This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.4 Duality of Linear Spaces
Everything has another side
Orthonormal basis allows to reduce any question on Hilbert space to a
question on sequence of numbers. This is powerful but sometimes heavy
technique. Sometime we need a smaller and faster tool to study
questions which are represented by a single number, for example to
demonstrate that two vectors are different it is enough to show that
there is a unequal values of a single coordinate. In such cases linear
functionals are just what we needed.
–Is it functional?
–Yes, it works!
4.1 Dual space of a normed space
Definition 1
A linear functional on a vector space
V is a linear mapping α:
V→ ℂ
(or
α:
V→ ℝ
in the real case), i.e.
| α(ax+by)=aα(x)+bα(y), for all
x,y∈ V and a,b∈ℂ.
|
Exercise 2
Show that α(0) is necessarily 0.
We will not consider any functionals but linear, thus below
functional always means linear functional.
Example 3
-
Let V=ℂn and ck, k=1,…,n be
complex numbers. Then
α((x1,…,xn))=c1x1+⋯+c2x2 is a linear
functional.
- On C[0,1] a functional is given by
α(f)=∫01 f(t) d t.
- On a Hilbert space H for any x∈ H a functional
αx is given by αx(y)=⟨ y,x
⟩.
Theorem 4
Let V be a normed space and α
is a linear
functional. The following are equivalent:
-
α is continuous (at any point of V).
-
α is continuous at point 0.
-
sup{| α(x) |: ||x||≤ 1}< ∞,
i.e. α is a bounded linear
functional.
Proof.
Implication
1 ⇒
2 is trivial.
Show 2 ⇒
3. By the
definition of continuity: for any
є>0 there exists δ>0 such that
||v||<δ implies
| α(v)−α(0) |<є . Take є=1
then | α(δ x) |<1 for all x with norm
less than 1 because ||δ x||< δ. But from
linearity of α the inequality | α(δ x) |<1
implies | α(x) |<1/δ<∞ for all
||x||≤ 1.
3 ⇒
1. Let mentioned supremum be M. For
any x, y∈ V such that x≠ y vector
(x−y)/||x−y|| has norm 1. Thus | α
((x−y)/||x−y||) |<M. By the linearity of α this
implies that | α
(x)−α(y) |<M||x−y||. Thus α is continuous.
□
Definition 5
The dual space X* of a normed space
X is the set of continuous linear functionals on X. Define a
norm on it by
|
| ⎪⎪
⎪⎪ | α | ⎪⎪
⎪⎪ | = | | ⎪
⎪ | α(x) | ⎪
⎪ | .
(23) |
Exercise 6
-
Show that the chain of inequalities:
| | ⎪⎪
⎪⎪ | α | ⎪⎪
⎪⎪ | ≤ | | | ⎪
⎪ | α(x) | ⎪
⎪ |
≤ | |
| | ≤ | ⎪⎪
⎪⎪ | α | ⎪⎪
⎪⎪ | .
|
Deduce that any of the mentioned supremums deliver the norm of
α. Which of them you will prefer if you need to show
boundedness of α? Which of them is better to use if
boundedness of α is given?
- Show that | α(x) |≤
||α||·||x|| for all x∈ X, α ∈ X*.
The important observations is that linear functionals form a normed
space as follows:
Exercise 7
-
Show that X* is a linear space with natural (point-wise) operations.
- Show that (23) defines a norm on X*.
Furthermeore, X* is always complete, regardless of properties of X!
Theorem 8
X* is a Banach space with the defined norm (even if X was
incomplete).
Proof.
Due to Exercise
7 we only need to show that
X* is complete. Let (α
n) be a Cauchy sequence in
X*, then for any
x∈
X scalars α
n(
x) form a
Cauchy sequence, since
| α
m(
x)−α
n(
x) |≤||α
m−α
n||·||
x||. Thus
the sequence has a limit and we define α by
α(
x)=lim
n→∞α
n(
x). Clearly
α is a linear functional on
X. We should show that it
is bounded and α
n→ α. Given
є>0 there exists
N such that
||α
n−α
m||<є for all
n,
m≥
N. If
||
x||≤ 1 then | α
n(
x)−α
m(
x) |≤
є, let
m→∞ then | α
n(
x)−α(
x) |≤
є, so
| | ⎪
⎪ | α(x) | ⎪
⎪ | ≤ | ⎪
⎪ | αn(x) | ⎪
⎪ | +є≤
| ⎪⎪
⎪⎪ | αn | ⎪⎪
⎪⎪ | + є,
|
i.e. ||α|| is finite and ||α
n−α||≤
є, thus α
n→α.
□
Definition 9
The kernel of linear functional α, write kerα, is the set
all vectors x∈ X such that α(x)=0.
Exercise 10
Show that
-
kerα is a subspace of X.
- If α≢0 then obviously kerα ≠ X.
Furthermore, if X has at least two linearly independent vectors
then kerα ≠ {0}, thus kerα is a
proper subspace of X.
- If α is continuous then kerα is closed.
Study one and get any other for free!
Hilbert spaces sale
4.2 Self-duality of Hilbert space
Lemma 11 (Riesz–Fréchet)
Let H be a Hilbert space and α
a continuous linear
functional on H, then there exists the unique y∈
H such
that α(
x)=⟨
x,
y
⟩
for all x∈
H. Also
||α||
H*=||
y||
H.
Proof.
Uniqueness: if ⟨
x,
y
⟩=⟨
x,
y′
⟩ ⇔
⟨
x,
y−
y′
⟩=0 for all
x∈
H
then
y−
y′ is self-orthogonal and thus is zero
(Exercise
1).
Existence: we may assume that α≢0 (otherwise take
y=0), then M=kerα is a closed proper subspace of
H. Since H=M⊕ M⊥, there exists a non-zero z∈
M⊥, by scaling we could get α(z)=1. Then for any
x∈ H:
| x=(x−α(x)z)+α(x)z, with
x−α(x)z∈ M, α(x)z∈ M⊥.
|
Because ⟨ x,z
⟩=α(x)⟨ z,z
⟩=α(x)||z||2
for any x∈ H we set y=z/||z||2.
Equality of the norms ||α||H*=||y||H follows from the
Cauchy–Bunyakovskii–Schwarz
inequality in the form α(x)≤ ||x||·||y|| and the
identity α(y/||y||)=||y||.
□
Example 12
On L2[0,1]
let
α(
f)=⟨
f,
t2
⟩=∫
01 f(
t)
t2 d t. Then
| | ⎪⎪
⎪⎪ | α | ⎪⎪
⎪⎪ | = | ⎪⎪
⎪⎪ | t2 | ⎪⎪
⎪⎪ | = | ⎛
⎜
⎜
⎝ | |
(t2)2 d t | ⎞
⎟
⎟
⎠ | | = | | .
|
Last modified: November 6, 2024.