A STUDY OF A CLOSED GRAPH THEOREM AND THEIR APPLICATIONS IN BANACH SPACES FOR CLOSED LINEAR TRANSFORMATIONS Tsegaye Kebede Irena1, Dr.Tadesse Zegeye2, Dr. Vasudevarao Kota3 1Lecturer, 3Assistant Professor, Department of Mathematics, Ambo University, Ambo, Ethiopia. 2Assistant 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
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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 there corresponds 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 , (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
(x y) (xo yo ) x xo y yo for any xo and yo . Hence, + is continuous. Consider . as a function from F X into X . Since
x o xo (x xo ) ( o )xo . x xo o . xo for any fixed o & xo . Hence, . is continuous. 1
_ n Example 1.3 (a) Let x (x1, x2 ,..., xn ), xi 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: _ p p x (x1, x2 , x3 ,...), xi : xi . i1
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_ _ p For x (x1, x2 , x3 ,...) , y (y1, y2 , y3 ,...) arbitrary elements of , define _ _ _ x y (x1 y1, x2 y2 , x3 y3 ,...) and x (x1,x2 ,x3 ,...) .Then, is a linear space.
Example 1.4 Let the space be defined as follows:
_ _ x (x1, x2 , x3 ,...), xi : x is bounded . 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 x y is in whenever x and y are both in and , are any scalars. If is also closed as a subset of , 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 ,
. : X 0, which satisfies the following conditions:
N1 : x 0 , and x 0 if and only if x 0, x X;
N2 : x x for all F,x X;
N3 : x y x y 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 A sequence 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 f n converges to f , we write f lim f n or f n f . n
Another way of formulating the convergence of f n to f is by noting that f n f if
f n f 0.
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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 belongs to Lp and f f 0. Convergence in L is nearly uniform n n p 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 E c .
Verification. Let f n be a sequence of functions in L .
Suppose f f 0 . Given 0 , there exists N such that n
inf M : mt : f n (t) f (t) M 0 for n N .
Thus mt : 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 .
Conversely, suppose there exists a set with and f n converges uniformly to f on . Given 0 , there exists such that f (t) f (t) for and t E c . n 2 Thus t : f n (t) f (t) E for . 2 Hence, for . That is, f f 0 . n 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 in the space such that f n f .
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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 x in and show that it converges to a point in . n n1
Example 1.14 For 1 p , we denote by p the space of all sequences such that 1
p . 1 Prove that a normed linear space p is complete.
(n) p Verification. Let 1 p and let be a Cauchy sequence in . Given 0, there
(n) (m) p p exists N such that for n,m N.
(n) (m) p p (n) In particular, for n,m N and each . Thus for each , is Cauchy in so it converges to some .
k (n) p p Consider . Then for each k and each n N . So 1
(n) p p for n N .
(n) p p (n) Thus for n N . So and 0 .
Example 1.17 Let X and Y are Banach spaces. Show that the product space X Y , with the norm defined by (x, y) x y , x, y X Y is Banach.
Verification. To show that X Y is complete, let (xn , yn ) be a Cauchy sequence in .
Since xn , yn xm , ym xn xm , yn ym xn xm yn ym , we have 172 | P a g e
xn xm xn , yn xm , ym and
y y x , y x , y for all n,m 1. n m n n m m
These show that xn and yn are Cauchy sequences in X and Y , respectively. Since X and Y are Banach, they converge.
Let xn x and yn y . Then,
xn , yn x, y xn x, yn y
xn x yn 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 T(x y) (x) (y) 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 X and for each scalar, .
_ 2 2 Example 1.20 Let X Y . For each x x1, x2 , x3 , in define
_ x2 x3 T x 0, x1 , , , .. 2 3 Then T is a linear operator on 2 .
_ _
Verification. Let x x1, x2 , x3 , and y y1, y2 , y3 , be arbitrary elements of and let , be scalars. Then
_ _
x y x1,x2 ,x3 , y1,y2 ,y3 ,
x1 y1,x2 y2 ,x3 y3 ,
_ _ x y x y x y and T( x y) 0, 1 1 , 2 2 , 3 3 , 1 2 3
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x2 x3 y2 y3 0,x1, , , 0, y1, , , 2 3 2 3
_ _ T(x) T(y) 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 : X Y 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 0 and then .
(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 T(0) 0). But T is one-to-one. So, x 0. Conversely, assume that whenever Tu 0 , then u 0. So, let Tx Ty. Then, Tx Ty 0 and by the linearity of T , T(x y) 0 . By hypothesis, x y 0 which implies x y . 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 T : X Y be a linear operator. Then T is said to be bounded if there exists some constant K 0 such that for all x X,
Tx K x , (1.1) the constant K is called a bound for T and in this case, T is called a bounded linear operator. 174 | P a g e
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 xn converging to x ,
Txn Tx.
Theorem 1.25 Let and be normed linear spaces and let be a linear operator. Then the following statements are equivalent: (i) is continuous;
(ii) is continuous at 0. That is, if xn is a sequence in X such that xn 0 as n
then Txn 0 in Y as n ; (iii) 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 be a sequence in X such that xn 0. By continuity of T , we have
Txn T(0) . But T is a linear operator so that T(0) 0 . Hence, is continuous at 0.
(ii)(iii) We are given that if is any sequence in X such that xn 0 as n , then
T(xn ) 0. 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 xn X , xn 0 such that
Txn n xn .
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Tx This implies that n 1. n xn
xn Now, define the sequence un . Clearly, un 0. For, n xn
xn 1 xn 1 un 0 0. n xn n xn n
However, Tun does not converge to 0. For,
T(xn ) T(xn ) Tun 0 1, n xn n xn 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
y 1. 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. x x Consider the vector u : 1 2 . Clearly, u D , and so, by condition (iv) , (i.e.,T(D) is x1 x2
x x bounded), there exists a constant K 0 such that T(u) K . i.e., T( 1 2 ) K , or x1 x2
Tx1 Tx2 K x1 x2 .
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 . K 1
So, if x1 x2 we obtain, K Tx Tx K x x K ( ) 1 2 1 2 K 1
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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 T is continuous at xo , then it must be continuous everywhere. (iii) If is continuous at , then it is bounded.
Proof. (i) Suppose T is a bounded linear operator. Then Tx Ty T x y for all x, y X with x y . T
Thus, is uniformly continuous and so is continuous.
Conversely, to show that continuity implies boundedness, the proof will be by contradiction. Thus, suppose is continuous but not bounded. The negation of being bounded is that for any natural number n, however large, there is some point, say xn X, xn 0 , such that
Txn n xn .
Tx This implies that n 1. n xn
xn 1 Consider now the vectors un with norm un . n xn n 1 Clearly, u 0 because u 0 0 . n n n
Since is linear and continuous, this implies T(un ) T(0) 0.
xn Txn But Tun 0 T( ) 1 and so Tun does not converge to 0. This contradicts n xn n xn the hypothesis that T is continuous. Thus our assumption that was not bounded must be false when is continuous and so continuity implies boundedness.
Thus for linear maps, continuity and boundedness are equivalent.
(ii) By the hypothesis, if xn xo , then T(xn ) T(xo ).
To prove continuity everywhere, we must show, for any y X , if yn y , then
T(yn ) T(y). 177 | P a g e
To this end, suppose yn y X and consider
T(yn ) T(yn y xo y xo )
T(yn y xo ) T(y xo ) .
Since yn y xo xo ,
T(yn y xo ) T(xo ) and T(yn ) T(y). Hence, T is continuous everywhere.
(iii) Suppose now that is a linear operator that is continuous at xo . Then there is a 0 such that Tx Txo 1 for all x such that x xo . z For any z in X with z 0, set w , where 0 . z
Then Tz Tw T(w x ) T(x ) and Tz T(w x ) T(x ) 1, since z o o z o o
w xo xo w .
Consequently, Tz 1 z and T is bounded.
Example 1.27 Show that if Tn T and xn x, then Tn xn Tx .
Verification. Suppose Tn T and xn x . Then,
Tn T 0 and xn x 0.
Since Tn xn Tx Tn xn Txn Txn Tx Tn T xn T xn x and xn is bounded.
This implies Tn xn 0 . That is, Tn xn Tx .
Thus kerT is closed.
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 .
Note that from definition 1.30 we obtain immediately that Tx T x for each x X and
that for every 0, there exists x X , x 0 such that Tx ( T ) x
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Theorem 1.31 Let B(X,Y) be the family of all bounded linear maps from X to Y . Then we have the following: For arbitrary T B(X,Y),
Tx T sup Tx sup Tx sup x 1 x 1 x 0 x
Proof. Since T is bounded and linear, there exists K 0 such that for all x X , Tx K x .
If x 1, then Tx K x K and so,Tx : x 1 is a bounded set in , so its
“sup” exists and sup Tx K for any K such that x 1
Tx K x x X.
Taking the infimum over all such K's we obtain that sup Tx inf K 0 : Tx K x for all x X T x 1
Hence, sup Tx T . (1.2) x 1
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).
x Let u , then u 1 and Tu T . x Consequently, we obtain from inequality (1.2) that
x T sup Tx sup Tx sup T( T x 1 x 1 x 0 x Since 0 is arbitrary, we get that Tx T sup Tx sup Tx sup T . x 1 x 1 x 0 x
So that all these inequalities are equal.
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Lemma 2.1 Let T be a continuous linear transformation of the Banach space X onto the Banac space Y . Then the image by of the unit sphere in 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 GT , is defined by
GT (x,Tx) : x X. Observe that is a subset of X Y and that
(x, y)GT if and only if Tx y . Example 2.4: Let X 0,1,Y , and T :0,1 be defined by
2 Tx x , x 0,1. Then, the graph of T , is given by
2 GT (x,Tx) : x0,1 (x, x ) : x0,1. 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 T : D Y is called closed if for every convergent sequence xn of points of D , where xn x X say, such that Txn is a convergent sequence of points of Y , where Txn 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 GT and show that its limit is in .
So, let (x ,Tx ) be an arbitrary sequence in and suppose (x ,Tx ) (x, y) as n . n n n n
This implies that xn x and Txn y as n .
Observe that the map is continuous on 0,1. Thus, xn x Txn Tx as n .
By uniqueness of limit, Tx y so that (x, y) GT .
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Hence, GT 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 T : D Y is called closed if and only if its graph is a closed subspace.
T Proof: Suppose X DY is a closed linear map.
To show that GT is closed, we must show that any limit point of GT is actually a member of .
Let (x, y) be a limit point of . Since this is so, there must be a sequence of points of
, (xn ,Txn ) , where xn D , converging to (x, y). This is equivalent to requiring that
(xn x,Txn y) 0 or xn x Txn y 0 , which implies that
xn x and Txn y . Since T is closed, we can say that this implies
x D and y Tx , and, in view of this, we can write
(x, y) (x,Tx)GT . Therefore, is closed.
Conversely, suppose is closed and that
xn x , xn D for all n , and Txn y . We must show that x D and y Tx .
___
The condition implies that (xn ,Txn ) (x, y) GT .
___
Since is closed, GT GT , and we have that
(x, y) GT . By definition of this means that
x D and y Tx . Therefore, T is closed. Theorem 2.10: Let T be a bounded linear map such that
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T X DY where D is a closed subspace of X . Then T is closed.
Proof. Suppose xn is a sequence of points of D such that
xn x and Txn y . Since is closed, the limit of the sequence must be in D . That is, x D. The continuity of implies that
Txn Tx . By the uniqueness of limit, we have Tx y . Hence, 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 C0,1 Y with the sup norm and D consists of all those functions that have a continuous derivative; that is, D f X / f ' C0,1 We take the linear map T to be differentiation, Tf f ' for any f D . Then, (i) T is linear (ii) is closed; and (iii) T is not bounded.
Verification. (i) Let f , g D and , be scalars. Then, T(f g) (f g)' (f )'(g)'f 'g'Tf Tg 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 and Tf f ' y . n n n
' Then, Tfn y sup (Tfn )(t) y(t) sup f n (t) y(t) 0 as n . t0,1 t0,1 But convergence with respect to the sup norm is uniform convergence.
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' ' So, f n Tfn y uniformly and y(t) lim f n (t) n Since the convergence is uniform, we have
t t t ' ' y(s)ds lim fn (s)ds lim fn (s)ds lim( fn (t) fn (0)) f (t) f (0) . 0 0 n n 0 n
t Hence, f (t) f (0) y(s)ds 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,
f D and Tf y . Therefore, T is closed. (iii) To show that T is not bounded, consider the sequence of points of D , f (t) t n n for n 1,2,.
n1 In this case Tfn nt . We now observe that, with the sup norm,
n n1 f n sup t 1 and Tfn sup nt n for all n . t0,1 t0,1
Hence, Tfn n n f n . Since Tf Tf T sup sup n supn f 0 f n f n n 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 , and suppose there is a constant C such that 2 x C x 1 2 for all x X . Then the norms are equivalent. That is, there is a second constant C' such that x C' x 2 1 for all x X
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Proof. The identity map of ( X , ) onto ( X , ) is a one-to-one continuous linear 2 1 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 , GT is closed. Then, T is a bounded (continuous) operator. Proof. Define a new norm in by x x Tx . 2 1 1
Then is complete in the norm . For if x x 0 , then x x 0 and 2 p q 2 p q 1
Tx Tx 0 . Hence by the completeness of and we have x X and y Y such p q 1 that x x 0 and Tx y 0. By the hypothesis of our theorem y Tx . Hence, p 1 p 1 x x 0 , and X is complete with respect to . Now since is complete in both p 2 2 norms and , there is C' such that 1 2 x x Tx C' x . 2 1 1 1 Thus Tx C' x , 1 1 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 ,GT is closed (or equivalently, T is a closed map).
To see this, suppose xn is a sequence in such that xn x and Txn y . Continuity of
T implies Txn Tx, and uniqueness of limit gives Tx y .
Hence, (x, y) (x,Tx)GT and so T is a closed linear map. Example 2.15 Let S be a linear subspace of C0,1 that is closed as a subspace of L2 0,1
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(a) Show that S is a closed subspace of C0,1. (b) Show that there is a constant M such that for all f S we have
f f and f M f . 2 2
Verification. (a) Let f n be a sequence in S converging to f in C0,1. That is,
f f 0 . Then since f f f f , we have f f 0. Thus . n n 2 n n 2 f S Hence S is closed as a subspace of .
1 1 2 2 (b) For any f S , we have f f 2 f 2 f . Since is closed in both 2 and L2 0,1, it is complete in both norms. Thus there exists M such that
f M f . 2 Some Applications of the Closed Graph Theorem
Example 2.17 Prove that if k k is convergent whenever k , then k1 k1
sup k . k1
Consider the map T : 1 defined by
T(1,2 ,) (11,11 22 ,11 22 33 ,) .
n That is, T(1,2 ,3 ,) k k . k1 n1 Then T is (i) Well defined; (ii) Linear; (iii) Bounded.
Verification. (i) x 1 . Then by hypothesis . k k1 k k k k1 k1
n This implies the sequence S converges and so is bounded. n n1 k k k1 n1
n That is, sup Sn sup kk . That means, Tx . n1 n1 k1 Hence, T is well defined.
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1 (ii) Let x 1,2 ,3 ,... and y 1,2 ,3 ,... be arbitrary elements of 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 ),...
= 11 11,11 11 22 22 ,...
=11,11 22 ,... 11,11 22 ,...
=11,11 22 ,... 11,11 22 ,... =Tx Ty Hence, T is a linear operator.
(iii) To prove T is bounded; we use the closed graph theorem.
Let x n be an arbitrary sequence in 1 and x 1 such that n n1 k n1 k k1
xn Txn in 1 and in x z k k1 It suffices to show that Tx z . Now,
n x x x x 1 0 . n n k k k1
i Tx z Tx z sup n 0 as n . n n k k k i1 k1
Therefore, for each i 1,2,3,
i i n n k (k k ) sup k . k k 0 as . k1 k1,2,...,i k1
i i Thus, for each i , Tx n converges to in . n n1 k k k k k1 n1 k1 i1
But Tx also converges (by hypothesis) to z . n n1 k k1
i So by the uniqueness of limit, i k k , i 1,2,... k1
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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 is bounded (continuous). Furthermore, Tx T . x . That is, 1 sup i kk T . k . i k1 k1
In particular, for each i 1,2,, and for all x in 1 we have k k1
i k k T . k (2.1) k1 k1
We now consider x k defined by
__ __ 0 if i 0; __ sign( i ) if i k __ k where __ . sign( ) i i if 0 0, otherwise __ i i
i Then, k k i for each i and k 1. k1 k1
Using this in (2.1), now yields i T for all i and
So, supi T . i1
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 and some 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).
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[4]. H.L.Royden, Real Analysis, 3rd edition, Macmillan, New York (July, 1987).
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[6]. V.K.Krishnan, Textbook of Functional Analysis: A Problem-Oriented Approach, Prentice-Hall of
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