A function α: [0,\mathrm{\infty})\to [0,1) is said to be an \mathcal{MT}function (or ℛfunction) [12, 25–30] if {lim\hspace{0.17em}sup}_{s\to {t}^{+}}\alpha (s)<1 for all t\in [0,\mathrm{\infty}). It is obvious that if α: [0,\mathrm{\infty})\to [0,1) is a nondecreasing function or a nonincreasing function, then α is an \mathcal{MT}function. So the set of \mathcal{MT}functions is a rich class.
Recently, Du [28] first proved the following characterizations of \mathcal{MT}functions which are quite useful for proving our main results.
Theorem D ([[28], Theorem 2.1])
Let \alpha :[0,\mathrm{\infty})\to [0,1) be a function. Then the following statements are equivalent.

(a)
α is an \mathcal{MT}function.

(b)
For each t\in [0,\mathrm{\infty}), there exist {r}_{t}^{(1)}\in [0,1) and {\epsilon}_{t}^{(1)}>0 such that \alpha (s)\le {r}_{t}^{(1)} for all s\in (t,t+{\epsilon}_{t}^{(1)}).

(c)
For each t\in [0,\mathrm{\infty}), there exist {r}_{t}^{(2)}\in [0,1) and {\epsilon}_{t}^{(2)}>0 such that \alpha (s)\le {r}_{t}^{(2)} for all s\in [t,t+{\epsilon}_{t}^{(2)}].

(d)
For each t\in [0,\mathrm{\infty}), there exist {r}_{t}^{(3)}\in [0,1) and {\epsilon}_{t}^{(3)}>0 such that \alpha (s)\le {r}_{t}^{(3)} for all s\in (t,t+{\epsilon}_{t}^{(3)}].

(e)
For each t\in [0,\mathrm{\infty}), there exist {r}_{t}^{(4)}\in [0,1) and {\epsilon}_{t}^{(4)}>0 such that \alpha (s)\le {r}_{t}^{(4)} for all s\in [t,t+{\epsilon}_{t}^{(4)}).

(f)
For any nonincreasing sequence {\{{x}_{n}\}}_{n\in \mathbb{N}} in [0,\mathrm{\infty}), we have 0\le {sup}_{n\in \mathbb{N}}\alpha ({x}_{n})<1.

(g)
α is a function of contractive factor; that is, for any strictly decreasing sequence {\{{x}_{n}\}}_{n\in \mathbb{N}} in [0,\mathrm{\infty}), we have 0\le {sup}_{n\in \mathbb{N}}\alpha ({x}_{n})<1.
Let us recall the concept of \mathcal{MT}cyclic contractions introduced first by Du and Lakzian [12].
Definition 3.1 [12]
Let A and B be nonempty subsets of a metric space (X,d). If a map T:A\cup B\to A\cup B satisfies

(MT1)
T(A)\subseteq B and T(B)\subseteq A;

(MT2)
there exists an \mathcal{MT}function α: [0,\mathrm{\infty})\to [0,1) such that
d(Tx,Ty)\le \alpha (d(x,y))d(x,y)+(1\alpha (d(x,y)))dist(A,B)\phantom{\rule{1em}{0ex}}\text{for any}x\in A\text{and}y\in B,
then T is called an \mathcal{MT}cyclic contraction with respect to α on A\cup B.
The following example shows that there exists an \mathcal{MT}cyclic contraction which is not a cyclic contraction.
Example 3.1 [12]
Let X=\{{v}_{1},{v}_{2},{v}_{3},\dots \} be a countable set and \{{\tau}_{n}\} be a strictly increasing convergent sequence of positive real numbers. Denote by {\tau}_{\mathrm{\infty}}:={lim}_{n\to \mathrm{\infty}}{\tau}_{n}. Then {\tau}_{2}<{\tau}_{\mathrm{\infty}}. Let d:X\times X\to [0,\mathrm{\infty}) be defined by d({v}_{n},{v}_{n})=0 for all n\in \mathbb{N} and d({v}_{n},{v}_{m})=d({v}_{m},{v}_{n})={\tau}_{m} if m>n. Then d is a metric on X. Set A=\{{v}_{1},{v}_{3},{v}_{5},\dots \}, B=\{{v}_{2},{v}_{4},{v}_{6},\dots \}. Define a map T:A\cup B\to A\cup B by
T{v}_{n}\stackrel{\mathrm{def}}{=}\{\begin{array}{ll}{v}_{2}& \text{if}n=1,\\ {v}_{n1}& \text{if}n1\end{array}
for n\in \mathbb{N} and define \alpha :[0,\mathrm{\infty})\to [0,1) by
\alpha (t)\stackrel{\mathrm{def}}{=}\{\begin{array}{ll}\frac{{\tau}_{n1}}{{\tau}_{n}}& \text{if}t={\tau}_{n}\text{for some}n\in \mathbb{N}\text{with}n2,\\ 0& \text{otherwise}.\end{array}
Then T is an \mathcal{MT}cyclic contraction with respect to α, but not a cyclic contraction on A\cup B.
The following result tells us the relation between an \mathcal{MT}cyclic contraction and a Caristitype cyclic map.
Theorem 3.1 Let A and B be nonempty subsets of a metric space (X,d) and T:A\cup B\to A\cup B be an \mathcal{MT}cyclic contraction with respect to α. Then there exist a bounded below function f:A\cup B\to \mathbb{R} and a nondecreasing function \phi :\mathbb{R}\to (0,+\mathrm{\infty}) such that T is a Caristitype cyclic map dominated by f and φ.
Proof Denote {T}^{0}=I (the identity mapping). Let x\in A\cup B be given. Define a sequence {\{{w}_{n}\}}_{n\in \mathbb{N}} in A\cup B by {w}_{1}=x and {w}_{n}=T{w}_{n}={T}^{n1}x for n\in \mathbb{N}. Clearly, the condition (MT2) implies that T satisfies
d(Tx,Ty)\le d(x,y)\phantom{\rule{1em}{0ex}}\text{for any}x\in A\text{and}y\in B.
So from the last inequality we deduce
d({w}_{n+1},{w}_{n+2})\le d({w}_{n},{w}_{n+1})\phantom{\rule{1em}{0ex}}\text{for all}n\in \mathbb{N}.
Hence the sequence \{d({w}_{n},{w}_{n+1})\} is nonincreasing in [0,\mathrm{\infty}). Since α is an \mathcal{MT}function, by (g) of Theorem D, we obtain
0\le \underset{n\in \mathbb{N}}{sup}\alpha (d({w}_{n},{w}_{n+1}))<1.
Since x\in A\cup B is arbitrary, we can define a new function \beta :A\cup B\to [0,1) by
\beta (x):=\underset{n\in \mathbb{N}}{sup}\alpha (d({w}_{n},{w}_{n+1}))=\underset{n\in \mathbb{N}}{sup}\alpha \left(d({T}^{n1}x,{T}^{n}x)\right)\phantom{\rule{1em}{0ex}}\text{for}x\in A\cup B.
Clearly, for each x\in A\cup B, we have
\beta (Tx)\le \beta (x)
(3.1)
and
\alpha \left(d({T}^{n1}x,{T}^{n}x)\right)\le \beta (x)\phantom{\rule{1em}{0ex}}\text{for all}n\in \mathbb{N}.
(3.2)
Let x\in A\cup B be given. Without loss of generality, we may assume x\in A. Then Tx\in B. By (MT2), we get
d(Tx,{T}^{2}x)dist(A,B)\le \alpha (d(x,Tx))(d(x,Tx)dist(A,B))
and hence
d(x,Tx)\alpha (d(x,Tx))(d(x,Tx)dist(A,B))\le d(x,Tx)d(Tx,{T}^{2}x)+dist(A,B).
(3.3)
By exploiting inequalities (3.1), (3.2) and (3.3), we obtain
\begin{array}{rcl}d(x,Tx)dist(A,B)& \le & \frac{1}{1\alpha (d(x,Tx))}d(x,Tx)\frac{1}{1\alpha (d(x,Tx))}d(Tx,{T}^{2}x)\\ \le & \frac{1}{1\beta (x)}d(x,Tx)\frac{1}{1\beta (Tx)}d(Tx,{T}^{2}x).\end{array}
Let \phi :\mathbb{R}\to [0,+\mathrm{\infty}) and f:A\cup B\to \mathbb{R} be defined by
\phi (t)=1\phantom{\rule{1em}{0ex}}\text{for}t\in \mathbb{R}
and
f(x)=\frac{1}{1\beta (x)}d(x,Tx)\phantom{\rule{1em}{0ex}}\text{for}x\in A\cup B,
respectively. Then φ is a nondecreasing function and f is a bounded below function. Clearly, f(x)<+\mathrm{\infty} for all x\in A\cup B. From (3.3), we obtain
d(x,Tx)dist(A,B)\le \phi (f(x))(f(x)f(Tx))\phantom{\rule{1em}{0ex}}\text{for all}x\in A\cup B,
which means that T is a Caristitype cyclic map dominated by f and φ. □
Theorem 3.2 [12]
Let A and B be nonempty subsets of a metric space (X,d) and T:A\cup B\to A\cup B be an \mathcal{MT}cyclic contraction with respect to α. Then there exists a sequence {\{{x}_{n}\}}_{n\in \mathbb{N}} such that
\underset{n\to \mathrm{\infty}}{lim}d({x}_{n},T{x}_{n})=\underset{n\in \mathbb{N}}{inf}d({x}_{n},T{x}_{n})=dist(A,B).
Proof Applying Theorem 3.1, there exist a bounded below function f:A\cup B\to \mathbb{R} and a nondecreasing function \phi :\mathbb{R}\to (0,+\mathrm{\infty}) such that T is a Caristitype cyclic map dominated by f and φ. Let u\in A\cup B be given. Let {\{{x}_{n}\}}_{n\in \mathbb{N}}\subset A\cup B be defined by {x}_{1}=u and {x}_{n+1}=T{x}_{n} for n\in \mathbb{N}. Applying Theorem 2.1, we have
\underset{n\to \mathrm{\infty}}{lim}d({x}_{n},{x}_{n+1})=dist(A,B).
(3.4)
Since the condition (MT2) implies that T satisfies
d(Tx,Ty)\le d(x,y)\phantom{\rule{1em}{0ex}}\text{for any}x\in A\text{and}y\in B,
we know that the sequence \{d({x}_{n},{x}_{n+1})\} is nonincreasing in [0,\mathrm{\infty}). By (3.4), we get
dist(A,B)=\underset{n\to \mathrm{\infty}}{lim}d({x}_{n},{x}_{n+1})=\underset{n\in \mathbb{N}}{inf}d({x}_{n},{x}_{n+1})\ge 0.
The proof is completed. □
Theorem 3.3 [12]
Let A and B be nonempty subsets of a metric space (X,d) and T:A\cup B\to A\cup B be an \mathcal{MT}cyclic contraction with respect to α. For a given {x}_{1}\in A, define an iterative sequence {\{{x}_{n}\}}_{n\in \mathbb{N}} by {x}_{n+1}=T{x}_{n} for n\in \mathbb{N}. Suppose that \{{x}_{2n1}\} has a convergent subsequence in A, then there exists v\in A such that d(v,Tv)=dist(A,B).
Proof Applying Theorem 3.1, there exist a bounded below function f:A\cup B\to \mathbb{R} and a nondecreasing function \phi :\mathbb{R}\to (0,+\mathrm{\infty}) such that T is a Caristitype cyclic map dominated by f and φ. Let u={x}_{1}\in A be given. Let {\{{x}_{n}\}}_{n\in \mathbb{N}}\subset A\cup B be defined by {x}_{n+1}=T{x}_{n} for n\in \mathbb{N}. Since the condition (MT2) implies the condition (H2) as in Theorem 2.2, all the assumptions of Theorem 2.2 are satisfied. By applying (a) of Theorem 2.2, there exists v\in A such that d(v,Tv)=dist(A,B). □
Remark 3.1 ([[3], Proposition 3.2])
(i.e., Theorem 1.1) is a special case of Theorem 3.3.
Finally, applying Theorem 2.1, we can establish a new Caristitype fixed point theorem without assuming that the dominated functions possess the lower semicontinuity property.
Theorem 3.4 Let M be a nonempty subset of a metric space (X,d), f:M\to (\mathrm{\infty},+\mathrm{\infty}] be a proper and bounded below function, \phi :\mathbb{R}\to (0,+\mathrm{\infty}) be a nondecreasing function and T:M\to M be a selfmap on X. Suppose that T is of Caristi type on M dominated by φ and f, that is,
d(x,Tx)\le \phi (f(x))(f(x)f(Tx))\phantom{\rule{1em}{0ex}}\mathit{\text{for each}}x\in M.
(3.5)
Then there exists a sequence {\{{x}_{n}\}}_{n\in \mathbb{N}} in M such that {\{{x}_{n}\}}_{n\in \mathbb{N}} is Cauchy.
Moreover, if (X,d) is complete and M is closed in X, and one of the following conditions is satisfied:

(D1)
T is continuous on M;

(D2)
T is closed, that is, GrT=\{(x,y)\in M\times M:y=Tx\}, the graph of T, is closed in M\times M;

(D3)
T he map g:X\to [0,\mathrm{\infty}) defined by g(x)=d(x,Tx) is l.s.c.
Then the mapping T admits a fixed point in X, and for any u\in X with f(u)<+\mathrm{\infty}, the sequence {\{{T}^{n}u\}}_{n\in \mathbb{N}} converges to a fixed point of T.
Proof Let A=B=M. Then we have A\cup B=M, T(A)\subseteq B, T(B)\subseteq A and dist(A,B)=0. So (3.5) implies
d(x,Tx)dist(A,B)\le \phi (f(x))(f(x)f(Tx))\phantom{\rule{1em}{0ex}}\text{for all}x\in A\cup B=M.
Hence T a Caristitype cyclic map dominated by f and φ on A\cup B. Since f is proper, there exists u\in M=A\cup B such that f(u)<+\mathrm{\infty}. Let {\{{x}_{n}\}}_{n\in \mathbb{N}}\subset M be defined by {x}_{1}=u and {x}_{n+1}=T{x}_{n} for n\in \mathbb{N}. By applying Theorem 2.1, we have

(a)
f({x}_{n+1})\le f({x}_{n})<+\mathrm{\infty} for each n\in \mathbb{N},

(b)
d({x}_{n},{x}_{n+1})\le \phi (f({x}_{n}))(f({x}_{n})f({x}_{n+1})) for all n\in \mathbb{N},

(c)
{lim}_{n\to \mathrm{\infty}}d({x}_{n},{x}_{n+1})=0.
Since φ is nondecreasing, by (a), we have
\phi (f({x}_{n}))\le \phi (f({x}_{1}))\phantom{\rule{1em}{0ex}}\text{for all}n\in \mathbb{N}.
(3.6)
Since f is bounded below and the sequence {\{f({x}_{n})\}}_{n\in \mathbb{N}} is nonincreasing in [0,+\mathrm{\infty}),
\gamma :=\underset{n\to \mathrm{\infty}}{lim}f({x}_{n})=\underset{n\in \mathbb{N}}{inf}f({x}_{n})\text{exists}.
(3.7)
For m>n with m,n\in \mathbb{N}, taking into account (3.5), (3.6) and (3.7), we get
d({x}_{n},{x}_{m})\le \sum _{j=n}^{m1}d({x}_{j},{x}_{j+1})\le \phi (f({x}_{1}))(f({x}_{n})\gamma ).
Let {\lambda}_{n}=\phi (f({x}_{1}))(f({x}_{n})\gamma ), n\in \mathbb{N}. Then
sup\{d({x}_{n},{x}_{m}):m>n\}\le {\lambda}_{n}\phantom{\rule{1em}{0ex}}\text{for each}n\in \mathbb{N}.
(3.8)
Since {lim}_{n\to \mathrm{\infty}}f({x}_{n})=\gamma, {lim}_{n\to \mathrm{\infty}}{\lambda}_{n}=0. From (3.8) we obtain
\underset{n\to \mathrm{\infty}}{lim}sup\{d({x}_{n},{x}_{m}):m>n\}=0,
which proves that {\{{x}_{n}\}}_{n\in \mathbb{N}} is a Cauchy sequence in M.
Moreover, assume that (X,d) is complete and M is closed in X. So (M,d) is a complete metric space. By the completeness of M, there exists {v}_{u}\in M such that {x}_{n}\to {v}_{u} as n\to \mathrm{\infty}. We claim {v}_{u}\in \mathcal{F}(T). If (D1) holds, since T is continuous on M, {x}_{n+1}=T{x}_{n} for each n\in \mathbb{N} and {x}_{n}\to {v}_{u} as n\to \mathrm{\infty}, we get
{v}_{u}=\underset{n\to \mathrm{\infty}}{lim}{x}_{n}=\underset{n\to \mathrm{\infty}}{lim}{x}_{n+1}=\underset{n\to \mathrm{\infty}}{lim}T{x}_{n}=T\left(\underset{n\to \mathrm{\infty}}{lim}{x}_{n}\right)=T{v}_{u},
If (D2) holds, since T is closed, {x}_{n+1}=T{x}_{n} for each n\in \mathbb{N} and {x}_{n}\to {v}_{u} as n\to \mathrm{\infty}, we have T{v}_{u}={v}_{u}. Finally, assume that (D3) holds. Since {lim}_{n\to \mathrm{\infty}}d({x}_{n},{x}_{n+1})=0, we obtain
\begin{array}{rcl}d({v}_{u},T{v}_{u})& =& g({v}_{u})\\ \le & \underset{n\to \mathrm{\infty}}{lim\hspace{0.17em}inf}g({x}_{n})\\ =& \underset{n\to \mathrm{\infty}}{lim}d({x}_{n},{x}_{n+1})=0,\end{array}
we obtain d({v}_{u},T{v}_{u})=0 and hence T{v}_{u}={v}_{u}. This completes the proof. □