Linear Functionals

Linear Functional

Suppose that

V

is a vector space over a field

K \in {ℝ, ℂ}

, then a linear functional is a linear operator

f : V \to K

Note that in the case of the reals or complex we have the norm as the absolute value. Also we say that a linear functional is real if $F = ℝ$ and say that it is complex if $F = ℂ$

Bounded Linear Functional

We say that a linear functional

f

is bounded when it is a bounded linear operator

The above equates to saying that there exists some $c \in ℝ^{+}$ such that for all $x \in dom (f), | f (x) | \leq c ‖ x ‖$

Dual Space

Suppose that

X

is a normed vector space over a field

F

then we define the the dual space to be the set of all bounded linear functionals, denoted by

X^{*}

The Dual Space Is a Normed Vector Space

Given a vector space

X

then

X^{*}

is a normed vector space using the operator norm.

TODO: Add the proof here.

The Dual Space Is a Banach Space

Given a vector space

X

then

X^{*}

is a banach space

TODO: Add the proof here.

What's interesting is that the above holds whether or not $X$ is a banach space.

The Dual Space of Rn Is Rn

The dual space of

{(ℝ)}^{n}

is isomorphic to

ℝ^{n}

TODO: Add the proof here.

The Dual of Lp Is Lq

The dual space of

l^{p}

l^{q}

where

q \in (1, \infty)

and

\frac{1}{p} + \frac{1}{q} = 1

A Schauder basis for

l^{p}

(e_{k})

, where

e_{k} = (δ_{k i})

as in the preceding example. Then every

x \in l^{p}

has a unique representation (9)

x = \sum_{k = 1}^{\infty} ξ_{k} e_{k}

Given any

f

in the dual space of

l^{p}

, since it is

f

is linear and bounded we have:

f (x) = \sum_{k = 1}^{\infty} ξ_{k} γ_{k} γ_{k} = f (e_{k})

Let

q

be the conjugate of

p

(cf. 1.2-3) and consider

x_{n} = (ξ_{k}^{(n)})

with

ξ_{k}^{(n)} = {\begin{matrix} {| γ_{k} |}^{q} / γ_{k} & if k ≦ n and γ_{k} \neq 0 \\ 0 & if k > n or γ_{k} = 0 \end{matrix}

Thus we have:

f (x_{n}) = \sum_{k = 1}^{\infty} ξ_{k}^{(n)} γ_{k} = \sum_{k = 1}^{n} {| γ_{k} |}^{4}

We also have, using (11) and

(q - 1) p = q

\begin{matrix} f (x_{n}) ≦ ‖ f ‖ ‖ x_{n} ‖ & = ‖ f ‖ {(\sum {| ξ_{k}^{(n)} |}^{p})}^{1 / p} \\ = ‖ f ‖ {(\sum {| γ_{k} |}^{(q - 1) p})}^{1 / p} \\ = ‖ f ‖ {(\sum {| γ_{k} |}^{q})}^{1 / p} \end{matrix}

(sum from 1 to

n

). Together,

f (x_{n}) = \sum {| γ_{k} |}^{q} \leq ‖ f ‖ {(\sum {| γ_{k} |}^{q})}^{1 / p}

Dividing by the last factor and using

1 - 1 / p = 1 / q

, we get

{(\sum_{k = 1}^{n} {| γ_{k} |}^{q})}^{1 - 1 / p} = {(\sum_{k = 1}^{n} {| γ_{k} |}^{q})}^{1 / q} \leq ‖ f ‖ .

The Dual Space of l1 Is l Infinity

As per title.

We first recall that a Schauder basis for

l^{1}

is given by

e_{k} = (δ_{k j})

, that is

e_{k}

has 1 in the kth place and zeros everywhere else, so that every

x \in l^{1}

has a unique representation of the form

x = \sum_{k = 1}^{\infty} c_{k} e_{k} .

We consider any

f

in the dual space of

l^{1}

. Since

f

is linear and bounded then we know

f (x) = \sum_{k = 1}^{\infty} c_{k} f (e_{k})

By construction we note that

‖ e_{k} | = 1

and therefore we have:

| f (e_{k}) | = \leq {‖ f ‖}_{op} ‖ e_{k} ‖ = {‖ f ‖}_{op}

Therefore we can conclude that

\sup_{k} | f (e_{k}) | \leq {‖ f ‖}_{op}

so then

{(f (e_{k}))}_{k} \in l^{\infty}

. But for every

b = (β_{k}) \in l^{\infty}

then

g

defined by

g (x) = \sum_{k = 1}^{\infty} c_{k} β_{k}

where

(c_{n}) \in l^{1}

is linear, additionally it is bounded:

| g (x) | \leq \sum_{k = 1}^{\infty} | c_{k} β_{k} | \leq \sup_{i} | β_{i} | \sum_{k = 1}^{\infty} | c_{k} | = ‖ x ‖ \sup_{i} | β_{i} |

Therefore we can conclude that

g

is in the dual of

l^{1}

. We will now show that the norm of

f

is the norm on the space

l^{\infty}

| f (x) | = | \sum_{k = 1}^{\infty} c_{k} f (e_{k}) | \leq \sup_{j} | f (e_{k}) | \sum_{k = 1}^{\infty} | c_{k} | = ‖ x ‖ \sup_{j} | f (e_{j}) |

Therefore when we take this supremum over all

x

on the unit circle we attain that:

{‖ f ‖}_{op} \leq \sup_{j} | f (e_{k}) |

But earlier in the proof we had the other direction of the above inequality so we have

{‖ f ‖}_{op} = \sup_{j} | f (e_{j}) |

which is the

l^{\infty}

norm, that is we have

{‖ f ‖}_{op} = {‖ y ‖}_{\infty}

where

y = (f (e_{k})) \in l^{\infty}

, showing that the mapping

f \mapsto (y)

is a bijective linear mapping of

l^{1}

onto

l^{\infty}

therefore it is an isomorphism, as needed.

Hahn Banach

Let

X

be a real vector space,

p

a sublinear functional on

𝒳, ℳ

a subspace of

𝒳

, and

f

a linear functional on

ℳ

such that

f (x) \leq p (x)

for all

x \in ℳ

. Then there exists a linear functional

F

𝒳

such that

F (x) \leq p (x)

for all

x \in 𝒳

and

F | ℳ = f

We begin by showing that if

x \in X \ 2 M, f

can be extended to a linear functional

g

ℳ + ℝ x

satisfying

g (y) \leq p (y)

there. If

y_{1}, y_{2} \in ℳ

, we have

f (y_{1}) + f (y_{2}) = f (y_{1} + y_{2}) \leq p (y_{1} + y_{2}) \leq p (y_{1} - x) + p (x + y_{2}),

f (y_{1}) - p (y_{1} - x) \leq p (x + y_{2}) - f (y_{2}) .

Hence

\sup {f (y) - p (y - x) : y \in ℳ} \leq \inf {p (x + y) - f (y) : y \in ℳ} .

Let

α

be any number satisfying

\sup {f (y) - p (y - x) : y \in 𝒩} \leq α \leq \inf {p (x + y) - f (y) : y \in ℳ}

and define

g : ℳ + ℝ x \to ℝ

g (y + λ x) = f (y) + λ α

. Then

g

is clearly linear, and

g | ℳ = f

, so that

g (y) \leq p (y)

for

y \in ℳ

. Moreover, if

λ > 0

and

y \in ℳ

g (y + λ x) = λ [f (y / λ) + α] \leq λ [f (y / λ) + p (x + (y / λ)) - f (y / λ)] = p (y + λ x),

whereas if

λ = - μ < 0

g (y + λ w) - μ [f (y / μ) - a] \leq μ [f (y / μ) - f (y / μ) + p ((y / μ) - x)] = p (y + λ x)

Thus

g (z) \leq p (z)

for all

z \in ℳ + ℝ x

. Evidently the same reasoning can be applied to any linear extension

F

f

satisfying

F \leq p

on its domain, and it shows that the domain of a maximal linear extension satisfying

F \leq p

must be the whole space

X

. But the family

ℱ

of all tinear extensions

F

f

satisfying

F \leq p

is partially ordered by inclusion (maps from subspaces of

X

ℝ

being regarded as subsets of

X \times ℝ

). Since the union of any increasing family of subspaces of

X

is agath a subspace, one casily sees that the union of a linearly ordered subfamily of

ℱ

lies in

ℱ

. The proof is therefore completed by invoking Zom's lemma.

5.6 The Hahn-Banach Theorem.

5.7 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) | \leq p (x)

for

x \in ℳ

. Then there exists a complex linear functional

F

X

such that

| F (x) | \leq p (x)

for all

x \in 𝒳

and

F | ℳ = f

. Proof. Let

u = Re f

. By Theorem 5.6 there is a real linear extension

U

u

X

such that

| U (x) | \leq p (x)

for all

x \in X

. Let

F (x) = U (x) - i U (i x)

as in Proposition 5.5. Then

F

is a complex linear extension of

f

, and as in the proof of Proposition 5.5, if

α = sgn F (x)

we have

| F (x) | = α F (x) = F (α x) = U (α x) \leq p (α x) = p (x)

5.8 Theorem. Let

x

be a normed vector space. a. If

ℳ

is a closed subspace of

\dot{X}

and

x \in \dot{X} \ ℳ

, there exists

f \in X

such that

f (x) \neq 0

and

f | ℳ = 0

. In fact, if

δ = \inf_{y \in ℳ} ‖ x - y ‖ . f

can be taken to satisfy

‖ f ‖ = 1

and

f (x) = δ

. b. If

x \neq 0 \in X

, there exists

f \in X^{*}

such that

‖ f ‖ = 1

and

f (x) = ‖ x ‖

. c. The bounded linear functionals on

X

separate points. d. If

x \in X

, define

\hat{x} : X^{+} \to ℂ

\hat{x} (f) = f (x)

. Then the map

x \mapsto \hat{x}

is a linear isometry from

X

into

X^{* *}

(the dual of

X^{*}

). Proof. To prove (a), define

f

ℳ + ℂ x

f (y + λ x) = λ δ (y \in ℳ, λ \in ℂ)

. Then

f (x) = δ, f | ℳ = 0

, and for

λ \neq 0, | f (y + λ x) | = | λ | δ \leq | λ | ‖ λ^{- 1} y + x | | =

‖ y + λ x ‖

. Thus the Hahn-Banach theorem can be applied, with

p (x) = ‖ x ‖

and

ℳ

replaced by

M + ℂ x

. (b) is the spectal case of (a) with

M = {0}

, and (c) follows immediately: if

x \neq y

, there exists

f \in X^{+}

with

f (x - y) \neq 0

, i.e.,

f (x) \neq f (y)

. As for (d), obviously

\hat{x}

is a linear functional on

X^{*}

and the map

x \mapsto \hat{x}

is linear. Moroever,

| \hat{x} (f) | = | f (x) | \leq ‖ f ‖ ‖ x ‖

, so

‖ \hat{x} ‖ \leq ‖ x ‖

. On the other hand, (b) implies that

‖ \hat{2} ‖ \geq ‖ x ‖

. With notation as in Theorem 5.8d, let

\hat{X} = {\hat{x} : x \in X}

. Since

x^{*}

is always complete, the closure

\hat{X}

\hat{X}

X^{* *}

is a Banach space, and the map

x \mapsto \hat{x}

embeds

x

into

\overline{X}

as a dense subspace.

\overline{x}

is called the completion of

X

. In particular, if

X

is itself a Banach space then

\overline{x} = \hat{x}

. If

x

is finite-dimensional, then of course

\hat{x} = x^{* *}

, since these spaces have the same dimension. For infinite-dimensional Banach spaces it may or may not happen that

\hat{x} = x \cdot *

; if it does,

x

is called reflexive. The examples of Banach spaces we have examined so far are not reflexive except in trivial cases where they turn out to be finite-dimensional. We shall prove some cases of this assertion and present examples of reflexive Banach spaces in later sections. Usually we shall identify

\hat{x}

with

x

and thus regard

x^{* *}

as a superspace of

x

; reflexivity then means that

x^{* *} = x

We already know (Background Information): (Folland)Theorem 5.8 - Let

𝒳

be a normed vector space. a.) If

M

is a closed subspace of

𝒳

and

x \in 𝒳 \ M

, there exists

f \in 𝒳^{*}

such that

f (x) \neq 0

and

f (M) = {0}

. In fact, if

δ = \inf_{y \in M} ‖ x - y ‖, f

can be taken to satisfy

‖ f ‖ = 1

and

f (x) = δ

. Question: Let

X

be a normed vector space. a.) If

M

is a closed subspace and

x \in X \ M

then

M + ℂ x

is closed. (Use Theorem 5.8a.)). b.) Every finite-dimensional subspace of

X

is closed. Proof a.) - Let

M

be a proper closed subspace of

X

and let

x \in X \ M

. There exists

f \in X^{*}

such that

f (x) \neq 0

and

f (M) = {0}

. Let

{u_{n} + a_{n} x}_{1}^{\infty}

be a sequence in

M + K x

that converges to

y \in X

. Then

f (y) = f (\lim_{n \to \infty} (u_{n} + a_{n} x)) = \lim_{n \to \infty} f (u_{n} + a_{n} x) = \lim_{n \to \infty} a_{n} f (x)

since

f

is continuous, so

{a_{n}}_{1}^{\infty}

converges to

a : = f (y) / f (x)

, which implies that

{a_{n} x}_{1}^{\infty}

converges to

a x

. Therefore

{u_{n}}_{1}^{\infty} = {(u_{n} + a_{n} x) - a_{n} x}_{1}^{\infty} \to (y - a x)

which lies in

M

because

M

is closed. It follows that

y \in M + K x

, which shows that

M + K x

is closed. Proof b.) - Let

Y

be a finite dimensional subspace of

X

. Let

{e_{1}, \dots, e_{n}}

be a basis of

Y

. Now, note that

M = {0}

is a closed subspace of

X

. So, by item a.),

M + C e_{1}

is a closed subspace of

X

. Again, by item a.) we have that

M + ℂ e_{1} + ℂ 𝕖 e_{2}

is a closed subspace of

X

. Repeating the argument enough times, we have that

M + ℂ 𝕖 e_{1} + \dots + ℂ e_{n}

is a closed subspace of

X

. But

M + ℂ 𝕖 e_{1} + \dots + ℂ e_{n} = Y

. So we have proved that

Y

is a closed subspace of

X