3.
In general case one uses Zorn’s lemma.
Let
F
be the family of linear
extensions,
F
, of
f
to a subspace
M
F
with
F
(
x
)
≤
p
(
x
). It is partially ordered by
inclusion:
F
1
≤
F
2
if
{
(
x, F
1
(
x
)) :
x
∈
M
F
1
} ⊂ {
(
x, F
2
(
x
) :
x
∈
M
F
2
}
. Since the
union of increasing family of subspaces of
X
is a subspace, one has that the union
of totally ordered subfamilies of
F
lies in
F
.
Hence by Zorn’s lemma
F
has a
maximal element. It is defined on the whole space since otherwise it would admit
an extension and, hence, would not be the maximal element of
F
.
Remark 12.4.
1. Step 2 is presented here with the main reason to demonstrate
that, for separable spaces, the theorem can be proved without using the Zorn
lemma.
2. If
p
is a seminorm/norm then
f
(
x
)
≤
p
(
x
)
⇔ |
f
(
x
)
| ≤
p
(
x
)
.
In fact, in this case,
|
f
(
x
)
|
=
±
f
(
x
) =
f
(
±
x
) and
f
(
x
)
≤
p
(
x
)
⇐ −
f
(
x
)
≤
p
(
−
x
) =
p
(
x
) giving the statement.
68
Theorem 12.5.
(
The complex Hahn-Banach Theorem
) Let
X
be a complex
vector space,
p
a seminorm on
X
,
M
a subspace of
X
, and
f
a complex linear
functional on
M
such that
|
f
(
x
)
| ≤
p
(
x
)
for
x
∈
M
. Then there exists a complex
linear functional
F
on
X
such that
|
F
(
x
)
| ≤
p
(
x
)
for all
x
∈
X
and
F
(
x
) =
f
(
x
)
for all
x
∈
M
.
Proof.
Consider
X
and
M
as vector spaces over reals and denote them by
X
R
and
M
R
respectively. Note that
X
R
(
M
R
) and
X
(
M
respectively) are different as
linear spaces but coincides as sets. Clearly
p
is a sublinear functional on
X
R
.
Let
u
= Re
f
. Then
u
is a real linear functional on
M
R
satisfying the condition
|
u
(
x
)
| ≤
p
(
x
)
and hence
u
(
x
)
≤
p
(
x
).
By the real Hahn-Banach theorem there exists a real linear functional
U
on
X
R
such that
U
(
x
)
≤
p
(
x
), for all
x
∈
X
R
and
U
(
x
) =
u
(
x
) for all
x
∈
M
R
.
Clearly
−
U
(
x
) =
U
(
−
x
)
≤
p
(
−
x
) =
p
(
x
) and hence
|
U
(
x
)
| ≤
p
(
x
)
, x
∈
X
R
.
Define a functional
F
on
X
by letting
F
(
x
) =
U
(
x
)
−
iU
(
ix
)
.
F
is a complex linear functional:
F
is real linear since so is
U
and
F
(
ix
) =
U
(
ix
)
−
iU
(
−
x
) =
U
(
ix
) +
iU
(
x
) =
i
(
U
(
x
)
−
iU
(
ix
)) =
iF
(
x
)
, x
∈
X.
F
is an extension of
f
to
X
, i.e.
F
(
x
) =
f
(
x
),
x
∈
M
: as
Im
f
(
x
) =
−
Re(
if
(
x
)) =
−
Re(
f
(
ix
)) =
−
u
(
ix
)
we have
F
(
x
) =
u
(
x
)
−
iu
(
ix
) = Re
f
(
x
) +
i
Im
f
(
x
) =
f
(
x
)
.
It remains to prove that
|
F
(
x
)
| ≤
p
(
x
) for all
x
∈
X
. Assume contrary that for
some
x
0
∈
X
,
|
F
(
x
0
)
|
> p
(
x
0
). Write
F
(
x
0
) =
|
F
(
x
0
)
|
e
iφ
. Let
y
0
=
e
−
iφ
x
0
. Then
U
(
y
0
) = Re
F
(
y
0
) = Re[
e
−
iφ
F
(
x
0
)] =
|
F
(
x
0
)
|
> p
(
x
0
) =
p
(
y
0
)
.
A contradiction that gives the statement.
Corollary 12.6.
Let
X
be a normed space and let
M
be its subspace. Then for
every linear continuous functional
f
defined on
M
there exists a functional
F
∈
X
∗
such that
F
(
x
) =
f
(
x
)
for all
x
∈
M
and
∥
F
∥
=
∥
f
∥
.
Proof.
Let
p
(
x
) =
∥
f
∥∥
x
∥
,
x
∈
X
. We have
|
f
(
x
)
| ≤
p
(
x
) for any
x
∈
M
. By the
complex Hahn-Banach theorem there exists a linear functional
F
on
X
such that
|
F
(
x
)
| ≤
p
(
x
) =
∥
f
∥∥
x
∥
for all
x
∈
X
and
F
|
M
=
f
. Hence,
F
is bounded and
∥
F
∥ ≤ ∥
f
∥
. As
∥
F
∥
= sup
{|
F
(
x
)
|
:
∥
x
∥ ≤
1
, x
∈
X
} ≥
sup
{|
F
(
x
)
|
:
∥
x
∥ ≤
1
, x
∈
M
}
=
∥
f
∥
, we have
∥
F
∥
=
∥
f
∥
.
69
12.1
Corollaries of the Hahn-Banach Theorem
Theorem 12.7.
Let
X
be a normed space.
