Note.
Functions in
L
1
(
E
) are often called the essentially bounded.
Example.
C
[
a, b
]
✓
L
1
[
a, b
], since all continuous functions are measurable and all
f
2
C
[
a, b
] are bounded and thus
k
f
k
L
1
[
a,b
]
sup
x
2
[
a,b
]

f
(
x
)

<
1
. Note that
C
(
R
)
6✓
L
1
(
R
). All truly bounded measurable functions are in
L
1
.
Exercise.
Prove
k
f
k
1
= inf
{
↵
2
R
:
m
{
x
:

f
(
x
)

>
↵
}
= 0
}
Note.
This implies that if
β
>
k
f
k
1
, then take
k
f
k
1
<
↵
<
β
with
m
{
x
:

f
(
x
)

>
↵
}
, so
m
{
x
:

f
(
x
)

>
β
}
= 0.
Theorem.
If
k
f
k
1
= 0 then
f
= 0 a.e. .
Proof.
By above,
m
{
x
:

f
(
x
)

>
1
n
}
= 0 for all
n
2
N
. Then
1
[
n
=1
⇢
x
:

f
(
x
)

>
1
n
=
{
x
:

f
(
x
)

6
= 0
}
thus
m
{
x
:

f
(
x
)

6
= 0
}
= 0 by
σ
subadditivity, so
f
= 0 a.e. .
Note.
Similarly,
m
{
x
:

f
(
x
)

>
k
f
k
1
}
= 0.
Also
R
E

f

k
f
k
1
m
(
E
), since
Z
E

f

=
Z
E
\
{
x
:

f
(
x
)

k
f
k
1
}

f

+
Z
E
\
{
x
:

f
(
x
)

>
k
f
k
1
}

f

k
f
k
1
m
(
E
\
{
x
:

f
(
x
)

k
f
k
1
}
) + 0
k
f
k
1
m
(
E
)
Definition.
Let 1
p
1
. Take
q
2
[1
,
1
] such that
1
p
+
1
q
= 1 where
1
1
= 0.
q
is called the
conjugate index
of
p
.
Example.
Examples of conjugate pairs: (1
,
1
)
,
(
1
,
1)
,
(2
,
2)
,
(4
,
4
3
).
Theorem
(Holder’s Inequality)
.
Let 1
p, q
1
be conjugate indices, so
1
p
+
1
q
= 1.
Then for all measurable
functions
f, g
,
Z

fg

k
f
k
p
k
g
k
q
.
In particular, if
f
2
L
p
and
g
2
L
q
, then
fg
2
L
1
and
k
fg
k
1
k
f
k
p
k
g
k
q
.
Note.
This generalizes CauchySchwarz for
L
2
:

h
f, g
i

h
f, f
i
1
/
2
h
g, g
i
1
/
2
=
k
f
kk
g
k
.
For
L
2
,
h
f, g
i
=
R
f
g
, so
h
f, f
i
=
R

f

2
=
k
f
k
2
L
2
, so CS is equivalently:
Z
f
g
k
f
k
L
2
k
g
k
L
2
.
Proof of Holder’s Inequality.
We first consider the
p
= 1
, q
=
1
case. If RHS =
1
or
k
f
k
1
= 0 or
k
g
k
1
= 0, we are
done. In particular, we can assume
k
g
k
1
<
1
. Let
A
=
{
x
:

g
(
x
)

>
k
g
k
1
}
, so
m
(
A
) = 0. Then
Z

fg

=
Z
A

fg

+
Z
A
C

fg

0 +
Z
A
C
k
g
k
1

f

=
k
g
k
1
Z
A
C

f

k
g
k
1
Z

f

=
k
g
k
1
k
f
k
1
.