Fixed point theory in symmetric spaces with applications to probabilistic spaces

The basic reference for this paper is the book by Schweizer and Sklar [11]. The reader can consult the references for more recent and specialized results. The purpose of this paper is to provide some important fixed point theorems in very general probabilistic structures. To do this we first prove some fixed point theorems for symmetric spaces. These theorems should be of interest to analysts. Then we show how to use these by proving that very general probabilistic structures admit a compatible symmetric or semi-metric such that the distribution functions F_x,y is related to the symmetric d by d(x,y)<t if and only if F_x,y(t)>1−t where t>0.

Definition 1. A symmetric 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; and

• 2.

  d(x,y)=d(y,x).

Let d be a symmetric on a set X and for ε>0 and any x∈X, let $\text{B(x,ε)={y∈X:}\text{d(x,y)<ε}}$. 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 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). There are several concepts of completeness in this setting. A sequence is d-Cauchy if it satisfies the usual metric condition.

Definitions 2–5. Let (X,d) be a symmetric space.

• 1.

  (X,d) is S-complete if for every d-Cauchy sequence {x_n}, there exists x in X with $\text{lim}\text{d(x}{}_{\mathrm{n}}\text{,x)=0}$.

• 2.

  (X,d) is d-Cauchy complete if for every d-Cauchy sequence {x_n}, there exists x in X with $\text{lim}\text{x}{}_{\mathrm{n}}\text{=x}$ with respect to t(d).

• 3.

  $\text{f}\text{:}\text{X→X}$ is d-continuous if $\text{lim}\text{d(x}{}_{\mathrm{n}}\text{,x)=0}$ implies $\text{lim}\text{d(fx}{}_{\mathrm{n}}\text{,fx)=0}$.

• 4.

  $\text{f}\text{:}\text{X→X}$ is t(d)-continuous if $\text{lim}\text{x}{}_{\mathrm{n}}\text{=x}$ with respect to t(d) implies $\text{lim}\text{f(x}{}_{\mathrm{n}}\text{)=f(x)}$ with respect to t(d).

Definition 6. Let (X,t) be a topological space and $\text{d}\text{:}\text{X×X→[0,∞)}$ with d(x,y)=0 if and only if x=y. Then (X,t) is a d-complete topological space if ∑_n=1^∞d(x_n,x_n+1)<∞ implies there exists x in X with $\text{lim}\text{x}{}_{\mathrm{n}}\text{=x}$ with respect to the topology t.

The following two axioms were given by Wilson [17]. Let (X,d) be a symmetric space.

• (W.3) Given $\text{{x}{}_{\mathrm{n}}\text{},}\text{x}$ and y in $\text{X,}\text{d(x}{}_{\mathrm{n}}\text{,x)→0}$ and d(x_n,y)→0 imply that x=y.

• (W.4) Given {x_n},{y_n} and an x in $\text{X,}\text{d(x}{}_{\mathrm{n}}\text{,x)→0}$ and d(x_n,y_n)→0 imply that d(y_n,x)→0.

Note that for a semi-metric d, if t(d) is Hausdorff, then (W.3) holds. In 3, 4, 5 many metric space fixed point theorems were extended to d-complete topological spaces. The idea of completeness in d-complete topological spaces generalizes completeness in metric and quasi-metric spaces. In [9], Jachymski et al. state that, in a semi-metric setting, this concept of completeness is rather strong. They choose to use the above concept of d-Cauchy complete. They give an example of a fixed point free Banach contraction on a Hausdorff d-Cauchy complete semi-metric space. They show that Banach’s fixed point theorem holds if d is also bounded. Their work clearly motivates the following theorem. They also prove that a d-Cauchy complete semi-metric space that satisfies (W.4) is a d-complete topological space. This gives another class of d-complete topological spaces. In [6], it is shown that the F-complete topological spaces of Choudhury [1] are d-complete topological spaces. Thus, many fixed points theorems already given for d-complete topological spaces hold for F-complete topological spaces.

Theorem 1. Let d be a bounded symmetric (semi-metric) for X that satisfies (W.3). Suppose (X,d) is S-complete (d-Cauchy complete) and $\text{f}\text{:}\text{X→X}$ is d-continuous (t(d)-continuous). Then f has a fixed point if and only if there exists α∈(0,1) and a d-continuous (t(d)-continuous) function $\text{g}\text{:}\text{X→X}$ which commutes with f and satisfies$\text{g(X)⊂f(X) and d(gx,gy)≤αd(fx,fy)}\text{for all}\text{x,y∈X.}$Indeed, f and g have a unique common fixed point if (1) holds.

Proof. Suppose f(a)=a for some a∈X, Put g(x)=a for all x∈X. The proof that (1) holds is just as in the metric space case. Just compute or see [10].

Suppose there exist α and g so that (1) holds. Let $\text{M=}\text{sup}\text{{d(x,y)}\text{:}\text{x,y∈X}}$. Let x₀∈X. Choose x₁ such that g(x₀)=f(x₁). In general, choose x_n such that f(x_n)=g(x_n−1). We show that$\text{d(f(x}{}_{\mathrm{n}}\text{),f(x}{}_{\mathrm{n+m}}\text{))≤α}{}^{\mathrm{n}}\text{M.}$Now$\text{d(f(x}{}_{\mathrm{n}}\text{),f(x}{}_{\mathrm{n+m}}\text{))=d(g(x}{}_{\mathrm{n-1}}\text{),}\text{g(x}{}_{\mathrm{n+m-1}}\text{))≤α}{}^1\text{d(f(x}{}_{\mathrm{n-1}}\text{),f(x}{}_{\mathrm{n+m-1}}\text{))≤⋯≤α}{}^{\mathrm{n}}\text{d(f(x}{}_0\text{),f(x}{}_{\mathrm{m}}\text{))≤α}{}^{\mathrm{n}}\text{M.}$Clearly, {f(x_n)} is a d-Cauchy sequence and the S-completeness of (X,d) gives an x in X with $\text{lim}\text{d(f(x}{}_{\mathrm{n}}\text{),x)=0}$. g is d-continuous implies $\text{lim}\text{d(gf(x}{}_{\mathrm{n}}\text{),g(x))=0}$. Now f(x_n)=g(x_n−1) so that $\text{lim}\text{d(g(x}{}_{\mathrm{n}}\text{),x)=0}$. f is d-continuous gives lim d(fg(x_n),f(x))=0. Since fg=gf, $\text{lim}\text{d(fg(x}{}_{\mathrm{n}}\text{),f(x))=}\text{lim}\text{d(fg(x}{}_{\mathrm{n}}\text{),g(x))=0}$. Condition (W.3) gives f(x)=g(x). Also, fg(x)=gf(x). Thus, f(fx)=f(gx)=g(fx)=g(gx) and d(gx,g(gx))≤αd(fx,f(gx))=αd(gx,g(gx)) implies g(x)=g(gx). Hence g(x)=g(gx)=f(gx) so gx is a common fixed point of f and g.

If x=f(x)=g(x) and y=f(y)=g(y), then (1) gives d(x,y)=d(gx,gy)≤αd(fx,fy)=αd(x,y) or x=y.

For d a semi-metric, d(x_n,x)→0 if and only if x_n→x in the topology t(d). Thus, for a t(d)-continuous function f,x_n→x implies f(x_n)→f(x) and the above proof holds without change. Also, note that B(x,ε) is a t(d)-neighborhood of x, and (W.3) holds if t(d) is a Hausdorff topology.

Corollary 1. Let d be a bounded symmetric (semi-metric) for X that satisfies (W.3). Suppose (X,d) is S-complete (d-Cauchy complete) and f and g are commuting d-continuous (t(d)-continuous) mappings from X to X and g(X)⊂f(X). If there exists α∈(0,1) and a positive integer k such that d(g^kx,g^ky)≤αd(fx,fy) for all x,y∈X, then f and g have a unique common fixed point.

Proof. The proof of Corollary 10 of [4] carries over without change.

Letting f=i, the identity, gives a generalization of Banach’s theorem. If we also let k=1, we get Banach’s theorem in this new setting, which was given in [9] as Proposition 1.

Corollary 2. Let d be a bounded symmetric (semi-metric) for X that satisfies (W.3). Let n be a positive integer and let k>1. If g is a surjective d-continuous (t(d)-continuous) self map of X such that d(gⁿx,gⁿy)≥kd(x,y) for all x,y∈X, then g has a unique fixed point.

Proof. The proof of Corollary 11 of [4] carries over without change since f=g²ⁿ is also d-continuous.

We now look at applications to probabilistic spaces. A real-valued function f defined on the set of real numbers is a distribution function if it is nondecreasing, left continuous, with inff=0 and $\text{sup}\text{f=1}$. H denotes the distribution function defined by H(x)=0 if x≤0, and H(x)=1 for x>0.

Definition 7. Let X be a set and $\text{F}$ a function on X×X such that $\text{F}\text{(x,y)=F}{}_{\mathrm{x,y}}$ is a distribution function. Consider the following conditions:

I. F_x,y(0)=0 for all $\text{x,y}\text{in}\text{X}$.

II. F_x,y=H if and only if x=y.

III. F_x,y=F_y,x.

IV. If F_x,y(ε)=1 and F_y,z(δ)=1, then F_x,z(ε+δ)=1.

IV_m. F_x,z(ε+δ)≥T(F_x,y(ε),F_y,z(δ)).

If $\text{F}$ satisfies I and II, then it is called a PPM-structure on X and the pair $\text{(X,}\text{F}\text{)}$ is called a PPM space. An $\text{F}$ satisfying III is said to be symmetric. A symmetric PPM-structure $\text{F}$ satisfying IV is a probabilistic metric structure (PM-structure) and the pair $\text{(X,}\text{F}\text{)}$ is a probabilistic metric space (PM-space).

Definition 8. A Menger space is a PM-space that satisfies IV_m, where T is a 2-place function on the unit square satisfying:

(1) T(0,0)=0 and T(a,1)=a.

(2) T(a,b)=T(b,a).

(3) If a≤c and b≤d, then T(a,b)≤T(c,d).

(4) T(T(a,b),c)=T(a,T(b,c)).

T is called a t-norm.

Most fixed point theorems have been proved for Menger spaces satisfying additional conditions. Let $\text{(X,}\text{F}\text{)}$ be a PPM-space. For ε,λ>0 and x in X, let N_x(ε,λ)={y:F_x,y(ε)>1−λ}. A T₁ topology $\text{t(}\text{F}\text{)}$ on X is obtained as follows: $\text{U∈t(}\text{F}\text{)}$ if for each x in U, there exists ε>0 such that N_x(ε,ε)⊂U. Now N_x(ε,ε) may not be a $\text{t(}\text{F}\text{)}$ neighborhood of x. If it is, then $\text{t(}\text{F)}$ is said to be topological. The following condition is a reasonable generalization of a metric contraction to PPM-spaces. Let 0<k<1.$\text{(C)}\text{For}\text{t>0,F}{}_{\mathrm{x,y}}\text{(t)>1−t}\text{implies}\text{F}{}_{\mathrm{fx,fy}}\text{(kt)>1−kt.}$A self-map f satisfying (C) is called an H-contraction [2]. If the metric space (X,d) is made into a PM-space in the usual way; that is, F_x,y(t)=H(t−d(x,y)), then any metric contraction is an H-contraction.

Let $\text{(X,}\text{F}\text{)}$ be a symmetric PPM-space. Then $\text{t(}\text{F}\text{)}$ is a T₁-topology for X. Set$\text{d(x,y)=}\text{0}\text{if}\text{y∈N}{}_{\mathrm{x}}\text{(ε,ε)}\text{for all}\text{ε>0.}\text{sup}\text{{ε:}\text{y∉N}{}_{\mathrm{x}}\text{(ε,ε),0<ε<1}}\text{otherwise.}$

Clearly, d(x,x)=0. Since $\text{⋂{N}{}_{\mathrm{x}}\text{(ε,ε):}\text{ε>0}={x}}$, for $\text{x≠y,}\text{d(x,y)=}\text{sup}\text{{ε}\text{:}\text{F}{}_{\mathrm{x,y}}\text{(ε)≤1−ε}}$. Note that d is a symmetric for X. We refer to this d in what follows.

Definition 9. Let $\text{(X,}\text{F}\text{)}$ be a symmetric PPM-space. A sequence {x_n} is a fundamental sequence if for all t>0. The space is complete if for every fundamental sequence {x_n} there exists x in X such that for all t>0.

The following theorem generalizes Theorem 2 of [8]. It is used, along with Theorem 1 of this paper, to give very general fixed point theorems for symmetric PPM-spaces. These theorems greatly improve results in [8] and in many other papers.

Theorem 2. Let $\text{(X,}\text{F}\text{)}$ be a symmetric PPM-space. Define d as above. Then

• 1.

  d(x,y)<t if and only if F_x,y(t)>1−t.

• 2.

  d is a compatible symmetric for $\text{t(}\text{F}\text{)}$.

• 3.

  If $\text{f}\text{:}\text{X→X}$ and 0<k<1,f is an H-contraction if and only if d(fx,fy)≤kd(x,y).

• 4.

  $\text{(X,}\text{F}\text{)}$ is complete if and only if (X,d) is S-complete.

• 5.

  If $\text{t(}\text{F}\text{)}$ is topological, d is a semi-metric.

Proof. The details for (1) and (3) are essentially the same as given for Theorem 2 of [8]. We given them for completeness. Observe that if $\text{t<r,}\text{N}{}_{\mathrm{x}}\text{(t,t)⊂N}{}_{\mathrm{x}}\text{(r,r)}$. Also, $\text{∩{N}{}_{\mathrm{x}}\text{(ε,ε):}\text{0<ε<1}={x}}$. For, if $\text{x≠y,}\text{F}{}_{\mathrm{x,y}}\text{≠H}$. Thus, there exists ε>0 such that F_x,y(ε)=δ where 0<δ<1. Set r=1−δ and let $\text{s=}\text{min}\text{{ε,}\text{r}}$. Then F_x,y(s)≤F_x,y(ε)=δ=1−r≤1−s gives y∉N_x(s,s).

(1) If $\text{1<t,}\text{d(x,y)≤1<t}$ and also F_x,y(t)≥0>1−t. Suppose d(x,y)<t≤1. Choose δ such that d(x,y)<δ<t≤1. Then y∈N_x(δ,δ) and F_x,y(t)≥F_x,y(δ)>1−δ>1−t. For, if we assume y∉N_x(δ,δ), then $\text{d(x,y)=}\text{sup}\text{{ε>0}\text{:}\text{y∉N}{}_{\mathrm{x}}\text{(ε,ε),0<ε<1}≥δ}$, a contradiction. Conversely, suppose F_x,y(t)>1−t where 0<t≤1. Then y∈N_x(t,t). If y∉N_x(ε,ε) for all $\text{ε<t,}\text{F}{}_{\mathrm{x,y}}\text{(t)=}\text{lim}{}_{\mathrm{\epsilon \to t-}}\text{F}{}_{\mathrm{x,y}}\text{(ε)≤}\text{lim}{}_{\mathrm{\epsilon \to t-}}\text{(1−ε)=1−t}$, a contradiction. Thus, there exists 0<ε<t such that y∈N_x(ε,ε). Hence d(x,y)≤ε<t.

Clearly, (2) follows from (1).

(3) Suppose d(fx,fy)≤kd(x,y) and F_x,y(t)>1−t. Then d(x,y)<t and d(fx,fy)<kt. From (1), F_fx,fy(kt)>1−kt. If (C) holds, let ε>0 be given. Let t=d(x,y)+ε. d(x,y)=t−ε<t gives F_x,y(t)>1−t, and F_fx,fy(kt)>1−kt follows from (C). Thus, d(fx,fy)<kt=k(d(x,y)+ε)=kd(x,y)+kε. Since ε>0 was arbitrary, d(fx,fy)≤kd(x,y).

(4) Let {x_n} be a fundamental sequence. Then lim_n,m→∞F_xn,xm(t)=1 for every t>0. Fix ε>0 and 0<t<ε. Choose N such that F_xn,xm(t)>1−ε for all n,m>N. Now F_xn,xm(ε)≥F_xn,xm(t)>1−ε for all n,m>N. From (1), d(x_n,x_m)<ε for all n,m>N. Thus, {x_n} is a d-Cauchy sequence. If {x_n} is a d-Cauchy sequence, fix t>0 and let 0<ε<t. Choose N such that d(x_n,x_m)<ε for all n,n≥N. Then 1−ε<F_xn,xm(ε)≤F_xn,xm(t)≤1<1+ε for all n,m≥N. Thus, {x_n} is a fundamental sequence. Replacing x_m by x in the above argument, we have lim_n→∞F_xn,x(t)=1 for all t>0 if and only if lim_n→∞d(x_n,x)=0. This proves (4).

(5) This follows from (1). Note that for d a semi-metric, d(x_n,x)→0 if and only if x_n→x with respect to $\text{t(}\text{F}\text{)}$.

In the proof of (4) in Theorem 2, it was shown that F_xn,x(t)→1 for all t>0 if and only if d(x_n,x)→0. Thus, the condition (W.3), defined earlier, is equivalent, for d as in Theorem 2, to the following condition:We need the analogue in $\text{(X,}\text{F}\text{)}$ of a d-continuous function defined earlier. $\text{f}\text{:}\text{(X,}\text{F}\text{)→(X,}\text{F}\text{)}$ is $\text{F}$-continuous if F_xn,x(t)→1 for all t>0 implies F_fxn,fx(t)→1 for all t>0. Clearly, this is equivalent to $\text{f}\text{:}\text{(X,d)→(X,d)}$ is S-continuous, if d is as in Theorem 2. Note that if $\text{t(}\text{F}\text{)}$ is topological, d is a semi-metric and f is $\text{F}$-continuous is equivalent to f is $\text{t(}\text{F}\text{)}$ continuous. Also, f is d-continuous is equivalent to f is t(d) continuous.

Theorem 3. Let $\text{(X,}\text{F}\text{)}$ be a complete symmetric PPM-space that satisfies P(3). Suppose $\text{f}\text{:}\text{X→X}$ is $\text{F}$-continuous. Then f has a fixed point in X if and only if there exists an α in (0,1) and an $\text{F}$-continuous function $\text{g}\text{:}\text{X→X}$ which commutes with f and satisfies:

(2)$\text{g(X)⊂f(X)}$ and F_fx,fy(t)>1−t implies F_gx,gy(αt)>1−αt, for every t>0.

Indeed, if (2) holds, f and g have a unique common fixed point.

Proof. Let d be as in Theorem 2. Now d is a bounded compatible symmetric for $\text{t(}\text{F}\text{)}$, and f and g are S-continuous functions from (X,d) into (X,d). Theorem 2 also gives (X,d) is S-complete and d(x,y)<t if and only if F_x,y(t)>1−t. Suppose $\text{g}\text{:}\text{X→X}$ such that (2) holds. Let ε>0 be given, and set t=d(fx,fy)+ε. Then d(fx,fy)<t gives F_fx,fy(t)>1−t and (2) yields F_gx,gy(αt)>1−αt. Hence d(gx,gy)<αt=αd(fx,fy)+αε. Since ε was arbitrary, d(gx,gy)≤αd(fx,fy). We noted above that $\text{F}$-satisfies P(3) is equivalent to d satisfies (W.3). We now apply Theorem 1.

Corollary 3. Let $\text{(X,}\text{F}\text{)}$ be a complete symmetric PPM-space that satisfies P(3). Suppose f and g are commuting $\text{F}$-continuous mappings from X to X and g(X)⊂f(X). If there exists α∈(0,1) and a positive integer k such that for t>0

(3) F_fx,fy(t)>1−t implies F_g^kx,g^ky(αt)>1−αt, then f and g have a unique common fixed point.

Proof. The proof of Theorem 3 shows that (3) translates to$\text{d(g}{}^{\mathrm{k}}\text{x,g}{}^{\mathrm{k}}\text{y)≤αd(fx,fy).}$The result follows from Corollary 1.

When k=1 and f=I, Corollary 3 gives the analogue of Banach’s fixed point theorem for PPM-spaces.

Corollary 4. Let n be a positive integer and let k>1. If g is an $\text{F}$-continuous onto self map of a complete symmetric PPM-space $\text{(X,}\text{F}\text{)}$ that satisfies P(3) such thatthen g has a unique fixed point.

Proof. Let f(x)=g²ⁿ(x)=gⁿ(gⁿx) and α=(1/k). For $\text{t>0,}\text{t=k(t/α)=kt}{}_1\text{>0}$. F_{g²ⁿx,g²ⁿy}(kt₁)>1−kt₁ implies F_gⁿx,gⁿy(t₁)>1−t₁ or F_{g²ⁿx,g²ⁿy}(t)>1−t implies F_gⁿx,gⁿy(αt)>1−αt. Now g is onto implies g(X)⊂f(X). Since 0<α<1, Corollary 3 gives g²ⁿ and gⁿ have a unique common fixed point x. If gⁿy=y,g²ⁿy=gⁿy so x=y.

Remarks. We could have obtained Theorem 1 as a corollary of our next theorem. However, the proof is much more difficult and, for many applications, Theorem 1 will suffice. This allows one to proceed more rapidly to applications in probabilistic structures. The next theorem involves a function Q. Various conditions on Q have been studied by many different authors. Let $\text{Q}\text{:}\text{R}{}^{\mathrm{+}}\text{→}\text{R}{}^{\mathrm{+}}$ satisfy the following:

• 1.

  Q is nondecreasing on $\text{R}{}^{\mathrm{+}}$,

• 2.

  0<Q(s)<s for each $\text{s∈}\text{R}{}^{\mathrm{+}}$, and

• 3.

  lim_nQⁿ(s)=0 for each $\text{x∈}\text{R}{}^{\mathrm{+}}$.

  As a consequence of the above properties, Q also satisfies the following condtions:

• 4.

  Q(0)=0,

• 5.

  for any $\text{a,b,∈}\text{R}{}^{\mathrm{+}}\text{,}\text{Q(}\text{max}\text{(a,b))=}\text{max}\text{(Q(a),Q(b))}$, and

• 6.

  Q(s)≤s for each $\text{s∈}\text{R}{}^{\mathrm{+}}$.

Proof of (d). From the definition of $\text{Q,}\text{0≤Q(0)}$. From (a) and (b), Q(0)≤Q(s)<s for each s>0. Combining these inequalities yields 0≤Q(0)<s. Now take the limit as s→0 to obtain Q(0)=0.

Proof of (e). Without loss of generality we may assume that a≤b. Then Q(max(a,b))=Q(b)=max(Q(a),Q(b)), by (a).

Proof of (f). If s>0, then, from $\text{(b),}\text{Q(s)<s}$. If s=0, then, from (d), Q(0)=0. Therefore, in all cases we have Q(s)≤s.

Theorem 4. Let d be a bounded symmetric (semi-metric) for X that satisfies (W.3). Suppose that (X,d) is S-complete (d-Cauchy complete) and $\text{S,T}\text{:}\text{X→X}$ 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 $\text{A,B}\text{:}\text{X→X}$ such that A commutes with S,B commutes with T and satisfy$\text{A(X)⊂T(X),}\text{B(X)⊂S(X),}\text{and}\text{d(Ax,By)≤Q(}\text{max}\text{{d(Sx,Ty),d(Sx,Ax),d(Ty,By)}).}$If (4) 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. Define Ax=Bx=a for each x∈X. Then the conditions of Eq. (4)are satisfied.

To prove the converse, suppose there exists a Q satisfying (a)–(c) so that (4) holds. Let x₀∈X. Choose x₁ so that Ax₀=Tx₁,x₂ so that Bx₁=Sx₂, and, in general, define x_n so that Tx_2n+1=Ax_2n and Sx_2n+2=Bx_2n+1. Define {y_n} by y_2n=Sx_2n,y_2n+1=Tx_2n+1, and t_n≔d(y_n,y_n+1).

Lemma 1. {t_n} is nonincreasing in n, and $\text{lim}\text{t}{}_{\mathrm{n}}\text{=0}$.

Proof. Consider t₁. If t₁=0, then we shall show that t_n=0 for all n>0. Suppose that t₂≠0. Then, from (4),$\text{t}{}_2\text{=d(y}{}_2\text{,y}{}_3\text{)=d(Sx}{}_2\text{,Tx}{}_3\text{)=d(Bx}{}_1\text{,Ax}{}_2\text{)=d(Ax}{}_2\text{,Bx}{}_1\text{)≤Q(}\text{max}\text{{d(Sx}{}_2\text{,Tx}{}_1\text{),}\text{d(Sx}{}_2\text{,Ax}{}_2\text{),d(Tx}{}_1\text{,Bx}{}_1\text{)})=Q(}\text{max}\text{{t}{}_1\text{,t}{}_2\text{,t}{}_1\text{})=Q(t}{}_2\text{)<t}{}_2\text{,}$a contradiction. Therefore t₂=0.

Suppose that t_k=0 for 1≤k≤n. Then t_n+1=0. For, if t_n+1≠0, then, a repeat of the above argument will yield t_n+1≤Q(t_n+1)<t_n+1, a contradiction. Therefore t_n+1=0.

Suppose that t₁>0. Then, from the above,$\text{t}{}_2\text{=d(y}{}_2\text{,y}{}_3\text{)≤Q(}\text{max}\text{{t}{}_1\text{,t}{}_2\text{}).}$The assumption that t₂≥t₁ leads to the contradiction t₂≤Q(t₂)<t₂. Therefore t₂<t₁.

Now consider t₂. Using the above arguments for t₁, the assumption that t₂=0 leads to t_n=0 for all n>1, and the assumption t₂≠0 leads to t₃<t₂. The cases for general n are established in the same manner.

Since {t_n} is a nonincreasing nonnegative sequence, it has a limit t≥0. We shall assume that t₁<0, since, if t₁=0,t is automatically zero. Then, from Eq. (4),$\text{t}{}_{\mathrm{2n}}\text{=d(y}{}_{\mathrm{2n}}\text{,y}{}_{\mathrm{2n+1}}\text{)=d(Sx}{}_{\mathrm{2n}}\text{,Tx}{}_{\mathrm{2n+1}}\text{)=d(Bx}{}_{\mathrm{2n-1}}\text{,Ax}{}_{\mathrm{2n}}\text{)=d(Ax}{}_{\mathrm{2n}}\text{,Bx}{}_{\mathrm{2n-1}}\text{)≤Q(}\text{max}\text{{d(Sx}{}_{\mathrm{2n}}\text{,Tx}{}_{\mathrm{2n-1}}\text{),d(Sx}{}_{\mathrm{2n}}\text{,Ax}{}_{\mathrm{2n}}\text{),d(Tx}{}_{\mathrm{2n-1}}\text{,Bx}{}_{\mathrm{2n-1}}\text{)})=Q(}\text{max}\text{{t}{}_{\mathrm{2n-1}}\text{,t}{}_{\mathrm{2n}}\text{,t}{}_{\mathrm{2n-1}}\text{})=Q(t}{}_{\mathrm{2n-1}}\text{).}$$\text{t}{}_{\mathrm{2n+1}}\text{=d(y}{}_{\mathrm{2n+1}}\text{,y}{}_{\mathrm{2n+2}}\text{)=d(Tx}{}_{\mathrm{2n+1}}\text{,Sx}{}_{\mathrm{2n+2}}\text{)=d(Ax}{}_{\mathrm{2n}}\text{,Bx}{}_{\mathrm{2n+1}}\text{)≤Q(}\text{max}\text{{d(Sx}{}_{\mathrm{2n}}\text{,Tx}{}_{\mathrm{2n+1}}\text{),d(Sx}{}_{\mathrm{2n}}\text{,Ax}{}_{\mathrm{2n}}\text{),d(Tx}{}_{\mathrm{2n+1}}\text{,Bx}{}_{\mathrm{2n+1}}\text{)})=Q(}\text{max}\text{{t}{}_{\mathrm{2n}}\text{,t}{}_{\mathrm{2n}}\text{,t}{}_{\mathrm{2n+1}}\text{})=Q(t}{}_{\mathrm{2n}}\text{).}$Thus, for all n,$\text{t}{}_{\mathrm{n}}\text{≤Q(t}{}_{\mathrm{n-1}}\text{).}$Treating (5) as a recursion formula, it follows that$\text{t}{}_{\mathrm{n}}\text{≤Q(Q(t}{}_{\mathrm{n-2}}\text{))=Q}{}^2\text{(t}{}_{\mathrm{n-2}}\text{)≤⋯Q}{}^{\mathrm{n-1}}\text{(t}{}_1\text{).}$Taking the limit of the above as n→∞ and using (c) yields t≤0. Therefore t=0.

Lemma 2. {y_n} is Cauchy.

Proof. Define $\text{M=}\text{sup}\text{{d(x,y)}\text{:}\text{x,y∈X}<∞}$, since d is a bounded symmetric. Fix ε>0. Choose N so that n≥N implies that t_2n−2<ε. Then choose N₁>N so that n≥N₁ implies that Q^2(n−N)(M)<ε. For n>N₁,m>0,$\text{d(y}{}_{\mathrm{2n}}\text{,y}{}_{\mathrm{2n+2m-1}}\text{)=d(Sx}{}_{\mathrm{2n}}\text{,Tx}{}_{\mathrm{2n+2m-1}}\text{)=d(Bx}{}_{\mathrm{2n-1}}\text{,Ax}{}_{\mathrm{2n+2m-2}}\text{)=d(Ax}{}_{\mathrm{2n+2m-2}}\text{,Bx}{}_{\mathrm{2n-1}}\text{)≤Q(}\text{max}\text{{d(Sx}{}_{\mathrm{2n+2m-2}}\text{,Tx}{}_{\mathrm{2n-1}}\text{),d(Sx}{}_{\mathrm{2n+2m-2}}\text{,Ax}{}_{\mathrm{2n+2m-2}}\text{),d(Tx}{}_{\mathrm{2n-1}}\text{,Bx}{}_{\mathrm{2n-1}}\text{)})=Q(}\text{max}\text{{d(y}{}_{\mathrm{2n+2m-2}}\text{,y}{}_{\mathrm{2n-1}}\text{),t}{}_{\mathrm{2n+2m-2}}\text{,t}{}_{\mathrm{2n-1}}\text{})=Q(}\text{max}\text{{d(y}{}_{\mathrm{2n+2m-2}}\text{,y}{}_{\mathrm{2n-1}}\text{),t}{}_{\mathrm{2n-1}}\text{})=}\text{max}\text{{Q(d(y}{}_{\mathrm{2n+2m-2}}\text{,y}{}_{\mathrm{2n-1}}\text{)),Q(t}{}_{\mathrm{2n-1}}\text{)}≤}\text{max}\text{{Q(d(y}{}_{\mathrm{2n+2m-2}}\text{,y}{}_{\mathrm{2n-1}}\text{)),ε},}$$\text{d(y}{}_{\mathrm{2n+2m-2}}\text{,y}{}_{\mathrm{2n-1}}\text{)=d(Sx}{}_{\mathrm{2n+2m-2}}\text{,Tx}{}_{\mathrm{2n-1}}\text{)=d(Bx}{}_{\mathrm{2n+2m-3}}\text{,Ax}{}_{\mathrm{2n-2}}\text{)=d(Ax}{}_{\mathrm{2n-2}}\text{,Bx}{}_{\mathrm{2n+2m-3}}\text{)≤Q(}\text{max}\text{{d(Sx}{}_{\mathrm{2n-2}}\text{,Tx}{}_{\mathrm{2n+2m-3}}\text{),d(Sx}{}_{\mathrm{2n-2}}\text{,Ax}{}_{\mathrm{2n-2}}\text{),d(Tx}{}_{\mathrm{2n+2m-3}}\text{,Bx}{}_{\mathrm{2n+2m-3}}\text{)})=Q(}\text{max}\text{{d(y}{}_{\mathrm{2n-2}}\text{,y}{}_{\mathrm{2n+2m-3}}\text{),t}{}_{\mathrm{2n-2}}\text{,t}{}_{\mathrm{2n+2m-3}}\text{})=Q(}\text{max}\text{{d(y}{}_{\mathrm{2n-2}}\text{,y}{}_{\mathrm{2n+2m-3}}\text{),t}{}_{\mathrm{2n-2}}\text{})=}\text{max}\text{{Q(d(y}{}_{\mathrm{2n-2}}\text{,y}{}_{\mathrm{2n+2m-3}}\text{)),Q(t}{}_{\mathrm{2n-2}}\text{)}≤}\text{max}\text{{Q(d(y}{}_{\mathrm{2n-2}}\text{,y}{}_{\mathrm{2n+2m-3}}\text{)),ε}.}$Substituting Eq. (7)into Eq. (6)yields$\text{d(y}{}_{\mathrm{2n}}\text{,y}{}_{\mathrm{2n+2m-1}}\text{)≤}\text{max}\text{{Q(}\text{max}\text{{Q(d(y}{}_{\mathrm{2(n-1)}}\text{,y}{}_{\mathrm{2(n-1)+2m-1}}\text{,ε}),ε}=}\text{max}\text{{Q}{}^2\text{(d(y}{}_{\mathrm{2(n-1)}}\text{,y}{}_{\mathrm{2(n-1)+2m-1}}\text{)),Q(ε),ε}=}\text{max}\text{{Q}{}^2\text{(d(y}{}_{\mathrm{2(n-1)}}\text{,y}{}_{\mathrm{2(n-1)+2m-1}}\text{)),ε}}$

Treating Eq. (8)as a recursion formula one obtains$\text{d(y}{}_{\mathrm{2n}}\text{,y}{}_{\mathrm{2n+2m-1}}\text{)≤}\text{max}\text{{Q}{}^2\text{(d(y}{}_{\mathrm{2(n-1)}}\text{,y}{}_{\mathrm{2(n-1)+2m-1}}\text{)),ε}≤}\text{max}\text{{Q}{}^2\text{[}\text{max}\text{{Q}{}^2\text{(d(y}{}_{\mathrm{2(n-2)}}\text{,y}{}_{\mathrm{2(n-2)+2m-1}}\text{)),ε}],ε}=}\text{max}\text{{Q}{}^4\text{(d(y}{}_{\mathrm{2(n-2)}}\text{,y}{}_{\mathrm{2(n-2)+2m-1}}\text{)),Q}{}^2\text{(ε),ε}=}\text{max}\text{{Q}{}^4\text{(d(y}{}_{\mathrm{2(n-2)}}\text{,y}{}_{\mathrm{2(n-2)+2m-1}}\text{)),ε}≤⋯≤}\text{max}\text{{Q}{}^{\mathrm{2(n-N)}}\text{(d(y}{}_{\mathrm{2N}}\text{,y}{}_{\mathrm{2N+2m-1}}\text{)),ε}≤}\text{max}\text{{Q}{}^{\mathrm{2(n-N)}}\text{(M),ε}<ε,}$and$\text{lim}\text{n}\text{d(y}{}_{\mathrm{2n}}\text{,y}{}_{\mathrm{2n+2m-1}}\text{)=0.}$Now consider d(y_n,y_n+m).

Case I. n,m even. Then$\text{d(y}{}_{\mathrm{2n}}\text{,y}{}_{\mathrm{2n+2m}}\text{)≤d(y}{}_{\mathrm{2n}}\text{,y}{}_{\mathrm{2n+2m-1}}\text{)+t}{}_{\mathrm{2n+2m-1}}\text{→0}\text{as}\text{n→∞,}\text{using Eq. (10)}$

Case II. n even, m odd. Then, from Eq. (10),$\text{d(y}{}_{\mathrm{2n}}\text{,y}{}_{\mathrm{2n+2m-1}}\text{)→0}\text{as}\text{n→∞.}$

Case III. n odd, m even. Then, using Eq. (10),$\text{d(y}{}_{\mathrm{2n+1}}\text{,y}{}_{\mathrm{2m+2n+1}}\text{)≤t}{}_{\mathrm{2n}}\text{+d(y}{}_{\mathrm{2n}}\text{,y}{}_{\mathrm{2n+2m+1}}\text{)→0}\text{as}\text{n→∞.}$

Case IV. n,m odd. Then, from Eq. (10),$\text{d(y}{}_{\mathrm{2n+1}}\text{,y}{}_{\mathrm{2n+2m}}\text{)≤t}{}_{\mathrm{2n}}\text{+d(y}{}_{\mathrm{2n}}\text{,y}{}_{\mathrm{2m+2n-1}}\text{)+t}{}_{\mathrm{2m+2n-1}}\text{→0}\text{as}\text{n→∞.}$

Therefore {y_n} is Cauchy. Since (X,d) is S-complete, there exists an x in X with $\text{lim}\text{d(y}{}_{\mathrm{n}}\text{,x)=0}$. In particular, $\text{lim}\text{d(Sx}{}_{\mathrm{2n}}\text{,x)=}\text{lim}\text{d(y}{}_{\mathrm{2n}}\text{,x)=0}$ and $\text{lim}\text{d(Tx}{}_{\mathrm{2n-1}}\text{,x)=}\text{lim}\text{d(y}{}_{\mathrm{2n-1}}\text{,x)=0}$. Since $\text{Tx}{}_{\mathrm{2n-1}}\text{=Ax}{}_{\mathrm{2n-2}}\text{,}\text{lim}\text{d(Ax}{}_{\mathrm{2n}}\text{,x)=0}$. The d-continuity of A and S implies that $\text{lim}\text{d(ASx}{}_{\mathrm{2n}}\text{,Ax)=0}$ and $\text{lim}\text{d(SAx}{}_{\mathrm{2n}}\text{,Sx)=0}$. But ASx_2n=SAx_2n, since A and S commute. Condition (W.3) then yields Sx=Ax.

The d-continuity of B and T implies that $\text{lim}\text{d(BTx}{}_{\mathrm{2n-1}}\text{,Bx)=0}$ and $\text{lim}\text{d(TBx}{}_{\mathrm{2n-1}}\text{,Tx)=0}$. Since B and T commute, BTx_2n−1=TBx_2n−1 and, by (W.3), Tx=Bx.

Using Eq. (4),$\text{d(Ax,Bx)≤Q(}\text{max}\text{{d(Sx,Tx),d(Sx,Ax),d(Tx,Bx)})=Q(}\text{max}\text{{d(Ax,Bx),0,0})=Q(d(Ax,Bx)),}$which implies that Ax=Bx. Thus x is a common coincidence point of A,B,S, and T. Hence AAx=ABx=BAx=BBx=BTx=TBx=TTx=TAx=ASx=SAx=SSx.

From Eq. (4),$\text{d(Ax,BAx)≤Q(}\text{max}\text{{d(Sx,TAx),d(Sx,Ax),d(TAx,BAx)})=Q(}\text{max}\text{{d(Ax,BAx),0,0})=Q(d(Ax,BAx))}$and Ax=BAx. Thus Ax is a fixed point of B.

Since TAx=AAx=SAx=BAx, Ax is a common fixed point of A,B,S, and T.

To prove uniqueness, suppose that y is also a common fixed point of A,B,S, and T.

Then, from Eq. (4)$\text{d(Ax,y)=d(AAx,By)≤Q(}\text{max}\text{{d(SAx,Ty),d(SAx,AAx),d(Ty,By)})=Q(}\text{max}\text{{d(AAx,By),0,0})=Q(d(Ax,y)),}$which implies that Ax=y, and the common fixed point is unique.

Corollary 5. Let d be a bounded symmetric (semi-metric) for X that satisfies (W.3). Suppose that (X,d) is S-complete (d-Cauchy complete) and {A,S} and {B,T} are commuting d-continuous (t(d)-continuous) self mappings of X and A(X)⊂T(X),B(X)⊂S(X). If there exists a Q satisfying (a)–(c) and positive integers r, s, t and u such that$\text{d(A}{}^{\mathrm{r}}\text{x,B}{}^{\mathrm{s}}\text{y)≤Q(}\text{max}\text{{d(S}{}^{\mathrm{t}}\text{x,T}{}^{\mathrm{u}}\text{y),d(S}{}^{\mathrm{t}}\text{x,A}{}^{\mathrm{r}}\text{x),d(T}{}^{\mathrm{u}}\text{y,B}{}^{\mathrm{s}}\text{y)})}$for each x,y∈X, then A,B,S, and T have a unique common fixed point.

Proof. Define $\text{f=A}{}^{\mathrm{r}}\text{,}\text{g=B}{}^{\mathrm{s}}\text{,}\text{h=S}{}^{\mathrm{t}}\text{,k=T}{}^{\mathrm{u}}$. Then, from Theorem 4, f, g, h and k have a unique common fixed point p; i.e., A^rp=B^sp=S^tp=T^up=p. Since A and S commute, this implies that S(A^rp)=A^r(Sp)=Sp, S(S^tp)=S^t(Sp)=Sp, so that Sp is a common fixed point of A^r and S^t. Similarly, Tp is a common fixed point of B^s and T^u.

Using Eq. (4),$\text{d(Sp,Tp)=d(A}{}^{\mathrm{r}}\text{Sp,B}{}^{\mathrm{s}}\text{Tp)≤Q(}\text{max}\text{{d(S}{}^{\mathrm{t}}\text{Sp,T}{}^{\mathrm{u}}\text{Tp),d(S}{}^{\mathrm{t}}\text{Sp,A}{}^{\mathrm{r}}\text{Sp),d(T}{}^{\mathrm{u}}\text{Tp,B}{}^{\mathrm{s}}\text{Tp))=Q(}\text{max}\text{{d(Sp,Tp),0,0})=Q(d(Sp,Tp)),}$implies that Sp=Tp. Therefore, Sp is a common fixed point of A^r, B^s, S^t, and T^u. Uniqueness implies that p=Sp=Tp. A similar argument leads to p=Ap=Bp. Therefore p is a common fixed point of A, B, S, and T.

The proof of uniqueness is the same as that of Theorem 4.

Setting A=B, or S=T, yields two corollaries of Theorem 4 for three maps.

Corollary 6. Let d be a bounded symmetric (semi-metric) for X that satisfies (W.3). Suppose that (X,d) is S-complete (d-Cauchy complete) and $\text{S,T}\text{:}\text{X→X}$ 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 a d-continuous (t(d)-continuous) function $\text{A}\text{:}\text{X→X}$ such that A commutes with S and satisfies$\text{A(X)⊂T(X),}\text{d(Ax,Ay)≤Q(}\text{max}\text{{d(Sx,Ty),d(Sx,Ax),d(Ty,Ay)})}$If Eq. (11)holds then A, S, and T have a unique common fixed point.

Corollary 7. Let d be a bounded symmetric (semi-metric) for X that satisfies (W.3). Suppose that (X,d) is S-complete (d-Cauchy-complete) and $\text{S}\text{:}\text{X→X}$ is d-continuous (t(d)-continuous). Then S has a fixed point if and only if there exists a function Q satisfying (a)–(c) and d-continuous (t(d)-continuous) functions $\text{A,B}\text{:}\text{X→X}$ such that A and B commute with S and satisfy$\text{A(X)⊂S(X),B(X)⊂S(X),}\text{d(Ax,By)≤Q(}\text{max}\text{{d(Sx,Sy),d(Sx,Ax),d(Sy,By)}).}$If Eq. (12)holds then A, S, and T have a unique common fixed point.

If one adds condition (W.4), then one can replace the commuting condition in Theorem 4 with compatibility.

In this setting, two maps A and S are said to be compatible if, for each sequence of values {x_n} such that $\text{lim}\text{Ax}{}_{\mathrm{n}}\text{=}\text{lim}\text{Sx}{}_{\mathrm{n}}\text{=t∈X}$, one has $\text{lim}\text{d(ASx}{}_{\mathrm{n}}\text{,SAx}{}_{\mathrm{n}}\text{)=0}$.

Theorem 5. Let d be a bounded symmetric (semi-metric) for X that satisfies (W.3) and (W.4). Suppose that (X,d) is S-complete (d-Cauchy complete) and $\text{S,T}\text{:}\text{X→X}$ are d-continuous (t(d)-continuous). Then A and S have a common fixed point if and only if there exists a Q satisfying (a)–(c) and d-continuous (t(d)-continuous) functions $\text{A,B}\text{:}\text{X→X}$ which are such that {A,S} and {B,T} are pairwise compatible and satisfy$\text{A(X)⊂T(X),B(X)⊂S(X),}\text{d(Ax,}\text{By)≤Q(}\text{max}\text{{d(Sx,Ty),d(Sx,Ax),d(Ty,By)}).}$If Eq. (13)holds then A, B, S, and T have a unique common fixed point.

Proof. The proof that Eq. (13)holds remains unchanged.

As in the proof of Theorem 4, {y_n} converges to a point x in X. In particular, $\text{lim}\text{d(Sx}{}_{\mathrm{2n}}\text{,x)=}\text{lim}\text{d(y}{}_{\mathrm{2n}}\text{,x)=0}$ and $\text{lim}\text{d(Tx}{}_{\mathrm{2n-1}}\text{,x)=}\text{lim}\text{d(y}{}_{\mathrm{2n-1}}\text{,x)=0}$. Since $\text{Tx}{}_{\mathrm{2n-1}}\text{=Ax}{}_{\mathrm{2n-2}}\text{,}\text{lim}\text{d(Ax}{}_{\mathrm{2n}}\text{,x)=0}$. The d-continuity of A and S implies that $\text{lim}\text{d(ASx}{}_{\mathrm{2n}}\text{,Ax)=0}$ and $\text{lim}\text{d(SAx}{}_{\mathrm{2n}}\text{,Sx)=0}$. Since A and S are compatible, $\text{lim}\text{d(ASx}{}_{\mathrm{2n}}\text{,SAx}{}_{\mathrm{2n}}\text{)=0}$. By (W.4), Ax=Sx. Since compatible maps commute at coincidence points, the balance of the proof of Theorem 4 applies.

There is no analog of Corollary 1 for compatible maps, since the proof requires that the maps commute. However, Corollaries 2 and 3 have analogs for the situation in which the maps are pairwise compatible.

Theorem 6. Let $\text{(X,}\text{F}\text{)}$ be a complete symmetric PPM-space that satisfies P(3). Suppose that $\text{f}\text{:}\text{X→X}$ is $\text{F}$-continuous. Then f has a fixed point in X if and only if there exists a Q satisfying (a)–(c) and $\text{lim}\text{ε→}\text{o}\text{Q(t+ε)=Q(t)}$ for each t>0 and an $\text{F}$-continuous function $\text{g}\text{:}\text{X→X}$ which commutes with f and satisfies:$\text{g(X)⊂f(X)}\text{and}\text{F}{}_{\mathrm{fx,fy}}\text{(t)>1−t}\text{implies}\text{F}{}_{\mathrm{gx,gy}}\text{(Q(t))>1−Q(t)}$for every t>0. Indeed, if (14) holds, f and g have a unique common fixed point.

Proof. Let d be as in Theorem 2. Then d is a bounded compatible symmetric for $\text{t(}\text{F}\text{)}$, and f and g are S-continuous selfmaps of (X,d). Also from Theorem 2, (X,d) is S-complete and d(x,y)<t if and only if F_x,y(t)>1−t. Let $\text{g}\text{:}\text{X→X}$ be such that Eq. (14)holds. For any ε>0 set t=max{d(fx,fy)+ε,d(fx,gx),d(fy,gy)}. Then t>0 and d(fx,fy)+ε≤t, or d(fx,fy)≤t−ε<t. Thus F_fx,fy(t)>1−t, and Eq. (14)implies that F_gx,gy(Q(t))>1−Q(t). Therefore d(gx,gy)<Q(t)=Q(max{d(fx,fy)+ε,d(fx,gx),d(fy,gy)}=max{Q(d(fx,fy))+ε),Q(d(fx,gx)),Q(d(fy,gy))}. Since ε is aribitrary, it follows that Eq. (4)is satisfied. As noted above, the condition that $\text{F}$ satisfies P(3) is equivalent to d satisfying (W.3). Now apply Theorem 4.

Corollary 8. Let $\text{(X,}\text{F}\text{)}$ be a complete symmetric PPM-space that satisfies P(3). Suppose that f and g are commuting $\text{F}$-continuous self mappings of X and g(X)⊂f(X). If there exists an α∈(0,1) and positive integers r and s such thatthen f and g have a unique common fixed point.

Proof. The proof of Theorem 6 shows that (15) translates into (4). The result now follows from Corollary 5.

Next we look at an application to random normed structures.

Definition 10. X is a linear space over R or C and for each $\text{x∈X,}\text{F}{}_{\mathrm{x}}$ is a distribution function. Let $\text{F}\text{={F}{}_{\mathrm{x}}\text{:}\text{x∈X}}$ and let T be a t-norm. The triple $\text{(X,}\text{F}\text{,T)}$ is a random normed space if the following conditions hold:

• (R1) F_x(0)=0 for all x in X.

• (R2) F_x=H if and only if x=0.

• (R3) F_λx(μ)=F_x(μ/|λ|) for all μ, all λ≠0 and all x in X.

• (R4) F_x+y(μ+v)≥T(F_x(μ),F_y(v)) for all x,y in X and all μ≥0,v≥0.

• (R5) T(μ,v)≥max{μ+v−1,0} for all $\text{μ,v}\text{in}\text{[0,1]}$.

A random normed space is a Menger space if we set G_x,y=F_x−y. It is also known that, if the t-norm T is continuous, X is a topological linear space using the usual topology $\text{t(}\text{F}\text{)}$.

Definition 11. Suppose X is a linear space and, for each x in X, we have a distribution function F_x such that (R1), (R2), (R3a), and (R3b) hold.

• (R3a) If 0<|α|<1,F_αx(ε)≥F_x(ε).

• (R3b) For any x and any ε>0,F_αx(ε)→1 as α→0.

Condition (R3) implies (R3a) and (R3b). Let G_x,y=F_x−y and $\text{G}\text{={G}{}_{\mathrm{x,y}}\text{:}\text{x,y∈X}}$. It is easily shown that $\text{(X,}\text{G}\text{)}$ is a symmetric PPM-space. Thus, the previous fixed point theorems and corollaries hold for RPN-spaces. Note that the contractive condition in Theorem 3 becomes g(X)⊂f(X) and for t>0,$\text{F}{}_{\mathrm{fx-fy}}\text{(t)>1−t}\text{implies}\text{F}{}_{\mathrm{gx-gy}}\text{(αt)>1−αt.}$We only state Banach’s fixed point theorem in this setting. It is clearly a corollary of the above results.

Theorem 7. Let f be a self map of a complete RPN-space $\text{(X,}\text{F}\text{)}$ that satisfies P(3). Assume 0<k<1. If f satisfies F_x−y(t)>1−t implies F_fx−fy(kt)>1−kt for all t>0, then f has a unique fixed point.

Note that a RPN-space is much more general than a random normed linear space. Other results are given for RPN-spaces in [7]. In particular, a necessary condition is given for a RPN-space to be a metrizable topological linear space. Let $\text{f}\text{:}\text{X→X}$ where $\text{(X,}\text{F}\text{)}$ is a symmetric PPM-space or some other probabilistic space. Suppose a condition involving f and the distribution functions F_x,y arise from some applied problem and one needs to know if this implies f has a fixed point. Use the compatible distance function d with d(x,y)<t if and only if F_x,y(t)>1−t to translate the condition to one involving d. This is usually easier, since fixed point theory in metric spaces is well developed. If d is only a symmetric, Theorems 1 and 4 show it is possible to obtain fixed point theorems.