ISSN 2307-7743 http://scienceasia.asia
FIXED POINT THEOREMS FOR CONTRACTIVE MAPPINGS OF INTEGRAL TYPE
ROBAB HAMLBARANI HAGHI*, NEGAR BAKHSHI SADABADI
Department of Mathematics, Payame Noor University, P.O. Box 19395-3697, Tehran, Iran
*Email: [email protected]
Abstract. In this paper, we define a class of functions and prove fixed point theorems under this class for contractive mappings of integral type in complete metric space. Examples are included. Finally, we discuss the application of our main results in the research of functional equations.
1. Introduction
Branciari [5] was the first to study the existence of fixed points for the contractive map-pings of integral type. He established a nice integral version of the Banach contraction principle [3].
Afterwards, many authors continued the study of Branciari [5] and obtained many fixed point theorems for several classes of contractive mappings of integral type.see,e.g [6, 10, 13]. Throughout this paper, we assume thatR+ = [0,∞) and
Φ1 = {ϕ: ϕ : R+ → R+is Lebesgue integrable, summable on each compact subset of R+
andR0φ(t)dt>0 for each >0},
Φ2 = {ϕ : ϕ : R+ → R+satisfies thatlim inf
n→∞ ϕ(an) > 0 if and only iflim infn→∞ (an) > 0 for each{an} ⊆R+},
Φ3 ={ϕ:ϕ:R+ →R+ is nondecreasing continuous andϕ(t) = 0 if and only if t = 0},
Φ4 ={ϕ:ϕ:R+ →[0,1) satisfy thatlim sup
s→t
ϕ(s)<1 for each t >0}, The following lemmas play important roles in this paper.
Lemma 1.1. [9]Let ϕ∈Φ1 and {rn}n∈N be a nonnegative sequence with lim
n→∞rn =a. Then lim
n→∞
Z rn
0
ϕ(t)dt=
Z a
0
ϕ(t)dt.
Keywords: contractive mappings of integral type, directed graph, functional equation.
c
2019 Science Asia
Lemma 1.2. [9] Let ϕ∈Φ1 and {rn}n∈N be a nonnegative sequence. Then
lim
n→∞
Z rn
0
ϕ(t)dt= 0
if and only if lim
n→∞rn = 0.
Lemma 1.3. [9] Let ϕ∈Φ2. Then ϕ(t)>0 if and only if t >0.
2. Main Results
In this section we show the existence and uniqueness of fixed points for contractive map-pings of integral type. Now we give the following definition.
Definition 2.1. Let ζ be the set of functions g : (0,∞)3 → (0,∞) satisfying the following
conditions:
(1) g is continuous,
(2) if R0vϕ(t)dt <R0g(u,u,v)ϕ(t)dt, then R0vϕ(t)dt <R0uϕ(t)dt, (3) if R0uϕ(t)dt ≤ kRg(u,0,0)
0 ϕ(t)dt or
Ru
0 ϕ(t)dt ≤ k
Rg(0,0,u)
0 ϕ(t)dt for all k ∈ (0,1),
then u= 0.
As examples, the following functions belong to ζ (1) Ifg(a, b, c) = max{a, b, c}, then g ∈ζ.
Proof. obviously, g is continuous. IfR0vϕ(t)dt < R0max{u,u,v}ϕ(t)dt = max{Ru
0 ϕ(t)dt,
Rv
0 ϕ(t)dt}, then we have
Rv
0 ϕ(t)dt <
Ru
0 ϕ(t)dt.
Also if
Z u
0
ϕ(t)dt≤k
Z max{0,0,u}
0
ϕ(t)dt =kmax{
Z 0
0
ϕ(t)dt,
Z u
0
ϕ(t)dt}=k
Z u
0
ϕ(t)dt,
so(1−k)R0uϕ(t)dt≤0, then u= 0. (2) g(a, b, c) = a+2b,
(3) g(a, b, c) = a.
Theorem 2.2. Let f be a mapping from a complete metric space(X, d) into itself satisfying
Z d(f x,f y)
0
ϕ(t)dt≤α(M(x, y))
Z M(x,y)
0
ϕ(t)dt, x, y ∈X,
where M(x, y) =g(d(x, y), d(x, f x), d(y, f y)) and ϕ∈Φ1 and α :R+ →[0,1) is a function
with
lim sup
s→t
α(s)<1, t >0.
Proof. Letx∈X , n∈N and xn+1 =f xn and letdn =d(f xn, f xn+1). Since ϕ:R+→R+
we get Rdn0
0 ϕ(t)dt>0, for all n0 ∈ N. We show that dn ≤dn−1. On the contrary, suppose
that there existsn0 ∈Nsuch that dn0 > dn0−1. We have
Z dn0−1
0
ϕ(t)dt≤
Z dn0
0
ϕ(t)dt=
Z d(f xn0,f xn0+1)
0
ϕ(t)dt
≤α(M(xn0, xn0+1))
Z M(xn0,xn0+1)
0
ϕ(t)dt, (2.1)
where
M(xn0, xn0+1) =g(d(xn0, xn0+1), d(xn0, f xn0), d(xn0+1, f xn0+1)),
from2.1
Z dn0−1
0
ϕ(t)dt≤
Z dn0
0
ϕ(t)dt<
Z g(dn0−1,dn0−1,dn0)
0
ϕ(t)dt,
so from definition 2.1, we have
Z dn0−1
0
ϕ(t)dt≤
Z dn0
0
ϕ(t)dt<
Z dn0−1
0
ϕ(t)dt,
which is impossible. So the sequence{dn}n∈N is the nonnegative non increasing which implies
that there exists a constant c≥0 such that limn→∞dn=c. Supposec >0, then Z dn
0
ϕ(t)dt=
Z d(f xn,f xn+1)
0
ϕ(t)dt≤α(M(xn, xn+1))
Z M(xn,xn+1)
0
ϕ(t)dt, (2.2)
where M(xn, xn+1) = g(d(xn, xn+1), d(xn, xn+1), d(xn+1, xn+2)). Then
Z dn
0
ϕ(t)dt<
Z g(d(xn,xn+1),d(xn,xn+1),d(xn+1,xn+2))
0
ϕ(t)dt=
Z g(dn−1,dn−1,dn)
0
ϕ(t)dt,
so from definition 2.1, we have
Z dn
0
ϕ(t)dt<
Z dn−1
0
ϕ(t)dt,
taking n → ∞
Z c
0
ϕ(t)dt<
Z c
0
ϕ(t)dt,
that is a contradiction. Thenc= 0. Now, we show that{xn}is a cauchy sequence. Contrary,
suppose that {xn} is not a cauchy sequence. Then there existsδ >0 such that for allk >0
there exists m(k)> n(k)> k with d(xmk, xnk)≥δ and d(xmk−1, xnk)< δ. We have
δ ≤d(xmk, xnk) ≤ d(xmk, xmk−1) +d(xmk−1, xnk)
≤ d(xmk, xmk−1) +δ,
since lim
k→∞d(xmk, xmk−1) = 0, by letting k → ∞, we have klim→∞d(xmk, xnk) =δ. So
Z δ
0
ϕ(t)dt = lim sup
k→∞
Z d(f xmk,f xnk)
0
ϕ(t)dt
≤ lim sup
k→∞
(α(M(xmk, xnk)))
Z M(xmk,xnk)
0
ϕ(t)dt
= lim sup
k→∞
(α(M(xmk, xnk)))
Z g(δ,0,0)
0
ϕ(t)dt, (2.4)
whereM(xmk, xnk) =g(d(xmk, xnk), d(xmk, xmk+1), d(xnk, xnk+1)). From definition2.1,δ = 0.
That is a contradiction. Hence {xn}is a cauchy sequence. Since (X, d) is a complete metric
space, there exists a∈X such that xn→a. We have Z d(xn+1,f(a))
0
ϕ(t)dt≤α(M(xn, a))
Z M(xn,a)
0
ϕ(t)dt,
where
M(xn, a) =g(d(xn, a), d(xn, f(xn)), d(a, f(a))).
Letting n→ ∞
Z d(a,f a)
0
ϕ(t)dt≤ lim
n→∞α(M(xn, a))
Z g(0,0,d(a,f a))
0
ϕ(t)dt,
From definition 2.1, d(a, f a) = 0. Now, we prove uniqueness, suppose that f has another fixed point b ∈X with a 6=b. It follows that
Z d(a,b)
0
ϕ(t)dt=
Z d(f a,f b)
0
ϕ(t)dt≤α(M(a, b))
Z M(a,b)
0
ϕ(t)dt.
Where M(a, b) = g(d(a, b), d(a, f(a)), d(b, f(b))). So
Z d(a,b)
0
ϕ(t)dt≤α(g(d(a, b),0,0))
Z g(d(a,b),0,0)
0
ϕ(t)dt,
hence d(a, b) = 0. That is, a=b.
Example 2.3. Let X = [12,1] be endowed with euclidean metric d = |.|. Assume that f : X →X and ϕ:R+ →R+ and α : R+ → [0,1) are defined by f(x) = x
2 and ϕ(t) = 2t
and
α(t) =
1 3 +
t2
2, t∈[0,1] 1
2t, t∈(1,3]
1
t12
, t∈(3,∞)
also put g : (0,∞)3 → (0,∞) with g(a, b, c) = max{a, b, c}. Clearly (X, d) is a complete
metric space and (ϕ, α)∈Φ1×Φ4. Let x, y ∈X and x > y. We have
M(x, y) = g(|x−y|,|x−f x|,|y−f y|) = g(|x−y|,|x−x 2|,|y−
(1) if M(x, y) = |x−y| then
Z d(f x,f y)
0
ϕ(t)dt =
Z x−2y
0
2tdt= |x−y|
2
4
≤ (1 3 +
1
2|x−y|
2)|x−y|2
= α(M(x, y))
Z M(x,y)
0
ϕ(t)dt.
(2) if M(x, y) = |x
2| then
Z d(f x,f y)
0
ϕ(t)dt =
Z x−2y
0
2tdt= |x−y|
2
4
≤ (1 3 +
1 8|x|
2)|x|2
4
= α(M(x, y))
Z M(x,y)
0
ϕ(t)dt.
So all the conditions of theorem 2.2 are satisfied. Then f has a unique fixed point in X. We see that 0∈X is the unique fixed point of f.
In theorem 2.2we chose g(a, b, c) =a, we can obtain the following theorem.
Theorem 2.4. [9]Let f be a mapping from a complete metric space (X, d) into itself satis-fying
Z d(f x,f y)
0
ϕ(t)dt≤α(d(x, y))
Z d(x,y)
0
ϕ(t)dt, x, y ∈X,
where φ∈Φ1 and α:R+ →[0,1) is a function with
lim sup
s→t
α(s)<1, t >0.
Then f has a unique fixed pointa ∈X such that lim
n→∞f
n(x) = a for each x∈X.
Theorem 2.5. Let f be mapping from a complete metric space (X, d) into itself satisfying
Ψ(
Z d(f x,f y)
0
ϕ(t)dt)≤Ψ(
Z M(x,y)
0
ϕ(t)dt)−φ(
Z M(x,y)
0
ϕ(t)dt), x, y ∈X,
where M(x, y) = g(d(x, y), d(x, f x), d(y, f y)) and ϕ∈ Φ1 and φ ∈Φ2,Ψ∈ Φ3. Then f has
a unique fixed point.
Proof. Let x ∈ X, n ∈ N and xn+1 = f xn and dn = d(f xn, f xn+1). First we show that
dn0 > dn0−1. Since ϕ∈Φ1, then
Rdn0
0 ϕ(t)dt >0. We have
Ψ(
Z dn0−1
0
ϕ(t)dt) < Ψ(
Z dn0
0
ϕ(t)dt) = Ψ(
Z d(f xn0,f xn0+1)
0
ϕ(t)dt)
≤ Ψ(
Z M(xn0,xn0+1)
0
ϕ(t)dt)−φ(
Z M(xn0,xn0+1)
0
ϕ(t)dt) (2.5)
where M(xn0, xn0+1) =g(d(xn0, xn0+1), d(xn0, xn0+1), d(xn0+1, xn0+2)). sinceψ ∈Φ3, then
Z dn0−1
0
ϕ(t)dt<
Z dn0
0
ϕ(t)dt<
Z g(dn0−1,dn0−1,dn0)
0
ϕ(t)dt,
from definition2.1
Z dn0−1
0
ϕ(t)dt<
Z dn0
0
ϕ(t)dt<
Z dn0−1
0
ϕ(t)dt,
that is a contradiction. So {dn} is nonnegative and nonincreasing, which means that there
exists a constant c≥0 such that lim
n→∞dn=c. Suppose c >0. It follows that
Ψ(
Z dn
0
ϕ(t)dt) = Ψ(
Z d(f xn,f xn+1)
0
ϕ(t)dt)≤Ψ(
Z M(xn,xn+1)
0
ϕ(t)dt)−φ(
Z M(xn,xn+1)
0
ϕ(t)dt), (2.6)
where M(xn, xn+1) = g(d(xn, xn+1), d(xn, xn+1), d(xn+1, xn+2)). sinceψ ∈Φ3, then
Z dn
0
ϕ(t)dt<
Z g(dn−1,dn−1,dn)
0
ϕ(t)dt,
from definition2.1
Z dn
0
ϕ(t)dt<
Z dn−1
0
ϕ(t)dt,
letting n→ ∞, we have
Z c
0
ϕ(t)dt<
Z c
0
ϕ(t)dt,
which is impossible. Hence c = 0. Now we show that {xn} is a cauchy sequence. Suppose
on the contrary that {xn} is not a cauchy sequence. So there exists δ >0 such that for all
k >0 there exists m(k)> n(k)> k with d(xmk, xnk) ≥δ and d(xmk−1, xnk)< δ. Similar to
the theorem 2.2, we can get lim
k→∞d(xmk, xnk) = δ. Then
Ψ(
Z d(f xmk,f xnk)
0
ϕ(t)dt)≤Ψ(
Z M(xmk,xnk)
0
ϕ(t)dt)−φ(
Z M(xmk,xnk)
0
ϕ(t)dt),
where M(xmk, xnk) = g(d(xmk, xnk), d(xmk, xmk+1), d(xnk, xnk+1)). So
Ψ(
Z d(f xmk,f xnk)
0
ϕ(t)dt)<Ψ(
Z g(d(xmk,xnk),d(xmk,xmk+1),d(xnk,xnk+1))
0
ϕ(t)dt),
since ψ ∈Φ3
Z d(f xmk,f xnk)
0
ϕ(t)dt<
Z g(d(xmk,xnk),d(xmk,xmk+1),d(xnk,xnk+1))
0
since g is continuous and by letting n→ ∞
Z δ
0
ϕ(t)dt<
Z g(δ,0,0)
0
ϕ(t)dt.
From definition 2.1, δ= 0. That is a contradiction. Then {xn} is a cauchy sequence. Since
(X, d) is a complete metric space, there exists a∈X, such thatxn →a. We infer that
Ψ(
Z d(xn+1,f a)
0
ϕ(t)dt)≤Ψ(
Z M(xn,a)
0
ϕ(t)dt)−φ(
Z M(xn,a)
0
ϕ(t)dt),
(2.7)
Where M(xn, a) = g(d(xn, a), d(xn, xn+1), d(a, f a)). Then
Ψ(
Z d(xn+1,f a)
0
ϕ(t)dt)<Ψ(
Z g(d(xn,a),d(xn,xn+1),d(a,f a))
0
ϕ(t)dt),
since ψ is nondecreasing
Z d(xn+1,f a)
0
ϕ(t)dt<
Z g(d(xn,a),d(xn,xn+1),d(a,f a))
0
ϕ(t)dt,
letting n→ ∞
Z d(a,f a)
0
ϕ(t)dt<
Z g(0,0,d(a,f a))
0
ϕ(t)dt,
from definition 2.1, d(a, f a) = 0. Finally, we show the uniqueness. Suppose that f has another fixed point b∈X with a6=b. It follows that
Ψ(
Z d(a,b)
0
ϕ(t)dt) = Ψ(
Z d(f a,f b)
0
ϕ(t)dt)≤Ψ(
Z M(a,b)
0
ϕ(t)dt)−Φ(
Z M(a,b)
0
ϕ(t)dt), (2.8)
where
M(a, b) =g(d(a, b), d(a, f a), d(b, f b)).
Hence from 2.8 we deduce that Ψ(
Z d(a,b)
0
ϕ(t)dt)<Ψ(
Z g(d(a,b),d(a,f a),d(b,f b))
0
ϕ(t)dt),
then
Z d(a,b)
0
ϕ(t)dt<
Z g(d(a,b),0,0)
0
ϕ(t)dt,
hence d(a, b) = 0, and this completes the proof.
Example 2.6. Let X = [12,1] be endowed with euclidean metric d = |.|. Assume that f :X →X and ϕ, φ,Ψ :R+→R+ are defined by f(x) = x
2,
ϕ(t) =
( t
2, x∈[0,1]
1, x∈(1,∞) φ(t) =
( t2
4, x∈[0,1]
t2
and
Ψ(t) =
(
t, x∈[0,1]
t2+1
2 , x∈(1,∞)
Put g(a, b, c) = max{a, b, c}. Clearly (X, d) is a complete metric space and (ϕ, φ,Ψ) ∈ Φ1×Φ2×Φ3. Let x, y ∈X and x > y. We have
M(x, y) =g(|x−y|,|x−f x|,|y−f y|) =g(|x−y|,|x 2|,|
y 2|), we have two cases
(1) If M(x, y) =|x−y|, then Ψ(
Z d(f x,f y)
0
ϕ(t)dt) = Ψ(
Z 12|x−y|
0
ϕ(t)dt) = Ψ(|x−y|
2
16 ) =
|x−y|2
16
≤ |x−y|
2
4 −
|x−y|4
16 = Ψ(
|x−y|2
4 )−φ(
|x−y|2
4 )
= Ψ(
Z |x−y|
0
ϕ(t)dt)−φ(
Z |x−y|
0
ϕ(t)dt)
= Ψ(
Z d(x,y)
0
ϕ(t)dt)−φ(
Z d(x,y)
0
ϕ(t)dt).
(2) If M(x, y) =|x
2|, then
Ψ(
Z d(f x,f y)
0
ϕ(t)dt) = Ψ(
Z 12|x−y|
0
t 2dt) =
|x−y|2
16 ≤ |x|2
16 − |x|4
1024
= Ψ(
Z |x2|
0
ϕ(t)dt)−φ(
Z |x2|
0
ϕ(t)dt)
Thus by theorem 2.4, f has a unique fixed point in X. We see that 0 is the fixed point of f. In theorem 2.5we chose g(a, b, c) =a, we can obtain the following theorem.
Theorem 2.7. [9]Let f be a mapping from a complete metric space (X, d) into itself satis-fying
Z d(f x,f y)
0
ϕ(t)dt≤ψ(
Z d(x,y)
0
ϕ(t)dt)−φ(
Z d(x,y)
0
ϕ(t)dt), x, y ∈X,
where ϕ ∈ Φ1, φ ∈ Φ2 and ψ ∈ Φ3. Then f has a unique fixed point a ∈ X such that
lim
n→∞f
n(x) =a for each x∈X.
3. Fixed point theorem endowed with a graph
In 2007, Jachymski [8] introduced the concept of G-contraction on a metric space endowed with graph G. Later on many authors undertook further investigations in this direction [1, 2, 4, 12].
that Ghas no parallel edges. Now we can identifyG with the pair(v(G), E(G)). The graph G can be converted to a weighted graph by assigning to each edge a weight equal to the distance between its vertices.
LetG−1 denote conversion of the graphGobtained from graphGby reversing the direction of edges. Thus we have v(G−1) = v(G)andE(G−1) = {(x, y)∈X×X; (y, x)∈E(G)}. The
letter G¯ denotes the undirected graph obtained from G by ignoring the direction of edges. It is convenient to treat G¯ as a directed graph for which the set of its edges is symmetric. That is E( ¯G) = E(G)∪E(G−1).
If x, y are vertices in a graph G, then a path in G from x to y of length l is a sequence {xi}li=0 of l+ 1 vertices such that x0 =x, xl=y and (xi−1, xi)∈E(G)for i= 1, ..., l.
A graphG is called connected if there is a path between any two vertices ofG. Gis weakly connected if G¯ is connected. The space (X, d) is said to have property P if Whenever a sequence {xn} in X, convergence to x and (xn, xn+1) ∈ E(G) for n ∈ N, then there exists
a subsequence {xnk} such that (xnk, x) ∈ E(G) for all k ∈ N. G is transitive if for any
two vertices x and y that are connected by a directed finite path, we have (x, y) ∈ E(G). Suppose that F ixf be all fixed points of f.
Theorem 3.1. Let (X, d) be a complete metric space endowed with a graph G and let f be a self mapping on X and (ϕ, φ,Ψ) ∈Φ1×Φ2×Φ3. Suppose that the following is satisfied
(1) for all x, y ∈X, (x, y)∈E(G) then (f x, f y)∈E(G), (2) there exists x0 ∈X such that (x0, f x0)∈E(G),
(3) G is transitive and has the property P, (4) for all (x, y)∈E(G)
Z d(f x,f y)
0
ϕ(t)dt≤α(M(x, y))
Z M(x,y)
0
ϕ(t)dt,
where M(x, y) =g(d(x, y), d(x, f x), d(y, f y)).
thenf has a fixed point. Also suppose that Gf is weakly connected, then f has a unique fixed
point, where v(Gf) = F ixf and E(Gf)⊆E(G).
Proof. Letxn+1 =f xn. Since(x0, x1)∈E(G), so(f x0, f x1)∈E(G)that is(x1, x2)∈E(G).
By this we have (xn, xn+1)∈E(G). Since G is transitive we deduce that (xmk, xnk)∈E(G)
for mk, nk ∈ N. Similar to the proof of theorem 2.2, we can show that {xn} is a cauchy
sequence. From completeness of(X, d), there existsa∈X such thatxn →a. From property
P, there exists {xnk} such that (xnk, a)∈E(G). So
Z d(f xnk+1,f a)
0
ϕ(t)dt≤α(M(xnk, a))
Z M(xnk,a)
0
ϕ(t)dt,
where M(xnk, a) = g(d(xnk, a), d(a, f a), d(xnk, f xnk)). Similar to the proof of theorem 2.2,
we show thatf a=a.
So there exists a path{xi}li=0 of l+ 1vertices with x0 =a andxl =b and (xi, xi+1)∈E( ¯Gf)
that is (a, b)∈E( ¯Gf). Then Z d(a,b)
0
ϕ(t)dt=
Z d(f a,f b)
0
ϕ(t)dt≤α(M(a, b))
Z M(a,b)
0
ϕ(t)dt,
where M(a, b) =g(d(a, b), d(a, f a), d(b, f b)). Hence
Z d(a,b)
0
ϕ(t)dt≤α(g(d(a, b),0,0))
Z g(d(a,b),0,0)
0
ϕ(t)dt,
from definition2.1, d(a, b) = 0. That is a=b.
Corollary 3.2. Let (X, d,) be a complete partially ordered metric space and let f be a self-mapping on X. Suppose that
(1) f is nondecreasing,
(2) there exists x0 ∈X such that x0 f x0,
(3) for all x, y ∈ X with x y and ϕ∈ Φ1 and α :R+ →[0,1) with lim sups→tα(s)<
1, t >0
Z d(f x,f y)
0
ϕ(t)dt≤α(M(x, y))
Z M(x,y)
0
ϕ(t)dt,
where M(x, y) =g(d(x, y), d(x, f x), d(y, f y)).
(4) if nondecreasing sequence {xn} convergence to x, then xnx.
Then f has a fixed point.
Proof. Define a graphGbyv(G) =XandE(G) ={(x, y)∈X×X;xy}. So all conditions
of Theorem 3.1is satisfied and then f has a fixed point.
4. Application
In this section, we apply Theorem 2.2, to show the existence for the solution of the functional equations arising in dynamic programming.
Throughout this section, suppose that(X,||.||)and(Y,||.||0)be real banach spaces andS ⊆X be the state space andD⊆Y be the decision space. Assume thatB(S)be the set of all real-valued bounded functions onS. We equipB(S)with the metricd(f, g) = supx∈S|f(x)−g(x)| forf ∈B(S). Note that(B(S), d(f, g))is a Banach space. Consider the following functional equation
f(x) = infopty∈D{A(x, y, f(a(x, y))), q(x, y) +B(x, y, f(b(x, y)))}, x∈S
(4.1)
Lemma 4.1. [11] let C be a set, p and q : C → R be mappings such that opty∈Cp(y),
opty∈Cq(y) and supy∈C|p(y)−q(y)| are bounded. Then
|opty∈Cp(y)−opty∈Cq(y)| ≤sup y∈C
|p(y)−q(y)|.
Lemma 4.2. [11] Let a, b, c and d be in R. Then
|opt{a, b} −opt{c, d}| ≤max{|a−c|,|b−d|}.
Theorem 4.3. Let u, v : S → R , A, B :S ×D×R → R and T : S×D → S satisfy the following conditions
(1) q, A, B are bounded,
(2) R0|A(x,y,u(T(x,y)))−A(x,y,v(T(x,y)))|ϕ(t)dt≤α(M(u, v))R0M(u,v)ϕ(t)dt or
Z |B(x,y,u(T(x,y)))−B(x,y,v(T(x,y)))|
0
ϕ(t)dt≤α(M(u, v))
Z M(u,v)
0
ϕ(t)dt,
where M(u, v) = g(d(u, v), d(u, Hu), d(v, Hv)) and Hz(x) = infopty∈D{A(x, y, z(T(x, y))), q(x, y) + B(x, y, z(T(x, y)))} for all x ∈ S and
z ∈B(S).
Then the functional equation 4.1 has a unique solution. Proof. From assumption 1, there exists k > 0such that
sup{|A(x, y, t)|,|B(x, y, t)|,|q(x, y)|}< k, (x, y, t)∈S×D×R, (4.2)
also from4.2and lemma 4.1, we deduce thatH is the mapping onB(S). By theorem 12.34 in [14], we obtain that for each >0, there exists δ >0 such that
Z
C
ϕ(t)dt < , ∀C ⊆[0,2M] with m(C)≤δ (4.3)
where m(C)is the Lebesgue measure of C. Put
c(x, y, z) = opty∈D{A(x, y, z(T(x, y))), q(x, y) +B(x, y, z(T(x, y)))}
Letx∈S. For any u, v ∈B(S), there exists y, h∈D with
Hu(x)≤c(x, y, u), Hu(x)> c(x, y, u)−δ
Hv(x)≤c(x, y, v), Hv(x)> c(x, y, v)−δ,
so by using lemma4.2, we have
Hu(x)−Hv(x) < c(x, y, u)−c(x, h, v) +δ
< max{|A(x, y, u)−A(x, h, v)|,|B(x, y, u)−B(x, h, v)|+δ, (4.4)
and
Hu(x)−Hv(x) > c(x, y, u)−c(x, h, v)−δ
Hence
|Hu(x)−Hv(x)|<max{|A(x, y, u)−A(x, h, v)|,|B(x, y, u)−B(x, h, v)|+δ, (4.6)
put
A1 =A(x, y, u), A2 =A(x, h, v), B1 =B(x, y, u), B2 =B(x, h, v),
so assumption2 combining with 4.6 and 4.3 implies that
Z |Hu(x)−Hv(x)|
0
ϕ(t)dt ≤
Z max{|A1−A2|,|B1−B2|}+δ
0
ϕ(t)dt
= max{
Z |A1−A2|+δ
0
ϕ(t)dt,
Z |B1−B2|+δ
0
ϕ(t)dt}
= max{
Z |A1−A2|
0
ϕ(t)dt+
Z |A1−A2|+δ
|A1−A2|
ϕ(t)dt,
Z |B1−B2|
0
ϕ(t)dt+
Z |B1−B2|+δ
|B1−B2|
ϕ(t)dt}
≤ max{
Z |A1−A2|
0
ϕ(t)dt,
Z |B1−B2|
0
ϕ(t)dt}+ max{
Z |A1−A2|+δ
|A1−A2|
ϕ(t)dt,
Z |B1−B2|+δ
|B1−B2|
ϕ(t)dt}
≤ α(M(u, v))
Z M(u,v)
0
ϕ(t)dt+, (4.7)
letting →0+ implies that
Z d(Hu(x),Hv(x))
0
ϕ(t)dt ≤α(M(u, v))
Z M(u,v)
0
ϕ(t)dt.
Thus by theorem 2.2, the equation has a solution.
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