A STUDY OF A CLOSED GRAPH THEOREM AND THEIR APPLICATIONS IN BANACH SPACES FOR CLOSED LINEAR TRANSFORMATIONS 1 2 3 Tsegaye Kebede Irena , Dr.Tadesse Zegeye , Dr. Vasudevarao Kota 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 168 | P a g e 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 169 | P a g e _ _ 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 . p n Another way of formulating the convergence of f n to f is by noting that f n f if f n f 0. 170 | P a g e 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 . 171 | P a g e 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 yn ym xn , yn xm , ym for all n,m 1. 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 .