We first give the following lemma and then we state the main results of this paper. We recall that if \(A_{0}\) (or \(B_{0}\)) is a nonempty subset, then A and B are nonempty subsets.
Lemma 2.1
Let
\((X, F,\Delta) \)
be a complete probabilistic Menger space such that Δ is a t-norm of H-type, and
\(A,B \subseteq X \)
be such that
\(A_{0} \)
is a nonempty closed set. If
\(T : A\rightarrow B \)
is a proximal contraction mapping such that
\(T(A_{0} )\subseteq B_{0}\), then there exists a unique
\(x\in A_{0} \)
such that
\(F_{x,Tx}(t)=F_{A,B}(t) \)
for all
\(t>0\).
Proof
Since \(A_{0} \) is nonempty and \(T(A_{0} )\subseteq B_{0}\), there exist \(x_{1},x_{0}\in A_{0}\) such that \(F_{x_{1},Tx_{0}}(t)=F_{A,B}(t) \). Since \(Tx_{1}\in B_{0} \), there exists \(x_{2}\in A_{0}\) such that \(F_{x_{2},Tx_{1}}(t)=F_{A,B}(t) \). Continuing this process, we obtain a sequence \(( x_{n})\subseteq A_{0} \) such that \(F_{x_{n+1},Tx_{n}}(t)=F_{A,B}(t) \) for all \(n\in\mathbb {N} \) and \(t>0\). Since for all \(n\in\mathbb{N} \),
$$F_{x_{n},Tx_{n-1}}(t)=F_{A,B}(t)= F_{x_{n+1},Tx_{n}}(t) \quad (t>0) $$
and T is a proximal contraction, we have
$$F_{x_{n+1},x_{n}}(t)\geq F_{x_{n},x_{n-1}}\biggl(\frac{t}{\alpha}\biggr)\quad (0< \alpha< 1, t>0). $$
Therefore, by Lemma 1.1, \(( x_{n}) \) is a Cauchy sequence and so converges to some \(x\in A_{0} \). Again by the assumption \(T(A_{0} )\subseteq B_{0}\), \(Tx \in B_{0}\). Then there exists an element \(u\in A_{0} \) such that \(F_{u,Tx}(t)=F_{A,B}(t)\) for all \(t>0\). Since for all \(n\in\mathbb{N} \),
$$F_{u,Tx}(t)=F_{A,B}(t)=F_{x_{n+1},Tx_{n}}(t)\quad (t>0), $$
by the hypothesis we have
$$F_{u,x_{n+1}}(t)\geq F_{x,x_{n}}\biggl(\frac{t}{\alpha}\biggr)\geq F_{x,x_{n}}(t) \quad (t>0). $$
Letting \(n\rightarrow\infty\) shows that \(x_{n}\rightarrow u \) and thus \(x =u\), so \(F_{x,Tx}(t)=F_{A,B}(t) \). If there exists another element y such that \(F_{y,Ty}(t)=F_{A,B}(t) \), then by the hypothesis we have \(F_{x,y}(t)\geq F_{x,y}(\frac{t}{\alpha}) \), which means that \(x=y\). □
Proposition 2.2
Let
\((X, F,\Delta) \)
be a probabilistic Menger space, and
\(A,B \subseteq X \)
be such that
\(A_{0} \)
is a nonempty set. Suppose that
\(T : A\rightarrow B \)
is a proximal contraction mapping such that
\(T(A_{0} )\subseteq B_{0}\)
and
\(g : A\rightarrow A \)
is an isometry mapping such that
\(A_{0}\subseteq g(A_{0}) \). Denote
\(G = g(A)\)
and
$$G_{0} =\bigl\lbrace z\in G : \exists y\in B \textit{ s.t. } \forall t \geq0, F_{z,y}(t)=F_{G,B}(t)\bigr\rbrace . $$
Then
\(Tg^{-1}\)
is a proximal contraction, and
\(G_{0}=A_{0} \).
Proof
Since \(G\subseteq A \), \(F_{G,B}(t)\leq F_{A,B} (t)\) for all \(t>0 \). Assume that \(x\in A_{0} \subseteq g(A_{0})\). Then \(x=g(x^{\prime}) \) for some \(x^{\prime} \in A_{0}\), and so there exists \(y\in B \) such that \(F_{A,B}(t)=F_{g(x^{\prime}),y}(t)\leq F_{G,B}(t) \) for all \(t>0 \). Thus, \(F_{A,B}(t)=F_{G,B}(t) \) for all \(t>0 \). Now we show that \(Tg^{-1} \) is a proximal contraction. To this end, suppose that \(u,v,x,y\in G \) are such that
$$F_{u,Tg^{-1}x}(t)=F_{G,B}(t)=F_{A,B}(t)=F_{v,Tg^{-1}y}(t) \quad (t>0). $$
By the hypothesis we have
$$F_{u,v}(t)\geq F_{g^{-1}x,g^{-1}y}\biggl(\frac{t}{\alpha } \biggr)=F_{gg^{-1}x,gg^{-1}y}\biggl(\frac{t}{\alpha}\biggr)=F_{x,y}\biggl( \frac{t}{\alpha }\biggr) \quad (t>0) $$
for some \(\alpha\in(0,1) \). Therefore, \(Tg^{-1}\) is a proximal contraction. If \(x\in G_{0} \), then \(x\in G\subseteq A\), and there exists \(y\in B \) such that \(F_{x,y}(t)=F_{G,B}(t)=F_{A,B}(t)\) for all \(t>0 \), so that \(x\in A_{0} \). If \(x\in A_{0} \subseteq A\), then there exists \(y\in B \) such that \(F_{x,y}(t)=F_{A,B}(t)=F_{G,B}(t)\) for all \(t>0 \). On the other hand, by the hypothesis \(x\in G \), and therefore \(G_{0}=A_{0} \). □
Corollary 2.3
Let the hypotheses of Lemma
2.1
be satisfied. Suppose that
\(T : A\rightarrow B \)
is a proximal contraction mapping such that
\(T(A_{0} )\subseteq B_{0}\)
and
\(g: A\rightarrow A\)
is an isometry mapping such that
\(A_{0}\subseteq g(A_{0}) \). Then there exists a unique
\(x\in A_{0} \)
such that
\(F_{gx,Tx}(t)=F_{A,B}(t) \).
Proof
By Proposition 2.2, \(Tg^{-1}:G=g(A)\rightarrow B\) is proximal contraction, and \(Tg^{-1}(G_{0})=Tg^{-1}(A_{0})\subseteq T(A_{0})\subseteq B_{0}\). Now by Lemma 2.1 there exists a unique \(x'\in A_{0} \) such that \(F_{x',Tg^{-1}x'}(t)=F_{A,B}(t) \). Since \(A_{0}\subseteq g(A_{0})\), there exists \(x\in A_{0} \) such that \(x'=g(x)\), so that \(F_{g(x),Tx}(t)=F_{A,B}(t) \). Note that g is an injective mapping, therefore, by Lemma 2.1, x is unique, and hence the result follows. □
Theorem 2.4
Let
\((X, \nu,\Delta_{m}) \)
be a probabilistic Banach space, \(A,B \subseteq X \)
be such that
A
is a convex set, \(A_{0}\)
be a nonempty compact set, and
B
be a bounded convex set. Suppose that
\(T : A\rightarrow B \)
is a continuous affine and proximal nonexpansive mapping such that
\(T(A_{0} )\subseteq B_{0}\)
and
\(g : A\rightarrow A \)
is an isometry mapping such that
\(A_{0}\subseteq g(A_{0}) \). Then there exists an element
\(x\in A_{0} \)
such that
\(\nu _{gx-Tx}(t)=\nu_{A-B}(t) \)
for all
\(t>0\).
Proof
Fix \(z\in A_{0} \) and \(i \in(0,1)\). We define the mapping \(T_{i}:A\rightarrow B \) by
We show that \(T_{i} \) is a proximal contraction. Let \(u,v,x,y\in A \) be such that
$$\nu_{u-T_{i}x}(t)=\nu_{A-B}(t)=\nu_{v-T_{i}y}(t)\quad (t>0). $$
Since T is an affine mapping, we have
$$\nu_{u-T((1-i)z+ix)}(t)=\nu_{A-B}(t)=\nu_{v-T((1-i)z+iy)}(t)\quad (t>0). $$
So by the hypothesis we have
$$\begin{aligned} \nu_{u-v}(t) & \geq\nu_{(1-i)z+ix-(1-i)z-iy}(t) \\ &= \nu_{i(x-y)}(t)=\nu_{x-y}\biggl(\frac{t}{i}\biggr)\quad (t>0). \end{aligned}$$
Hence, \(T_{i} \) is a proximal contraction. Let \(x\in A_{0} \), so that \(Tx\in B_{0} \) and \(Tz\in B_{0} \). Therefore, there exist \(u,v\in A_{0} \) such that
$$\nu_{u-Tx}(t)=\nu_{A-B}(t)=\nu_{v-Tz}(t)\quad (t>0). $$
Put \(y=iu+(1-i)v\in A \). Then
$$\begin{aligned} \nu_{y-T_{i}x}(t)&= \nu_{iu+(1-i)v-(1-i)Tz-iTx}(t) \\ &= \nu_{i(u-Tx)+(1-i)(v-Tz)}(t) \\ &\geq \Delta_{m}\bigl(\nu_{i(u-Tx)} (it ),\nu_{(1-i)(v-Tz)} \bigl((1-i)t \bigr)\bigr) \\ &= \Delta_{m}\bigl( \nu_{u-Tx}(t),\nu_{v-Tz}(t)\bigr) \\ &= \Delta_{m}\bigl(\nu_{A-B}(t),\nu_{A-B}(t)\bigr)= \nu_{A-B}(t) \quad (t>0), \end{aligned}$$
and thus \(T_{i}(A_{0})\subseteq B_{0} \). By Corollary 2.3 there exists a unique \(x_{i}\in A_{0} \) such that \(\nu_{gx_{i}-T_{i}x_{i}}(t)=\nu _{A-B}(t) \) for all \(t>0 \). Fix \(j\in(0,1) \). Then
$$\begin{aligned} \nu_{gx_{i}-Tx_{i}}(t) & \geq\Delta_{m}\bigl(\nu_{gx_{i}-T_{i}x_{i}}(jt), \nu _{T_{i}x_{i}-Tx_{i}}\bigl((1-j)t\bigr)\bigr) \\ & =\Delta_{m}\bigl(\nu_{A-B}(jt),\nu_{(1-i)(Tz-Tx_{i})} \bigl((1-j)t\bigr)\bigr) \\ &= \Delta_{m} \biggl(\nu_{A-B}(jt),\nu_{Tz-Tx_{i}} \biggl( \frac {(1-j)t}{1-i} \biggr) \biggr) \\ &\geq\Delta_{m} \biggl(\nu_{A-B}(jt),D_{B} \biggl( \frac{(1-j)t}{1-i} \biggr) \biggr)\quad (t>0). \end{aligned}$$
Now letting \(i\rightarrow1 \), we obtain
$$\lim_{i\rightarrow1}\nu_{gx_{i}-Tx_{i}}(t)\geq\Delta_{m} \bigl(\nu_{A-B}(jt),1\bigr)=\nu _{A-B}(jt)\quad \bigl(\forall j \in(0,1), t>0\bigr). $$
Then letting \(j\rightarrow1 \), we have
$$\lim_{i\rightarrow1}\nu_{gx_{i}-Tx_{i}}(t)=\nu_{A-B}(t) \quad (t>0). $$
So we can create a sequence \((x_{n})\) in \(A_{0}\) such that
$$\nu_{gx_{n}-Tx_{n}}(t)\rightarrow\nu_{A-B}(t) \quad (t>0). $$
Since \(A_{0}\) is compact, the sequence \((x_{n})\) has a subsequence \((x_{n_{k}})\) such that \(x_{n_{k}}\rightarrow x\in A_{0}\). By Remark 1.5, g is continuous mapping, and so \(g-T\) is a continuous mapping by Remark 1.8. Indeed, since \(\Delta_{m}\) is a continuous t-norm, \(p\rightarrow \nu_{P}\) is continuous ([27], Chapter 12), and we get
$$\nu_{gx -Tx }(t)=\lim_{k\rightarrow\infty}\nu _{gx_{n_{k}}-Tx_{n_{k}}}(t)= \nu_{A-B}(t), $$
as required. □
Theorem 2.5
Let
\((X, F,\Delta) \)
be a complete probabilistic Menger space such that Δ is a t-norm of H-type, and
\((A,B) \)
be a pair of subsets of
X
with the weak
P-property such that
\(A_{0}\)
is a nonempty closed set. If
\(T:A\rightarrow B \)
is a contraction mapping such that
\(T(A_{0} )\subseteq B_{0}\), then there exists a unique
x
in
A
such that
\(F_{x,Tx}(t)=F_{A,B}(t) \)
for all
\(t>0 \).
Proof
It is a direct consequence of Remark 1.20 and Lemma 2.1. □
Clearly, the pair \((A, A) \) has the P-property, so we have the following result.
Corollary 2.6
Let
\((X, F,\Delta) \)
be a complete probabilistic Menger space such that Δ is a t-norm of H-type. Then every contraction self-mapping from each nonempty closed subset of
X
has a unique fixed point.
Theorem 2.7
Let
\((X, \nu,\Delta_{m}) \)
be a probabilistic Banach space, and
\((A,B) \)
be a semisharp proximinal pair of
X
such that
A
is a
p-star-shaped set, \(A_{0}\)
be a nonempty compact set, B
be a
q-star-shaped set, and let
\(\nu_{p - q}(t)=\nu_{A-B}(t)\)
for all
\(t>0\). If
\(T : A\rightarrow B \)
is a proximal nonexpansive mapping such that
\(T(A_{0} )\subseteq B_{0}\), then there exists an element
\(x\in A_{0} \)
such that
\(\nu_{x - Tx}(t)=\nu_{A-B}(t)\)
for all
\(t>0\).
Proof
For each integer \(i\geq1 \), define \(T_{i}:A_{0}\rightarrow B_{0} \) by
$$T_{i}(x)= \biggl(1-\frac{1}{i} \biggr) Tx +\frac{1}{i}q \quad ( x\in A_{0}). $$
Then by the hypothesis we have \(T_{i}(A_{0} )\subseteq B_{0}\). Next, we show that for each i, \(T_{i} \) is a proximal contraction with \(\alpha=1-\frac{1}{i}< 1 \). To do this, suppose that \(x,y,u,v,s,r\in A_{0} \) and \(t>0\) are such that
$$\nu_{u-T_{i}x}(t)=\nu_{v-T_{i}y}(t)=\nu_{A_{0}-B_{0}}(t)= \nu_{A-B}(t)=\nu _{s-Tx}(t)=\nu_{r-Ty}(t). $$
Now we define
$$u'= \biggl(1-\frac{1}{i} \biggr) s +\frac{1}{i}p\in A_{0},\qquad v'= \biggl(1-\frac{1}{i} \biggr) r + \frac{1}{i}p\in A_{0}, $$
so we have
$$\begin{aligned} \nu_{A-B}(t)&\geq\nu_{u'-T_{i}x}(t) =\nu_{ (1-\frac{1}{i} ) s +\frac{1}{i}p- (1-\frac{1}{i} ) Tx -\frac{1}{i}q}(t) \\ & =\nu_{ (1-\frac{1}{i} )(s-Tx)+\frac{1}{i}(p-q)}(t) \\ & \geq\Delta_{m} \biggl(\nu_{ (1-\frac{1}{i} )(s-Tx)} \biggl( t \biggl(1- \frac{1}{i} \biggr) \biggr) ,\nu_{\frac{1}{i}(p-q)} \biggl( t \biggl( \frac {1}{i} \biggr) \biggr) \biggr) \\ & =\Delta_{m}\bigl( \nu_{s-Tx}(t),\nu_{p-q}(t) \bigr) \\ & =\Delta_{m}\bigl(\nu_{A-B}(t),\nu_{A-B}(t)\bigr)= \nu_{A-B}(t). \end{aligned}$$
Hence, \(\nu_{u'-T_{i}x}(t)= \nu_{A-B}(t)\). Since \(\nu _{u-T_{i}x}(t)= \nu_{A-B}(t)\) and \((A,B) \) is a semisharp proximinal pair, we have \(u'=u\). By the same method we also have \(v'=v \). Since T is a proximal nonexpansive mapping, we have
$$\begin{aligned} \nu_{u-v}(t)&=\nu_{u'-v'}(t) =\nu_{ (1-\frac{1}{i} )(s-r)}(t) \\ &= \nu_{s-r} \biggl(\frac{t}{1-\frac{1}{i}} \biggr)\geq\nu_{x-y} \biggl(\frac {t}{1-\frac{1}{i}} \biggr). \end{aligned}$$
Therefore, \(T_{i} \) is a proximal contraction with \(\alpha=1-\frac {1}{i}< 1 \). By Lemma 2.1, for each \(i\geq1 \), there exists a unique \(u_{i}\in A_{0} \) such that \(\nu_{u_{i}-T_{i}u_{i}}(t)=\nu _{A_{0}-B_{0}}(t)= \nu_{A-B}(t)\). Since \(A_{0} \) is compact and \(( u_{i})\subseteq A_{0} \), without loss of generality, we can assume that \(u_{i} \) is a convergent sequence and \(u_{i}\rightarrow x \in A_{0} \).
For each \(i\geq1\), since \(T(u_{i})\in T(A_{0})\subseteq B_{0}\), there exists \(v_{i}\in A_{0} \) such that \(\nu_{v_{i}-Tu_{i}}(t)= \nu _{A-B}(t)\). So we have
$$\begin{aligned} \nu_{A-B}(t)&\geq \nu_{ (1-\frac{1}{i} )v_{i} +\frac {1}{i}p-T_{i}u_{i}}(t) \\ &= \nu_{ (1-\frac{1}{i} ) v_{i} +\frac{1}{i}p- (1-\frac {1}{i} ) Tu_{i} -\frac{1}{i}q}(t) \\ &\geq\Delta_{m} \biggl(\nu_{(1-\frac{1}{i})( v_{i}- Tu_{i})}\biggl( t\biggl(1- \frac {1}{i}\biggr)\biggr) ,\nu_{\frac{1}{i}(p-q)}\biggl( t\biggl( \frac{1}{i}\biggr) \biggr) \biggr) \\ &=\Delta_{m}\bigl( \nu_{v_{i}- Tu_{i}}(t),\nu_{p-q}(t)\bigr) \\ &= \Delta_{m}\bigl(\nu_{A-B}(t), \nu_{A-B}(t) \bigr)=\nu_{A-B}(t). \end{aligned}$$
Thus, \(\nu_{A-B}(t)= \nu_{ (1-\frac{1}{i} )v_{i} +\frac {1}{i}p-T_{i}u_{i}}(t)\). Since \((A,B) \) is a semisharp proximinal pair and \(\nu_{A-B}(t)=\nu _{u_{i}-T_{i}u_{i}}(t) \), we have \(u_{i}=(1-\frac{1}{i}) v_{i} +\frac{1}{i}p\), and so
$$\nu_{u_{i}-v_{i}}(t)=\nu_{\frac{1}{i}(v_{i}-p)}(t)=\nu_{v_{i}-p}(it) . $$
Since \(A_{0} \) is compact and \(( v_{i})\subseteq A_{0} \), without loss of generality, we can assume that \(v_{i} \) is a convergent sequence and \(v_{i}\rightarrow z \in A_{0} \). For every \(j\leq i\), we have
$$\nu_{u_{i}-v_{i}}(t)=\nu_{v_{i}-p}(it)\geq\nu_{v_{i}-p}(jt)\geq \Delta _{m} \biggl(\nu_{v_{i}-z}\biggl(\frac{j}{2}t \biggr),\nu_{z-p}\biggl(\frac{j}{2}t\biggr)\biggr). $$
Letting \(i\rightarrow\infty\), we have
$$\lim_{i\rightarrow\infty}\nu_{u_{i}-v_{i}}(t)\geq\nu_{z-p} \biggl(\frac {j}{2}t\biggr) \quad (\forall j\geq1). $$
Now letting \(j \rightarrow\infty\), we have
$$\lim_{i\rightarrow\infty}\nu_{u_{i}-v_{i}}(t)\geq\lim_{j \rightarrow\infty} \nu_{z-p}\biggl(\frac{j}{2}t\biggr)=1. $$
Therefore, \(\nu_{u_{i}-v_{i}}(t)\rightarrow1 \), so that \(z=\lim_{i\rightarrow\infty}v_{i}= \lim_{i\rightarrow\infty}u_{i}=x\). Since \(Tx\in B_{0} \), there must exist \(u\in A_{0} \) such that \(\nu _{A-B}(t)=\nu_{u-Tx}(t)\). Since we know that \(\nu_{A-B}(t)=\nu _{v_{i}-Tu_{i}}(t)\) and T is a proximal nonexpansive mapping, it follows that \(\nu _{v_{i}-u}(t)\geq\nu_{u_{i}-x}(t) \rightarrow1\). This implies that \(u= \lim_{i\rightarrow\infty}v_{i}=x\) and then \(\nu_{A-B}(t)=\nu _{x-Tx}(t) \), as required. □
Theorem 2.8
Let
\((X,\nu,\Delta_{m}) \)
be a probabilistic Banach space, \((A,B) \)
be a semisharp proximinal pair of
X
with the weak
P-property such that
A
is a
p-star-shaped set, \(A_{0} \)
be a nonempty compact set, B
be a
q-star-shaped set, and let
\(\nu_{p - q}(t)=\nu_{A-B}(t)\)
for all
\(t>0\). If
\(T:A\rightarrow B \)
is a nonexpansive mapping such that
\(T(A_{0} )\subseteq B_{0}\), then
T
has a best proximity point in
\(A_{0} \).
Proof
It is a direct consequence of Remark 1.20 and Theorem 2.7. □
Proposition 2.9
Let
\((X, F,\Delta) \)
be a probabilistic Menger space, and
\(A,B \subseteq X \)
be such that
\(A_{0} \)
is a nonempty set. Suppose that
\(T : A\rightarrow B \)
is a proximal nonexpansive mapping such that
\(T(A_{0} )\subseteq B_{0}\)
and
\(g : A\rightarrow A \)
is an isometry mapping such that
\(A_{0}\subseteq g(A_{0}) \). Denote
\(G = g(A)\)
and
$$G_{0} =\bigl\lbrace z\in G : \exists y\in B \textit{ s.t. } \forall t \geq0, F_{z,y}(t)=F_{G,B}(t)\bigr\rbrace . $$
Then
\(Tg^{-1}\)
is a proximal nonexpansive, and
\(G_{0}=A_{0} \).
Proof
The result follows by using a similar argument as in the proof of Proposition 2.2. □
The following theorem is an immediate consequence of Theorem 2.7 and Proposition 2.9.
Theorem 2.10
Let
\((X, \nu,\Delta_{m}) \)
be a probabilistic Banach space, \((A,B) \)
be a semisharp proximinal pair of
X
such that
A
is a
p-star-shaped set, \(A_{0}\)
be a nonempty compact set, B
be a
q-star-shaped set, and let
\(\nu_{p - q}(t)=\nu_{A-B}(t)\)
for all
\(t>0\). If
\(T : A\rightarrow B \)
is a proximal nonexpansive mapping such that
\(T(A_{0} )\subseteq B_{0}\)
and
\(g : A\rightarrow A \)
is an isometry mapping such that
\(A_{0}\subseteq g(A_{0}) \), then there exists an element
\(x\in A_{0} \)
such that
\(\nu_{gx - Tx}(t)=\nu_{A-B}(t)\)
for all
\(t>0\).
Corollary 2.11
Let
\((X, \nu,\Delta_{m}) \)
be a probabilistic Banach space, and let
\((A,B) \)
be a pair of convex subsets of
X
with the
P-property such that
\(A_{0}\)
is a nonempty compact set. If
\(T : A\rightarrow B \)
is a nonexpansive mapping such that
\(T(A_{0} )\subseteq B_{0}\)
and
\(g : A\rightarrow A \)
is an isometry mapping such that
\(A_{0}\subseteq g(A_{0}) \), then there exists an element
\(x\in A_{0} \)
such that
\(\nu _{gx-Tx}(t)=\nu_{A-B}(t) \)
for all
\(t>0\).
In Corollary 2.11, if \(g(x)=x\), then we have the following corollary.
Corollary 2.12
With the hypotheses of the previous corollary, if
\(T:A\rightarrow B \)
is a nonexpansive mapping such that
\(T(A_{0} )\subseteq B_{0}\), then
T
has a best proximity point.
In Corollary 2.12, if \(A=B\), then we have the following corollary.
Corollary 2.13
If
A
is a nonempty, compact, and convex subset of a probabilistic Banach space
\((X, \nu,\Delta_{m}) \)
and
\(T:A\rightarrow A \)
is a nonexpansive mapping, then
T
has a fixed point.
In the following, we give some examples that defend our main results.
Example 2.14
Let \(X=\mathbb{R}^{2} \), \(A=\lbrace(0,y) : y\in\mathbb{R}\rbrace\) and \(B=\lbrace(1,y): y\in\mathbb{R}\rbrace\). Suppose that \(T :A\rightarrow B\) is defined by \(T(0,y)= (1,\frac {y}{4} ) \), \(g :A\rightarrow A\) is defined by \(g(0,y)=(0,-y) \), and \(F_{(x,x'),(y,y')}(t)=\frac{t}{t+|x-y|+|x'-y'|} \). It is easy to see that \((X,F,\Delta_{m})\) is a complete probabilistic Menger space, \(F_{A,B}(t)=\frac{t}{t+1} \), \(A_{0} =A\), \(B_{0}=B \), \(T(A_{0})\subseteq B_{0} \), and
$$F_{g(0,x),g(0,y)}(t)=F_{(0,-x),(0,-y)}(t)=\frac{t}{t+|x-y|}=F_{(0,x),(0,y)}(t). $$
If \((0,u),(0,x), (0,v),(0,y)\in A \) are such that
$$\frac{t}{t+1+|u-\frac{x}{4}|} = F_{(0,u),T(0,x)}(t)=F_{A,B}(t)= F_{(0,v),T(0,y)}(t)=\frac{t}{t+1+|v-\frac{y}{4}|}, $$
then \(u= \frac{x}{4}\) and \(v= \frac{y}{4}\), so that
$$F_{(0,u),(0,v)}(t)=F_{(0,\frac{x}{4}),(0, \frac{y}{4})}(t)=\frac {t}{t+\frac{1}{4}|x-y|}=F_{(0,x),(0,y)} \biggl(\frac{t}{\frac{1}{4}} \biggr). $$
Therefore, all the hypothesis of Corollary 2.3 are satisfied, and we also have
$$F_{(0,0),T(0,0)}(t)=F_{(0,0),(1,0)}(t)=\frac{t}{t+1}=F_{A,B}(t). $$
Example 2.15
Let \(X=\mathbb{R} \), \(A=[0,2] \) and \(B=[3,5] \). For every \(x\in X \), define \(\nu_{x} (t)=\frac{t}{t+|x|}\). It is easy to see that \((X, \nu,\Delta_{m}) \) is a probabilistic Banach space, \(\nu_{A-B}(t)=\frac{t}{t+1} \), \(A_{0}=\lbrace2\rbrace\), and \(B_{0}=\lbrace3\rbrace\). For every \(x\in A \), define \(T:A\rightarrow B \) by \(Tx=5-x \) and let g be the identity mapping. Clearly, T is a continuous affine and proximal nonexpansive mapping, and \(T(A_{0})=\{T(2)\}=\{3\}=B_{0} \). Therefore, all the hypotheses of Theorem 2.4 are satisfied, and also we have
$$\nu_{2-T2}(t)=\nu_{2-3}(t)=\frac{t}{t+1}= \nu_{A-B}(t). $$
The following example shows that the weak P-property of the pair \((A, B) \) cannot be removed from Theorem 2.5.
Example 2.16
Let \(X =\mathbb{R} \), \(A =\lbrace-10,10\rbrace\), \(B = \lbrace-2, 2\rbrace\), and \(F_{p,q}(t)=\frac{t}{t+|p-q|} \). Clearly, \((X, F,\Delta_{m}) \) is a complete probabilistic Menger space. Then \(A_{0} = A \), \(B_{0} = B \), and \(F_{A,B}(t)=\frac{t}{t+8} \). Let \(T : A \rightarrow B\) be a mapping given by \(T (-10) = 2 \) and \(T (10) = -2 \). It is easy to see that for \(\alpha=\frac{1}{5} \), T is a contraction mapping with \(T (A_{0})\subseteq B_{0}\). The mapping T does not have any best proximity point because \(F_{x,Tx}(t)=\frac{t}{t+12} < \frac{t}{t+8}= F_{A,B}(t)\) for all \(x \in A \). It should be noted that the pair \((A, B) \) does not have the weak P-property.
Example 2.17
Let \(X=\mathbb{R} \), \(A=[0,1] \), and \(B=[2,3] \). For every \(x\in X \), define \(\nu_{x}(t)=\frac{t}{t+|x|} \). It is easy to see that \((X, \nu,\Delta_{m}) \) is a probabilistic Banach space, A is 1-star-shaped set, B is 2-star-shaped set,
$$\nu_{A-B}(t)=\sup_{x\in A, y\in B}\nu_{x-y}(t)= \frac{t}{t+1},\qquad A_{0}=\lbrace1\rbrace,\qquad B_{0}= \lbrace2\rbrace, $$
and
$$\nu_{1-2}(t)=\frac{t}{t+|1-2|}=\frac{t}{t+1}= \nu_{A-B}(t). $$
Also, \((A,B) \) is a semisharp proximinal pair. Now for each \(x\in A \), define \(T:A\rightarrow B \) by \(Tx=3-x \). If \(u,v,x,y\in A \), then
$$\frac{t}{t+|u-3+x|} =\nu_{u-Tx}(t)=\nu_{A-B}(t)= \nu_{v-Ty}(t)=\frac {t}{t+|v-3+y|}, $$
so that \(u= x=1 \) and \(v= y=1 \). Thus,
$$\nu_{u-v}(t)=1=\nu_{x-y}(t). $$
So T is a proximal nonexpansive, and \(T(A_{0})=B_{0}\). Therefore, all the hypotheses of Theorem 2.7 are satisfied, and we also have
$$\nu_{1-T1}(t)=\nu_{1-2}(t)=\frac{t}{t+1}= \nu_{A-B}(t). $$
Example 2.18
Let \(X=\mathbb{R}^{2} \), \(A=\lbrace(x,0) : 0\leq x \leq1\rbrace\), \(B_{1}=\lbrace(x,y): x+y=1, -1\leq x \leq0\rbrace\), \(B_{2}=\lbrace(x,1): 0\leq x \leq1\rbrace\), \(B=B_{1}\cup B_{2}\), and \(\nu_{(x,x')}(t)=\frac{t}{t+ |x|+|x'|} \). It is easy to see that \((X, \nu,\Delta_{m}) \) is a probabilistic Banach space, \(\nu_{A-B}(t)= \frac{t}{t+1} \), B is not convex but is a \((0,1) \)-star-shaped set, and A is \((0,0)\)-star-shaped set. Clearly, \(A_{0}=A\) and \(B_{0}=B_{2} \). So
$$\nu_{(0,0)-(0,1)}(t)=\frac{t}{t+ |0|+|1|}=\frac{t}{t+1}= \nu_{A-B}(t), $$
and \((A,B) \) is a semisharp proximinal pair. Suppose that \(T:A\rightarrow B \) is defined by
$$ T(x,0)= \left\{ \textstyle\begin{array}{l@{\quad}l} (0,1), & x= 0, \\ (\sin x,1), & x\neq0, \end{array}\displaystyle \right. $$
and \((u,0),(v,0),(x,0),(y,0)\in A \) are such that
$$\nu_{(u,0)-T(x,0)}(t)=\nu_{A-B}(t)=\frac{t}{t+1}= \nu_{(v,0)-T(y,0)}(t). $$
If \(x=y=0 \), then \(u=v=0 \), and therefore
$$\nu_{(u,0)-(v,0)}(t)=\nu_{(0,0)-(0,0)}(t)=1=\nu_{(x,0)-(y,0)}(t). $$
If \(x,y\neq0\), then \(u=\sin x\), \(v=\sin y \), and therefore
$$\begin{aligned} \nu_{(u,0)-(v,0)}(t)&=\nu_{(\sin x,0)-( \sin y,0)}(t)=\frac{t}{t+|\sin x-\sin y|} \\ &\geq\frac{t}{t+| x- y|} \\ &=\nu_{(x,0)-(y,0)}(t). \end{aligned}$$
If \(x=0\) and \(y\neq0\), then \(u=0 \) and \(v=\sin y \), and therefore
$$\nu_{(u,0)-(v,0)}(t)=\nu_{(0,0)-(\sin y,0)}(t)=\frac{t}{t+|\sin y|}\geq \frac{t}{t+| y|}\geq\nu_{(0,0)-(y,0)}(t). $$
If \(x\neq0\) and \(y= 0\), then \(u=\sin x \) and \(v=0 \), and therefore
$$\nu_{(u,0)-(v,0)}(t)=\nu_{(\sin x,0)-(0,0)}(t)=\frac{t}{t+|\sin x|}\geq \frac{t}{t+| x|}\geq\nu_{(x,0)-(0,0)}(t). $$
Hence, T is proximal nonexpansive, and \(T(A_{0})\subseteq B_{2}=B_{0}\), so all the hypotheses of Theorem 2.7 are satisfied, and we also have
$$\nu_{(0,0)-T(0,0)}(t)=\nu_{(0,0)-(0,1)}(t)=\frac{t}{t+1}= \nu_{A-B}(t). $$
Example 2.19
Let \(X=\mathbb{R} \), \(A=[0,1] \), \(B=[\frac{15}{8},2] \), and \(\nu _{x}(t)=\frac{t}{t+|x|} \). Clearly, \((X,\nu,\Delta_{m})\) is a probabilistic Banach space, \(\nu_{A-B}(t)=\frac{t}{t+\frac{7}{8}} \), the pair \((A,B)\) has the P-property, \(A_{0}=\lbrace1\rbrace\), and \(B_{0}=\lbrace\frac{15}{8}\rbrace\). If \(Tx=-\frac{1}{8}x+2 \), then \(T(A_{0})=\{T(1)\}=\{\frac{15}{8}\}=B_{0} \). Let \(x,y\in A\). Then we have
$$\nu_{Tx-Ty}(t)=\nu_{-\frac{1}{8}(x-y)}(t)=\nu_{x-y}(8t)\geq \nu_{x-y}(t). $$
Therefore, all the hypotheses of Corollary 2.12 are satisfied, and hence T has a best proximity point, and we also have
$$\nu_{1-T1}(t)=\nu_{1-\frac{15}{8}}(t)=\frac{t}{t+\frac{7}{8}}= \nu_{A-B}(t). $$