Now we prove our first main result.
Theorem 2.1
Suppose
A
and
B
are nonempty closed subsets of a complete metric space
X
with
\(A_{0}\neq\emptyset\). Let
\(T:A\to B\)
satisfy (1.3) together with the following assertions:
-
(i)
\(T(A_{0})\subseteq B_{0}\)
and
\((A,B)\)
satisfies the weak
P-property,
-
(ii)
T
is proximal
\(\alpha^{+}\)-admissible,
-
(iii)
there exist
\(x_{0}, x_{1} \in A_{0}\)
such that
$$d(x_{1},Tx_{0})=d(A,B) \quad \textit{and} \quad \alpha(x_{0},x_{1})\geq0, $$
-
(iv)
T
is continuous, or
-
(v)
A
is
α-regular, that is, if
\(\{x_{n}\}\)
is a sequence in
A
such that
\(\alpha(x_{n},x_{n+1})\geq0\)
and
\(x_{n}\to x\in A\)
as
\(n\to\infty\), then
\(\alpha(x_{n}, x)\geq0\)
for all
\(n\in\mathbb{N}\).
Then
\(\operatorname {Bpp}(T)\)
is nonempty.
Proof
Since \(T(A_{0})\subseteq B_{0}\), we have \(x_{2}\in A_{0}\) such that
$$d(x_{2},Tx_{1})=d(A,B). $$
As T satisfies (iii) and is proximal \(\alpha^{+}\)-admissible, we obtain \(\alpha(x_{1},x_{2})\geq0\). That is,
$$d(x_{2},Tx_{1})=d(A,B), \qquad\alpha(x_{1},x_{2}) \geq0. $$
Again, since \(T(A_{0})\subseteq B_{0}\), there exists \(x_{3}\in A_{0}\) such that
$$d(x_{3},Tx_{2})=d(A,B). $$
Thus we have
$$d(x_{2},Tx_{1})=d(A,B),\qquad d(x_{3},Tx_{2})=d(A,B), \qquad \alpha(x_{1},x_{2})\geq0. $$
Again since T is proximal \(\alpha^{+}\)-admissible, so \(\alpha(x_{2},x_{3})\geq0\). Hence,
$$d(x_{3},Tx_{2})=d(A,B),\qquad \alpha(x_{2},x_{3}) \geq0. $$
We continue this process, to get
$$ d(x_{n+1},Tx_{n})=d(A,B), \qquad \alpha(x_{n+1},x_{n})\geq 0 \quad \mbox{for all } n\in\mathbb{N} \cup\{0\}. $$
(2.1)
By using the above observations we can write
$$ \begin{aligned} \frac{1}{2}d^{*}(x_{n-1},Tx_{n-1})&= \frac {1}{2}\bigl[d(x_{n-1},Tx_{n-1})-d(A,B)\bigr] \\ &\leq\frac{1}{2}\bigl[d(x_{n-1},x_{n})+d(x_{n},Tx_{n-1})-d(A,B) \bigr] \\ &=\frac{1}{2}d(x_{n-1},x_{n}) \\ &\leq d(x_{n},x_{n-1}). \end{aligned} $$
That is,
$$\frac{1}{2}d^{*}(x_{n-1},Tx_{n-1})\leq d(x_{n},x_{n-1}). $$
Now from (1.3) we get
$$ d(Tx_{n-1},Tx_{n})\leq\alpha(x_{n-1},x_{n})+d(Tx_{n-1},Tx_{n}) \leq \psi\bigl(M(x_{n-1},x_{n})\bigr). $$
(2.2)
By a simple calculation we obtain (see for details [2, 5]),
$$ \begin{aligned}[b] M(x_{n-1},x_{n})={}& \max \biggl\{ d(x_{n-1},x_{n}),\frac {d(x_{n-1},Tx_{n-1})+d(x_{n},Tx_{n})}{2}-d(A,B), \\ &{} \frac{d(x_{n-1},Tx_{n})+d(x_{n},Tx_{n-1})}{2}-d(A,B) \biggr\} \\ \leq{}& \max\bigl\{ d(x_{n-1},x_{n}),d(x_{n},x_{n+1}) \bigr\} . \end{aligned} $$
(2.3)
By the weak P-property and (2.1) one obtains
$$d(x_{n},x_{n+1})\leq d(Tx_{n-1},Tx_{n}) \quad \mbox{for all } n\in\mathbb{N}. $$
Equations (2.2) and (2.3) imply that
$$ d(x_{n},x_{n+1})\leq\psi\bigl(M(x_{n-1},x_{n}) \bigr)\leq \psi\bigl(\max\bigl\{ d(x_{n-1},x_{n}),d(x_{n},x_{n+1}) \bigr\} \bigr)\quad \mbox{for all } n\in\mathbb{N}. $$
(2.4)
If \(x_{n_{0}}=x_{n_{0}+1}\) for some \(n_{0}\in \mathbb{N}\), from (2.1) one obtains
$$d(x_{n_{0}}, Tx_{n_{0}})=d(x_{n_{0}+1}, Tx_{n_{0}})=d(A,B), $$
that is, \(x_{n_{0}}\in \operatorname {Bpp}(T)\). Thus, we suppose that
$$ d(x_{n+1},x_{n})>0\quad \mbox{for all } n\in \mathbb{N}\cup\{0\}. $$
(2.5)
If, \(\max\{d(x_{n-1},x_{n}),d(x_{n},x_{n+1})\}=d(x_{n},x_{n+1})\), then (2.4) implies
$$ d(x_{n},x_{n+1})\leq\psi\bigl(d(x_{n},x_{n+1}) \bigr)< d(x_{n},x_{n+1}), $$
which is a contradiction. Thus,
$$ d(x_{n},x_{n+1})\leq \psi\bigl(d(x_{n-1},x_{n}) \bigr)< d(x_{n-1},x_{n}) \quad \mbox{for all } n\in \mathbb{N}. $$
(2.6)
Applying the monotonicity of ψ, by induction, it follows from (2.6),
$$d(x_{n},x_{n+1})\leq\psi^{n}\bigl(d(x_{1},x_{0}) \bigr)\quad \mbox{for all } n\in \mathbb{N}\cup\{0\}. $$
Suppose ϵ is any positive real number. Then there exists \(N\in\mathbb{N}\) such that
$$\sum_{n\geq N}\psi^{n}\bigl(d(x_{0},x_{1}) \bigr)< \epsilon\quad \text{for all } n \in\mathbb{N}. $$
If \(m,n\in\mathbb{N}\) with \(m>n\geq N\). We apply the triangle inequality to get
$$d(x_{n},x_{m})\leq\sum_{k=n}^{m-1}d(x_{k},x_{k+1}) \leq\sum_{k=n}^{m-1}\psi^{k} \bigl(d(x_{0},x_{1})\bigr) < \sum _{n\geq N}\psi^{n}\bigl(d(x_{0},x_{1}) \bigr)< \epsilon. $$
Consequently \(\lim_{m,n,\to+\infty}d(x_{n},x_{m})=0\), which implies \(\{x_{n}\}\) is Cauchy sequence. By completeness of X, \(x_{n}\to z\in X\). If (iv) holds, then \(Tx_{n}\to Tz\) as \(n\to\infty\) and
$$d(A,B)=\lim_{n\to\infty}d(x_{n+1},Tx_{n})=d(z,Tz), $$
as required. Next, assume that (v) holds. Then \(\alpha(x_{n},z)\geq0\).
If the following inequalities hold:
$$ \frac{1}{2}d^{*}(x_{n},Tx_{n})> d(x_{n},z) \quad \text{and}\quad \frac{1}{2}d^{*}(x_{n+1},Tx_{n+1})> d(x_{n+1},z), $$
for some \(n\in\mathbb{N}\), then by using (2.6) and definition of \(d^{*}\), we obtain the following contradiction:
$$\begin{aligned} d(x_{n},x_{n+1}) \leq& d(x_{n},z)+d(x_{n+1},z) \\ < &\frac{1}{2}\bigl[d^{*}(x_{n},Tx_{n})+d^{*}(x_{n+1},Tx_{n+1}) \bigr] \\ \leq&\frac {1}{2}\bigl[d(x_{n},Tx_{n})+d(x_{n+1},Tx_{n+1})-2d(A,B) \bigr] \\ =&d(x_{n},x_{n+1}). \end{aligned}$$
Consequently,, for any \(n\in\mathbb{N}\), either
$$ \frac{1}{2}d^{*}(x_{n},Tx_{n})\leq d(x_{n},z) \quad \text{or}\quad \frac{1}{2}d^{*}(x_{n+1},Tx_{n+1}) \leq d(x_{n+1},z) $$
holds. Thus, we may pick a subsequence \(\{x_{n_{k}}\}\) of \(\{x_{n}\}\) such that
$$\frac{1}{2}d^{*}(x_{n_{k}},Tx_{n_{k}})\leq d(x_{n_{k}},z)\quad \mbox{and}\quad \alpha (x_{n_{k}},x_{n_{k}+1}) \geq0 $$
for all \(k\in\mathbb{N}\). By (1.3) we get
$$ d(Tx_{n_{k}},Tz)\leq\psi\bigl(M(x_{n_{k}},z)\bigr). $$
(2.7)
Notice that
$$\begin{aligned} M(x_{n_{k}},z) =&\max \biggl\{ d(x_{n_{k}},z), \frac {d(x_{n_{k}},Tx_{n_{k}})+d(z,Tz)}{2}-d(A,B), \\ &{}\frac{d(x_{n_{k}},Tz)+d(z,Tx_{n_{k}})}{2}-d(A,B) \biggr\} \\ \leq&\max \biggl\{ d(x_{n_{k}},z), \frac{d(x_{n_{k}},x_{n_{k}+1})+d(x_{n_{k}+1},Tx_{n_{k}})+d(z,Tz)}{2}-d(A,B), \\ &{}\frac {d(x_{n_{k}},z)+d(z,Tz)+d(z,x_{n_{k}+1})+d(x_{n_{k}+1},Tx_{n_{k}})}{2}-d(A,B) \biggr\} \\ =&\max \biggl\{ d(x_{n_{k}},z), \frac{d(x_{n_{k}},x_{n_{k}+1})+d(A,B)+d(z,Tz)}{2}-d(A,B), \\ &{}\frac {d(x_{n_{k}},z)+d(z,Tz)+d(z,x_{n_{k}+1})+d(A,B)}{2}-d(A,B) \biggr\} , \end{aligned}$$
which implies
$$\begin{aligned} \lim_{k\to\infty}M(x_{n_{k}},z) \leq \frac{d(z,Tz)-d(A,B)}{2}. \end{aligned}$$
(2.8)
Further,
$$\begin{aligned} d(z,Tz) &\leq d(z,x_{n_{k}+1})+d(x_{n_{k}+1},Tx_{n_{k}})+d(Tx_{n_{k}},Tz) \\ &\leq d(z,x_{n_{k}+1})+d(A,B)+d(Tx_{n_{k}},Tz), \end{aligned}$$
which gives
$$\begin{aligned} d(z,Tz)-d(z,x_{n_{k}+1})-d(A,B)\leq d(Tx_{n_{k}},Tz). \end{aligned}$$
(2.9)
As \(k\to\infty\) in (2.9) we deduce
$$\begin{aligned} d(z,Tz)-d(A,B)\leq\lim_{k\to\infty}d(Tx_{n_{k}},Tz). \end{aligned}$$
(2.10)
Therefore from (2.7), (2.8), and (2.10)
$$\begin{aligned} d(z,Tz)-d(A,B)&\leq\lim_{k\to\infty}d(Tx_{n_{k}},Tz) \\ &\leq \psi\Bigl(\lim_{k\to\infty}M(x_{n_{k}},z)\Bigr)\leq \psi \biggl(\frac{d(z,Tz)-d(A,B)}{2}\biggr). \end{aligned}$$
(2.11)
Now, if \(d(z,Tz)-d(A,B)>0\), then we get
$$ d(z,Tz)-d(A,B)\leq \psi\biggl(\frac{d(z,Tz)-d(A,B)}{2}\biggr)< \frac{d(z,Tz)-d(A,B)}{2}, $$
(2.12)
a contradiction. Hence, \(d(z,Tz)=d(A,B)\) as desired. □
Example 2.1
Suppose \(X=\mathbb{R}^{2}\) is equipped with the metric
$$d \bigl((p_{1},p_{2}),(q_{1},q_{2}) \bigr) = \vert p_{1} - q_{1}\vert +\vert p_{2} - q_{2}\vert , $$
for all \((p_{1},p_{2}),(q_{1},q_{2}) \in X\). Let \(A_{1}=\{(p,q) | p=1, 0\leq q\leq\frac{1}{2}\} \), \(A_{2}=\{(p,q) |p=4, q\geq5\}\), \(A_{3}=\{(p,q) p=5, q\geq4\}\), \(A_{4}=\{(p,q) | p=3,q\geq3\}\) and \(A=A_{1}\cup A_{2}\cup A_{3}\cup A_{4}\). Further define \(B_{1}=\{(p,q)| p= \frac{1}{2}, \frac{1}{2}\leq q\leq1\}\), \(B_{2}=\{(p,q)| p=0, q\leq4\}\), \(B_{3}=\{(P,q)| p=4, q\leq0\}\), and \(B=B_{1}\cup B_{2}\cup B_{3}\).
Note that \(d(A,B)=1\), \(A_{0}=\{(p,q) | p=1, 0\leq q\leq \frac{1}{2}\}\), and \(B_{0}=\{(p,q)| p= \frac{1}{2}, \frac{1}{2}\leq q\leq1\}\). Let, for \(x_{1}=(1, u_{1}), x_{2}=(1, u_{2})\in A_{0}\) and \(y_{1}=(\frac{1}{2}, v_{1}), y_{2}=(\frac{1}{2}, v_{2})\in B_{0}\), us have \(d(x_{1},y_{1})=d(A,B)=1\) and \(d(x_{2},y_{2})=d(A,B)=1\). Then
$$\frac{1}{2}+\vert u_{1}-v_{1}\vert =1 $$
and
$$\frac{1}{2}+\vert u_{2}-v_{2}\vert =1, $$
and so \(\vert u_{1}-v_{1}\vert =\frac{1}{2}\) and \(\vert u_{2}-v_{2}\vert =\frac{1}{2}\). Since \(v_{1},v_{2}\geq u_{1}, u_{2}\), we have \(v_{1}=\frac{1}{2}+u_{1}\) and \(v_{2}=\frac{1}{2}+u_{2}\). This shows that \(d(x_{1},x_{2})\leq d(y_{1},y_{2})\). So \((A,B)\) satisfy the weak P-property. Let \(T:A\to B\) be defined by
$$T(p_{1},p_{2})= \textstyle\begin{cases} (\frac{1}{2},\frac{1}{2}) & \text{if } p_{1}= p_{2} , \\ (p_{1},0) & \text{if } p_{1}< p_{2} , \\ (0,p_{2}) & \text{if } p_{1}>p_{2}. \end{cases} $$
Notice that \(T(A_{0})\subseteq B_{0}\).
Define the functions \(\psi:[0,+\infty)\rightarrow[0,+\infty)\) and \(\alpha: A\times A\to[-\infty,\infty)\) by
$$\psi(t)=\frac{8}{9}t\quad \mbox{and} \quad \alpha(p,q)= \textstyle\begin{cases} 0, & \mbox{if } p,q\in\{(1,0),(4,5),(5,4)\},\\ -\infty, & \mbox{otherwise.} \end{cases} $$
Assume that \(\frac{1}{2}d^{*}(p,Tp)\leq d(p,q)\) and \(\alpha(p,q)\geq0\), for \(p,q\in A\). Then
$$\textstyle\begin{cases} p=(1,0),\qquad q=(4,5) &\mbox{or} \\ p=(1,0),\qquad q=(5,4) & \mbox{or} \\ q=(1,0),\qquad p=(4,5) & \mbox{or} \\ q=(1,0),\qquad p=(5,4). & \end{cases} $$
Since \(d(Tp,Tq)=d(Tq,Tp)\) and \(M(p,q)=M(q,p)\) for all \(p,q\in A\), we can suppose that
$$(p,q)=\bigl((1,0),(4,5)\bigr) \quad \mbox{or}\quad (p,q)=\bigl((1,0),(5,4)\bigr). $$
Now, we discuss the following cases:
-
(i)
if \((p,q)=((1,0),(4,5))\), then
$$d\bigl(T(1,0),T(4,5)\bigr)=4\leq7=\frac{7}{8} \cdot8=\psi\bigl(d \bigl((1,0),d(4,5)\bigr)\bigr)\leq \psi\bigl(M(p,q)\bigr); $$
-
(ii)
if \((p,q)=((1,0),(5,4))\), then
$$d\bigl(T(1,0),T(5,4)\bigr)=4\leq\frac{7}{8} \cdot8=\psi\bigl(d \bigl((1,0),(5,4)\bigr)\bigr)\leq\psi \bigl(M(p,q)\bigr). $$
Consequently, we have
$$ \frac{1}{2}d^{*}(p,Tp)\leq d(p,q)\quad \Rightarrow\quad d(Tp,Tq)\leq \psi \bigl(M(p,q)\bigr). $$
Thus all the assumptions of Theorem 2.1 are satisfied and \(\operatorname {Bpp}(T)=\{(1,0)\}\).
The next result can be deduced easily from Theorem 2.1.
Theorem 2.2
Let
X, A, \(A_{0}\), and
B
be as in Theorem
2.1. Assume that
\(T:A\to B\)
satisfies the assertions (i)-(v) in Theorem
2.1
and
$$ \alpha(p,q)+d(Tp,Tq)\leq\psi\bigl(M(p,q)\bigr) $$
holds for all
\(p,q\in A\). Then
\(\operatorname {Bpp}(T)\)
is nonempty.
If \(\alpha=0\) on A, in Theorem 2.1, we obtain the following new result.
Corollary 2.1
Suppose
X, A, \(A_{0}\), and
B
are as in Theorem
2.1
and
\(T:A\to B\)
satisfies the following assumptions:
-
(i)
\(T(A_{0})\subseteq B_{0}\)
and
\((A,B)\)
satisfies the weak
P-property,
-
(ii)
for all
\(p,q\in A\)
with
\(\frac{1}{2}d^{*}(p,Tp)\leq d(p,q)\)
we have
$$ d(Tp,Tq)\leq\psi\bigl(M(p,q)\bigr). $$
Then
\(\operatorname {Bpp}(T)\)
is nonempty.