FIXED POINT THEOREMS IN GENERALIZED B-METRIC SPACES HASSEN AYDI 1 , STEFAN CZERWIK
1
2
University of Dammam, Department of Mathematics. College of Education of Jubail, P.O: 12020, Industrial Jubail 31961. Saudi Arabia. Department of Medical Research, China Medical University Hospital, China Medical University, Taichung, Taiwan. 2 Institute of Mathematics, Silesian University of Technology, Kaszubska 23, 44-100 Gliwice, Poland Abstract. In the paper we present fixed point theorems in generalized b-metric spaces both for linear and nonlinear contractions, generalizing several existing results.
1. Introduction and Preliminaries The idea of b-metric space is due is due to the second author of this paper (for details see [2] and [3]). Now we extend this idea in the following way. Let X be a set (nonempty). A function d : X × X → [0, ∞] is said to be a generalized b-metric (briefly gbm) on X, provided that for x, y, z ∈ X the following conditions hold true: (a) d(x, y) = 0 if and only if x = y, (b) d(x, y) = d(y, x), (c) d(x, y) ≤ s[d(x, z) + d(z, y)], where s ≥ 1 is a fixed constant. Then (X, d) is called a generalized b-metric space with generalized b-metric d. As usual, by N, N0 , R+ we denote the set of all natural numbers, the set of all nonnegative integers or the set of all nonnegative real numbers, respectively. If f : X → X, by f n we denote the n-th iterate of f : f 0 (x) = x,
x ∈ X;
f n+1 = f ◦ f n .
Here the symbol ϕ ◦ f denotes the function ϕ[f (x)] for x ∈ X. 2. Linear quasi-contractions We start with the following theorem Theorem 2.1. Let (X, d) be a complete generalized b-metric space. Assume that T : X → X is continuous and satisfies the condition d(T (x), T 2 (x)) ≤ αd(x, T (x)),
(1)
for x ∈ X, such that d(x, T (x)) < ∞, and αs = q < 1. Let x ∈ X be a arbitrarily fixed. Then the following alternative holds: either (A) for every nonnegative integer n ∈ N0 , d(T n (x), T n+1 (x)) = ∞, or Key words and phrases. fixed point, Geraghty type contraction, metric-like space.
[email protected] 1 ,
[email protected] 2 . 1
(2)
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H. AYDI, S. CZERWIK
(B) there exists an k ∈ N0 such that d(T k (x), T k+1 (x)) < ∞. In (B) (i) the sequence {T m (x)} is a Cauchy sequence in X; (ii) there exists a point u ∈ X such that lim d(T m (x), u) = 0 and
m→∞
T (u) = u.
Proof. From (1) we get (in case (B)) d(T k+1 (x), T k+2 (x)) ≤ αd(T k (x), T k+1 (x)) < ∞ and by induction d(T k+n (x), T k+n+1 (x)) ≤ αn d(T k (x), T k+1 (x)),
n = 0, 1, 2, . . . .
(3)
Consequently, for n, v ∈ N0 , by (3) we obtain d(T k+n (x), T k+n+v (x)) ≤ sd(T k+n (x), T k+n+1 (x)) + . . . + sv−1 d(T k+n+v−2 (x), T k+n+v−1 (x)) + sv d(T k+n+v−1 (x), T k+n+v (x)) ≤ sαn d(T k (x), T k+1 (x)) + . . . + sv−1 αn+v−2 d(T k (x), T k+1 (x)) + αn+v−1 sv d(T k (x), T k+1 (x)) ≤ sαn [1 + (sα) + . . . + (sα)v−1 ]d(T k (x), T k+1 (x)) ∞ X n ≤ sα (sα)m d(T k (x), T k+1 (x)) m=0
sαn d(T k (x), T k+1 (x)). ≤ 1 − sα Finally, d(T k+n (x), T k+n+v (x)) ≤
sαn d(T k (x), T k+1 (x)) 1 − sα
(4)
for n, v ∈ N0 . By (4) it follows that {T n (x)} is a Cauchy sequence of elements of X. Since X is complete, so there exists u ∈ X with lim d(T n (x), u) = 0.
n→∞
Because T is continuous with respect to d (see the assumptions), therefore u = lim T n+1 (x) = T ( lim T n (x)) = T (u), n→∞
n→∞
and u is a fixed point of T , which ends the proof.
Remark 2.1. In the space X, T may have more than one fixed point (for consider T (x) = x). Remark 2.2. If additionally d is a continuous function (as a function of one variable), then we have the estimation (see (4)): d(T k+n (x), u) ≤
sαn d(T k (x), T k+1 (x)). 1 − sα
(5)
Remark 2.3. A function d may not be continuous even as a function of one variable (e.g. see [10]).
FIXED POINT THEOREMS IN GENERALIZED B-METRIC SPACES
Remark 2.4. An operator T satisfying (1) may not be continuous (see [2]). In fact, if T satisfies α d(T (x), T (y)) ≤ [d(x, T (x)) + d(y, T (y))], x, y ∈ X, 2 then T satisfies also (1).
3
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Remark 2.5. Theorem 2.1 generalizes theorems Diaz, Margolis ([6]), Luxemburg ([8], [9]), Banach ([1]), Czerwik, Kr´ol ([4]) and other (see also [11], [12], [13], [14], [3], [7]). 3. Nonlinear contractions In this section we present the following result. Theorem 3.1. Assume that (X, d) is a complete generalized b-metric space and T : X → X satisfies the condition d(T (x), T (y)) ≤ ϕ[d(x, y)] (7) for x, y ∈ X, d(x, y) < ∞, where ϕ : [0, ∞) → [0, ∞) is nondecreasing and lim ϕn (z) = 0
n→∞
f or z > 0.
(8)
Let x ∈ X be arbitrarily fixed. Then the following alternative holds: either (C) for every nonnegative integer n ∈ N0 d(T n (x), T n+1 (x)) = ∞, or (D)
there exists an k ∈ N0 such that d(T k (x), T k+1 (x)) < ∞.
In (D) (iii) (iv)
the sequence {T m (x)} is a Cauchy sequence in X; there exists a point u ∈ X such that lim d(T n (x), u) = 0 and
n→∞
(v) (vi)
T (u) = u;
u is the unique fixed point of T in B := {t ∈ X : d(T k (x), t) < ∞}; for every t ∈ B, lim d(T n (t), u) = 0. n→∞
If, moreover, d is continuous (with respect to one variable) and ∞ X
sk ϕk (t) < ∞
f or
t > 0,
k=1
then for t ∈ B m
d(T (t), u) ≤
∞ X
sk+1 ϕm+k [d(t, T (t))],
m ∈ N0 .
(9)
k=0
Proof. The proof will consist with a few steps. 0 1 Let’s take x ∈ X and ε > 0. Take n ∈ N such that ε ϕn (ε) < , 2s n n m and put F = T , α = ϕ and xm = F (x) for m ∈ N. Then for all x, y ∈ X such that d(x, y) < ∞, one gets d(F (x), F (y)) ≤ ϕn [d(x, y)] = α[d(x, y)]. (10)
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H. AYDI, S. CZERWIK
20 One can prove that (B, d) is a complete b-metric space. Clearly, T k (x), T k+1 (x) ∈ B. 30 Now we observe that T : B → B. For it if t ∈ B i.e. d(T k (x), t) < ∞, then d(T k (x), T (t)) ≤ s[d(T k (x), T k+1 (x)) + d(T k+1 (x), T (t))] ≤ s ε1 + ϕ[d(T k (x), t)] ≤ s[ε1 + ε2 ] < ∞, where Also F : B → B. ε1 , ε2 some positive numbers. m 0 4 For t ∈ B we have {F (t)} ⊂ B, for all m ∈ N0 . We verify that {F m (t)} is a Cauchy sequence. In fact, putting ym = F m (t), m ∈ N0 , we get d(F (t), F 2 (t)) ≤ α[d(t, F (t))] and by induction d(F m (t), F m+1 (t)) ≤ αm [d(t, F (t))] i.e. d(ym , ym+1 ) ≤ αm [d(t, F (t))], whence d(ym , ym+1 ) → 0 as m → ∞. Let m be such that
ε . 2s Then for every z ∈ K(ym , ε) := {y ∈ X : d(ym , y) ≤ ε} we obtain d(ym , ym+1 )
0. Then T has exactly one fixed point u ∈ X, and lim d(T n (x), u) = 0
n→∞
for each x ∈ X. If, moreover, d is continuous (with respect to one variable) and the series of iterates ∞ X
sk ϕk (t) < ∞
f or
t > 0,
k=1
then for z ∈ X and m ∈ N0 m
d(T (z), u) ≤
∞ X
sk+1 ϕm+k [d(z, T (z))].
(11)
k=0
Remark 3.1. Corollary is contained in [2] for s = 2 (see also [3]). Remark 3.2. It would be interesting to consider a convergence of series of iterates: ∞ X ϕn (t). n=1
For example, one of the sufficient conditions is the following ϕn+1 (t) = q(t) < 1, lim sup n ϕ (t) n→∞ Let’s also note another one such conditions: ln ϕn (t) lim inf − = α(t) > 1, n→∞ ln n For more details see [5].
t > 0.
t > 0.
(12)
(13)
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H. AYDI, S. CZERWIK
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