We construct a sequence starting from
\(x_{0} \in X\). If
\(x_{0} \in Tx _{0}\), then
\(x_{0}\) is a fixed point of
T and the proof is completed. Suppose that
\(x_{0} \notin Tx_{0}\). Because
\(Tx_{0}\) is a compact subset of
X, then
\(D(x_{0},Tx_{0})>0\) and we can choose
\(x_{1} \in Tx_{0}\) such that
\(d(x_{0},x_{1})=D(x_{0},Tx_{0})\). If
\(x_{1} \in Tx_{1}\), then
\(x_{1}\) is a fixed point of
T, and subsequently, the proof is completed. Assume that
\(x_{1} \notin Tx_{1}\), then it is clear that
\(D(x_{1},Tx_{1})>0\) because
\(Tx_{1}\) is a compact subset of
X. On the other hand, from
\(D(x_{1},Tx_{1}) \leq H(Tx_{0},Tx_{1})\) and
\((\varPhi _{1})\), we obtain
$$\phi \bigl(D(x_{1},Tx_{1})\bigr) \leq \phi \bigl(H(Tx_{0},Tx_{1})\bigr). $$
It follows from (
2.1) and Remark
2.4 that
$$ \begin{aligned}[b] \phi \bigl(D(x_{1},Tx_{1}) \bigr) &\leq \phi \bigl(H(Tx_{0},Tx_{1})\bigr) \leq \psi \bigl(\phi \bigl(M(x _{0},x_{1})\bigr)\bigr) \\ &=\psi \biggl(\phi \biggl(\frac{D(x_{0},Tx_{0})D(x_{0},Tx_{1})+D(x _{1},Tx_{1})D(x_{1},Tx_{0})}{\max \{D(x_{0},Tx_{1}),D(x_{1},Tx_{0})\}}\biggr)\biggr) \\ &=\psi \bigl(\phi \bigl(D(x_{0},Tx_{0})\bigr)\bigr) \\ &< \phi \bigl(D(x_{0},Tx_{0})\bigr). \end{aligned} $$
(3.1)
Since
\(Tx_{1}\) is a compact subset of
X, we can choose
\(x_{2} \in Tx_{1}\) such that
\(d(x_{1},x_{2})=D(x_{1},Tx_{1})\). Then from (
3.1) we get
$$ \phi \bigl(d(x_{1},x_{2})\bigr)=\phi \bigl(D(x_{1},Tx_{1}) \bigr)< \phi \bigl(D(x_{0},Tx_{0})\bigr)= \phi \bigl(d(x_{0},x_{1})\bigr). $$
(3.2)
It follows from (
3.2) and
\((\varPhi _{1})\) that
$$d(x_{1},x_{2}) \leq d(x_{0},x_{1}). $$
We continue constructing the sequence similarly. If
\(x_{2} \in Tx_{2}\), then this proof is done. Thus, we assume that
\(x_{2} \notin Tx_{2}\). Then
\(D(x_{2},Tx_{2})>0\) since
\(Tx_{2}\) is a compact subset of
X, and from
\(D(x_{2},Tx_{2}) \leq H(Tx_{1},Tx_{2})\), we have
$$ \begin{aligned}[b] \phi \bigl(D(x_{2},Tx_{2}) \bigr) &\leq \phi \bigl(H(Tx_{1},Tx_{2})\bigr) \leq \psi \bigl(\phi \bigl(M(x _{1},x_{2})\bigr)\bigr) \\ &=\psi \biggl(\phi \biggl(\frac{D(x_{1},Tx_{1})D(x_{1},Tx_{2})+D(x _{2},Tx_{2})D(x_{2},Tx_{1})}{\max \{D(x_{1},Tx_{2}),D(x_{2},Tx_{1})\}}\biggr)\biggr) \\ &=\psi \bigl(\phi \bigl(D(x_{0},Tx_{0})\bigr)\bigr) \\ &< \phi \bigl(D(x_{1},Tx_{1})\bigr). \end{aligned} $$
(3.3)
In addition, the compactness of
\(Tx_{2}\) implies that there exists
\(x_{3} \in Tx_{2}\) such that
\(d(x_{2},x_{3})=D(x_{2},Tx_{2})\). Then from (
3.3) we get
$$ \phi \bigl(d(x_{2},x_{3})\bigr)=\phi \bigl(D(x_{2},Tx_{2}) \bigr)< \phi \bigl(D(x_{1},Tx_{1})\bigr)= \phi \bigl(d(x_{1},x_{2})\bigr). $$
(3.4)
It follows from (
2.4) and
\((\varPhi _{1})\) that
$$d(x_{2},x_{3}) \leq d(x_{1},x_{2}). $$
By induction, we obtain a sequence
\(\{x_{n}\}_{n \in \mathbb{N}_{0}}\) satisfying
$$ x_{n+1} \in Tx_{n},x_{n+1} \notin Tx_{n+1}, \quad d(x_{n},x_{n+1})=D(x _{n},Tx_{n})>0, $$
(3.5)
and
$$d(x_{n},x_{n+1}) \leq d(x_{n-1},x_{n}), $$
for all
\(n \in \mathbb{N}\). Therefore the sequence
\(\{d(x_{n},x_{n+1}) \}_{n \in \mathbb{N}_{0}}\) is a positive and non-increasing sequence, and hence
$$\lim_{n \to \infty } d(x_{n},x_{n+1}) \geq 0. $$
Now, we claim that
$$\lim_{n \to \infty } d(x_{n},x_{n+1})=0. $$
In fact, from (
3.5) and
\((\varPhi _{1})\), by using (
2.1), we get
$$\begin{aligned} 0 &\leq \phi \bigl(d(x_{n},x_{n+1}) \bigr) = \phi \bigl(D(x_{n},Tx_{n})\bigr) \\ &\leq \phi \bigl(H(Tx_{n-1},Tx_{n})\bigr) \leq \psi \bigl( \phi \bigl(M(x_{n-1},x_{n})\bigr)\bigr) \\ &= \psi \biggl( \phi \biggl( \frac{D(x_{n-1},Tx_{n-1})D(x_{n-1},Tx_{n})+D(x_{n},Tx _{n})D(x_{n},Tx_{n-1})}{\max \{D(x_{n-1},Tx_{n}),D(x_{n},Tx_{n-1})\}}\biggr)\biggr) \\ &=\psi \bigl(\phi \bigl(D(x_{n-1},Tx_{n-1})\bigr)\bigr) \leq \psi ^{2}\bigl(\phi \bigl(D(x_{n-2},Tx _{n-2})\bigr) \bigr) \\ &\leq \psi ^{3}\bigl(\phi \bigl(D(x_{n-3},Tx_{n-3}) \bigr)\bigr) \leq \cdots \\ &\leq \psi ^{n}\bigl(\phi \bigl(D(x_{0},Tx_{0}) \bigr)\bigr). \end{aligned} $$
From
\((\varPsi _{2})\) we have
$$\lim_{n \to \infty } \psi ^{n}\bigl(\phi \bigl(D(x_{0},Tx_{0}) \bigr)\bigr) =0. $$
By using the sandwich theorem, we get
$$\lim_{n \to \infty }\phi \bigl(d(x_{n},x_{n+1}) \bigr)=0. $$
Therefore, from
\((\varPhi _{2})\) we obtain
$$\lim_{n \to \infty }d(x_{n},x_{n+1})=0 $$
and hence
$$ \lim_{n \to \infty }D(x_{n},Tx_{n})=0. $$
(3.6)
Now, we claim that
$$\lim_{n,m \to \infty }d(x_{n},x_{m})=0. $$
Arguing by contradiction, we assume that there exists a
\(\varepsilon >0\) for which we can seek two sequences
\(\{p(n)\}_{n=1}^{\infty }\) and
\(\{q(n)\}_{n=1}^{\infty }\) of natural numbers such that, for all
\(n\in \mathbb{N}\),
\(p(n)\) is the smallest index for which
$$ p(n)>q(n)>n, \quad\quad d(x_{p(n)},x_{q(n)})\geq \varepsilon , \quad\quad d(x_{p(n)-1},x _{q(n)})< \varepsilon . $$
(3.7)
Thus, for all
\(n\in \mathbb{N}\), by using the triangle inequality, we have
$$ \varepsilon \leq d(x_{p(n)},x_{q(n)}) \leq D(x_{p(n)},Tx_{q(n)})+D(Tx _{q(n)},x_{q(n)}). $$
(3.8)
It follows from (
3.6) and (
3.8) and by using the sandwich theorem again, we have
$$\liminf_{n\to \infty } D(x_{p(n)},Tx_{q(n)})\geq \varepsilon . $$
Thus, there exists
\(n_{1}\in \mathbb{N}\), such that
$$D(x_{p(n)},Tx_{q(n)})>\frac{\varepsilon }{2}, $$
for all
\(n>n_{1}\).
On the other hand, from (
3.9) we know that
$$\begin{aligned} 0 &\leq \frac{D(x_{p(n)},Tx_{p(n)})D(x_{p(n)},Tx_{q(n)})+ D(x_{q(n)},Tx _{q(n)})D(x_{q(n)},Tx_{p(n)})}{\max \{D(x_{p(n)},Tx_{q(n)}),D(Tx_{p(n)},x _{q(n)})\}} \\ &=\frac{D(x_{p(n)},Tx_{p(n)})D(x_{p(n)},Tx_{q(n)})}{ \max \{D(x_{p(n)},Tx_{q(n)}),D(Tx_{p(n)},x_{q(n)})} \\ &\quad {}+\frac{D(x _{q(n)},Tx_{q(n)})D(x_{q(n)},Tx_{p(n)})}{\max \{D(x_{p(n)},Tx_{q(n)}),D(Tx _{p(n)},x_{q(n)})} \\ &\leq D(x_{p(n)},Tx_{p(n)})+D(x_{q(n)},Tx_{q(n)}). \end{aligned} $$
Let
\(n \to \infty \) in the above inequality and by taking (
3.6) into account, we obtain
$$\lim_{n\to \infty } \frac{D(x_{p(n)},Tx_{p(n)})D(x_{p(n)},Tx_{q(n)})+ D(x_{q(n)},Tx_{q(n)})D(x_{q(n)},Tx_{p(n)})}{\max \{D(x_{p(n)},Tx_{q(n)}),D(Tx _{p(n)},x_{q(n)})\}}=0. $$
So, there exists
\(n_{3}\in \mathbb{N}\) such that
$$\frac{D(x_{p(n)},Tx_{p(n)})D(x_{p(n)},Tx_{q(n)})+ D(x_{q(n)},Tx_{q(n)})D(x _{q(n)},Tx_{p(n)})}{\max \{D(x_{p(n)},Tx_{q(n)}),D(Tx_{p(n)},x_{q(n)}) \}}< \frac{\varepsilon }{2}, $$
for all
\(n>n_{3}\)
In combination with (
3.12) and (
3.13) we get
$$\phi \biggl(\frac{\varepsilon }{2}\biggr)< \phi \biggl(\frac{D(x_{p(n)},Tx_{p(n)})D(x _{p(n)},Tx_{q(n)})+ D(x_{q(n)},Tx_{q(n)})D(x_{q(n)},Tx_{p(n)})}{ \max \{D(x_{p(n)},Tx_{q(n)}),D(Tx_{p(n)},x_{q(n)})\}}\biggr) \leq \phi \biggl(\frac{ \varepsilon }{2}\biggr), $$
for all
\(n>\max \{n_{1},n_{2},n_{3}\}\), which is a contradiction. Hence
$$\lim_{n,m\rightarrow \infty }d(x_{n},x_{m})=0. $$
Therefore, we conclude that
\(\{x_{n}\}_{n=1}^{\infty }\) is a Cauchy sequence in
X. Since
\((X,d)\) is a complete metric space, so there exists
\(x^{*}\in X\) such that
$$\lim_{n\to \infty }d\bigl(x_{n},x^{*}\bigr)=0 $$
and
$$ \lim_{n\to \infty }D\bigl(x_{n+1},Tx^{*}\bigr)=d \bigl(x^{*},Tx^{*}\bigr). $$
(3.14)
Now, we claim that
\(x^{*}\in Tx^{*}\).
From (b), we get
$$ \max \bigl\{ D\bigl(x_{n},Tx^{*}\bigr),D\bigl(x^{*},Tx_{n} \bigr)\bigr\} >0. $$
(3.15)
Since
T is a generalized multivalued Khan-type
\((\psi ,\phi )\)-contraction, from (
3.15) we obtain
$$ \begin{aligned}[b] \phi \bigl(D\bigl(x_{n+1},Tx^{*} \bigr)\bigr) &\leq \phi \bigl(H\bigl(Tx_{n},Tx^{*}\bigr)\bigr) \leq \psi \bigl( \phi \bigl(M\bigl(x_{n},x^{*}\bigr)\bigr) \bigr) \\ &=\psi \biggl(\phi \biggl(\frac{D(x_{n},Tx_{n})D(x_{n},Tx ^{*})+D(x^{*},Tx^{*})D(x^{*},Tx_{n})}{\max \{D(x_{n},Tx^{*}),D(Tx_{n},x ^{*})\}}\biggr)\biggr) \\ &< \phi \biggl(\frac{D(x_{n},Tx_{n})D(x_{n},Tx^{*})+D(x^{*},Tx ^{*})D(x^{*},Tx_{n})}{\max \{D(x_{n},Tx^{*}),D(Tx_{n},x^{*})\}} \biggr). \end{aligned} $$
(3.16)
On the other hand, in combination with (
3.6), (
3.14) and (
3.15), we get
$$\lim_{n\rightarrow \infty }\frac{D(x_{n},Tx_{n})D(x_{n},Tx^{*})+D(x ^{*},Tx^{*})D(x^{*},Tx_{n})}{\max \{D(x_{n},Tx^{*}),D(Tx_{n},x^{*})\}}=0. $$
Thus, taking
\(D(x^{*},Tx^{*})>0\) into account, there exists
\(n_{4} \in \mathbb{N}\), such that
$$\frac{D(x_{n},Tx_{n})D(x_{n},Tx^{*})+D(x^{*},Tx^{*})D(x^{*},Tx_{n})}{ \max \{D(x_{n},Tx^{*}),D(Tx_{n},x^{*})\}}< \frac{1}{2}D\bigl(x^{*},Tx^{*} \bigr), $$
for all
\(n>n_{4}\). And this implies that
$$\phi \biggl(\frac{D(x_{n},Tx_{n})D(x_{n},Tx^{*})+D(x^{*},Tx^{*})D(x^{*},Tx _{n})}{\max \{D(x_{n},Tx^{*}),D(Tx_{n},x^{*})\}}\biggr)\leq \phi \biggl(\frac{1}{2}D\bigl(x ^{*},Tx^{*}\bigr)\biggr). $$
It follows from (
3.16) that
$$\phi \bigl(D\bigl(x_{n+1},Tx^{*}\bigr)\bigr)< \phi \biggl( \frac{1}{2}D\bigl(x^{*},Tx^{*}\bigr)\biggr), $$
for all
\(n>n_{4}\).