Volume 62, Number 2, February 1977. FIXED POINT THEOREMS FOR MAPPINGS. WITH A CONTRACTIVE ITERATE AT A POINT1. JANUSZ MATKOWSKI.
PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 62, Number 2, February
1977
FIXED POINT THEOREMS FOR MAPPINGS WITH A CONTRACTIVE ITERATE AT A POINT1 JANUSZ MATKOWSKI Abstract. Let (X,d) be a complete metric space, T: X -» X, and a: [0, oo) -* [0, oo) be nondecreasing with respect to each variable. Suppose that for the function y(t) = a(t,t,t,2t,2t), the sequence of iterates y" tends to 0 in [0, oo) and \imt^,x(t - y(t)) = oo. Furthermore, suppose that for each x G X there exists a positive integer n = n(x) such that for all y € X, d(T"x,Tny)
< a(d(x,Tnx),d(x,Tny),d(x,y),d(T,'x,y),d(Tny,y)).
Under these assumptions our main result states that T has a unique fixed point. This generalizes an earlier result of V. M. Sehgal and some recent results of L. Khazanchi and K. Iseki.
1. For a function y: [0, oo) -* [0, oo) denote by y", n = 0, 1, ..., iterate of y. Before stating the main result we prove the following.
the nth
Lemma . Suppose that y: [0, oo) -» [0, oo) is nondecreasing. Then for every t > 0, limn^xyn(t) = 0 implies y(t) < /.
Proof. Suppose that for some i0 > 0 we have y(t0) > t0. Hence, by the monotonicity of y, y"('o) > t0forn = 1,2,.... This proves the lemma. Remark 1. Note that for every right continuous function y: [0, oo) -* [0, oo)
such that y(t) < t for t > 0, lim,,^ y"(t) = 0. Theorem
1. Let (X,d) be a complete metric space, T: X -» X, a: [0, oo)
-» [0, oo), and let y(t) = a(t, t, t, 2t, 2i)for t > 0. Suppose that Io. a is nondecreasing with respect to each variable, 20.linW>(/-Y(í))= oo, 3°.limn^0Oyn(i) = 0, t>0, 4°. for every x E X, there exists a positive integer n = n(x) such that for all
y ex, d(T"x,Tny)
< a(d(x,Tnx),d(x,T"y),d(x,y),d(Tnx,y),d(T"y,y)).
Then T has a unique fixed point a E X and for each x E X, lim^^.^ Tkx = a.
Proof.
First we shall show that for every x E X, the orbit {T'x}°l0 is
bounded. Received by the editors January 7, 1975 and, in revised form, May 19, 1975.
AMS (MOS) subject classifications(1970).Primary 54H25, 47H50. Key words and phrases. Fixed point, orbit, complete metric, Cauchy sequence. 1 The author is indebted to the referee for his valuable remarks. © American Mathematical Society 1977
344
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FIXED POINT THEOREMS
345
To prove this assertion we fix an x E X, an integer s, 0 < i < n = n(x), and we put
uk = d(x,Tkn+sx),
fe-0,1,
...,
h = max(u0,d(x,T"x)).
By 2° there is a c, c > h, such that
t - y(t) > h,
t> c.
From the choice of c we have w0 < c. Suppose that there exists a positive integer j such that u, > c. Evidently, we may assume that u¡ < c for i < j. Hence, by the triangle inequality, d(Tnx,T(j-x)n+sx) d(TJn+sx,T(j-x)n+sx)
< d(x,T"x)
+ uj_x < 2uj,
< uj + uj_x < 2uj.
Now, using 4° and Io, we get
uj = d(x, TJ"+sx) < d(Tnx, TnT^-x^n+sx) + d(x, T"x) < a(uj,Uj,Uj,2Uj,2uj)
+ h = y(uj) + h,
i.e. U: — y(uj) < h which together with h- > c contradicts to the choice of c. Therefore u, < c for 7" = 0, 1, ..., and, consequently, the orbit {T'x}°^0 is
bounded. Take anr06l
(0
and put n0 = n(x0). Define {xk} as follows
**+l = T"kxk,
nk = n(xk), k = 0, 1, ....
Evidently, {xk} is a subsequence of the orbit {T'x0}°i0. We shall prove that {xk} is a Cauchy sequence. Let k and 1 be positive integers. From (1) we have
xk+i = r*-'+"
•+***.
Denoting j0 = «¿+/_i + • • • + nk, we can write d{xk,xk+i) = d{xk,Tsxk).
For the simplicity of the notations we put r, = dixk_x, T'xk_x). Denote by i, that of the numbers s0, nk_x, s0 + nk_x for which tS]has the greatest value. Thus, by the triangle inequality, we have diT"^xk_x,TsoXk_x)
< tnki + tso < 2ts¡,
diT">->+*oXk_x,T*oXk_x < V|+jo
+ tso < 2tsi.
Now 4° and 1° imply License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
346
JANUSZ MATKOWSKI
d(xk,T**xk) = d(T"^xk_x,T"^Tsoxk_x)
xk_x)).
Repeating this procedure, we can find positive integers Sj,j=
1, ...,
k — 1,
such that d(xk_j,TsJxk_j)
< y(d(xk_j_x,Ts^xk_j_x)).
Hence, since y is nondecreasing, we obtain d(xk,xk+i)
< y^xo^xo))
< yk(M),
where M denotes the diameter of the orbit {T'x0}^0. By 3°, limk^aoyk(M) = 0. This proves that [xk] is a Cauchy sequence. By the completeness of X there is an a = lim^^ xk, a E X. We shall show that for n = n(a), T"a = a. For an indirect proof suppose that e = d(T"a,a) > 0. Using the argument of the preceding part of the proof, we see that
lim d(T"xk,xk)
= 0.
Therefore, by the Lemma, there exists a k0 such that d(a,xk) < lie - y(e)),
d(T"xk,xk)
< \(e - y(e)),
k > kQ.
Hence we get e = d(Tna,a)
< d(Tna,Tnxk)
+ d(T"xk,xk)
+ d(xk,a)
< a(d(a, T"a),d(a, Tnxk),d(a,xk),d(T"a,xk),d(T"xk,xk))
+ & - y(e)l Since d(a,T"xk) < d(a,xk) + d(xk,T"xk), d(a, xk), it follows that for k > kQ, d(a, T"xk) < \(e - y(£)) < e,
d(T"a,xk)
d(Tna,xk)
< d(T"a,a)
+
< 2e.
Therefore, by Io, e < a(e, e, e, 2e, 2e) + \(e - y(e)) = j(e + y(e)) < e,
which is a contradiction.
Consequently,
T"a = a.
Suppose that there is a point b E X, b ¥= a, such that T"b = b with n = n(a). Then by 4° and the Lemma
d(a,b) = d(T"a,T"b) < a(0,d(a,b),d(a,b),d(a,b),0) < y(d(a,b))
0. Then for every x G X — {p) there exists a positive integer n — n(x) such that for all y E X, d(T"x, Tny) < ad(x, T"x). Using this remark one can easily construct an example of a mapping T satisfying all the conditions of Theorem 3 and such that for every n, T" is discontinuous.
References 1. L. F. Guseman, Fixed point theorems for mappings with a contractive iterate at a point, Proc.
Amer. Math. Soc. 26 (1970),615-618.MR 42 #919. 2. K. Iseki, A generalization of Sehgal-Khazanchi's fixed point theorems, Math. Seminar Notes
Kobe Univ. 2 (1974),1-9. 3. L. Khazanchi, Results on fixed points in complete metric space, Math. Japonicae (to appear). 4. V. M. Sehgal, A fixed point theorem for mappings with a contractive iterate, Proc. Amer. Math.
Soc. 23 (1969),631-634.MR 40 #3531. Lubertowicza
3/1, 43-300 Btelsko-Biala,
Poland
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