Two general fixed point theorems for asymptotically regular self-mappings on a metric space X which satisfy the contractive condition (1) below are proved. Our results extend and generalize results of Sharma and Yuel [4] and Guay and Singh [3]

Lj. Ćirić

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Mathematical Communications 10(2005), 111-114 111Fixed points of asymptotically regular mappingsLjubomir C´iric´∗Abstract. Two general fixed point theorems for asymptoticallyregular self-mappings on a metric space X which satisfy the contractivecondition (1) below are proved. Our results extend and generalize resultsof Sharma and Yuel [4] and Guay and Singh [3].Key words: fixed point, asymptotically regular mappingAMS subject classifications: 54H25Received April 25, 2005 Accepted September 13, 20051. IntroductionBanach fixed point theorem and its applications are well known. Many authors haveextended this theorem, introducing more general contractive conditions, which im-ply the existence of a fixed point. Almost all of conditions imply the asymptoticregularity of the mappings under consideration. So the investigation of the asymp-totically regular maps plays an important role in fixed point theory.Sharma and Yuel [4] and Guay and Singh [3] were among the first who used theconcept of asymptotic regularity to prove fixed point theorems for a wider class ofmappings than a class of mappings introduced and studied by C´iric´ [2].The purpose of this short paper is to study a wide class of asymptotically reg-ular mappings which possess fixed points in complete metric spaces. Our resultsgeneralize and unify the results of Sharma and [4] and Guay and Singh [3].2. Main ResultsBrowder and Petryshyn [1] defined the following notion.Definition 1. A selfmapping T on a metric space (X, d) is said to be asymp-totically regular at a point x in X, ifd(T nx, T nTx)→ 0 as n→∞, (1)where T nx denotes the n-th iterate of T at x.Let R+ be the set of nonnegative reals and let Fi : R+ → R+ be functionssuch that Fi(0) = 0 and Fi is continuous at 0 (i = 1, 2).∗University of Belgrade, Serbia and Montenegro, e-mail: lciric@afrodita.rcub.bg.ac.yu112 Lj. C´iric´Our main result is the following theorem.Theorem 1. Let (X, d) be a complete metric space and T a selfmapping on Xsatisfying the following condition:d(Tx, T y) ≤ a1F1[min{d(x, Tx), d(y, T y)}] + a2F2[d(x, Tx) · d(y, T y)]+a3d(x, y) + a4[d(x, Tx) + d(y, T y)] + a5[d(x, T y) + d(y, Tx)] (2)for all x, y in X, where ai = ai(x, y) (i = 1, 2, 3, 4, 5) are nonnegative functionssuch that for arbitrarily fixed K > 0 and 0 < λ1 < 1, 0 < λ2 < 1:a1(x, y), a2(x, y) ≤ K, (3)a4(x, y) + a5(x, y) ≤ λ1, (4)a3(x, y) + 2a5(x, y) ≤ λ2. (5)If T is asymptotically regular at some x0 in X, then T has a unique fixed point inX and at this point T is continuous.Proof. We show that {xn} is a Cauchy sequence, where xn = T nx0. Denotedn = d(xn, xn+1). (6)Using the triangle inequality, from (2) we haved(xn, xm) ≤ dn + d(Txn, Txm) + dm ≤ dn + dm+a1F1[min{dn, dm}] + a2F2(dn · dm)+a3d(xn, xm) + a4(dn + dm) + a5[d(xn, Txm) + d(xm, Txn)],where ai = ai(xn, xm). Using again the triangle inequality, we getd(xn, xm) ≤ (a3 + 2a5)d(xn, xm) + (1 + a4 + a5)(dn + dm)+a1F1[min{dn, dm}] + a2F2(dn · dm).Hence, because of (3), (4) and (5), we obtain(1 − λ2)d(xn, xm) ≤ (1 + λ1)(dn + dm) +KF1[min{dn, dm}] +KF2(dn · dm). (7)Since T is asymptotically regular and F1 and F2 are continuous at zero, taking thelimit as m tends to infinity we obtain(1 − λ2) limn>m→∞ d(xn, xm) ≤ 0, (8)which implies that {xn} is a Cauchy sequence.Since X is complete, there is some u in X such thatlimxn = u. (9)Now we show that u is a unique fixed point of T . Suppose that d(u, Tu) > 0. From(2) we haved(u, Tu) ≤ d(u, Txn) + d(Txn, Tu) ≤ d(u, xn+1)+a1F1[min{dn, d(u, Tu)}] + a2F2[dn · d(u, Tu)]+a3d(xn, u) + a4[dn + d(u, Tu)] + a5[d(xn, Tu) + d(u, xn+1)],Fixed points of asymptotically regular mappings 113where ai = ai(xn, u). Using the triangle inequality we getd(u, Tu) ≤ (1 + a5)d(u, xn+1) + a1F1[min{dn, d(u, Tu)}]+a2F2[dn · d(u, Tu)] + (a3 + a5)d(xn, u) + a4dn + (a4 + a5)d(u, Tu).Therefore, from (3), (4) and (5),d(u, Tu) ≤ λ1d(u, Tu) + (1 + λ2)d(u, xn+1) + λ2d(u, xn)+K · F1[min{dn, d(u, Tu)}] +KF2[dn · d(u, Tu)].Taking the limit we get d(u, Tu) ≤ λ1d(u, Tu) < d(u, Tu), a contradiction. There-fore, d(u, Tu) = 0; hence Tu = u.To prove the uniqueness of u, let us suppose that u and v are two fixed pointsof T . From (2), with ai = ai(u, v),d(u, v) = d(Tu, T v) ≤ a1F1(0) + a2F2(0) + a3d(u, v) + a4 · 0 + 2a5d(u, v)= (a3 + 2a5)d(u, v).Hence, because of (5),(1 − λ2)d(u, v) ≤ 0, (10)which implies v = u.To prove that T is continuous at u, suppose that xn → u = Tu. Then from (2),d(Txn, u) = d(Txn, Tu) ≤ a1 · F1(0) + a2F2(0)+a3d(xn, u) + a4d(xn, Txn)+a5[d(xn, u) + d(Txn, u)]≤ (a3 + a4 + a5)d(xn, u) + (a4 + a5)d(u, Txn),where ai = ai(xn, u). Hence, using (4) and (5),(1− λ1)d(Txn, u) ≤ (λ1 + λ2)d(xn, u). (11)Letting n go to infinity, we obtain(1− λ1) limn→∞ d(Txn, u) ≤ 0, (12)which implies that limn→∞ Txn = u. ✷Remark 1. The contractive condition considered by Sharma and Yuel [4] isdefined as follows:d(Tx, T y) ≤ αd(y, T y)[1 + d(x, Tx)]1 + d(x, y)+ βd(x, y), (13)where α, β are nonnegative reals, satisfyingα < 1, β < 1. (14)114 Lj. C´iric´By symmetry of d, it is clear that (13) impliesd(Tx, T y) ≤ αmin{d(x, Tx), d(y, T y)}+ d(x, Tx) · d(y, T y)1 + d(x, y)+ βd(x, y). (15)Our condition (2) becomes (15) ifa1(x, y) = a2(x, y) =11 + d(x, y), F1(t) = F1(t) = α · t,a3(x, y) = β, a4(x, y) = a5(x, y) = 0,and clearly (3) and (5) becomes a1 = a2 ≤ 1 and β < 1, respectively.Remark 2. The contractive condition, introduced and considered by Guay andSingh [3], is defined as follows:d(Tx, T y) ≤ pd(x, y) + q[d(x, Tx) + d(y, T y)] + r[d(x, Tx) + d(y, T y)], (16)where p, q and r are fixed real numbers such that q + r < 1, p+ 2r < 1. It is clearthat our condition (2) becomes (16), if a1 = 0, a2 = 0 and a3 = p, a4 = q anda5 = r.Remark 3. The example of Sharma and Yuel [4] shows that the assumption ofasymptotically regularity in the above theorems cannot be dropped.The following theorem generalizes Theorem 2 of Sharma and Yuel [4].Theorem 2. Let (X, d) be a metric space, not necessarily complete, and let Tbe as in Theorem 1. If the sequence of iterates {T nx0} at some x0 has a subsequenceconverging to a point u in X, then u is the unique fixed point of T , and {T nx0}also converges to u and T is continuous at u.Proof. As shown in the proof of Theorem 1, {T nx0} is a Cauchy sequence.Since it contains a subsequence converging to u, limT nx0 = u. The rest of theresult follows by the same method of proof as in Theorem 1. ✷References[1] F.E.Browder, W.V.Petrysyn, The solution by iteration of nonlinearfunctional equation in Banach spaces, Bull. Amer. Math. Soc. 72(1966), 571-576.[2] Lj.B. C´iric´, Generalized contractions and fixed point theorems, Publ. Inst.Math. (Beograd) 12(26)(1971), 19-26.[3] M.D.Guay, K. L. Singh, Fixed points of asymptotically regular mappings,Mat. Vesnik 35(1983), 101-106.[4] P.L. Sharma A.K.Yuel, Fixed point theorems under asymptotic regularityat a point, Math. Sem. Notes 35(1982), 181-190.

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