Fixed point theorems in symmetric spaces and applications to probabilistic spaces

Nonlinear Analysis 72 (2010) 527–532

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Fixed point theorems in symmetric spaces and applications to probabilistic spaces Yan Shen ∗ , Min Lu School of Mathematics and Statistics, Nanjing Audit University, Nanjing, 210029, PR China

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Article history: Received 13 March 2009 Accepted 24 June 2009 Keywords: Symmetric space Semi-metric space PPM-space Common fixed point

abstract Several fixed point theorems for symmetric spaces are proved. Then these theorems are used in symmetric PPM-space to prove and generalize Theorem 6 of [T.L. Hicks, B.E. Rhoades, Fixed point theory in symmetric spaces with applications to probabilistic spaces, Nonlinear Anal. 36 (1999) 331–344]. © 2009 Elsevier Ltd. All rights reserved.

1. Introduction In [1], Hicks and Rhoades proved a number of fixed point theorems in a symmetric space, and applied these theorems to the very general structure of probability, and got the corresponding fixed point theorems in a symmetric PPM-space. Unfortunately, in the proof of Theorem 4 and Theorem 5 of [1], the triangle inequality for d was abusively used, for d is only a symmetric (semi-metric) space and in general symmetric (semi-metric) space the triangle inequality does not necessarily hold. To repair this mistake, Dorel Mihet replaced the condition (W .3) with the condition (W ) so that the conclusion of the theorem can be proved [2]. Here

(W .3) : (W ) :

lim d(xn , x) = 0 and lim d(xn , y) = 0 imply that x = y.

n→∞

n→∞

lim d(xn , yn ) = 0 and lim d(yn , zn ) = 0 imply that lim d(xn , zn ) = 0.

n→∞

n→∞

n→∞

However, the condition (W ) is stronger than (W .3) and in a symmetric PPM-space, the compatible metric d does not necessarily satisfy (W ). Here d is defined by d(x, y) =

0, sup{ε : y 6∈ Nx (ε, ε),  > 0},



if y ∈ Nx (ε, ε) for all  > 0 otherwise.

Therefore, Theorem 3 of [2] cannot be applied to a symmetric PPM-space. Thus, it seems worthwhile to give a proper proof of Theorem 6 of [1]. In addition, as Ar (X ) ⊂ T u (X ) and Bs (X ) ⊂ S t (X ) should not be deduced from A(X ) ⊂ T (X ) and B(X ) ⊂ S (X ), Corollary 5 and Corollary 8 of [1] are not obviously tenable. The purpose of this paper is to prove Theorem 6 and Corollary 8 of [1] and generalize Theorem 3 of [1].

∗

Corresponding author. E-mail addresses: [email protected] (Y. Shen), [email protected] (M. Lu).

0362-546X/$ – see front matter © 2009 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2009.06.102

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Y. Shen, M. Lu / Nonlinear Analysis 72 (2010) 527–532

2. Fixed point theorems in symmetric spaces Definition 1. A symmetric space on a set X is a real-valued function d on X × X such that (1) d(x, y) ≥ 0 and d(x, y) = 0 if and only if x = y; (2) d(x, y) = d(y, x). Let d be a symmetric space on a set X and for  > 0 and any x ∈ X , let B(x, ε) = {y ∈ X : d(x, y) < ε}. A topology t (d) on X is given by U ∈ t (d) if and only if for each x ∈ U, B(x, ε) ⊂ U for some ε > 0. A symmetric space d is a semi-metric if for each x ∈ X and each ε > 0, B(x, ε) is a neighborhood of x in the topology t (d). A sequence is d-Cauchy if it satisfies the usual metric condition. Definition 2–5. Let (X , d) be a symmetric space. (2) (X , d) is S-complete if for every d-Cauchy sequence {xn }, there exists an x in X with limn→∞ d(xn , x) = 0; (3) (X , d) is d-Cauchy complete if for every d-Cauchy sequence {xn }, there exists an x in X with limn→∞ xn = x with respect to t (d); (4) f : X → X is d-continuous if limn→∞ d(xn , x) = 0 implies limn→∞ d(f (xn ), f (x)) = 0; (5) f : X → X is t (d)-continuous if limn→∞ xn = x with respect to t (d) implies limn→∞ f (xn ) = f (x) with respect to t (d). If d is a semi-metric on X ,then limn→∞ d(xn , x) = 0 is equivalent to limn→∞ xn = x with respect to t (d), f is d-continuous is equivalent to f is t (d)-continuous, (X , d) is S-complete is equivalent to (X , d) is d-Cauchy complete. The next theorem involves a function Q . Let Q : [0, +∞) → [0, ∞) satisfy the following: (a) Q is nondecreasing on [0, +∞); (b) 0 < Q (s) < s for each s > 0; (c) limn→∞ Q n (s) = 0 for each s > 0. Theorem 1. Let d be a bounded symmetric (semi-metric) space for X that satisfies (W .3). Suppose that (X , d) is S-complete (d-Cauchy complete) and f : X → X is d-continuous (t (d)-continuous). Then f has a fixed point if and only if there exists a function Q satisfying (a)–(c) and a d-continuous (t (d)-continuous) function g : X → X which commutes with f and satisfies g (X ) ⊂ f (X ) and there exists a positive integer k such that d(gx, gy) ≤ Q [max{d(f k x, f k y), d(f k x, gx), d(f k y, gy)}]

for all x, y ∈ X .

(1)

If (1) holds, then f and g have a unique common fixed point. Proof. Suppose f (a) = a for some a ∈ X , Put g (x) = a for all x ∈ X . Then the conditions of the theorem are satisfied. To prove the converse, suppose there exists a Q satisfying (a)–(c) so that (1) holds. Let M = sup{d(x, y)|x, y ∈ X }.Let x0 ∈ X . Choose x1 such that g (x0 ) = f (x1 ). In general, choose xn such that g (xn−1 ) = f (xn ). We show that 1. d(f s (xn ), f l (xn+1 )) ≤ Q n (M ) for all positive integers s, l, and n. d(f s (x1 ), f l (x2 )) = d(f s−1 gx0 , f l−1 gx1 ) = d(gf s−1 x0 , gf l−1 x1 )

≤ Q [max{d(f k+s−1 x0 , f k+l−1 x1 ), d(f k+s−1 x0 , gf s−1 x0 ), d(f k+l−1 x1 , gf l−1 x1 )}] ≤ Q (M ). Suppose that d(f s (xn ), f l (xn+1 )) ≤ Q n (M ) for all positive integers s, l. Then d(f s (xn+1 ), f l (xn+2 )) = d(f s−1 gxn , f l−1 gxn+1 ) = d(gf s−1 xn , gf l−1 xn+1 )

≤ Q [max{d(f k+s−1 xn , f k+l−1 xn+1 ), d(f k+s−1 xn , gf s−1 xn ), d(f k+l−1 xn+1 , gf l−1 xn+1 )}] ≤ Q [max{Q n (M ), d(f k+l−1 xn+1 , f l xn+2 )}]. If d(f

k+l−1

d(f

(2)

xn+1 , f xn+2 ) > Q (M ), then

k+l−1

l

n

xn+1 , f l xn+2 ) = d(f k+l−2 gxn , f l−1 gxn+1 ) = d(gf k+l−2 xn , gf l−1 xn+1 )

≤ = ≤ =

Q [max{d(f 2k+l−2 xn , f k+l−1 xn+1 ), d(f 2k+l−2 xn , gf k+l−2 xn ), d(f k+l−1 xn+1 , gf l−1 xn+1 )}] Q [max{d(f 2k+l−2 xn , f k+l−1 xn+1 ), d(f k+l−1 xn+1 , f l xn+2 )}] Q [max{Q n (M ), d(f k+l−1 xn+1 , f l xn+2 )}] Q [d(f k+l−1 xn+1 , f l xn+2 )] < d(f k+l−1 xn+1 , f l xn+2 )

a contradiction. Therefore d(f k+l−1 xn+1 , f l xn+2 ) ≤ Q n (M ).

(3)

Y. Shen, M. Lu / Nonlinear Analysis 72 (2010) 527–532

529

Substituting Eq. (3) into Eq. (2) yields d(f s (xn+1 ), f l (xn+2 )) ≤ Q [Q n (M )] = Q n+1 (M ). 2. d(f s (xn ), f l (xn+m )) ≤ Q n (M ) for all positive integers s, l, m and n. d(f s (x1 ), f l (x1+m )) = d(f s−1 gx0 , f l−1 gxm ) = d(gf s−1 x0 , gf l−1 xm )

≤ Q [max{d(f k+s−1 x0 , f k+l−1 xm ), d(f k+s−1 x0 , gf s−1 x0 ), d(f k+l−1 xm , gf l−1 xm )}] ≤ Q (M ). Suppose that d(f s (xn ), f l (xn+m )) ≤ Q n (M ) for all positive integers s, l, m. Then d(f s (xn+1 ), f l (xn+1+m )) = d(f s−1 gxn , f l−1 gxn+m ) = d(gf s−1 xn , gf l−1 xn+m )

≤ Q [max{d(f k+s−1 xn , f k+l−1 xn+m ), d(f k+s−1 xn , gf s−1 xn ), d(f k+l−1 xn+m , gf l−1 xn+m )}] = Q [max{d(f k+s−1 xn , f k+l−1 xn+m ), d(f k+s−1 xn , f s xn+1 ), d(f k+l−1 xn+m , f l xn+m+1 )}]. From 1, d(f k+l−1 (xn+m ), f l (xn+m+1 )) ≤ Q n+m (M ), thus d(f s (xn+1 ), f l (xn+1+m )) ≤ Q [max{Q n (M ), Q n+m (M )}] = Q [Q n (M )] = Q n+1 (M ). In particular d(fxn , fxn+m ) ≤ Q n (M ). Since limn→∞ Q n (M ) = 0, {fxn } is a d-Cauchy sequence and the S-completeness of (X , d) gives an x in X with limn→∞ d(fxn , x) = 0. Since g (xn ) = f (xn+1 ), limn→∞ d(gxn , x) = limn→∞ d(fxn+1 , x) = 0. The d-continuity of f and g implies limn→∞ d(gfxn , gx) = 0 and limn→∞ d(fgxn , fx) = 0. But fgxn = gfxn since f and g commute. Condition (W .3) then yields fx = gx. Thus, f s x = f s−1 gx = gf s−1 x = gf s−2 gx = · · · = g s x

for each positive integer s.

Using Eq. (1) d(gx, ggx) ≤ Q [max{d(f k x, f k gx), d(f k x, gx), d(f k gx, ggx)}]

= Q [max{d(g k x, g k+1 x), d(g k x, gx), d(g k+1 x, g 2 x)}]. Let max{d(g k x, g k+1 x), d(g k x, gx), d(g k+1 x, g 2 x)} = d(g s1 x, g t1 x). Here s1 ∈ {k, k + 1}, t1 ∈ {k + 1, 1, 2}, s1 and t1 are positive integers. Then d(gx, ggx) ≤ Q [d(g s1 x, g t1 x)]. d(g s1 x, g t1 x) = d(gg s1 −1 x, gg t1 −1 x)

≤ Q [max{d(f k g s1 −1 x, f k g t1 −1 x), d(f k g s1 −1 x, g s1 x), d(f k g t1 −1 x, g t1 x)}] = Q [max{d(g k+s1 −1 x, g k+t1 −1 x), d(g k+s1 −1 x, g s1 x), d(g k+t1 −1 x, g t1 x)}]. Let max{d(g k+s1 −1 x, g k+t1 −1 x), d(g k+s1 −1 x, g s1 x), d(g k+t1 −1 x, g t1 x)} = d(g s2 x, g t2 x). Here s2 ∈ {k + s1 − 1, k + t1 − 1}, t2 ∈ {k + t1 − 1, s1 , t1 }, s2 and t2 are positive integers. Then d(g s1 x, g t1 x) ≤ Q [d(g s2 x, g t2 x)]. Hence d(gx, ggx) ≤ Q [d(g s1 x, g t1 x)] ≤ Q 2 [d(g s2 x, g t2 x)].

(4)

Treating Eq. (4) as a recursion formula one obtains d(gx, ggx) ≤ Q n [d(g sn x, g tn x)] ≤ Q n (M ). Since limn→∞ Q n (s) = 0, d(gx, ggx) = 0, which implies that gx = ggx and fgx = gfx = ggx = gx. Thus gx is a common fixed point of f and g. To prove uniqueness, suppose that y is also a common fixed point of f and g. If y 6= gx, then d(gx, y) = d(ggx, gy) ≤ Q [max{d(f k gx, f k y), d(f k gx, ggx), d(f k y, gy)}]

= Q [max{d(gx, y), d(gx, gx), d(y, y)}] = Q [d(gx, y)] < d(gx, y) a contradiction. Therefore y = gx, and the common fixed point is unique. Corollary 1. Let d be a bounded symmetric (semi-metric) space for X that satisfies (W .3). Suppose that (X , d) is S-complete (d-Cauchy complete) and f : X → X is d-continuous (t (d)-continuous). Then f has a fixed point if and only if there exists a function Q satisfying (a) − (c ) and a d-continuous (t (d)-continuous) function g : X → X which commutes with f and satisfies g (X ) ⊂ f (X ) and there exist positive integers k, l, such that d(g l x, g l y) ≤ Q [max{d(f k x, f k y), d(f k x, g l x), d(f k y, g l y)}] for all x, y ∈ X . If (5) holds, then f and g have a unique common fixed point.

(5)

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Proof. Define h = g l . Then, from Theorem 1, h, f have a unique common fixed point x0 , i.e., f (x0 ) = x0 , g l (x0 ) = x0 . If d(gx0 , x0 ) > 0, then d(gx0 , x0 ) = d(gg l x0 , g l x0 ) = d(g l gx0 , g l x0 )

≤ Q [max{d(f k gx0 , f k x0 ), d(f k gx0 , g l gx0 ), d(f k x0 , g l x0 )}] = Q [max{d(gx0 , x0 ), d(gx0 , gx0 ), d(x0 , x0 )}] = Q [d(gx0 , x0 )] < d(gx0 , x0 ) a contradiction. Therefore d(gx0 , x0 ) = 0, i.e., gx0 = x0 . So that x0 is a common fixed point of f and g. Since a common fixed point of f and g is also a common fixed point of f and g l , the common fixed point of f and g is unique. The following corollary slightly generalizes Theorem 1 of [1]. Corollary 2. Let d be a bounded symmetric (semi-metric) space for X that satisfies (W .3). Suppose that (X , d) is S-complete (d-Cauchy complete) and f : X → X is d-continuous (t (d)-continuous). Then f has a fixed point if and only if there exists an α ∈ (0, 1) and a d-continuous (t (d)-continuous) function g : X → X which commutes with f and satisfies g (X ) ⊂ f (X ) and there exist positive integers k, l, such that d(g l x, g l y) ≤ α d(f k x, f k y) for all x, y ∈ X .

(6)

If (6) holds, then f and g have a unique common fixed point. Proof. Suppose f (a) = a for some a ∈ X , Put g (x) = a for all x ∈ X . Then the conditions of the corollary are satisfied. To prove the converse, suppose there exists an α ∈ (0, 1) so that (6) holds. Let Q (s) = α s for each s ∈ [0, +∞). Then Q satisfies (a)–(c) and d(g l x, g l y) ≤ α d(f k x, f k y) = Q [d(f k x, f k y)] ≤ Q [max{d(f k x, f k y), d(f k x, g l x), d(f k y, g l y)}]. Hence f and g have a unique common fixed point. In the proof of Theorem 4 of [1], the reason why the triangle inequality for d needs to be used is the absence of any kind of ‘‘contraction’’ for S and T in the hypotheses of the theorem. In the following corollary we add certain conditions of ‘‘contraction’’ to S and T in Theorem 4 of [1], so that the theorem can be proved. Corollary 3. Let d be a bounded symmetric (semi-metric) space for X that satisfies (W .3). Suppose that (X , d) is S-complete (d-Cauchy complete) and S , T : X → X are d-continuous (t (d)-continuous). Then S and T have a common fixed point if and only if there exists a function Q satisfying (a)–(c) and d-continuous (t (d)-continuous) functions A, B : X → X such that A commutes with S , B commutes with T and satisfy A(X ) ⊂ S (X ), B(X ) ⊂ T (X ), and d(Ax, By) ≤ Q [max{d(Sx, Ty), d(Sx, Ax), d(Ty, By)}], d(Ax, Ay) ≤ Q [max{d(Sx, Sy), d(Sx, Ax), d(Sy, Ay)}], d(Bx, By) ≤ Q [max{d(Tx, Ty), d(Tx, Bx), d(Ty, By)}] for all x, y ∈ X .

(7)

If (7) holds, then A, B, S and T have a unique common fixed point. Proof. Let a ∈ X be a common fixed point of S and T . Put Ax = Bx = a for all x ∈ X . Then A, B : X → X satisfy the conditions of the corollary. From Theorem 1, S and A have a unique common fixed point a ∈ X , T and B have a unique common fixed point b ∈ X , i.e., Aa = a, Sa = a, Bb = b, Tb = b. If a 6= b, then d(a, b) = d(Aa, Bb) ≤ Q [max{d(Sa, Tb), d(Sa, Aa), d(Tb, Bb)}]

= Q [max{d(a, b), d(a, a), d(b, b)}] = Q (d(a, b)) < d(a, b) a contradiction. Hence a = b. Therefore, a is a unique common fixed point of A, B, S and T . 3. Fixed point theorems in a PPM-space We now look at applications to probabilistic spaces. The following definitions appeared in [1] and [3]. A real-valued function f defined on the set of real numbers is a distribution function if it is nondecreasing, left continuous, with inf f = 0 and sup f = 1. H denotes the distribution function defined by H (t ) = 0 if t ≤ 0, and H (t ) = 1 for t > 0. Definition 6. Let X be a nonempty set and F a function on X × X such that F (x, y) = Fx,y is a distribution function. Consider the following conditions:

Y. Shen, M. Lu / Nonlinear Analysis 72 (2010) 527–532

I. II. III. IV.

531

Fx,y (0) = 0 for all x, y in X ; Fx,y = H if and only if x = y; Fx,y = Fy,x ; If Fx,y (ε) = 1 and Fy,z (δ) = 1, then Fx,z (ε + δ) = 1.

If F satisfies I and II, then it is called a PPM-structure on X and the pair (X , F ) is called a PPM-space. An F satisfying III is said to be symmetric. A symmetric PPM-structure F satisfying IV is a probabilistic metric structure (PM-structure) and the pair (X , F ) is a probabilistic metric space (PM-space). Let (X , F ) be a PPM-space. For ε, λ > 0, and x in X , let Nx (ε, λ) = {y ∈ X : Fx,y (ε) > 1 − λ}. A T1 topology t (F ) on X is obtained as follows: U ∈ t (F ) if for each x in U, there exists an ε > 0 such that Nx (ε, ε) ⊂ U. Nx (ε, ε) may not be a t (F ) neighborhood of x. If it is, then t (F ) is said to be topological. Let (X , F ) be a symmetric PPM-space. A sequence {xn } is a fundamental sequence if limn,m→∞ Fxn ,xm (t ) = 1 for all t > 0. The space is complete if for every fundamental sequence {xn } there exists an x in X such that limn→∞ Fxn ,x (t ) = 1 for all t > 0. Let (X , F ) be a symmetric PPM-space. Set d(x, y) =

0, sup{ε : y 6∈ Nx (ε, ε),  > 0},



if y ∈ Nx (ε, ε) for all ε > 0 otherwise.

(8)

d is a bounded symmetric space for X . The following lemma was proved in [1]. Lemma 1. Let (X , F ) be a symmetric PPM-space. Define d as in (8). Then (1) (2) (3) (4) (5)

d(x, y) < t if and only if Fx,y (t ) > 1 − t; d is a compatible symmetric space for t (F ); Fxn ,x (t ) → 1 for all t > 0 if and only if d(xn , x) → 0; (X , F ) is complete if and only if (X , d) is S-complete; t (F ) is topological if and only if d is a semi-metric.

Let (X , F ) be a symmetric PPM-space. f : (X , F ) → (X , F ) is F -continuous if Fxn ,x (t ) → 1 for all t > 0 implies Ffxn ,fx (t ) → 1 for all t > 0. This is equivalent to f : (X , d) → (X , d) is d-continuous. The condition (W .3) is equivalent to the following condition:

(P .3) Fxn ,x (t ) → 1 and Fxn ,y (t ) → 1 for all t > 0 implies x = y. Lemma 2. Let (X , F ) be a symmetric PPM-space. Define d as in (8). f and g are self mappings of X . k and l are positive integers, Q : [0, +∞) → [0, +∞) satisfies (a)–(c), and Q is right continuous. Suppose that for each t > 0, Ff k x,f k y (t ) > 1 − t, implies Fg l x,g l y [Q (t )] > 1 − Q (t ), then d(g l x, g l y) ≤ Q [max{d(f k x, f k y), d(f k x, g l x), d(f k y, g l y)}] Proof. Let max{d(f k x, f k y), d(f k x, g l x), d(f k y, g l y)} = t0 . Then we shall show that d(g l x, g l y) ≤ Q (t0 ). Suppose that d(g l x, g l y) > Q (t0 ) = limt →t + Q (t ). Choose ε > 0 so that Q (t0 + ε) < d(g l x, g l y). d(f k x, f k y) ≤ t0 < t0 + ε gives 0

Ff k x,f k y (t0 + ε) > 1 − (t0 + ε). This leads to Fg l x,g l y [Q (t0 + ε)] > 1 − Q (t0 + ε), so that d(g l x, g l y) < Q (t0 + ε), a contradiction. Therefore d(g l x, g l y) ≤ Q (t0 ). Theorem 2. Let (X , F ) be a complete symmetric PPM-space that satisfies (P .3). Suppose that f : X → X is F -continuous. Then f has a fixed point in X if and only if there exists a function Q , which is right continuous and satisfies (a)–(c), and an F -continuous function g : X → X , which commutes with f and satisfies: g (X ) ⊂ f (X ) and there exist positive integers k, l such that Ff k x,f k y (t ) > 1 − t implies Fg l x,g l y [Q (t )] > 1 − Q (t ) for all x, y ∈ X and t > 0.

(9)

If (9) holds, f and g have a unique common fixed point. Proof. Suppose f (a) = a for some a ∈ X . Put g (x) = a for all x ∈ X . Then the conditions of the theorem are satisfied. To prove the converse, define d as in (8). Then d is a bounded symmetric space on X . The condition that F satisfies (P .3) is equivalent to d satisfies (W .3). Also from Lemma 1, (X , d) is S-complete. f and g are d-continuous. From Lemma 2, (5) holds.We now apply Corollary 1. The following theorem slightly generalizes Theorem 3 of [1].

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Theorem 3. Let (X , F ) be a complete symmetric PPM-space that satisfies (P .3). Suppose that f : X → X is F -continuous. Then f has a fixed point in X if and only if there exists an α ∈ (0, 1) and an F -continuous function g : X → X which commutes with f and satisfies: g (X ) ⊂ f (X ) and there exist positive integers k, l such that Ff k x,f k y (t ) > 1 − t implies Fg l x,g l y (α t ) > 1 − α t

for all x, y ∈ X and t > 0.

(10)

If (10) holds, f and g have a unique common fixed point. Proof. Suppose f (a) = a for some a ∈ X , Put g (x) = a for all x ∈ X . Then the conditions of the theorem are satisfied. To prove the converse, define d as in (8). We show that for every x, y ∈ X , d(g l x, g l y) ≤ α d(f k x, f k y). Suppose that d(g l x, g l y) > α d(f k x, f k y). Choose u, such that d(g l x, g l y) > u > α d(f k x, f k y). From d(f k x, f k y) < ,F ( u ) > 1 − αu . Thus, Fg l x,g l y (u) > 1 − u. Hence d(g l x, g l y) < u, a contradiction. We now apply Corollary 2. α f k x ,f k y α u

References [1] T.L. Hicks, B.E. Rhoades, Fixed point theory in symmetric spaces with applications to probabilistic spaces, Nonlinear Anal. 36 (1999) 331–344. [2] D. Mihet, A note on a paper of Hicks and Rhoades, Nonlinear Anal. 65 (2006) 1411–1413. [3] T.L. Hicks, Fixed point theory in probabilistic metric spaces II, Math. Japonica 44 (3) (1996) 487–493.