48
BRUCE K. DRIVER
†
4.
The Riemann Integral
In this short chapter, the Riemann integral for Banach space valued functions
is de
f
ned and developed. Our exposition will be brief, since the Lebesgue integral
and the Bochner Lebesgue integral will subsume the content of this chapter. The
following simple “Bounded Linear Transformation” theorem will often be used here
andintheseque
ltode
f
ne linear transformations.
Theorem 4.1
(B
.L
.T
.Th
eo
r
em
)
.
Suppose that
Z
is a normed space,
X
is a
Banach space, and
S
⊂
Z
is a dense linear subspace of
Z.
If
T
:
S
→
X
is a
bounded linear transformation (i.e. there exists
C<
∞
such that
k
Tz
k
≤
C
k
z
k
for all
z
∈
S
)
,
then
T
hasaun
iqueex
tens
iontoane
lemen
t
¯
T
∈
L
(
Z, X
)
and this
extension still satis
f
es
°
°
¯
Tz
°
°
≤
C
k
z
k
for all
z
∈
¯
S
.
Exercise 4.1.
Prove Theorem 4.1.
For the remainder of the chapter, let
[
a, b
]
be a
f
xed compact interval and
X
be
a Banach space. The collection
S
=
S
([
a, b
]
,X
)
of
step functions,
f
:[
a, b
]
→
X,
consists of those functions
f
which may be written in the form
(4.1)
f
(
t
)=
x
0
1
[
a,t
1
]
(
t
)+
n
−
1
X
i
=1
x
i
1
(
t
i
,t
i
+1
]
(
t
)
,
where
π
≡
{
a
=
t
0
<t
1
<
···
<t
n
=
b
}
is a partition of
[
a, b
]
and
x
i
∈
X.
For
f
as
in Eq. (4.1), let
(4.2)
I
(
f
)
≡
n
−
1
X
i
=0
(
t
i
+1
−
t
i
)
x
i
∈
X.
Exercise 4.2.
Show that
I
(
f
)
is well de
f
ned, independent of how
f
is represented
as a step function. (
Hint:
show that adding a point to a partition
π
of
[
a, b
]
does
not change the right side of Eq. (4.2).) Also verify that
I
:
S
→
X
is a linear
operator.
Proposition 4.2
(Riemann Integral)
.
The linear function
I
:
S
→
X
extends
uniquely to a continuous linear operator
¯
I
from
¯
S
(the closure of the step functions
inside of
c
∞
([
a, b
]
,X
))
to
X
and this operator satis
f
es,
(4.3)
k
¯
I
(
f
)
k
≤
(
b
−
a
)
k
f
k
∞
for all
f
∈
¯
S
.
Furthermore,
C
([
a, b
]
,X
)
⊂
¯
S
⊂
c
∞
([
a, b
]
,X
)
and for
f
∈
,
¯
I
(
f
)
may be computed
as
(4.4)
¯
I
(
f
)= l
im

π

→
0
n
−
1
X
i
=0
f
(
±
π
i
)(
t
i
+1
−
t
i
)
where
π
≡
{
a
=
t
0
<t
1
<
···
<t
n
=
b
}
denotes a partition of
[
a, b
]
,

π

=max
{
t
i
+1
−
t
i

:
i
=0
,...,n
−
1
}
is the mesh size of
π
and
±
π
i
may be chosen
arbitrarily inside
[
t
i
,t
i
+1
]
.
Proof.
Taking the norm of Eq. (4.2) and using the triangle inequality shows,
(4.5)
k
I
(
f
)
k
≤
n
−
1
X
i
=0
(
t
i
+1
−
t
i
)
k
x
i
k
≤
n
−
1
X
i
=0
(
t
i
+1
−
t
i
)
k