© Wuhan University 2025

0 Introduction

Fixed point results are widely used for finding solutions to various differential equations and integral equations. Since the valuable Banach contraction principle was created by Polish mathematician Banach in 1922^[1], many fixed point theorems about different contractions were discussed, such as the Kannan-type fixed point theorem^[2]. There are lots of other important contractions which were given by Reich^[3-4], Ciric^[5], Chatterjea^[6], Hardy and Rogers^[7], Bianchini^[8], and so on. Among them, the Kannan-type contraction gets more attention, since the Kannan-type fixed point theorems describe the relationship between the contraction and the completeness of metric spaces^[9].

Some other more generalized spaces were given by many researchers, which extended the structure of metric spaces, such as b-metric spaces^[10-11] and extended b-metric spaces^[12]. Numerous fixed point theorems were proved on these spaces. Whereafter, the notion of controlled metric-type space was introduced in Ref. [13], as a generation of extended b-metric space. In 2020, Mlaiki et al^[14] initiated a double double-controlled metric-type space. Later, a more general concept named double-controlled metric-like space appeared in Refs. [15-16], which was a sharp extension of all the types of metric spaces mentioned above. The authors established some fixed point results in this space.

Recently, Górnicki^[17] and Garai et al^[18] obtained some meaningful results for Kannan-type contractive self-mappings in metric spaces. In this paper, we prove some fixed point theorems about extended Kannan contraction in double controlled metric-like spaces, by introducing the concepts of bounded compactness, orbital compactness, weak orbital continuity, orbital completeness and asymptotic regularity in double controlled metric-like spaces. As compared to the existing results from Refs. [15,17-19], our main results weaken the compactness and completeness of the spaces, the orbital continuity and the $k$-continuity of the mappings. Furthermore, we prove that the completeness in double controlled metric-like spaces is necessary if the extended Kannan-type contractions have a fixed point in the set M. Besides these progresses, we also give some examples to illustrate that our new notions and theorems are genuine improvements and generalizations of the corresponding results in the literature.

1 Preliminaries

First, we recall some necessary definitions in double-controlled metric-like spaces.

Definition 1^[15] Assume a set $M\ne \mathrm{\varnothing }$ and $\alpha ,\beta :M\times M\to [\mathrm{1},+\mathrm{\infty })$ be two noncomparable functions. The function $\zeta :M\times M\to [\mathrm{0},+\mathrm{\infty })$ satisfies for all $f,g,h\in M:$$\begin{array}{l}\mathrm{1})\zeta (f,g)=\mathrm{0}\Rightarrow f=g,\\ \mathrm{2})\zeta (f,g)=\zeta (g,f),\\ \mathrm{3})\zeta (f,g)\le \alpha (f,h)\zeta (f,h)+\beta (h,g)\zeta (h,g).\end{array}$

In this case, $(M,\zeta )$ is called a double controlled metric-like space (DCMLS).

Note that if the condition 1) is replaced by $\zeta (f,g)=\mathrm{0}$ if and only if $f=g$, then it is a double controlled metric-type space (DCMTS)^[14]. Moreover, if the triangle inequality is $\zeta (f,g)\le \alpha (f,h)\zeta (f,h)+\alpha (h,g)\zeta (h,g)$, then it is a controlled metric-type space (CMTS)^[13]. Next, if this inequality is further written as $\zeta (f,g)\le \alpha (f,g)[\zeta (f,h)+\zeta (h,g)]$, then it is an extended b-metric space (EbMS)^[12]. Therefore, a double controlled metric-like space must be a double controlled metric-type space, a controlled metric-type space, an extended b-metric space, a b-metric space and a metric space, but not the converse. The following example greatly explores their relationships.

Example 1 Let $M={\mathbb{R}}^{+}$ and $\zeta :M\times M\to {\mathbb{R}}^{+}$ be defined as

$\zeta (f,g)=\{\begin{array}{ll}\mathrm{0}\Rightarrow & f=g,\\ \frac{\mathrm{1}}{\mathrm{4}},& \mathrm{i}\mathrm{f}f=g=\mathrm{0},\\ \frac{\mathrm{1}}{f+\mathrm{1}},& \mathrm{i}\mathrm{f}f\ge \mathrm{1}\mathrm{a}\mathrm{n}\mathrm{d}g\in [\mathrm{0,1}),\\ \frac{\mathrm{1}}{g+\mathrm{1}},& \mathrm{i}\mathrm{f}g\ge \mathrm{1}\mathrm{a}\mathrm{n}\mathrm{d}f\in [\mathrm{0,1}),\\ \mathrm{1},& \mathrm{o}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{w}\mathrm{i}\mathrm{s}\mathrm{e}.\end{array}$

The functions $\alpha ,\beta :M\times M\to [\mathrm{1},+\mathrm{\infty })$ are considered as

$\alpha (f,g)=\{\begin{array}{l}\mathrm{1},f,g\ge \mathrm{1},\\ f+\mathrm{1},\mathrm{o}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{w}\mathrm{i}\mathrm{s}\mathrm{e},\end{array}$

and$\beta (f,g)=\{\begin{array}{l}\mathrm{1},f,g<\mathrm{1},\\ \mathrm{2}(f+g),\mathrm{o}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{w}\mathrm{i}\mathrm{s}\mathrm{e}.\end{array}$

Obviously, 1) and 2) of Definition 1 are satisfied. Let's prove that condition 3) is also satisfied.

Case 1 If $f\ge \mathrm{1},\mathrm{0}\le g<\mathrm{1}$, then

$\zeta (f,g)=\frac{\mathrm{1}}{f+\mathrm{1}}\le \{\begin{array}{l}\mathrm{1}\times \mathrm{1}+\mathrm{2}(h+g)\cdot \frac{\mathrm{1}}{h+\mathrm{1}},h\ge \mathrm{1},\\ (f+\mathrm{1})\cdot \frac{\mathrm{1}}{f+\mathrm{1}}+\mathrm{1}\times \mathrm{1},\mathrm{0}\le h<\mathrm{1}.\end{array}$

Case 2 If $\mathrm{0}\le f<\mathrm{1},g\ge \mathrm{1},$ then

$\zeta (f,g)=\frac{\mathrm{1}}{g+\mathrm{1}}\le \{\begin{array}{ll}(f+\mathrm{1})\cdot \frac{\mathrm{1}}{h+\mathrm{1}}+\mathrm{2}(h+g)\cdot \mathrm{1},& h\ge \mathrm{1},\\ (f+\mathrm{1})\cdot \mathrm{1}+\mathrm{2}(h+g)\cdot \frac{\mathrm{1}}{g+\mathrm{1}},& \mathrm{0}\le h<\mathrm{1}.\end{array}$

Case 3 If $f,g\ge \mathrm{1},$ then

$\zeta (f,g)=\mathrm{1}\le \{\begin{array}{ll}\mathrm{1}\times \mathrm{1}+\mathrm{2}(h+g)\cdot \mathrm{1},& h\ge \mathrm{1},\\ (f+\mathrm{1})\cdot \frac{\mathrm{1}}{f+\mathrm{1}}+\mathrm{2}(h+g)\cdot \frac{\mathrm{1}}{g+\mathrm{1}},& \mathrm{0}\le h<\mathrm{1}.\end{array}$

Case 4 If $f,g<\mathrm{1},f\ne \mathrm{0}$ or $g\ne \mathrm{0}$, then

$\zeta (f,g)=\mathrm{1}\le \{\begin{array}{ll}(f+\mathrm{1})\cdot \frac{\mathrm{1}}{h+\mathrm{1}}+\mathrm{2}(h+g)\cdot \frac{\mathrm{1}}{h+\mathrm{1}},& h\ge \mathrm{1},\\ (f+\mathrm{1})\cdot \mathrm{1}+\mathrm{1}\times \mathrm{1},& \mathrm{0}\le h<\mathrm{1}.\end{array}$

Case 5 If $f=g=\mathrm{0}$, then

$\zeta (f,g)=\frac{\mathrm{1}}{\mathrm{4}}\le (f+\mathrm{1})\cdot \frac{\mathrm{1}}{\mathrm{4}}+\mathrm{1}\times \frac{\mathrm{1}}{\mathrm{4}},h=\mathrm{0},$

if $h\ne \mathrm{0}$, it also holds.

Thus, the condition 3) of Definition 1 is satisfied in the five cases. Therefore, $(M,\zeta )$ is a DCMLS. But, $\zeta (\mathrm{0,0})\ne \mathrm{0}$ shows that $(M,\zeta )$ is not a DCMTS. Furthermore, $(M,\zeta )$ is not an extended b-metric space or a controlled metric-type space.

Definition 2^[15] Let $(M,\zeta )$ be a DCMLS and $\left\{f_n\right\}$ be a sequence in M. Then,

(i) $\left\{f_n\right\}$ converges to $f$ (denoted $f_n\to f$) if and only if

$\underset{n\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\zeta (f_n,f)=\zeta (f,f)$;

(ii) $\left\{f_n\right\}$ is a $\zeta$-Cauchy if and only if

$\underset{n,m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\zeta (f_n,f_m)\mathrm{e}\mathrm{x}\mathrm{i}\mathrm{s}\mathrm{t}\mathrm{s}\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{f}\mathrm{i}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{e}$;

(iii) $(M,\zeta )$ is a complete space if for every $\zeta$-Cauchy sequence $\left\{f_n\right\}$, there exists $f\in M$ such that

$\underset{n\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\zeta (f_n,f)=\zeta (f,f)=\underset{n,m\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\zeta (f_n,f_m).$

2 Bounded Compactness and $\mathit{\tau }$-Orbital Compactness

The concepts of bounded compactness and $\tau$-orbital compactness were studied in usual metric spaces in Ref. [18], which were significant in weakening the condition of compactness. Moreover, Jaggi^[20] and Pant et al^[21] gave the notions of $f_{\mathrm{0}}$-orbital continuity and weak orbital continuity in metric spaces, respectively, which generalized the concepts of orbital continuity and $k$-continuity. In the following, we introduce the notions of extended Kannan-type contraction, bounded compactness, $\tau$-orbital compactness, $f_{\mathrm{0}}$-orbital continuity and weak orbital continuity in the framework of DCMLS. Of course, they are true in controlled metric spaces and extended b-metric spaces, etc.

Definition 3   Let $(M,\zeta )$ be a DCMLS. The mapping $\tau :M\to M$ is said to be an extended Kannan-type contraction if it satisfies

$\zeta (\tau f,\tau g)<a\zeta (f,\tau f)+b\zeta (g,\tau g),$(1)

for all $f,g\in M$ with $f\ne g$, where $a+b=\mathrm{1}$ and $a,b\in (\mathrm{0,1})$.

Definition 4   Let $(M,\zeta )$ be a DCMLS and $\tau$ be a self-mapping on $M$. Let $f\in M$ and $O_{\tau }(f)=\{f,\tau f,{\tau }^{\mathrm{2}}f,{\tau }^{\mathrm{3}}f,\cdots \}.$

1) The space $(M,\zeta )$ is said to be boundedly compact if every bounded sequence in $M$ has a convergent subsequence.

2) The set $M$ is said to be $\tau$-orbitally compact set if every sequence in $O_{\tau }(f)$ has a convergent subsequence for all $f\in M$.

3) The mapping $\tau$ is said to be $f_{\mathrm{0}}$-orbitally continuous for some $f_{\mathrm{0}}\in M$ if its restriction to the set $\overline{O(\tau ,f_{\mathrm{0}})}$ is continuous, i.e., $\tau :\overline{O(\tau ,f_{\mathrm{0}})}\to M$ is continuous; here, $\overline{O(\tau ,f_{\mathrm{0}})}$ represents the closure of $O(\tau ,f_{\mathrm{0}})$. Moreover, $\tau$ is said to be orbitally continuous if it is $f_{\mathrm{0}}$-orbitally continuous for all $f_{\mathrm{0}}\in M$.

4) The mapping $\tau$ is said to be weakly orbitally continuous if the set

$\{g\in M:u=\underset{i\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}{\tau }^{n_i}g\mathrm{i}\mathrm{m}\mathrm{p}\mathrm{l}\mathrm{i}\mathrm{e}\mathrm{s}\tau u=\underset{i\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\tau {\tau }^{n_i}g\}$

is nonempty whenever the set $\{f\in M:u=\underset{i\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}{\tau }^{n_i}f\}$ is nonempty for $u\in M$.

Given the above definitions, compactness of a set implies bounded compactness and $\tau$-orbital compactness of this set, but not the converse. In the same way, orbital continuity and $k$-continuity of $\tau$ imply weak orbital continuity and $f_{\mathrm{0}}$-orbital continuity, but the converse need not be true. The next example fully illustrates this point.

Example 2 1) Let $M={\mathbb{R}}^{+}$ and $\zeta :M\times M\to {\mathbb{R}}^{+}$ by

$\zeta (f,g)=\{\begin{array}{ll}\mathrm{1},& f=g=\mathrm{0},\\ |f-g|,& \mathrm{o}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{w}\mathrm{i}\mathrm{s}\mathrm{e}.\end{array}$

The functions $\alpha ,\beta :M\times M\to [\mathrm{1},\mathrm{\infty })$ are written as

$\alpha (f,g)=\{\begin{array}{ll}\frac{\mathrm{1}}{f},& g=\mathrm{0},f\in (\mathrm{0,1}],\\ \frac{\mathrm{1}}{g},& f=\mathrm{0},g\in (\mathrm{0,1}],\\ \mathrm{m}\mathrm{a}\mathrm{x}\{\frac{\mathrm{1}}{f},\frac{\mathrm{1}}{g}\},& f,g\in (\mathrm{0,1}],\\ \mathrm{1},& \mathrm{o}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{w}\mathrm{i}\mathrm{s}\mathrm{e},\end{array}$

and $\beta (f,g)=\{\begin{array}{ll}\frac{\mathrm{1}}{f}+\frac{\mathrm{1}}{g},& f,g\in (\mathrm{0,1}],\\ \mathrm{1},& \mathrm{o}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{w}\mathrm{i}\mathrm{s}\mathrm{e}.\end{array}$

Then, $(M,\zeta )$ is a DCMLS by the detailed calculations. Define the mappings ${\tau }_{\mathrm{1}},{\tau }_{\mathrm{2}}:M\to M$ as

$\begin{array}{l}{\tau }_{\mathrm{1}}(f)=\frac{f}{{(n+\mathrm{1})}^{\mathrm{2}}},n\le f<n+\mathrm{1},n\in \mathbb{N},\\ {\tau }_{\mathrm{2}}(f)=\mathrm{2}f+\mathrm{1},\end{array}$

for all $f\in M$. This implies that $M$ is ${\tau }_{\mathrm{1}}$-orbitally compact and boundedly compact but not ${\tau }_{\mathrm{2}}$-orbitally compact.

2) Let $M=[\mathrm{0,2})$. The metric $\zeta$ and the functions $\alpha ,\beta$ are defined the same as above and $\tau :M\to M$. It is obvious that $M$ is $\tau$ -orbitally compact but not complete.

3) Let $M=[\mathrm{0,2})$. The metric $\zeta$ and the functions $\alpha ,\beta$ are defined the same as above. The mapping $\tau :M\to M$ is given as

$\tau (f)=\{\begin{array}{ll}\frac{\mathrm{1}+f}{\mathrm{2}},& \mathrm{0}\le f<\mathrm{1},\\ \mathrm{2},& \mathrm{1}\le f\le \mathrm{2}.\end{array}$

It is not difficult to verify that $\tau$ is weakly orbitally continuous. If $f=\mathrm{2}$, then ${\tau }^n(\mathrm{2})\to \mathrm{2},$ $\tau ({\tau }^n(\mathrm{2}))\to \mathrm{2}=\tau (\mathrm{2})$. However, $\tau$ is not orbitally continuous, since ${\tau }^n(\mathrm{0})\to \mathrm{1}$, while $\tau ({\tau }^n(\mathrm{0}))\to \mathrm{1}\ne \tau (\mathrm{1})$. Likewise, we can prove that $\tau$ is not $k$-continuous for any integer $k\ge \mathrm{1}$, that is,

${\tau }^{k-\mathrm{1}}({\tau }^n(\mathrm{0}))\to \mathrm{1},{\tau }^k({\tau }^n(\mathrm{0}))\to \mathrm{1}\ne \tau (\mathrm{1}).$

Hence, the conditions of $k$-continuity and orbital continuity are stronger than the weak orbital continuity for the mapping $\tau$.

Theorem 1   Let $(M,\zeta )$ be a boundedly compact DCMLS and $\tau :M\to M$ be an extended Kannan-type contraction. The mapping $\tau$ is $f_{\mathrm{0}}$-orbitally continuous for some $f_{\mathrm{0}}\in M$ or weakly orbitally continuous, then $\tau$ has a unique fixed point $f^{\mathrm{*}}\in M$ and for each $f\in M$ the iterated sequence ${\tau }^{n}f$ converges to $f^{\mathrm{*}}$, i.e., $\tau$ is a Picard operator.

Proof   For arbitrary $f_{\mathrm{0}}\in M$, set $f_n=\tau f_{n-\mathrm{1}}={\tau }^n{f}_{\mathrm{0}},n\ge \mathrm{1}.$ Suppose $f_n\ne f_{n+\mathrm{1}}$ for all $n\in \mathbb{N}$. Indeed, if for some $n\in \mathbb{N},f_n=f_{n+\mathrm{1}}=\tau f_n$, then $f_n$ is the fixed point of $\tau$. Denote $u_n=\zeta (f_n,f_{n+\mathrm{1}})$ for each $n\in \mathbb{N}$. By (1), we have $u_n=\zeta (f_n,f_{n+\mathrm{1}})=\zeta (\tau f_{n-\mathrm{1}},\tau f_n)<a\zeta (f_{n-\mathrm{1}},f_n)+b\zeta (f_n,f_{n+\mathrm{1}})$$=a{u}_{n-\mathrm{1}}+b{u}_n.$ As $a+b=\mathrm{1}$ and $a,b\in (\mathrm{0,1})$, we have $(\mathrm{1}-b)u_n<a{u}_{n-\mathrm{1}}=(\mathrm{1}-b)u_{n-\mathrm{1}},$ which gives that $u_n<u_{n-\mathrm{1}}$ for all $n\in \mathbb{N}$. Repeat this process, we conclude that $\mathrm{0}<\cdots <u_n<u_{n-\mathrm{1}}<\cdots <u_{\mathrm{0}}=\zeta (f_{\mathrm{0}},f_{\mathrm{1}}).$

So, $\underset{n\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}{u}_n=u^{\mathrm{*}}$ for some $u^{\mathrm{*}}\ge \mathrm{0}$. It follows that $\zeta (f_n,f_m)=\zeta (\tau f_{n-\mathrm{1}},\tau f_{m-\mathrm{1}})<a\zeta (f_{n-\mathrm{1}},\tau f_{n-\mathrm{1}})+b\zeta (f_{m-\mathrm{1}},\tau f_{m-\mathrm{1}})=$

$a{u}_{n-\mathrm{1}}+b{u}_m<(a+b)u_{\mathrm{0}}=u_{\mathrm{0}},$ for all $n,m\in \mathbb{N}$, which shows $\left\{f_n\right\}$ is a bounded sequence in $M$. By the fact that $M$ is boundedly compact, the sequence $\left\{f_n\right\}$ has a convergent subsequence $\{f_{n_i}\}$ and $f^{\mathrm{*}}\in M$ such that $f_{n_i}\to f^{\mathrm{*}}$ $\mathrm{a}\mathrm{s}i\to \mathrm{\infty }$.

If $\tau$ is $f_{\mathrm{0}}$-orbitally continuous, then $\tau f_{n_i}\to \tau f^{\mathrm{*}}.$ If $\tau$ is weakly orbitally continuous, then ${\tau }^n{f}_{\mathrm{0}}\to f^{\mathrm{*}}$ for each $f_{\mathrm{0}}\in M$. By weak orbital continuity of $\tau$, we know ${\tau }^n{g}_{\mathrm{0}}\to f^{\mathrm{*}}$ and ${\tau }^{n+\mathrm{1}}{g}_{\mathrm{0}}\to \tau f^{\mathrm{*}}$ for some $g_{\mathrm{0}}$ in $M$. Next, let us prove that $f^{\mathrm{*}}=\tau f^{\mathrm{*}}$. Assume $u^{\mathrm{*}}>\mathrm{0}$. We have $\mathrm{0}<u^{\mathrm{*}}=\underset{i\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\zeta (f_{n_i},\tau f_{n_i})=\zeta (f^{\mathrm{*}},\tau f^{\mathrm{*}}).$

Moreover, $\mathrm{0}<u^{\mathrm{*}}=\underset{i\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}{u}_{n_i}=\underset{i\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\zeta (\tau f_{n_i},{\tau }^{\mathrm{2}}{f}_{n_i})=\zeta (\tau f^{\mathrm{*}},$

${\tau }^{\mathrm{2}}{f}^{\mathrm{*}})<a\zeta (f^{\mathrm{*}},\tau f^{\mathrm{*}})+b\zeta (\tau f^{\mathrm{*}},{\tau }^{\mathrm{2}}{f}^{\mathrm{*}}),$ which means $u^{\mathrm{*}}=\zeta (\tau f^{\mathrm{*}},$

${\tau }^{\mathrm{2}}{f}^{\mathrm{*}})<\zeta (f^{\mathrm{*}},\tau f^{\mathrm{*}})=u^{\mathrm{*}},$ a contradiction. Thus, $u^{\mathrm{*}}=\mathrm{0}$ and $f^{\mathrm{*}}=\tau f^{\mathrm{*}}$, i.e., $f^{\mathrm{*}}$ is a fixed point of $\tau$. For the uniqueness of $f^{\mathrm{*}}$, if $\tau f=f$ for some $f\in M$, then $\zeta (f,f^{\mathrm{*}})=\zeta (\tau f,\tau f^{\mathrm{*}})<a\zeta (f,\tau f)+b\zeta (f^{\mathrm{*}},\tau f^{\mathrm{*}})=\mathrm{0},$ lead to a contradiction.

For $\tau$ is a Picard operator, since $a+b=\mathrm{1}$ and $a,b\in (\mathrm{0,1})$, we can infer that

$\begin{array}{l}\zeta (f_{n+\mathrm{1}},f^{\mathrm{*}})=\zeta (\tau f_n,\tau f^{\mathrm{*}})<a\zeta (f_n,\tau f_n)+b\zeta (f^{\mathrm{*}},\tau f^{\mathrm{*}})\\ =a\zeta (f_n,f_{n+\mathrm{1}})+b\zeta (f^{\mathrm{*}},f^{\mathrm{*}})=a\zeta (f_n,f_{n+\mathrm{1}})\to \mathrm{0}(n\to \mathrm{\infty }).\end{array}$

Hence, $f_{n+\mathrm{1}}={\tau }^n{f}_{\mathrm{0}}\to f^{\mathrm{*}}(n\to \mathrm{\infty })$, i.e., $\tau$ is a Picard operator.

Theorem 2   Let $(M,\zeta )$ be a $\tau$-orbitally compact DCMLS and $\tau :M\to M$ be an extended Kannan-type contraction. The mapping $\tau$ is $f_{\mathrm{0}}$-orbitally continuous for some $f_{\mathrm{0}}\in M$ or weakly orbitally continuous, then $\tau$ has a unique fixed point $f^{\mathrm{*}}\in M$ and for each $f\in M$ the iterated sequence ${\tau }^{n}f$ converges to $f^{\mathrm{*}}$, i.e., $\tau$ is a Picard operator.

Proof   The proof is similar to Theorem 1. We first get the sequence $f_n=\tau f_{n-\mathrm{1}}={\tau }^n{f}_{\mathrm{0}},n\ge \mathrm{1}.$ Suppose $f_n\ne f_{n+\mathrm{1}}$ for all $n\in \mathbb{N}$. Indeed, if for some $n\in \mathbb{N},f_n=f_{n+\mathrm{1}}=\tau f_n$, then $f_n$ is the fixed point of $\tau$. Without loss of generality, we assume that $f_n\ne f_{n+\mathrm{1}}$ for all $n\in \mathbb{N}$, then $u_n\to u^{\mathrm{*}}\ge \mathrm{0}\left(n\to \mathrm{\infty }\right)$ by using analogous arguments as above. Because $M$ is $\tau$-orbitally compact, there is a convergent subsequence $\{f_{n_i}\}$ of $\left\{f_n\right\}$ and $f^{\mathrm{*}}\in M$ such that $f_{n_i}\to f^{\mathrm{*}}\mathrm{a}\mathrm{s}i\to \mathrm{\infty }$. By $f_{\mathrm{0}}$-orbital continuity or weak orbital continuity of $\tau$, we obtain $\tau f_{n_i}\to \tau f^{\mathrm{*}}$. The rest proof is similar to Theorem 1.

Remark 1   The main theorems in our paper are extensions of the results in Refs. [17-18]. The results in these literature always rely strongly on the compactness of the metric spaces. Moreover, Theorem 2.2 in Ref. [17] needs the continuity of the mapping and Theorem 2.1, 2.2 in Ref. [18] need orbital continuity or $k$-continuity. However, the continuity condition is weakened by weak orbital continuity and the compactness of the spaces is relaxed by bounded compactness and orbital compactness in our results. In addition, we extend $a=b=\frac{\mathrm{1}}{\mathrm{2}}$ to $a,b\in (\mathrm{0,1})$. As a consequence, we have more conveniences for applications in the future.

From the above two theorems, we immediately obtain the following several assertions.

Corollary 1   Let $(M,\zeta )$ be a boundedly compact DCMTS and $\tau :M\to M$ be an extended Kannan-type contraction. The mapping $\tau$ is $f_{\mathrm{0}}$-orbitally continuous for some $f_{\mathrm{0}}\in M$ or weakly orbitally continuous, then $\tau$ has a unique fixed point $f^{\mathrm{*}}\in M$ and for each $f\in M$ the iterated sequence ${\tau }^{n}f$ converges to $f^{\mathrm{*}}$, i.e., $\tau$ is a Picard operator.

Corollary 2   Let $(M,\zeta )$ be a $\tau$-orbitally compact DCMTS and $\tau :M\to M$ be an extended Kannan-type contraction. The mapping $\tau$ is $f_{\mathrm{0}}$-orbitally continuous for some $f_{\mathrm{0}}\in M$ or weakly orbitally continuous, then $\tau$ has a unique fixed point $f^{\mathrm{*}}\in M$ and for each $f\in M$ the iterated sequence ${\tau }^{n}f$ converges to $f^{\mathrm{*}}$, i.e., $\tau$ is a Picard operator.

Corollary 3   Let $(M,\zeta )$ be a boundedly compact CMTS and $\tau :M\to M$ be an extended Kannan-type contraction. The mapping $\tau$ is $f_{\mathrm{0}}$-orbitally continuous for some $f_{\mathrm{0}}\in M$ or weakly orbitally continuous, then $\tau$ has a unique fixed point $f^{\mathrm{*}}\in M$ and for each $f\in M$ the iterated sequence ${\tau }^{n}f$ converges to $f^{\mathrm{*}}$, i.e., $\tau$ is a Picard operator.

Corollary 4   Let $(M,\zeta )$ be a $\tau$-orbitally compact CMTS and $\tau :M\to M$ be an extended Kannan-type contraction. The mapping $\tau$ is $f_{\mathrm{0}}$-orbitally continuous for some $f_{\mathrm{0}}\in M$ or weakly orbitally continuous, then $\tau$ has a unique fixed point $f^{\mathrm{*}}\in M$ and for each $f\in M$ the iterated sequence ${\tau }^{n}f$ converges to $f^{\mathrm{*}}$, i.e., $\tau$ is a Picard operator.

Corollary 5   Let $(M,\zeta )$ be a boundedly compact EbMS and $\tau :M\to M$ be an extended Kannan-type contraction. The mapping $\tau$ is $f_{\mathrm{0}}$-orbitally continuous for some $f_{\mathrm{0}}\in M$ or weakly orbitally continuous, then $\tau$ has a unique fixed point $f^{\mathrm{*}}\in M$ and for each $f\in M$ the iterated sequence ${\tau }^{n}f$ converges to $f^{\mathrm{*}}$, i.e., $\tau$ is a Picard operator.

Corollary 6   Let $(M,\zeta )$ be a $\tau$-orbitally compact EbMS and $\tau :M\to M$ be an extended Kannan-type contraction. The mapping $\tau$ is $f_{\mathrm{0}}$-orbitally continuous for some $f_{\mathrm{0}}\in M$ or weakly orbitally continuous, then $\tau$ has a unique fixed point $f^{\mathrm{*}}\in M$, and for each $f\in M$, the iterated sequence ${\tau }^{n}f$ converges to $f^{\mathrm{*}}$, i.e., $\tau$ is a Picard operator.

There is an example in which the mapping $\tau$ has a fixed point since it satisfies our results. However, the space is incomplete and the mapping is not continuous, which means that our results are meaningful.

Example 3 Let $M=(-\mathrm{1,1}]\bigcup \left\{\mathrm{2}\right\}$ and define the metric $\zeta :M\times M\to {\mathbb{R}}^{+}$ by

$\zeta (f,g)=\{\begin{array}{ll}k,& f=g=\mathrm{0},\\ k|f-g|,& \mathrm{o}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{w}\mathrm{i}\mathrm{s}\mathrm{e}\end{array}$

for all $f,g\in M$, where $k>\mathrm{0}$. The functions $\alpha ,\beta :M\times M\to [\mathrm{1},+\mathrm{\infty })$ are written as

$\alpha (f,g)=\{\begin{array}{ll}\frac{\mathrm{1}}{|g|}+\mathrm{1},& f=\mathrm{0},g\ne \mathrm{0},\\ \mathrm{1},& \mathrm{o}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{w}\mathrm{i}\mathrm{s}\mathrm{e},\end{array}$

and $\beta (f,g)=\{\begin{array}{ll}|g|+\mathrm{1},& f=\mathrm{0},g\ne \mathrm{0},\\ \mathrm{1},& \mathrm{o}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{w}\mathrm{i}\mathrm{s}\mathrm{e}.\end{array}$

Furthermore, define the mapping $\tau :M\to M$ by:

$\tau f=\{\begin{array}{ll}\frac{f}{\mathrm{5}},& f\ne \mathrm{1},\\ \mathrm{0},& f=\mathrm{1}.\end{array}$

Obviously, $\tau$ is not continuous but $f_{\mathrm{0}}$-orbitally continuous for $f_{\mathrm{0}}\in (-\mathrm{1,1})\bigcup \{\mathrm{2}\}$. Moreover, $(M,\zeta )$ is an incomplete but $\tau$-orbitally compact DCMLS. Take $a=\frac{\mathrm{1}}{\mathrm{4}},b=\frac{\mathrm{3}}{\mathrm{4}}$. There are the following two cases for $f\ne g.$

(i) If $f=\mathrm{1}$ and $g\ne \mathrm{1}$, then

$\begin{array}{l}\zeta (\tau f,\tau g)=k\left|\frac{g}{\mathrm{5}}-\mathrm{0}\right|<\frac{\mathrm{1}}{\mathrm{4}}k\left|\mathrm{1}-\mathrm{0}\right|+\frac{\mathrm{3}}{\mathrm{4}}k\left|g-\frac{g}{\mathrm{5}}\right|\\ =a\zeta (f,\tau f)+b\zeta (g,\tau g).\end{array}$

On the other hand, if $f\ne \mathrm{1}$ and $g=\mathrm{1}$, it is obviously true by the symmetry of the double controlled metric.

(ii) If $f,g\in M$ with $f\ne \mathrm{1},g\ne \mathrm{1}$ and $f\ne g$, then

$\begin{array}{l}\zeta (\tau f,\tau g)=k\left|\frac{f}{\mathrm{5}}-\frac{g}{\mathrm{5}}\right|\le k\left|\frac{f}{\mathrm{5}}\right|+k\left|\frac{g}{\mathrm{5}}\right|<\frac{\mathrm{1}}{\mathrm{4}}k\left|f-\frac{f}{\mathrm{5}}\right|\\ +\frac{\mathrm{3}}{\mathrm{4}}k\left|g-\frac{g}{\mathrm{5}}\right|=a\zeta (f,\tau f)+b\zeta (g,\tau g).\end{array}$

Therefore, the mapping $\tau$ has a unique fixed point in $M$ by Theorem 1. Since $\tau (\mathrm{0,0})=k>\mathrm{0}$, the space is not a DCMTS (or CMTS, EbMS etc.). Furthermore, $a=b=\frac{\mathrm{1}}{\mathrm{2}}$ is a special case of $a+b=\mathrm{1},$$a,b\in (\mathrm{0,1})$. Hence, the existing theorems in the literature are not applicable here.

3 Asymptotic Regularity and Orbital Completeness

In the following, we obtain the fixed point theorems of generalized contractive mapping in orbitally complete DCMLS, under the condition of asymptotic regularity. The continuity of the mapping is relaxed. At first, we give the definitions of asymptotic regularity and $\tau$-orbital completeness in DCMLS.

Definition 5   Let $(M,\zeta )$ be a DCMLS. The mapping $\tau :M\to M$ is said to be asymptotically regular, if $\underset{n\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\zeta ({\tau }^{n+\mathrm{1}}f,{\tau }^{n}f)=\mathrm{0}$ for all $f\in M$.

Definition 6   Let $(M,\zeta )$ be a DCMLS. The space $(M,\zeta )$ is said to be $\tau$-orbitally complete, if every $\zeta$-Cauchy sequence which is contained in $O_{\tau }(f)$ for some $f\in M$ converges in $M$. Every complete DCMLS is $\tau$-orbitally complete for any $\tau$, but a $\tau$-orbitally complete DCMLS needs not be complete.

The following theorem is obtained under a generalized version of extended Kannan-type contraction, where $a,b\in [\mathrm{0},+\mathrm{\infty })$.

Theorem 3   Let $\tau :M\to M$ be an asymptotically regular mapping on the $\tau$-orbitally complete DCMLS. The mapping $\tau$ is $f_{\mathrm{0}}$-orbitally continuous for some $f_{\mathrm{0}}\in M$ or weakly orbitally continuous and satisfies

$\zeta (\tau f,\tau g)\le a\zeta (f,\tau f)+b\zeta (g,\tau g),$(2)

for all $f,g\in M$, where $a,b\in [\mathrm{0},+\mathrm{\infty })$. Then, $\tau$ has a unique fixed point $f^{\mathrm{*}}\in M$ and for each $f\in M$ the iterated sequence ${\tau }^{n}f$ converges to $f^{\mathrm{*}}$, i.e.,$\tau$ is a Picard operator.

Proof   For arbitrary $f_{\mathrm{0}}\in M$, set $f_n=\tau f_{n-\mathrm{1}}={\tau }^n{f}_{\mathrm{0}},n\ge \mathrm{1}.$ Suppose $f_n\ne f_{n+\mathrm{1}}$ for all $n\in \mathbb{N}$. Indeed, if for some $n\in \mathbb{N},f_n=f_{n+\mathrm{1}}=\tau f_n$, then $f_n$ is the fixed point of $\tau$. Thus, we consider that $f_n\ne f_{n+\mathrm{1}}$ for any $n\in \mathbb{N}$. For all $m>n$, we have $\zeta (f_n,f_m)=\zeta (\tau f_{n-\mathrm{1}},\tau f_{m-\mathrm{1}})\le a\zeta (f_{n-\mathrm{1}},\tau f_{n-\mathrm{1}})+b\zeta (f_{m-\mathrm{1}},$

$\tau f_{m-\mathrm{1}})=a\zeta (f_{n-\mathrm{1}},f_n)+b\zeta (f_{m-\mathrm{1}},f_m).$

According to asymptotic regularity, $\underset{n\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\zeta (f_{n-\mathrm{1}},f_n)=\mathrm{0}$, so $\left\{f_n\right\}$ is a $\zeta$-Cauchy in $M$. Since $(M,\zeta )$ is $\tau$-orbitally complete, there exists $f^{\mathrm{*}}\in M$ such that $f_n\to f^{\mathrm{*}}\mathrm{a}\mathrm{s}n\to \mathrm{\infty }$. We shall prove $\tau f^{\mathrm{*}}=f^{\mathrm{*}}$, i.e., $f^{\mathrm{*}}$ is the fixed point of $\tau$.

If $\tau$ is $f_{\mathrm{0}}$-orbitally continuous, then $f_n={\tau }_n{f}_{\mathrm{0}}\to f^{\mathrm{*}}$for each $f_{\mathrm{0}}\in M$.

If $\tau$ is weakly orbitally continuous, then ${\tau }^n{f}_{\mathrm{0}}\to f^{\mathrm{*}}$ for each $f_{\mathrm{0}}\in M$. By virtue of weak orbital continuity of $\tau$, we know ${\tau }^n{g}_{\mathrm{0}}\to f^{\mathrm{*}}$ and ${\tau }^{n+\mathrm{1}}{g}_{\mathrm{0}}\to \tau f^{\mathrm{*}}$ for some $g_{\mathrm{0}}$ in $M$. It implies $f^{\mathrm{*}}=\tau f^{\mathrm{*}}$ since ${\tau }^{n+\mathrm{1}}{g}_{\mathrm{0}}\to f^{\mathrm{*}}$. $f^{\mathrm{*}}$ is a fixed point of $\tau$.

For the uniqueness of $f^{\mathrm{*}}$, if $g^{\mathrm{*}}\in M$, $\tau g^{\mathrm{*}}=g^{\mathrm{*}}$, then

$\zeta (f^{\mathrm{*}},g^{\mathrm{*}})=\zeta (\tau f^{\mathrm{*}},\tau g^{\mathrm{*}})\le a\zeta (f^{\mathrm{*}},\tau f^{\mathrm{*}})+b\zeta (g^{\mathrm{*}},\tau g^{\mathrm{*}})=\mathrm{0},$

which yields $f^{\mathrm{*}}=g^{\mathrm{*}}$. Hence, $f^{\mathrm{*}}$ is the unique fixed point of $\tau$ and $f_n={\tau }_n{f}_{\mathrm{0}}\to f^{\mathrm{*}}(n\to \mathrm{\infty })$.

Corollary 7   Let $\tau :M\to M$ be an asymptotically regular mapping on the $\tau$-orbitally complete DCMLS. The mapping $\tau$ is $f_{\mathrm{0}}$ -orbitally continuous for some $f_{\mathrm{0}}\in M$ or weakly orbitally continuous and satisfies

$\zeta (\tau f,\tau g)\le k[\zeta (f,\tau f)+\zeta (g,\tau g)]$(3)

for all $f,g\in M$, where $k\in [\mathrm{0},+\mathrm{\infty })$. Then, $\tau$ has a unique fixed point $f^{\mathrm{*}}\in M$, and for each $f\in M$, the iterated sequence ${\tau }^{n}f$ converges to $f^{\mathrm{*}}$, i.e., $\tau$ is a Picard operator.

Now, if $a\in [\mathrm{0},+\mathrm{\infty }),b\in (\mathrm{0,1})$ and the noncomparable functions $\alpha :M\times M\to [\mathrm{1},+\mathrm{\infty }),$ $\beta :M\times M\to [\mathrm{1},\frac{\mathrm{1}}{b})$, both the limits $\underset{n\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\alpha (f,f_n)$ and $\underset{n\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\beta (f,f_n)$ exist and are finite for any $f,f_{\mathrm{0}}\in M$ and $f_n={\tau }^n{f}_{\mathrm{0}},n\ge \mathrm{1}$, then the condition "$\tau$ is $f_{\mathrm{0}}$ -orbitally continuous for some $f_{\mathrm{0}}\in M$ or weakly orbitally continuous" in Theorem 3 can be deleted.

Theorem 4   Let $\tau :M\to M$ be an asymptotically regular mapping on the $\tau$-orbitally complete DCMLS. The mapping $\tau$ satisfies

$\zeta (\tau f,\tau g)\le a\zeta (f,\tau f)+b\zeta (g,\tau g),$(4)

for all $f,g\in M$, where $a\in [\mathrm{0},+\mathrm{\infty }),$$b\in (\mathrm{0,1})$. If for any $f,f_{\mathrm{0}}\in M$, $f_n={\tau }^n{f}_{\mathrm{0}}$ and the noncomparable functions $\alpha :M\times M\to [\mathrm{1},+\mathrm{\infty }),\beta :M\times M\to [\mathrm{1},\frac{\mathrm{1}}{b})$, both the limits $\underset{n\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\alpha (f,f_n)$ and $\underset{n\to \mathrm{\infty }}{\mathrm{l}\mathrm{i}\mathrm{m}}\beta (f,f_n)$ exist and are finite, then $\tau$ has a unique fixed point $f^{\mathrm{*}}\in M$, and for each $f\in M$, the iterated sequence ${\tau }^{n}f$ converges to $f^{\mathrm{*}}$, i.e., $\tau$ is a Picard operator.