contained in the set R of all reals. If corresponding to any
number  > 0, there exists a number  > 0 (depending only on G) such that
|x – y| < , x, y  A |f(x) – f(y)| < *
then we say that f is uniformly continuous on the subset A.
An immediate consequence of the definition of uniform continuity is that uniform continuity in
a set A implies pointwise continuity in A. This is proved in the following theorem.
Theorem 4: If a function f is uniformly continuous in a set A, then it is continuous in A.
Proof: Since f is uniformly continuous in A, given a positive number E, there corresponds a
positive number 6 such that
|x – y| < ; x, y  A  |f(x) – f(y)| <  ... (7)
Let a be any point of A. In the above result (1), take y = a. Then we get,
|x – a| < ; x  A  |f(x) – f(a)| < 

which shows that f is continuous at ‘a’. Since ‘a’ is any point of A, it follows that f is continuous
in A.
Now we consider some examples.

Example: Show that the function f : R

f(x) = x V x R,
is uniformly continuous on R
Solution: For a given  > 0, 6 can be chosen to be  itself so that
|x – y| < 6 = G  |f(x) – f(y)| = |x – y| < .

Example: Show that the function f : R – R given by

2
f(x) = x " x R
is not uniformly continuous on R.

164 LOVELY PROFESSIONAL UNIVERSITY