Square Root Functional Equation on Positive Cones 1 Introduction

Home , Functional equation

Analysis in Theory and Applications DOI: 10.4208/ata.2013.v29.n4.2 Anal. Theory Appl., Vol. 29, No. 4 (2013), pp. 333-341

Square Root Functional Equation on Positive Cones

H. Rezaei∗ Department of Mathematics, College of Sciences, Yasouj University, Yasouj-75914-74831, Iran Received 27 March 2012; Accepted (in revised version) 15 October 2013 Available online 31 December 2013

Abstract. A square root functional equation on positive cones of C∗-algebras is intro- duced and its solution and Hyers-Ulam-Rassias stability are investigated. Key Words: Hyers-Ulamstability, ﬁxed point, additive functional equation. AMS Subject Classiﬁcations: 39B52, 47H10

1 Introduction

The stability theory of functional equation is originated from the well-known Ulam’s problem [1] concerning the stability of homomorphisms in metric groups: Let (G, ) be ∗ a group and (X, ) be a metric group. Does for every ε > 0 there exist δ > 0 such that if · f : G X satisfies → d( f (x y), f (x) f (y))< δ for x,y G, ∗ · ∈ then a homomorphism h : G X exists with d( f (x),h(x)) < ε for x G? Hyers [2] gave → ∈ a first affirmative partial answer to the question of Ulam for Banach spaces. Hyers’s Theorem was generalized by Aoki [3] for additive mappings and by Th. M. Rassias [4] for linear mappings by considering an unbounded Cauchy difference. Theorem 1.1 (Th. M. Rassias). Consider two Banach spaces E ,E , and let f : E E be a 1 2 1 → 2 mapping such that f (tx) is continuous in t for each fixed x. Assume that there exist θ 0 and ≥ p [0,1), such that ∈ f (x+y) f (x) f (y) k − − k θ for any x,y E . x p + y p ≤ ∈ 1 k k k k Then there exists a unique linear mapping T : E E such that 1 → 2 f (x) T(x) 2θ k − k for any x E . x p ≤ 2 2p ∈ 1 k k −

∗Corresponding author. Email address: [email protected] (H. Rezaei) http://www.global-sci.org/ata/ 333 c 2013 Global-Science Press 334 H. Rezaei / Anal. Theory Appl., 29 (2013), pp. 333-341

The paper of Th. M. Rassias [4] has provided a lot of inﬂuence in the development of what we call generalized Hyers-Ulam stability or Hyers-Ulam-Rassias stability of functional equations. A generalization of the Th. M. Rassias theorem was obtained by Gavruta [5] by using a general controb function in place of the unbounded Cauchy difference in the spirit of Th. M. Rassias’s approach. Following the innovative approach of the Th. M. Rassias theorem [4], J. M. Rassias [6] replaced the factor x p + y p by k k k k x p y q for p,q R with p+q =1. The stability problem of several functional equations k k k k ∈ has been extensively investigated by a number of authors and there are many interesting results concerning this problem (see [7–10]). Let A be a C -algebra and a A be a self- ∗ ∈ adjoint element, i.e., a=a∗. Then a is said to be positive if it is of the form a=bb∗ for some a A. The set of positive elements of A is denoted by A+. Note that A+ is a closed convex ∈ cone (see [11]). Moreover, It is well-known that for a positive element a and a positive integer n there exists a unique positive element x A+ such that a = xn. In this case, we ∈ denote x by √n a. In the following some preliminary properties of A+ are listed [11]:

Theorem 1.2. Suppose that A isaC∗-algebra.

(i) A+ is closed in A, (ii) ax A+ if x A+ and a 0, ∈ ∈ ≥ (iii) x+y A+ if x,y A+, ∈ ∈ (iv) xy A+ if x,y A+ and xy = yx, ∈ ∈ (v) x A+ and x A+, then x =0. ∈ − ∈ + + Let A and B be two C∗-algebra and A and B be the corresponding positive cones. We introduce the following pair of functional equations

f (x) f (y)= f (y) f (x), (1.1) ( f (ax+by)= a2 f (x)+2ab f (x) f (y)+b2 f (y), p for every x,y A+ and f :A+ B+. Here a, b are two nonnegative real scalars that a+b=0. ∈ → 6 By part (iv) of Theorem 1.2, the first equation of condition (1.1) is needed for the second equation of (1.1) to be well-defined. Note that the function f : A+ B+ by f (x)= cx2, → c 0, is a solution of the functional equation (1.1). Applying (1.1) for x = y =0, x =0, and ≥ y = 0, separately, gives f (0)= 0; f (ax)= a2 f (x), and f (by)= b2 f (y), respectively. Hence (1.1) can be modified by the following:

2 f (ax+by)= f (ax)+2 f (ax) f (by)+ f (by)= f (ax)+ f (by) , q q q and consequently, f (ax+by)= f (ax)+ f (by). q q q H. Rezaei / Anal. Theory Appl., 29 (2013), pp. 333-341 335

For this reason, the Eq. (1.1) is named by square root functional equation. Considering the solution of Cauchy additive mapping it is easy to see that if a function f :R+ R+ satisﬁes → the above equation it must be of the form f (x)= cx2 for some scalar c 0. ≥ In the present paper, using the direct method and ﬁxed point method, we prove the Hyers-Ulam stability of the functional equation (1.1) in positive cones of C∗-algebras.

2 Hyers-Ulam-Rassias stability of (1.1): direct method

In what follows a,b are two nonnegative real scalars which a+b = 0, A,B are two + + 6 C∗-algebras with corresponding positive cones A ,B respectively. Using the direct method, we prove the stability of the positive functional equation (1.1). Theorem 2.1. Let h :A+ A+ [0,+∞) be a function with this property that × → h (a+b)x,(a+b)y γ :=sup : x,y A+, h(x,y) =0 <(a+b)2 (2.1) h(x,y) ∈ 6 and f :A+ B+ be a function satisfying → f (x) f (y)= f (y) f (x), (2.2) f (ax+by) a2 f (x) 2ab f (x) f (y) b2 f (y) h(x,y), ( k − − − k≤ for all x,y A+. Then there exists a unique mappingp T :A+ B+ such that ∈ → h(x,x) f (x) T(x) , x A+, (2.3) k − k≤ (a+b)2 γ ∈ − and T(x)T(y)= T(y)T(x), (2.4) ( T(ax+by)= a2 T(x)+2ab T(x)T(y)+b2T(y). for every x,y A+. p ∈ Proof. Let β = a+b, then by (2.1), γ < β2 and

h(βx,βy) γh(x,y), x,y A+. ≤ ∈ Applying the second equation of (2.2) for y = x, we observe that f satisﬁes the following functional inequality f (βx) β2 f (x) h(x,x), x A+. k − k≤ ∈ Let q (x) := β 2n f (βn x) for every n 1 and x A+. Then replacing x A by βn x in the n − ≥ ∈ ∈ above inequality to see

f (βn+1x) β2 f (βn x) h(βn x,βn x), x A+. k − k≤ ∈ 336 H. Rezaei / Anal. Theory Appl., 29 (2013), pp. 333-341

Thus

2n 2 n+1 2n n q (x) q (x) = β− − f (β x) β− f (β x) k n+1 − n k k − k 2n 2 n+1 2 n =β− − f (β x) β f (β x) k − k 2n 2 n n β− − h(β x,β x), ≤ which in the last inequality, the contractive property of h is used. Hence,

2 2 n + q (x) q (x) β− (γβ− ) h(x,x), x A , (2.5) k n+1 − n k≤ ∈ and γβ 2 <1, so the sequence q (x) is a Cauchy sequence for each x A+. Since B+ is − { n } ∈ closed in the complete metric space B, there exists a limit function T(x) := limn ∞ qn(x) → in B+. Now by induction on n, we prove that

n 1 − 2 2 i qn(x) f (x) ∑ β− (γβ− ) h(x,x), (2.6) k − k≤ i=0 for any n N and x X. Fix x A+, note that ∈ ∈ ∈ 2 q (x) f (x) = β− f (βx) f (x) k 1 − k k − k 2 2 2 β− f (βx) β f (x) β− h(x,x). ≤ k − k≤ Now suppose (2.6) holds, then by (2.5) and (2.6), we obtain

q (x) f (x) q (x) q (x) + q (x) f (x) k n+1 − k≤k n+1 − n k k n − k n 1 2 2 n − 2 2 i β− (γβ− ) h(x,x)+ ∑ β− (γβ− ) h(x,x) ≤ i=0 n 2 2 i =∑ β− (γβ− ) h(x,x). i=0

Lending n +∞ in (2.6), we get → β 2h(x,x) h(x,x) T(x) f (x) − = , x A+. k − k≤ 1 (γβ 2) β2 γ ∈ − − − We prove T satisﬁes (2.4). To see this, replacing x,y in the ﬁrst equation of (2.2) with βnx, βny, gives f (βn x) f (βny)= f (βny) f (βn x), x,y A+, ∈ and so q (x)q (y)= q (y)q (x), x,y A+. n n n n ∈ H. Rezaei / Anal. Theory Appl., 29 (2013), pp. 333-341 337

Taking n +∞, yields the second equation of (2.4). On the other hand, replacing x,y in → the second equation of (2.2) with βn x, βny, gives

f (βn(ax+by)) a2 f (βn x) 2ab f (βn x) f (βny) b2 f (βny) h(βn x,βny). − − − ≤ q

Hence by deﬁnition of qn we obtain

q (ax+by) a2q (x) 2ab q (x)q (y) b2q (y) n − n − n n − n q 2n n 2 2n n n n n 2 2n n = β− f (β (ax+by)) a β− f (β x) 2ab β 4 f ( β x) f (β y) b β− f (β y) − − − − q 2n n 2 n n n 2 n β− f (β (ax+by)) a f (β x) 2ab f (β x) f (β y) b f (β y) ≤ − − − 2n n n 2n n q2 n β− h(β x,β y) β− γ h(x,x)=(γβ− ) h(x,x). ≤ ≤ Therefore,

2 2 2 n q (ax+by) a q (x) 2ab q (x)q (y) b q (y) (γβ− ) h(x,x), n − n − n n − n ≤ q for every x ,y A+ and n N. Considering 0 < γβ 2 < 1, and taking n +∞ in the last ∈ ∈ − → inequality; implies the second equation of (2.4). We now want to prove that T is a unique function satisfying (2.3) and (2.4). Assume + + that there exists another one, denoted by T′ :A B satisfying (2.3) and (2.4). Use (2.4) 2 2 → for y = x to get Tβx = β Tx and T′βx = β T′x and more generality

n 2n n 2n T(β x)= β Tx and T′(β x)= β T′x.

Hence,

2n n 2n n Tx = β− T(β x) and T′x = β− T′(β x), (2.7) for every x A+ and n N. By the triangle inequality, (2.3) and (2.7), we obtain ∈ ∈ 2n n 2n n 2n n n T(x) T′(x) = β− T(β x) β− T′(β x) β− T(β x) T′(β x) k − k k − k≤ k − k 2n n n n n β− ( T(β x) f (β x) + f (β x) T′(β x) ) ≤ k − k k − k 2 n n 2n n n 2n β− h(β x,β x) 2 β− h(β x,β x) β− 2 2 2β− 2 ≤ 1 γβ− ≤ 1 γβ− 2 −n − 2 (γβ− ) h(x,x) 2β− , ≤ 1 γβ 2 − − for every x A+ and n N. Since, 0 < γβ 2 < 1, letting n +∞, we get T(x)= T (x) for ∈ ∈ − → ′ all x A+. ∈ 338 H. Rezaei / Anal. Theory Appl., 29 (2013), pp. 333-341

Recall that if X is a vector space, a function h: X X [0,+∞) is called homogeneous × → of order p whenever h(λx,λy)= λ p h(x,y) for every scalar λ and every x,y X. In this | | ∈ case h (a+b)x,(a+b)y γ :=sup = a+b p. h(x,y) | | n o This led to the following corollary:

Corollary 2.1. Let h : A+ A+ [0,+∞) be a homogeneous function of order p < 2. If × → f :A+ B+ is a function satisfying (2.2) then there exists a unique mapping T :A+ B+ → → satisfying (2.4) and

h(x,x) f (x) T(x) , x A+. k − k≤ a+b 2 a+b p ∈ | | −| | Since the function h(x,y)= θ( x p + y p) is a homogenous of order p, we have the k k k k following corollary:

Corollary 2.2. Assume that f :A+ B+ is a function satisfying → f (x) f (y)= f (y) f (x), f (ax+by) a2 f (x) 2ab f (x) f (y) b2 f (y) θ( x p + y p), ( k − − − k≤ k k k k p for some θ 0 and p < 2, and for every x,y A+. Then there exists a unique mapping ≥ ∈ T :A+ B+ satisfying (2.4) and → h(x,x) f (x) A(x) , x A+. k − k≤ a+b 2 a+b p ∈ | | −| | 3 Hyers-Ulam-Rassias stability of (1.1): ﬁxed point method

In 2003, V. Radu proved the Hyers-Ulam-Rassias stability of the additive Cauchy equation by using the fixed point method (see [12] or [13]). In this section, using some ideas from [12] or [13], we prove Theorem 2.1 by the fixed point method. Before state it we need to introduce one of the fundamental results of the fixed point theory. For the proof, we refer to [14].

Theorem 3.1. Let (Ω,d) be a generalized complete metric space. Assume that Λ : Ω Ω is a → strictly contractive operator with the Lipschitz constant L <1, i.e.,

d(Λg,Λh) Ld(g,h) ≤ for all g,h Ω. If there exists a nonnegative integer n such that d(Λn0+1 f ,Λn0 f )<+∞ for some ∈ 0 f Ω, then the following statements are true: ∈ H. Rezaei / Anal. Theory Appl., 29 (2013), pp. 333-341 339

(i) The sequence Λn f converges to a fixed point A of Λ. { } (ii) A is the unique fixed point of Λ in Ω = g Ω : d(Λn0 f ,g)<+∞ . ∗ { ∈ } (iii) If g X∗, then ∈ 1 d(g,A) d(Λg,g). ≤ 1 L − Now we are ready to prove Theorem 2.1 is by the fixed point method:

Theorem 3.2. Under the assumptions of Theorem 2.1, let h : A+ A+ [0,+∞) be a function × → satisfying (2.1), and f : A+ B+ be a function satisfying (2.2). Then there exists a unique → mapping T :A+ B+ satisfying (2.3) and (2.4). → Proof. Letting y = x in (2.2), we get

f (βx) β2 f (x) h(x,x) k − k≤ for every x A+. Consider the set ∈ Ω := g :A+ B+ : g(x)g(y)= g(y)g(x) for all x,y A+ { → ∈ } and deﬁne the generalized metric d on Ω by

d(g ,g )=inf µ (0,+∞) : g (x) g (x) µh(x,x) for all x A+ 1 2 ∈ k 1 − 2 k≤ ∈ g (x) g (x) =sup k 1 − 2 k : x A+, h(x,x) =0 . h(x,x) ∈ 6 n o It is easy to show that (Ω,d) is a complete metric space (see [15]). Now we consider the linear mapping Λ : Ω Ω by → 1 Λg(x)= g(βx) β2 for all x A+. For given g ,g Ω, ∈ 1 2 ∈ 1 1 Λg (x) Λg (x) = g (βx) g (βx) k 1 − 2 k β2 1 − β2 2

1 g (βx) g (βx) ≤ β2 k 1 − 2 k 1 d(g ,g )h(x,x) ≤ β2 1 2 for every x A+. By deﬁnition of d, ∈ γ d(Λg ,Λg ) d(g ,g ). 1 2 ≤ β2 1 2 340 H. Rezaei / Anal. Theory Appl., 29 (2013), pp. 333-341

Hence Λ is indeed a contraction. Note that 1 f (x) Λ f (x) = f (x) f (βx) k − k − β2

1 1 β2 f (x) f (β x) h(x,x) ≤ β2 k − k≤ β2 for all x A+ whence d(Λ f , f ) β 2 < +∞. By the preceding theorem, there exists a ∈ ≤ − mapping T :A+ B+ satisfying the following conditions: → (i) T(x)T(y)= T(y)T(x) for all x,y A+. ∈ (ii) T is a ﬁxed point of Λ, i.e.,

1 T(βx)=(ΛT)(x)= T(x), β2

whence T(βx)= β2 T(x) for all x A+. Moreover, T is a unique ﬁxed point of Λ in ∈ the set Ω∗ := g Ω : d( f ,g)<+∞ , { ∈ } which implies that

f (x) T(x) d( f ,T)h(x,x), x A+. k − k≤ ∈ (iii) d(Λn f ,T) 0 as n +∞, deducing that T(x)=lim β n f (βn x). → → n − (iv) By (iii) of the preceding theorem we conclude that

1 1 2 1 d( f ,A) d( f ,Λ f )< β− = , ≤ 1 γβ 2 1 γβ 2 β2 γ − − − − − deducing the inequality

h(x,x) f (x) T(x) d( f ,T)h(x,x) , x A+. k − k≤ ≤ β2 γ ∈ − From here we proceed exactly as in the proof of Theorem 2.1 to show that T :A+ B+ is → indeed a unique mapping satisfying (2.4).

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