1. Chapter 5
Limits and Continuity

2. Section 5.4
Uniform Continuity

3. For a function f : D  to be continuous on D, it is required that
for every x0  D and for every  > 0 there exists a  > 0 such that
| f (x) – f (x0)| <  whenever | x – x0 | <  and x  D.
Note the order of the quantifiers:  may depend on both  and the point x0.
If it happens that, given  > 0, there is a  > 0 that works for all x0 in D,
then f is said to be uniformly continuous.
Definition 5.4.1
Let f :D  We say that f is uniformly continuous on D if
for every  > 0 there exists a  > 0 such that | f (x) – f ( y)| < 
whenever | x – y | <  and x, y  D .

4. Example 5.4.2*
We claim that the function f (x) = 3x is uniformly continuous on .
Given any  > 0, we want to make | f (x) – f ( y)| <  by making x sufficiently close to y.
We have | f (x) – f ( y)| = | 3x – 3y| = 3| x – y|.
So we may take  =  /3.
Then whenever | x – y | <  we have
| f (x) – f ( y)| = 3| x – y| < 3 = .
We conclude that f is uniformly continuous on .

5. Example 5.4.4 The function f (x) = x2 is not uniformly continuous on .
Let’s look at this graphically.
Given a point x1 close to 0 and an
 -neighborhood about f (x1),
the required  can be fairly large.
f (x)
f (x) = x2
But as x increases, the value
of  must decrease.
f (x2)
This means the continuity is
not uniform.
 -neighborhood
f (x1) x1 x2 x

6. Example 5.4.4 The function f (x) = x2 is not uniformly continuous on .
As a prelude to proving this, let’s write out the statement of uniform continuity
and its negation.
The function f is uniformly continuous on D if
  > 0   > 0 such that  x, y  D, | x – y | <  implies | f (x) – f ( y)| <  .
So the function f fails to be uniformly continuous on D if
  > 0 such that
  > 0,
 x, y  D such that
| x – y | <  and | f (x) – f ( y)|  .
Now for the proof.
Suppose we take  = 1.
(Any  > 0 would work.)
We must show that given any  > 0, there exist x, y in such that | x – y | <  and
| f (x) – f ( y)| = | x2 – y2| = | x + y |  | x – y |  1.
For any x, if we let y = x +  /2, then | x – y | =  /2 <  .
Thus to make
1  | x + y |  | x – y | = | x + y |  ( /2),
we need to have | x + y |  2/ .
This prompts us to let x = 1/ .
Here is the formal proof.

7. Example 5.4.4
The function f (x) = x2 is not uniformly continuous on .
Proof:
Let  = 1.
Then given any  > 0, let x = 1/ and y = 1/ +  /2.
Then | x – y | =  /2 <  , but
| f (x) – f ( y)| = | x2 – y2| = | x + y |  | x – y |
Thus f is not uniformly continuous on 

8. Example 5.4.5*
The function f (x) = x2 is uniformly continuous on D if D is a bounded set.
For example, let D = [– 3, 3].
Then | x + y |  6.
So given  > 0, if  =  /6 and | x – y | <  , we have
| f (x) – f ( y)| = | x2 – y2| = | x + y |  | x – y |
 6| x – y | < 6 =  .
This is a special case of the following theorem.

9. Theorem 5.4.6
Suppose f : D  is continuous on a compact set D. Then f is uniformly
continuous on D.
Proof:
Let  > 0 be given. Since f is continuous on D, f is continuous at each x  D.
Thus for each x  D, there exists a  x > 0 such that
| f (x) – f ( y)| <  /2 whenever | x – y | <  x and y  D.
Now the family of neighborhoods F
is an open cover of D.
Since D is compact, F contains a finite subcover.
That is, there exist x1, …, xn in D such that
If x, y  D with | x – y | <  , then it can be shown
(in the book) that there is some index i such that
Now let  = min
Open cover of D
It follows that | f (x) – f (xi)| <  /2 and | f ( y) – f (xi)| <  /2,
so that | f (x) – f ( y)|  | f (x) – f (xi)| + | f (xi) – f ( y)|
<  /2 +  /2 = .  D

10. How does uniform continuity relate to sequences?
Recall that the continuous image of a convergent sequence need not be convergent
if the limit of the sequence is not in the domain of the function.
For example, let f (x) = 1/x and xn = 1/n.
Then (xn) converges to 0, but since f (xn) = n for all n,
the sequence ( f (xn)) diverges to + .
But with uniform continuity we have the following:
Theorem 5.4.8
Let f : D  be uniformly continuous on D and suppose that (xn) is a Cauchy
sequence in D. Then ( f (xn)) is a Cauchy sequence.
Proof:
Given any  > 0, since f is uniformly continuous on D there exists a  > 0 such that
| f (x) – f ( y)| <  whenever | x – y | <  and x, y  D.
Since (xn) is Cauchy, there
exists N  such that | xn – xm | <  whenever m, n  N.
Thus for m, n  N we have | f (xn) – f ( xm)| <  , so ( f (xn)) is Cauchy. 

11. Using Theorem 5.4.8 we can derive a useful test to determine if a function is uniformly
continuous on a bounded open interval.
We say that a function : E  is an extension of f : D  if D  E and
f (x) = (x) for all x  D. f ~
Theorem 5.4.9
A function f : (a, b)  is uniformly continuous on (a, b) iff it can be extended
to a function that is continuous on [a, b]. f ~
Proof:
If f can be extended to a function that is continuous on the compact set [a, b],
then is uniformly continuous on [a, b] by Theorem f ~
It follows that (and hence f ) is also uniformly continuous on the subset (a, b). f ~

12. Theorem 5.4.9
A function f : (a, b)  is uniformly continuous on (a, b) iff it can be extended
to a function that is continuous on [a, b]. f ~
Proof:
Conversely, suppose that f is uniformly continuous on (a, b).
We claim that lim x  a f (x) and lim x  b f (x) both exist as real numbers.
To see this, let (sn) be a sequence in (a, b) that converges to a.
Then (sn) is Cauchy, so Theorem implies that ( f (sn)) is also Cauchy.
Theorem then implies that ( f (sn)) converges to some real number, say p.
It follows (Theorem ) that lim x  a f (x) = p.
Similarly, we have
lim x  b f (x) = q, for some real number q.
Now define : [a, b]  by f ~
Then is an extension of f . f ~
(x) =
f (x) if a < x < b,
p if x = a,
q if x = b. f ~
Since f is continuous on (a, b), so is f ~
But is also continuous at a and b, so is continuous on [a, b].  f ~