Application to Lipschitzian and Integral Systems via a Quadruple Coincidence Point in Fuzzy Metric Spaces

Ci a ion: Hammad, H.A.; De la Sen,
M. Applica ion o Lipschi zian and
In eg al Sys ems ia a Quad uple
Coincidence Poin in Fuzzy Me ic
Spaces. Ma hema ics 2022,10, 1905.
h ps://doi.o g/10.3390/
ma h10111905
Academic Edi o s: Sal ado
Romague a and Manuel Sanchis
Recei ed: 5 May 2022
Accep ed: 31 May 2022
Published: 2 June 2022
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ma hema ics
A icle
Applica ion o Lipschi zian and In eg al Sys ems ia a
Quad uple Coincidence Poin in Fuzzy Me ic Spaces
Hasanen A. Hammad 1,2,* and Manuel De la Sen 3
1Depa men o Ma hema ics, Unaizah College o Sciences and A s, Qassim Uni e si y,
Bu aydah 52571, Saudi A abia
2Depa men o Ma hema ics, Facul y o Science, Sohag Uni e si y, Sohag 82524, Egyp
3Ins i u e o Resea ch and De elopmen o P ocesses, Depa men o Elec ici y and Elec onics,
Facul y o Science and Technology, Uni e si y o he Basque Coun y, 48940 Leioa, Bizkaia, Spain;
[email p o ec ed]
*Co espondence: [email p o ec ed]
Abs ac :
In his pape , he esul s o a quad uple coincidence poin (QCP) a e es ablished o
commu ing mapping in he se ing o uzzy me ic spaces (FMSs) wi hou using a pa ially o de ed
se . In addi ion, se e al ela ed esul s a e p esen ed in o de o gene alize some o he p io indings in
his a ea. Finally, o suppo and enhance ou heo e ical ideas, non- i ial examples and applica ions
o inding a unique solu ion o Lipschi zian and in eg al quad uple sys ems a e discussed.
Keywo ds:
quad uple coincidence poin ; commu ing mapping; Lipschi zian mappings; an in eg al
equa ion; uzzy me ic spaces
MSC: 47H10; 54H25
1. In oduc ion
Fixed-poin (FP) heo y has many applica ions, no only in nonlinea analysis and i s
ends—including solu ions o di e en ial and in eg al equa ions, unc ional equa ions a is-
ing om dynamic p og amming, opologies, and dynamic sys ems—bu also in economics,
game heo y, biological sciences, compu e sciences, chemis y, e c. [1–4].
The e is no doub ha he s udy o uzzy se s is ex emely impo an o hei many
applica ions, such as in he con ol o ill-de ined, complex, and non-linea sys ems. I
is mo e common o ind solu ions o con ol p oblems ha a e di icul o sol e wi h
he classical con ol heo y. Fuzzy se heo y is becoming an inc easingly impo an ool,
especially in he apidly e ol ing discipline o a i icial in elligence, such as in expe
sys ems and neu al ne wo ks. I c ea es comple ely new oppo uni ies o he applica ion
o uzzy se s in chemical enginee ing [5–8].
The concep o uzzy se s was ini ia ed by Zadeh [
9
] in 1965. Many ma hema icians
used hese se s o in oduce in e es ing concep s in o he ield o ma hema ics, such as uzzy
logic, uzzy di e en ial equa ions, and uzzy me ic spaces. I is known ha an FMS is
an impo an gene aliza ion o an o dina y me ic space whe e he opological de ini ions
a e ex ended, and he e a e possible applica ions in se e al a eas. Many ma hema icians
ha e conside ed his p oblem in many ways. Fo example, he au ho s o [
10
] modi ied
he concep o an FMS ha was ini ia ed by K amosil and Michalek [
11
] and de ined he
Hausdo opology o an FMS. Fo mo e de ails abou his idea, we ad ise he eade
o see [12–17].
In 2011, he coupled ixed-poin (FP) [
18
] esul was ex ended o a ipled FP in pa ially
o de ed me ic spaces by Be inde and Bo cu [
19
]. Using hese spaces, hey in oduced
exci ing esul s o ipled FP heo ems. Fo mo e de ails, see [20–24].
Ma hema ics 2022,10, 1905. h ps://doi.o g/10.3390/ma h10111905 h ps://www.mdpi.com/jou nal/ma hema ics
Ma hema ics 2022,10, 1905 2 o 16
In he se ing o FMSs, coupled FP esul s we e p esen ed and some impo an heo-
ems we e gi en by Zhu and Xiao [
25
] and Hu [
26
]. Elagan e al. [
27
] s udied he exis ence
o an FP in a locally con ex opology gene a ed by uzzy n−no med spaces.
Mo i a ed by he esul s o he no ions o coupled and ipled FPs in pa ially o de ed
me ic spaces, Ka apina [
28
] sugges ed he concep o a quad uple FP and p o ed some
ela ed consequences o FPs in he same spaces.
Based on he las wo pa ag aphs, in his publica ion, a QCP is conside ed, and some
new and ele an FP esul s in FMSs a e epo ed. Ou pape ’s s eng h is de e mined
by wo ac o s. Fi s , we can adap i o comple e me ic spaces (CMSs) so as o achie e
Ka apina ’s esul s [
28
] (in non- uzzy se s). So, ou pape co e s and uni ies a la ge numbe
o ou comes in he same di ec ion. Secondly, we can apply he heo e ical conclusions o
Lipschi zian and in eg al quad uple sys ems in o de o disco e a unique solu ion. Finally,
non- i ial examples a e men ioned and discussed.
2. P elimina ies
He ea e , we will e e o
ζ
as a non-emp y se ,
Ω(ρ
,
σ
,
τ
,
υ)
as
Ωρστυ
,
Ψ(ρ
,
σ
,
κ)
as
Ψρσ(κ), and ω(ρ,σ)as ωρσ.
The usual me ic space is a non-emp y se
ζ
equipped wi h a unc ion
ω:ζ×ζ→R+
such ha o all ρ,σ,τ∈ζ, he ollowing condi ions a e ue:
•ωρσ ≥0,
•ωρσ =0 i ρ=σ,
•ωρσ ≤ωρτ +ωτσ.
The pai (ζ,ω)is called an MS.
A mapping
k:ζ→ζ
on an MS
(ζ,ω)
is called Lipschi zian i he e is
≥
0 such ha
ωkρkσ≤ ωρσ,∀ρ,σ∈ζ.
The smalles cons an
—deno ed by
k
— ha sa is ies he abo e inequali y is called
he Lipschi z cons an o
k
. I is clea ha a Lipschi zian mapping (LM) is a con ac ion
wi h k<1.
Theo em 1
([
29
])
.
Le
(ζ,ω)
be a comple e MS and le
Q:ζ→ζ
be a con ac ion mapping, ha
is, he ollowing inequali y is ue:
ω(Qx,Qy)≤kω(Qx,Qy), o all x,y∈ζ,
whe e
k∈[
0, 1
)
. Then,
Q
has a unique FP
x∗
in
ζ
. Mo eo e , o
x0∈ζ
, he sequence
(Qnx0)n∈N
con e ges o x∗.
Fo examples on LMs, le
ζ=R
and le
ki:ζ→ζ
be de ined by
k1(ρ) = Λ
,
k2(ρ) = µρ
,
k3(ρ) = cos ρ
,
k4(ρ) = 1
1+ρ
,
k5(ρ) = 1
(1+ρ)2
, and
k6(ρ) = a csin ρ
, whe e
Λ,µ∈R.
De ini ion 1
([
30
])
.
A mapping
?:[
0, 1
]2→[
0, 1
]
is called a
κ
-no m i i is nondec easing in
bo h a gumen s, associa i e, commu a i e, and has 1as iden i y. Fo all
`∈[
0, 1
]
, he sequence
{?m`}∞
m=1
is induc i ely de ined by
?1`=`
,
?m`=?m−1`? `
. A iangula no m
?
is o
Υ
- ype i
{?m`}∞
m=1
is equicon inuous a
`=
1, ha is, o each
e∈(
0, 1
)
, he e is
κ∈(
0, 1
)
such ha i `∈(1−κ, 1], hen ?m` > 1−e o each m ∈N.
The mos amous con inuous
κ
-no m o he
Υ
- ype is
?=min
, which sa is ies
min(`1
,
`2)≥
`1`2 o all `1,`2∈[0, 1].
The esul s below include a wide ange o κ-no ms o he Υ- ype.
Ma hema ics 2022,10, 1905 3 o 16
Lemma 1
([
30
])
.
Assume ha
?
is a
κ
-no m and
$∈(
0, 1
]
is a eal numbe . De ine
?$
by
ρ?$σ=ρ?σ
i
max{ρ
,
σ} ≤
1
−$
, and
ρ?$σ=min{ρ
,
σ}
i
max{ρ
,
σ}>
1
−$
. Then,
?$
is a κ-no m o he Υ- ype.
De ini ion 2
([
11
])
.
Le
ζ6=∅
be an a bi a y se , le
?
be a con inuous
κ
-no m, and le
Ψ:
ζ×ζ×[
0,
∞)→[
0, 1
]
be a uzzy se . We say ha
(ζ,Ψ,?)
is an FMS i he unc ion
Ψ
sa is ies
he hypo heses below o each ρ,σ,τ∈ζ, and κ,µ>0 :
( ms 1)
Ψρσ(0) = 0;
( ms 2)
Ψρσ(κ) = 1⇔ρ=σ;
( ms 3)
Ψρσ(κ) = Ψσρ(κ);
( ms 4)
Ψρσ(.):[0, ∞)→[0, 1]is le con inuous;
( ms 5)
Ψρσ(κ)?Ψστ(µ)≤Ψρτ(κ+µ).
He e, we also conside
(ζ,Ψ)
an FMS unde
?
, and we will only conside he FMS ha e i ies:
(D)
lim
κ→∞Ψρσ(κ) = 1,∀ρ,σ∈ζ.
Lemma 2 ([12]).On he in ini e se [0, ∞),Ψρσ(.)is a non-dec easing unc ion.
De ini ion 3
([
10
])
.
Assume ha
(ζ,Ψ)
is an FMS unde some
κ
-no m; a sequence
{ρm} ⊂ ζ
is called:
•
Con e gen o
ρ∈ζ
, and we w i e
lim
m→∞ρm=ρ
i , o e e y
e>
0,
κ>
0, he e is
m0∈N
such ha Ψρmρ(κ)>1−e o all m ≥m0.
•
A Cauchy sequence i , o e e y
e>
0,
κ>
0, he e is
m0∈N
such ha
Ψρmρj(κ)>
1
−e
o all m,j≥m0.
•An FMS is called comple e i e e y Cauchy sequence is con e gen .
De ini ion 4
([
11
])
.
We say ha a unc ion
k:ζ→ζ
de ined on an FMS is con inuous a
ρ0∈ζ
i
lim
m→∞kρm=kρ0
o any
{ρm} ∈ ζ
such ha
lim
m→∞ρm=ρ0
. As is amilia , o
ρ0∈ζ
, we will
deno e k−1(ρ0) = {ρ∈ζ:kρ=ρ0}.
Rema k 1
([
11
])
.
I
`1≤`2
, hen
ρ`1≥ρ`2
p o ided ha
ρ∈[
0, 1
]
and
`1
,
`2∈(
0,
∞)
. This ac
will be exp essed he e as ollows: 0< `1≤`2≤1implies ha Ψρσ(κ)`1≥Ψρσ(κ)`2≥Ψρσ(κ).
Fo any
κ
-no m
?
, i is ob ious ha
?≤min
. So, i
(ζ,Ψ)
is an FMS ia
min
, hen
(ζ,Ψ)is an FMS unde any κ-no m.
In he examples below, we only de ine Ψρσ(κ) o κ>0 and ρ6=σ.
Example 1
([
10
])
.
Fo
κ>
0and
ρ6=σ
, we de ine an FMS in di e en ways om an MS
(ζ,ω)
as ollows:
•Ψω
ρσ(κ) = κ
κ+ωρσ •Ψe
ρσ(κ) = e−ωρσ
κ•Ψo
ρσ(κ) = 0, i κ≤ωρσ,
1, i κ>ωρσ.
I is ob ious ha , unde he p oduc
?=
.,
(ζ,Ψω)
is an FMS, which is called he s anda d
FMS on
(ζ,ω)
. In addi ion,
(ζ,Ψω)
,
(ζ,Ψe)
, and
(ζ,Ψo)
a e FMSs unde
min
. This is a s anda d
me hod o seeing he MS (ζ,ω)as an FMS, hough i is no as well known.
Mo eo e , (ζ,ω)is a CMS i (ζ,Ψω),(ζ,Ψe), o (ζ,Ψo)is a comple e FMS.
3. Main Resul s
We begin his sec ion wi h he ollowing simple de ini ion.
De ini ion 5. Assume ha Ω:ζ4→ζand k:ζ→ζa e wo mappings.
•We say ha Ωand ka e commu ing i kΩρστυ =Ωkρkσkτkυ,∀`,σ,ρ,υ∈ζ.
Ma hema ics 2022,10, 1905 4 o 16
•We say ha (ρ,σ,τ,υ)∈ζ4is a QCP o Ωand ki
Ωρστυ =kρ,Ωστυρ =kσ,Ωτυρσ =kτand Ωυρστ =kυ.
Theo em 2.
Assume ha
?
is a
κ
-no m o he
Υ
- ype such ha
µ?κ≥µκ
o all
µ
,
κ∈[
0, 1
]
.
Suppose ha
(ζ,Ψ,?)
is a comple e FMS and
Ω:ζ4→ζ
,
k:ζ→ζ
a e wo mappings such ha
(a)
Ωζ4⊆k(ζ),
(b)
kis con inuous,
(c)
kis commu ing wi h Ω,
(d)
o all ρ,σ,τ,υ,b
ρ,bσ,b
τ,b
υ∈ζ,
ΨΩρστυΩb
ρbσb
τb
υ(κ )≥Ψkρkb
ρ(κ)`1?Ψkσkbσ(κ)`2?Ψkτkb
τ(κ)`3?Ψkυkb
υ(κ)`4, (1)
whe e
∈(
0, 1
)
and
`1
,
`2
,
`3
,
`4
a e eal numbe s in
[
0, 1
]
such ha
`1+`2+`3+`4≤
1.
Then, he ollowing conclusions hold.
(1)
The e is a unique ρ∈ζsuch ha ρ=kρ=Ωρρρρ. In pa icula ,
(2)
The e is a leas a QCP o he mappings
k
and
Ω
; mo eo e , in he case o
Ω=ρ0
, he e
is a cons an on
ζ4
. This holds only i he in e se o he mapping
k
exis s and i sa is ies
.k−1(ρ0) = {ρ0}; hen, we ha e
(3)
(ρ,ρ,ρ,ρ)is a unique QCP o kand Ω.
No e ha , o a oid he uniden i ied quan i y 0
0
, we conside he e
Ψkρkb
ρ(κ)0=
1 o
all κ>0 and all ρ,b
ρ∈ζ.
P oo . We di ide he p oo in o wo cases:
Case 1.
When
Ω⊆ζ
is cons an , ha is, he e is
ρ0∈ζ
such ha , o all
ρ,σ,τ,υ∈ζ,
Ωρστυ =ρ0
. Since
Ω
and
k
a e commu ing, one can w i e
kρ0=kΩρστυ =Ωkρkσkτkυ=ρ0
.
The e o e,
ρ0=kρ0=Ωρ0ρ0ρ0ρ0
and
(ρ0
,
ρ0
,
ρ0
,
ρ0)
is a QCP o
Ω
and
k
. On he o he
hand, assume ha
k−1(ρ0) = {ρ0}
and
(ρ,σ,τ,υ)∈ζ4
is ano he QCP o
Ω
and
k
.
Then,
kρ=Ωρστυ =ρ0
, so
ρ∈k−1(ρ0) = {ρ0}
. In he same manne , we can w i e
ρ=σ=τ=υ=ρ0; hence, (ρ0,ρ0,ρ0,ρ0)is a unique QCP o Ωand k.
Case 2.
Assume ha
Ω∈ζ
is no cons an ; o his, le
(`1,`2,`3,`4)6= (
0, 0, 0, 0
)
. In
his case, we conside
j
and
m
o be non-nega i e in ege s and
κ∈[
0,
∞)
. This case is
di ided in o i e s eps.
S 1
. De i ing ou sequences
{ρm}
,
{σm}
,
{τm}
, and
{υm}
: Suppose ha
ρ0
,
σ0
,
τ0
,
υ0
a e a bi a y poin s in
ζ
. As
Ωζ4⊆k(ζ)
, we can selec
ρ1
,
σ1
,
τ1
,
υ1∈ζ
so ha
kρ1=Ωρ0σ0τ0υ0,kσ1=Ωσ0τ0υ0ρ0
,
kτ1=Ωτ0υ0ρ0σ0
and
kυ1=Ωυ0ρ0σ0τ0
. Again, wi h
Ωζ4⊆k(ζ)
, we can selec
ρ2
,
σ2
,
τ2
,
υ2∈ζ
so ha
kρ2=Ωρ1σ1τ1υ1
,
kσ2=Ωσ1τ1υ1ρ1
,
kτ2=Ωτ1υ1ρ1σ1
, and
kυ2=Ωυ1ρ1σ1τ1
. Con inuing wi h he same scena io, we can cons uc
{ρm}
,
{σm}
,
{τm}
, and
{υm}
so ha o
m≥
0,
kρm+1=Ωρmσmτmυm
,
kσm+1=Ωσmτmυmρm
,
kτm+1=Ωτmυmρmσm, and kυm+1=Ωυmρmσmτm.
S 2
.
{ρm}
,
{σm}
,
{τm}
, and
{υm}
a e Cauchy sequences. Fo
m≥
0 and all
κ>
0,
we de ine
Ξm(κ) = Ψkρmkρm+1(κ)?Ψkσmkσm+1?Ψkτmkτm+1?Ψkυmkυm+1.
Ξmis a non-dec easing unc ion and κ−κ ≤κ≤κ
, so we ge
Ξm(κ−κ )≤Ξm(κ)≤Ξmκ
, o all κ>0 and m≥0. (2)
I ollows om (1) ha , o all m∈Nand all κ≥0,
Ψkρmkρm+1(κ)=ΨΩρm−1σm−1τm−1υm−1Ωρmσmτmυm(κ)
≥Ψkρm−1kρmκ
`1?Ψkσm−1kσmκ
`2
?Ψkτm−1kτmκ
`3?Ψkυm−1kυmκ
`4;
(3)
Ma hema ics 2022,10, 1905 5 o 16
Ψkσmkσm+1(κ)=ΨΩσm−1τm−1υm−1ρm−1Ωσmτmυmρm(κ)
≥Ψkσm−1kσmκ
`1?Ψkτm−1kτmκ
`2
?Ψkυm−1kυmκ
`3?Ψkρm−1kρmκ
`4;
(4)
Ψkτmkτm+1(κ)=ΨΩτm−1υm−1ρm−1σm−1Ωτmυmρmσm(κ)
≥Ψkτm−1kτmκ
`1?Ψkυm−1kυmκ
`2
?Ψkρm−1kρmκ
`3?Ψkσm−1kσmκ
`4;
(5)
Ψυmkυm+1(κ)=ΨΩυm−1ρm−1σm−1τm−1Ωυmρmσmτm(κ)
≥Ψkυm−1kυmκ
`1?Ψkρm−1kρmκ
`2
?Ψkσm−1kσmκ
`3?Ψkτm−1kτmκ
`4;
(6)
I ollows om (3)–(6) and Rema k 1 ha
Ψkρmkρm+1(κ)
≥Ψkρm−1kρmκ
`1?Ψkσm−1kσmκ
`2?Ψkρm−1kρmκ
`3?Ψkυm−1kυmκ
`4
≥Ψkρm−1kρmκ
?Ψkσm−1kσmκ
?Ψkρm−1kρmκ
?Ψkυm−1kυmκ

=Ξm−1(κ
);
Ψkσmkσm+1(κ)
≥Ψkσm−1kσmκ
`1?Ψkτm−1kτmκ
`2?Ψkυm−1kυmκ
`3?Ψkρm−1kρmκ
`4
≥Ψkσm−1kσmκ
?Ψkτm−1kτmκ
?Ψkυm−1kυmκ
?Ψkρm−1kρmκ

=Ξm−1(κ
);
Ψkτmkτm+1(κ)
≥Ψkτm−1kτmκ
`1?Ψkυm−1kυmκ
`2?Ψkρm−1kρmκ
`3?Ψkσm−1kσmκ
`4
≥Ψkτm−1kτmκ
?Ψkυm−1kυmκ
?Ψkρm−1kρmκ
?Ψkσm−1kσmκ

=Ξm−1(κ
);
and
Ψυτmkυm+1(κ)
≥Ψkυm−1kυmκ
`1?Ψkρm−1kρmκ
`2?Ψkσm−1kσmκ
`3?Ψkτm−1kτmκ
`4
≥Ψkυm−1kυmκ
?Ψkρm−1kρmκ
?Ψkσm−1kσmκ
?Ψkτm−1kτmκ

=Ξm−1(κ
).
This p o es ha , o all κ>0 and all m≥0,
Ψkρmkρm+1(κ),Ψkσmkσm+1(κ),Ψkτmkτm+1(κ),Ψυτmkυm+1(κ)≥Ξm−1(κ
)≥Ξm−1(κ). (7)
Pu ing κ− κ ins ead o κ, we ob ain, o all κ>0 and all m≥0, ha

Ma hema ics 2022,10, 1905 6 o 16
Ψkρmkρm+1(κ− κ),Ψkσmkσm+1(κ− κ),Ψkτmkτm+1(κ− κ),Ψυτmkυm+1(κ− κ)
≥Ξm−1(κ− κ).(8)
Since ?is commu a i e and ?≥., using (3)–(6), we deduce ha
Ξm(κ) = Ψkρmkρm+1(κ)?Ψkσmkσm+1?Ψkτmkτm+1?Ψkυmkυm+1
≥Ψkρm−1kρmκ
`1?Ψkσm−1kσmκ
`2?Ψkτm−1kτmκ
`3?Ψkυm−1kυmκ
`4
?Ψkρm−1kρmκ
`2?Ψkσm−1kσmκ
`3?Ψkτm−1kτmκ
`4?Ψkυm−1kυmκ
`1
?Ψkρm−1kρmκ
`3?Ψkσm−1kσmκ
`4?Ψkτm−1kτmκ
`1?Ψkυm−1kυmκ
`2
?Ψkρm−1kρmκ
`4?Ψkσm−1kσmκ
`1?Ψkτm−1kτmκ
`2?Ψkυm−1kυmκ
`3
=Ψkρm−1kρmκ
`1?Ψkρm−1kρmκ
`2?Ψkρm−1kρmκ
`3?Ψkρm−1kρmκ
`4
?Ψkσm−1kσmκ
`2?Ψkσm−1kσmκ
`3?Ψkσm−1kσmκ
`4?Ψkρm−1kρmκ
`2
?Ψkτm−1kτmκ
`3?Ψkτm−1kτmκ
`4?Ψkτm−1kτmκ
`1?Ψkτm−1kτmκ
`2
?Ψkυm−1kυmκ
`4?Ψkυm−1kυmκ
`1?Ψkυm−1kυmκ
`2?Ψkυm−1kυmκ
`3.
I ollows ha
Ξm(κ)≥Ψkρm−1kρmκ
`1.Ψkρm−1kρmκ
`2.Ψkρm−1kρmκ
`3.Ψkρm−1kρmκ
`4
?Ψkσm−1kσmκ
`2.Ψkσm−1kσmκ
`3.Ψkσm−1kσmκ
`4.Ψkρm−1kρmκ
`2
?Ψkτm−1kτmκ
`3.Ψkτm−1kτmκ
`4.Ψkτm−1kτmκ
`1.Ψkτm−1kτmκ
`2
?Ψkυm−1kυmκ
`4.Ψkυm−1kυmκ
`1.Ψkυm−1kυmκ
`2.Ψkυm−1kυmκ
`3
=Ψkρm−1kρmκ
`1+`2+`3+`4?Ψkσm−1kσmκ
`1+`2+`3+`4
?Ψkτm−1kτmκ
`1+`2+`3+`4?Ψkυm−1kυmκ
`1+`2+`3+`4
≥Ψkρm−1kρmκ
?Ψkσm−1kσmκ
?Ψkτm−1kτmκ
?Ψkυm−1kυmκ

=Ξm−1(κ
)
By using (2), one can w i e
Ξm(κ)≥Ξm−1(κ
)≥Ξm−1(κ)≥Ξm−1(κ−κ ),∀κ>0, and m≥1. (9)
By con inuing in he same manne , we ha e
Ξm(κ)≥Ξm−1(κ
)≥Ξm−2(κ
2)≥... ≥Ξ0(κ
m),∀κ>0, and m≥1,
which leads o ind ha o all κ>0
lim
m→∞Ξm(κ)≥lim
m→∞Ξ0(κ
m) = 1⇒lim
m→∞Ξm(κ) = 1. (10)
Ma hema ics 2022,10, 1905 7 o 16
F om (7) and (9), we ha e
Ψkρmkρm+1(κ),Ψkσmkσm+1(κ),Ψkτmkτm+1(κ),Ψυτmkυm+1(κ)≥Ξm(κ)≥Ξm−1(κ−κ ). (11)
A e ha , we will p o e ha , o all κ>0 and all m, ≥1,
Ψkρmkρm+ (κ),Ψkσmkσm+ (κ),Ψkτmkτm+ (κ),Ψυτmkυm+ (κ)≥? Ξm−1(κ−κ ). (12)
We can show his by induc ion in
≥
1 as ollows: Inequali y (12) holds i
=
1 o all
m≥
1 and all
κ>
0 by (11). Assume ha (12) is ue o all
m≥
1 and all
κ>
0 o some
.
Now, we p o e he ela ion o
+
1. I ollows om (1), he induc ion assump ion, and
?≥. ha
Ψkρm+1kρm+ +1( κ)
=ΨΩρmσmτmυmΩρm+ σm+ τm+ υm+ ( κ)
≥Ψkρmkρm+ (κ)`1?Ψkσmkσm+ (κ)`2?Ψkτmkτm+ (κ)`3?Ψkυmkυm+ (κ)`4
≥(? Ξm−1(κ−κ ))`1?(? Ξm−1(κ−κ ))`2?(? Ξm−1(κ−κ ))`3?(? Ξm−1(κ−κ ))`4
≥(? Ξm−1(κ−κ ))`1.(? Ξm−1(κ−κ ))`2.(? Ξm−1(κ−κ ))`3.(? Ξm−1(κ−κ ))`4
=(? Ξm−1(κ−κ ))`1+`2+`3+`4≥? Ξm−1(κ−κ ).
Simila ly, we a i e a
Ψkρm+1kρm+ +1( κ),Ψkσm+1kσm+ +1( κ),Ψkτm+1kτm+ +1( κ),Ψkυm+1kυm+ +1( κ)≥? Ξm−1(κ−κ ).
F om De ini ion 2( ms 5), (8), and he induc ion assump ion, we ge
Ψkρm+1kρm+ +1(κ)=Ψkρm+1kρm+ +1(κ−κ +κ )
≥Ψkρmkρm+1(κ−κ )?Ψkρm+1kρm+ +1(κ )
≥Ξm−1(κ−κ )?(? Ξm−1(κ−κ ))
=? +1Ξm−1(κ−κ ).
In addi ion, he same esul holds i we conside
Ψkσm+1kσm+ +1(κ)
,
Ψkτm+1kτm+ +1(κ)
,
and
Ψkυm+1kυm+ +1(κ)
. This leads o (12) being ue. This allows us o p o e ha
{kρm}
is
Cauchy. Assume ha
κ>
0 and
ε∈(
0, 1
)
a e gi en. F om his assump ion, as
?
is a
κ
-no m
o he
Υ
- ype, he e is
ϕ∈(
0, 1
)
such ha
? `1>
1
−ε
o all
`1∈(
1
−ϕ
, 1
]
and o all
≥1. F om (10), limm→∞Ξm(κ) = 1, so he e is m0∈Nsuch ha
Ξm(κ−κ )>1−ϕ,∀m≥m0.
Hence, by (12), we ha e
Ψkρmkρm+ (κ),Ψkσmkσm+ (κ),Ψkτmkτm+ (κ),Ψυτmkυm+ (κ)>1−ε,∀m≥m0and ≥1.
Thus,
{kρm}
is a Cauchy sequence. Simila ly,
{kσm}
,
{kτm}
, and
{kυm}
a e also
Cauchy sequences.
S 3
. P o ing ha
Ω
and
k
ha e a QCP: As
ζ
is comple e, he e a e
ρ
,
σ
,
τ
,
υ∈ζ
such ha
lim
m→∞kρm=ρ, lim
m→∞kσm=σ, lim
m→∞kτm=τand lim
m→∞kυm=υ.
The con inui y o kimplies ha
lim
m→∞kkρm=kρ, lim
m→∞kkσm=kσ, lim
m→∞kkτm=kτ, and lim
m→∞kkυm=kυ.
Ma hema ics 2022,10, 1905 8 o 16
The commu a i i y o Ωand kleads o
kkρm+1=kΩ(ρm,σm,τm,υm)=Ω(kρm,kσm,kτm,kυm).
By (1), we ge
Ψkkρm+1Ωρστυ (κ )=ΨΩkρmσmτmυmΩρστυ (κ )
≥Ψkkρmkρ(κ)`1?Ψkkσmkσ(κ)`2?Ψkkτmkτ(κ)`3?Ψkkυmkυ(κ)`4
≥Ψkkρmkρ(κ)?Ψkkσmkσ(κ)?Ψkkτmkτ(κ)?Ψkkυmkυ(κ).
(13)
As m→∞, in (13), we ind ha
lim
m→∞kkρm+1=Ωρστυ =kρ.
Simila ly, we deduce ha
Ωστυρ =kσ
,
Ωτυρσ =kτ
,
Ωυρστ =kυ
. This shows ha
(ρ,σ,τ,υ)is a QCP o Ωand k.
kρ=Ωρστυ,kσ=Ωστυρ,kτ=Ωτυρσ,kυ=Ωυρστ. (14)
S 4
. Showing ha
ρ=Ωρστυ
,
σ=Ωστυρ
,
τ=Ωτυρσ
, and
υ=Ωυρστ
: F om
S ipula ion (1), we ge
Ψkρkσm+1(κ )
=ΨΩρστυΩσmτmυmρm(κ )
≥Ψkρkσm(κ)`1?Ψkσkτm(κ)`2?Ψkτkυm(κ)`3?Ψkυkρm(κ)`4;
(15)
Ψkσkτm+1(κ )
=ΨΩστυρΩτmυmρmσm(κ )
≥Ψkσkτm(κ)`1?Ψkτkυm(κ)`2?Ψkυkρm(κ)`3?Ψkρkσm(κ)`4;
(16)
Ψkτkυm+1(κ )
=ΨΩτυρσΩυmρmσmτm(κ )
≥Ψkτkυm(κ)`1?Ψkυkρm(κ)`2?Ψkρkσm(κ)`3?Ψkσkτm(κ)`4;
(17)
Ψkυkρm+1(κ )
=ΨΩυρστ Ωρmσmτmυm(κ )
≥Ψkυkρm(κ)`1?Ψkρkσm(κ)`2?Ψkσkτm(κ)`3?Ψkτkυm(κ)`4.
(18)
We se
∇m(κ ) = Ψkρkσm(κ )?Ψkσkτm(κ )?Ψkτkυm(κ )?Ψkυkρm(κ )
o all
κ>0
and m≥0. I ollows om (15)–(18) ha
∇m+1(κ ) = Ψkρkσm+1(κ )?Ψkσkτm+1(κ )?Ψkτkυm+1(κ )?Ψkτkυm+1(κ )
≥Ψkρkσm(κ)`1?Ψkσkτm(κ)`2?Ψkτkυm(κ)`3?Ψkυkρm(κ)`4
?Ψkσkτm(κ)`1?Ψkτkυm(κ)`2?Ψkυkρm(κ)`3?Ψkρkσm(κ)`4
?Ψkτkυm(κ)`1?Ψkυkρm(κ)`2?Ψkρkσm(κ)`3?Ψkσkτm(κ)`4
?Ψkυkρm(κ)`1?Ψkρkσm(κ)`2?Ψkσkτm(κ)`3?Ψkτkυm(κ)`4
=Ψkρkσm(κ)`1?Ψkρkσm(κ)`4?Ψkρkσm(κ)`3?Ψkρkσm(κ)`2
?Ψkσkτm(κ)`2?Ψkσkτm(κ)`1?Ψkσkτm(κ)`4?Ψkσkτm(κ)`3
?Ψkτkυm(κ)`3?Ψkτkυm(κ)`2?Ψkτkυm(κ)`1?Ψkτkυm(κ)`4
?Ψkυkρm(κ)`4?Ψkυkρm(κ)`3?Ψkυkρm(κ)`2?Ψkυkρm(κ)`1,
Ma hema ics 2022,10, 1905 9 o 16
which implies ha
∇m+1(κ )≥Ψkρkσm(κ)`1.Ψkρkσm(κ)`4.Ψkρkσm(κ)`3.Ψkρkσm(κ)`2
?Ψkσkτm(κ)`2.Ψkσkτm(κ)`1.Ψkσkτm(κ)`4.Ψkσkτm(κ)`3
?Ψkτkυm(κ)`3.Ψkτkυm(κ)`2.Ψkτkυm(κ)`1.Ψkτkυm(κ)`4
?Ψkυkρm(κ)`4.Ψkυkρm(κ)`3.Ψkυkρm(κ)`2.Ψkυkρm(κ)`1
=Ψkρkσm(κ)`1+`2+`3+`4?Ψkσkτm(κ)`1+`2+`3+`4
?Ψkτkυm(κ)`1+`2+`3+`4?Ψkυkρm(κ)`1+`2+`3+`4
≥Ψkρkσm(κ)?Ψkσkτm(κ)?Ψkτkυm(κ)?Ψkυkρm(κ)=∇m(κ).
This implies ha
∇m+1(κ )≥ ∇m(κ)
o all
m≥
0 and all
κ>
0. Repea ing his p ocess,
∇m(κ)≥ ∇m−1(κ
)≥ ∇m−2(κ
2)≥... ≥ ∇0(κ
m),∀κ>0 and m≥1. (19)
F om (15)–(19), we conclude ha
Ψkρkσm+1(κ )≥Ψkρkσm(κ)`1?Ψkσkτm(κ)`2?Ψkτkυm(κ)`3?Ψkυkρm(κ)`4
≥ ∇m(κ)≥ ∇0(κ
m);(20)
Ψkσkτm+1(κ )≥Ψkσkτm(κ)`1?Ψkτkυm(κ)`2?Ψkυkρm(κ)`3?Ψkρkσm(κ)`4
≥ ∇m(κ)≥ ∇0(κ
m);(21)
Ψkτkυm+1(κ )≥Ψkτkυm(κ)`1?Ψkυkρm(κ)`2?Ψkρkσm(κ)`3?Ψkσkτm(κ)`4
≥ ∇m(κ)≥ ∇0(κ
m);(22)
Ψkυkρm+1(κ )≥Ψkυkρm(κ)`1?Ψkρkσm(κ)`2?Ψkσkτm(κ)`3?Ψkτkυm(κ)`4
≥ ∇m(κ)≥ ∇0(κ
m).(23)
Thus,
Ψkρkσm+1(κ ),Ψkσkτm+1(κ ),Ψkτkυm+1(κ ),Ψkυkρm+1(κ )≥ ∇0(κ
m),∀κ>0 and m≥1.
Taking he limi as
m→∞
in (20)–(23) and using
limm→∞∇0(κ
m) =
1, o all
κ>
0,
we ge
limm→∞kρm=kυ
,
limm→∞kσm=kρ
,
limm→∞kτm=kσ
, and
limm→∞kυm=kτ
.
This shows, oge he wi h (14), ha
Ωρστυ =kρ=lim
m→∞kσm=σ,Ωστυρ =kσ=lim
m→∞kτm=τ,
Ωτυρσ =kτ=lim
m→∞kυm=υ,Ωυρστ =kυ=lim
m→∞kρm=ρ.
S 5
. We shall p o e ha
ρ=σ=τ=υ
. We se
Π(κ) = Ψρσ(κ)?Ψστ(κ)?Ψτυ(κ)?
Ψυρ(κ) o all κ>0. Then, acco ding o (1), we can w i e
Ψρσ(κ )=ΨΩρστυΩστυρ (κ )≥Ψkρkσ(κ)`1?Ψkσkτ(κ)`2?Ψkτkυ(κ)`3?Ψkυkρ(κ)`4
=Ψστ(κ)`1?Ψτυ(κ)`2?Ψυρ(κ)`3?Ψρσ(κ)`4;(24)
Ψστ(κ )=ΨΩστυρΩτυρσ (κ )≥Ψkσkτ(κ)`1?Ψkτkυ(κ)`2?Ψkυkρ(κ)`3?Ψkρkσ(κ)`4
=Ψτυ(κ)`1?Ψυρ(κ)`2?Ψρσ(κ)`3?Ψστ(κ)`4;(25)
Ψτυ(κ )=ΨΩτυρσΩυρστ (κ )≥Ψkτkυ(κ)`1?Ψkυkρ(κ)`2?Ψkρkσ(κ)`3?Ψkσkτ(κ)`4
=Ψυρ(κ)`1?Ψρσ(κ)`2?Ψστ(κ)`3?Ψτυ(κ)`4;(26)
Ψυρ(κ )=ΨΩυρστ Ωρστυ (κ )≥Ψkυkρ(κ)`1?Ψkρkσ(κ)`2?Ψkσkτ(κ)`3?Ψkτkυ(κ)`4
=Ψρσ(κ)`1?Ψστ(κ)`2?Ψτυ(κ)`3?Ψυρ(κ)`4.(27)
Ma hema ics 2022,10, 1905 16 o 16
In o med Consen S a emen : No applicable.
Da a A ailabili y S a emen : No applicable.
Acknowledgmen s:
The au ho s a e g a e ul o he Spanish Go e nmen and he Eu opean Commis-
sion o G an IT1207-19.
Con lic s o In e es : The au ho s decla e no con lic o in e es .
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