Page 171 - DMTH401_REAL ANALYSIS

Unit 12: Properties of Continuous Functions

Solution: Let  be any positive number. Let  > 0 be any arbitrary positive number. Choose Notes
x > / and y = x + /2. Then


|x – y| = < .
2
2
2
(fix) – f(y)| = |x – y | = |x + y||x – y|
æ  ö æ  ö 
= ç ÷ |x + y| = ç ÷ 2x +
è 2 ø è 2 ø 2

 æ 2  ö  2
> ç + ÷ =  + > 
2 è  2 ø 4
That is whatever  > 0 we choose, there exist real numbers x, y such that |x – y| <  but |f(x) –
f(y)|> G which proves that f is not uniformly continuous.
But we know that f is a continuous function on R.

Example: In the above example if we restrict the domain of f to be the closed interval
[–1, 1], then show that f is uniformly continuous on [–1, 1].
8
Solution: Given E > 0, choose  < . If |x – y| <  and x, y [–1, 1],
2
then using the triangle inequality for || we get,
|f(x) – f(y)| = |x – y | = |x + y||x – y|
2
2
<  (|x| + |y|)
 2 (since |x|1, |y| 1)
You should be able to solve the following exercises:

Task 1. Show that f(x) = x , n > 1 is not uniformly continuous on R even though for
n
each a > 1, it is a continuous function on R.
1
2. Show that the function f(x) = for 0 < x < 1 is continuous for every x but not
x
uniformly on ]0, 1[.
1
3. Show that the function f(x) = sin is not uniformly continuous on the interval ]0, 1[
x
even though it is continuous in that interval.

4. Show that f(x) = cx where c is a fixed non-zero real number is a uniformly continuous
function on R.

We have seen that the function defined by f(x) = 1/x on the open interval ]0, 1 [ is not uniformly
continuous on ]0, 1[ even though it is a continuous function on ]0, 1[. Similarly the function f
defined as f(x) = x is continuous on the entire real line R but is not uniformly continuous on R.
2
However, if we restrict the domain of this function to the bounded closed interval [–1, 1], then it
is uniformly continuous. This property is not a special property of the function f, where f(x) = x 2

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