≤
t
−
t
0

max
{
F
(
y, s
)

:(
)
∈
R
}
≤
δM
≤
r
so (
I
(
z
)(
t
)
,t
)
∈
R
for all
t
∈
[
t
0
−
δ, t
0
+
δ
]. Thus,
I
:
S
→
S
.
Given two functions
x, z
∈
S
and
t
∈
[
t
0
−
δ, t
0
+
δ
],

I
(
z
)(
t
)
−
I
(
x
)(
t
)

=
⎩
⎩
⎩
⎩
y
0
+
Z
t
t
0
F
(
z
(
s
)
,s
)
ds
−
y
0
−
Z
t
t
0
F
(
x
(
s
)
)
ds
⎩
⎩
⎩
⎩
=
⎩
⎩
⎩
⎩
Z
t
t
0
(
F
(
z
(
s
)
)
−
F
(
x
(
s
)
))
ds
⎩
⎩
⎩
⎩
t
−
t
0

sup
{
F
(
z
(
s
)
)
−
F
(
x
(
s
)
)

:
s
∈
[
t
0
−
δ, t
0
+
δ
]
}
≤
δK
sup
{
z
(
s
)
−
x
(
s
)

:
s
∈
[
t
0
−
δ, t
0
+
δ
]
}
≤
k
z
−
x
k
∞
2
Therefore,
k
I
(
z
)
−
I
(
x
)
k
∞
≤
k
z
−
x
k
∞
2
,so
I
is a contraction. Since
S
is a complete metric space,
I
has a
unique Fxed point
y
∈
S
. Therefore, for all
t
∈
[
t
0
−
δ, t
0
+
δ
], we have
y
(
t
)=
y
0
+
Z
t
t
0
F
(
y
(
s
)
)
ds
F
is continuous, so the ±undamental Theorem of Calculus implies that
y
0
(
t
F
(
y
(
t
)
)
for all
t
∈
(
t
0
−
δ, t
0
+
δ
). Since we also have
y
(
t
0
y
0
+
Z
t
0
t
0
F
(
y
(
s
)
)
ds
=
y
0
y
(restricted to (
t
0
−
δ, t
0
+
δ
)) is a solution of the initial value problem (1).
On the other hand, suppose that ˆ
y
is any solution of the initial value problem (1) on (
t
0
−
δ, t
0
+
δ
). It
is easy to check that (ˆ
y
(
s
)
)
∈
R
for all
s
∈
(
t
0
−
δ, t
0
+
δ
), so we have

F
(ˆ
y
(
s
)
)
≤
M
;
3