Rectangular cone bmetric spaces over Banach algebra and contraction principle
George et al. Fixed Point Theory and Applications
Rectangular cone bmetric spaces over Banach algebra and contraction principle
Reny George 3 4
Hossam A Nabwey 2 4
R Rajagopalan 4
Stojan Radenovic´ 0 1
KP Reshma 5
0 Faculty of Mathematics and Statistics, Ton Duc Thang University , Ho Chi Minh City , Vietnam
1 Nonlinear Analysis Research Group, Ton Duc Thang University , Ho Chi Minh City , Vietnam
2 Department of Basic Engineering Sciences, Faculty of Engineering, Menofia University , Menofia , Egypt
3 Department of Mathematics and Computer Science, St. Thomas College , Bhilai, Chhattisgarh , India
4 Department of Mathematics, College of Science, Prince Sattam bin Abdulaziz University , AlKharj, Kingdom of Saudi Arabia
5 Department of Mathematics, Rungta College of Engineering and Technology , Bhilai, Chhattisgarh , India
Rectangular cone bmetric spaces over a Banach algebra are introduced as a generalization of metric space and many of its generalizations. Some fixed point theorems are proved in this space and proper examples are provided to establish the validity and superiority of our results. An application to solution of linear equations is given which illustrates the proper application of the results in spaces over Banach algebra.
fixed points; cone rectangular bmetric space; rectangular metric space; rectangular bmetric space

For a given cone P ⊂ A and x, y ∈ A, we say that x y if and only if y – x ∈ P. Note that
is a partial order relation defined on A. For more details on the basic concepts of a Banach
algebra, solid cone, unit element e, zero element θ , invertible elements in Banach algebra
etc. the reader may refer to [–].
For basic properties of Banach algebra and spectral radius refer to [, ].
In what follows A will always denote a Banach algebra, P a solid cone in A and e the unit
element of A.
Definition . ([]) Let P be a solid cone in a Banach space E. A sequence {un} ⊂ P is said
to be a csequence if for each c θ there exists a natural number N such that un c for
all n > N .
Remark . For more on csequences see [, , , ].
Lemma . ([]) Let E be a Banach space.
(i) If a, b, c ∈ E and a b c, then a c.
(ii) If θ a c for each c θ , then a = θ .
3 Main results
In this section first we introduce the definition of a rectangular cone bmetric space over
a Banach algebra (in short RCbMSBA) and furnish examples to show that this concept
is more general than that of CMSBA and CbMSBA. We then define convergence and a
Cauchy sequence in a RCbMSBA and then prove fixed point results in this space.
Definition . Let χ be a nonempty set and drcb : χ × χ → A be such that for all x, y, u, v ∈
χ , x = u, v = y:
(RCbM) θ drcb(x, y) and drcb(x, y) = θ if and only if x = y;
(RCbM) drcb(x, y) = drcb(y, x);
(RCbM) there exist s ∈ P, e s such that drcb(x, y) s[drcb(x, u) + drcb(u, v) + drcb(v, y)].
Then drcb is called a rectangular cone bmetric on χ and (χ , drcb) is called a rectangular
cone bmetric space over a Banach algebra (in short RCbMSBA) with coefficient s. If s = e
we say that (χ , drcb) is a rectangular cone metric space over a Banach algebra (in short
RCMSBA).
In the above definition if condition RCbM is replaced with
(CbM) drcb(x, y) s[drcb(x, z) + drcb(z, y)] for all x, y, z ∈ χ ,
then (χ , drcb) is a CbMSBA as defined in [].
Note that every CMSBA is a CbMSBA and CbMSBA is a RCbMSBA but the converse
is not necessarily true. Inspired by [, ] we furnish the following examples, which will
establish our claim.
Example . Let A = {a = (ai,j)× : ai,j ∈ R, ≤ i, j ≤ }, a = ≤i,j≤ ai,j,
P = {a ∈ A : ai,j ≥ , ≤ i, j ≤ } be a cone in A. Let χ = B ∪ N, where B = { n : n ∈ N}.
otherwise.
Then (χ , drcb) is a RCbMSBA over A with coefficient s = . But it is not possible to
find s ∈ P, e s satisfying condition CbM and so (χ , d) is not a CbMSBA over a Banach
algebra A.
Example . Let χ = [, ] and let A = CR (χ ). For α = (f , g) and β = (u, v) in A, we define
α.β = (f .u, g.v) and α = max( f , g ) where f = supx∈χ f (x). Then A is a Banach
algebra with unit e = (, ), zero element θ = (, )and P = {(f , g) ∈ A : f (t) ≥ , g(t) ≥ ,
t ∈ χ } a cone in A. Consider drcb : χ × χ → A given by
⎧⎪ drcb(x, y)(t) = (, )
⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨ drcb(x, y)(t) = (c + d.t, a + bt)
if x = y;
if x, y ∈ B = [, ) and
a, b, c, d are some fixed real
numbers;
⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩ ddrrccbb((xx,, yy))((tt)) == ((nx –(cy+d(c.t+), dn.t()a, +x b–ty))(a + b.t)) iyoft∈xhe{=rw,n(i}ns;e≥. ) ∈ B and
Clearly (χ , drcb) is a RCbMSBA over A with s = (, ). Again it is not possible to find a real
number s ∈ P, e s satisfying condition CbM and so (χ , drcb) is not a CbMSBA over a
Banach algebra A.
For any a ∈ χ , the open sphere with center a and radius λ
θ is given by
Bλ(a) = b ∈ χ : drcb(a, b) ≺ λ .
Let U = {Y ⊆ χ : ∀x ∈ Y , ∃r θ , such that Br(x) ⊆ Z}. Then U defines the rectangular
bmetric topology for the RCbMSBA (χ , drcb).
The definitions of convergent sequence, Cauchy sequence, csequence and
completeness in RCbMSBA are along the same lines as for CbMSBA given in [] and so we omit
these definitions.
Remark . We refer to Example . for the following:
(i) Open balls in RCbMSBA need not be an open set. For example Bλ( ) with
λ = is not open because open balls with center are not contained in Bλ( ).
(ii) The limit of a sequence in RCbMSBA is not unique. For instance { n } converges to
and .
(iii) Every convergent sequence in RCbMSBA need not be Cauchy. For
d( n , n+p ) = θ as n → ∞, so { n } is not a Cauchy sequence.
(iv) A RCbMSBA need not be Hausdorff, as it is impossible to find r, r
Br () ∩ Br () = φ.
such that
Theorem . Let (χ , drcb) be a complete RCbMSBA over A with θ
there exist λ ∈ P, r(λ) < such that
s and T : χ → χ . If
for all x, y ∈ χ , then T has a unique fixed point.
Proof Let x ∈ χ be arbitrary. Consider the iterative sequence defined by xn+ = Txn for all
n ≥ . We divide the proof into three cases.
Case : Let r(λ) ∈ [, s ) (s > ). If xn = xn+ then xn is fixed point of T . Moreover, for any
x ∈ X the iterative sequence {T nx} (n ∈ N) converges to the fixed point. So, suppose that
xn = xn+ for all n ≥ . Setting drcb(xn, xn+) = dn, it follows from (.) that
drcb(xn, xn+) = drcb(Txn–, Txn)
dn
λdn– ≺ dn–,
λd(xn–, xn),
dn
λnd.
Repeating this process we obtain
We consider d(xn, xn+p) in two cases.
(.)
(.)
(.)
(.)
i.e. the sequence {dn} is strictly decreasing and from this it follows that dn = dm whenever
n = m. Continuing this process we get
Again setting dn∗ = drcb(xn, xn+) for any n ∈ N, using (.) we get
Since r(λ) < s , we have r(sλ) ≤ sr(λ) ≤ sr(λ).r(λ) < s < and so e – sλ is invertible and
If p is odd, say m + , then using (.) as well as the fact that dn = dm whenever n = m
we obtain
sλn e + sλ + sλ + · · · d + sλn+ e + sλ + sλ + · · · d
+ smλn+md
=
∞
i=
sλ i sλnd +
sλ i sλn+d
∞
i=
= e – sλ –sλnd + e – sλ –sλn+d
= λn e – sλ –sd[e + λ].
Since r(λ) < s < , using Lemma . of [], it is easy to see that λn is a csequence. Again
using Proposition . of [], λn(e – (sλ))–sd[ + λ] → θ as n → ∞, and so it follows that,
for any c ∈ A with θ c, there exists a natural number N such that, for any n > N, we
have
λn e – sλ –sd[ + λ]
If p is even say m, using (.) and (.) as well as the fact that dn = dm whenever n = m
we obtain
drcb(xn, xn+m) s drcb(xn, xn+) + drcb(xn+, xn+) + drcb(xn+, xn+m)
s[dn + dn+]
+ s drcb(xn+, xn+) + drcb(xn+, xn+) + drcb(xn+, xn+m)
s[dn + dn+] + s[dn+ + dn+] + s[dn+ + dn+] + · · ·
+ sm–[dm– + dm–] + sm–drcb(xn+m–, xn+m)
s λnd + λn+d + s λn+d + λn+d + s λn+d + λn+d + · · ·
+ sm– λm–d + λm–d + sm–λn+m–d∗
sλn e + sλ + sλ + · · · d + sλn+ e + sλ + sλ + · · · d
+ sm–λn+m–d∗
∞
i=
∞
i=
sλ i sλnd +
sλ i sλn+d + sm–λn+m–d∗
e – sλ –sλnd + e – sλ –sλn++d + sm–λn+m–d∗
e – sλ –sλnd[e + λ] + sm–λn+m–d∗
Note that r(λ) < s < and so using Lemma . of [], it is easy to see that λn is a csequence.
Again using Proposition . of [], (e – (sλ))–sd[e + λ]λn + ( es – λ)–d∗λn → θ as n → ∞
and so it follows that, for any c ∈ A with θ c, there exists a natural number N such that,
for any n > N, we have
e – sλ –sd[e + λ]λn +
Let N = Max{N, N}. Then for all n ≥ N we have
e – λ
s
–
d∗λn
c.
drcb(xn, xn+p)
c.
nl→im∞ xn = u.
Thus {xn} is a Cauchy sequence and since (χ , drcb) is complete, we can find u ∈ χ such that
Since dn = dm whenever n = m there exists k ∈ N such that drcb(u, Tu) = {dk, dk+, . . .}. Then
for any n > k
(.)
(.)
(.)
drcb(u, Tu) s drcb(u, xn) + drcb(xn, xn+) + drcb(xn+, Tu)
= s drcb(u, xn) + dn + drcb(Txn, Tu)
s drcb(u, xn) + dn + λdrcb(xn, u)
s (e + λ)drcb(xn, u) + λnd → θ
as n → ∞,
i.e. Tu = u. Now if Tv = v and drcb(u, v) = θ then using (.) one can easily deduce that
drcb(u, v) = , and so the fixed point is unique.
Case : Let r(λ) ∈ [ s , ) (s > ). In this case, we have r(λ)n → as n → ∞, and so there
exists n ∈ N such that r(λ)n < s . Note that r(λn ) ≤ r(λ)n < s . Also by (.),
drcb T n x, T n y = drcb T T n–x , T T n–y
λdrcb T n–x , T n–y
= λdrcb T T n–x , T T n–y
λdrcb T n–x , T n–y
· · ·
λno drcb(x, y).
Thus by case , T n has a unique fixed point u∗ ∈ X. Now we have
T n Tu∗ = T n+ u∗ = T T n u∗ = Tu∗,
(.)
i.e. Tu∗ is also a fixed point of T n . Hence, by the uniqueness of the fixed point of T n we get
Tu∗ = u∗. Now suppose Tu = u and Tv = v. Then T n u = T n–(Tu) = T n–u = · · · = Tu = u
and T n v = T n–(Tv) = T n–v = · · · = Tv = v. By the uniqueness of the fixed point of T n
we get u = v.
Case : s = . The proof follows from case .
Remark . In an open problem in [] the authors have asked whether it is possible to
increase the range of λ in Theorem . of [] from (, s ) to (, ). Since every rectangular
bmetric space is a RCbMSBA, Theorem . gives a positive answer to the question posed
by the authors.
Definition . Let (χ , drcb) be a RCbMSBA, θ s and T : χ → χ . Then T is called a
weak Kannan contraction iff there exist L, λ ∈ P such that ≤ r(λ) < s+ , and
drcb(Tu, Tv)
λ drcb(u, Tu) + drcb(v, Tv) + L.α(u, v) ∀u, v ∈ χ
(.)
and α(u, v) = drcb(u, Tv) or drcb(v, Tu).
Theorem . Let (χ , drcb) be a complete RCbMSBA with θ s, and T : χ → χ be a
mapping. If T is a weak Kannan contraction mapping then T has a fixed point. Further if
L < or drcb(Tx, Ty)
L∗. drcb(x, Tx) + d(y, Ty)
(.)
for some L∗ ∈ P, then the fixed point is unique.
Proof Let x ∈ χ be arbitrary. Consider the iterative sequence defined by xn+ = Txn for all
n ≥ . Let drcb(xn, xn+) = dn and suppose α(x, y) = drcb(x, Ty). It follows from (.) that
drcb(xn, xn+) = drcb(Txn–, Txn),
dn
λ[dn– + dn].
If α(x, y) = drcb(x, Ty)
drcb(xn, xn+) = drcb(Txn–, Txn),
Thus in both cases
dn
dn
dn
λ[dn– + dn].
λ[dn– + dn],
(e – λ)–λdn– = βdn–,
where β = (e – λ)–λ. Repeating this process we obtain
dn
βnd.
(.)
Also, for α(x, y) = drcb(x, Ty)
drcb(xn, xn+) = drcb(Txn–, Txn+) = drcb(Txn+, Txn–)
λ[dn– + dn+] + L.dn
λ[dn– + dn+] + L.dn;
for α(x, y) = drcb(y, Tx)
where η = (λ(e + β) + Lβ)d ∈ P. Thus we have
Note that r(λ) < and so (e – λ)– is invertible and (e – λ)– = i∞= λi. Therefore r(β) =
r((e – λ)–λ) = ( i∞= λi) ≤ i∞= r(λ)i = –r(rλ(λ)) (as r(λ) < ). Thus we have r(β) < s . Therefore
r sβ = sr β ≤ sr(β)r(β) = s <
so e – sβ is invertible and
e – sβ – =
sβ i.
∞
i=
We will analyze drcb(xn, xn+p) as follows: For some odd p say m +
drcb(xn, xn+m+) s dn + dn+ + drcb(xn+, xn+m+)
(.)
(.)
s[dn + dn+] + s dn+ + dn+ + drcb(xn+, xn+m+)
s[dn + dn+] + s[dn+ + dn+] + s[dn+ + dn+] + · · · + smdn+m
s βnd + βn+d + s βn+d + βn+d + s βn+d + βn+d + · · ·
+ smβn+md
sβn + sβ + sβ + · · · d + sβn+ + sβ + sβ + · · · d
=
∞
Note that r(β) < s < and using Lemma . of [], βn is a csequence. Again using
Proposition . of [], (e – (sβ))–sβnd + (e – (sβ))–sβn+d → θ as n → ∞. It follows that, for
any c ∈ A with θ c, there exists N ∈ N such that, for any n > N, we have
e – sβ –sβnd + e – sβ –sβn+d
(.)
For some even p, say m,
drcb(xn, xn+m) s dn + dn+ + drcb(xn+, xn+m)
s[dn + dn+] + s dn+ + dn+ + drcb(xn+, xn+m)
s[dn + dn+] + s[dn+ + dn+] + s[dn+ + dn+] + · · ·
sβn + sβ + sβ + · · · d + sβn+ + sβ + sβ + · · · d
+ sm–ηβm–βn–d
since r(β) < s .
∞
i=
∞
i=
(.)
(.)
(.)
d(xn, xn+p)
c.
Note that r(β) < s < and using Lemma . of [], βn– is a csequence. Again using
Proposition . of [] (e – (sβ))–sββn–d + (e – (sβ))–sββn–d + η( es – β)–βn–d → θ as
n → ∞. It follows that, for any c ∈ A with θ c, there exists N ∈ N such that, for any
n > N,
drcb(xn, xn+m)
e – sβ –sββn–d + e – sβ –sββn–d
+ η e – β
s
Let N = Max{N, N}. Then for all n ≥ N we have
Thus {xn} is a Cauchy sequence and by completeness of (χ , drcb) there exists u ∈ χ such
that
Since dn = dm whenever n = m there exists k ∈ N such that d(u, Tu) = {dk, dk+, . . .}. Then
for any n > k
Note that r(sλ) ≤ sr(λ) < < and so e – sλ is invertible. Also, r(β) < s < . Hence
using Lemma . of [], βn is a csequence and use of Proposition . of [] gives
s(e – λs)–[drcb(u, xn) + (e + λ)βnd] + L.drnb(u, xn+) → θ as n → ∞. It follows that, for
c ∈ A and θ c, there exists N ∈ N, such that, for any n > N,
drcb(u, Tu) s(e – λs)– drcb(u, xn) + ( + λ)βnd + L.drcb(u, xn+)
c,
(.)
i.e. Tu = u. Uniqueness follows easily from (.).
Theorem . Let (χ , drcb) be a complete RCbMSBA with s ≥ and T : χ → χ be a
mapping. If there exists λ ∈ P such that ≤ r(λ) < s+ , and
for all x, y ∈ χ then T has a unique fixed point.
(.)
Proof Note that (.) implies (.) and (.). Hence the result follows from
Theorem ..
Corollary . Theorem . of [] and Theorem . of [].
Proof Note that, for k ∈ P with r(k) < ,
where r(k) < and r(k) < . Thus T satisfies conditions (.) and (.) of Theorem
., with s = . Since every CMSBA is a RCbMSBA with s = , the proof follows from
Theorem ..
Corollary . Theorem . of [] and Theorem . of [].
Proof Since every CMSBA is a RCbMSBA with s = , the proof follows from
Theorem ..
Example . Let A = {a = (ai,j)× : ai,j ∈ R, ≤ i, j ≤ }, a = maxi
j= ai,j, P = {a ∈
A : ai,j ≥ , ≤ i, j ≤ } be a cone in A. Let χ = A ∪ B, where A = [, ] and B = [, ]. Let
drcb : χ × χ → A be given by
if u = v;
if u, v ∈ A – {, , , , , };
if u = n (n ≥ ) ∈ A and v ∈ {, };
otherwise.
drcb(, ) = drcb( , ) = drcb( , ) =
⎧ drcb(, ) = drcb( , ) = drcb( , ) =
⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪
drcb(, ) = drcb( , ) = drcb( , ) =
drcb(, ) = drcb( , ) = drcb( , ) =
drcb(, ) = drcb( , ) = drcb( , ) =
.. .. ;
.. .. ;
.. .. ;
. . ;
. .
.. .. ;
drcb(u, v) =
⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪ dddrrrcccbbb(((uuu,,, vvv))) === u–nnvnnu–v
u–v u–v
Then (χ , drcb) is a RCbMSBA over A with s = . However, for u = n and v = m it is
impossible to find s ∈ P, e
s such that drcb( n , m )
N
s(drcb( n , ) + drcb(, m )) for all n, m ∈ ,
drcb( , ) =
. .
. .
drcb( , ) + d( , ) =
.. .. . Define T by
and so (χ , drcb) is not a CbMSBA over A. Also (χ , drcb) is not a CMSBA over A as
⎧
⎪ ,
⎪⎩ ,
u ∈ B ∪ { };
u ∈ A – {D ∪ }.
Tu = ⎨ – u, u ∈ D = { n : n ≥ , n = };
drcb(Tu, Tv) = drcb( , ) =
drcb(v, Tv) = drcb( , ) =
(.).
Then T satisfies condition (.) For α(x, y) = θ then it is enough if we take L sufficiently
large. If α(u, v) = θ , we proceed as follows.
Case (i): u ∈ B ∪ { }, v ∈ D, drcb(u, Tv) = drcb(u, – v); drcb(v, Tu) = drcb(v, ); α(u, v) = θ
iff u + v = or v = . Since v ∈ D, v = . u + v = only at u = and v = . Then
.. .. ; drcb(u, Tu) = drcb( , ) =
.. .. ;
.. .. . Clearly we can find λ =
k
k
with k ∈ (, ) satisfying
Case (ii): u ∈ B ∪ { }, v ∈ A – {D ∪ }, drcb(u, Tv) = drcb(u, ); drcb(v, Tu) = drcb(v, );
α(u, v) = only at u = or v = . But u = . Let v = and u ∈ B ∪ { }. Then
drcb(Tu, Tv) = drcb( , ) =
.. .. ; drcb(v, Tv) = drcb( , ) =
.. .. ; drcb(u, Tu) =
x–  x– 
drcb(x, ) = x–  x–  ; when u = , drcb(u, Tu) =
ist k ∈ (, ) such that λ = k satisfying (.).
k
Case (iii): u ∈ D, v ∈ A – {D ∪ }, drcb(u, Tv) = drcb(u, ); drcb(v, Tu) = drcb(v, – u);
α(u, v) = at u = and u + v = . At u = , drcb(Tu, Tv) = ( , ) = . Hence condition
(.) is satisfied. At u = n and v = – n , drcb(Tu, Tv) = drcb( – n , ) =
drcb(u, Tu) = drcb( n , – n ) =
 n –   n – 
 n –   n –  ; drcb(v, Tv) = drcb( – n , ) =
 – n   – n 
 – n   – n  ;
 – n   – n 
 – n   – n  ;
.. .. . Clearly there
exNote that drcb(u, Tu) + drcb(v, Tv) =  n –  +  – n  =  – n  = drcb(Tu, Tv). Hence
(.) is satisfied with λ =
.
Other cases follow similarly. Indeed condition (.) is satisfied. Note that drcb(u, Tu) +
drcb(v, Tv) = θ for any u, v ∈ χ and so T satisfies (.) for sufficiently large L∗. Theorem .
is thus applicable and Fix(T ) = { }. However, condition (.) is not satisfied at u = and
v = as drcb(Tu, Tv) = drcb( , ) = .. .. [drcb(u, Tu) + drcb(v, Tv)] = [drcb( , ) +
drcb( , )] = .. .. . Hence Theorem . is not applicable.
Example . Let χ = [, ] and drcb(x, y) = x – y. Let Tx = x for all x, y ∈ χ . Then
Theorem . is applicable on T and Fix(T ) = {}. However, Corollary . is not applicable
on T . If we take X = [, ] then T satisfies (.) but neither L < nor T satisfy (.).
and are two fixed points of T .
Now, we will apply Theorem . to study the existence and uniqueness of solutions of a
system of linear equations.
Consider the following system of linear equations:
⎧⎪ ax + ax + · · · + anxn = b,
⎪⎪⎨ ax + ax + · · · + anxn = b,
.
.
⎪⎪⎪⎩ .anx + anx + · · · + annxn = bn,
C
with aij, bi (i = · · · n, j = · · · n) ∈ .
Theorem . If
n
i=,i=j aij +  – ajj < , then (.) has a unique solution.
Proof Consider the Banach algebra A = {a = (aij)n×n : aij ∈ C, ≤ i, j ≤ n}, with e being the
identity matrix of order n, multiplication defined as ordinary matrix multiplication and
a = in= aij. Let P = {a ∈ A : ai,j ≥ , ≤ i, j ≤ } be a cone in A. Let χ = C . Define
drcb : χ × χ → A by
(.)
⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪ α ⎛⎝ xxxn –––...yyyn xxxn –––...yyyn ......... xxxn–––...yyyn ⎞⎠
drcb(x, y) = ⎨⎪ ⎛ p x–y p x–y ... p x–y ⎞
⎪⎪⎪⎪⎪⎪⎪ ⎜⎜⎝ p x...–y p x...–y ... p x...–y⎟⎟⎠
p xn–yn p xn–yn ... p xn–yn
⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩ ⎝⎛ xxxn –––...yyyn xxxn –––...yyyn ......... xxxn–––...yyyn ⎠⎞
if x, y ∈ X,
x – y ∈ [ – , ] × [ – , ] × · · · × [ – , ],
α is the largest integer such that
α ∈ {, , , , , , , , } and
maxi xi – yi < α ;
if x = (x, x, . . . , xn) ∈ X:
x + x + · · · + xn = p ∈ N and
y = (y, y, . . . , yn) ∈ N × N × · · · × N;
otherwise.
Then (χ , drcb) is a RCbMSBA over A with s = . Let T : χ → X be defined by
Tx = (I – A)x + B for all x ∈ χ ,
where An×n, χsn× and Bn× are the coefficient matrices of (.).
Then the system of linear equation (.) is equivalent to x = Tx. We will show that T
satisfies (.). Let x, y ∈ χ .
Case . x – y ∈ [ – , ] × [ – , ] × · · · × [ – , ] and α is the largest integer such that
maxi{xi – yi : i = , , . . . , n} < α , α ∈ {, , , , , , , , }. Then Tx = (γ, γ, . . . , γn);
γi = jn=,j=i aijxj + ( – aii)xi, Ty = (η, η, . . . , ηn); ηi = jn=,j=i aijxj + ( – aii)xi and Tx –
Ty = (λ, λ, . . . , λn) where λi = jn=,j=i aij(xj – yj) + ( – aii)(xi – yi) and λi = γi – ηi =
n
j=,j=i aijxj – yj +  – aiixi – yi < β , β ≥ α. Thus we have
x – y . . . x – y ⎞
x – y . . . x – y⎟
... ... ⎟⎟⎟⎠ .
xn – yn . . . xn – yn
Case . x – y ∈/ [ – , ] × [ – , ] × · · · × [ – , ]. Then we have
⎛ ( – a)
⎜ a
drcb(Tx, Ty) = ⎜⎜⎝⎜ ...
an
⎛ x – y
× ⎜⎜⎜⎜⎝ x –... y
xn – yn
an
Thus in both cases we have d(Tx, Ty) ≤ γ .d(x, y), where
.
.
.
an
a
an
. . .
an
⎞
the system of linear equations (.) has a unique solution.
Theorem . If
n
j=,j=i aij +  – aii < , then the conclusion of Theorem . still holds.
Proof Let A and P be as in the proof of Theorem . and a =
Let drcb : χ × χ → A be given by
j= aij. Let χ = C .
n
drcb(x, y) =
⎧ ⎛ x–y x–y ... x–y ⎞
⎨ ⎝ x –...y x –...y ... x –...y⎠
⎩ xn–yn xn–yn ... xn–yn
∀x, y ∈ X.
Tx = (I – A)x + B for all x ∈ X,
Then (χ , drcb) is a RCbMSBA over A with s = . Define the self map T of χ by
where An×n, Xn× and Bn× are the coefficient matrices of (.).
Then the system of linear equations (.) is the problem x = Tx. We will show that T
satisfies (.). Let x, y ∈ χ . Then
⎜
drcb(Tx, Ty) = ⎜⎜
⎜
⎝
⎛ ( – a)
a
.
.
.
an
a
. . .
an
an
⎞
an ⎟
... ⎟⎟⎟ drcb(x, y) = γ .drcb(x, y)
⎠
. . . ( – ann)
with γ = I – A and r(γ ) ≤ γ =
n
i=,i=j aij +  – ajj < . Thus T satisfies (.) and so by
Theorem . the system of linear equations (.) has a unique solution.
Acknowledgements
This project is supported by Deanship of Scientific research at Prince Sattam bin Abdulaziz University, Al kharj, Kingdom
of Saudi Arabia, under International Project Grant No. 2016/01/6714. The authors are thankful to the learned reviewers for
their valuable suggestions which helped in bringing this paper in its present form.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript.
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