Now, we are in a position to state and prove our main result.
Proof.
Denote by
F
the operator associated with the righthand side of (1). Then, (1) takes the form
(14)
x
=
F
x
,
where
(15)
F
x
=
a
+
F
H
x
,
(
H
x
)
(
t
)
=
(
G
x
)
(
t
)
·
(
U
x
)
(
t
)
,
(
U
x
)
(
t
)
=
β
Γ
(
α
)
∫
0
t
s
β

1
u
(
t
,
s
,
x
(
s
)
)
(
t
β

s
β
)
1

α
d
s
,
t
∈
R
+
.
Here,
F
and
G
are the superposition operators, generated by the functions
f
=
f
(
t
,
x
)
and
g
=
g
(
t
,
x
)
involved in (1), defined by
(16)
(
F
x
)
(
t
)
=
f
(
t
,
x
(
t
)
)
,
(17)
(
G
x
)
(
t
)
=
g
(
t
,
x
(
t
)
)
,
respectively, where
x
=
x
(
t
)
is an arbitrary function defined on
R
+
(see [39]).
Solving (1) is equivalent to finding a fixed point of the operator
F
defined on the space
B
C
(
R
+
)
.
For convenience, we divide the proof into several steps.
Step 1 (
F
x
is continuous on
R
+
). To prove the continuity of the function
F
x
on
R
+
it suffices to show that if
x
∈
B
C
(
R
+
)
, then
U
x
is continuous function on
R
+
, thanks to
(
h
1
)
,
(
h
2
)
, and
(
h
3
)
. For this purpose, take an arbitrary
x
∈
B
C
(
R
+
)
and fix
ɛ
>
0
and
T
>
0
. Assume that
t
1
,
t
2
∈
[
0
,
T
]
are such that

t
2

t
1

≤
ɛ
. Without loss of generality we can assume that
t
2
>
t
1
. Then we get
(18)

(
U
x
)
(
t
2
)

(
U
x
)
(
t
1
)

=

β
Γ
(
α
)
∫
0
t
2
s
β

1
u
(
t
2
,
s
,
x
(
s
)
)
(
t
2
β

s
β
)
1

α
d
s

β
Γ
(
α
)
∫
0
t
1
s
β

1
u
(
t
1
,
s
,
x
(
s
)
)
(
t
1
β

s
β
)
1

α
d
s

≤

β
Γ
(
α
)
∫
0
t
2
s
β

1
u
(
t
2
,
s
,
x
(
s
)
)
(
t
2
β

s
β
)
1

α
d
s

β
Γ
(
α
)
∫
0
t
1
s
β

1
u
(
t
2
,
s
,
x
(
s
)
)
(
t
2
β

s
β
)
1

α
d
s

+

β
Γ
(
α
)
∫
0
t
1
s
β

1
u
(
t
2
,
s
,
x
(
s
)
)
(
t
2
β

s
β
)
1

α
d
s

β
Γ
(
α
)
∫
0
t
1
s
β

1
u
(
t
1
,
s
,
x
(
s
)
)
(
t
2
β

s
β
)
1

α
d
s

+

β
Γ
(
α
)
∫
0
t
1
s
β

1
u
(
t
1
,
s
,
x
(
s
)
)
(
t
2
β

s
β
)
1

α
d
s

β
Γ
(
α
)
∫
0
t
1
s
β

1
u
(
t
1
,
s
,
x
(
s
)
)
(
t
1
β

s
β
)
1

α
d
s

≤
β
Γ
(
α
)
∫
t
1
t
2
s
β

1

u
(
t
2
,
s
,
x
(
s
)
)

(
t
2
β

s
β
)
1

α
d
s
+
β
Γ
(
α
)
∫
0
t
1
s
β

1
[

u
(
t
2
,
s
,
x
(
s
)
)

u
(
t
1
,
s
,
x
(
s
)
)

]
(
t
2
β

s
β
)
1

α
d
s
+
β
Γ
(
α
)
∫
0
t
1
s
β

1

u
(
t
1
,
s
,
x
(
s
)
)

mm
×
[
(
t
1
β

s
β
)
α

1

(
t
2
β

s
β
)
α

1
]
d
s
.
Let us denote
(19)
ω
d
T
(
u
,
ɛ
)
=
sup
{

u
(
t
2
,
s
,
y
)

u
(
t
1
,
s
,
y
)

:
s
,
t
1
,
t
2
∈
[
0
,
T
]
,
m
t
1
≥
s
,
t
2
≥
s
,

t
2

t
1

≤
ɛ
,
m
y
∈
[

d
,
d
]
;
d
≥
0

u
(
t
2
,
s
,
y
)

u
(
t
1
,
s
,
y
)

}
;
then we obtain
(20)

(
U
x
)
(
t
2
)

(
U
x
)
(
t
1
)

≤
β
Γ
(
α
)
×
∫
t
1
t
2
s
β

1
[

u
(
t
2
,
s
,
x
(
s
)
)

u
(
t
2
,
s
,
0
)

+

u
(
t
2
,
s
,
0
)

]
(
t
2
β

s
β
)
1

α
d
s
+
β
Γ
(
α
)
∫
0
t
1
s
β

1
ω
∥
x
∥
T
(
u
,
ɛ
)
(
t
2
β

s
β
)
1

α
d
s
+
β
Γ
(
α
)
×
∫
0
t
1
s
β

1
[

u
(
t
1
,
s
,
x
(
s
)
)

u
(
t
1
,
s
,
0
)

+

u
(
t
1
,
s
,
0
)

]
mm
×
[
(
t
1
β

s
β
)
α

1

(
t
2
β

s
β
)
α

1
]
d
s
≤
β
Γ
(
α
)
∫
t
1
t
2
s
β

1
[
l
(
t
2
)
Φ
(

x
(
s
)

)
+
u
*
(
t
2
)
]
(
t
2
β

s
β
)
1

α
d
s
+
ω
∥
x
∥
T
(
u
,
ɛ
)
Γ
(
α
+
1
)
[
t
2
α
β

(
t
2
β

t
1
β
)
α
]
+
β
Γ
(
α
)
∫
0
t
1
s
β

1
[
l
(
t
1
)
Φ
(

x
(
s
)

)
+
u
*
(
t
1
)
]
mm
×
[
(
t
1
β

s
β
)
α

1

(
t
2
β

s
β
)
α

1
]
d
s
≤
[
l
(
t
2
)
Φ
(
∥
x
∥
)
+
u
*
(
t
2
)
]
Γ
(
α
+
1
)
(
t
2
β

t
1
β
)
α
+
ω
∥
x
∥
T
(
u
,
ɛ
)
Γ
(
α
+
1
)
t
2
α
β
+
l
(
t
1
)
Φ
(
∥
x
∥
)
+
u
*
(
t
1
)
Γ
(
α
+
1
)
[
t
1
α
β

t
2
α
β
+
(
t
2
β

t
1
β
)
α
]
.
Thus
(21)
ω
T
(
U
x
,
ɛ
)
≤
1
Γ
(
α
+
1
)
{
2
ε
α
β
[
l
^
(
T
)
Φ
(
∥
x
∥
)
+
u
^
(
T
)
]
+
T
α
β
ω
∥
x
∥
T
(
u
,
ɛ
)
}
,
where
(22)
l
^
(
T
)
=
max
{
l
(
t
)
:
t
∈
[
0
,
T
]
}
,
u
^
(
T
)
=
max
{
u
*
(
t
)
:
t
∈
[
0
,
T
]
}
.
In view of the uniform continuity of the function
u
on
[
0
,
T
]
×
[
0
,
T
]
×
[

∥
x
∥
,
∥
x
∥
]
we have that
ω
∥
x
∥
T
(
u
,
ɛ
)
→
0
as
ɛ
→
0
. From the above inequality we infer that the function
U
x
is continuous on the interval
[
0
,
T
]
for any
T
>
0
. This yields the continuity of
U
x
on
R
+
and, consequently, the function
F
x
is continuous on
R
+
.
Step 2 (
F
x
is bounded on
R
+
). In view of our hypotheses for arbitrary
x
∈
B
C
(
R
+
)
and for a fixed
t
∈
R
+
we have
(23)

(
F
x
)
(
t
)

≤

a
(
t
)
+
f
(
t
,
β
g
(
t
,
x
(
t
)
)
Γ
(
α
)
∫
0
t
s
β

1
u
(
t
,
s
,
x
(
s
)
)
(
t
β

s
β
)
1

α
d
s
)

≤
∥
a
∥
+
β
Γ
(
α
)
m
(
t
)
[

g
(
t
,
x
(
t
)
)

g
(
t
,
0
)

+

g
(
t
,
0
)

]
×
∫
0
t
s
β

1
[

u
(
t
,
s
,
x
(
s
)
)

u
(
t
,
s
,
0
)

+

u
(
t
,
s
,
0
)

]
(
t
β

s
β
)
1

α
d
s
+

f
(
t
,
0
)

≤
∥
a
∥
+
f
*
+
β
m
(
t
)
[
n
(
t
)
∥
x
∥
+

g
(
t
,
0
)

]
Γ
(
α
)
×
∫
0
t
s
β

1
[
l
(
t
)
Φ
(

x
(
s
)

)
+
u
*
(
t
)
]
(
t
β

s
β
)
1

α
d
s
≤
∥
a
∥
+
f
*
+
m
(
t
)
[
n
(
t
)
∥
x
∥
+

g
(
t
,
0
)

]
Γ
(
α
+
1
)
×
[
l
(
t
)
Φ
(
∥
x
∥
)
+
u
*
(
t
)
]
t
α
β
=
∥
a
∥
+
f
*
+
1
Γ
(
α
+
1
)
[
ϕ
(
t
)
∥
x
∥
Φ
(
∥
x
∥
)
mm
+
ψ
(
t
)
∥
x
∥
+
ξ
(
t
)
Φ
(
∥
x
∥
)
+
η
(
t
)
]
.
Hence,
F
x
is bounded on
R
+
, thanks to hypothesis
(
h
5
)
.
Step 3 (
F
maps the ball
B
r
0
into itself). Steps
2
and
3
allow us to conclude that the operator
F
transforms
B
C
(
R
+
)
into itself. Moreover, from the last estimate we have
(24)
∥
F
x
∥
≤
∥
a
∥
+
f
*
+
1
Γ
(
α
+
1
)
[
ϕ
*
∥
x
∥
Φ
(
∥
x
∥
)
+
ψ
*
∥
x
∥
+
ξ
*
Φ
(
∥
x
∥
)
+
η
*
]
.
From the last estimate with hypothesis
(
h
6
)
we deduce that there exists
r
0
>
0
such that the operator
F
maps
B
r
0
into itself.
Step 4 (an estimate of
F
with respect to the quantity
c
). Let us take a nonempty set
X
⊂
B
r
0
. Then, for arbitrary
x
,
y
∈
X
and for a fixed
t
∈
R
+
, we obtain
(25)

(
F
x
)
(
t
)

(
F
y
)
(
t
)

≤
β
m
(
t
)
Γ
(
α
)

g
(
t
,
x
(
t
)
)
∫
0
t
s
β

1
u
(
t
,
s
,
x
(
s
)
)
(
t
β

s
β
)
1

α
d
s
m

g
(
t
,
y
(
t
)
)
∫
0
t
s
β

1
u
(
t
,
s
,
y
(
s
)
)
(
t
β

s
β
)
1

α
d
s

≤
β
m
(
t
)

g
(
t
,
x
(
t
)
)

g
(
t
,
y
(
t
)
)

Γ
(
α
)
×
∫
0
t
s
β

1

u
(
t
,
s
,
x
(
s
)
)

(
t
β

s
β
)
1

α
d
s
+
β
m
(
t
)

g
(
t
,
y
(
t
)
)

Γ
(
α
)
×
∫
0
t
s
β

1

u
(
t
,
s
,
x
(
s
)
)

u
(
t
,
s
,
y
(
s
)
)

(
t
β

s
β
)
1

α
d
s
≤
β
m
(
t
)
n
(
t
)

x
(
t
)

y
(
t
)

Γ
(
α
)
×
∫
0
t
s
β

1
[

u
(
t
,
s
,
x
(
s
)
)

u
(
t
,
s
,
0
)

+

u
(
t
,
s
,
0
)

]
(
t
β

s
β
)
1

α
d
s
+
β
m
(
t
)
[
n
(
t
)

y
(
t
)

+

g
(
t
,
0
)

]
Γ
(
α
)
×
∫
0
t
s
β

1
l
(
t
)
Φ
(

x
(
s
)

y
(
s
)

)
(
t
β

s
β
)
1

α
d
s
≤
β
m
(
t
)
n
(
t
)

x
(
t
)

y
(
t
)

Γ
(
α
)
×
∫
0
t
s
β

1
[
l
(
t
)
Φ
(

x
(
s
)

)
+
u
*
(
t
)
]
(
t
β

s
β
)
1

α
d
s
+
β
m
(
t
)
[
n
(
t
)

y
(
t
)

+

g
(
t
,
0
)

]
Γ
(
α
)
×
∫
0
t
s
β

1
l
(
t
)
Φ
(

x
(
s
)

+

y
(
s
)

)
(
t
β

s
β
)
1

α
d
s
≤
β
m
(
t
)
n
(
t
)
l
(
t
)
(

x
(
t
)

+

y
(
t
)

)
Γ
(
α
)
×
∫
0
t
s
β

1
Φ
(

x
(
s
)

)
(
t
β

s
β
)
1

α
d
s
+
β
m
(
t
)
n
(
t
)
u
*
(
t
)

x
(
t
)

y
(
t
)

Γ
(
α
)
×
∫
0
t
s
β

1
(
t
β

s
β
)
1

α
d
s
+
β
m
(
t
)
n
(
t
)
l
(
t
)

y
(
t
)

Γ
(
α
)
×
∫
0
t
s
β

1
Φ
(

x
(
s
)

+

y
(
s
)

)
(
t
β

s
β
)
1

α
d
s
+
β
m
(
t
)
l
(
t
)

g
(
t
,
0
)

Γ
(
α
)
×
∫
0
t
s
β

1
Φ
(

x
(
s
)

+

y
(
s
)

)
(
t
β

s
β
)
1

α
d
s
≤
2
β
m
(
t
)
n
(
t
)
l
(
t
)
r
0
Φ
(
r
0
)
Γ
(
α
)
∫
0
t
s
β

1
(
t
β

s
β
)
1

α
d
s
+
β
m
(
t
)
n
(
t
)
u
*
(
t
)
diam
X
(
t
)
Γ
(
α
)
∫
0
t
s
β

1
(
t
β

s
β
)
1

α
d
s
+
β
m
(
t
)
n
(
t
)
l
(
t
)
r
0
Φ
(
2
r
0
)
Γ
(
α
)
∫
0
t
s
β

1
(
t
β

s
β
)
1

α
d
s
+
β
m
(
t
)
l
(
t
)

g
(
t
,
0
)

Φ
(
2
r
0
)
Γ
(
α
)
∫
0
t
s
β

1
(
t
β

s
β
)
1

α
d
s
≤
2
ϕ
(
t
)
r
0
Φ
(
r
0
)
Γ
(
α
+
1
)
+
ψ
(
t
)
Γ
(
α
+
1
)
diam
X
(
t
)
+
ϕ
(
t
)
r
0
Φ
(
2
r
0
)
Γ
(
α
+
1
)
+
ξ
(
t
)
Φ
(
2
r
0
)
Γ
(
α
+
1
)
.
Hence, we can easily deduce the following inequality:
(26)
diam
(
F
X
)
(
t
)
≤
2
ϕ
(
t
)
r
0
Φ
(
r
0
)
Γ
(
α
+
1
)
+
ψ
(
t
)
Γ
(
α
+
1
)
diam
X
(
t
)
+
ϕ
(
t
)
r
0
Φ
(
2
r
0
)
Γ
(
α
+
1
)
+
ξ
(
t
)
Φ
(
2
r
0
)
Γ
(
α
+
1
)
.
Now, taking into account hypothesis
(
h
5
)
we obtain
(27)
c
(
F
X
)
≤
q
c
(
X
)
,
where
q
=
(
ϕ
*
Φ
(
r
0
)
+
ψ
*
)
/
Γ
(
α
+
1
)
≥
ψ
*
/
Γ
(
α
+
1
)
. Obviously, in view of hypothesis
(
h
6
)
we have that
q
<
1
.
Step 5 (an estimate of
F
with respect to the modulus of continuity
ω
0
∞
). Take arbitrary numbers
ɛ
>
0
and
T
>
0
. Choose a function
x
∈
X
and take
t
1
,
t
2
∈
[
0
,
T
]
such that

t
2

t
1

≤
ɛ
. Without loss of generality we can assume that
t
2
>
t
1
. Then, taking into account our hypotheses and (21), we have
(28)

(
F
x
)
(
t
2
)

(
F
x
)
(
t
1
)

≤

a
(
t
2
)

a
(
t
1
)

+
m
(
t
2
)

(
G
x
)
(
t
2
)
(
U
x
)
(
t
2
)

(
G
x
)
(
t
1
)
(
U
x
)
(
t
2
)

+
m
(
t
2
)

(
G
x
)
(
t
1
)
(
U
x
)
(
t
2
)

(
G
x
)
(
t
1
)
(
U
x
)
(
t
1
)

+

f
(
t
2
,
(
G
x
)
(
t
1
)
(
U
x
)
(
t
1
)
)
m

f
(
t
1
,
(
G
x
)
(
t
1
)
(
U
x
)
(
t
1
)
)

≤
ω
T
(
a
,
ɛ
)
+
β
m
(
t
2
)

g
(
t
2
,
x
(
t
2
)
)

g
(
t
1
,
x
(
t
1
)
)

Γ
(
α
)
×
∫
0
t
2
s
β

1
[

u
(
t
2
,
s
,
x
(
s
)
)

u
(
t
2
,
s
,
0
)

+

u
(
t
2
,
s
,
0
)

]
(
t
2
β

s
β
)
1

α
d
s
+
m
(
t
2
)
[

g
(
t
1
,
x
(
t
1
)
)

g
(
t
1
,
0
)

+

g
(
t
1
,
0
)

]
Γ
(
α
+
1
)
×
{
2
ε
α
β
[
l
^
(
T
)
Φ
(
∥
x
∥
)
+
u
^
(
T
)
]
+
T
α
β
ω
∥
x
∥
T
(
u
,
ɛ
)
}
+

f
(
t
2
,
(
G
x
)
(
t
1
)
(
U
x
)
(
t
1
)
)
m

f
(
t
1
,
(
G
x
)
(
t
1
)
(
U
x
)
(
t
1
)
)

≤
ω
T
(
a
,
ɛ
)
+
β
m
(
t
2
)
[
n
(
t
2
)

x
(
t
2
)

x
(
t
1
)

+
ω
g
T
(
ɛ
)
]
Γ
(
α
)
×
∫
0
t
2
s
β

1
[
l
(
t
2
)
Φ
(

x
(
s
)

)
+
u
*
(
t
2
)
]
(
t
2
β

s
β
)
1

α
d
s
+
m
(
t
2
)
[
n
(
t
1
)

x
(
t
1
)

+

g
(
t
1
,
0
)

]
Γ
(
α
+
1
)
×
{
2
ε
α
β
[
l
^
(
T
)
Φ
(
∥
x
∥
)
+
u
^
(
T
)
]
+
T
α
β
ω
∥
x
∥
T
(
u
,
ɛ
)
}
+

f
(
t
2
,
(
G
x
)
(
t
1
)
(
U
x
)
(
t
1
)
)
m

f
(
t
1
,
(
G
x
)
(
t
1
)
(
U
x
)
(
t
1
)
)

≤
ω
T
(
a
,
ɛ
)
+
t
2
α
β
Γ
(
α
+
1
)
m
(
t
2
)
×
[
n
(
t
2
)
ω
T
(
x
,
ɛ
)
+
ω
g
T
(
ɛ
)
]
[
l
(
t
2
)
Φ
(
r
0
)
+
u
*
(
t
2
)
]
+
m
^
(
T
)
[
n
(
t
1
)
r
0
+
g
^
(
T
)
]
Γ
(
α
+
1
)
×
{
2
ε
α
β
[
l
^
(
T
)
Φ
(
r
0
)
+
u
^
(
T
)
]
+
T
α
β
ω
∥
x
∥
T
(
u
,
ɛ
)
}
+

f
(
t
2
,
(
G
x
)
(
t
1
)
(
U
x
)
(
t
1
)
)
m

f
(
t
1
,
(
G
x
)
(
t
1
)
(
U
x
)
(
t
1
)
)

≤
ω
T
(
a
,
ɛ
)
+
[
ϕ
(
t
2
)
Φ
(
r
0
)
+
ψ
(
t
2
)
]
Γ
(
α
+
1
)
ω
T
(
x
,
ɛ
)
+
T
α
β
ω
g
T
(
ɛ
)
Γ
(
α
+
1
)
m
^
(
T
)
[
l
^
(
T
)
Φ
(
r
0
)
+
u
^
(
T
)
]
+
m
^
(
T
)
[
n
^
(
T
)
r
0
+
g
^
(
T
)
]
Γ
(
α
+
1
)
×
{
2
ε
α
β
[
l
^
(
T
)
Φ
(
r
0
)
+
u
^
(
T
)
]
+
T
α
β
ω
∥
x
∥
T
(
u
,
ɛ
)
}
+

f
(
t
2
,
(
G
x
)
(
t
1
)
(
U
x
)
(
t
1
)
)
m

f
(
t
1
,
(
G
x
)
(
t
1
)
(
U
x
)
(
t
1
)
)

≤
ω
T
(
a
,
ɛ
)
+
[
ϕ
*
Φ
(
r
0
)
+
ψ
*
]
Γ
(
α
+
1
)
ω
T
(
x
,
ɛ
)
+
T
α
β
ω
g
T
(
ɛ
)
Γ
(
α
+
1
)
m
^
(
T
)
[
l
^
(
T
)
Φ
(
r
0
)
+
u
^
(
T
)
]
+
m
^
(
T
)
[
n
^
(
T
)
r
0
+
g
^
(
T
)
]
Γ
(
α
+
1
)
×
{
2
ε
α
β
[
l
^
(
T
)
Φ
(
r
0
)
+
u
^
(
T
)
]
+
T
α
β
ω
∥
x
∥
T
(
u
,
ɛ
)
}
+

f
(
t
2
,
(
G
x
)
(
t
1
)
(
U
x
)
(
t
1
)
)
m

f
(
t
1
,
(
G
x
)
(
t
1
)
(
U
x
)
(
t
1
)
)

.
In the last estimates, we have denoted by
(29)
ω
g
T
(
ɛ
)
=
sup
{

g
(
t
2
,
x
)

g
(
t
1
,
x
)

:
t
1
,
t
2
∈
[
0
,
T
]
,
mm

t
2

t
1

≤
ɛ
,
x
∈
[

r
0
,
r
0
]
}
,
n
^
(
T
)
=
max
{
n
(
t
)
:
t
∈
[
0
,
T
]
}
,
m
^
(
T
)
=
max
{
m
(
t
)
:
t
∈
[
0
,
T
]
}
,
g
^
(
T
)
=
max
{

g
(
t
,
0
)

:
t
∈
[
0
,
T
]
}
.
Hence,
(30)
ω
T
(
F
x
,
ɛ
)
≤
ω
T
(
a
,
ɛ
)
+
[
ϕ
*
Φ
(
r
0
)
+
ψ
*
]
Γ
(
α
+
1
)
ω
T
(
x
,
ɛ
)
+
T
α
β
ω
g
T
(
ɛ
)
Γ
(
α
+
1
)
m
^
(
T
)
[
l
^
(
T
)
Φ
(
r
0
)
+
u
^
(
T
)
]
+
m
^
(
T
)
[
n
^
(
T
)
r
0
+
g
^
(
T
)
]
Γ
(
α
+
1
)
×
{
2
ε
α
β
[
l
^
(
T
)
Φ
(
r
0
)
+
u
^
(
T
)
]
+
T
α
β
ω
∥
x
∥
T
(
u
,
ɛ
)
}
+
sup
t
1
,
t
2
∈
[
0
,
T
]
,
∥
x
∥
≤
r
0

f
(
t
2
,
(
G
x
)
(
t
1
)
(
U
x
)
(
t
1
)
)
mm

f
(
t
1
,
(
G
x
)
(
t
1
)
(
U
x
)
(
t
1
)
)

.
Since the function
f
(
t
,
y
)
is uniformly continuous on the set
[
0
,
T
]
×
[

H
,
H
]
, the function
g
(
t
,
x
)
is uniformly continuous on the set
[
0
,
T
]
×
[

r
0
,
r
0
]
and the function
u
(
t
,
s
,
x
)
is uniformly continuous on the set
[
0
,
T
]
×
[
0
,
T
]
×
[

r
0
,
r
0
]
, where
(31)
H
=
sup
{
∫
0
t
1
s
β

1

u
(
t
1
,
s
,
x
(
s
)
)

(
t
1
β

s
β
)
1

α
β

g
(
t
1
,
x
(
t
1
)
)

Γ
(
α
)
m
×
∫
0
t
1
s
β

1

u
(
t
1
,
s
,
x
(
s
)
)

(
t
1
β

s
β
)
1

α
d
s
:
t
1
∈
[
0
,
T
]
,
m
∥
x
∥
≤
r
0
β

g
(
t
1
,
x
(
t
1
)
)

Γ
(
α
)
∫
0
t
1
s
β

1

u
(
t
1
,
s
,
x
(
s
)
)

(
t
1
β

s
β
)
1

α
}
;
we have
(32)
sup
{

f
(
t
2
,
y
)

f
(
t
1
,
y
)

:
t
1
,
t
2
∈
[
0
,
T
]
,

t
2

t
1

≤
ɛ
,
mi

y

≤
H
}
⟶
0
as
ɛ
⟶
0
.
It is easy to see that
H
<
∞
because
u
(
t
,
s
,
x
)
is bounded on
[
0
,
T
]
×
[
0
,
T
]
×
[

r
0
,
r
0
]
,
g
(
t
,
x
)
is bounded on
[
0
,
T
]
×
[

r
0
,
r
0
]
, and
(
β
/
Γ
(
α
)
)
∫
0
t
1
(
s
β

1
/
(
t
1
β

s
β
)
1

α
)
d
s
≤
T
α
β
/
Γ
(
α
+
1
)
.
Therefore, from the last estimate we derive the following one:
(33)
ω
0
T
(
F
X
)
≤
q
ω
0
T
(
X
)
.
Hence we have
(34)
ω
0
∞
(
F
X
)
≤
q
ω
0
∞
(
X
)
.
Step 6 (
F
is contraction with respect to the measure of noncompactness
μ
). From (27) and (34) and the definition of the measure of noncompactness
μ
given by formula (8), we obtain
(35)
μ
(
F
X
)
≤
q
μ
(
X
)
.
Step 7. We construct a nonempty, bounded, closed, and convex set
Y
on which we will apply a fixed point theorem. In the sequel let us put
B
r
0
1
=
Conv
F
(
B
r
0
)
,
B
r
0
2
=
Conv
F
(
B
r
0
1
)
, and so on. In this way we have constructed a decreasing sequence of nonempty, bounded, closed, and convex subsets
(
B
r
0
n
)
of
B
r
0
such that
F
(
B
r
0
n
)
⊂
B
r
0
n
for
n
=
1,2
,
…
. Hence, in view of (35) we obtain
(36)
μ
(
B
r
0
n
)
≤
q
n
μ
(
B
r
0
)
,
for
any
n
=
1,2
,
3
,
…
.
This implies that
lim
n
→
∞
μ
(
B
r
0
n
)
=
0
. Hence, taking into account Definition 1 we infer that the set
Y
=
⋂
n
=
1
∞
B
r
0
n
is nonempty, bounded, closed, and convex subset of
B
r
0
. Moreover,
Y
∈
ker
μ
. Also, the operator
F
maps
Y
into itself.
Step 8 (
F
is continuous on the set
Y
). Let us fix a number
ɛ
>
0
and take arbitrary functions
x
,
y
∈
Y
such that
∥
x

y
∥
≤
ɛ
. Using the fact that
Y
∈
ker
μ
and keeping in mind the structure of sets belonging to
ker
μ
we can find a number
T
>
0
such that for each
z
∈
Y
and
t
≥
T
we have that

z
(
t
)

≤
ɛ
. Since
F
maps
Y
into itself, we have that
F
x
,
F
y
∈
Y
. Thus, for
t
≥
T
we get
(37)

(
F
x
)
(
t
)

(
F
y
)
(
t
)

≤

(
F
x
)
(
t
)

+

(
F
y
)
(
t
)

≤
2
ɛ
.
On the other hand, let us assume
t
∈
[
0
,
T
]
. Then we obtain
(38)

(
F
x
)
(
t
)

(
F
y
)
(
t
)

≤
β
m
(
t
)
n
(
t
)

x
(
t
)

y
(
t
)

Γ
(
α
)
×
∫
0
t
s
β

1
[
l
(
t
)
Φ
(

x
(
s
)

)
+
u
*
(
t
)
]
(
t
β

s
β
)
1

α
d
s
+
β
m
(
t
)
[
n
(
t
)

y
(
t
)

+

g
(
t
,
0
)

]
Γ
(
α
)
×
∫
0
t
s
β

1
l
(
t
)
Φ
(

x
(
s
)

y
(
s
)

)
(
t
β

s
β
)
1

α
d
s
≤
[
m
(
t
)
n
(
t
)
l
(
t
)
Φ
(
r
0
)
+
m
(
t
)
n
(
t
)
u
*
(
t
)
]
ɛ
β
Γ
(
α
)
×
∫
0
t
s
β

1
(
t
β

s
β
)
1

α
d
s
+
[
m
(
t
)
n
(
t
)
l
(
t
)
r
0
+
m
(
t
)
l
(
t
)

g
(
t
,
0
)

]
Φ
(
ɛ
)
β
Γ
(
α
)
×
∫
0
t
s
β

1
(
t
β

s
β
)
1

α
d
s
≤
ϕ
(
t
)
Φ
(
r
0
)
+
ψ
(
t
)
Γ
(
α
+
1
)
ɛ
+
ϕ
(
t
)
r
0
+
ξ
(
t
)
Γ
(
α
+
1
)
Φ
(
ɛ
)
≤
ϕ
*
Φ
(
r
0
)
+
ψ
*
Γ
(
α
+
1
)
ɛ
+
ϕ
*
r
0
+
ξ
*
Γ
(
α
+
1
)
Φ
(
ɛ
)
.
Now, taking into account (37) and (38) and hypothesis
(
h
5
)
we conclude that the operator
F
is continuous on the set
Y
.
Step 9 (application of Schauder fixed point principle). Linking all aboveobtained facts about the set
Y
and the operator
F
:
Y
→
Y
and using the classical Schauder fixed point principle we deduce that the operator
F
has at least one fixed point
x
in the set
Y
. Obviously the function
x
=
x
(
t
)
is a solution of the quadratic integral equation (1). Moreover, since
Y
∈
ker
μ
we have that all solutions of (1) belonging to
B
r
0
are asymptotically stable in the sense of Definition 2. This completes the proof.