HACETTEPE JOURNAL OF MATHEMATICS AND STATISTICS

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1 HACETTEPE UNIVERSITY FACULTY OF SCIENCE TURKEY HACETTEPE JOURNAL OF MATHEMATICS AND STATISTICS A Bimonthly Publication Volume 42 Issue ISSN

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3 HACETTEPE JOURNAL OF MATHEMATICS AND STATISTICS Volume 42 Issue 1 February 2013 A Peer Reviewed Journal Published Bimonthly by the Faculty of Science of Hacettepe University Abstracted/Indexed in SCI-EXP, Journal Citation Reports, Mathematical Reviews, Zentralblatt MATH, Current Index to Statistics, Statistical Theory & Method Abstracts, SCOPUS, Tübitak-Ulakbim. ISSN This Journal is typeset using L A TEX.

4 Hacettepe Journal of Mathematics and Statistics Cilt 42 Sayı 1 (2013) ISSN KÜNYE YAYININ ADI: HACETTEPE JOURNAL OF MATHEMATICS AND STATISTICS YIL : 2013 SAYI : 42-1 AY : Şubat YAYIN SAHİBİNİN ADI : H. Ü. Fen Fakültesi Dekanlığı adına Prof. Dr. Kadir Pekmez SORUMLU YAZI İŞL. MD. ADI : Prof. Dr. Yücel Tıraş YAYIN İDARE MERKEZİ ADRESİ : H. Ü. Fen Fakültesi Dekanlığı YAYIN İDARE MERKEZİ TEL. : YAYININ TÜRÜ : Yaygın BASIMCININ ADI : Hacettepe Üniversitesi Hastaneleri Basımevi. BASIMCININ ADRESİ : Sıhhıye, ANKARA. BASIMCININ TEL. : BASIM TARİHİ - YERİ : - ANKARA

5 Hacettepe Journal of Mathematics and Statistics A Bimonthly Publication Volume 42 Issue 1 (2013) ISSN EDITORIAL BOARD EDITOR : Yücel Tıraş ASSOCIATE EDITOR : Lawrence M. Brown MEMBERS Gary F. Birkenmeier (University of Louisiana at Lafayette, USA.) G. C. L. Brümmer (University of Cape Town, South Africa) Cem Kadılar (Hacettepe University, Ankara, Turkey) Varga Kalantarov (Koç University, Istanbul, Turkey) Vladimir Levchuk (Siberian Federal University, Russia) Cihan Orhan (Ankara University, Ankara, Turkey) Ivan Reilly (University of Auckland, New Zealand) Patrick Smith (Glasgow University, United Kingdom) Bülent Saraç (Hacettepe University, Ankara, Turkey) Alexander P. Šostak (University of Latvia, Riga, Latvia) Ağacık Zafer (Middle East Technical University, Ankara, Turkey) Published by Hacettepe University Faculty of Science

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7 Mathematics CONTENTS E. Kılıç and P. Stănică General Approach in Computing Sums of Products of Binary Sequences S. Ebru Daṣ Dynamics of a Nonlinear Rational Difference Equation Süleyman Solak and Mustafa Bahşi On the Spectral Norms of Toeplitz Matrices with Fibonacci and Lucas Numbers Mujahid Abbas, Talat Nazır and B. E. Rhoades Fixed Points of Multivalued Mapping Satisfying Ciric Type Contractive Conditions in G-Metric Spaces Hasan Atik Cofibration Category and Homotopies of Three Crossed Complexes A. Dilek Güngör Maden, Ivan Gutman and A. Sinan Çevik Bounds for Resistance Distance Spectral Radius Xianping Liu, Dajing Xiang, K. P. Shum and Jianming Zhan Soft Rings Related to Fuzzy Set Theory Derya Keskin Tütüncü and Sultan Eylem Toksoy Absolute Co-Supplement and Absolute Co-Coclosed Modules Yiqiu Du and Yu Wang A Result on Generalized Derivations in Prime Rings Statistics Ahmet Kara Dynamics of Performance in Higher Education an Applied Stochastic Model and a Case Study

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9 MATHEMATICS

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11 Hacettepe Journal of Mathematics and Statistics Volume 42 (1) (2013), 1 7 GENERAL APPROACH IN COMPUTING SUMS OF PRODUCTS OF BINARY SEQUENCES E. Kılıç, P. Stănică Received 10 : 02 : 2010 : Accepted 16 : 12 : 2011 Keywords: Abstract In this paper we find a general approach to find closed forms of sums of products of arbitrary sequences satisfying the same recurrence with different initial conditions. We apply successfully our technique to sums of products of such sequences with indices in (arbitrary) arithmetic progressions. It generalizes many results from literature. We propose also an extension where the sequences satisfy different recurrences. Second order recurrences; Sums; Products 2000 AMS Classification: 11B37, 11C20 1. Introduction We consider a generic (nondegenerate, that is, δ = p 2 4q 0) binary recurrence satisfying (1.1) X n+1 = px n qx n 1, n Z with some initial conditions. Let α, β be the roots of the equation x 2 px + q = 0, and so, α + β = p, αβ = q, δ = α β. We associate the companion Lucas sequence L n which also satisfies (1.1) together with L 0 = 2, L 1 = p, and so L n = α n + β n. Let {U n (j) } p j=1 be a set of p binary sequences, all of which will satisfy the recurrence (1.1) with some initial conditions, such that the Binet formula for these sequences is where A j = U (j) 1 U (j) n = A jα n + B jβ n, n Z, (j) U 0 β δ, B j = U (j) 0 (j) α U 1. δ TOBB Economics and Technology University, MathematicsDepartment, Sogutozu, Ankara, Turkey. ekilic@etu.edu.tr Naval Postgraduate School, Department of Applied Mathematics, Monterey, CA 93943, USA. pstanica@nps.edu

12 2 E. Kiliç, P. Stănică For easy notation, we will denote the recurrence {X n} given by (1.1) by {X n (p, q, a, b)} where a = X 0 and b = X 1 are initial conditions of it. Several authors investigated products of two terms of a sequence or products of two sequences, and also, the sums of these products. As a first example, note that the sum of square terms of Fibonacci numbers [7, 8, 12] is n = F nf n+1. i=1 F 2 i The sum of products of variable subscripted terms of certain second order recurrences have been considered by several authors. For example (see [11]) n F if i+2 =F 2n+1F 2n+2 1, i=1 n F if i+1 =F2n+1 2 1, i=1 n F 2i 1F 2i+3 = ( 3F2n+2 2 2F2n n 1 ) /5. i=1 Certainly, the classical Fibonacci, F n and Pell numbers P n are F n = X n (1, 1, 0, 1) and P n = X n (2, 1, 0, 1). Generalizations of the above sums by taking different recurrences and their variable subscripted terms have also been studied. For example, in [9], the author found n i=1 FiPi. Melham [10] looked at the sum of the squares of the sequence {X n (2, 1, 0, 1)}. Recently, in [2, 3, 4, 5, 6], the authors gave several formulas for sums of squares of even and odd Fibonacci, Lucas and Pell-Lucas numbers, and their sums of products of even and odd subscripted terms. Also the authors of [1] established several formulas for sums and alternating sums of products of certain subscripted terms of recurrences {X n (p, q, 0, 1)} and {X n (p, q, 2, p)}. It is our goal in this paper to propose a general approach for the theory of closed forms for sums of products of nondegenerate second-order recurrent sequences, thus generalizing many of these kind of results that the reader can find scattered throughout the literature. 2. Main Results Let P(n) be the power set of {1, 2,..., n}, that is the set of all subsets of {1, 2,..., n}. Given a sequence of p functions f j(i), j = 1,..., p, for all M P(p), we let F M (i) = l M f l(i), F (i) = 0, and for simplicity, F (i) = F {1,...,p} (i) = p l=1 f l(i). Let us define a set of twisted product sequences, indexed by the sets M P(p), in the following way: for a set M P(p), we let W n (M) be a (M-twisted product) rational sequence satisfying (1.1) with the Binet formula W n (M) = A j B k α n + B j A k β n. j M k M Further, we use M = {1, 2,..., p} \ M, for the complement of the set M in {1, 2,..., p}. We shall first show that W n (M) is a rational sequence, even more precise that W n (M) 1 δ 2p 1 Z. j M k M 2.1. Lemma. For any integer n, the twisted product sequences satisfy W ( M) n = q n W (M) n. Proof. Straightforward using the Binet formula.

13 General Approach in Computing Sums of Products of Binary Sequences Theorem. For p, n Z, we have W (M) n 1 δ 2p 1 Z. Proof. We will prove the claim by induction. First, we let p = 2, and consider two sequences U n = A 1α n + B 1β n, V n = A 2α n + B 2β n (for simplicity of notations). We write the superscript sets as {a,...} instead of ({a,...}). The associated twisted product sequences are W {1,2} n = A 1A 2α n + B 1B 2β n, W {1} n = A 1B 2α n + B 1A 2β n, W {2} n = A 2B 1α n + A 1B 2β n, W n = B 1B 2α n + A 1A 2β n. Since our index n runs through the entire set of integers, by Lemma 2.1, it will be sufficient to consider only the case of W n {1,2}, and W n {1}. First, using the expressions for A 1, A 2, B 1, B 2 in terms of initial conditions of U n, V n, and simplifying, we get Further, δ 2 (A 1A 2 + B 1B 2) = 2U 1V 1 p(u 0V 1 + U 1V 0) + (p 2 2q)U 0V 0 Z δ 2 (A 1A 2α + B 1B 2β) = pu 1V 1 2q(U 1V 0 + U 0V 1) + pqu 0V 0 Z. δ 2 (A 1B 2 + B 1A 2) = p(u 1V 0 + V 1U 0) 2qU 0V 0 2U 1V 1 Z δ 2 (A 1B 2α + B 1A 2β) = (p 2 2q)U 1V 0 p(u 0V 0q + U 1V 1) + 2U 0V 1q Z. 1 Now, let U n Z and, from the induction step, assume that V n Z. As before, δ 2p 1 writing δ 2 W {1,2} 0, δ 2 W {1,2} 1, δ 2 W {1} 0, δ 2 W {1} 1 in terms of U 0, U 1, V 0, V 1, we see that each 1 term in these expressions contains only one factor based on either V 0, or V 1 δ 2p 1 Z, and therefore W {1,2} i, W {1} i 1 δ2p Z, i = 0, 1. Certainly, since the initial terms of the twisted product sequences are in 1 (M) δ2p Z, so is W n. We show now our general approach to finding sums of products of recurrences Theorem. Given a set of p functions f j(i), j = 1,..., p, such that f j(i) f l (i) is a function of j, l only and it does not depend on i, we have n p i=0 j=1 U (j) f j (i) = 1 2 n M P(p) i=0 q F (i) F M (i) W (M) 2F M (i) F (i). Proof. First, we associate to every set M P(p) a bit string ɛ of length p in the usual manner (a 1 bit appears in the bit string if and only if its corresponding position appears in M, otherwise the bit is 0). For ɛ Z p 2, we let wt(ɛ) to be the Hamming weight of the bit string ɛ, that is, the number of 1 s in its expression, and supp(ɛ) = {i 1 < i 2 <... < i wt(ɛ) } to be the support of ɛ (the positions where 1 s appear in ɛ). Certainly, supp(ɛ) {1, 2,..., p}.

14 4 E. Kiliç, P. Stănică Next, we compute the product p j=1 U (j) f j (i) = = p (A ) jα f j (i) + B jβ f j (i) j=1 p A ɛ j j B1 ɛ j j ɛ=(ɛ 1,...,ɛ p) Z p j=1 2 = 1 2 ɛ Z p 2 + j supp(ɛ) k supp(ɛ) j supp(ɛ) k supp(ɛ) = M P(p) j M k M = 1 2 (αβ) j M f j (i) α ɛ j f j (i) β (1 ɛ j )f j (i) A jb k α j supp(ɛ) f j (i) β k supp(ɛ) f k(i) A k B jβ j supp(ɛ) f j (i) α k supp(ɛ) f k(i) A jb k α j M f j (i) j M f j (i) j M k M A k B jβ j M f j (i) j M f j (i) M P(p) q F (i) F M (i) W (M) 2F M (i) F (i), from which our theorem follows easily. Obviously, if the sum n i=0 qf (i) F M (i) W (M) 2F M (i) F (i) can be simplified, then the previous theorem takes quite an attractive form. The rest of the paper is devoted in finding various functions f j for which such a sum can be computed. Many papers are investigating sums of products of very few recurrences (mostly two) where the indices are very specific linear functions. We will attack this case in its full generality here and solve it completely, by taking f j to be arbitrary linear functions. Let W n be our generic sequence satisfying (1.1) such that W n = Aα n + Bβ n, and recall that L n = α n + β n is the companion Lucas sequence Lemma. For a, b, c, d Z, we have the generating function n i=0 x a+bi W c+di = x a q d x b(n+2) W c+dn x b(n+1) W c+d(n+1) x b q d W c d + W c. x 2b q d x b L d + 1

15 General Approach in Computing Sums of Products of Binary Sequences 5 Proof. Using Binet formula for W n, we obtain n x bi W c+di = Aα c i=0 n i=0 n (x b α d ) i + Bβ c (x b β d ) i = Aα c (x b α d ) n+1 1 x b α d 1 i=0 + Bβ c (x b β d ) n+1 1 x b β d 1 = Axb(n+2) β d α c+d(n+1) Ax b(n+1) α c+d(n+1) Ax b α c β d + Aα c +Bx b(n+2) α d β c+d(n+1) Bx b(n+1) β c+d(n+1) Bx b β c α d + Bβ c x 2b (αβ) d x b (α d + β d ) + 1 = qd x b(n+2) (Aα c+dn + Bβ c+dn ) x b(n+1) (Aα c+d(n+1) + Bβ c+d(n+1) ) q d x b (Aα c d + Bβ c d ) + (Aα c + Bβ c ) x 2b q d x b L d + 1 = qd x b(n+2) W c+dn x b(n+1) W c+d(n+1) x b q d W c d + W c. x 2b q d x b L d + 1 Taking W n = u n = X n (p, q, 0, 1), we reach at the following result: n i=0 ( 1) i u r+4i = ( 1) n υ 4n+r+2 + u r 2 v 2 where v n = X n (p, q, 2, p). One can also find this result in [1, Lemma 5]. Let f j(i) = a j + b ji be linear functions. Under these conditions, p F (i) F M (i) = j + b ji) j=1(a (a j + b ji) j M = a j + j M j M b j i = a ( M) + b ( M) i, where we use the notations a ( M) = j M aj a (M) = j M aj, b(m) = j M bj. Further, and b( M) = j M bj. We shall also use 2F M (i) F (i) = (a j + b ji) j + b ji) j M j M(a ( ) ( ) = a (M) a ( M) + b (M) b ( M) i. Applying Lemma 2.4 with x := q, a := a ( M), b := b ( M), c := a (M) a ( M), d := b (M) b ( M), and using Theorem 2.3 we obtain our next result.

16 6 E. Kiliç, P. Stănică 2.5. Theorem. Given a set of linear functions f j(i) = a j + b ji, and binary sequences satisfying (1.1) with some initial conditions, we have U (j) k n p i=0 j=1 U (j) f j (i) = A Particular Case M P(p) (M) +b ( M) (n+1) qb W (M) a(m) a q (M) a ( M) +n(b (M) b ( M) ) q b( M) (n+1) W (M) a (M) a ( M) +(n+1)(b (M) b ( M) ) q b(m) W (M) (M) + W a (M) +a ( M) b (M) b ( M) q b(m) q b( M) L b (M) b ( M) + 1 a (M) a ( M) To understand our general result better, we shall consider now a particular case of two binary recurrences, which is the case most often encountered in literature. Let U n, V n be two binary recurrent sequences satisfying (1.1) with some initial conditions. The Binet formula indicates that U n = A 1α n + B 1β n, V n = A 2α n + B 2β n,. where A 1 = U 0β U 1, β α B1 = U 1 U 0 α, β α A2 = V 0β V 1, β α B2 = V 1 V 0 α As before, we take the twisted products W n {1,2}, W n {1} β α., satisfying (1.1), with initial conditions W {1,2} 0 = A 1A 2+B 1B 2, W {1,2} 1 = A 1A 2α+B 1B 2β, W {1} 0 = A 1B 2+B 1A 2, W {1} 1 = n = A 1B 2α n + B 1A 2β n. A 1B 2α + B 1A 2β, so that W {1,2} n = A 1A 2α n + B 1B 2β n, and W {1} From Theorem 2.2 we know that W {1} n, W {1,2} n 1 δ 2 Z. We next consider the example f 1(i) = r + ki, f 2(i) = s + ki Theorem. Let k, r, s be fixed integers. We have n U r+ki V s+ki = q s W {1} q k(n+1) 1 r s q k 1 i=0 Proof. First, (3.1) + q2k W {1,2} r+s+2kn W {1,2} r+s+2k(n+1) q2k W {1,2} r+s 2k + W {1,2} r+s. q 2k L 2k + 1 U r+ki V s+ki = (A 1α r+ki + B 1β r+ki )(A 2α s+ki + B 2β s+ki ) = (A 1A 2α r+s+2ki + B 1B 2β r+s+2ki ) + (A 1B 2α r+ki β s+ki + A 2B 1α s+ki β r+ki ) = W {1,2} r+s+2ki + qs+ki (A 1B 2α r s + A 2B 1β r s ) = W {1,2} r+s+2ki + qs+ki W {1} r s. In the notations of Theorem 2.5, the previous product will be 1 ( q f 1(i)+f 2 (i) W F (i) + q f2(i) W {1} f 2 1 (i) f 2 (i) + qf 1(i) W {2} ) +W {1,2} f 1 (i)+f 2 (i) f 2 (i) f 1 (i) = W {1,2} f 1 (i) + qf 2(i) W {1} f 1 (i) f 2 (i).

17 General Approach in Computing Sums of Products of Binary Sequences 7 Using (3.1), we separate the sum n i=0 U r+kiv s+ki into two sums. First, n n (3.2) q s+ki W {1} r s = qs W {1} r s (q k ) i = q s W {1} q k(n+1) 1 r s, q k 1 i=0 i=0 (we could have also used Lemma 2.4 with x := q, a = s, b = k and c = r s, d = 0). Next, using Lemma 2.4 with x := q, a = b = 0 and c = r + s, d = 2k, we get n W t+li = ql W t+ln W t+l(n+1) q l W t l + W t. q l L l + 1 i=0 and the second sum becomes n W {1,2} r+s+2ki = q2k W {1,2} r+s+2kn W {1,2} r+s+2k(n+1) q2k W {1,2} r+s 2k + W {1,2} r+s, q 2k L 2k + 1 i=0 which finishes the proof of our theorem. If we take u n = X n (p, q, 0, 1) and v n = X n (p, q, 2, p) (p 0, p 2 4q 0), then by required arrangements, we obtain for k = 2 n u r+2iv s+2i = v4n+r+s+2 vr+s 2 p (n + 1) qr u s r p (p 2 4q) i=0 which is the main result of [1, Theorem 1]. References [1] Belbachir H. and Bencherif, F. Sums of products of generalized Fibonacci and Lucas numbers, Arxiv: v1, [2] Cerin, Z. Sums of products of generalized Fibonacci and Lucas numbers, Demons. Math., 42 (2), , [3] Cerin, Z. On Sums of Products of Horadam Numbers, Kyungpook Math. J., 49, , [4] Cerin, Z. and Gianella G.M. On sums of squares of Pell-Lucas numbers, Integers 6 A15, 16 pp., [5] Cerin, Z. Alternating sums of Fibonacci products, Atti Semin. Mat. Fis. Univ. Modena Reggio Emilia 53 (2), , [6] Cerin, Z. Some alternating sums of Lucas numbers, Cent. Eur. J. Math. 3 (1), 1 13, [7] Kılıç, E. Sums of the squares of terms of sequence {u n}, Proc. Indian Acad. Sci. 118 (1), 27 41, [8] Koshy, T. Fibonacci and Lucas numbers with applications, Pure and Appl. Math., Wiley- Interscience, New York, [9] Mead, D.G. Problem B-67, Fibonacci Quart. 3 (4), , [10] Melham, R.S. On sums of powers of terms in a linear recurrence, Portugal. Math. 56 (4), , [11] Rao, K.S. Some properties of Fibonacci numbers, The Amer. Math. Monthly, 60, , [12] Vajda, S. Fibonacci & Lucas numbers, and the golden section, John Wiley & Sons, Inc., New York, 1989.

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19 Hacettepe Journal of Mathematics and Statistics Volume 42 (1) (2013), 9 14 DYNAMICS OF A NONLINEAR RATIONAL DIFFERENCE EQUATION S. Ebru Daṣ Received 22 : 08 : 2010 : Accepted 20 : 12 : 2011 Keywords: Abstract In this paper, we investigate the dynamical properties of the following nonlinear difference equation: x n+1 = xa nx n 2x n 3 + x nx n 2x a n x a nx n 3 + x nx a n 3 + 1, n = 0, 1,... Difference equation, Stability, Global Stability, Oscillation AMS Classification: 39A Introduction Recently, there has been a great interest in studying the qualitative behavior of rational difference equations. Berenhaut et al.[4] has showed that the unique positive equilibrium ȳ = 1 of the difference equation: y n = y n k + y n m 1 + y n k y n m, n = 0, 1,... is globally asymptotically stable. Chen et al.[5] investigated the dynamical properties of the following fourth-order nonlinear difference equation: x n+1 = xa n 2 + x n 3, n = 0, 1,... x a n 2xn with nonnegative initial conditions and a [0, 1). Das [6] investigate the qualitative behavior of the following fourth-order difference equation: x n+1 = xn 1xa n 2 + x n 1x a n x a n 2 + xa n 3 + 1, n = 0, 1,... Yıldız Technical University, Department of Mathematics, Istanbul, Turkey. eyeni@yildiz.edu.tr

20 10 S. E. Daṣ where a (0, ) and the initial conditions x 3, x 2, x 1, x 0 (0, ). For more work, see [1, 2, 3, 7, 8, 9, 10]. To be motivated by the above studies, in this paper, we consider the following nonlinear difference equation: (1.1) x n+1 = xa nx n 2x n 3 + x nx n 2x a n x a nx n 3 + x nx a n 3 + 1, n = 0, 1,... where a (0, ) and the initial conditions are arbitrary positive real numbers. It is easy to see that the positive equilibrium x = 1 of Eq.(1.1) satisfies x = (2 x a+2 +1)/(2 x a+1 +1). In the following, we state some main definitions used in this paper Definition. A positive semi-cycle of a solution {x n} n= 3 of Eq.(1.1) consists of a string of terms {x l, x l+1,..., x m} all greater than or equal to the equilibrium x, with l 3 and m < such that either l = 3 or l > 3 and x l 1 < x and either m = or m < and x m+1 < x A negative semi-cycle of a solution {x n} n= 3 of Eq.(1.1) consists of a string of terms {x l, x l+1,..., x m} all less than x, with l 3 and m < such that either l = 3 or l > 3 and x l 1 x and either m = or m < and x m+1 x The length of a semi-cycle is the number of the total terms contained in it Definition. A solution {x n} n= 3 of Eq.(1.1) is said to be eventually trivial if x n is eventually equal to x = 1 ; Otherwise the solution is said to be nontrivial. A solution {x n} n= 3 of Eq.(1.1) is said to be eventually positive (negative) if x n is eventually greater (less) than x = Three Lemmas Before to draw a qualitatively clear picture for the positive solutions of Eq.(1.1), we first establish three basic lemmas which will play a key role in the proof of our main results Lemma. A positive solution {x n} n= 3 of Eq.(1.1) is eventually equal to 1 if and only if (2.1) (x 2 1)(x 1 1)(x 0 1) = 0 Proof. Assume that (2.1) holds. Then according to Eq.(??), it is easy to see that the following conclusions hold: (i) if x 2 = 1, then x n = 1 for n 40 (ii) if x 1 = 1, then x n = 1 for n 40 (ii) if x 0 = 1, then x n = 1 for n 40

21 Dynamics of a Nonlinear Rational Difference Equation 11 Conversely, assume that (2.2) (x 2 1)(x 1 1)(x 0 1) 0 Then one can show that (2.3) x n 1 for any n 1 Assume the contrary that for some N 1, (2.4) x N = 1 and that x n 1 for 2 n N 1 It is easy to see that (2.5) 1 = x N = xa N 1x N 3x N 4 + x N 1x N 3x a N x a N 1 xn 4 + xn 1xa N which implies (x a N 1x N 4 + x N 1x a N 4)(x N 3 1) = 0. Obviously, this contradicts (2.3) Remark. If the initial conditions do not satisfy Eq.(1.1), then, for any solution x n of Eq.(1.1), x n 1 for n 3. Here, the solution is a nontrivial one Lemma. Let {x n} n= 3 be a nontrivial positive solution of Eq.(1.1). Then the following conclusions are true for n 0: (a) (x n+1 1)(x n 2 1) > 0 (b) (x n+1 x n 2)(x n 2 1) < 0 Proof. It follows in light of Eq.(1.1) that x n+1 1 = (xa nx n 3 + x nx a n 3)(x n 2 1) x a nx n 3 + x nx a n 3 + 1, n = 0, 1,... (1 x n 2) x n+1 x n 2 =, n = 0, 1,... x a nx n 3 + x nx a n from which inequalities (a) and (b) follow Lemma. (i) If x 2, x 1, x 0 > 1, then {x n} n= 3 has a positive semi-cycle with an infinite number of terms and it monotonically tends to the positive equilibrium point x = 1. (ii) If x 2, x 1, x 0 < 1, then {x n} n= 3 has a negative semi-cycle with an infinite number of terms and it monotonically tends to the positive equilibrium point x = 1. Proof. (i) If x 2, x 1, x 0 > 1, from Lemma 2.3.(a) and (b), for n 3 1 < x 3k 2 <... < x 4 < x 1 < x 2 1 < x 3k 1 <... < x 5 < x 2 < x 1 1 < x 3k <... < x 6 < x 3 < x 0, k = 0, 1,... Clearly, {x n} n= 3 has a positive semi-cycle with an infinite number of terms and monotonically decreasing for n 0. So the limit (2.6) lim n xn = L exists and finite. Taking the limits on both sides of Eq.(1.1), we have L = 2La L a+1 + 1

22 12 S. E. Daṣ we can easily see that {x n} n= 3 tends to the positive equilibrium point x = 1. (ii) If x 2, x 1, x 0 < 1, from Lemma 2.3.(a) and (b), for n 2 x 2 < x 1 < x 4 <... < x 3k 2 < 1 x 1 < x 2 < x 5 <... < x 3k 1 < 1 x 0 < x 3 < x 6 <... < x 3k < 1 k = 0, 1,... Therefore, {x n} n= 3 has a negative semi-cycle with an infinite number of terms and monotonically increasing for n 0. So the limit (2.7) lim n xn = M exists and finite. Taking the limits on both sides of Eq.(1.1), we have M = 2M a M a So, {x n} n= 3 tends to the positive equilibrium point x = Main Results and their proofs First we analyze the structure of the semi-cycles of nontrivial solutions of Eq.(1.1). Here we confine us to consider the situation of the strictly oscillatory solution of Eq.(1.1) Theorem. Let {x n} n= 3 be a strictly oscillatory solution of Eq.(1.1). Then the rule for the lengths of positive and negative semi-cycles of this solution to successively occur is , 1, 2 +, 1, 2 +, 1,... or... 2, 1 +, 2, 1 +, 2, 1 +,.... Proof. By Lemma 2.3.(a) and (b), one can see that the length of a positive semi-cycle is not larger than 2 and the length of a negative semi-cycle is at most 2. Based on the strictly oscillatory character of the solution, we see, for some p 0, that one of the following two cases must occur: Case1. x p 2 > 1, x p 1 < 1 and x p > 1 Case2. x p 2 > 1, x p 1 < 1 and x p < 1 If Case 1. Occurs, it follows from Lemma 2.3.(a) that x p+1 > 1, x p+2 < 1, x p+3 > 1, x p+4 > 1, x p+5 < 1, x p+6 > 1, x p+7 > 1, x p+8 < 1,... It means that the rule of the lengths of positive and negative semi-cycles of the solution of Eq.(1.1) to occur successively is , 1, 2 +, 1, 2 +, 1,.... If Case 2. Occurs, it follows from Lemma 2.3.(a) that x p+1 > 1, x p+2 < 1, x p+3 < 1, x p+4 > 1, x p+5 < 1, x p+6 < 1, x p+7 > 1, x p+8 < 1, x p+9 < 1,... It means that the rule of the lengths of positive and negative semi-cycles of the solution of Eq.(1.1) to occur successively is... 2, 1 +, 2, 1 +, 2, Therefore, the proof is complete. Now we present the global asymptotically stable results for Eq.(1.1) Theorem. Assume that a (0, ). Then the positive equilibrium of Eq.(1.1) is globally asymptotically stable.

23 Dynamics of a Nonlinear Rational Difference Equation 13 Proof. We should prove that the positive equilibrium point x of Eq.(1.1) is both locally asymptotically stable and globally attractive. The linearized equation of Eq.(1.1) about the positive equilibrium point x = 1 is y n+1 = 0.y n + 2.yn yn 3, n = 0, 1,... 3 By virtue of [7, Remark 1.3.7], x is locally asymptotically stable. It remains to verify that every positive solution {x n} n= 3 of Eq.(1.1) converges to 1 as n. Namely, we want to prove (3.1) lim n xn = 1 If the solution is nonoscillatory about the positive equilibrium point x of Eq.(1.1), then from Lemma 2.1 and Lemma 2.4, the solution is either equal to 1 or eventually positive or negative one which has an infinite number of terms and monotonically tends to the positive equilibrium point x of Eq.(1.1), and so Eq.(3.1) holds.therefore, it suffices to prove that Eq.(3.1) holds for the solution to be strictly oscillatory. Consider now {x n} to be strictly oscillatory about the positive equilibrium point x of Eq.(1.1). By virtue of Theorem 3.1, one understands that the rule for the lengths of positive and negative semi-cycles which occur successively is , 1, 2 +, 1, 2 +, 1,... or..., 2, 1 +, 2, 1 +, 2, 1 +,... Now, we investigate the case where the rule for the lengths of positive and negative semi-cycles which occur successively is , 1, 2 +, 1,... For simplicity, we denote by {x t, x t+1} + the terms of a positive semi-cycle of length two, followed by {x t+2} the terms of a negative semi-cycle with length one,followed by {x t+3, x t+4} + the terms of a positive semi-cycle of length two, followed by {x t+5} the terms of a negative semi-cycle with length one,and so on. Namely, the rule for the lengths of positive and negative semi-cycles to occur successively can be periodically expressed as follows for n = 0, 1,...: {x t+6n, x t+6n+1} +, {x t+6n+2}, {x t+6n+3, x t+6n+4} +, {x t+6n+5} then the following results can be easily observed: (3.2) 1 < x t+6n+4 < x t+6n+1 (3.3) 1 < x t+6n+6 < x t+6n+3 < x t+6n (3.4) x t+6n+2 < x t+6n+5 < 1 It follows from 3.2 that {x t+6n+1} n=0 is decreasing with lower bound 1. So the limit lim n xt+6n+1 = L exists and finite. Accordingly, by view of 3.2, we obtain lim n xt+6n+4 = L Also, it is easy to see from 3.3 that {x t+6n} n=0 is decreasing with lower bound 1. So the limit lim n xt+6n = M

24 14 S. E. Daṣ exists and finite. By view of 3.4, we obtain lim xt+6n+3 = lim n n xt+6n+6 = M Lastly, from 3.4 that {x t+6n+2} n=0 is increasing with upper bound 1. So the limit lim n xt+6n+2 = N exists and finite. By view of 3.4, we obtain lim n xt+6n+5 = N Taking the limits on both sides of x t+6n+6 = xa t+6n+5x t+6n+3x t+6n+2 + x t+6n+5x t+6n+3x a t+6n x a t+6n+5 xt+6n+2 + xt+6n+5xa t+6n one has, M = (2MN a+1 + 1)/(2N a+1 + 1), which gives rise to M = 1. Similarly, taking the limits on both sides of x t+6n+5 = xa t+6n+4x t+6n+2x t+6n+1 + x t+6n+4x t+6n+2x a t+6n x a t+6n+4 xt+6n+1 + xt+6n+4xa t+6n one has, N = (2NL a+1 + 1)/(2L a+1 + 1), which gives rise to N = 1. Lastly, taking the limits on both sides of x t+6n+4 = xa t+6n+3x t+6n+1x t+6n + x t+6n+3x t+6n+1x a t+6n + 1 x a t+6n+3 xt+6n + xt+6n+3xa t+6n + 1 one has, L = (2LM a+1 + 1)/(2M a+1 + 1), which gives rise to L = 1. So we can see that lim x t+6n+k = 1, k = 0, 1,..., 6 n For..., 2, 1 +, 2, 1 +, 2, 1 +,... can be similarly shown. References [1] Agarwal R.P. Difference Equations and Inequalities, Marcel Dekker, Newyork, NY, USA, 1st Edition, [2] Agarwal R.P. Difference Equations and Inequalities, Marcel Dekker, Newyork, NY, USA, 2nd Edition, [3] Bayram, M. and Das, S. E.Global Asymptotic Stability of a Nonlinear Recursive Sequence, Int. Math. For., 5(22), , [4] Berenhaut, K.S., Foley, J.D. and Stevic, S. The Global Attractivity of the Rational Difference Equation, Appl. Math. Lett., 20, 54-58, [5] Chen, Y. and Li, X. Dynamical Properties in a Fourth-order Nonlinear Difference Equation, Adv. Diff. Equ., ID , 9 pages, [6] Das, S. E. Qualitative Behavior of a Fourth-order Rational Difference Equation, in review, [7] Das, S. E. Global Asymptotic Stability for a Fourth-order Rational Difference Equation, Int. Math. For., 5(32), , [8] Kocic, V. L. and Ladas, G. Global Behavior of Nonlinear Difference Equations of Higher Order with Applications, Vol. 256 of Mathematics and its Applications, Kluwer Academic Publishers, Dordrecht, The Netherlands, [9] Li, X. and Zhu, D. Global asymptotic stability of a nonlinear recursive sequence, Compt.Math.Appl., 17, , [10] Li, X. Qualitative properties for a fourth-order rational difference equation, J.Math.Anal.Appl., 311, , 2005.

25 Hacettepe Journal of Mathematics and Statistics Volume 42 (1) (2013), ON THE SPECTRAL NORMS OF TOEPLITZ MATRICES WITH FIBONACCI AND LUCAS NUMBERS Süleyman Solak and Mustafa Bahşi Received 08 : 03 : 2010 : Accepted 02 : 01 : 2012 Keywords: Abstract This paper is concerned with the work of the authors [M.Akbulak and D. Bozkurt, on the norms of Toeplitz matrices involving Fibonacci and Lucas numbers, Hacettepe Journal of Mathematics and Statistics, 37(2), (2008), 89-95] on the spectral norms of the matrices: A = [F i j] and B = [L i j], where F and L denote the Fibonacci and Lucas numbers, respectively. Akbulak and Bozkurt have found the inequalities for the spectral norms of n n matrices A and B, as for us, we are finding the equalities for the spectral norms of matrices A and B. Spectral norm, Toeplitz matrix, Fibonacci number, Lucas Number AMS Classification: 15A60, 15A15, 15B05, 11B Introduction and Preliminaries The matrix T = [t ij] n 1 i,j=0 is called Toeplitz matrix such that tij = tj i. In Section 2, we calculate the spectral norms of Toeplitz matrices (1) A = [F j i] n 1 i,j=0 and (2) B = [L j i] n 1 i,j=0 where F k and L k denote k-th the Fibonacci and Lucas numbers, respectively. Now we start with some preliminaries. Let A be any n n matrix. The spectral norm of the matix A is defined as A 2 = max 1 i n λi (AH A) where A H is the conjugate transpose of matrix A. For a square matrix A, the square roots of the eigenvalues N.E. University, A.K. Education Faculty, 42090, Meram, Konya-TURKEY. ssolak42@yahoo.com Corresponding Author. Aksaray University, Education Faculty, Aksaray-TURKEY. mhvbahsi@yahoo.com

26 16 S. Solak, M. Bahşi of A H A are called singular values of A. Generally, we denote the singular values as σ n = { λ i : λ i is eigenvalue of matrix A H A }. Moreover, the spectral norm of matrix A is the maximum singular value of matrix A. The equation det(a λi) = 0 is known as the characteristic equation of matrix A and the left-hand side known as the characteristic polynomial of matrix A. The solutions of characteristic equation are known as the eigenvalues of matrix. Fibonacci and Lucas numbers are the numbers in the following sequences, respectively: 0, 1, 1, 2, 3, 5, 8, 13, 21,... and 2, 1, 3, 4, 7, 11, 18, 29, 47,... in addition, these numbers are defined backwards by 0, 1, 1, 2, 3, 5, 8, 13, 21,... and 2, 1, 3, 4, 7, 11, 18, 29, 47, Main Results 2.1. Theorem. Let the matrix A be as in (1). Then the singular values of A are σ 1,2 = { Fn, if n is even F 2 n 1, if n is odd and σ m = 0, where m=3,4,...,n. Proof. From matrix multiplication AA H = [ n 1 ] n 1 F k i F k j. k=0 i,j=0 By using mathematical induction principle on n, we have n 1 F n 1F n (i+j) + F if j, if n is odd F k i F k j = k=0 F nf n (i+j+1), if n is even. Since the singular values of matrix A are the square roots of the eigenvalues of matrix AA H, we must find the roots of characteristic equation λi AA H = 0, for this there are two cases. Case I: If n is odd, since AA H characteristic equation: = [ ] n 1 F n 1F n (i+j) + F if j, in this case the i,j=0 λi AA H = λ F n 1F n Fn 1 2 F n 1F 1 Fn 1 2 λ F n 1F n 2 F 1F 1 F n 1F 0 F 1F 1 n... F n 1F 1 F n 1F 0 F 1 nf 1 λ F n 1F n+2 F 1 nf 1 n = 0. Let e [(i, j), r, k] be an elementary row operation, where e [(i, j), r, k] is addition of k times of addition of ith and jth rows to rth row. Firstly, we apply e [(i + 1, i + 2), i, 1], (i = 1, 2,..., n 2). Secondly, we add proper times of first n 2 rows to (n 1)th row and then to nth row, so we have

27 On the Spectral Norms of Toeplitz Matrices with Fibonacci λi AA H = λ λ λ λ λ λ = λ λ λ λ Fn λ Fn = λ n 2 ( λ F 2 n + 1 ) 2 = 0. Hence, the singular values of the matrix A are σ 1,2 = F 2 n 1, σ m = 0, where m = 3, 4,..., n. Case II : If n is even, since AA H = [ ] n 1 F nf n (i+j+1), the characteristic equation: i,j=0 λ F nf n 1 F nf n 2 F nf 1 F nf 0 F nf n 2 λ F nf n 3 F nf 0 F nf 1 λi AA H =.... F nf 1 F nf 0 λ F nf 3 n F nf 2 n F nf 0 F nf 1 F nf 2 n λ F nf 1 n If we apply elemanter row operations in Case I to the determinant given above, we have = 0. λi AA H = λ λ λ λ λ λ = λ λ λ λ Fn λ Fn 2 = λ n 2 ( λ F 2 n) 2 = 0. In that case, the singular values of the matrix A are σ 1,2 = F 2 n, σ m = 0, where m = 3, 4,..., n. Thus the proof is completed Corollary. Let the matrix A be as in (1), then A 2 = { Fn, if n is even F 2 n 1, if n is odd. Proof. The proof is trivial from Theorem Theorem. Let the matrix B be as in (2). Then the singular values of B are { Ln ± 1, if n is odd σ 1,2 = F 2 and σ n 1, if n is even m = 0, where m = 3, 4,..., n. Proof. From matrix multiplication BB H = [ n 1 ] n 1 L k i L k j. k=0 i,j=0

28 18 S. Solak, M. Bahşi By using mathematical induction principle on n, we have n 1 F n (i+j+1) L n 1 + F n (i+j+2) L n+2 5F if j, if n is odd L k i L k j = k=0 5F nf n (i+j+1), if n is even. Firstly, we must find the roots of characteristic equation λi BB H = 0, for this there are two cases. Case I: If n is odd, since BB H = [ F n (i+j+1) L n 1 + F n (i+j+2) L n+2 5F if j ] n 1 i,j=0, in this case the characteristic equation: λi BB H = λ F n 1L n 1 F n 2L n+2 F 0L n 1 F 1L n+2 F n 2L n 1 F n 3L n+2 F 1L n 1 F 2L n+2 + 5F 1F 1 n.. F 0L n 1 F 1L n+2 λ F 1 nl n 1 F nl n+2 + 5F 1 nf 1 n = 0. If we apply elementary row operations in Case I of Theorem 2.1 to the determinant given above, we have λi BB H = λ λ λ λ λ λ = λ λ λ λ a 1 2F n 3L n F n 1L n λ a 2 = λ n 2 [ λ 2 ( (L n 1) 2 + (L n + 1) 2) λ + (L 2 n 1) 2] = 0 where a 1 = (L n 1) 2 (2F n 2 2)L n and a 2 = (L n + 1) 2 + (2F n 2 2)L n. Hence, the singular values of the matrix B are σ 1,2 = L n ± 1, σ m = 0, where m = 3, 4,..., n. Case II: If n is even, since BB H = [ ] n 1 5F nf n (i+j+1), in this case the characteristic i,j=0 equation: λi BB H = λ 5F nf n 1 5F nf n 2 5F nf 0 5F nf n 2 λ 5F nf n 3 5F nf F nf 0 5F nf 1 λ 5F nf 1 n = 0. If we apply elementary row operations in Case I of Theorem 2.1 to the determinant given above, we have

29 On the Spectral Norms of Toeplitz Matrices with Fibonacci λi BB H = λ λ λ λ λ λ = λ λ λ λ L 2 n λ L 2 n + 4 = λ n 2 ( λ L 2 n + 4 ) 2 = 0. Hence, the singular values of the matrix B are σ 1,2 = L 2 n 4, σ m = 0, where m = 3, 4,..., n. Thus the proof is completed Corollary. Let the matrix B be as in (2), then B 2 = { Ln + 1, if n is odd L 2 n 4, if n is even. Proof. The proof is trivial from Theorem 2.3. References [1] Akbulak, M., and Bozkurt, D. On the norms of Toeplitz matrices involving Fibonacci and Lucas numbers, Hacettepe Journal of Mathematics and Statistic, 37 (2), 89-95, 2008.

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31 Hacettepe Journal of Mathematics and Statistics Volume 42(1) (2013), FIXED POINTS OF MULTIVALUED MAPPING SATISFYING CIRIC TYPE CONTRACTIVE CONDITIONS IN G-METRIC SPACES Mujahid Abbas, Talat Nazır and B. E. Rhoades Received 12 : 10 : 2011 : Accepted 10 : 01 : 2012 Keywords: spaces. Abstract In this paper, study of necessary conditions for existence of fixed point of multivalued mappings satisfying Ciric type contractive conditions in the setting of generalized metric spaces is initiated. Examples to support our results are presented. Since every symmetric generalized metric reduces to an ordinary metric, we give a new example of a nonsymmetric generalized metric to justify the study of fixed point theory in generalized metric spaces. Multivalued mappings, fixed point, non symmetric, generalized metric 2000 AMS Classification: 47H10, 54H25, 54E Introduction and Preliminaries The study of fixed points of mappings satisfying certain contractive conditions has been at the center of rigorous research activity. Mustafa and Sims [10] generalized the concept of a metric space. Based on the notion of generalized metric spaces, Mustafa et al. ([9, 11, 12]) obtained some fixed point theorems for mappings satisfying different contractive conditions. Abbas and Rhoades [1] motivated the study of common fixed point theory in generalized metric spaces. Recently, Saadati et al. [14] proved some fixed Department of Mathematics, COMSATS Institute of Information Technology, 22060, Abbottabad, Pakistan. (M. Abbas) ABBAS@lums.edu.pk (T. Nazır) talt@ciit.net.pk Department of Mathematics, Indiana University, Bloomington, IN (B. E. Rhoades) rhoades@indiana.edu

32 22 M. Abbas, T. Nazır, B. E. Rhoades point results for contractive mappings in partially ordered G metric spaces. Abbas et al. [2] obtain some periodic point results in generalized metric spaces. The aim of this paper is to prove various fixed points results for multivalued mappings taking closed values in generalized metric spaces. It is worth mentioning that our results do not rely on the notion of continuity of the mappings involved therein. Our results extend and unify various comparable results in ([4], [5] and [13]). Consistent with Mustafa and Sims [10], the following definitions and results will be needed in the sequel Definition. Let X be a nonempty set. Suppose that a mapping G : X X X R + satisfies: (a) G(x, y, z) = 0 if x = y = z; (b) 0 < G(x, y, z) for all x, y X, with x y; (c) G(x, x, y) G(x, y, z) for all x, y, z X, with y z; (d) G(x, y, z) = G(p{x, y, z}), where p is a permutation of x, y, z (symmetry); (e) G(x, y, z) G(x, a, a) + G(a, y, z) for all x, y, z, a X. Then G is called a G metric on X and (X, G) is called a G metric space Definition. A sequence {x n} in a G metric space X is: (i) a G Cauchy sequence if, for any ε > 0, there is an n 0 N ( the set natural number ) such that for all n, m, l n 0, G(x n, x m, x l ) < ε, (ii) a G Convergent sequence if, for any ε > 0, there is an x X and an n 0 N, such that for all n, m n 0, G(x, x n, x m) < ε. A G metric space on X is said to be G complete if every G Cauchy sequence in X is G convergent in X. It is known that {x n} G converges to x X if and only if G(x m, x n, x) 0 as n, m [10] Proposition. [10] Let X be a G metric space. Then the following are equivalent: (1) {x n} is G convergent to x. (2) G(x n, x n, x) 0 as n. (3) G(x n, x, x) 0 as n. (4) G(x n, x m, x) 0 as n, m Definition. A G metric on X is said to be symmetric if G(x, y, y) = G(y, x, x) for all x, y X Proposition. Every G metric on X will define a metric d G on X by d G(x, y) = G(x, y, y) + G(y, x, x), x, y X. For a symmetric G metric space d G(x, y) = 2G(x, y, y), x, y X. However, if G is not symmetric, then the following inequality holds: 3 G(x, y, y) dg(x, y) 3G(x, y, y), x, y X. 2 Now we give an example of non-symmetric G metric Example. Let X = {1, 2, 3}, G : X X X R +, be defined as

33 Fixed Points of Multivalued Mapping Satisfying Ciric Type (x, y, z) G(x, y, z) (1, 1, 1), (2, 2, 2), (3, 3, 3) 0 (1, 1, 2), (1, 2, 1), (2, 1, 1), (2, 2, 3), (2, 3, 2), (3, 2, 2), (1, 1, 3), (1, 3, 1), (3, 1, 1), 1 (1, 2, 2), (2, 1, 2), (2, 2, 1), (2, 3, 3), (3, 2, 3), (3, 3, 2) (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1), 2 (1, 3, 3), (3, 1, 3), (3, 3, 1) Note that G satisfies all of the axioms of a generalized metric but G(1, 1, 3) G(1, 3, 3). Therefore G is not a symmetric on X. Let X be a G metric space. We denote by P (X) the family of all nonempty subsets of X, and by P cl (X) the family of all nonempty closed subsets of X. A point x in X is called a fixed point of a multivalued mapping T : X P cl (X) provided x T x. The collection of all fixed point of T is denoted by F ix(t ). 2. Fixed Point Theorems Kannan [4] proved a fixed point theorem for a single valued self mapping T of a metric space X satisfying the property d(t x, T y) h{d(x, T x) + d(y, T y)} for all x, y in X and for a fixed h [0, 1 ). Cirić [3] proved a fixed point theorem for a 2 single valued self mapping T of a metric space X satisfying the property d(t x, T y) ad(x, y) + bd(x, T x) + cd(y, T y) + e[d(x, T y) + d(y, T x)] for all x, y in X and for a fixed a, b, c, e 0 with a + b + c + 2e < 1. Latif and Beg [5] introduced the notion of a K multivalued mapping, which is the extension of Kannan mappings, to multivalued mappings. Continuing in this direction, Rus et al. [13] coined the term R multivalued mapping, which is a generalization of a K multivalued mapping. In this section, we obtain some fixed point theorems for a multivalued mapping satisfying Ciric type contractive conditions on generalized metric spaces without using the continuity condition Theorem. Let X be a complete G metric space and T : X P cl (X). If for each x, y X, u x T (x) there exist u y T (y) such that (2.1) G(u x, u y, u y) h max{g(x, y, y), G(x, u x, u x), G(y, u y, u y), where h [0, 1), then T has a fixed point. 1 [G(x, uy, uy) + G(y, ux, ux)]}, 2 Proof. Let x 0 be an arbitrary point of X, and x 1 T (x 0). Then there exists an x 2 T (x 1) such that G(x 1, x 2, x 2) h max{g(x 0, x 1, x 1), G(x 0, x 1, x 1), G(x 1, x 2, x 2), 1 [G(x0, x2, x2) + G(x1, x1, x1)]} 2 = h max{g(x 0, x 1, x 1), G(x 1, x 2, x 1 2), [G(x0, x2, x2)]}. 2

34 24 M. Abbas, T. Nazır, B. E. Rhoades But, from property (e) of Definition 1.1, G(x 0, x 2, x 2) 2 and we now have 1 [G(x0, x1, x1) + G(x1, x2, x2)] 2 max{g(x 0, x 1, x 1), G(x 1, x 2, x 2)}, (2.2) G(x 1, x 2, x 2) h max{g(x 0, x 1, x 1), G(x 1, x 2, x 2)}. If G(x 1, x 2, x 2) = 0, then, by (1) of proposition 1 in [10], x 1 = x 2. Then x 2 T (x 1) = T (x 2), and x 2 is a fixed point of T. If G(x 1, x 2, x 2) 0, then (2.2) becomes G(x 1, x 2, x 2) hg(x 0, x 1, x 1). Continuing this process, we obtain a sequence {x n} in X, that is for x n T (x n 1), there exists x n+1 T (x n) such that G(x n, x n+1, x n+1) h max{g(x n 1, x n, x n), G(x n 1, x n, x n), G(x n, x n+1, x n+1), (G(x n 1, x n+1, x n+1) + G(x n, x n, x n))/2} = h max{g(x n 1, x n, x n), G(x n, x n+1, x n+1), (G(x n 1, x n+1, x n+1))/2} = h max{g(x n 1, x n, x n), G(x n, x n+1, x n+1)}. Without loss of generality we may assume that x n x n+1 for each n, since, otherwise, it follows that x n+1 is a fixed point of T. Thus we have G(x n, x n+1, x n+1) hg(x n 1, x n, x n)... h n G(x 0, x 1, x 2). For any m > n 1, repeated use of property (e) gives G(x n, x m, x m) G(x n, x n+1, x n+1) G(x m 1, x m, x m) [h n + h n h m 1 ]G(x 0, x 1, x 1) hn G(x0, x1, x1), 1 h and so G(x n, x m, x m) 0 as n, m. Hence {x n} is a G Cauchy sequence. By the G completeness of X, there exist a u X such that {x n} converges to u. Let n N be given. Then, for each x n T (x n 1), there exists a u n T (u) such that Now, if implies that G(u n, u n, u) G(x n, u n, u n) + G(x n, x n, u) h max{g(x n 1, u, u), G(x n 1, x n, x n), G(u, u n, u n), 1 [G(xn 1, un, un) + G(u, xn, xn)]} + G(xn, xn, u) 2 h max{g(x n 1, u, u), G(x n 1, x n, x n), G(u n, u n, u), 1 [G(xn 1, u, u) + G(un, un, u) + G(xn, xn, u)]} + G(xn, xn, u). 2 max{g(x n 1, u, u), G(x n 1, x n, x n), G(u n, u n, u), 1 [G(xn 1, u, u) + G(un, un, u) + G(u, xn, xn)]} 2 = G(x n 1, u, u), G(u n, u n, u) hg(x n 1, u, u) + G(x n, x n, u).

35 Fixed Points of Multivalued Mapping Satisfying Ciric Type Taking limit as n, implies G(u n, u n, u) 0, and u n u. If then max{g(x n 1, u, u), G(x n 1, x n, x n), G(u n, u n, u), 1 [G(xn 1, u, u) + G(un, un, u) + G(xn, xn, u)]} 2 = G(x n 1, x n, x n), G(u n, u n, u) hg(x n 1, x n, x n) + G(x n, x n, u) hg(x n 1, u, u) + 2G(x n, x n, u). On letting limit n, implies G(u n, u n, u) 0, and u n u. In case then max{g(x n 1, u, u), G(x n 1, x n, x n), G(u n, u n, u), 1 [G(xn 1, u, u) + G(un, un, u) + G(xn, xn, u)]} 2 = G(u n, u n, u), G(u n, u n, u) hg(u n, u n, u) + G(x n, x n, u) which further implies that G(u n, u n, u) 1 G(xn, xn, u). 1 h Taking the limit as n, implies G(u n, u n, u) 0, gives u n u. Finally, if max{g(x n 1, u, u), G(x n 1, x n, x n), G(u n, u n, u), 1 [G(xn 1, u, u) + G(un, un, u) + G(xn, xn, u)]} 2 = 1 [G(xn 1, u, u) + G(un, un, u) + G(xn, xn, u)], 2 then G(u n, u n, u) h [G(xn 1, u, u) + G(un, un, u) + G(xn, xn, u)] + G(xn, xn, u) 2 h 2 G(xn 1, u, u) G(u, un, un) + 3 G(xn, xn, u), 2 which further implies G(u n, u n, u) hg(x n 1, u, u) + 3G(x n, x n, u). Taking the limit as n, implies that G(u n, u n, u) 0. Thus u n u as n. Since u n T (u) and T (u) is closed, it follows that u T (u). The following corollary generalizes Theorem 3.1 of Rus et al. [13] to G metric spaces Corollary. Let X be a complete G metric space and T : X P cl (X). If for each x, y X, u x T (x), there exists a u y T (y) such that (2.3) G(u x, u y, u y) a 1G(x, y, y) + a 2G(x, x, y) + a 3G(x, u x, u x) + a 4G(x, x, u x) + a 5G(y, u y, u y) + a 6G(y, y, u y),

36 26 M. Abbas, T. Nazır, B. E. Rhoades where a i 0 for i = 1, 2,..., 6 and a 1 + a 3 + a 5 + 2(a 2 + a 4 + a 6) < 1, then T has a fixed point. Proof. Note that (2.3) implies that G(u x, u y, u y) h max{g(x, y, y), G(x, u x, u x), G(y, u y, u y), G(x, x, y), 2 where h = a 1 + a 3 + a 5 + 2(a 2 + a 4 + a 6) < 1. Which further implies that G(x, x, ux), 2 G(y, y, uy) }, 2 G(u x, u y, u y) h max{g(x, y, y), G(x, u x, u x), G(y, u y, u y)}, and the result follows from Theorem Example. Let X = [0, ) and G(x, y, z) = max{ x y, y z, z x } be a symmetric G metric on X. Define T : X P cl (X) as T x = [0, x 6 ]. Now for case x = y, u x T x. Take u y = 0, then G(u x, u y, u y) = u x x (0) ( 5x 6 ) (x) 1 12 (x y) (x y) (x ux) (x ux) (y uy) + 1 (y uy) 12 = a 1G(x, y, y) + a 2G(x, x, y) + a 3G(x, u x, u x) +a 4G(x, x, u x) + a 5G(y, u y, u y) + a 6G(y, y, u y). Thus (2.3) is satisfied with a 1 + a 3 + a 5 + 2(a 2 + a 4 + a 6) = Now when x < y, u x T x. Take u y = 0, then G(u x, u y, u y) = u x x (0) ( 5x 6 ) (x) 1 12 (y x) (y x) (x ux) (x ux) (y uy) + 1 (y uy) 12 = a 1G(x, y, y) + a 2G(x, x, y) + a 3G(x, u x, u x) +a 4G(x, x, u x) + a 5G(y, u y, u y) + a 6G(y, y, u y). Thus (2.3) is satisfied with a 1 + a 3 + a 5 + 2(a 2 + a 4 + a 6) =

37 Fixed Points of Multivalued Mapping Satisfying Ciric Type Finally for, y < x, u x T x. Take u y = y 6, then G(u x, u y, u y) = u x u y u x + u y 1 (x + y) (x y) ( 5x 6 ) ( 5y 6 ) 1 12 (x y) (x y) (x ux) (x ux) (y uy) + 1 (y uy) 12 = a 1G(x, y, y) + a 2G(x, x, y) + a 3G(x, u x, u x) +a 4G(x, x, u x) + a 5G(y, u y, u y) + a 6G(y, y, u y). Thus (2.3) is satisfied with a 1 + a 3 + a 5 + 2(a 2 + a 4 + a 6) = Hence all conditions of Corollary 2.2 are satisfied. Moreover, T has a fixed point Corollary. Let X be a complete G metric space and T : X P cl (X). If for each x, y X, u x T (x), there exist u y T (y) such that (2.4) G(u x, u y, u y) αg(x, y, y) + βg(x, u x, u x) + γg(y, u y, u y), where α, β, γ 0 and α + β + γ < 1, then T has a fixed point Example. Let X = {0, 1} and a nonsymmetric G metric from X to R + be define as Define T : X P cl (X) as T (0) = T (1) = {0, 1}. G(0, 0, 0) = G(1, 1, 1) = 0, G(0, 0, 1) = G(0, 1, 0) = G(1, 0, 0) = 0.5, G(0, 1, 1) = G(1, 0, 1) = G(1, 1, 0) = 1. Now if x = 0, y = 0, u x T (0). Then two cases arise. When u x = 0, take u y = 0 T (y), then G(u x, u y, u y) = G(0, 0, 0) = 0 When u x = 1, take u y = 1 T (y), then = 1 8 (0) (0) (0) = αg(0, 0, 0) + βg(0, 0, 0) + γg(0, 0, 0) = αg(x, y, y) + βg(x, u x, u x) + γg(y, u y, u y). G(u x, u y, u y) = G(1, 1, 1) = 0 < 1 8 (0) (1) (1) Thus (2.4) is satisfied with α + β + γ = 7 8. = αg(0, 0, 0) + βg(0, 1, 1) + γg(0, 1, 1) = αg(x, y, y) + βg(x, u x, u x) + γg(y, u y, u y).

38 28 M. Abbas, T. Nazır, B. E. Rhoades For case x = 0, y = 1, u x T (0). Then for u x = 0, take u y = 0 T (y), then G(u x, u y, u y) = 0 < 1 8 (1) (0) (0.5) And when u x = 1, take u y = 0 T (y), then = αg(0, 1, 1) + βg(0, 0, 0) + γg(1, 0, 0) = αg(x, y, y) + βg(x, u x, u x) + γg(y, u y, u y). G(u x, u y, u y) = G(1, 0, 0) = 0.5 < 1 8 (1) (1) (0.5) Thus (2.4) is satisfied with α + β + γ = 7 8. = αg(0, 1, 1) + βg(0, 1, 1) + γg(1, 0, 0) = αg(x, y, y) + βg(x, u x, u x) + γg(y, u y, u y). For case x = 1, y = 0, u x T (1). Then for u x = 0, take u y = 0, we have G(u x, u y, u y) = 0 < 1 8 (0.5) (0.5) (0) = αg(1, 0, 0) + βg(1, 0, 0) + γg(0, 0, 0) and when u x = 1, again by taking u y = 1, we have = αg(x, y, y) + βg(x, u x, u x) + γg(y, u y, u y). G(u x, u y, u y) = G(1, 1, 1) = 0 < 1 8 (0.5) (0) (1) Thus (2.4) is satisfied with α + β + γ = 7 8. = αg(1, 0, 0) + βg(1, 1, 1) + γg(0, 1, 1) = αg(x, y, y) + βg(x, u x, u x) + γg(y, u y, u y). Finally for x = 1, y = 1, u x T (1), then for the case u x = 0, take u y = 0 T (1), we have G(u x, u y, u y) = 0 < 1 8 (0) (0.5) (0.5) And if u x = 1, take u y = 1 T (1), implies = αg(1, 1, 1) + βg(1, 0, 0) + γg(1, 0, 0) = αg(x, y, y) + βg(x, u x, u x) + γg(y, u y, u y). G(u x, u y, u y) = G(1, 1, 1) = 0 = 1 8 (0) (0) (0) = αg(1, 1, 1) + βg(1, 1, 1) + γg(1, 1, 1) = αg(x, y, y) + βg(x, u x, u x) + γg(y, u y, u y). Thus (2.4) is satisfied with α + β + γ = 7. Hence all the conditions of Corollary 2.4 are 8 satisfied and F ix(t ) 0. The following corollary generalizes Theorem 4.1 of Latif and Beg [5] to G metric Spaces.