L
p
SPACES FOR
0
< p <
1
KEITH CONRAD
1.
Introduction
In a ﬁrst course in functional analysis, a great deal of time is spent with Banach spaces,especially the interaction between such spaces and their dual spaces. Banach spaces are aspecial type of topological vector space, and there are important topological vector spacesthat do not lie in the Banach category, such as the Schwartz spaces.The most fundamental theorem about Banach spaces is the HahnBanach theorem, whichlinks the srcinal Banach space with its dual space. What we want to illustrate here is awide collection of topological vector spaces where the HahnBanach theorem has no obviousextension because the dual space is
zero
. The model for a topological vector space withzero dual space will be
L
p
[0
,
1] when 0
< p <
1. After proving the dual of this space is
{
0
}
,we’ll see how to make the proof work for other
L
p
spaces, with 0
< p <
1. The argumenteventually culminates in a pretty theorem from measure theory (Theorem 4.2) that can beunderstood at the level of a ﬁrst course on measures and integration.2.
Banach Spaces and Beyond
In this section, to provide some context, we recall some basic classes of vector spaces thatare important in analysis.Throughout, our vector spaces are real vector spaces. All that we say would go throughwith minimal change to complex vector spaces.
Deﬁnition 2.1.
A
norm
on a vector space
V
is a function
·
:
V
→
R
satisfying
ã 
v
 ≥
0, with equality if and only if
v
= 0,
ã 
v
+
w
 ≤ 
v

+

w

for all
v
and
w
in
V
.
ã 
cv

=

c

v

for all scalars
c
and
v
∈
V
.Given a norm on a vector space, we get a metric by
d
(
v,w
) =

v
−
w

.
Example 2.2.
On
R
n
, we have the supnorm

x

sup
= max
1
≤
i
≤
n

x
i

and the
L
2
norm

x

2
=
x
·
x
= (
ni
=1

x
i

2
)
1
/
2
. The metric coming from the
L
2
norm is the usual notion of distance on
R
n
. The metric on
R
n
coming from the supnorm has balls that are actuallycubes. These two norms give rise to the same topology on
R
n
. Actually, all norms on
R
n
yield the same topology.
Example 2.3.
The space
C
[0
,
1] of continuous realvalued functions on [0
,
1] has the supnorm

f

sup
= sup
x
∈
[0
,
1]

f
(
x
)

and the
L
2
norm

f

2
= (
10

f
(
x
)

2
d
x
)
1
/
2
. While functionsthat are close in the supnorm are close in the
L
2
norm, the converse is false: a functionwhose graph is close to the
x
axis except for a tall thin spike is near 0 in the
L
2
norm butnot in the supnorm.
Deﬁnition 2.4.
A
Banach space
is a vector space
V
equipped with a norm
·
such that,with respect to the metric deﬁned by
d
(
v,w
) =

v
−
w

,
V
is complete.
1
2 KEITH CONRAD
Example 2.5.
Under either norm in Example 2.2,
R
n
is a Banach space.
Example 2.6.
In the supnorm,
C
[0
,
1] is a Banach space (convergence in the supnorm isexactly the concept of uniform convergence). But in the
L
2
norm,
C
[0
,
1] is not a Banachspace. That is,
C
[0
,
1] is not complete for the
L
2
norm.
Deﬁnition 2.7.
For a Banach space
V
, its
dual space
is the space of continuous linearfunctionals
V
→
R
, and is denoted
V
∗
.Continuity is important. We do not care about arbitrary linear functionals (as in linearalgebra), but only those that are continuous. One of the important features of a Banachspace is that we can use continuous linear functionals to separate points.
Theorem 2.8.
Let
V
be a Banach space. For each nonzero
v
∈
V
, there is a
ϕ
∈
V
∗
such that
ϕ
(
v
)
= 0
. Thus, given distinct
v
and
w
in
V
, there is a
ϕ
∈
V
∗
such that
ϕ
(
v
)
=
ϕ
(
w
)
.
Theorem 2.8 is a special case of the HahnBanach theorem, and can be found in any texton functional analysis. Even this special case can’t be proved in a constructive way (when
V
is inﬁnitedimensional). Its general proof depends on the axiom of choice.
Example 2.9.
On
C
[0
,
1], the evaluation maps
e
a
:
f
→
f
(
a
), for
a
∈
R
, are linear functionals. Since

f
(
a
)
 ≤ 
f

sup
, each
e
a
is continuous for the supnorm. If
f
= 0 in
C
[0
,
1],there is some
a
such that
f
(
a
)
= 0, and then
e
a
(
f
) =
f
(
a
)
= 0
.
Using the supnorm topology on
C
[0
,
1], the dual space
C
[0
,
1]
∗
is the space of boundedBorel measures (or RiemannStieltjes integrals) on [0
,
1], with the evaluation maps
e
a
corresponding to point masses.
Deﬁnition 2.10.
A
topological vector space
is a (real) vector space
V
equipped with aHausdorﬀ topology in which addition
V
×
V
→
V
and scalar multiplication
R
×
V
→
V
are continuous.Note the Hausdorﬀ condition is included in the deﬁnition. We won’t be meeting anynonHausdorﬀ spaces.
Example 2.11.
The usual topology on
R
n
makes it a topological vector space. So doesthe discrete topology on
R
n
. The usual topology on
R
n
is the only nondiscrete Hausdorﬀ topology that makes it a topological vector space. (This is not trivial. Try the case of dimension 1.)
Example 2.12.
Any Banach space is a topological vector space.
Deﬁnition 2.13.
A subset of a vector space is called
convex
if, for any
v
and
w
in thesubset, the line segment
tv
+ (1
−
t
)
w
, for 0
≤
t
≤
1, is in the subset.More generally, if a subset is convex and
v
1
,...,v
m
are in the subset, then any weightedsum
mi
=1
c
i
v
i
with
c
i
≥
0 and
mi
=1
c
i
= 1 is in the subset. In particular, the subsetcontains the average (1
/m
)
mi
=1
v
i
.
Deﬁnition 2.14.
A topological vector space is called
locally convex
if the convex open setsare a base for the topology: given any open set
U
around a point, there is a convex openset
C
containing that point such that
C
⊂
U
.
Example 2.15.
Any Banach space is locally convex, since all open balls are convex. Thisfollows from the deﬁnition of a norm.
L
p
SPACES FOR 0
< p <
1 3
Since topological vector spaces are homogeneous (we can use addition to translate neighborhoods around one point to neighborhoods around other points), the locally convex condition can be checked by focusing at the srcin: the open sets around 0 need to contain abasis of convex open sets.
Example 2.16.
The space
C
(
R
) of continuous realvalued functions on
R
does not havethe supnorm over all of
R
as a norm: a continuous function on
R
could be unbounded.But
C
(
R
) can be made into a locally convex topological vector space as follows. For eachpositive integer
n
, deﬁne a “seminorm”
·
n
by

f

n
= sup

x
≤
n

f
(
x
)

.
This is just like a norm, except it might assign value 0 to a nonzero function. That is,a function could vanish on [
−
n,n
] without vanishing everywhere. Of course, if we take
n
large enough, a nonzero continuous function will have nonzero
n
th seminorm, so thetotal collection of seminorms
·
n
as
n
varies, rather than one particular seminorm, letsus distinguish diﬀerent functions from each other. Using these seminorms, deﬁne a basicopen set around
g
∈
C
(
R
) to be all functions close to
g
in a ﬁnite number of seminorms:
{
f
∈
C
(
R
) :

f
−
g

n
1
≤
ε,...,

f
−
g

n
r
≤
ε
}
for a choice of ﬁnitely many seminorms and an
ε
. These sets are a basis for a topologythat makes
C
(
R
) a locally convex topological vector space.
Deﬁnition 2.17.
When
V
is a topological vector space, its
dual space
V
∗
is the space of continuous linear functionals
V
→
R
.Theorem 2.8 generalizes to all locally convex spaces, as follows.
Theorem 2.18.
Let
V
be a locally convex topological vector space. For any distinct
v
and
w
in
V
, there is a
ϕ
∈
V
∗
such that
ϕ
(
v
)
=
ϕ
(
w
)
.Proof.
See the chapters on locally convex spaces in [3] or [9].
Let’s meet some topological vector spaces that are
not
locally convex.
Example 2.19.
Let
L
1
/
2
[0
,
1] be the measurable
f
: [0
,
1]
→
R
such that
10

f
(
x
)

1
/
2
d
x
is ﬁnite, with functions equal almost everywhere identiﬁed. (We need to make such anidentiﬁcation, since integration does not distinguish between functions that are changed ona set of measure 0.)The function
d
(
f,g
) =
10

f
(
x
)
−
g
(
x
)

1
/
2
d
x
is a metric on
L
1
/
2
[0
,
1]. The topology
L
1
/
2
[0
,
1] obtains from this metric is
not
locally convex. To see why, consider any open ballaround 0:
f
∈
L
1
/
2
[0
,
1] :
10

f
(
x
)

1
/
2
d
x < R
.
We will show, for any
ε >
0, that the
ε
ball around 0 contains functions whose average liesoutside the ball of radius
R
. That violates the meaning of local convexity.Pick
ε >
0 and
n
≥
1. Select
n
disjoint intervals in [0
,
1] (they need not cover all of [0
,
1]).Call them
A
1
,A
2
,...,A
n
. Set
f
k
= (
ε/µ
(
A
k
))
2
χ
A
k
, where
µ
is Lebesgue measure (so
µ
(
A
k
)is simply the length of
A
k
). Then
10

f
k
(
x
)

1
/
2
d
x
=
ε
, so every
f
k
is at distance
ε
from
4 KEITH CONRAD
0. However, since the
f
k
’s are supported on disjoint sets, their average
g
n
= (1
/n
)
nk
=1
f
k
satisﬁes
10

g
n
(
x
)

1
/
2
d
x
= 1
n
1
/
2
n
k
=1
10

f
k
(
x
)

1
/
2
d
x
=
n
1
/
2
ε.
Taking
n
large enough (depending on
ε
), we see the distance between
g
n
and 0 in
L
1
/
2
[0
,
1]can be made as large as desired.
Example 2.20.
For 0
< p <
1, let
L
p
[0
,
1] be the set of measurable functions
f
: [0
,
1]
→
R
such that
10

f
(
x
)

p
d
x <
∞
, with functions equal almost everywhere identiﬁed. Thefunction
d
(
f,g
) =
10

f
(
x
)
−
g
(
x
)

p
d
x
is a metric on
L
p
[0
,
1]. With the inherited metrictopology,
L
p
[0
,
1] is not locally convex, by exactly the same argument as in the previousexample (where
p
= 1
/
2). Indeed, in the previous construction, replace the exponent 2in the deﬁnition of
f
k
with 1
/p
, and at the end you’ll get
10

g
n
(
x
)

p
d
x
=
n
1
−
p
ε
. Since1
−
p >
0, the distance between
g
n
and 0 can be made arbitrarily large with suitable choiceof
n
. In fact, what this means is that the
only
convex open set in
L
p
[0
,
1] is the whole space.This method of constructing topological vector spaces that are not locally convex is dueto Tychonoﬀ [10, pp. 768–769], whose actual example was the sequence space
1
/
2
=
{
(
x
i
) :
i
≥
1
√
x
i
<
∞}
rather than the function space
L
1
/
2
[0
,
1].One can’t push a result like Theorem 2.18 to all topological vector spaces, as the nextresult [4] vividly illustrates.
Theorem 2.21.
For
0
< p <
1
,
L
p
[0
,
1]
∗
=
{
0
}
. That is, the only continuous linear map
L
p
[0
,
1]
→
R
is
0
.Proof.
We argue by contradiction. Assume there is
ϕ
∈
L
p
[0
,
1]
∗
with
ϕ
= 0. Then
ϕ
hasimage
R
(a nonzero linear map to a onedimensional space is surjective), so there is some
f
∈
L
p
[0
,
1] such that

ϕ
(
f
)
 ≥
1.Using this choice of
f
, map [0
,
1] to
R
by
s
→
s
0

f
(
x
)

p
d
x.
This is continuous, so there is some
s
between 0 and 1 such that(2.1)
s
0

f
(
x
)

p
d
x
= 12
10

f
(
x
)

p
d
x >
0
.
Let
g
1
=
fχ
[0
,s
]
and
g
2
=
fχ
(
s,
1]
, so
f
=
g
1
+
g
2
and

f

p
=

g
1

p
+

g
2

p
. So
10

g
1
(
x
)

p
d
x
=
s
0

f
(
x
)

p
d
x
= 12
10

f
(
x
)

p
d
x,
hence
10

g
2
(
x
)

p
d
x
=
12
10

f
(
x
)

p
d
x
. Since

ϕ
(
f
)
 ≥
1,

ϕ
(
g
i
)
 ≥
1
/
2 for some
i
. Let
f
1
= 2
g
i
, so

ϕ
(
f
1
)
 ≥
1 and
10

f
1
(
x
)

p
d
x
= 2
p
10

g
i
(
x
)

p
d
x
= 2
p
−
1
10

f
(
x
)

p
d
x
. Note2
p
−
1
<
1.Iterate this to get a sequence
{
f
n
}
in
L
p
[0
,
1] such that

ϕ
(
f
n
)
 ≥
1 and
d
(
f
n
,
0) =
10

f
n
(
x
)

p
d
x
= (2
p
−
1
)
n
10

f
(
x
)

p
d
x
→
0
,
a contradiction of continuity of
ϕ
.