# Alignment Transport Between Ultracold Polar Molecules

J. Smucker  Department of Physics and Astronomy, Stony Brook University, Stony
Brook, 11790 USA<EMAIL_ADDRESS>J. Pérez-Ríos  Department of
Physics and Astronomy, Stony Brook University, Stony Brook, 11790 USA

###### Abstract

We propose an array of ultracold polar molecules as a platform to study
alignment transport between molecules. We envision a polar molecule being
aligned with an intense off-resonant laser field whose alignment migrates to a
nearby molecule due to dipole-dipole interactions. Our results show that the
transport of the alignment is due to a complex interplay between electric
field-driven excitations and dipole-dipole interactions. All mechanisms for
transfer are elucidated and analyzed, and using NaCs, a prototype molecule, we
find that the time for alignment transfer is $\mathcal{O}(10\mu\rm s)$, which
makes the phenomena readily observable in the lab.

††preprint: APS/123-QED

Light-matter interactions are key in many fields of physics, especially in
laser induced alignment of molecules [1] within atomic, molecular and optical
physics. When an intense off resonant laser hits a molecule, the molecule will
align with the laser polarization axis. In this scenario, there are two
competing time scales at play: the rotational period and the pulse duration.
When the pulse duration is shorter than the rotational period, entering in the
so-called non-adiabatic limit, the alignment experiences revivals which
periodicity is controlled by the rotational constant of the molecule [2, 3, 4,
5, 6, 7, 8, 9]. This effect is seen in isolated molecules as well as in
molecules in a quantum solvent, such as helium droplets. In principle, in the
case of multiple molecules the alignment could transfer to nearby molecules
because of dipole-dipole interactions, which we hypothesize could be observed
in arrays of ultracold polar molecules.

Ultracold polar molecules are getting attention due to their prospective
applications in quantum information sciences, many-body physics, cold and
ultracold chemistry and quantum metrology [10, 11, 12, 13, 14, 15, 16, 17].
For instance, arrays of ultracold polar molecules offer a unique scenario to
study many-body physics such as exciton dynamics [18], quantum synchronization
[19], other many-body phenomena [20, 21, 22], and even as a quantum simulator
of complex spin Hamiltonians [23, 24, 25, 26]. Indeed, alignment transport
could be better explored in arrays of ultracold polar molecules. As we will
discuss, the transport of the alignment is a complicated process involving
multiple intermediate states. This complexity combined with the possibility of
expanding to systems of more dipoles and different geometries offer a rich new
topic to be explored.

Figure 1: Schematic of alignment transport. Fig: 1a shows the states of the
first molecule before and after the laser pulse. Fig: 1b is a representation
of the two molecule system. Fig: 1c is a sketch of the level diagram of the
two dipole system. The laser pulse excites the first molecule to the $j=2$
state and this excitation is passed to the second molecule via dipole-dipole
interactions.

In this letter, we study alignment transfer between two dipoles in an array
assisted by the dipole-dipole interaction, as shown schematically in panel (b)
of Fig: 1. First, a laser pulse excites a molecule, populating the $j=2$
rotational state and inducing the molecules alignment, as shown in panel (a).
After this, the presence of a constant electric field plus the dipole-dipole
interaction transfers the alignment to the second dipole, as shown in panel
(c). Our results show that the transport mechanism is very involved and can be
efficiently controlled based on molecular and laser field properties. We will
use a combination of techniques to understand the underlying quantum mechanics
behind transferring the excitation created by the laser pulse. This
understanding will allow us to narrow down the number of states that must be
included in our calculations, speeding up our calculations and helping speed
up future calculations of larger systems. Many states are required to
accurately predict the transport of the alignment because multiple pathways
are involved. The effects of the distance between the dipoles, the strength of
the dipoles, the rotational constant, and the properties of the laser pulse
are all discussed.

In the case of ultracold molecules, the kinetic energy is not enough to excite
the vibrational degrees of freedom, and hence it is a good approximation to
treat the molecule as a rigid rotor. In this case, the Hamiltonian of the
system consists of four different parts:

$H(t)=H_{1,\text{rot}}+H_{2,\text{rot}}+H_{E}+H_{dd}+H_{Lsr}(t).$ (1)

$H_{i,\text{rot}}$ is the rotational Hamiltonian of the ith molecule.
$H_{i,\text{rot}}=B_{0,i}\hbar^{2}j_{i}(j_{i}+1)$, where $B_{0,i}$ is the ith
molecule’s equilibrium rotational constant. $H_{E}=-\bm{\mu}\cdot\bm{E}$ is
the Hamiltonian due to the electric field with $\bm{\mu}$ being the dipole of
the molecule and $\bm{E}$ being the electric field. $H_{Lsr}$(t) is the
Hamiltonian due to the laser pulse. $H_{dd}$ is the Hamiltonian due to the
dipole-dipole interaction. We assume that the z-component of the angular
momentum $m$ is always zero for both molecules, allowing our states to be
represented in Dirac notation as $|j_{1},j_{2}\rangle$ with $j_{1}$ and
$j_{2}$ representing the total angular momentum quantum numbers for the two
dipoles. We also assume our initial state is the $|0,0\rangle$ state.

The dipole-dipole Hamiltonian can be represented as [27]:

$H_{dd}=\frac{\mu^{2}}{R^{3}}(1-3\cos^{2}{\alpha})\cos{\theta_{1}}\cos{\theta_{2}},$
(2)

assuming the two dipoles have the same dipole moment. $R$ is the distance
between the dipoles and $\theta_{i}$ is the angle of the corresponding
$i^{th}$ molecule, shown in Fig: 1b. We set $\alpha=\pi/2$. The laser pulse’s
Hamiltonian for a symmetric top or linear molecule is [28, 8]:

$H_{Lsr}=-\frac{(E_{Lsr}\epsilon(t))^{2}}{4}(\alpha_{yy}+\alpha_{\Delta}\cos^{2}{\theta}),$
(3)

where $E_{Lsr}$ is the strength of the laser pulse, $\epsilon(t)$ is the laser
pulse profile and $\alpha_{\Delta}=\alpha_{zz}-\alpha_{yy}$. $\alpha_{yy}$ and
$\alpha_{zz}$ are the polarizabilities of the molecule being hit by the laser
in the $y$ and $z$ directions respectively. Since we are assuming our molecule
is symmetric around the z-axis, $\alpha_{yy}=\alpha_{xx}$. Here, we chose NaCs
as our molecule with a rotational constant of $0.058$ cm-1 [29, 30], a
polarizability in the parallel direction of $\alpha_{zz}=669.26$ a.u and a
polarizability in the perpendicular direction of $\alpha_{yy}=328.25$ a.u. The
polarizabilities were calculated using the MP2 level of theory as implemented
in the Molpro quantum chemistry package [31], at the experimental equilibrium
distance [30] using the AVQZ basis set Na [32] and def2-QZVPP for Cs [33].
These polarizabilities only impact the effect the laser pulse has on the
population of the rotational state and do not impact the alignment transport
process. We can ignore the interaction between the laser’s electric field and
the molecules dipole moment since the laser pulses we consider contain
multiple cycles of the electric field causing this interaction to average to
approximately zero[28, 8]. We assume a Gaussian pulse:
$\epsilon(t)=e^{-2\log{2}~{}t^{2}/\tau^{2}}$ with $\tau$ being a time constant
that changes the width. For our simulations we set $\tau=1/3$ ps, and hence,
we are in the non-adiabatic limit, where revivals in the alignment are
expected. We also used a peak laser pulse intensity of $3\times 10^{13}$ W/cm2
and a constant electric field strength of $0.5$ kV/cm unless otherwise stated.

We delay the peak of the laser pulse by $2$ ps, as shown in panel (a) of Fig.
1. We solve the time dependent Schrödinger equation numerically up to $4$ ps
(we set this time to be $t=0$), at this point the laser field is effectively
zero. Since $H_{Lsr}(t)$ is the only part of the Hamiltonian that explicitly
depends on time, the Hamiltonian can be treated as time independent after $4$
ps allowing for the time dependent Schrödinger equation to be solved using
separation of variables from this point on. We take the state at $4$ ps from
the numerical solution to the Schrödinger equation and use it as an initial
state for the separation of variables solution. We found that by including $3$
rotational states per molecule, the results are roughly the same as
simulations that take into account many more rotational states.

Fig: 2a shows the result of a simulation with two NaCs molecules separated by
a distance of $100$ nm. This figure also shows the population of the
$|2,0\rangle$ and the $|0,2\rangle$ states as a function of time. As expected,
the alignment on dipole 1 shows revivals. However, due to the presence of the
second dipole the lowest alignment of dipole 1 corresponds to the largest
alignment of dipole 2, thus, showing how the alignment hops between the two
dipoles. The rapid oscillations of the alignment, highlighted in Fig: 2b, are
mostly due to the energy gap between the $|0,0\rangle$ and the $|2,0\rangle$
states. This short period is therefore around $\pi/3B_{0}$ in atomic units.

Figure 2: The top curves in panel (a) show the alignment as a function of time
for two NaCs molecules separated by a distance of $100$ nm. The lower curves
in Fig: 2a show the population of the $|2,0\rangle$ and the $|0,2\rangle$
states as a function of time. Panel (b) highlights the rapid oscillations of
the alignment. Panel (c) shows a schematic view of the most important pathways
for the transfer of the population between the $|2,0\rangle$ and the
$|0,2\rangle$ states. The text above the arrows show the mechanism allowing
for these individual transitions.

It is clear from Fig: 2a that the transport of the alignment excitation
corresponds to a transfer of the population from the $|2,0\rangle$ state to
the $|0,2\rangle$ state. These states are not coupled directly via the dipole-
dipole interaction, meaning the process must involve at least one intermediate
states.

After an extensive study on the role of the rotational states, we found the
minimum set of rotational states leading to the same results as using the
complete manifold. The relevant states are: $|0,0\rangle$, $|1,0\rangle$,
$|2,0\rangle$, $|0,1\rangle$, $|1,1\rangle$, $|2,1\rangle$, $|0,2\rangle$,
$|1,2\rangle$, $|2,2\rangle$. Given these states, there are three dominant
pathways for the alignment excitation to transfer. These pathways are shown in
Fig: 2c. The middle pathway in Fig: 2c contains only one intermediate state
but the dipole-dipole interactions are much slower than the E-field
interactions. Therefore, this pathway is less dominant then the other too.
Since two of the three pathways depend on the E-field, increasing the E-field
will increase the rate of alignment transport slightly. Although, too large of
an E-field will mix all the states too efficiently and drastically change the
underlying physics.

It is not immediately clear why the $|2,2\rangle$ state is relevant as it does
not appear in any of the most relevant pathways. By removing the $|2,2\rangle$
state, the rate of alignment transport increases noticeably. When setting the
dipole-dipole interaction between the $|2,2\rangle$ and the $|1,1\rangle$
states to zero, the alignment transport rate does not change. This indicates
that it is the E-field interactions with the $|2,1\rangle$ and $|1,2\rangle$
states that make the $|2,2\rangle$ state relevant. Therefore, the
$|2,2\rangle$ state slows down the rate of alignment transport by diverting
part of the $|2,1\rangle$ and $|1,2\rangle$ states’ populations. Similarly, by
setting the dipole-dipole interaction between the $|1,1\rangle$ and the
$|0,2\rangle$ state to zero (effectively eliminating the pathway shown in the
middle of Fig: 2c) the rate of alignment transport increases. Again the
$|1,1\rangle$ state and the slower middle pathway in Fig: 2c divert part of
the population from taking the other two faster pathways.

Figure 3: The period of the alignment transport as a function of dipole
strength and distance. The rotational constant dependence is show in the
inset. These calculations were preformed with an E-field strength of $0.25$
kV/cm.

Using Fourier analysis the period of the transport of the alignment can be
calculated from our simulations. From this we can find the dipole strength,
distance and rotational constant dependence of the period of the alignment
excitation. This dependence is shown in Fig: 3. The results for the dipole
dependence and distance dependence. shown in Fig: 3, were fit to the function:

$T=a\frac{R^{3}}{\mu^{4}},$ (4)

where $a$ is the fit parameter, $R$ is the distance between the dipoles and
$\mu$ is the dipole strength. This function fit to the results from the
simulations with an R-Squared value of $0.999957$. This function offered the
best fit when compared to functions involving other powers of $R$ and $\mu$.
Unsurprisingly, the period depends on $R^{3}$ as the strength of the dipole-
dipole interaction also depends on $R^{3}$. As we can tell from the relevant
pathways shown in Fig: 2b, the dipole strength $\mu$ dependence is more
complex since both the E-field interactions and the dipole-dipole interactions
depend on $\mu$. E-field interactions include a single power of $\mu$ while
the dipole-dipole interactions involve $\mu^{2}$. Since the more dominate
pathways (the top pathway and bottom pathways in Fig: 2b) both involve two
E-field interactions and one dipole-dipole interaction their contributions to
the period go as $\mu^{4}$, which is the period of the alignment transport is
proportional to $\mu^{-4}$. The inset in Fig: 3 shows the rotational constant
dependence which fits to a line with an R-Squared value of $0.999977$.

Up until this point our study has focused on relatively low energy laser
pulses. Assuming that the system starts completely in the $|0,0\rangle$ state,
when the peak of the pulse is small enough the population remains mostly in
the $|0,0\rangle$ state with part of the population being pumped into the
$|2,0\rangle$ state. When the energy of the laser pulse is too great the
system starts to be pumped into the $|4,0\rangle$ state. The $|4,0\rangle$
state transfers its population to the $|0,4\rangle$ state (through similar
multi-state pathways as the $|2,0\rangle$ and $|0,2\rangle$ pathways) but this
process takes orders of magnitudes longer. For example, a system of two NaCs
molecules separated by a distance of $100$ nm has a predicted period for
complete alignment transport of $249$ s if the $|4,0\rangle$ state is
populated, compared to the $65$ $\mu$s it takes when the $|4,0\rangle$ state
isn’t populated. Part of the alignment will still transfer to the observer
molecule through the $|2,0\rangle$ and $|0,2\rangle$ pathways with each
iteration of this process resulting in more and more of the total alignment
excitation being transported. As long as $|4,0\rangle$ state isn’t
significantly populated the strength of the laser pulse will only affect the
magnitude of the alignment and will not affect the period of the alignment
transport. The overall shape of the laser pulse should not have an impact on
the alignment as long as the same amount of total energy is being pumped into
the system. This is true until the laser pulse’s duration approaches the
period of the short oscillations due to the electric field, i.e., we are in
the non-adiabatic regime.

Due to extensive studies of laser cooling and trapping polar molecules[10,
15], experimental studies of alignment transport should be feasible. Table 1
has the relevant data for a few common polar molecules as well as the
predicted period of the alignment transport. The estimated periods are on the
same time scale as estimates for CNOT gates using polar molecules [10]. The
ideal molecule has a large dipole moment and a small rotational constant.

Table 1: The periods for alignment transport for various different diatomic molecules. For all simulations the static electric field strength was $0.5$ kV/cm and the peak laser intensity was $3\times 10^{13}$ W/cm2. The distances between the molecules was set to $100$ nm. The dipole strengths, the equilibrium rotational constants and the polarizabilities were predicted from MP2 calculations using AVQZ basis set for Li, Na and K, and the def2-QZVPP basis set for Rb and Cs. Species |  | $B_{0}$
---
cm-1
| $\mu$
---
(Debye)
| $\alpha_{zz}$
---
(a.u)
| $\alpha_{yy}$
---
(a.u)
| Period
---
($\mu$s)
NaCs | 0.058 | 4.69 | 669.26 | 328.25 | 65.43
LiCs | 0.1907 | 5.5 | 586.41 | 292.93 | 308.38
RbCs | 0.015 | 1.24 | 1040.98 | 447.22 | 926.75
LiRb | 0.2186 | 4.1 | 500.41 | 266.53 | 1280.82
KRb | 0.037 | 0.6 | 711.35 | 385.43 | 88021.2

Despite many studies of alignment and of dipole-dipole interactions, studies
combining these two effects are minimal. Our results show that ultracold polar
molecules can find new applications in the study of alignment transport.
Similarly, our findings indicate that the transport time occurs on time scales
relevant to quantum gates and can be readily observed in the lab. The relevant
states and most important pathways for the alignment transport have been
predicted from our simulations. The period of the alignment transport has been
determined to depend on the distance cubed and the dipole strength to the
negative fourth power. This marks only an introductory study into alignment
transport, as a large portion of parameter space is still yet to be explored.
Of particular interest is the study of systems involving many dipoles, which
drastically increases the parameter space. Preliminary results show that the
transport time is reduced when introducing a third dipole in the system,
indicating that in an array of ultracold polar molecules, the alignment could
efficiently transfer among them. This will be the topic of future studies.

The authors acknowledge the support of the Simons Foundation.

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# New directions in fixed point theory in $G$-metric spaces and applications
to mappings contracting perimeters of triangles

Mohamed Jleli111Department of Mathematics, College of Science, King Saud
University, Riyadh 11451, Saudi Arabia; E-mail address<EMAIL_ADDRESS>Cristina Maria Păcurar222Faculty of Mathematics and Computer Science,
Transilvania University of Braşov, 50 Iuliu Maniu Blvd., Braşov, Romania;
E-mail address<EMAIL_ADDRESS>Bessem Samet333Department of
Mathematics, College of Science, King Saud University, Riyadh 11451, Saudi
Arabia; E-mail address<EMAIL_ADDRESS>
###### Abstract

We are concerned with the study of fixed points for mappings $T:X\to X$, where
$(X,G)$ is a $G$-metric space in the sense of Mustafa and Sims. After the
publication of the paper [Journal of Nonlinear and Convex Analysis. 7(2)
(2006) 289–297] by Mustafa and Sims, a great interest was devoted to the study
of fixed points in $G$-metric spaces. In 2012, the first and third authors
observed that several fixed point theorems established in $G$-metric spaces
are immediate consequences of known fixed point theorems in standard metric
spaces. This observation demotivated the investigation of fixed points in
$G$-metric spaces. In this paper, we open new directions in fixed point theory
in $G$-metric spaces. Namely, we establish new versions of the Banach, Kannan
and Reich fixed point theorems in $G$-metric spaces. We point out that the
approach used by the first and third authors [Fixed Point Theory Appl. 2012
(2012) 1–7] is inapplicable in the present study. We also provide some
interesting applications related to mappings contracting perimeters of
triangles.

2010 Mathematics Subject Classification: 47H10, 54E50, 47S20.
Key words and phrases: Fixed point, $G$-metric, mappings contracting
perimeters of triangles.

##  1 Introduction

The notion of $G$-metric spaces was introduced by Mustafa and Sims in [8].
After the publication of this work, a great attention was accorded to the
study of fixed points in such spaces, see e.g. [1, 2, 5, 7, 9, 14, 16, 18] and
the references therein. In [4] (see also [17]), Jleli and Samet remarked that
several fixed point theorems in $G$-metric spaces can be deduced immediately
from fixed point theorems in (standard) metric spaces. Namely, it was shown
that, if $G:X\times X\times X\to\mathbb{R}^{+}$ is a $G$-metric on $X$, then
the mapping $\delta:X\times X\to\mathbb{R}^{+}$ defined by
$\delta(x,y)=\max\\{G(x,x,y),G(y,x,x)\\}$ is a metric on $X$. Moreover, if
$(X,G)$ is a complete $G$-metric space, then $(X,\delta)$ is a complete metric
space. From this observation, it was proved that several contractions studied
in complete $G$-metric spaces can be reduced to standard contractions in the
complete metric space $(X,\delta)$. For instance, if $(X,G)$ is a complete
$G$-metric space and $T:X\to X$ is a mapping satisfying the inequality
$G(Tx,Ty,Tz)\leq\lambda G(x,y,z)$ for all $x,y,z\in X$, where
$\lambda\in(0,1)$ is a constant, then (see [7]) $T$ admits a unique fixed
point. Taking $z=y$, the above inequality reduces to $G(Tx,Ty,Ty)\leq\lambda
G(x,y,y)$ for all $x,y\in X$. Replacing $(x,y,z)$ by $(y,x,x)$, we obtain
$G(Ty,Tx,Tx)\leq\lambda G(y,x,x)$ for all $x,y\in X$. Consequently, we get
$\delta(Tx,Ty)\leq\lambda\delta(x,y)$ for all $x,y\in X$, and the Banach fixed
point theorem (in metric spaces) applies. After the publications of the papers
[4, 17], the attention paid to the study of fixed point theorems in $G$-metric
spaces was attenuated.

In this paper, we open new directions in fixed point theory on $G$-metric
spaces. Namely, we establish new extensions of Banach, Kannan and Reich fixed
point theorems in $G$-metric spaces. We point out that the used approach in
[4, 17] cannot be used to establish our obtained results. Moreover, motivated
by the recent paper [11], we provide (among other applications) an application
to the study of fixed points contracting perimeters of triangles.

The rest of the paper is organized as follow. In Section 2, we briefly recall
some basic notions and properties related to $G$-metric spaces. Our main
results are stated and proved in Section 3. Finally, some applications are
provided in Section 4.

Throughout this paper, the following notations are used:
$\mathbb{R}^{+}=[0,\infty)$, $X$ denotes a nonempty set, $|X|$ denotes the
cardinal of $X$, and for a mapping $T:X\to X$, the set of fixed points of $T$
is denoted by $\operatorname{Fix}(T)$.

##  2 Preliminaries

Let us recall briefly some basic notions related to $G$-metric spaces. For
more details, we refer to Mustafa and Sims [8]. Throughout this paper, $X$
denotes a nonempty set and $\mathbb{R}^{+}=[0,\infty)$.

Let $G:X\times X\times X\to\mathbb{R}^{+}$ be a given mapping. We say that $G$
is a $G$-metric on $X$, if for all $x,y,z,w\in X$, we have

  * (P1)

$G(x,y,z)=0$ if and only if $x=y=z$;

  * (P2)

If $x\neq y$, then $G(x,x,y)>0$;

  * (P3)

$G(x,y,z)=G(\sigma(x,y,z))$ for every permutation
$\sigma:\\{x,y,z\\}\to\\{x,y,z\\}$;

  * (P4)

If $y\neq z$, then $G(x,x,y)\leq G(x,y,z)$;

  * (P5)

$G(x,y,z)\leq G(x,w,w)+G(w,y,z)$.

If the above conditions are satisfied, then $(X,G)$ is called a $G$-metric
space.

Prototype examples of $G$-metrics are

$G(x,y,z)=d(x,y)+d(y,z)+d(x,z),\quad x,y,z\in X$ (2.1)

and

$G(x,y,z)=\max\left\\{d(x,y),d(y,z),d(x,z)\right\\},\quad x,y,z\in X,$ (2.2)

where $d$ is a (standard) metric on $X$.

Let $\\{x_{n}\\}$ be a sequence in $X$ and $x\in X$. We say that $\\{x_{n}\\}$
is $G$-convergent to $x$, if
$\displaystyle\lim_{n,m\to\infty}G(x_{n},x_{m},x)=0$. The following assertions
are equivalent:

  * (A1)

$\\{x_{n}\\}$ is $G$-convergent to $x$;

  * (A2)

$\displaystyle\lim_{n\to\infty}G(x_{n},x_{n},x)=0$;

  * (A3)

$\displaystyle\lim_{n\to\infty}G(x_{n},x,x)=0$.

Notice that, if $\\{x_{n}\\}$ is $G$-convergent to $x$ and $\\{x_{n}\\}$ is
$G$-convergent to $y$, then $x=y$. This can be easily seen by (P5), (A2) and
(A3).

We say that $\\{x_{n}\\}$ is $G$-Cauchy, if
$\displaystyle\lim_{n,m\to\infty}G(x_{n},x_{m},x_{m})=0$.

If every $G$-Cauchy sequence in $X$ is $G$-convergent to an element of $X$,
then $(X,G)$ is called a complete $G$-metric space.

The following two lemmas will be useful later.

######  Lemma 2.1.

Let $d$ be a metric on $X$ and $G:X\times X\times X\to\mathbb{R}^{+}$ be the
$G$-metric on $X$ defined by (2.1). The following assertions are equivalent:

  * (i)

$(X,d)$ is a complete metric space;

  * (ii)

$(X,G)$ is a complete $G$-metric space.

###### Proof.

Assume first that $(X,d)$ is a complete metric space. Let $\\{x_{n}\\}\subset
X$ be a $G$-Cauchy sequence. For all $n,m$, we have

$\displaystyle d(x_{n},x_{m})$ $\displaystyle\leq 2d(x_{n},x_{m})$
$\displaystyle=d(x_{n},x_{m})+d(x_{m},x_{m})+d(x_{n},x_{m})$
$\displaystyle=G(x_{n},x_{m},x_{m}).$

Since $\\{x_{n}\\}$ is a $G$-Cauchy sequence, we have
$\displaystyle\lim_{n,m\to\infty}G(x_{n},x_{m},x_{m})=0$, which implies by the
above inequality that $\displaystyle\lim_{n,m\to\infty}d(x_{n},x_{m})=0$, that
is, $\\{x_{n}\\}$ is a Cauchy sequence in the metric space $(X,d)$. From the
completeness of $(X,d)$, we deduce the existence of $x\in X$ such that

$\lim_{n\to\infty}d(x_{n},x)=0.$ (2.3)

On the other hand, we have

$\displaystyle G(x_{n},x,x)$ $\displaystyle=d(x_{n},x)+d(x,x)+d(x_{n},x)$
$\displaystyle=2d(x_{n},x),$

which implies by (2.3) that $\displaystyle\lim_{n\to\infty}G(x_{n},x,x)=0$,
that is, $\\{x_{n}\\}$ is $G$-convergent to $x$. This shows that $(X,G)$ is a
complete $G$-metric space.

Assume now that $(X,G)$ is a complete $G$-metric space. Let
$\\{x_{n}\\}\subset X$ be a Cauchy sequence in the metric space $(X,d)$. From
the identity

$G(x_{n},x_{m},x_{m})=\frac{1}{2}d(x_{n},x_{m}),$

we deduce that $\displaystyle\lim_{n,m\to\infty}G(x_{n},x_{m},x_{m})=0$, that
is, $\\{x_{n}\\}$ is $G$-Cauchy. Consequently, there exists $x\in X$ such that

$\lim_{n\to\infty}G(x_{n},x,x)=0.$

Then, from the identity

$d(x_{n},x)=\frac{1}{2}G(x_{n},x,x),$

we deduce that $\displaystyle\lim_{n\to\infty}d(x_{n},x)=0$, which shows that
$(X,d)$ is a complete metric space. ∎

######  Lemma 2.2 (see [4]).

Let $d$ be a metric on $X$ and $G:X\times X\times X\to\mathbb{R}^{+}$ be the
$G$-metric on $X$ defined by (2.2). The following assertions are equivalent:

  * (i)

$(X,d)$ is a complete metric space;

  * (ii)

$(X,G)$ is a complete $G$-metric space.

##  3 Main results

###  3.1 Banach’s fixed point theorem in $G$-metric spaces

Our first main result is an extension of the Banach fixed point theorem to
$G$-metric spaces.

######  Theorem 3.1.

Let $(X,G)$ be a complete $G$-metric space with $|X|\geq 3$. Let $T:X\to X$ be
a mapping satisfying the following conditions:

  * (I)

For all $x\in X$, $T(Tx)\neq x$, provided $Tx\neq x$;

  * (II)

There exists $\lambda\in(0,1)$ such that for all pairwise distinct points
$x,y,z\in X$, we have

$G(Tx,Ty,Tz)\leq\lambda G(x,y,z).$

Then $\operatorname{Fix}(T)\neq\emptyset$ and $|\operatorname{Fix}(T)|\leq 2$.

###### Proof.

Let us first prove that $\operatorname{Fix}(T)\neq\emptyset$. Let $u_{0}\in X$
be fixed and consider the Picard sequence $\\{u_{n}\\}\subset X$ defined by

$u_{n+1}=Tu_{n},\quad n\geq 0.$

If $u_{n}=u_{n+1}$ for some $n$, then $u_{n}\in\operatorname{Fix}(T)$ and the
result is proved. So, we may suppose that

$u_{n}\neq u_{n+1},\quad n\geq 0,$

which implies by (I) that $u_{n}\neq T(Tu_{n})$ ($=u_{n+2}$). Consequently,
$u_{n},u_{n+1}$ and $u_{n+2}$ are pairwise distinct points for every $n\geq
0$. Then, making use of (II), we obtain

$\displaystyle G(u_{1},u_{2},u_{3})=G(Tu_{0},Tu_{1},Tu_{2})\leq\lambda
G(u_{0},u_{1},u_{2}),$ $\displaystyle
G(u_{2},u_{3},u_{4})=G(Tu_{1},Tu_{2},Tu_{3})\leq\lambda
G(u_{1},u_{2},u_{3})\leq\lambda^{2}G(u_{0},u_{1},u_{2}),$
$\displaystyle\vdots$ $\displaystyle
G(u_{n},u_{n+1},u_{n+2})=G(Tu_{n-1},Tu_{n},Tu_{n+1})\leq\lambda
G(u_{n-1},u_{n},u_{n+1})\leq\lambda^{n}G(u_{0},u_{1},u_{2}),$

that is,

$G(u_{n},u_{n+1},u_{n+2})\leq\tau_{0}\lambda^{n},\quad n\geq 0,$ (3.1)

where $\tau_{0}=G(u_{0},u_{1},u_{2})>0$ (by (P1)). We now show that
$\\{u_{n}\\}$ is $G$-Cauchy. First, by (P3), (P4) and using that
$u_{n},u_{n+1}$ and $u_{n+2}$ are pairwise distinct points, we obtain

$\displaystyle G(u_{n},u_{n+1},u_{n+1})$
$\displaystyle=G(u_{n+1},u_{n+1},u_{n})$ $\displaystyle\leq
G(u_{n+1},u_{n},u_{n+2})$ $\displaystyle=G(u_{n},u_{n+1},u_{n+2}),$

which implies by (3.1) that

$G(u_{n},u_{n+1},u_{n+1})\leq\tau_{0}\lambda^{n},\quad n\geq 0.$ (3.2)

We now use (3.2) and (P5) to obtain that for all $n<m$,

$\displaystyle G(u_{n},u_{m},u_{m})$ $\displaystyle\leq
G(u_{n},u_{n+1},u_{n+1})+G(u_{n+1},u_{m},u_{m})$ $\displaystyle\leq
G(u_{n},u_{n+1},u_{n+1})+G(u_{n+1},u_{n+2},u_{n+2})+G(u_{n+2},u_{m},u_{m})$
$\displaystyle\vdots$ $\displaystyle\leq
G(u_{n},u_{n+1},u_{n+1})+G(u_{n+1},u_{n+2},u_{n+2})+\cdots+G(u_{m-1},u_{m},u_{m})$
$\displaystyle\leq\tau_{0}\left(\lambda^{n}+\lambda^{n+1}+\cdots+\lambda^{m-1}\right)$
$\displaystyle=\frac{\tau_{0}}{1-\lambda}\lambda^{n}\left(1-\lambda^{m-n}\right)$
$\displaystyle\leq\frac{\tau_{0}}{1-\lambda}\lambda^{n},$

which shows that (since $0<\lambda<1$)

$\lim_{n,m\to\infty}G(u_{n},u_{m},u_{m})=0,$

that is, $\\{u_{n}\\}$ is a $G$-Cauchy sequence. Now, the completeness of
$(X,G)$ yields the existence of an element $u^{*}\in X$ such that
$\\{u_{n}\\}$ is $G$-convergent to $u^{*}$. On the other hand, if there exists
$k$ such that for all $n\geq k$, we have $u_{n}=u^{*}$, this contradicts the
fact that $u_{n},u_{n+1}$ and $u_{n+2}$ are pairwise distinct points for all
$n$. Consequently, we can extract a subsequence $\\{u_{n(k)}\\}_{k}$ of
$\\{u_{n}\\}$ such that $u_{n(k)}\neq u^{*}$ for all $k$. In order to simplify
writing, we always denote the sequence $\\{u_{n(k)}\\}_{k}$ by $\\{u_{n}\\}$
with $u_{n}\neq u^{*}$ for all $n$. We now show that
$u^{*}\in\operatorname{Fix}(T)$. Indeed, by (P3) and (P5), we have

$\displaystyle G(u^{*},u^{*},Tu^{*})$ $\displaystyle\leq
G(u^{*},u_{n},u_{n})+G(u_{n},u^{*},Tu^{*})$
$\displaystyle=G(u^{*},u_{n},u_{n})+G(u^{*},u_{n},Tu^{*})$ $\displaystyle\leq
G(u^{*},u_{n},u_{n})+G(u^{*},u_{n+1},u_{n+1})+G(u_{n+1},u_{n},Tu^{*})$
$\displaystyle=G(u^{*},u_{n},u_{n})+G(u^{*},u_{n+1},u_{n+1})+G(Tu_{n},Tu_{n-1},Tu^{*}),$

which implies by (II) and the fact that $u_{n},u_{n-1}$ and $u^{*}$ are
pairwise distinct points that

$G(u^{*},u^{*},Tu^{*})\leq
G(u^{*},u_{n},u_{n})+G(u^{*},u_{n+1},u_{n+1})+\lambda G(u_{n},u_{n-1},u^{*}).$
(3.3)

On the other hand, by the definition of the $G$-convergence and (A2), we have

$\lim_{n\to\infty}\left[G(u^{*},u_{n},u_{n})+G(u^{*},u_{n+1},u_{n+1})+\lambda
G(u_{n},u_{n-1},u^{*})\right]=0,$

which implies by (3.3) together with (P1) that $G(u^{*},u^{*},Tu^{*})=0$ and
$u^{*}=Tu^{*}$. This proves that $\operatorname{Fix}(T)\neq\emptyset$.

Suppose now that $v_{i}$, $i=1,2,3$, are pairwise distinct fixed points of
$T$. Then, making use of (P1) and (II), we get

$0<G(v_{1},v_{2},v_{3})=G(Tv_{1},Tv_{2},Tv_{3})\leq\lambda
G(v_{1},v_{2},v_{3})$

and we reach a contradiction with $0<\lambda<1$. This shows that
$|\operatorname{Fix}(T)|\leq 2$. The proof of Theorem 3.1 is then completed. ∎

######  Remark 3.2.

We point out that the condition $x,y,z\in X$ are pairwise distinct in
assumption (II) of Theorem 3.1 is essential. Otherwise, as we mentioned in
Section 1, using the approach in [4], (II) reduces to

$\delta(Tx,Ty)\leq\lambda\delta(x,y),\quad x,y\in X,$

where $\delta:X\times X\to\mathbb{R}^{+}$ is the metric on $X$ defined by

$\delta(x,y)=\max\\{G(x,y,y),G(y,x,x)\\},\quad x,y\in X,$ (3.4)

and the existence of a (unique) fixed point of $T$ will be an immediate
consequence of the Banach fixed point theorem in metric spaces.

We provide below an example to illustrate our obtained result.

######  Example 3.3.

Let $X=\\{A,B,C\\}\subset\mathbb{R}^{2}$, where

$A=\left(\frac{7}{8},\frac{\sqrt{15}}{8}\right),\quad B=(1,0),\quad C=(0,0).$

We introduce the mappings $T_{1},T_{2}:X\to X$ defined by

$T_{1}A=A,\quad T_{1}B=B,\quad T_{1}C=A$

and

$T_{2}A=B,\quad T_{2}B=A,\quad T_{2}C=A.$

Consider now the $G$-metric on $X$ given by

$G(u,v,w)=\max\left\\{\|u-v\|,\|v-w\|,\|u-w\|\right\\},\quad u,v,w\in X,$

where $\|\cdot\|$ denotes the Euclidean norm in $\mathbb{R}^{2}$.

Clearly, the mapping $T_{1}$ satisfies condition (I) of Theorem 3.1.
Furthermore, we have

$\displaystyle G(T_{1}A,T_{1}B,T_{1}C)$ $\displaystyle=G(A,B,A)$
$\displaystyle=\|A-B\|$ $\displaystyle=\frac{1}{2}$

and

$\displaystyle G(A,B,C)$
$\displaystyle=\max\left\\{\|A-B\|,\|B-C\|,\|A-C\|\right\\}$
$\displaystyle=\max\left\\{\frac{1}{2},1,1\right\\}$ $\displaystyle=1,$

which shows that condition (II) of Theorem 3.1 holds for all
$\lambda\in\left[\frac{1}{2},1\right)$. On the other hand,
$\operatorname{Fix}(T_{1})=\\{A,B\\}$, which confirms the obtained result
given by Theorem 3.1.

Remark also that the mapping $T_{2}$ satisfies condition (II) of Theorem 3.1.
This can be easily seen observing that
$G(T_{2}A,T_{2}B,T_{2}C)=\|A-B\|=G(T_{1}A,T_{1}B,T_{1}C)$. On the other hand,
we have $T_{2}A\neq A$ and $T_{2}(T_{2}A)=T_{2}(B)=A$, which shows that
condition (I) of Theorem 3.1 is not satisfied. Furthermore, we have
$\operatorname{Fix}(T)=\emptyset$, which shows that in the absence of
condition (I), the result of Theorem 3.1 is not true.

###  3.2 Kannan’s fixed point theorem in $G$-metric spaces

The Kannan fixed point theorem [6] can be stated as follows: Let $(X,d)$ be a
complete metric space and $T:X\to X$ be a mapping satisfying the inequality

$d(Tx,Ty)\leq\lambda\left[d(x,Tx)+d(y,Ty)\right]$

for every $x,y\in X$, where $\lambda\in\left(0,\frac{1}{2}\right)$ is a
constant. Then $T$ admits a unique fixed point.

Our second main result is an extension of the above result from metric spaces
to $G$-metric spaces.

######  Theorem 3.4.

Let $(X,G)$ be a complete $G$-metric space with $|X|\geq 3$. Let $T:X\to X$ be
a mapping satisfying the following conditions:

  * (I)

For all $x\in X$, $T(Tx)\neq x$, provided $Tx\neq x$;

  * (II)

There exists $\lambda\in\left(0,\frac{1}{3}\right)$ such that for all pairwise
distinct points $x,y,z\in X$, we have

$G(Tx,Ty,Tz)\leq\lambda\left[G(x,Tx,Tx)+G(y,Ty,Ty)+G(z,Tz,Tz)\right].$

Then $\operatorname{Fix}(T)\neq\emptyset$ and $|\operatorname{Fix}(T)|\leq 2$.

###### Proof.

We first show that $T$ admits at least one fixed point. Let $u_{0}\in X$ be
fixed and consider the Picard sequence $\\{u_{n}\\}\subset X$ defined by

$u_{n+1}=Tu_{n},\quad n\geq 0.$

As in the proof of Theorem 3.1, we may suppose that $u_{n}\neq u_{n+1}$ for
all $n\geq 0$, which implies by (I) that $u_{n},u_{n+1}$ and $u_{n+2}$ are
pairwise distinct points for every $n\geq 0$. Then, making use of (II), we
obtain

$\displaystyle G(u_{1},u_{2},u_{3})$ $\displaystyle=$ $\displaystyle
G(Tu_{0},Tu_{1},Tu_{2})$ $\displaystyle\leq$
$\displaystyle\lambda\left[G(u_{0},u_{1},u_{1})+G(u_{1},u_{2},u_{2})+G(u_{2},u_{3},u_{3})\right],$

which yields

$(1-\lambda)G(u_{1},u_{2},u_{3})\leq\lambda\left[G(u_{0},u_{1},u_{1})+G(u_{2},u_{3},u_{3})\right].$
(3.5)

We now use (P3), (P4), (P5) and the fact that $u_{n},u_{n+1}$ and $u_{n+2}$
are pairwise distinct points to get

$G(u_{0},u_{1},u_{1})=G(u_{1},u_{1},u_{0})\leq
G(u_{1},u_{0},u_{2})=G(u_{0},u_{1},u_{2})$ (3.6)

and

$G(u_{2},u_{3},u_{3})=G(u_{3},u_{3},u_{2})\leq
G(u_{3},u_{1},u_{2})=G(u_{1},u_{2},u_{3}).$ (3.7)

Then, it follows from (3.5), (3.6) and (3.7) that

$(1-2\lambda)G(u_{1},u_{2},u_{3})\leq\lambda G(u_{0},u_{1},u_{2}),$

which implies (since $1-2\lambda>0$) that

$G(u_{1},u_{2},u_{3})\leq kG(u_{0},u_{1},u_{2}),$

where $k=\frac{\lambda}{1-2\lambda}$. Notice that by $0<\lambda<\frac{1}{3}$,
we have $0<k<1$. Next, by induction, we obtain easily that

$G(u_{n},u_{n+1},u_{n+2})\leq k^{n}\tau_{0},\quad n\geq 0,$

where $\tau_{0}=G(u_{0},u_{1},u_{2})>0$. Proceeding as in the proof of Theorem
3.1, we obtain that $\\{u_{n}\\}$ is a $G$-Cauchy sequence. Due to the
completeness of $(X,G)$, there exists $u^{*}\in X$ such that $\\{u_{n}\\}$ is
$G$-convergent to $u^{*}$. By the proof of Theorem 3.1, without restriction of
the generality, we may suppose that $u_{n}\neq u^{*}$ for all $n$. We now show
that $u^{*}\in\operatorname{Fix}(T)$. By (P3) and (P4), we have (see the proof
of Theorem 3.1)

$G(u^{*},u^{*},Tu^{*})\leq
G(u^{*},u_{n},u_{n})+G(u^{*},u_{n+1},u_{n+1})+G(Tu_{n},Tu_{n-1},Tu^{*}),$

which implies by (II) and the fact that $u_{n},u_{n-1}$ and $u^{*}$ are
pairwise distinct points that

$\displaystyle G(u^{*},u^{*},Tu^{*})$ $\displaystyle\leq
G(u^{*},u_{n},u_{n})+G(u^{*},u_{n+1},u_{n+1})$
$\displaystyle\quad+\lambda\left[G(u_{n},u_{n+1},u_{n+1})+G(u_{n-1},u_{n},u_{n})+G(u^{*},Tu^{*},Tu^{*})\right]$
$\displaystyle\leq G(u^{*},u_{n},u_{n})+G(u^{*},u_{n+1},u_{n+1})$
$\displaystyle\quad+\lambda\left[G(u_{n},u_{n+1},u_{n+1})+G(u_{n-1},u_{n},u_{n})+2G(u^{*},u^{*},Tu^{*})\right],$

that is,

$\displaystyle(1-2\lambda)G(u^{*},u^{*},Tu^{*})$ $\displaystyle\leq
G(u^{*},u_{n},u_{n})+G(u^{*},u_{n+1},u_{n+1})$
$\displaystyle\quad+\lambda\left[G(u_{n},u_{n+1},u_{n+1})+G(u_{n-1},u_{n},u_{n})\right].$

Since $1-2\lambda>0$, passing to the limit as $n\to\infty$ in the above
inequality, $G(u^{*},u^{*},Tu^{*})=0$, which yields $u^{*}=Tu^{*}$. This shows
that $\operatorname{Fix}(T)\neq\emptyset$.

Suppose now that $v_{i}$, $i=1,2,3$, are pairwise distinct fixed points of
$T$. Then, making use of (II), we get

$\displaystyle G(v_{1},v_{2},v_{3})$ $\displaystyle=G(Tv_{1},Tv_{2},Tv_{3})$
(3.8)
$\displaystyle\leq\lambda\left[G(v_{1},Tv_{1},Tv_{1})+G(v_{2},Tv_{2},Tv_{2})+G(v_{3},Tv_{3},Tv_{3})\right]$
$\displaystyle=\lambda\left[G(v_{1},v_{1},v_{1})+G(v_{2},v_{2},v_{2})+G(v_{3},v_{3},v_{3})\right].$

On the other hand, we have by (P4) that

$\displaystyle G(v_{1},v_{1},v_{1})\leq G(v_{1},v_{1},v_{2})\leq
G(v_{1},v_{2},v_{3})$ $\displaystyle G(v_{2},v_{2},v_{2})\leq
G(v_{2},v_{2},v_{3})\leq G(v_{2},v_{3},v_{1})=G(v_{1},v_{2},v_{3})$
$\displaystyle G(v_{3},v_{3},v_{3})\leq G(v_{3},v_{3},v_{2})\leq
G(v_{3},v_{2},v_{1})=G(v_{1},v_{2},v_{3}).$

Then, by (3.8), we obtain

$(1-3\lambda)G(v_{1},v_{2},v_{3})\leq 0,$

and we reach a contradiction with $0<\lambda<\frac{1}{3}$ and
$G(v_{1},v_{2},v_{3})>0$. This shows that $|\operatorname{Fix}(T)|\leq 2$ and
the proof of Theorem 3.4 is completed. ∎

######  Remark 3.5.

Notice that, if conditions (II) of Theorem 3.4 holds for every $x,y,z\in X$,
then following the approach in [4], the existence of a (unique) fixed point of
$T$ can be deduced immediately from Ćirić’s fixed point theorem in metric
spaces [3]. Namely, in this case, taking $z=y$, (II) reduces to

$G(Tx,Ty,Ty)\leq\lambda\left[G(x,Tx,Tx)+2G(y,Ty,Ty)\right],\quad x,y\in X.$

Replacing $y$ by $x$ and $z$ by $y$, (II) reduces to

$G(Tx,Tx,Ty)\leq\lambda\left[2G(x,Tx,Tx)+G(y,Ty,Ty)\right],\quad x,y\in X.$

The above two inequalities yield

$\delta(Tx,Ty)\leq k\max\left\\{\delta(x,Tx),\delta(y,Ty)\right\\},$

where $k=3\lambda$ and $\delta$ is the metric on $X$ defined by (3.4). Since
$k=3\lambda\in(0,1)$, then by Ćirić’s fixed point theorem, $T$ admits a
(unique) fixed point.

######  Example 3.6.

Let $X=\left\\{a,b,c\right\\}\subset\mathbb{R}$, where

$a=0,\quad b=\frac{1}{5},\quad c=1.$

We introduce the mappings $T_{1},T_{2}:X\to X$ defined by

$T_{1}a=a,\quad T_{1}b=b,\quad T_{1}c=a$

and

$T_{2}a=b,\quad T_{2}b=a,\quad T_{2}c=a.$

Consider now the $G$-metric on $X$ given by

$G(u,v,w)=\max\left\\{|u-v|,|v-w|,|u-w|\right\\},\quad u,v,w\in X.$

Clearly, the mapping $T_{1}$ satisfies condition (I) of Theorem 3.4.
Furthermore, we have

$G(T_{1}a,T_{1}b,T_{1}c)=G(a,b,a)=|a-b|=\frac{1}{5}$

and

$\displaystyle G(a,T_{1}a,T_{1}a)+G(b,T_{1}b,T_{1}b)+G(c,T_{1}c,T_{1}c)$
$\displaystyle=G(a,a,a)+G(b,b,b)+G(c,a,a)$ $\displaystyle=G(c,a,a)$
$\displaystyle=|a-c|=1,$

which shows that condition (II) of Theorem 3.4 holds for all
$\lambda\in\left(0,\frac{1}{3}\right)$. On the other hand,
$\operatorname{Fix}(T_{1})=\\{a,b\\}$, which confirms the obtained result
given by Theorem 3.4.

Remark also that the mapping $T_{2}$ satisfies condition (II) of Theorem 3.4.
Namely, we have

$G(T_{2}a,T_{2}b,T_{2}c)=G(b,a,a)=|b-a|=\frac{1}{5}$

and

$\displaystyle G(a,T_{2}a,T_{2}a)+G(b,T_{2}b,T_{2}b)+G(c,T_{2}c,T_{2}c)$
$\displaystyle=G(a,b,b)+G(b,a,a)+G(c,a,a)$ $\displaystyle=2|a-b|+|c-a|$
$\displaystyle=\frac{2}{5}+1=\frac{7}{5},$

which shows that condition (II) of Theorem 3.4 holds for every
$\lambda\in\left[\frac{1}{7},\frac{1}{3}\right)$. On the other hand, we have
$T_{2}a\neq a$ and $T_{2}(T_{2}a)=T_{2}b=a$, which shows that condition (I) of
Theorem 3.4 is not satisfied. Furthermore, we have
$\operatorname{Fix}(T)=\emptyset$, which shows that in the absence of
condition (I), the result of Theorem 3.4 is not true.

###  3.3 The Reich fixed point theorem in $G$-metric spaces

The following fixed point result was obtained by Reich [15]: Let $(X,d)$ be a
complete metric space and $T:X\to X$ be a mapping satisfying the inequality

$d(Tx,Ty)\leq a_{1}d(x,Tx)+a_{2}d(y,Ty)+a_{3}d(x,y)$

for every $x,y\in X$, where $a_{i}$ ($i=1,2,3$) are nonnegative constants with
$0<\sum_{i=1}^{3}a_{i}<1$. Then $T$ admits a unique fixed point.

We now extend the Reich fixed point theorem to $G$-metric spaces.

######  Theorem 3.7.

Let $(X,G)$ be a complete $G$-metric space with $|X|\geq 3$. Let $T:X\to X$ be
a mapping satisfying the following conditions:

  * (I)

For all $x\in X$, $T(Tx)\neq x$, provided $Tx\neq x$;

  * (II)

There exist nonnegative constants $a_{i}$ ($i=1,2,3,4$) with
$0<\sum_{i=1}^{4}a_{i}<1$ such that for all pairwise distinct points $x,y,z\in
X$, we have

$G(Tx,Ty,Tz)\leq
a_{1}G(x,Tx,Tx)+a_{2}G(y,Ty,Ty)+a_{3}G(z,Tz,Tz)+a_{4}G(x,y,z).$

Then $\operatorname{Fix}(T)\neq\emptyset$ and $|\operatorname{Fix}(T)|\leq 2$.

###### Proof.

Let us first show that $T$ admits a fixed point. Let $u_{0}\in X$ be fixed,
but arbitrarily chosen and consider the Picard sequence $\\{u_{n}\\}\subset X$
defined by

$u_{n+1}=Tu_{n},\quad n\geq 0$

.

As in the proof of Theorem 3.1, we can assume that $u_{n}\neq u_{n+1}$ for all
$n\geq 0$, and thus, by (I) we obtain that $u_{n},u_{n+1},u_{n+2}$ are
pairwise distinct for all $n\geq 0$. Then, by (II), we have

$\displaystyle G(u_{1},u_{2},u_{3})$ $\displaystyle=G(Tu_{0},Tu_{1},Tu_{2})$
$\displaystyle\leq
a_{1}G(u_{0},Tu_{0},Tu_{0})+a_{2}G(u_{1},Tu_{1},Tu_{1})+a_{3}G(u_{2},Tu_{2},Tu_{2})+a_{4}G(u_{0},u_{1},u_{2})$

so

$\displaystyle G(u_{1},u_{2},u_{3})\leq
a_{1}G(u_{0},u_{1},u_{1})+a_{2}G(u_{1},u_{2},u_{2})+a_{3}G(u_{2},u_{3},u_{3})+a_{4}G(u_{0},u_{1},u_{2})$
(3.9)

By (P3) and (P4), since $u_{n},u_{n+1},u_{n+2}$ are pairwise distinct for all
$n\geq 0$ we have

$G(u_{0},u_{1},u_{1})\leq G(u_{0},u_{1},u_{2})$ (3.10)
$G(u_{1},u_{2},u_{2})\leq G(u_{0},u_{1},u_{2})$ (3.11)
$G(u_{2},u_{3},u_{3})\leq G(u_{1},u_{2},u_{3}).$ (3.12)

By (3.9), (3.10), (3.11) and (3.12) it follows that

$\displaystyle(1-a_{3})G(u_{1},u_{2},u_{3})\leq(a_{1}+a_{2}+a_{4})G(u_{0},u_{1},u_{2}),$

and since $1-a_{3}>0$ we obtain

$G(u_{1},u_{2},u_{3})\leq\dfrac{a_{1}+a_{2}+a_{4}}{1-a_{3}}G(u_{0},u_{1},u_{2}).$

Let $k=\dfrac{a_{1}+a_{2}+a_{4}}{1-a_{3}}$. Since $\sum_{i=1}^{4}a_{i}<1$
notice that $k<1$ and we have

$G(u_{1},u_{2},u_{3})\leq k\tau_{0},$

where $\tau_{0}=G(u_{0},u_{1},u_{2})>0$. By induction, we get

$G(u_{n},u_{n+1},u_{n+2})\leq k^{n}\tau_{0}.$

As in the proof of Theorem 3.1, we obtain that $\\{u_{n}\\}$ is a G-Cauchy
sequence and by completeness of $(X,G)$, there exists $u^{*}\in X$ such that
$\\{u_{n}\\}$ is G-convergent to $u^{*}$. We will show that $u^{*}\in Fix(T)$.
By (P5), we have

$\displaystyle G(u^{*},Tu^{*},Tu^{*})$ $\displaystyle\leq
G(u^{*},u_{n},u_{n})+G(u_{n},Tu^{*},Tu^{*})$
$\displaystyle=G(u^{*},u_{n},u_{n})+G(Tu_{n-1},Tu^{*},Tu^{*}),$

which implies, by (II) and the fact that $u_{n}$, $u_{n-1}$ and $x^{*}$ are
pairwise distinct that

$\displaystyle G(u^{*},Tu^{*},Tu^{*})$ $\displaystyle\leq
G(u^{*},u_{n},u_{n})+a_{1}G(u_{n-1},Tu_{n-1},Tu_{n-1})$
$\displaystyle+a_{2}G(u^{*},Tu^{*},Tu^{*})+a_{3}G(u^{*},Tu^{*},Tu^{*})+a_{4}G(u_{n-1},u^{*},u^{*}),$

so we have

$\displaystyle G(u^{*},Tu^{*},Tu^{*})$ $\displaystyle\leq
G(u^{*},u_{n},u_{n})+a_{1}G(u_{n-1},u_{n},u_{n})+a_{4}G(u_{n-1},u^{*},u^{*}).$

Since $1-(a_{2}+a_{3})>0$, passing to limit as $n\to\infty$ in the above
inequality, we obtain $G(u^{*},Tu^{*},Tu^{*})=0$, which yields $u^{*}=Tu^{*}$,
and thus Fix$(T)\neq\emptyset$.

Now suppose that $v_{i}$, $i=1,2,3$, are pairwise distinct points of $T$.
Then, by (II) we have

$\displaystyle G(v_{1},v_{2},v_{3})$ $\displaystyle=G(Tv_{1},Tv_{2},Tv_{3})$
$\displaystyle\leq
a_{1}G(v_{1},Tv_{1},Tv_{1})+a_{2}G(v_{2},Tv_{2},Tv_{2})+a_{3}G(v_{3},Tv_{3},Tv_{3})+a_{4}G(v_{1},v_{2},v_{3})$
$\displaystyle=a_{1}G(v_{1},v_{1},v_{1})+a_{2}G(v_{2},v_{2},v_{2})+a_{3}G(v_{3},v_{3},v_{3})+a_{4}G(v_{1},v_{2},v_{3}),$

that is

$\displaystyle(1-a_{4})G(v_{1},v_{2},v_{3})$ $\displaystyle\leq
a_{1}G(v_{1},v_{1},v_{1})+a_{2}G(v_{2},v_{2},v_{2})+a_{3}G(v_{3},v_{3},v_{3})$
$\displaystyle=0.$

Thus we obtain $a_{4}\geq 1$ which is a contradiction. This shows that
$|\operatorname{Fix}(T)|\leq 2$ and the proof of Theorem 3.7 is completed.

∎

######  Remark 3.8.

If condition (II) of Theorem 3.7 holds for every $x,y,z\in X$, then following
the approach in [4], the existence of a (unique) fixed point of $T$ can be
deduced immediately from Ćirić’s fixed point theorem in metric spaces [3] .
Namely, taking $z=y$, (II) reduces to

$\displaystyle G(Tx,Ty,Ty)$ $\displaystyle\leq
a_{1}G(x,Tx,Tx)+a_{2}G(y,Ty,Ty)+a_{3}G(y,Ty,Ty)+a_{4}G(x,y,y)$
$\displaystyle\leq
a_{1}\max\\{G(x,Tx,Tx),G(x,x,Tx)\\}+(a_{2}+a_{3})\max\\{G(y,Ty,Ty),G(y,y,Ty)\\}$
$\displaystyle\quad+a_{4}\max\\{G(x,y,y),G(y,x,x)\\}$

Replacing $y$ by $x$ and $z$ by $y$, (II) reduces to

$\displaystyle G(Tx,Tx,Ty)$ $\displaystyle\leq
a_{1}G(x,Tx,Tx)+a_{2}G(x,Tx,Tx)+a_{3}G(y,Ty,Ty)+a_{4}G(x,x,y)$
$\displaystyle\leq(a_{1}+a_{2})\max\\{G(x,Tx,Tx),G(x,x,Tx)\\}+a_{3}\max\\{G(y,Ty,Ty),G(y,y,Ty)\\}$
$\displaystyle\quad+a_{4}\max\\{G(x,y,y),G(y,x,x)\\}.$

Then, it follows from the above two inequalities that

$\delta(Tx,Ty)\leq
A\max\left\\{\delta(x,Tx),\delta(y,Ty),\delta(x,y)\right\\}$

for every $x,y\in X$, where $A=\sum_{i=1}^{4}a_{i}<1$ and $\delta$ is the
metric on $X$ defined by (3.4), and Ćirić’s fixed point theorem applies.

######  Example 3.9.

Let $\lambda\in\left(0,\frac{1}{4}\right)$ be fixed. We consider the set
$X_{\lambda}=\left\\{a,b,c,d\right\\}\subset\mathbb{R}$, where

$a=0,\quad b=\frac{2\lambda}{2\lambda-1},\quad c=1,\quad d=2.$

We introduce the mapping $T:X\to X$ defined by

$Ta=a,\quad Tb=b,\quad Tc=b,\quad Td=b.$

Consider now the $G$-metric on $X$ given by

$G(u,v,w)=\max\left\\{|u-v|,|v-w|,|u-w|\right\\},\quad u,v,w\in X.$

Remark that $T(Tc)=Tb=b\neq c$ and $T(Td)=Tb=b\neq d$, which shows that
condition (I) of Theorem 3.7 is satisfied.

We now show that condition (II) of Theorem 3.7 is satisfied for all pairwise
distinct points $x,y,z\in X$. By symmetry, we only study the cases

$(x,y,z)\in\left\\{(a,b,c),(a,b,d),(a,c,d),(b,c,d)\right\\}.$

$\bullet$ The case $(x,y,z)=(a,b,c)$: In this case, we have

$G(Tx,Ty,Tz)=G(Ta,Tb,Tc)=G(a,b,b)=|a-b|=-b$

and

$\displaystyle\lambda\left[G(x,Tx,Tx)+G(y,Ty,Ty)+G(z,Tz,Tz)+G(x,y,z)\right]$
$\displaystyle=\lambda\left[G(a,Ta,Ta)+G(b,Tb,Tb)+G(c,Tc,Tc)+G(a,b,c)\right]$
$\displaystyle=\lambda\left[G(a,a,a)+G(b,b,b)+G(c,b,b)+G(a,b,c)\right],$

that is,

$\displaystyle\lambda\left[G(x,Tx,Tx)+G(y,Ty,Ty)+G(z,Tz,Tz)+G(x,y,z)\right]$
$\displaystyle=\lambda\left[G(c,b,b)+G(a,b,c)\right]$
$\displaystyle=\lambda\left[|c-b|+\max\left\\{|a-b|,|b-c|,|a-c|\right\\}\right]$
$\displaystyle=\lambda\left[\frac{1}{1-2\lambda}+\max\left\\{\frac{2\lambda}{1-2\lambda},\frac{1}{1-2\lambda},1\right\\}\right]$
$\displaystyle=\lambda\left[\frac{1}{1-2\lambda}+\frac{1}{1-2\lambda}\right]$
$\displaystyle=\frac{2\lambda}{1-2\lambda}$ $\displaystyle=-b.$

Consequently, we obtain

$G(Tx,Ty,Tz)=\lambda\left[G(x,Tx,Tx)+G(y,Ty,Ty)+G(z,Tz,Tz)+G(x,y,z)\right].$

$\bullet$ The case $(x,y,z)=(a,b,d)$: In this case, we have

$G(Tx,Ty,Tz)=G(Ta,Tb,Td)=G(a,b,b)=|a-b|=-b$

and

$\displaystyle\lambda\left[G(x,Tx,Tx)+G(y,Ty,Ty)+G(z,Tz,Tz)+G(x,y,z)\right]$
$\displaystyle=\lambda\left[G(a,Ta,Ta)+G(b,Tb,Tb)+G(d,Td,Td)+G(a,b,d)\right]$
$\displaystyle=\lambda\left[G(a,a,a)+G(b,b,b)+G(d,b,b)+G(a,b,d)\right],$

that is,

$\displaystyle\lambda\left[G(x,Tx,Tx)+G(y,Ty,Ty)+G(z,Tz,Tz)+G(x,y,z)\right]$
$\displaystyle=\lambda\left[G(d,b,b)+G(a,b,d)\right]$
$\displaystyle=\lambda\left[|d-b|+\max\left\\{|a-b|,|b-d|,|a-d|\right\\}\right]$
$\displaystyle=\lambda\left[\frac{2-2\lambda}{1-2\lambda}+\max\left\\{\frac{2\lambda}{1-2\lambda},\frac{2-2\lambda}{1-2\lambda},2\right\\}\right]$
$\displaystyle=\lambda\left[\frac{2-2\lambda}{1-2\lambda}+\frac{2-2\lambda}{1-2\lambda}\right]$
$\displaystyle=\frac{4\lambda(1-\lambda)}{1-2\lambda}$
$\displaystyle=-2(1-\lambda)b.$

Since $-2(1-\lambda)b\leq-b$, then it holds that

$G(Tx,Ty,Tz)\leq\lambda\left[G(x,Tx,Tx)+G(y,Ty,Ty)+G(z,Tz,Tz)+G(x,y,z)\right].$

$\bullet$ The case $(x,y,z)=(a,c,d)$: In this case, we have

$G(Tx,Ty,Tz)=G(Ta,Tc,Td)=G(a,b,b)=|a-b|=-b$

and

$\displaystyle\lambda\left[G(x,Tx,Tx)+G(y,Ty,Ty)+G(z,Tz,Tz)+G(x,y,z)\right]$
$\displaystyle=\lambda\left[G(a,Ta,Ta)+G(c,Tc,Tc)+G(d,Td,Td)+G(a,c,d)\right]$
$\displaystyle=\lambda\left[G(a,a,a)+G(c,b,b)+G(d,b,b)+G(a,c,d)\right],$

that is,

$\displaystyle\lambda\left[G(x,Tx,Tx)+G(y,Ty,Ty)+G(z,Tz,Tz)+G(x,y,z)\right]$
$\displaystyle=\lambda\left[G(c,b,b)+G(d,b,b)+G(a,c,d)\right]$
$\displaystyle=\lambda\left[|c-b|+|d-b|+\max\left\\{|a-c|,|c-d|,|a-d|\right\\}\right]$
$\displaystyle=\lambda\left[\frac{1}{1-2\lambda}+\frac{2-2\lambda}{1-2\lambda}+\max\left\\{1,1,2\right\\}\right]$
$\displaystyle=\frac{\lambda(5-6\lambda)}{1-2\lambda}$
$\displaystyle=-\frac{(5-6\lambda)}{2}b.$

Since $-b\leq-\frac{(5-6\lambda)}{2}b$, we also obtain

$G(Tx,Ty,Tz)\leq\lambda\left[G(x,Tx,Tx)+G(y,Ty,Ty)+G(z,Tz,Tz)+G(x,y,z)\right].$

$\bullet$ The case $(x,y,z)=(b,c,d)$: In this case, we have

$G(Tx,Ty,Tz)=G(Tb,Tc,Td)=G(b,b,b)=0,$

which implies immediately that

$G(Tx,Ty,Tz)\leq\lambda\left[G(x,Tx,Tx)+G(y,Ty,Ty)+G(z,Tz,Tz)+G(x,y,z)\right].$

Consequently, condition (II) of Theorem 3.7 holds for all pairwise distinct
points $x,y,z\in X$ with $a_{1}=a_{2}=a_{3}=a_{4}=\lambda$. Notice that since
$\lambda\in\left(0,\frac{1}{4}\right)$, then $\sum_{i=1}^{4}a_{i}<1$.
Furthermore, we have $\operatorname{Fix}(T)=\\{a,b\\}$, which confirms the
result given by Theorem 3.7.

##  4 Applications

Some applications of our obtained results are provided in this section.

###  4.1 Mappings contracting perimeters of triangles

Petrov [11] introduced an interesting class of mappings $T:X\to X$, where
$(X,d)$ is a metric space, which can be characterized as mappings contracting
perimeters of triangles.

######  Definition 4.1 (Petrov [11]).

Let $(X,d)$ be a metric space with $|X|\geq 3$. We shall say that $T:X\to X$
is a mapping contracting perimeters of triangles on $X$, if there exists
$\lambda\in(0,1)$ such that the inequality

$d(Tx,Ty)+d(Ty,Tz)+d(Tz,Tx)\leq\lambda[d(x,y)+d(y,z)+d(z,x)],$ (4.1)

holds for all three pairwise distinct points $x,y,z\in X$.

The following result due to Petrov [11] is an immediate consequence of our
Theorem 3.1.

######  Corollary 4.2.

Let $(X,d)$, $|X|\geq 3$, be a complete metric space and let the mapping
$T:X\to X$ satisfies the following two conditions:

  * (I)

For all $x\in X$, $T(Tx)\neq x$, provided $Tx\neq x$;

  * (II)

$T$ is a mapping contracting perimeters of triangles on $X$.

Then $\operatorname{Fix}(T)\neq\emptyset$ and $|\operatorname{Fix}(T)|\leq 2$.

###### Proof.

Consider the $G$-metric on $X$ defined by (2.1). By Lemma 2.1, since $(X,d)$
is a complete metric space, then $(X,G)$ is a complete $G$-metric space.
Furthermore, (4.1) is equivalent to

$G(Tx,Ty,Tz)\leq\lambda G(x,y,z)$

for all three pairwise distinct points $x,y,z\in X$. Then, applying Theorem
3.1, the desired result follows. ∎

Further results related to mappings contracting perimeters of triangles can be
found in [10, 12, 13].

###  4.2 Further applications

Further interesting consequences can be deduced from our obtained results. Let
$(X,d)$ be a metric space with $|X|\geq 3$. We introduce the class of mappings
$T:X\to X$ satisfying the inequality

$\max\left\\{d(Tx,Ty),d(Ty,Tz),d(Tz,Tx)\right\\}\leq\lambda\max\left\\{d(x,y),d(y,z),d(x,z)\right\\}$
(4.2)

for all three pairwise distinct points $x,y,z\in X$, where $\lambda\in(0,1)$
is a constant.

######  Corollary 4.3.

Let $(X,d)$, $|X|\geq 3$, be a complete metric space and let the mapping
$T:X\to X$ satisfies the following two conditions:

  * (I)

For all $x\in X$, $T(Tx)\neq x$, provided $Tx\neq x$;

  * (II)

(4.2) holds for all three pairwise distinct points $x,y,z\in X$.

Then $\operatorname{Fix}(T)\neq\emptyset$ and $|\operatorname{Fix}(T)|\leq 2$.

###### Proof.

Consider the $G$-metric on $X$ defined by (2.2). By Lemma 2.2, since $(X,d)$
is a complete metric space, then $(X,G)$ is a complete $G$-metric space.
Furthermore, (4.2) is equivalent to

$G(Tx,Ty,Tz)\leq\lambda G(x,y,z)$

for all three pairwise distinct points $x,y,z\in X$. Then, applying Theorem
3.1, the desired result follows. ∎

We now introduce the class of mappings $T:X\to X$ satisfying the inequality

$d(Tx,Ty)+d(Ty,Tz)+d(Tx,Tz)\leq\lambda\left[d(x,Tx)+d(y,Ty)+d(z,Tz)\right]$
(4.3)

for all three pairwise distinct points $x,y,z\in X$, where
$\lambda\in\left(0,\frac{1}{3}\right)$ is a constant.

######  Corollary 4.4.

Let $(X,d)$, $|X|\geq 3$, be a complete metric space and let the mapping
$T:X\to X$ satisfies the following two conditions:

  * (I)

For all $x\in X$, $T(Tx)\neq x$, provided $Tx\neq x$;

  * (II)

(4.3) holds for all three pairwise distinct points $x,y,z\in X$.

Then $\operatorname{Fix}(T)\neq\emptyset$ and $|\operatorname{Fix}(T)|\leq 2$.

###### Proof.

Consider the $G$-metric defined by (2.1). By Lemma 2.1, since $(X,d)$ is a
complete metric space, then $(X,G)$ is a complete $G$-metric space. Moreover,
(4.3) is equivalent to

$G(Tx,Ty,Tz)\leq\dfrac{\lambda}{2}[G(x,Tx,Tx)+G(y,Ty,Ty)+G(z,Tz,Tz)]$

for all three pairwise distinct points $x,y,z\in X$. Then, since
$\lambda<\dfrac{1}{3}$ applying Theorem 3.4, the desired result follows. ∎

Consider now the class of mappings $T:X\to X$ satisfying the inequality

$\displaystyle d(Tx,Ty)+d(Ty,Tz)+d(Tx,Tz)$ (4.4) $\displaystyle\leq
a_{1}d(x,Tx)+a_{2}d(y,Ty)+a_{3}d(z,Tz)+a_{4}\left(d(x,y)+d(y,z)+d(x,z)\right)$

for all three pairwise distinct points $x,y,z\in X$, where $a_{i}$
($i=1,2,3,4$) are nonnegative constants with $0<\sum_{i=1}^{4}a_{i}<1$.

######  Corollary 4.5.

Let $(X,d)$, $|X|\geq 3$, be a complete metric space and let the mapping
$T:X\to X$ satisfies the following two conditions:

  * (I)

For all $x\in X$, $T(Tx)\neq x$, provided $Tx\neq x$;

  * (II)

(4.4) holds for all three pairwise distinct points $x,y,z\in X$.

Then $\operatorname{Fix}(T)\neq\emptyset$ and $|\operatorname{Fix}(T)|\leq 2$.

###### Proof.

Consider the $G$-metric defined by (2.1). By Lemma 2.1, since $(X,d)$ is a
complete metric space, then $(X,G)$ is a complete $G$-metric space. Moreover,
(4.4) is equivalent to

$G(Tx,Ty,Tz)\leq\dfrac{a_{1}}{2}G(x,Tx,Tx)+\dfrac{a_{2}}{2}G(y,Ty,Ty)+\dfrac{a_{3}}{2}G(z,Tz,Tz)+a_{4}G(x,y,z)$

for all three pairwise distinct points $x,y,z\in X$. Then, since
$\sum_{i=1}^{4}a_{i}<1$, applying Theorem 3.7, the desired result follows. ∎

We finally consider the class of mappings $T:X\to X$ satisfying the inequality

$\displaystyle\max\left\\{d(Tx,Ty),d(Ty,Tz),d(Tx,Tz)\right\\}$ (4.5)
$\displaystyle\leq
a_{1}d(x,Tx)+a_{2}d(y,Ty)+a_{3}d(z,Tz)+a_{4}\max\left\\{d(x,y),d(y,z),d(x,z)\right\\}$

for all three pairwise distinct points $x,y,z\in X$, where $a_{i}$
($i=1,2,3,4$) are nonnegative constants with $0<\sum_{i=1}^{4}a_{i}<1$.

######  Corollary 4.6.

Let $(X,d)$, $|X|\geq 3$, be a complete metric space and let the mapping
$T:X\to X$ satisfies the following two conditions:

  * (I)

For all $x\in X$, $T(Tx)\neq x$, provided $Tx\neq x$;

  * (II)

(4.5) holds for all three pairwise distinct points $x,y,z\in X$.

Then $\operatorname{Fix}(T)\neq\emptyset$ and $|\operatorname{Fix}(T)|\leq 2$.

###### Proof.

Consider the $G$-metric defined by (2.2). By Lemma 2.2, since $(X,d)$ is a
complete metric space, then $(X,G)$ is a complete $G$-metric space. Moreover,
(4.5) is equivalent to

$G(Tx,Ty,Tz)\leq
a_{1}G(x,Tx,Tx)+a_{2}G(y,Ty,Ty)+a_{3}G(z,Tz,Tz)+a_{4}G(x,y,z)$

for all three pairwise distinct points $x,y,z\in X$. Then, applying Theorem
3.7, the desired result follows. ∎

### Data availability statement

No new data were created or analyzed in this study.

### Conflict of interest

The authors declare that they have no competing interests.

### Funding

The third author is supported by Researchers Supporting Project number
(RSP2024R4), King Saud University, Riyadh, Saudi Arabia.

### Authors’ contributions

All authors contributed equally and significantly in writing this article. All
authors read and approved the final manuscript.

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