A FIXED POINT THEOREM FOR MAPPINGS WITH A CONTRACTIVE ITERATE V. M. SEHGAL 1. Let (X, d) he a metric satisfies the condition (1) space and /: X—>X a mapping. If / d(f(x),f(y))^kd(x,y) for all x, yEX and for some k<l, then/is called a contraction. A well-known theorem of Banach [l] states that for X complete, each contraction mapping/ has a unique fixed point u, and the successive approximations {/"(xo)} converge to u for each Xo£A. In this paper, we investigate mappings which are not necessarily contractions and prove the following result. Theorem. Let (X, d) be a complete metric space, and f: X—*X a continuous mapping satisfying the condition: there exists a k < 1 such that for each x£A, there is a positive integer n(x) such that for all yEX (2) d(fnM(y),f^(x)) ^ kd(y, x). Then f has a unique fixed point u and /"(xo)—»w for each XoEX. Lemma. If f: X—>X be any mapping satisfying the condition of the above theorem then for each x£A, r(x) =sup„ d(f"(x), x) is finite. Proof. Let x£A and let l(x) = ma.x{d(fk(x), x): k = 1, 2, ■ • ■ , n(x)}. If n is a positive integer, <n^(s + l)-n(x), and there exists an integer 5^0 such that s-n(x) d(f"(x), x) ^ <*(/»<*> ■/»-»<*>(*), fnM(x)) + d(fn^(x), x) ^ kd(f"-"M(x), x) + l(x) g /(*) + */(*) + k2l(x) +-h g l(x)/(l - k) Hence r(x) =sup„ Proof xi=/m°(xo) k'l(x) for all n ^ 0. d(fn(x), x) is finite. of the theorem. and inductively Let x0£A be arbitrary. Let m0 = n(x0), w,- = w(x,), xt+i =/•"•'(x,). We show Received by the editors February 12, 1969. 631 License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use that 632 V. M. SEHGAL the sequence {x„} is a convergent [December sequence. By routine calculation we have d(xn+1, Xn) = d(fm^-f^(Xn-x),f^-1(Xn-x)) ^ kd(f^(xn-x), *»-l) ^ • ■■ g knd(fm»(Xo), Xo). Therefore, m>n, it follows by lemma that m—X d(xn+1, xn)^knr(x0). Thus, for Un d(xm, xn) g 2^1 d(xi+i, Xi) g-r(x0) i=n —>0 t as ra —> oo. & The sequence {x„} is therefore Cauchy. Let x„—-*uEX. lif(u) 9^u, then, there exists a pair of disjoint closed neighborhoods U and V such that uEU, f(u) £ V and (3) P = Ini{d(x,y): xEU,yEV] Since/ is continuous, However, xnEU and/(x„)£ d(f(Xn), Xn) = d(fm^-f(xn_x),fmn'1(Xn-x)) >0. V for all n sufficiently ^ kd(f(xn-l), large. Xn-l) g • ■ • < knd(f(xo), x0)->0 as ra—>co, contradicting (3). Thus/(w)=M. The uniqueness of the fixed point follows To show that/'l(xo)—>u, set p* = max{d(fm(xo),u):m If ra is a sufficiently immediately from (2). = 0, 1, 2 • • • , (n(u) — 1)}. large integer, then n=r-n(u)+q, 0^q<n(u), r>0, and d(f(x0), u) = d(f'-"<-u)+"(xo),fnM(u)) g kd(f^-^-"^+"(xo), u) g • • • g kTd(fq(xo), u) g £rp*. Since ra—>=oimplies r—»°°, we have d(f"(xo), establishes the theorem. u)—>0 as ra—»°°. This 2. In this section we give an example of a continuous function / which satisfies condition (2), but is not a contraction. The author is thankful to the referee for showing that no iterate of/is a contraction. Bryant [2 ] considers a function / (not necessarily continuous) on a complete metric space X into itself for which there exists a^,0^^<l, and a positive integer w, such that d(fn(x), f"(y)) ^kd(x, y) for all x, yEX. That Bryant's condition is stronger than condition (2) is shown below. The author is grateful to Professor David R. Anderson for his help in the following example. License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use 1969] fixed point theorem for mappings Example. Let X be the closed unit interval metric. Write 633 [0, 1 ] with the usual x = u |~—,—~|u{o}, and let/: X—>A be defined as follows. For each » = 1, 2 • • • , let /:r±,_L"urj_,ri |_2» 2»-!j L2n+1 2»J be defined by »+ 2 / n + 3\ /(x) =-(x-)+ =- 1\1 2"-1/ —' ifx£-' 2" 1 r 3re + 5 L2"+1(w+2) in 2"~l J - r 1 3» + 5 1 if x E —>- 2B+1 |_2" 2"+1(« + 2)J and let/(0)=0. It is obvious that/ is a nondecreasing, continuous function on [0, l] with 0 as the only fixed point, and that / is not a contraction. If xE [l/2n, l/2n_1] and yEX, then by a routine examination of cases y£[l/2m, l/2m_1] for m^n and m^n, it is easy to verify that/ satisfies |/(x) -f(y) I ^ n—— \x-y\ for all y £ X. n+ 4 Therefore, if we choose £ = 1/2 in (2), then for each x£ [1/2", l/2n_1], n(x) may be taken asn + 3, whereas n(0) may be taken as any integer greater than one. To show that the condition in [2] is stronger than (2), let 0^&<1, and N (a natural number) be given; it will be shown that there exists x and y such that \fN(x)—fN(y)\>k\x n> (Nk/1— k))— 2. Since/* i=l, is uniformly —y\. continuous Choose and on fix [0, l] for 2, • • ■ , N, there is some 5>0 such that . . \x - y\1 < 8 => \Uf(x) - f(y) JWn . <- (n n+ N + 3 + N + 2)2»+"+! for i = l, 2, • ■ ■, N. Setting x = l/2n_1 and y any member of [l/2n, 1/2"-1] such that 0< |x-y| <5, it can be shown that/*(x) and f(y) are both members of License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use 634 V. M. SEHGAL r 3(ra + i) + 5 \_2n+i+1(n+ i+2)' 1 -I 2"+'-1J for i = 1,2, • • • , N. Thus i \f(x)-f(y)\ i ra + 2 , =—— |x-y|, w+ 3 \f(x) -f2(y) | = ^—4 | x-y| , • • • , |/*(x) -f»(y) | w+ 4 =- ra + 2 . . . x — y\ > k\ x — y\ . n + 2+ N ' ' ' ' The author acknowledges the help of Professor Reid in the preparation of this paper. A. T. Bharucha- References 1. S. Banach, Sur les operations dan les ensembles abstraits et leurs applications aux integrates, Fund. Math. 3 (1922), 133-181. 2. V. W. Bryant, A remark on a fixed point theorem for iterated mappings, Math. Monthly 75 (1968), 399^00. University of Wyoming License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use Amer.
© Copyright 2025 Paperzz