Seminar on Fixed Point Theory Cluj-Naoca, Volume 3, 2002, 317-322 htt://www.math.ubbcluj.ro/ nodeacj/journal.htm A GENERALIZATION OF JENSEN EQUATION FOR SET-VALUED MAPS DORIAN POPA Technical University Cluj-Naoca, Deartment of Mathematics Str. C. Daicoviciu nr.15, Romania E-mail: Poa.Dorian@math.utcluj.ro Abstract. It is given a reresentation of the solutions of a generalization of Jensen equation for set-valued mas with closed and convex values. Keywords: Set-valued ma, Jensen equation, additive. AMS Subject Classification: 39B52, 54C60. 1. Introduction Let X be a real vector sace, Y a real toological vector sace. We denote by 0 X, 0 Y the origin of X and Y, by P 0 (Y ) the collection of all nonemty subsets of Y and by CCl(Y ) the collection of all nonemty, closed and convex subsets of Y. For two nonemty subsets A and B of X (or Y ) and a real number t we define the sets A + B = {x x = a + b, a A, b B} (1) t A = {x x = ta, a A}. For any nonemty sets A, B X and any real numbers s, t the following relations holds s(a + B) = sa + sb (2) (s + t)a sa + ta. If A is a convex set and st 0 then (3) (s + t)a = sa + ta. Let K be a convex cone in X with 0 X K and a real number 0 < < 1. We are looking for set-valued solutions F : K P 0 (Y ) of the equation (4) F ((1 )x + y) = (1 )F (x) + F (y) + M where M P 0 (Y ). For = 1 2 and M = {0 Y } the equation (4) becomes the classical Jensen equation. It is well known that the solutions of classical Jensen equation for single valued mas are of the form F (x) = a(x) + k, where a : K Y is an additive function and k Y 317
318 DORIAN POPA is a constant [4]. Classical Jensen equation for set-valued mas was studied by Z. Fifer [4], K. Nikodem [6], [7]. They give a characterization of the solutions of Jensen equation for set-valued mas with comact and convex values in a toological vector sace. A generalization of Jensen equation was considered and studied by the author in [9]. The goal of this aer is to give a reresentation theorem of the solutions of the equation (4) if F has close and convex values. 2. Main results We start by roving an auxiliary lemma. Lemma 2.1. Let X, Y be real vector saces and K a convex cone in X containing the origin of X. If the set-valued ma F : K P 0 (Y ) satisfies the equation (4) then (5) F (x + y) + F (0 X ) = F (x) + F (y) for every x, y K. Proof. For x = y = 0 X in (4) we get (6) F (0 X ) = (1 )F (0 X ) + F (0 X ) + M and for x = 0 X, resectively y = 0 X in (4) we obtain (7) F ((1 )x) = (1 )F (x) + F (0 X ) + M, x K, (8) F (y) = (1 )F (0 X ) + F (y) + M, y K. Now let u, v K. By (4) we have ( ) u (9) F (u + v) = F (1 ) 1 + v = (1 )F and taking account of (6), (7), (8) we get ( ) u F (u+v)+f (0 X ) = (1 )F +F 1 = [ ( ) u (1 )F + F (0 X ) + M 1 and the lemma is roved. Consider in what follows the equation ( v ] + = F (u) + F (v) ( ) u + F 1 ( ) v + M ) +M +(1 )F (0 X )+F (0 X )+M = [ (1 )F (0 X ) + F (10) F (x + y) + C = F (x) + F (y) ( ) ] v + M = where F : K P 0 (Y ), K is a convex cone in X with 0 X K, C P 0 (Y ), X is a real vector sace and Y is a real toological vector sace. Lemma 2.2. Let C be a convex and sequentially comact subset of Y with 0 Y C. A set-valued ma F : K CCl(Y ) satisfies the equation (10) if and only if there exists an additive set-valued ma A : K CCl(Y ) such that for every x X. F (x) = A(x) + C
A GENERALIZATION OF JENSEN EQUATION FOR SET-VALUED MAPS 319 Proof. Suose that F : K CCl(Y ) satisfies the equation (10). It can be easily roved by induction that (11) F (nx) + (n 1)C = nf (x) for every n N and every x X. For n = 1 the relation (11) is obvious. Suose that the relation (11) holds for n N and we have to rove that (12) We have: F ((n + 1)x) + nc = (n + 1)F (x). F ((n + 1)x) + nc = F (nx + x) + C + (n 1)C = = F (nx) + F (x) + (n 1)C = nf (x) + F (x) = (n + 1)F (x). Now let x K. From (11) we get: 1 2 n F (2n x) + 2n 1 2 n C = F (x), n N. Denote A n (x) = 1 2 n F (2n x), n 0. The sequence of sets (A n (x)) n 0 is decreasing. Indeed: A n+1 (x) = 1 2 n+1 F (2n x + 2 n x) 1 2 n+1 (F (2n x + 2 n x) + C) = = 1 2 n+1 (F (2n x) + F (2 n x)) = 1 2 n+1 2F (2n x) = 1 2 n F (2n x) = A n (x). Put A(x) = n 0 A n (x) CCl(Y ). Prove that A(x). Let u F (x) fixed. From (12) it results that for every n N there exists a n A n (x) and c n C such that u = a n + 2n 1 2 n c n. The set C is sequentially comact, so there exists a subsequence (c nk ) k 0 of (c n ) n 0 convergent to c C and u = a nk + 2nk 1 1 2 n c nk, k 0. k It results that a nk u c as k. We show that u c n 0 A n (x). Suose that u c n 0 A n (x). Then there exists N such that u c A n. We have a nk A n for k and lim k a n k A n, because A n is closed, contradiction with u c A n. We rove that (13) A(x) + C = F (x), x X. Let u A(x) + C, u = a + c, a A(x), c C. It results that a A n (x) for every n 0 and let c n = 2n 1 2 n c 2n 1 2 n C. From the relation (12) it results that there exists b n F (x) such that a + c n = b n, n 0 and lim b n = a + c F (x), because n F (x) is closed. Hence A(x) + C F (x).
320 DORIAN POPA Now let b F (x). From (12) it results that for every n N there exists a n A n (x) and c n C such that b = a n + 2n 1 2 n c n. The sequence (c n ) n 0 has a subsequence (c nk ) k 0 convergent to c C, taking account of the sequential comactity of C. Hence the sequence (a n ) n 0 is convergent to b c A(x). Then b = (b c) + c A(x) + C. The relation F (x) A(x) + C is roved. It follows that the relation (13) is true. We rove that A is an additive set-valued ma. By the relation (10) and (13) we obtain A(x + y) + C + C = A(x) + A(y) + C + C and taking account of the cancellation law of Radström [2] it results that A(x + y) = A(x) + A(y) for every x, y X, hence A is an additive set-valued ma. If F (x) = A(x) + C, x K, where A : K CCl(Y ) is an additive set-valued ma, then we get: F (x + y) + C = A(x + y) + C + C = A(x) + A(y) + C + C = = (A(x) + C) + (A(y) + C) = F (x) + F (y) for every x, y X. The lemma is roved. Theorem 2.1. If a set-valued ma F : K CCl(Y ), with F (0 X ) sequentially comact set, satisfies the equation (4) then there exists an additive set-valued ma A : K ccl(y ) and a comact convex set B P 0 (Y ) such that (14) F (x) = A(x) + B for every x X. Proof. Suose that F satisfies the equation (1) and let α F (0 X ). The setvalued ma G : K CCl(Y ), given by the relation (15) G(x) = F (x) α, x X satisfies the equation (16) G((1 )x + y) = (1 )G(x) + G(y) + M and 0 Y G(0 X ). By Lemma 2.1 it results that G satisfies also the relation (17) G(x + y) + G(0 X ) = G(x) + G(y) for every x, y X and G(0 X ) is sequentially comact set. Then in view of Lemma 2.2 it results that there exists an additive set-valued ma A : K CCl(Y ) such that G(x) = A(x) + G(0 X ) for every x X. It follows that F (x) = A(x) + F (0 X ) for every x X. Denoting B = F (0 X ) the theorem is roved. References [1] J. Aczél, Lectures on functional equations and their alications, Academic Press, New York and London, 1966. [2] G. Beer, Toologies on closed and closed convex sets, Kluwer Academic Publishers, Dordrecht- Boston-London, 1993.
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