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Homework 6: solutions
1.
Problem 2, page 119.
We have

f
(
x
)
 ≤ k
f
k
∞
except possibly on a set of measure zero. Then
k
f
k
p
p
=
Z
1
0

f

p
≤
Z
1
0
k
f
k
p
∞
=
k
f
k
p
∞
,
thus
k
f
k
p
≤ k
f
k
∞
. On the other hand,
k
f
k
∞
= inf
{
M
:
m
{
x
∈
[0
,
1] :

f
(
x
)

> M
}
= 0
}
,
and therefore for any
² >
0 such that
k
f
k
∞

² >
0, the measure of the set
E
²
=
{
x
∈
[0
,
1] :

f
(
x
)

>
k
f
k
∞

²
}
is nonzero. Thus,
k
f
k
p
p
=
Z
1
0

f

p
>
Z
E
²
(
k
f
k
∞

²
)
p
= (
k
f
k
∞

²
)
p
m
(
E
²
)
.
Therefore,
k
f
k
∞

²
= (
k
f
k
∞

²
) lim
p
→∞
(
mE
²
)
1
/p
≤
liminf
p
→∞
k
f
k
p
≤
limsup
p
→∞
k
f
k
p
≤ k
f
k
∞
.
Since
²
is arbitrary, we obtain that
liminf
p
→∞
k
f
k
p
= limsup
p
→∞
k
f
k
p
=
k
f
k
∞
.
2.
Prove H¨older’s inequality for
p
= 1
, q
=
∞
. Show that equality holds if and
only if

g
(
x
)

=
k
g
k
∞
at almost every point
x
such that
f
(
x
)
6
= 0.
If
f
∈
L
1
[0
,
1]
, g
∈
L
∞
[0
,
1] then

g
(
x
)
 ≤ k
g
k
∞
except on a set of measure zero.
Thus,
Z
1
0

fg
 ≤
Z
1
0

f
 · k
g
k
∞
=
k
f
k
1
k
g
k
∞
.
Assume now that

g
(
x
)

=
k
g
k
∞
at almost every point
x
such that
f
(
x
)
6
= 0. Then
Z
1
0

fg

=
Z
{
x
:
f
(
x
)
6
=0
}

f
k
g
k
∞
=
k
f
k
1
k
g
k
∞
.
Conversely, assume that
Z
1
0

fg

=
k
f
k
1
k
g
k
∞
.
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This note was uploaded on 03/23/2010 for the course MATH 515 taught by Professor Staff during the Spring '08 term at Iowa State.
 Spring '08
 Staff

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