A STUDY OF A CLOSED GRAPH THEOREM AND THEIR APPLICATIONS IN BANACH SPACES FOR CLOSED LINEAR TRANSFORMATIONS.

A STUDY OF A CLOSED GRAPH THEOREM AND

THEIR APPLICATIONS IN BANACH SPACES FOR

CLOSED LINEAR TRANSFORMATIONS

Tsegaye Kebede Irena

1

,

Dr.Tadesse Zegeye

2

, Dr. Vasudevarao Kota

3

1

Lecturer, 3Assistant Professor, Department of Mathematics, Ambo University, Ambo, Ethiopia.

2

Assistant Professor, Department of Mathematics, Bahir Dar University, Bahir Dar, Ethiopia

I. INTRODUCTION

In this paper we deal with one of the fundamental theorems for Banach spaces, namely, the closed graph theorem and bounded linear operators and have proved some theorems concerning such operators. In applications, therefore, whenever one encounters a map, if one is able to prove that the map is linear and bounded, then one has those theorems at ones disposal. It turns out, however, that some of the important maps which are defined in applications, under certain conditions, are linear but not bounded. In particular, many maps defined in terms of differentiation (ordinary or partial) are discontinuous. Luckily, these operators often posses another property which, in a way, makes up for the fact that they are not bounded. These operators are closed (or have closed graphs ) with the definitions and some properties of closed linear maps, closed graphs and these will lead to one of the key theorems of functional analysis - the closed graph theorem.

A mapping from one topological space to another is called an open mapping if the image of each open set is open. Thus a one-to-one continuous open mapping is a homeomorphism. We shall show that a continuous linear transformation of one Banach space onto another is always an open mapping and use this to give criteria for the continuity of linear transformation.

Keywords: Linear spaces, Normed Linear spaces, Banch spaces, Linear operators,

Bounded linear operators, Closed Graphs

II. LINEAR (VECTOR) SPACES

Definition 1.1 A set X of elements is called a vector space or a linear space over a scalar field F (F or C if we have a function + on X X _to X and a function . on FX to

X that satisfy the following conditions: (i) xy  yx; x,yX;

(ii) (x y)z x(yz); x,y,zX;

(iii) There is a unique vector 0 in X such that x0x for all x in X ;

(iv) To each x in X therecorresponds a unique element x such that x(x)0; (v) (x y)xy; F, x,yX;

(vi) ()xxx; ,F , xX; (vii) (x)()x; ,F, xX; (viii) 1.x x for 1F, xX;

(ix) 0.x0 for 0F, xX.

We call + addition and . multiplication by scalars. If F is a real, then X is called a real linear space and if F is a complex, then X is called a complex linear space.

Example 1.2 Show that + is continuous function from XX into X and that . is a continuous function from FX into X.

Verification. Consider + as a function from XX into X . Since

o o

o

o y x x y y

x y

x )(  )    

( for any x_o and y_o.

Hence, + is continuous.

Consider . as a function from FX into X . Since

o o o

o o o

o

ox x x x x x x

x  ( ) (  ) .   .

          for any fixed o &xo.

Hence, . is continuous.

 

1

Example 1.3 (a) Let

   



 _{ }



 n i

n

x x x x

x ( ₁, ₂,..., ),

_

with addition and scalar multiplication defined component wise. Then, n is a linear space over .

(b) X  is not a linear space over the set of complex numbers,C

(c) The space p,1 p, let p be defined as follows:

. :

,...), , , (

1 3

2 1 _

   

 

  

 









i p i i

p

x x

x x x x

For ( 1, 2, 3,...)

_

x x x

x , ( 1, 2, 3,...)

_

y y y

y  arbitrary elements of p

 , define ,...)

, ,

( _{1 1 2 2 3 3}

_ _

y x y x y x y

x     and ( ₁, ₂, ₃,...)

_

x x x

x   

  .Then, p is a linear space.

Example 1.4 Let the space  be defined as follows:

   



 _{ }





bounded is

x x

x x x

x i

_ 3

2 1 _

: ,...),

, , (

 .

That is,  is the set of all bounded sequences. If addition and scalar multiplication are defined component wise as in example 1.3(c), then  is a linear space.

Definition 1.6 A nonempty subset S of a vector space X is a subspace or linear manifold if

y

x 

  is in S whenever x and y are both in S and , are any scalars. If S is also closed as a subset of X , then it is called a closed linear manifold.

 

2

Nor med Linear Spaces

Definition 1.7 Let X be a linear space over a field F, where F  or C. A norm on X is a real–valued function ,

 

  0, :

. X

which satisfies the following conditions: 0

:

1 x 

N , and x 0if and only if x 0,xX;

x x

N₂ :    for all F,xX;

y x y x

N₃ :    for arbitrary x,yX ( triangle inequality). A linear space with a norm defined on it is called a normed linear space.

[

2.1 Convergence and Completeness

The notion of convergence for a sequence of real numbers generalizes to give us a notion of convergence for sequences in a normed linear space.

Definition 1.11 Asequence f_n in a normed linear space is said to converge to an element

f in the space if given  0, there is an N such that for all nN, we have f  f_n  . If fn converges to f, we write n

n f

f

 

lim or fn  f .

Another way of formulating the convergence of f_n to f is by noting that f_n  f if 0

  f

Convergence in the space Lp, 1 p is often referred to as convergence in the mean of order p. Thus a sequence of functions f_n is said to converge to f in the mean of order

p if each f_n belongs to Lp and  0

p n

f

f . Convergence in L is nearly uniform convergence.

Just as for the case of sequences of real numbers, we say that a sequence f_n in a normed linear space is a Cauchy sequence if given  0, there exists N such that for all n,mN

we have f_n  f_m  .

It is easily verified that each convergent sequence is a Cauchy sequence.

Example 1.12 Let f_n be a sequence of functions in L. Prove that f_n converges to f in



L if and only if there is a set E of measure zero such that f_n converges to f uniformly on

c

E .

Verification. Let f_n be a sequence of functions in L. Suppose f_n  f _ 0. Given  0, there exists N such that







: : ( ) ( )  0





inf M mt f_n t f t M for n N. Thus m



t: f_n(t) f(t) 



0 for n N.

Let E



t: f_n(t) f(t) 



. Then mE0 and f_n converges uniformly to f on Ec. Conversely, suppose there exists a set E with mE0 and f_n converges uniformly to f

on Ec.

Given 0, there exists N such that

2 ) ( )

(t  f t  

fn for nN and

c

E t .

Thus t fn t f t E

   



 _{ }

2 ) ( ) (

:  for n N.

Hence, inf



M :m



t: f_n(t) f(t) M



0



 for nN. That is, f_n  f _ 0.

Definition 1.13 A normed linear space is called complete if every Cauchy sequence in the space converges; that is, if for each Cauchy sequence f_n in the space there is an element f

A complete normed linear space is called a Banach space.

In the case of the real line, every Cauchy sequence converges; that is, being a Cauchy sequence is sufficient to guarantee the existence of a limit. In the general case, however, this is not so.

To check or verify that a normed linear space X is complete, we take an arbitrary Cauchy sequence xn _n1 in X and show that it converges to a point in X .

Example 1.14 For 1 p, we denote by p

 the space of all sequences _

1

 

 such that







 

1

 

 p

.

Prove that a normed linear space p

 is complete.

Verification.Let 1 p and let _(n) be a Cauchy sequence in p. Given  0, there exists N such that



 n  m p p



( ) ( ) for n,m N.

In particular, _(n)_(m) p p for n,m N and each . Thus for each , _(n) is Cauchy in  so it converges to some _.

Consider _ . Then





 

k

p p n

1 ) (

  

 

 for each k and each n N. So



 n  p p



( ) for nN.

Thus (n) p for n N. So  p and (n)   0.

Example 1.17 Let X and Y are Banach spaces. Show that the product space XY, with the norm defined by

y x y

x, )  

( ,

 

x,y XY

is Banach.

Verification. To show that XY is complete, let (x_n,y_n) be a Cauchy sequence in

Y X .

x_n x_m 



x_n,y_n

 

 x_m,y_m



and

yn ym 



xn,yn

 

 xm,ym



for all n,m1.

These show that x_n and y_n are Cauchy sequences in X and Y , respectively. Since X

and Y are Banach, they converge. Let x_n x and yn  y. Then,



x_n,y_n

   

 x,y  x_n x,y_n y



 x_n x  y_n y 0

That is,



xn,yn

  

 x,y in XY. Hence, XY is a Banach space.

2.2

Linear Operators

Definition 1.19 Let X and Y be linear spaces over a scalar field F.

A mapping T:X Y is said to be a linear operator, a linear transformation, a linear map if )

( ) ( )

( x y x y

T    

for all x,y in X and all scalars ,F .

Definition (1.19) is equivalent to the following two conditions: )

(i T(xy)TxTy for all x,yX; and )

(ii T(x)T(x) for all xXand for each scalar,  .

Example 1.20 LetX 2 Y. For each 



1, 2, 3,



_

x x x

x in 2 define

   



 _

 ,

3 , 2 , ,

0 2 3

1

_ _{x x}

x x

T ..

Then T is a linear operator on 2

 .

Verification. Let 



1, 2, 3,



_

x x x

x and 



1, 2, 3,



_

y y y

y be arbitrary elements of 2

 and let , be scalars. Then

















 _{1, 2}, ₃, ₁, ₂, ₃,

_ _

y y y x

x x y

x        





x₁y₁,x₂ y₂,x₃y₃,



and 

  



    _



 ,

3 , 2 , 1 , 0 )

( 1 1 2 2 3 3

_

_ _{x y x y x y}

y x



  



 _

    



 _

 ,

3 , 2 , , 0 ,

3 , 2 , ,

0 2 3

1 3

2 1

y y y x

x

x     



( ) ( )

_ _

y T x

T 

 



and so T is a linear operator.

Proposition 1.21 Let X and Y be two linear spaces over a scalar field F, and let

Y X 

: be a linear operator. Then, )

(i (0)0; )

(ii The range of T, R(T)



yY/Tx y for some xX



is a linear subspace of Y ; )

(iii T is one-to-one if and only if T(x)0 implies x0.

Proof.(i) Since  is linear, we have T(x)T(x) for each xX and each scalar . Take  0and then (0)0.

(ii) We need to show that for y₁,y₂R(T) and , scalars, y₁y₂R(T).

Now, y1,y2R(T) implies that there exist x1,x2X such that T(x1) y1 and T(x2) y2.

Moreover, x₁x₂X since X is a linear space. Furthermore, by the linearity of T we have

T(x₁x₂)T(x₁)T(x₂)y₁y₂.

Hence,y₁y₂R(T), and so R(T) is a linear subspace of Y_.

(iii) Assume T is one- to one. Clearly Tx0T(x)T(0) since T is linear (and so ).

0 ) 0 ( 

T But T is one-to-one. So,x0.

Conversely, assume that whenever Tu0, then u0. So, let TxTy. Then, 0

 Ty

Tx and by the linearity of T, T(xy)0. By hypothesis, xy0 which implies

y x .

Hence, T is one-to-one.

2.3 Bounded Linear Operators

Definition 1.23 Let X _and Y be normed linear spaces over a scalar field F, and let

Y X

T:  be a linear operator. Then T is said to be bounded if there exists some constant 0



K such that for all xX,

We now turn our attention to linear maps that are continuous. The notion of continuity can be stated, for linear maps in several useful equivalent forms. We state these equivalent forms in the following theorem.

Theorem 1.24 X and Y be normed linear spaces and let T:X Y be a linear operator. Then T is continuous at the point x if and only if for every sequence x_n converging to x,

Tx Txn  .

Theorem 1.25 Let X and Y be normed linear spaces and let T:X Y be a linear operator. Then the following statements are equivalent:

)

(i T is continuous; )

(ii T is continuous at 0. That is, if x_n is a sequence in X such that x_n 0 as n

then Txn 0 in Y as n;

)

(iii T is bounded. That is, there exists a constant K0 such that for each ,

X

x Tx  K x _;

)

(iv If D



xX : x 1



is the closed unit disc in X , then T(D) is bounded. That is, there exists a constant M 0 such that Tx M for all xD.

Proof. (i)(ii) Let T:X Y be linear and continuous.

So, let x_n be a sequence in X such that xn 0. By continuity of T, we have )

0 (

T

Txn  . But T is a linear operator so that T(0)0.

Hence, T is continuous at 0.

(ii)(iii) We are given that if x_n is any sequence in X such that x_n 0 as n, then .

0 ) (xn 

T We want to prove that there exists a constant M 0 such that Tx  M x

for each xX.

Suppose for contradiction that there is no such M. Then, for any positive integer n, there exists some x_n X , x_n 0 such that

n

n n x

This implies that 1 n n x n Tx .

Now, define the sequence .

n n n x n x

u  Clearly, u_n 0. For,

0   1  1 0.

n x x n x n x u n n n n n

However, Tu_n does not converge to 0. For,

0  ( )  ( ) 1,

n n n n n x n x T x n x T Tu

and so Tun does not converge to 0 (even though un 0). This contradicts the hypothesis

that T is continuous at the origin.

Thus our assumption is false and so, (ii)(iii). (iii)(iv). Given that a linear map T is bounded.

Take yD arbitrary. Then, y 1. By(iii), Ty K y for some constant K0. But .

1



y So, Ty  K y  K.

Since y was arbitrarily chosen in D, it follows that for all y in D, Ty K. Now, take K M and we are done.

(iv)(i) Let x₁,x₂ be arbitrary elements of X. Assume first that x₁x₂ 0. Consider the vector

2 1 2 1 : x x x x u  

 . Clearly, uD , and so, by condition (iv), (i.e.,T(D) is

bounded), there exists a constant K 0 such that T(u) K. i.e., K x x x x T    ) ( 2 1 2

1 _{, or}

2 1 2

1 Tx K x x

Tx    .

If, on the other hand, x₁x₂ 0, this inequality clearly also holds. Thus, the inequality holds for all x₁,x₂ X.

Now, given any  0, choose

1   K   .

So, if x₁x₂  we obtain,

and hence T is (uniformly) continuous on X _.

Corollary 1.26 Let X and Y be normed linear spaces, and let T be a linear operator such that T:X Y. Then

)

(i T is a bounded linear operator if and only if it is continuous. )

(ii If Tis continuous at x_o, then it must be continuous everywhere. )

(iii If T is continuous at x_o, then it is bounded.

Proof. (i) Suppose T is a bounded linear operator. Then

   

Ty T x y

Tx for all x,yX with

T y

x   . Thus, T is uniformly continuous and so T is continuous.

Conversely, to show that continuity implies boundedness, the proof will be by contradiction. Thus, suppose T is continuous but not bounded. The negation of being bounded is that for any natural number n, however large, there is some point, say x_nX,x_n 0, such that

n

n n x

Tx  .

This implies that 1

n n

x n

Tx

.

Consider now the vectors

n n n

x n

x

u  with norm

n u_n  1 .

Clearly, u_n 0 because 0  1 0

n

un .

Since T is linearand continuous, this implies T(un)T(0)0.

But 0  ( )  1

n n n

n n

x n

Tx

x n

x T

Tu and so Tu_n does not converge to 0. This contradicts the hypothesis that T is continuous. Thus our assumption that T was not bounded must be false when T is continuous and so continuity implies boundedness.

Thus for linear maps, continuity and boundedness are equivalent.

(ii) By the hypothesis, if x_n x_o, then T(x_n)T(x_o).

To prove continuity everywhere, we must show, for any yX , if yn  y, then

). ( )

(y T y

To this end, suppose yn  yX and consider

) (

)

(y_n T y_n y x_o y x_o

T     

T(y_nyx_o)T(yx_o). Since yn yxo xo,

) ( )

(y_n y x_o T x_o

T    and T(y_n)T(y). Hence, T is continuous everywhere.

(iii) Suppose now that T is a linear operator that is continuous at x_o. Then there is a  0 such that TxTx_o 1 for all x such that xx_o  .

For any z in X with z0, set

z z

w , where 0  .

Then Tz Tw T(w x_o) T(x_o)

z     

and Tz  T(wx_o)T(x_o) 1,

z 

since

    

x x w

w _{o o} .

Consequently, Tz 1 z and T is bounded.

Example 1.27 Show that if T_n T and x_n x, then T_nx_n Tx.

Verification. Suppose Tn T and xn x. Then,

0

 T

T_n and x_n x 0.

Since T_nx_nTx  T_nx_n Tx_n  Tx_n Tx  T_n T x_n  T x_n x and x_n is bounded.

This implies T_nx_n 0. That is, T_nx_n Tx. Thus ker is closed. T

Definition 1.30 If T: X Y is a bounded linear maps from a normed linear space X in to a normed linear space Y, we define the norm of T by

T inf



K: Tx  K x



for each xX.

Theorem 1.31 Let B(X,Y) be the family of all bounded linear maps fromX to Y. Then we have the following:

For arbitrary TB(X,Y),

x Tx Tx

Tx T

x x

x 1 1 0

sup sup

sup

 

  



Proof. Since T is bounded and linear, there exists K 0 such that for all xX ,

x K Tx  .

If x 1, then Tx  K x  K and so,



Tx : x 1



is a bounded set in , so its “sup” exists and Tx K

x



1

sup for any K such that

x K

Tx  xX.

Taking the infimum over all such K' we obtain that s



K Tx K x

Tx

x

 



 inf 0:

sup

1

for all xX



 T

Hence, Tx T

x



1

sup . (1.2)

Conversely, from definition 1.30, we get that for every  0, there exists x_ in X ,x_ 0, such that

Tx_ (T ) x_ .

(Otherwise Tx_ (T ) x_ and T would no longer be the infimum). Let

  

x x

u  , then u_ 1 and Tu_  T . Consequently, we obtain from inequality (1.2) that



 



  

 

 

 x T

x T Tx

Tx T

x x

x

( sup sup

sup

0 1

1

Since  0 is arbitrary, we get that

T x

Tx Tx

Tx T

x x

x

 

 

 

1 1 0

sup sup

sup .

Lemma 2.1 Let T be a continuous linear transformation of the Banach space X onto the Banac space Y. Then the image by T of the unit sphere in X contains a sphere about the origin in Y.

2.4The Graph of a Mapping

Definition 2.3 Let X and Y be normed linear spaces and T:X Y be any map. Then, the

graph of T denoted by G_T, is defined by

GT 



(x,Tx): xX



.

Observe that G_T is a subset of XY and that

T

G y x, )

( if and only if Tx y.

Example 2.4: LetX 

 

0,1 ,Y , and T:

 

0,1  be defined by Txx2,x

 

0,1.

Then, the graph of T, G_T is given by

 



( , ):  0,1







( , 2): 

 

0,1



 x Tx x x x x

GT .

2.5 Closed Linear Maps and Closed Graphs

Definition 2.6: Let X and Y be normed linear spaces and let D be a subspace of X . Then the linear map

Y D

T: 

is called closed if for every convergent sequence x_n of points of D, where xn xX

say, such that Tx_n is a convergent sequence of points of Y, where Tx_n yY say, the following two conditions are satisfied:

(i) xD (ii) Tx y

Example 2.7: The graph of T given in example 2.4 is closed.

To see this, it suffices to take an arbitrary convergent sequence in G_T and show that its limit is in G_T.

So, let (x_n,Tx_n) be an arbitrary sequence in G_T and suppose (x_n,Tx_n)(x,y) as n. This implies that x_n x and Tx_n  y as n.

Observe that the map T is continuous on

 

0,1 . Thus, xn xTxn Tx as n.

Hence, G_T is closed subset of XY and so T is a closed map.

Proposition 2.9: Let X and Y be normed linear spaces and let D be a subspace of X . Then the linear map

Y D

T: 

is called closed if and only if its graph is a closed subspace.

Proof: SupposeX D Y

T



 is a closed linear map.

To show that G_T is closed, we must show that any limit point of G_T is actually a member of

T

G .

Let (x,y) be a limit point of G_T. Since this is so, there must be a sequence of points of

T

G ,(x_n,Tx_n), where x_n D, converging to (x,y). This is equivalent to requiring that 0

) ,

(x_n x Tx_n  y  or x_n x  Tx_n  y 0, which implies that

x

x_n  and Tx_n  y. Since T is closed, we can say that this implies

D

x _and yTx, and, in view of this, we can write

T

G Tx x y

x, )( , )

( .

Therefore, G_T is closed.

Conversely, suppose G_T is closed and that

x

x_n  , xnD

for all n, and Tx_n  y.

We must show that xD and yTx. The condition implies that

___

) , ( ) ,

(xn Txn  x y GT .

Since G_T is closed,

___

T

T G

G  , and we have that

T

G y x, )

( .

By definition of G_T this means that

D

x _and yTx. Therefore, T is closed.

Y D X

T

 

where D is a closed subspace of X . Then T is closed.

Proof. Suppose x_n is a sequence of points of D such that

x

x_n  and Tx_n  y.

Since D is closed, the limit of the sequence must be in D. That is, xD. The continuity of Timplies that

Tx Tx_n  . By the uniqueness of limit, we have

y Tx . Hence, T is closed.

To appreciate the importance of closed graph theorem, we first give an example of a map which is

)

(i Linear; (ii) Closed; and (iii) Not bounded

Example 2.12: Suppose T:DY, DX, where X C

 

0,1 Y _{with the} sup norm and

D consists of all those functions that have a continuous derivative; that is,

 



f X/ f' C0,1



D  

We take the linear map T to be differentiation,

'

f Tf 

for any f D. Then, )

(i T is linear (ii) T is closed; and (iii) T is not bounded.

Verification. (i) Let f,gD and , be scalars. Then,

Tg Tf

g f g f

g f g

f

T(  )(  )'( )'( )' ' ' 

Hence, T is a linear map.

(ii) To show that T is closed, suppose f_n is a sequence of points of D such that

f f_n 

and Tfn  fn  y

'

. Then,

 ( )( ) ( ) sup  ( ) ( ) 0

sup '

1 , 0 1

, 0

 

 

 

 

t y t f t

y t Tf y

Tf _n

t n

t

n as n.

So, fn Tfn  y

'

uniformly and y(t) lim f_n'(t)

n



Since the convergence is uniform, we have

) 0 ( ) ( )) 0 ( ) ( ( lim ) ( lim ) ( lim ) ( 0 0 ' '

0y s ds f s ds f s ds n fn t fn f t f

t t n n n n t            







.

Hence, f t  f 



t y s ds

0 ( )

) 0 ( )

(

So, f' is continuous because it is the limit of a uniformly convergent sequence of continuous function. This implies that f'(t) exists and f'(t) y(t) for all t

 

0,1 .

Thus,

D

f  _and Tf  y. Therefore, T is closed.

(iii)To show that T is not bounded, consider the sequence of points of D,

n

n t t

f ( ) for n1,2,.

In this case Tf_n ntn1.

We now observe that, with the sup norm,

  1

sup 1 , 0    n t n t

f and

 nt n

Tf n

t

n  

  1 1 , 0

sup for all n.

Hence, Tf_n nn f_n . Since      n f Tf f Tf T n n n n f sup sup sup 0

This shows that T is not bounded (or not continuous) linear map.

Proposition 2.13 Let X be a linear vector space that is complete in each of the norms

1

and ₂, and suppose there is a constant C such that

2 1 C x

x 

for all xX. Then the norms are equivalent. That is, there is a second constantC' such that

1 2 C' x

x 

Proof. The identity map of (

2

,

X ) onto (

1

,

X ) is a one-to-one continuous linear transformation and so must be an isomorphism by proposition 2.2. Therefore, the inverse mapping must be bounded.

2.6 The Closed Graph Theorem

Theorem 2.14 (The Closed Graph Theorem) Let X and Y be two Banach spaces. Let )

(i T:X Y be a linear map; )

(ii The graph of T, G_T is closed.

Then, T is a bounded (continuous) operator.

Proof. Define a new norm in X by

1 1

2 x Tx

x   .

Then X is complete in the norm ₂. For if 0

2   _q

p x

x , then 0

1 

 q

p x

x and

0

1 

 q

p Tx

Tx . Hence by the completeness of X and Y we have xX and yY such

that 0

1 

x

x_p and 0

1 

y

Tx_p . By the hypothesis of our theorem yTx. Hence, 0

2  x

x_p , and X is complete with respect to ₂. Now since X is complete in both norms

1 and 2, there is C' such that 1 1

1

2 x Tx C' x

x    . Thus

1 1 C' x

Tx  , and T is bounded.

Therefore, T is continuous.

Note that the converse of Closed Graph Theorem is also true; i.e., if X and Y are Banach spaces and T:X Y is such that TB(X,Y), then the graph of T ,G_T is closed (or equivalently, T is a closed map).

To see this, suppose x_n is a sequence in X such that x_n x and Tx_n  y. Continuity of

T implies Txn Tx, and uniqueness of limit gives Tx y.

Hence, (x,y)(x,Tx)G_T and so T is a closed linear map.

(a) Show that S is a closed subspace of C

 

0,1 .

(b) Show that there is a constant M such that for all f S we have



 f f

2 and f  M f 2.

Verification. (a) Let f_n be a sequence in S converging to f in C

 

0,1 . That is, 0

  f _

f_n . Then since f_n  f  f_n  f _

2 , we have fn  f 2 0. Thus f S .

Hence S is closed as a subspace of C

 

0,1 .

(b) For any f S , we have f 

   



f 



f _ 2  f _

1 2 2

1 2

2 . Since S is closed in both

 

0,1

C and L2

 

0,1 , it is complete in both norms. Thus there exists M such that

2

f M f _  .

Some Applications of the Closed Graph Theorem

Example 2.17 Prove that if





1

k k k

 is convergent whenever



    1 k k

 , then

   k k  1 sup .

Consider the map 1 

:

T defined by

) , , , ( ) , ,

(₁ ₂   ₁₁ ₁₁₂₂ ₁₁₂₂ ₃₃

T . That is,   



    1 1 3 2

1, , , )

( n n k k k

T      .

Then T is )

(i Well defined; (ii) Linear; (iii) Bounded.

Verification. (i)



        1 1 1 k k k k

x    . Then by hypothesis



    1 k k k  .

This implies the sequence

     



1 1 1 n n k k k n n

S   converges and so is bounded.

That is, 





   n k k k n n n S 1 1 1 sup

sup   . That means, Tx.

(ii) Let x



1,2,3,...



and y 



1,2,3,...



be arbitrary elements of 1

 and let , be scalars. Then

xy



₁,₂,₃,...

 

 ₁,₂,₃,...



=



₁ ₁,₂ ₂,₃₃,...



andT(xy)



₁(₁₁),₁(₁₁)₂(₂₂),...



=



₁₁₁₁,₁₁₁₁₂₂ ₂₂,...



=



₁₁,₁₁₂₂,...

 

 ₁₁,₁₁₂₂,...



=



₁₁,₁₁₂₂,...







₁₁,₁₁₂₂,...



=TxTy

Hence, T is a linear operator.

(iii) To prove T is bounded; we use the closed graph theorem.

Let 

 

1  _{n 1}

n k n

n

x  be an arbitrary sequence in 1 and x _k _k11 such that x_n Tx_n

 in 1 and  in 

x z  k _k1

It suffices to show that Txz. Now,

x

x_n  



      1 0 1 k k n k n x

x _   .

0 sup 1 1      



   i k k n k k i n

n z Tx z

Tx _    as n.

Therefore, for each i1,2,3,

0 . sup ) ( 1 ,..., 2 , 1 1    





   i k k n k k i k i k k n k

k     

 as n.

Thus, for each i,

     



1 1 1 n i k n k k n n

Tx   converges to

  



1 1 i i k k k

 in .

But Txn _n1 also converges (by hypothesis) to

 

 k _{k 1}

z  .

So by the uniqueness of limit,



  i k k k i 1  

That is,zTx, and so T is a closed linear map. Since 1

 and  are Banach spaces, it follows from the closed graph theorem that T is bounded (continuous).

Furthermore, Tx _  T . x _1. That is,







 



1 1

. sup

k k i

k

k

k T

i    .

In particular, for each i1,2,, and for all x _k _k1 in 1 we have







 



1 1

.

k k i

k

k

k T 

 (2.1)

We now consider x k defined by

  

 _



otherwise k i if sign _i

k

, 0

) (

__



 where

      

  

0 ; 0 0

)

( __

__ __

__ __

i i

i

i

i _if

if

sign _

 



 .

Then, _i

i

k

k

k 

 



1

for each i and









1

1

k k

 .

Using this in (2.1), now yields _i  T for all i and

So,  

 i T

i



1

sup .

III. CONCLUSION

we have studied linear operators, bounded linear operators, the graph of a mapping, closed linear operators and closed graphs, and one of the key theorems of functional analysis for Banach spaces, namely-the closed graph theorem andsome of its applications.

The closed graph theorem basically tells us when a closed linear map is bounded or continuous, and thus providing a means of recovering boundedness or continuity for the class of closed linear maps.

REFERENCES

[1]. A.Tayler, Introduction to Functional Analysis, 2nd edition,Wilay,New York (1958).

[2].C.E.Chidume, Applicable Functional Analysis, University of Nigeria, Nigeria (July, 2006).

[4]. H.L.Royden, Real Analysis,3rd edition, Macmillan, New York (July, 1987).

[5].W. Rudin, Principle of Mathematical Analysis, 3rd edition, McGraw-Hill, New York (1976).

[6].V.K.Krishnan, Textbook of Functional Analysis: A Problem-Oriented Approach, Prentice-Hall of