Introduction to Hilbert spaces

Huynh Nhut Truong - Dao Nhat Gia An

The Mathematics and Statistics Academic Club

Faculty Mathematics and Statistics, Ton Duc Thang University

September 3, 2024
 Plan of the talk

1. Normed spaces

2. Linear bounded operators

3. Dual spaces

4. Inner product spaces

5. Hilbert spaces and properties

Huynh Nhut Truong - Dao Nhat Gia An (TDTU) Introduction to Hilbert spaces September 3, 2024 2 / 30
 Normed spaces

Definition 1
Let X be a vector space. A mapping || · || : X → R is called a norm on X if with
for all x , y ∈ X and α ∈ R we have
1. ||x || ≥ 0
2. ||x || = 0 ⇐⇒ x = θ
3. ||αx || = |α| · ||x ||
4. ||x + y || ≤ ||x || + ||y || (triangle inequality)
The couple (X , || · ||) is called a normed space.

Huynh Nhut Truong - Dao Nhat Gia An (TDTU) Introduction to Hilbert spaces September 3, 2024 3 / 30
 Normed spaces

Example
1. In the space Rn and 0 < p < ∞. Then the norm of a vector
n
!1/p
X
n p
x = (x1 , x2 , ..., xn ) ∈ R is defined by ||x ||p = |xi |
i=1
And when p = 2 it is called Euclidean norm. Another common norm on Rn
||x ||∞ = max |xi |
1≤i≤n
2. Consider the space C [a, b] is the set of all continuous function on [a, b].
The norm of f ∈ C [a, b] is defined by ||f ||∞ = sup |f (x )|. Besided that
x ∈[a,b]
!1/p
Z b
we have another norm of its is defined by ||f ||p = |f (x )|p dx
a

Huynh Nhut Truong - Dao Nhat Gia An (TDTU) Introduction to Hilbert spaces September 3, 2024 4 / 30
 Equivalent norms

Definition 2 (Equivalent norms)

Let || · ||a and || · ||b be 2 norms on the vector space X . We say that || · ||a is a
equivalent to || · ||b if there exits α, β ∈ R such that

α||x ||a ≤ ||x ||b ≤ β||x ||a , ∀x ∈ X

We write || · ||a ∼ || · ||b

Example: Consider on the Rn . For any x ∈ Rn we have 3 norms as follow

n n
!1/p
X X
p
||x ||1 = |xi |, ||x ||p = |xi | , ||x ||∞ = max |xi |
1≤i≤n
i=1 i=1

Show that || · ||1 ∼ || · ||p ∼ || · ||∞

Huynh Nhut Truong - Dao Nhat Gia An (TDTU) Introduction to Hilbert spaces September 3, 2024 5 / 30
 Linear operators

Definition 3 (Linear operators)

Let X and Y be 2 vector spaces. An operator T : X → Y is called linear if with
for all x , y ∈ X and α ∈ R we have

T (x + y ) = T (x ) + T (y )
T (αx ) = αT (x )

Remark
If T : X → Y is linear then
• T (θX ) = θY
• T (αx + y ) = αT (x ) + T (y ) for all x , y ∈ X and α ∈ R
df
Example: Let T : C 1 [a, b] → R be defined by T (f ) = . Prove that T is
dt
1
linear (where C [a, b] is the set of all the functions which have the first
derivative are continuous on [a, b])

Huynh Nhut Truong - Dao Nhat Gia An (TDTU) Introduction to Hilbert spaces September 3, 2024 6 / 30
 Bounded operators

Definition 4 (Bounded operators)

Let X and Y be 2 normed spaces. A linear operator T : X → Y is called
bounded if there exists a constant C such that

||T (x )|| ≤ C · ||x ||, ∀x ∈ X

In the cases, Y = R then T is called bounded linear functional.

Example: Given X = C [0, 1] with norm ||x || = max |x (t)|. Prove that the
t∈[0,1]
mapping

T :X →X
Z 1
x 7→ T (x ) = cos(t)x (t)dt
0

is bounded (where C [a, b] is the set of all function are continuous on [a, b])

Huynh Nhut Truong - Dao Nhat Gia An (TDTU) Introduction to Hilbert spaces September 3, 2024 7 / 30
 Normed of an operator

Definition 5 (Normed of an operator)

Let X , Y be 2 normed spaces and T : X → Y be a linear bounded operator.
Then the norm of T is defined by

||T (x )||
||T || = sup = sup ||T (x )||
x ∈X ||x || x ∈X
||x ||̸=0 ||x ||≤1

Remark
The norm of T is the minimum value C ≥ 0 such that ||T (x )|| ≤ C ||x ||

Example: Let X = C [a, b] be a normed spaces with norm ||x || = sup |x (t)|.
t∈[a,b]
Z b
An operator T : X → R is defined by T (x ) = x (t)dt. Show that it is a
a
linear bounded operator and find ||T ||

Huynh Nhut Truong - Dao Nhat Gia An (TDTU) Introduction to Hilbert spaces September 3, 2024 8 / 30
 L(X , Y ) spaces

Definition 6 (L(X , Y ) spaces)

Let X and Y be 2 vector spaces. Then L(X , Y ) is the set of all linear bounded
operator T : X → Y
1. L(X , Y ) is a vector space with 2 algebraic operations as follow

(T1 + T2 )(x ) = T1 (x ) + T2 (x )
(αT1 )(x ) = αT1 (x )

with for all T1 , T2 ∈ L(X , Y ), x ∈ X , α ∈ R

2. L(X , Y ) is a normed space with the norm is defined by

||T (x )||
||T || = sup = sup ||T (x )||
x ∈X ||x || x ∈X
||x ||̸=0 ||x ||≤1

where T ∈ L(X , Y ), x ∈ X

Huynh Nhut Truong - Dao Nhat Gia An (TDTU) Introduction to Hilbert spaces September 3, 2024 9 / 30
 L(X , Y ) spaces

Example: Let (X , || · ||X ), (Y , || · ||Y ) and (Z , || · ||Z ) be 3 normed spaces. Let

T1 : X → Y T2 : Y → Z be 2 linear bounded operators. Show that
T2 ◦ T1 : X → Z is also a linear bounded operator and ||T1 ◦ T2 || ≤ ||T1 || · ||T2 ||

Huynh Nhut Truong - Dao Nhat Gia An (TDTU) Introduction to Hilbert spaces September 3, 2024 10 / 30
 Dual spaces

Definition 7 (Dual spaces)

Let X be a normed space. The set of all linear functionals on X constitutes a
normed space with norm defined by

|f (x )|
||f || = sup = sup |f (x )|
x ∈X ||x || x ∈X
||x ||̸=0 ||x ||≤1

This spaces is called the dual space of X and denoted by X ∗

Huynh Nhut Truong - Dao Nhat Gia An (TDTU) Introduction to Hilbert spaces September 3, 2024 11 / 30
 Inner product spaces

Definition 8 (Inner product)

Let X be a vector space over filed R. An operator ⟨·, ·⟩ : X 2 → R is said to be
an inner product on X if with for all x , y , z ∈ X and α ∈ R then we have
1. ⟨x , x ⟩ ≥ 0;
2. ⟨x , x ⟩ = 0 ⇐⇒ x = θ;
3. ⟨x , y ⟩ = ⟨y , x ⟩;
4. ⟨x + y , z⟩ = ⟨x , z⟩ + ⟨y , z⟩;
5. ⟨αx , y ⟩ = α ⟨x , y ⟩.
And the couple (X , ⟨·, ·⟩) is called an inner product space.

Definition 9
In a inner
p product space X , the norm of a vector x is defined by the formula
||x || = ⟨x , x ⟩.

Huynh Nhut Truong - Dao Nhat Gia An (TDTU) Introduction to Hilbert spaces September 3, 2024 12 / 30
 Inner product spaces

Theorem 10
Let X be a inner product space and for any x , y ∈ X then
1. Cauchy-Schwarz inequality

| ⟨x , y ⟩ | ≤ ||x || · ||y ||;

2. Parallelogram equality

||x + y ||2 + ||x − y ||2 = 2(||x ||2 + ||y ||2 ).

Huynh Nhut Truong - Dao Nhat Gia An (TDTU) Introduction to Hilbert spaces September 3, 2024 13 / 30
 Inner product spaces

Definition 11
Let X be an inner product space. For all x , y ∈ X and A, B are subsets of X we
say that
1. x and y are orthogonal if ⟨x , y ⟩ = 0 and we write x ⊥ y ;
2. x and A are orthogonal if ⟨x , a⟩ = 0, ∀a ∈ A and we write x ⊥ A;
3. A and B are orthogonal if ⟨a, b⟩ = 0, ∀a ∈ A, ∀b ∈ B and we write A ⊥ B.

Theorem 12 (Pythagorean theorem)

Given X be an inner product space and x , y ∈ X such that x ⊥ y then

||x + y ||2 = ||x ||2 + ||y ||2

Huynh Nhut Truong - Dao Nhat Gia An (TDTU) Introduction to Hilbert spaces September 3, 2024 14 / 30
 Inner product spaces

Definition 13 (Orthogonal Complement)

Let A be a nonempty subset of inner product space X . Then the orthogonal
complement of A, denoted by A⊥ , is defined by A⊥ = {x ∈ X : x ⊥ A}

Example: Let X be an inner product space and A ⊂ B ⊂ X . Show that

A ⊂ A⊥⊥ and B ⊥ ⊂ A⊥ .

Huynh Nhut Truong - Dao Nhat Gia An (TDTU) Introduction to Hilbert spaces September 3, 2024 15 / 30
 Hilbert spaces

Definition 14 (Hilbert spaces)

A Hilbert space is a complete space equipped with an inner product.

Example: The Euclidean space Rn is a Hilbert space with the inner product is
defined as follow
n
X
⟨x , y ⟩ = xi yi
i=1

Huynh Nhut Truong - Dao Nhat Gia An (TDTU) Introduction to Hilbert spaces September 3, 2024 16 / 30
 Projection onto closed convex set

Definition 15 (Convex set)

A subset C of Hilbert space H which is called convex if it satisfies condition
follow
αx + (1 − α)y ∈ C , ∀x , y ∈ C , ∀α ∈ [0, 1]

Definition 16 (Distance)
Let H be a Hilbert space. The distance from x to A is defined by

dist(x , A) = inf ||x − a||

a∈A

Huynh Nhut Truong - Dao Nhat Gia An (TDTU) Introduction to Hilbert spaces September 3, 2024 17 / 30
 Projection onto closed convex set

Theorem 17 (Projection onto closed convex set)

Let A be a closed convex subset of Hilbert space H. Then for all vector x ∈ H
there exists a unique vector y in A such that

||x − y || = dist(x , A)

And the vector y is called the projection of vector x onto the set A. We denote
y = PA x

Huynh Nhut Truong - Dao Nhat Gia An (TDTU) Introduction to Hilbert spaces September 3, 2024 18 / 30
 Proof of theorem 17

• Existence: Let (an ) be a sequence in A such that

dn = ||x − an || → d = inf ||x − a||

a∈A

We will show that (an ) is a Cauchy sequence. Indeed, the parallelogram

equality applied with α = x − am and β = x − an leads to

||α + β||2 + ||α − β||2 = 2(||α||2 + ||β||2 )

2 2
⇐⇒ ||2x − am − an || + ||−am + an || = 2(||x − am ||2 + ||x − an ||2 )
 2
am + an  2

 an − am
⇐⇒ 2 x − + −2 = 2(dm2 + dn2 )
2 2
2 2
am + an an − am
⇐⇒ 4 x − +4 = 2(dm2 + dn2 )
2 2
2 2
am + an an − am 1 2
⇐⇒ x− + = (d + dn2 )
2 2 2 m

Huynh Nhut Truong - Dao Nhat Gia An (TDTU) Introduction to Hilbert spaces September 3, 2024 19 / 30
 Proof of theorem 17

am + an am + an
But ∈ A since A is convex then x− ≥ d. It follows
2 2
that
2 2
an − am 1 2 an − am 1 2
d2 + ≤ (d + dn2 ) =⇒ ≤ (d + dn2 ) − d 2
2 2 m 2 2 m
(*)

Passing m, n → ∞ we obtain that

2
an − am 1 2
lim ≤ lim (dm + dn2 ) − d 2 = 0
m,n→∞ 2 m,n→∞ 2

Then lim ||an − am || = 0. Therefore (an ) is a Cauchy sequence in A

m,n→∞
then it is convergent i.e.

∃a ∈ A such that lim an = a with d = ||x − a||

n→∞

Huynh Nhut Truong - Dao Nhat Gia An (TDTU) Introduction to Hilbert spaces September 3, 2024 20 / 30
 Proof of theorem 17

• Uniqueness: Suppose that there a1 , a2 ∈ A such that

||x − a1 || = ||x − a2 || = d
a1 + a2
Since A is convex then ∈ A which as in (*) implies that
2
2
a2 − a1 1 2
0≤ ≤ (d + d 2 ) − d 2 = 0
2 2

Thus a2 − a1 = θ =⇒ a2 = a1

Huynh Nhut Truong - Dao Nhat Gia An (TDTU) Introduction to Hilbert spaces September 3, 2024 21 / 30
 Weakly convergences

Definition 18 (Weakly convergence)

Let H be a Hilbert space. A sequence (xn )n∈N ⊂ H∗ converges weakly to
x ∈ H∗ if

⟨xn , y ⟩ → ⟨x , y ⟩ as n → ∞, ∀y ∈ H

We then write xn ⇀ x .

Definition 19 (Strongly convergence)

Let H be a Hilbert space. A sequence (xn )n∈N ⊂ H∗ converges strongly to
x ∈ H∗ if

||xn − x || → 0 as n → ∞

We then write xn → x .

Huynh Nhut Truong - Dao Nhat Gia An (TDTU) Introduction to Hilbert spaces September 3, 2024 22 / 30
 Weakly convergences
Theorem 20
If (xn )n∈N is strongly convergence to x in Hilbert space H then (xn )n∈N is
weakly convergence to x .

Proof.
Suppose that (xn )n∈N is strongly convergence in Hilbert space H i.e.

||xn − x || → 0 as n → ∞

For any y ∈ H. Consider

| ⟨xn , y ⟩ − ⟨x , y ⟩ | = | ⟨xn − x , y ⟩ | ≤ ||xn − x || · ||y ||

Taking n → ∞ we obtain that

0 ≤ lim | ⟨xn , y ⟩ − ⟨x , y ⟩ | ≤ lim (||xn − x || · ||y ||) = 0

n→∞ n→∞

Then ⟨xn , y ⟩ → ⟨x , y ⟩. Therefore, xn ⇀ x .

Huynh Nhut Truong - Dao Nhat Gia An (TDTU) Introduction to Hilbert spaces September 3, 2024 23 / 30
 Weakly convergences

Theorem 21
Assume that (xn )n∈N be a sequence in Hilbert space H such that xn ⇀ x and
||xn || → ||x || as n → ∞ then xn → x as n → ∞.

Proof.
Let (xn )n∈N be a sequence in Hilbert space H.
Consider ||xn − x ||2 = ⟨xn − x , xn − x ⟩ = ||xn ||2 − 2 ⟨xn , x ⟩ + ||x ||2
Taking n → ∞ we have

lim ||xn − x ||2 = lim ||xn ||2 − 2 ⟨xn , x ⟩ + ||x ||2

n→∞ n→∞
= ||x || − 2 ⟨x , x ⟩ + ||x ||2
2

= 2||x ||2 − 2||x ||2

=0 (since ||xn || → ||x || and ⟨xn , x ⟩ → ⟨x , x ⟩ as n → ∞)

=⇒ lim ||xn − x || = 0. Thus xn → x as n → ∞

n→∞

Huynh Nhut Truong - Dao Nhat Gia An (TDTU) Introduction to Hilbert spaces September 3, 2024 24 / 30
 Weakly convergences

Theorem 22
The weak limit of a sequence in Hilbert space is unique.

Proof.
Suppose that there exists two vectors y , z in Hilbert space H such that xn ⇀ x1
and xn ⇀ x2 . For any y ∈ H we have

lim ⟨xn , y ⟩ = ⟨x1 , y ⟩ and lim ⟨xn , y ⟩ = ⟨x2 , y ⟩

n→∞ n→∞

Then

⟨x1 , y ⟩ − ⟨x2 , y ⟩ = lim (⟨xn , y ⟩ − ⟨xn , y ⟩)

n→∞
=⇒ ⟨x1 − x2 , y ⟩ = 0

Choose y = x1 − x2 we have ⟨x1 − x2 , x1 − x2 ⟩ = 0

=⇒ x1 − x2 = 0 =⇒ x1 = x2

Huynh Nhut Truong - Dao Nhat Gia An (TDTU) Introduction to Hilbert spaces September 3, 2024 25 / 30
 Riesz representation theorem

Theorem 23 (Riesz representation theorem)

Let H be a Hilbert space. Then for every bounded linear functional f on H∗
there exists a unique Riesz representative z ∈ H such that

f (x ) = ⟨x , z⟩ , ∀x ∈ H

Furthermore, ||f || = ||z||.

Huynh Nhut Truong - Dao Nhat Gia An (TDTU) Introduction to Hilbert spaces September 3, 2024 26 / 30
 Proof of Theorem 23
Let f be a linear bounded functional on Hilbert space H∗
• Case1: If f = 0 then f (x ) = 0, ∀x ∈ H
Beside that we choose z = θ then ⟨x , z⟩ = 0, ∀x ∈ H
Thus f (x ) = ⟨x , z⟩ = 0, ∀x ∈ H and ||f || = ||z|| = 0
• Case2: If f ̸= 0.
Existence: Let M = ker f = {x ∈ H| f (x ) = 0}. Clearly, M ̸= H since if
M = H then f (x ) = 0, ∀x ∈ H =⇒ f = 0 (contradiction). So M ̸= H
=⇒ M is proper, closed linear subspace of H
=⇒ There exists a nonzero vector η such that η ⊥ M =⇒ η ∈ M ⊥
We define ξ = f (x )η − f (η)x then f (ξ) = f (x )f (η) − f (η)f (x ) = 0
=⇒ ξ ∈ ker f = M
=⇒ ⟨ξ, η⟩ = 0
=⇒ ⟨f (x )η − f (η)x , η⟩ = 0
=⇒ f (x ) ⟨η, η⟩ = f (z0 ) ⟨x ,
η⟩ 
f (η) f (η)
=⇒ f (x ) = ⟨x , η⟩ = x , η
||η|| ||η||
f (η)
Set z = η then f (x ) = ⟨x , z⟩
||η||

Huynh Nhut Truong - Dao Nhat Gia An (TDTU) Introduction to Hilbert spaces September 3, 2024 27 / 30
 Proof of Theorem 23

Uniqueness: Assume that there exists two vectors z1 , z2 ∈ H such that

f (x ) = ⟨x , z1 ⟩ = ⟨x , z2 ⟩ , ∀x ∈ H
=⇒ ⟨x , z1 ⟩ − ⟨x , z2 ⟩ = 0
=⇒ ⟨x , z1 − z2 ⟩ = 0

Choose x = z1 − z2 we obtain that

⟨z1 − z2 , z1 − z2 ⟩ = 0 =⇒ z1 − z2 = θ =⇒ z1 = z2

Show that ||f|| = ||z||: We have

|f (x )| = | ⟨x , z⟩ | ≤ ||x || · ||z|| (Cauchy-Schwarz inequality)

Taking sup both sideds with x ∈ H such that ||x || ≤ 1 we obtain that

sup ≤ sup (||x || · ||z||) =⇒ ||f || ≤ ||z||

x ∈H x ∈H
||x |||f (x )|≤1 ||x ||≤1

Huynh Nhut Truong - Dao Nhat Gia An (TDTU) Introduction to Hilbert spaces September 3, 2024 28 / 30
 Proof of Theorem 23

Moreover, |f (x )| ≤ ||f || · ||x ||, ∀x ∈ H. Choose x = z we have

| ⟨z, z⟩ |
|f (z)| ≤ ||f || · ||z|| =⇒ ||f || ≥ = ||z||
||z||

Therefore, ||f || = ||z||

Huynh Nhut Truong - Dao Nhat Gia An (TDTU) Introduction to Hilbert spaces September 3, 2024 29 / 30
 THANK YOU FOR YOUR ATTENTION
SEE YOU LATER!

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