Completeness Notes by Dr E_Prempeh_Ghana

8/13/2019 Completeness Notes by Dr E_Prempeh_Ghana

1/18

1.1 CompletenesWe consider the concept of completeness in normed linear spaces. It is preferable to

study this concept in metric spaces which relates to normed linear spaces via:

, ( , )x y E ( , ) ,x y x y ,x y in ,E .

Recall: In a metric space , ,E a sequence 1n n

x

is called Cauchy if

, 0n mx x as ,n m and the metric space ,E is said to be complete if

every Cauchy sequence in ,E converges to an element in E. The general

procedure to verify that a metric space is complete is as follows:

i. Construct an element x which is used as the limit of the Cauchy sequence.ii. Prove that x is in the space under considerationiii. Prove that nx x as n (in the sense of the metric under

consideration)

To construct x mentioned in (i), one uses the fact that the given sequence 1n n

x

is

Cauchy to generate a Cauchy sequence in a complete space that is normally associated

with the normed linear space under study. This complete space is usually the real line

or the complex plane, . Once x is constructed, (ii) and (iii) are generally not too

difficult to complete.

1.2.1 Completeness ofn

andn

, 1,n n .

Euclidean space 2

,n and the Unitary space 2

,n are complete.

Proof: Euclidean space 2,n

.


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Step (i):

Let

1

22

1 2 21

: , ,..., :n

n

n i

i

x x x x x x

Let ( )1

m

mx

be any Cauchy sequence in

n , where

(1) (1) (1) (1)

1 2

(2) (2) (2) (2)

1 2

( ) ( ) ( ) ( )

1 2

, ,...,

, ,...,

.

.

.

, ,...,

.

.

.

n

n

n

n

m m m m n

n

x x x x

x x x x

x x x x

(1.1)

( )

1

m

mx

is Cauchy implies given 0, : ,N s t N

( ) ( ),s tx x

12

2( ) ( )

1

ns t

i i

i

x x

, ,s t N

2( ) ( ) 2

1

,n

s t

i i

i

x x

,s t N

2 2 2( ) ( ) ( ) ( ) ( ) ( ) 2

1 1 2 2 ...s t s t s t

n nx x x x x x ,s t N

( ) ( )1 1 ,

s tx x ,s t N

( ) ( )

2 2 ,s t

x x ,s t N


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.

.

.

( ) ( )s t

n nx x , ,s t N

These inequalities imply that each column in array (1.1) is Cauchy. Since each column

in (1.1) is a sequence in , and is complete, it follows that each column converges

to a point in . Without loss of generality, let

( )m

i ix x as m

See the following:

(1) (1) (1) (1)

1 2

( ) ( ) ( ) ( )

1 2

, ,...,.

.

.

, ,...,

n

m m m m

n

x x x x

x x x x

.. (1.2)

( 1x, 1x

, , nx

)

Now define 1 2: ( , ,..., )nx x x x

This completes the step (i)

Remark

Observe that the completeness of is used to obtain ix, 1,2,...,i n and from

' ,ix s x is defined.

Step (ii)

W.T.S:nx

,


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Since1 2

: ( , ,..., ),n

x x x x ix, 1,2,...,i n are real numbers nx , completing

step (ii).

Step (iii)

W.T.S:( )

mm

x x

From basic analysis, we have the following:

1 1

mm

x x

1 1 1 1 12

0, : , ,m

N m N x xn

( n from n )

2 2

mm

x x

2 2 2 2 12

0, : , ,m

N m N x xn

( n from n )

.

.

.

m

m

n nx x

1

2

0, : , ,m

n n n nN m N x xn

( n from n )

Choose 1max :ii nN N

Then m N

12

2( ) ( )

1

( , ) :n

m m

i i

i

x x x x

1 12 22 2

1

.n

i

nn n

i.e.( )( , )mx x , m N ( )

mm

x x

Sincenx

(by part (ii)) and the Cauchy Sequence ( )1

m

mx

is arbitrary, it follows

that 2

,n

is complete.

The proof ofn

is complete follows similarly (Exercise!!!)


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1.2.2 Completeness of

The Space is complete

Proof: Recall, 1 2: { , ,... : ,i ix x x x x K and1

: sup ii

x x

}.

Step (i)

Let ( )1

m

mx

be a Cauchy sequence in l

, where

(1) (1) (1) (1)

1 2 3

(2) (2) (2) (1)

1 2 3

( ) ( ) ( ) ( )

1 2 3

, , ,...

, , ,...

.

.

.

, , ,...

.

.

.

m m m m

x x x x

x x x x

x x x x

..(1.3)

( )1

m

mx

is Cauchy implies 0 , : ,N s t N

( ) ( ) ( ) ( ), :s t s tx x x x

,s t N , ( ) ( ) ( ) ( ) ( ) ( )1 1 2 2 3 3sup , , ,...s t s t s tx x x x x x

Then

( ) ( )

1 1

s tx x ,s t N

( ) ( )

2 2

s tx x ,s t N


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.

.

.

These inequalities imply that each column in array (1.3) is Cauchy. Each column is a

sequence of points in and so by the completeness of , each column converges to

a point in . Thus

( )m

i ix x as m for 1,2,3,...i

Let

(1) (1) (1) (1)

1 2 3

(2) (2) (2) (1)

1 2 3

( ) ( ) ( ) ( )

1 2 3

, , ,...

, , ,...

.

.

.

, , ,...

.

.

.

m m m m

x x x x

x x x x

x x x x

.. (1.4)

Define 1 2 3: ( , , ,...)x x x x

Step (ii):

W. T. S: x l

Now ( )1

m

m

x

implies ( )m mx K for each m , which implies

( )m

j mx K for

each j . Also from Step (i), we have( )

mm

j jx x

. So that

( ) ( ) ( ) ( ): m m m mj j j j j j j mx x x x x x x K


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This inequality holds for every j on the right hand side of the above inequality (which

is independent of j ). Hence 1

: j jx x

is a bounded sequence of real numbers. This

implies that x l

.

Step (iii):

From construction in Step (i),( )m

j jx x m N , so

( ) ( )1

, : supm mj jj

x x x x

m N

(since( )m

j jx x , so that ( ) ( )

1

sup mj jj

x x

)

This shows that( )mx x

in l

.

Since ( )1

m

mx

is arbitrary, is complete.

1.2.3 c is complete:

Proof: Recall that c is a subspace of l

.

To show that c is a complete space, it suffices to prove that c is a closed subset of .l

- A closed subset of a complete metric space is complete. To prove that c is closed, it

suffices to prove that c c . But cc , so it remains to prove that c c !

Let 1 2( , ,...)x x x c . This implies that there is a sequence in c such that( )mx x

as m . Using our usual array notation:


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(1) (1) (1) (1)

1 2 3

(2) (2) (2) (1)

1 2 3

( ) ( ) ( ) ( )

1 2 3

( ) ( ) ( ) ( )

1 2 3

, , ,...

, , ,...

.

.

.

, , ,...

.

.

.

, , ,...

.

.

.

r r r r

N N N N

x x x x

x x x x

x x x x

x x x x

We have: 0, 0,N ( ) :N ( )( , )3

mx x

m N

That is( ) ( )

1( , ) : max

3

m m

i ii

x x x x

This implies that:

( )

3

m

i ix x

i and m N .

In particular, for m N and i ,

( )

3

N

i ix x

. (1.5)

Since( ) ( ) ( )

1 2( , ,...)N N Nx x x c , it follows that

( )

1

N

i ix

is convergent and so it is

Cauchy. Every convergent sequence is Cauchy! Hence 1 1: ,N r s N

( ) ( )............(1.6)

3

N N

r sx x

Triangle inequality on the components of 1 2( , ,...)x x x c gives


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( ) ( ) ( ) ( )N N N N

r s r r r s s sx x x x x x x x

( ) ( ) ( ) ( )N N N N

r r r s s sx x x x x x

3 3 3

(using (1.6))

r sx x

This shows that the sequence 1i i

x

is a Cauchy Sequence of real numbers and hence,

converges. Thus1 2( , ,...)x x x c .

Since x was arbitrarily chosen, c c . Hence c c and c c c c , i.e. c is

equal to its closure and so c is a closed subspace of

. But

is complete. Therefore

c is complete.

1.2.4 (1 )p p is complete

Proof:

(1) (1) (1) (1)

1 2 3

(2) (2) (2) (1)

1 2 3

( ) ( ) ( ) ( )

1 2 3

, , ,...

, , ,...

.

.

.

, , ,...

.

.

.

m m m m

x x x x

x x x x

x x x x

( 1 ,x

2 ,x

3 ,x

)


10/18

Now ( )1

m

mx

is Cauchy in p implies a positive integer : , ,N r s N

( ) ( ),r sx x . This implies that

( ) ( ) ( ) ( )1

, .............(1.7)p pr s r s pj jj

x x x x

( ) ( )r sj jx x for each j .

Hence ( )1

m

j mx

is a Cauchy sequence of real numbers and therefore, converges to a

point of . Without loss of generality, let( ) *m

j jx x , for each j .

Define* * * *

1 2 3

( , , ,...)x x x x

Next, we show that* * * *

1 2 3( , , ,...) px x x x

From (1.7),( ) ( )

1

pr s p

j j

j

x x

, ,r s N

Then,( ) ( )

1

,k

pr s p

j j

j

x x

( 1,2,3,...)k

Let s (and recalling that( ) (*)s

j jx x ), we obtain for r N

( ) *

1

,k

pr p

j j

j

x x

( 1,2,3,...)k

Now as ,k it follows that r N

( ) *

1

...............(1.8)p

r p

j j

j

x x

Replace rby m N in (1.8) to obtain

( ) *

1

,p

m p

j j

j

x x

( ) *m px x . But by hypothesis,( )m

px . Since p is a vector space, we

have:


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* ( ) * ( )m m px x x x .

Finally, we show that ( )1

m

m

x

converges to

* * * *

1 2 3( , , ,...)x x x x

From (1.8),

1

( ) (*) ( ) *

1

,p

pm m

j j

j

x x x x

m N

This implies that( ) *

mm

x x

. Since ( )1

m

mx

was an arbitrary Cauchy sequence in p ,

it follows that p is complete.

1.2.5 ,C a b , endowed with sup norm is complete.

Proof: Let 1n n

f

be a Cauchy sequence in ,C a b , then for each ,t a b , a

positive integer :N 0 ,

,

,

: supn m n mC a bt C a b

f f f t f t

,m n N

Hence, for any fixed 0 ,t a b , 0 0n mf t f t ,m n . This shows that

0 1n nf t

is a Cauchy sequence of real numbers, since is complete, 0 1n nf t

converges to a real number, say, 0f t as n , that is

0 0n

nf t f t

.

This is the same as saying that the functionnf converges point-wise to the function f.

We prove that, this point-wise convergence is actually uniform convergence in

, ,t a b that is, 0 , * :N

,

sup nt a b

f t f t

*,n N

Given 0 , choose :2

n mN f f

,n m N . Then for n N ,


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n n m mf t f t f t f t f t f t

,

sup n m mt a b

f t f t f t f t

n m mf f f t f t By choosing m sufficiently large ( m may depend on t), each term on the right-hand-

side can be made less than2

so that

,

sup nt a b

f t f t

n N , hence the

convergence is uniform,

i.e. nf t f t , uniformly on ,a b .

Since the 'n

f s are continuous on ,a b and the convergence is uniform, it follows from

a well-known theorem in analysis (advanced calculus) that the limit function f t is

continuous on ,a b . Hence, ,f C a b .

This proves the completeness of ,C a b with the sup norm.

Definition 1.1.7

Any complete normed linear space is called a Banach space.

1.2.6. Examples of Incomplete normed linear spaces.

Example 1.1.8

Let ,X C a b , the space of all continuous realvalued functions on the closed and

bounded interval ,a b with norm given by

( )

b

af f t dt , ,t a b , ,f C a b .

It suffices to produce a Cauchy Sequence in ,a b which does not converge to an

element of ,C a b . Without loss of generality, let , 1,1a b .

Consider the function


13/18

0, 1,01

( ) , 0

11, 1

n

if t

f t nt if tn

if tn

drawn as

It is clear that 1n n

f

is a sequence of continuous realvalued functions defined on

1,1 .

We show that 1n n

f

is Cauchy!

Let m n (so that1 1

m n

). The graphs of nf t and mf t on the same axes is

shown as

1

11

(0, 1)

()

0


14/18

We need to show that,

0n m

m nf f

.

Now

1

1......(1.9)m n m nf f f t f t dt

Two methods are used to evaluate (1.9)

Method 1

1

1m n m nf f f t f t dt

= Area of OAB

,1 1 1 1

.1 02 2

m n

ABn m

Hence 1n n

f

is Cauchy in 1,1C .

Method 2

1

1m n m nf f f t f t dt

1 10 1

1 11 0(0 0) ( ) (1 ) (1 1)

m n

m n

dt mt nt nt dt dt

1

11

(0, 1)

()

0 1

() ()

()


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1 1

10( ) (1 )

m n

m

mt nt nt dt

11

2 2

102 2

nm

m

m n t ntt

2 2

1 1 1 1

2 2 2

m n n

m n n m m

1 1 1 1 1

2 2 2 2m m n m m (Since 1

n

m )

0 as ,m n

So also 1n nf

is Cauchy in 1,1C .

Next, we show that 1n n

f

converges to an element which is not in 1,1C .

Examining the earlier sketch, a possible candidate for the limit of nf t is

= 01 ifif1 00 1

Claim: 0n

nf f

.

Proof of Claim:

1

11

(0, 1) ()

()


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1

1n nf f f t f t dt

= Area shaded.

1 10

2

n

n

(Direct Computation can also be used)

However 1,1f C , since fis not continuous. Thus 1,1C is not complete.

Remark 1.19

The interval 1,1 which is employed in the above example is certainly not crucial.

Other intervals can easily be used. For example, the interval 0,4 is quite suitable.

Consider the space 0,4X C and use the function!!

(Exercise!!)

drawn as

and the limit!!

(Exercise!!)

2 +2

4

(0, 1)

0 2


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could be used as the limits of the Cauchy Sequence 1n n

f

.

Example 1.20

Let 3,3X C with 1

3 22

2 3:f f t dt

.

Then 3,3X C is not complete.

Consider the function nf t defined by

Let : 3,3 0,1f be defined by

(Exercise!!)

(0, 1)

0 2

3

, 0 33

(0, 1)

0


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Claim:

i. 1n n

f

is Cauchy Sequence in 3,3C

ii. nnf f

in the sense of the norm: 0nf f .

(Do the above as an exercise!)

Exercise

1. Let 2,2 .X C Show that

i. 21 2

:f f t dt

ii. 1

2 22

2 2:f f t dt

for arbitrary 2,2f C is not complete.

2. Let 0,5 .X C Show that

i.

5

1 0

:f f t dt

ii. 1

5 22

2 0:f f t dt

for arbitrary 0,5f C is not complete.

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Completeness Notes by Dr E_Prempeh_Ghana