A CHARACTERIZATION OF REFLEXIVITY OF NORMED ALMOST LINEAR SPACES

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1 Comm. Korean Math. Soc. 12 (1997), No. 2, pp A CHARACTERIZATION OF REFLEXIVITY OF NORMED ALMOST LINEAR SPACES SUNG MO IM AND SANG HAN LEE ABSTRACT. In [6] we proved that if a nals X is reflexive, then X = W X + V X. In this paper we show that, for a split nals X = W X + V X, X is reflexive if and only if V X and W X are reflexive. 1. Preliminaries G. Godini [3,4,5] introduced a normed almost linear space(nals), a concept which generalizes normed linear space. An example of a nals is the collection of all nonempty, bounded and convex subsets of a real normed linear space. In [6], we defined the notion of reflexivity of a nals. Also, we proved that if a nals X is reflexive then X is split as X = W X + V X. In this note, we characterizes the reflexivity of a nals X (without basis). First of all we recall some definitions and results which are needed in this paper. An almost linear space (als) isasetxtogether with two mappings s : X X X and m : R X X satisfying the conditions (L 1 ) (L 8 ) given below. For x, y X and λ R we denote s(x, y) by x + y and m(λ, x) by λx, when these will not lead to misunderstandings. Let x, y, z X and λ, µ R. (L 1 ) x + (y + z) = (x + y) + z; (L 2 ) x + y = y + x; (L 3 ) There exists an element 0 X such that x + 0 = x for each x X; (L 4 ) 1x = x; (L 5 )λ(x+y)=λx+λy;(l 6 )0x=0; (L 7 )λ(µx)=(λµ)x; (L 8 )(λ+µ)x = λx + µx for λ 0, µ 0.We denote 1x by x, if there is no confusion likely, and in the sequel x y means x + ( y). A nonempty subset Y of an als X is called an almost linear subspace of X, if for each y 1, y 2 Y and λ R, s(y 1, y 2 ) Y and m(λ, y 1 ) Y. An almost Received January 9, Revised March 18, Mathematics Subject Classification: Primary 46B99, Secondary 46A15. Key words and phrases: almost linear space, normed almost linear space, almost linear functional.

2 212 Sung Mo Im and Sang Han Lee linear subspace Y of X is called a linear subspace of X if s : Y Y Y and m : R Y Y satisfy all the axioms of a linear space. For an als X we introduce the following two sets; (1.1) V X ={x X:x x=0}, (1.2) W X ={x X:x= x}. Then, we have the following properties: (1) The set V X is a linear subspace of X, and it is the largest one. (2) The set W X is an almost linear subspace of X and W X ={x x:x X}. (3)An als X is a linear space V X = X W X ={0},andV X W X ={0}. Let X and Y be two almost linear spaces. A mapping T : X Y is called a linear operator if T (λ 1 x 1 + λ 2 x 2 ) = λ 1 T (x 1 ) + λ 2 T (x 2 ) for all λ i R and x i X, i = 1, 2. An isomorphism T of an als X onto an als Y is a bijective mapping which preserves the two algebraic operations of an als; that is, T : X Y is a bijective linear operator. Then Y is said to be isomorphic with X. The following is well known. PROPOSITION 1.1. Let T be a linear operator from an als X into an als Y. Then (1) T (V X ) V Y, T (W X ) W Y. (2) If X = V X + W X,thenT(X)=T(V X )+T(W X ). In particular, if T is an isomorphism, then Y = T (X) = V Y + W Y. Let X be an als. A function f : X R is called an almost linear functional if the conditions (1.3) (1.5) are satisfied. (1.3) f (x + y) = f (x) + f (y) (x,y X) (1.4) f (λx) = λ f (x) (λ 0, x X) (1.5) f (w) 0 (w W X ). A functional f : X R is called a linear functional on X if it satisfies (1.3), and (1.4) for each λ R. Then (1.5) is also satisfied. Note that an almost

3 A characterization of reflexivity 213 linear functional is not a linear operator from X to R, but a linear functional is a linear operator. Let X # be the set of all almost linear functionals defined on an als X. We define two operations s : X # X # X # and m : R X # X # as follows: s( f 1, f 2 )(x) = f 1 (x) + f 2 (x) (f 1, f 2 X # ), m(λ, f )(x) = f (λx) (λ R, f X # ) for all x X. Clearly, s( f 1, f 2 ) X #, m(λ, f ) X #,ands,msatisfy (L 1 ) (L 8 ) with 0 X # being the functional which is 0 at each x X. Therefore X # is an als. X # is called the algebraic dual space of an als X. We denote s( f 1, f 2 ) by f 1 + f 2 and m(λ, f ) by λ f. PROPOSITION 1.2 ([6]). Let X be an als. Then X # = W X # + V X #. PROPOSITION 1.3 ([3]). If f isan almostlinear functional on an als X,then f V X # if and only if f WX = 0. PROPOSITION 1.4. Let X be a split als as X = W X + V X. If f is an almost linear functional on X, then f W X # if and only if f VX = 0. PROOF. Suppose that f W X #. Then for each v V X we have f (v)+ f (v) = f (v)+( 1 f )(v) = f (v)+ f ( v) = f (v v) = f (0) = 0, since 1 f = f. Therefore f VX = 0. Conversely, suppose that f VX = 0andx =v+w Xwith v V X, w W X.Sincef(v) = f ( v) = 0, we have ( 1 f )(x) = f ( x) = f (w v) = f (w) + f ( v) = f (w) + f (v) = f (w + v) = f (x). Therefore f W X #.

4 214 Sung Mo Im and Sang Han Lee 2. Reflexivity of NALS A norm on an als X is a functional : X Rsatisfying the conditions (N 1 ) (N 3 ) below. Let x, y, z X and λ R. (N 1 ) x z x y + y z ;(N 2 ) λx = λ x ; (N 3 ) x =0iffx=0. Using (N 1 ) we get (2.1) x + y x + y (x,y X) (2.2) x y x y (x, y X). By the above axioms it follows that x 0 for each x X. An als X together with : X Rsatisfying (N 1 ) (N 3 ) is called a normed almost linear space (nals). When X is a nals, for f X #, we define, as in the case of a normed linear space, (2.3) f =sup{ f (x) : x X, x 1}, and let X ={f X # : f < }. Then X is a nals[4], called the dual space of X. We denote the dual space (X ) of X by X and call it the second dual space of X. For a nals X and f X, an equivalent formula for the norm of f is f (x) (2.4) f = sup f (x) =sup x =1 x 0 x, hence f (x) f x. In the theory of a normed linear space an important tool is the Hahn-Banach theorem. An analogous theorem is no longer true in a nals [3, 4.5. Example]. But we have the following Propositions. PROPOSITION 2.1 ([5]). Let (X, )be a nals. Then for each x X there exists f x X such that f x =1and f x (x) = x.

5 A characterization of reflexivity 215 PROPOSITION 2.2 ([4]). Let (X, )be a nals. Then for each f (W X ) there exists f 1 W X such that f 1 WX = f and f 1 = f. PROPOSITION 2.3 ([4]). Let (X, )be a nals and split as X = W X + V X. Then for each f (V X ) there exists f 1 V X such that f 1 VX = f and f 1 = f. PROPOSITION 2.4. For any x in a nals X, wehave { } f (x) x =sup f : f X, f 0. PROOF. For any x X, by Proposition 2.1, there exists f x X such that f x =1and f x (x)= x.so, we have x = f { } x(x) f (x) f x sup f : f X, f 0. From f (x) f x,wehave { } f (x) sup f : f X, f 0 x { } for each f X. Hence x =sup f (x) : f X, f 0. f An isomorphism T of a nals X onto a nals Y is a bijective linear operator T : X Y which preserves the norm, that is, for all x X, T (x) = x. Then X is called isomorphic with Y. For x X let Q x be the functional on X defined, as in the case of a normed linear space, by (2.5) Q x ( f ) = f (x) (f X ). Then Q x is an almost linear functional on X and (2.6) Q x x. Hence Q x is an element of X, by definition of X. This defines a mapping (2.7) C : X X by C(x) = Q x. C is called the canonical mapping of X into X. If the canonical mapping C of a nals X into X defined by (2.7) is an isomorphism, then X is said to be reflexive.

6 216 Sung Mo Im and Sang Han Lee PROPOSITION 2.5. For a nals X, the canonical mapping C defined by (2.7) is a linear operator and preserves the norm. PROOF. By (2.4)and Proposition 2.4, we have Q x ( f ) Q x =sup f 0 f f (x) = sup f 0 f for each x X. Hence C preserves the norm. Let x, y X and α R. For each f X, we have = x Q x+y ( f ) = f (x + y) = f (x) + f (y) = Q x ( f ) + Q y ( f ), Q αx ( f ) = f (αx) = (α f )(x) = (α Q x )( f ). Thus, C(x + y) = C(x) + C(y) and C(αx) = α C(x). Therefore C is a linear operator. THEOREM 2.6 ([6]). If a nals X is reflexive, then X = W X + V X. THEOREM 2.7. IfanalsXsplits as X = W X + V X, then (1) V X is isomorphic with (V X ), (2) W X is isomorphic with (W X ). PROOF. Since x VX (V X ) for each x V X, we can define an operator T : V X (V X ) by T (x ) = x VX for each x V X. For x, y V X and α, β R, we have T (α x + β y )(v) = (α x + β y )(v) = x (αv) + y (βv) = T (x )(αv) + T (y )(βv) = (α T (x ))(v) + (β T (y ))(v) = [α T (x ) + β T (y )](v) for each v V X. Hence T is a linear operator.

7 A characterization of reflexivity 217 If x y V X, then x (v) y (v) for some v V X by Proposition 1.3. So, T (x ) T (y ). Hence T is injective. For each v (V X ), there exists x V X such that x VX = v by Proposition 2.3. Hence T is surjective. For any v V X, v v VX = T(v ). Also, if x = v + w X, v V X,w W X with x 1, then v 1andv (x)=v (v). So we have v = sup{ v (x) : x X, x 1} sup{ v (v) : v V X, v 1} = sup{ T (v )(v) : v V X, v 1} = T(v ) Hence T preserves the norm. Therefore, V X is isomorphic with (V X ). Similarly, applying Proposition 1.4 and Proposition 2.2, we can show that an operator T : W X (W X ), T (x ) = x WX (x W X ), is an isomorphism. COROLLARY 2.8. IfanalsXsplits as X = W X + V X, then (1) V X is isomorphic with (V X ), (2) W X is isomorphic with (W X ). An arbitrary normedalmost linearsubspace of anals X need not be reflexive even if X is reflexive. But, we have the following result: THEOREM 2.9. IfanalsXis reflexive, then V X and W X are reflexive. PROOF. By Theorem 2.6, X = W X + V X since X is reflexive. Let C : X X be the canonical isomorphism, and let C : V X (V X ) be the canonical mapping. We will show that C is bijective. Let v (V X ). By Theorem 2.7, T : V X (V X ), T (v ) = v VX (v V X ), is an isomorphism. Since x VX (V X ) for each x X, we can define a functional v : X R by v (x ) = v (x VX ) for each x X. Then v V X. Since C is an isomorphism of X onto X, there exists v V X such that C(v) = v. For

8 218 Sung Mo Im and Sang Han Lee this v V X, C (v) = v. Indeed, for each v (V X ), there exists v V X such that v VX = v by Proposition 2.3. So, we have v (v ) = v (v VX ) = v (v ) = C(v)(v ) = v (v) = v (v) = C (v)(v ). Hence C is surjective. If v 1 v 2 in V X,thenC(v 1 ) C(v 2 ) in X since C is an isomorphism. Choose f X such that C(v 1 )( f ) C(v 2 )( f ), i.e, f(v 1 ) f (v 2 ). For this f X, f VX (V X ). And f VX (v 1 ) f VX (v 2 ). So, we have C (v 1 ) C (v 2 ). Hence C is injective. Therefore C is an isomorphism. Similarly, we can show that W X is reflexive. THEOREM Let X be a split nals as X = W X + V X. If V X and W X are reflexive, then X is reflexive. PROOF. Note that X = W X + V X and X = W X + V X. Let C : V X (V X ) and C : W X (W X ) be the canonical isomorphism, and let C : X X be the canonical map. We will show that C is bijective. Let v V X. By Proposition 1.3, we have v (x ) = v (v ) for each x = v + w X,v V X,w W X.And v V X (V X ). Recall that T : V X (V X ), T (v ) = v VX (v V X ), is an isomorphism. Define a functional v : (V X ) R by v (v VX ) = v (v ), for each v VX (V X ). Then v (V X ). Since C is an isomorphism of V X onto (V X ), there exists v V X such that C (v) = v. For this v V X, C(v) = v. Indeed, v (x ) = v (v ) = v (v VX ) = C (v)(v VX ) = v VX (v) = v (v) = x (v) = C(v)(x ) for each x = v + w X with v V X,w W X. Similarly, for each w W X, there exists w W X such that C(w) = w. Hence, for each x = v + w X with v V X, w W X, there exists x = v + w X with v V X,w W X such that C(x) = C(v) + C(w) = v + w = x. Hence C is surjective. If w 1 w 2 in W X,thenC (w 1 ) C (w 2 ) in (W X ) since C is an isomorphism. Choose f (W X ) such that C (w 1 )( f ) C (w 2 )( f ), i.e., f (w 1 ) f (w 2 ). By Proposition 2.2, there exists f 1 X such that f 1 WX = f and f 1 = f.for this f 1,wehaveC(w 1 )( f 1 ) C(w 2 )( f 2 ) since f 1 (w 1 )

9 A characterization of reflexivity 219 f 1 (w 2 ). Hence C(w 1 ) C(w 2 ). Similarly, C(v 1 ) C(v 2 ) for v 1 v 2 in V X. Therefore C is injective since C is a linear operator. From Theorem 2.9 and Theorem 2.10, we have the following theorem: THEOREM Let X be a split nals as X = W X + V X. Then X is reflexive if and only if V X and W X are reflexive. References [1] M.M.Day,Normed linear spaces, Berlin-Göttingen-Heidelberg, Springer-Verlag, [2] N. Dunford and J. Schwarz, Linear operators. Part I, Pure and applied Mathematics, 7, New York, London, Interscience, [3] G. Godini, An approach to generalizing Banach spaces: Normed almost linear spaces, Proceedings of the 12th Winter School on Abstract Analysis (Srni 1984). Suppl. Rend. Circ. Mat. Palermo II. Ser. 5 (1984), [4], A framework for best simultaneous approximation: Normed almost linear spaces, J. Approximation Theory 43 (1985), [5], On Normed Almost Linear Spaces, Math. Ann. 279 (1988), [6] S.H.Lee,Reflexivity of normed almost linear spaces, Comm. Korean Math. Soc. 10 (1995), Department of Mathematics Chungbuk National University Cheongju , Korea