Analysis with an introduction to proof 5th by steven lay ch05b

Chapter 5
Limits and Continuity

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 5.4, Slide 1


Section 5.4
Uniform Continuity

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 5.4, Slide 2


For a function f : D →

to be continuous on D, it is required that

for every x0 ∈ D and for every ε > 0 there exists a δ > 0 such that
| f (x) – f (x0)| < ε whenever | x – x0 | < δ and x ∈ D.
Note the order of the quantifiers: δ may depend on both ε and the point x0.

If it happens that, given ε > 0, there is a δ > 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 ε > 0 there exists a δ > 0 such that | f (x) – f ( y)| < ε
whenever | x – y | < δ and x, y ∈ D .

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 5.4, Slide 3


Example 5.4.2*

We claim that the function f (x) = 3x is uniformly continuous on

.

Given any ε > 0, we want to make | f (x) – f ( y)| < ε by making x sufficiently close to y.

We have

So we may take δ = ε /3.

| f (x) – f ( y)| = | 3x – 3y| = 3| x – y|.

Then whenever | x – y | < δ we have
| f (x) – f ( y)| = 3| x – y| < 3δ = ε.

We conclude that f is uniformly continuous on

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

.

Section 5.4, Slide 4


Example 5.4.4
2
The function f (x) = x is not uniformly continuous on

.

Let’s look at this graphically.
Given a point x1 close to 0 and an

ε -neighborhood about f (x1),

f (x)
f (x) = x

2

the required δ can be fairly large.

But as x increases, the value

f (x2)

of δ must decrease.

This means the continuity is

ε -neighborhood

not uniform.

f (x1)
x1

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x2

x

Section 5.4, Slide 5


Example 5.4.4
2
The function f (x) = x 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
∀ ε > 0 ∃ δ > 0 such that ∀ x, y ∈ D, | x – y | < δ implies | f (x) – f ( y)| < ε .

So the function f fails to be uniformly continuous on D if
∃ ε > 0 such that

∀ δ > 0,

(Any ε > 0 would work.)

Suppose we take ε = 1.

Now for the proof.

| x – y | < δ and | f (x) – f ( y)| ≥ ε.

∃ x, y ∈ D such that

We must show that given any δ > 0, there exist x, y in

such that | x – y | < δ and

2 2
| f (x) – f ( y)| = | x – y | = | x + y | ⋅ | x – y | ≥ 1.
For any x, if we let y = x + δ /2, then | x – y | = δ /2 < δ .
1 ≤ | x + y| ⋅ | x – y|
we need to have | x + y | ≥ 2/δ .

=

Thus to make

| x + y | ⋅ (δ /2),

This prompts us to let x = 1/δ .

Here is the formal proof.

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 5.4, Slide 6


Example 5.4.4
2
The function f (x) = x is not uniformly continuous on

.

Proof:
Let ε = 1.

Then given any δ > 0, let x = 1/δ and y = 1/δ + δ /2.
Then | x – y | = δ /2 < δ , but
2 2
| f (x) – f ( y)| = | x – y | = | x + y | ⋅ | x – y |

=
Thus f is not uniformly continuous on

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

1 1 δ  δ   2  δ 
+ +
>  ÷ ÷ = 1.

δ δ 2  2 ÷
  δ  2 

.♦

Section 5.4, Slide 7


Example 5.4.5*
2
The function f (x) = x is uniformly continuous on D if D is a bounded set.
For example, let D = [– 3, 3].

Then | x + y | ≤ 6.

So given ε > 0, if δ = ε /6 and | x – y | < δ , we have
2 2
| f (x) – f ( y)| = | x – y | = | x + y | ⋅ | x – y |
≤ 6| x – y | < 6δ = ε .

This is a special case of the following theorem.

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 5.4, Slide 8


Theorem 5.4.6
Suppose f : D →

is continuous on a compact set D. Then f is uniformly

continuous on D.

Proof:
Let ε > 0 be given. Since f is continuous on D, f is continuous at each x ∈ D.
Thus for each x ∈ D, there exists a δ > 0 such that

x

| f (x) – f ( y)| < ε /2

whenever | x – y | < δ

Now the family of neighborhoods F
Since D is compact, F contains a finite subcover.
That is, there exist x1, …, xn in D such that

{ δ x / 2, K , δ x / 2} .

Now let δ = min

1

n

and y ∈ D.

  δ 


=  N  x ; x ÷: x ∈ D 
  2 

 δ x1
D ⊆ N  x1 ;
2


is an open cover of D.

δ xn


÷∪ L ∪ N  xn ;
2




÷.


If x, y ∈ D with | x – y | < δ , then it can be shown

(in the book) that there is some index i such that
Open cover of D

x

| x − xi | < δ xi and | y − xi | < δ xi .

It follows that | f (x) – f (x )| < ε /2 and | f ( y) – f (x )| < ε /2,

i

i

so that | f (x) – f ( y)| ≤ | f (x) – f (x )| + | f (x ) – f ( y)|

i

i

< ε /2 + ε /2 = ε. ♦
D
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Section 5.4, Slide 9


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 ε > 0, since f is uniformly continuous on D there exists a δ > 0 such that
| f (x) – f ( y)| < ε whenever | x – y | < δ and x, y ∈ D.
Since (xn) is Cauchy, there
exists N ∈

such that | xn – xm | < δ whenever m, n ≥ N.

Thus for m, n ≥ N we have | f (xn) – f ( xm)| < ε , so ( f (xn)) is Cauchy. ♦

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 5.4, Slide 10


Using Theorem 5.4.8 we can derive a useful test to determine if a function is uniformly
continuous on a bounded open interval.
~
f
We say that a function

~
f

f (x) =

:E→

is an extension of f : D →

if D ⊆ E and

(x) for all x ∈ D.

Theorem 5.4.9
A function f : (a, b) →
~
f

to a function

is uniformly continuous on (a, b) iff it can be extended

that is continuous on [a, b].

Proof:
~
If f can be extended to a function
f
~
f
It follows that

then

that is continuous on the compact set [a, b],

is uniformly continuous on [a, b] by Theorem 5.4.6.

~
f f ) is also uniformly continuous on the subset (a, b).
(and hence

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Section 5.4, Slide 11


Theorem 5.4.9
A function f : (a, b) →
~
f

to a function

is uniformly continuous on (a, b) iff it can be extended

that is continuous on [a, b].

Proof:
Conversely, suppose that f is uniformly continuous on (a, b).
We claim that lim → f (x) and lim → f (x) both exist as real numbers.
x a
x b
To see this, let (sn) be a sequence in (a, b) that converges to a.
Then (sn) is Cauchy, so Theorem 5.4.8 implies that ( f (sn)) is also Cauchy.

Theorem 4.3.12 then implies that ( f (sn)) converges to some real number, say p.
It follows (Theorem 5.1.10) that lim → f (x) = p.Similarly, we have
x a
Now define

lim → f (x) = q, for some real number q.
x b

~
f (x) =

f (x)

if a < x < b,

p

if x = a,

q

But

by

~
Then

is an
f extension of f .

Since f is continuous on (a, b), so is

if x = b.

~
is falso continuous at a and b, so

~
: [a, b] →
f

.

~
f

~
is continuous on [a, b]. ♦ f

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Section 5.4, Slide 12