2
INTRODUCTION TO LEBESGUE INTEGRATION
21
so
f
2
L
p
1
[0
,
1]. To see that
L
p
1
[0
,
1]
6
=
L
p
2
[0
,
1], let
a
= 1
/p
2
, so
p
2
a
= 1 and since
p
1
< p
2
,
p
1
a <
1. Then note that
n
p
2
a
n
2
=
1
n
and
n
p
1
a
n
2
=
1
n
1+
δ
for some
δ
>
0. Now let
f
=
P
1
n
=2
n
a
χ
(
1
n
,
1
n
+
1
n
2
)
. Then
Z
1
0

f

p
1
=
1
X
n
=2
n
p
1
a
n
2
=
1
X
n
=2
1
n
1+
δ
<
1
Z
1
0

f

p
2
=
1
X
n
=2
n
p
2
a
n
2
=
1
X
n
=2
1
n
=
1
so
f
2
L
p
2
[0
,
1] but
f /
2
L
p
1
[0
,
1].
Now consider the 1
p
1
< p
2
=
1
case. Suppose
f
2
L
1
[0
,
1], so
k
f
k
1
<
1
. Let
A
=
{
x
2
[0
,
1] :

f
(
x
)

>
k
f
k
1
}
.
We proved in class that
m
(
A
) = 0. Notice that on [0
,
1]
\
A
,
f
k
f
k
1
. Thus
Z
1
0

f

p
1
=
Z
[0
,
1]
\
A

f

p
1
+
Z
A

f

p
1
Z
[0
,
1]
\
A
k
f
k
p
1
1
+ 0
k
f
k
p
1
1
<
1
so
f
2
L
p
1
[0
,
1]. Finally,
L
1
[0
,
1]
6
=
L
p
1
[0
,
1] since if
f
(
x
) =
1
x
2
, then
f
2
L
p
1
[0
,
1] but
f /
2
L
1
[0
,
1].
Minkowski’s Inequality
(Triangle Inequality)
.
If
f, g
are measurable, then
k
f
+
g
k
p
k
f
k
p
+
k
g
k
p
for 1
p
1
.
Remark.
If
f
=
g
a.e. (say
f
=
g
except on
E
with
m
(
E
) = 0), then
k
f
k
1
=
k
f
χ
E
+
f
χ
E
C
k
1
k
f
χ
E
k
1
+
k
f
χ
E
C
k
1
= 0+
k
f
χ
E
C
k
1
=
k
g
χ
E
C
k
1
=
k
g

g
χ
E
k
1
k
g
k
1
+
k
g
χ
E
}
1
=
k
g
k
1
and similarly
k
g
k
1
k
f
k
1
, so we have
k
f
k
1
=
k
g
k
1
.
Proof of Minkowski’s Inequality.
p
=
1
case left as exercise. For
p
= 1,
k
f
+
g
k
1
=
Z

f
+
g

Z
(

f

+

g

) =
Z

f

+
Z

g

=
k
f
k
1
+
k
g
k
1
.
So suppose 1
< p <
1
. Assume
f, g
2
L
p
(since if not, RHS =
1
, so we are done). Take
q
such that
1
p
+
1
q
= 1
)
1 +
p
q
=
p
)
q
+
p
=
pq
)
p
=
q
(
p

1). Now by Holder’s inequality,
Z

f
+
g

p

1

f

k

f
+
g

p

1
k
q
k
f
k
p
.
Notice that
k

f
+
g

p

1
k
q
q
=
Z

f
+
g

(
p

1)
q
=
Z

f
+
g

p
=
k
f
+
g
k
p
p
) k

f
+
g

p

1
k
q
=
k
f
+
g
k
p/q
p
Thus
Z

f
+
g

p

1

f

k

f
+
g

p

1
k
q
k
f
k
p
=
k
f
+
g
k
p/q
p
k
f
k
p
.
Similarly,
Z

f
+
g

p

1

g

k
f
+
g
k
p/q
p
k
g
k
p
.
Now
k
f
+
g
k
p
p
=
Z

f
+
g

p
=
Z

f
+
g

p

1
(

f
+
g

)
Z

f
+
g

p

1
(

f

+

g

) =
Z

f
+
g

p

1

f

+
Z

f
+
g

p

1

g

k
f
+
g
k
p/q
p
(
k
f
k
p
+
k
g
k
p
)