Chapter 4
Hilbert Spaces
4.1
Inner Product Spaces
• Inner Product Space. A complex vector space E is called an inner product space (or a pre-Hilbert space, or a unitary space) if there is a mapping
(·, ·) : E × E → C, called an inner product, that satisfies the conditions:
∀x, y, z ∈ E, ∀α ∈ C:
1. (x, x) ≥ 0
2. (x, x) = 0 if and only if x = 0
3. (x, y + z) = (x, y) + (x, z)
4. (x, αy) = α(x, y)
5. (x, y) = (y, x)
• Finite-dimensional spaces. Cn is the space of n-tuples x = (x1 , . . . , xn ) of
complex numbers. It is a Hilbert space with the inner product
(x, y) =
n
�
x̄ j y j
j=1
• Orthogonal Vectors. Let E be an inner product space. Two vectors x, y ∈ E
are orthogonal if (x, y) = 0.
• Notation. The orthogonality of the vectors x and y is denoted by
x⊥y
61
62
CHAPTER 4. HILBERT SPACES
• The relation ⊥ is symmetric, that is, if x ⊥ y then y ⊥ x.
• Examples.
• Spaces of sequences
• The spaces of sequences l0 (sequences with zero tails) and l2 (sequences
with finite || · ||2 norm) are 0inner product spaces with the inner product
(x, y) =
∞
�
x̄ j y j
j=1
• Spaces of functions
• The space C([a, b]) is an inner product space with the inner product
� b
f¯g
( f, g) =
a
• The space C0 (R) (continuous functions with compact support) is an inner
product space with the inner product
�
( f, g) =
f¯g
R
• Spaces of square integrable functions
• The space L2 ([a, b]) is inner product space with the inner product
� b
( f, g) =
f¯g
a
• The space L2 (R) is and inner product space with the inner product
�
( f, g) =
f¯g
R
• L2 ([a, b], µ) (space of square integrable functions with the measure µ) with
the inner product
�
( f, g) =
b
a
µ f¯g ,
where µ is an almost everywhere positive measurable function, µ > 0.
mathphys15.tex; October 28, 2015; 15:35; p. 60
4.1. INNER PRODUCT SPACES
63
• L2 (R, µ) (space of square integrable functions with the measure µ) with the
inner product
�
µ f¯g ,
( f, g) =
R
where µ > 0 almost everywhere.
• Let Ω be an open set in Rn (in particular, Ω can be the whole Rn ). The space
L2 (Ω) is the set of complex valued functions such that
�
| f |2 < ∞ .
Ω
It is an inner product space with the inner product
�
f¯g
( f, g) =
Ω
• L2 (Rn ) with the inner product
( f, g) =
�
Rn
f¯g
• Let Ω be an open set in Rn (in particular, Ω can be the whole Rn ) and V
be a finite-dimensional vector space. Let � , � be the inner product on the
vector space V. The space L2 (Ω, V, µ) is the set of vector valued functions
f = ( f 1 , . . . , f N ) on Ω such that
�
Ω
µ � f, f � =
�
µ
Ω
N
�
i=1
| f i |2 < ∞ .
It is an inner product space with the inner product
( f, g) =
�
Ω
µ � f, g� =
�
µ
Ω
N
�
f¯i gi
i=1
• Real Inner Product Spaces. The inner product in a real inner product
space is symmetric.
mathphys15.tex; October 28, 2015; 15:35; p. 61
64
CHAPTER 4. HILBERT SPACES
• Direct Sum of Inner Product Spaces. Let E1 and E2 be inner product
spaces. The direct sum E = E1 ⊕ E2 of E1 and E2 is an inner product space
of ordered pairs z = (x, y) with x ∈ E1 and y ∈ E2 with the inner product
defined by
(z1 , z2 )E = (x1 , x2 )E1 + (y1 , y2 )E2 .
• Tensor Products of Inner Product Spaces. Let E1 and E2 be inner product
spaces. For each ϕ1 ∈ E1 and ϕ2 ∈ E2 let ϕ1 ⊗ ϕ2 denote the conjugate
bilinear form on E1 × E2 defined by
(ϕ1 ⊗ ϕ2 )(ψ1 , ψ2 ) = (ψ1 , ϕ1 )E1 (ψ2 , ϕ2 )E2
where ψ1 ∈ E1 and ψ2 ∈ E2 .
• Let E = E1 ⊗ E2 be the set of finite linear combinations of such bilinear
forms. An inner product on E can be defined by
(ϕ ⊗ ψ, η ⊗ µ)E = (ϕ, η)E1 (ψ, µ)E2
(with ϕ, η ∈ E1 and ψ, µ ∈ E2 ) and extending by linearity on E.
• Fock Space. Let E be an inner product space. The space
E ⊗ ··· ⊗ E
F(E) = C ⊕∞
n=1 ��������������������
n
is called the Fock space over E.
4.1.1
Norm in an Inner Product Space.
Let E be an inner product space. The norm in E is a functional � · �: E → R
defined by
�
� x �= (x, x).
• Theorem
√ 4.1.1 Every inner product space is a normed space�with the norm
� x �= (x, x) and a metric space with the metric d(x, y) = (x − y, x − y).
• Theorem 4.1.2 Schwarz’s Inequality. Let E be an inner product space.
Then for any x, y ∈ E we have
|(x, y)| ≤� x � � y � .
mathphys15.tex; October 28, 2015; 15:35; p. 62
4.1. INNER PRODUCT SPACES
65
The equality |(x, y)| =� x � � y � holds if and only if x and y are linearly
dependent.
Proof:
1.
�
• Corollary 4.1.1 Triangle Inequality. Let E be an inner product space.
Then for any x, y ∈ E we have
� x + y �≤� x � + � y � .
Proof:
1.
�
• Theorem 4.1.3 Parallelogram Law. Let E be an inner product space.
Then for any x, y ∈ E we have
�
�
� x + y �2 + � x − y �2 = 2 � x �2 + � y �2
Proof:
1.
�
•
Theorem 4.1.4 Pythagorean Theorem. Let E be an inner product
space. If two vectors x, y ∈ E are orthogonal then
Proof:
� x + y �2 =� x �2 + � y �2 .
1.
�
• Examples.
mathphys15.tex; October 28, 2015; 15:35; p. 63
66
CHAPTER 4. HILBERT SPACES
4.2
Hilbert Spaces
• Hilbert Space. An inner product space is called a Hilbert space if it is
complete as a normed space.
• Examples.
• Cn is complete.
• Spaces of sequences
• The space l2 of square summable sequences is complete. Proved before.
• The space l0 of sequences with vanishing tails is not complete. Counterexample.
• Spaces of continuous functions
• C([a, b]) is not complete. Counterexample.
• C0 (R) (space of continuous functions with compact support) is not complete. Counterexample.
• Spaces of square integrable functions.
• Remark. For any f, g ∈ L2 (Ω) there holds
|( f, g)| ≤ || f g||1 ≤ || f ||2 ||g||2
• If Ω is compact then for any f ∈ L2 (Ω)
|| f ||1 ≤ [µ(Ω)]1/2 || f ||2
• Theorem 4.2.1
The space L2 ([a, b]) is complete.
Proof:
1. Take a Cauchy sequence fn in L2 so that
|| fn − fm ||2 → 0
as n, m → ∞.
mathphys15.tex; October 28, 2015; 15:35; p. 64
4.2. HILBERT SPACES
67
2. By Schwarz (or Hoelder) inequality
�
|| fn − fm ||1 ≤ (b − a)|| fn − fm ||2 → 0
as n, m → ∞.
3. So, fn is Cauchy in L1 .
4. Since L1 is complete it converges to some f ∈ L1 .
5. Then there is a subsequence f pn convergent to f (almost everywhere).
6. Clearly, the subsequence f pn is Cauchy in L2 ,
|| f pn − f pm ||2 → 0
as n, m → ∞.
7. Therefore, as m → ∞
as n → ∞.
|| f pn − f ||2 → 0
8. Therefore, f ∈ L2 .
9. Finally,
|| f − fn ||2 ≤ || f − f pn ||2 + || f pn − fn ||2 → 0.
10. Therefore, fn → f .
�
• Theorem 4.2.2 The space L2 (R) is complete.
Proof. Idea: Construct a subsequence of a Cauchy sequence in L2 (R) that
converges almost everywhere by a diagonal argument.
1. Let fn be a Cauchy sequence in L2 (R).
2. Then fn is Cauchy in L2 ([−1, 1]).
3. Then it has a subsequence f1,n that converges in [−1, 1] almost everywhere.
4. Since f1,n is Cauchy in L2 ([−2, 2]) it has a subsequence f2,n that converges in [−2, 2] almost everywhere.
5. Then the subsequence fn,n of fn converges in R almost everywhere to
some f .
mathphys15.tex; October 28, 2015; 15:35; p. 65
68
CHAPTER 4. HILBERT SPACES
6. Then, one shows that f ∈ L2 (R) and fn → f in L2 (R).
• Theorem 4.2.3 Let µ be a measurable function on [a, b] positive almost
everywhere. The space L2 ([a, b], µ) is complete.
√
Proof. Rescale functions by µ.
• More generally,
The space L2 (Ω, µ, V) is complete.
• Theorem 4.2.4
• Sobolev Spaces
• Let Ω be an open set in Rn (in particular, Ω can be the whole Rn ) and V a
finite-dimensional complex vector space.
• Let C m (Ω, V) be the space of complex vector valued functions that have
continuous partial derivatives of all orders less or equal to m.
• Let
α = (α1 , . . . , αn ),
α j ∈ Z+ , be a multiindex of nonnegative integers, αi ≥ 0, and let
|α| = α1 + · · · + αn .
• Define
Dα f =
∂|α|
f.
∂x1α1 · · · ∂xnαn
• Then f ∈ C m (V, Ω) iff ∀α, |α| ≤ m, ∀i = 1, . . . , N, ∀x ∈ Ω we have
|Dα fi (x)| < ∞ .
• The space H̃ m (Ω, V) is the space of complex vector valued functions such
that ∀α, |α| ≤ m,
Dα f ∈ L2 (Ω, V)
i.e. such that ∀α, |α| ≤ m,
�
Ω
dx �D f, D f � =
α
α
� �
N
Ω i=1
|Dα fi (x)|2 dx < ∞ .
mathphys15.tex; October 28, 2015; 15:35; p. 66
4.2. HILBERT SPACES
69
• It is an inner product space with the inner product
( f, g) =
� � �
α, |α|≤m
Ω
N
� � �
�
¯
D f, D g =
Dα f¯i Dα gi
α
α
α, |α|≤m
Ω i=1
• The Sobolev space H m (Ω, V) is the completion of the space H̃ m (Ω, V) defined above.
4.2.1
Strong and Weak Convergence
• Strong Convergence. A sequence (xn ) of vectors in an inner product space
E is strongly convergent to a vector x ∈ E if
lim � xn − x �= 0
n→∞
• Notation. xn → x.
• Theorem 4.2.5 Let E be an inner product space. Then for every y ∈ E the
linear functional ϕy : E → C defined by
ϕy (x) = (y, x)
∀x ∈ E
is continuous (and, therefore, bounded).
Proof: Use Schwarz inequality.
• Weak Convergence. A sequence (xn ) of vectors in an inner product space
E is weakly convergent to a vector x ∈ E if for any y ∈ E
lim (y, xn − x) = 0
n→∞
w
• Notation. xn → x
• Theorem 4.2.6 A strongly convergent sequence is weakly convergent to the
same limit.
Proof: Use Schwarz inequality.
• Converse is not true. Counterexample later.
mathphys15.tex; October 28, 2015; 15:35; p. 67
70
CHAPTER 4. HILBERT SPACES
• Theorem 4.2.7 Let (xn ) be a sequence in an inner product space E and
x ∈ E. Suppose that
w
1. xn → x and
2. � xn �→� x �.
Then xn → x.
Proof: Easy.
• Theorem 4.2.8 Let S be a subset of an inner product space E such that
span S is dense in E, (xn ) be a bounded sequence in E and x ∈ E. Suppose
that for any y ∈ S ,
lim (y, xn − x) = 0.
n→∞
w
Then xn → x.
Proof: Let z ∈ E.
Then there is ym ∈ span S such that ym → z.
Then
(xn − x, z) = (xn − x, ym ) + (xn − x, z − ym )
converges to zero as n, m → ∞.
w
• It is possible that xn → x but ||xn || does not converge to ||x||. But at least the
sequence ||xn || has to be bounded.
• Theorem 4.2.9 Weakly convergent sequences in a Hilbert space are bounded.
Proof: Read in functional analysis books.
4.3
Orthogonal and Orthonormal Systems
• Orthogonal and Orthonormal Systems. Let E be an inner product space.
A set S of vectors in E is called an orthogonal system if any pair of distinct
vectors in S is orthogonal to each other.
• An orthogonal system of unit vectors is an orthonormal system.
• Every orthogonal system can be made orthonormal.
mathphys15.tex; October 28, 2015; 15:35; p. 68
4.3. ORTHOGONAL AND ORTHONORMAL SYSTEMS
71
• Let S be a set of vectors in E. We say that x ⊥ S if x ⊥ y for any y ∈ S .
• If x ⊥ S then x ⊥ span S .
• The orthogonal complement of S is the space
S ⊥ = {x ∈ E | x ⊥ S }
• Theorem 4.3.1 Orthogonal systems are linearly independent.
Proof: Easy.
• Orthonormal Sequence. A sequence of vectors which is an orthonormal
system is an orthonormal sequence; then
(xn , xm ) = δnm
• Examples.
1. Canonical basis in l2 .
(en )i = (δin )
2. Fourier basis. The functions
1
fn (x) = √ einx
2π
are orthonormal in L2 ([−π, π]).
3. Legendre polynomials.
Pn (x) =
1 n 2
∂ (x
n
2 n! x
− 1)n
Show that the functions
fn =
�
n+
1
Pn
2
are orthonormal in L2 ([−1, 1]).
mathphys15.tex; October 28, 2015; 15:35; p. 69
72
CHAPTER 4. HILBERT SPACES
4. Hermite polynomials.
2
Hn = (−1)n e x ∂nx e−x
2
Show that the functions
1
fn (x) = √
2n n!π1/4
�
�
x2
exp −
Hn (x)
2
are orthonormal in L2 (R).
• Any orthogonal sequence can be always made orthonormal.
• Gram-Schmidt orthonormalization process. Any sequence of linearly
independent vectors can be made orthonormal.
1. Let (yn ) be a linearly independent sequence.
2. Let
zn =
yn
||yn ||
3. Recall that for any orthonormal sequence en the operator Pn defined
by
n−1
�
Pn = I −
|ek ��ek |
k=1
or
Pn x = x −
n−1
�
(ek , x)ek
k=1
is the projections to the orthogonal complement of span {e1 , . . . , en−1 }.
4. Then the sequence (en ) defined by
e1 = z1 ,
en = Pn zn = zn −
n−1
�
(ek , zn )ek
k=1
is orthonormal.
mathphys15.tex; October 28, 2015; 15:35; p. 70
4.3. ORTHOGONAL AND ORTHONORMAL SYSTEMS
4.3.1
73
Properties of Orthonormal Systems
• Theorem 4.3.2 Pythagorean Formula. Let E be an inner product space
N
and {xn }n=1
be an orthogonal set in E. Then
����2
����
N
N
�
��
�����
����
xn ������ =
� xn �2
�� n=1 ��
n=1
Proof: By induction.
• Theorem 4.3.3 Bessel’s Equality and Inequality. Let E be an inner prodN
uct space and {en }n=1
be an orthonormal set in E. Then ∀x ∈ E
����
����2
N
N
�
�
�
�
��
|xn |2 + ������ x −
xn en ������
� x �2 =
��
��
n=1
n=1
and
N
�
n=1
where xn = (en , x).
|xn |2 ≤� x �2
Proof:
• Remarks. Let (en ) be an orthonormal sequence in an inner product space
E.
• The complex numbers xn = (en , x) are called the generalized Fourier coefficients of x with respect to the orthonormal sequence (en ).
• An orthonormal sequence (en ) in E induces a mapping ϕ : E → l2 defined
by
ϕ(x) = x̃ = (xn )
• The sequence of Fourier coefficients x̃ = (xn ) is square summable, that is
x̃ ∈ l2 since for any x ∈ E
∞
�
n=1
|xn |2 ≤� x �2
and, therefore, this series converges.
mathphys15.tex; October 28, 2015; 15:35; p. 71
74
CHAPTER 4. HILBERT SPACES
• The expansion
x∼
∞
�
xn en
n=1
is called the generalized Fourier series of x.
• The question is whether the mapping ϕ is bijective and whether the Fourier
series converges.
• Theorem 4.3.4 Let (en ) be an orthonormal sequence in a Hilbert space
�
H and (xn ) be a sequence of complex numbers. Then the series ∞
n=1 xn en
�
∞
2
2
converges if and only if (xn ) ∈ l , that is, the series n=1 |xn | converges.
In this case
����2
����
∞
∞
�
����
�����
����
xn en ���� =
|xn |2
��
�� n=1
n=1
Proof: By Pythagorean theorem
����
����2
m
m
�
�����
����
����
xk ek ���� =
|xk |2
�� k=n
��
k=n
• Fourier series of any x ∈ H in a Hilbert space H converges.
• Fourier series of x may converge to a vector different from x!
• Example. Let (en ) be an orthonormal sequence. Then fn = e2n is also
�
orthonormal. Let x = e1 . Then xn = ( fn , x) = 0 and 0 = ∞
n=1 xn fn � x.
• Let (en ) be an orthonormal sequence in an inner product space E. The sequence of Fourier coefficients xn = (en , x) is square summable, and, therefore,
lim (en , x) = 0
∀x ∈ E
n→∞
• Thus, every orthonormal sequence weakly converges to zero.
• Orthonormal sequences are not strongly convergent since � en �= 1 ∀n ∈ Z+ .
mathphys15.tex; October 28, 2015; 15:35; p. 72
4.3. ORTHOGONAL AND ORTHONORMAL SYSTEMS
75
• Complete Orthonormal Sequence. Let E be an inner product space. An
orthonormal sequence (en ) in E is complete if ∀x ∈ E the Fourier series of
x converges to x, that is,
∞
�
x=
xn en ,
n=1
more explicitly,
����
����
n
�
����
��
xk ek ������ = 0
lim ���� x −
n→∞ ��
��
k=1
• Example. L2 ([−π, π]) does not imply pointwise convergence.
• Orthonormal Basis. Let E be an inner product space. An orthonormal system B in E is an orthonormal basis if for any x ∈ E there exists a unique
orthonormal sequence (en ) in B and a unique sequence (xn ) of nonzero complex numbers such that
∞
�
x=
xn en .
n=1
• Remarks.
• A complete orthonormal sequence in an inner product space is an orthonormal basis.
• Let E be an inner product space and (en ) be a complete orthonormal sequence. Then the set
S = span {en | n ∈ Z+ }
is dense in E.
• Theorem 4.3.5 Let H be a Hilbert space. An orthonormal sequence (en ) in
H is complete if and only if the only vector orthogonal to this sequence is
the zero vector, that is,
span {en |n ∈ Z+ }⊥ = {0}.
Proof: Easy. To prove converse, let y =
So, x = y.
�
n (en ,
x)en . Then (x − y, en ) = 0.
mathphys15.tex; October 28, 2015; 15:35; p. 73
76
CHAPTER 4. HILBERT SPACES
• Theorem 4.3.6 Parseval’s Formula. Let H be a Hilbert space. An orthonormal sequence (en ) in H is complete if and only if ∀x ∈ H
2
�x� =
∞
�
n=1
|xn |2
where xn = (en , x).
Proof: Use Bessel equality.
• Theorem 4.3.7 Let H1 and H2 be Hilbert spaces. If {ϕk } and {ψl } are orthonormal bases for H1 and H2 respectively, then {ϕk ⊗ψl } is an orthonormal
basis for the tensor product H1 ⊗ H2 .
• Examples. The sequence
1
fn (x) = √ einx ,
2π
n∈Z
is orthonormal and complete in L2 ([−π, π]). Proof later.
• The sequence
1
g0 = √ ,
2π
1
gn (x) = √ cos(nx),
2π
1
fn (x) = √ sin(nx),
2π
n ∈ Z+
is orthonormal and complete in L2 ([−π, π]). Proof later.
• The sequence
1
h0 = √ ,
π
hn (x) =
�
2
cos(nx),
π
n ∈ Z+
is orthonormal and complete in L2 ([0, π]). Proof later.
• The sequence
ψn (x) =
�
2
sin(nx),
π
n = 0, 1, 2, . . . ,
is orthonormal and complete in L2 ([0, π]). Proof later.
mathphys15.tex; October 28, 2015; 15:35; p. 74
4.4. SEPARABLE HILBERT SPACES
4.4
77
Separable Hilbert Spaces
• Separable Spaces. An infinite-dimensional Hilbert space is separable if it
contains a complete orthonormal sequence.
• Examples.
• L2 ([−π, π])
• l2
• Example (Non-separable Hilbert Space). Let H be the space of all complex valued functions f : R → C on R such that they vanish everywhere
except a countable number of points in R and
�
| f (x)|2 < ∞.
f (x)�0
Define the inner product by
(g, f ) =
�
ḡ(x) f (x).
f (x)g(x)�0
Let fn be an orthonormal sequence in H. Then there are non-zero functions
f such that ( f, fn ) = 0 for all n ∈ Z+ . Therefore, fn cannot be complete and
H is not separable.
• Theorem 4.4.1 Let H be a separable Hilbert space. Then H contains a
countable dense subset.
Proof:
1. Let (en ) be a complete orthonormal sequence in H.
2. Define the set
 n



�




(α
+
iβ
)e
|
α
,
β
∈
Q,
n
∈
Z
S =
k
k k
k k
+




k=1
3. Then S is countable.
4. Also, ∀x ∈ H,
lim �
n→∞
n
�
(ek , x)ek − x �= 0.
k=1
mathphys15.tex; October 28, 2015; 15:35; p. 75
78
CHAPTER 4. HILBERT SPACES
5. Therefore, S is dense in H.
�
• Theorem 4.4.2 Let H be a separable Hilbert space. Then every orthogonal
set S in H is countable.
Proof:
1. Let S be an orthogonal set in H.
2. Let
3. Then ∀x, y ∈ S 1 , x � y,
�
x
S1 =
|x∈S .
�x�
�
� x − y �2 = 2.
4. Consider the collection of balls Bε (x) with ε = 21/2 /3 for every x ∈ S 1 .
5. Then, for any x, y ∈ S 1 , if x � y, then the corresponding balls are
disjoint,
Bε (x) ∩ Bε (y) = ∅.
6. Since H is separable, it has a countable dense subset A.
7. Since A is dense in H it must have at least one point in every ball Bε (x).
8. Therefore, S 1 must be countable.
9. Thus S is countable.
�
• Unitary Linear Transformations. Let H1 and H2 be Hilbert spaces. A
linear map T : H1 → H2 is unitary if ∀x, y ∈ H1
(T (x), T (y))H2 = (x, y)H1 .
• Hilbert Space Isomorphism. Let H1 and H2 be Hilbert spaces. Then H1
is isomorphic to H2 if there exists a linear unitary bijection T : H1 → H2
(called a Hilbert space isomorphism).
• Remark. Every Hilbert space isomorphism has unit norm
� T �= 1.
mathphys15.tex; October 28, 2015; 15:35; p. 76
4.5. TRIGONOMETRIC FOURIER SERIES
79
• Theorem 4.4.3 Every infinite-dimensional separable Hilbert space is isomorphic to l2 .
Proof:
1. Let (en ) be a complete orthonormal sequence in H.
2. Let x ∈ H.
3. Let xn = (en , x).
4. This defines a linear bijection T : H → l2 by
T (x) = (xn ) .
5. Let x, y ∈ H.
6. Then
(T (x), T (y))l2 =
∞
�
x̄n yn
n=1
7. On another hand
∞
∞
∞
�
�
�
(x, y)H = (
xn en , y) =
x̄n (en , y) =
x̄n yn
n=1
n=1
n=1
8. Therefore T is unitary, and is, therefore, an isomorphism from H onto
l2 .
�
• Remarks.
• Isomorphism of Hilbert spaces is an equivalence relation.
• All separable infinite-dimensional Hilbert spaces are isomorphic.
4.5
Trigonometric Fourier Series
• Consider the Hilbert space L2 ([−π, π]).
• The sequence
1
ϕn (x) = √ einx ,
2π
2
is an orthonormal sequence in L ([−π, π]).
n∈Z
mathphys15.tex; October 28, 2015; 15:35; p. 77
80
CHAPTER 4. HILBERT SPACES
• Note that L2 ⊂ L1 . Consider the space L1 ([−π, π]).
• Identify the elements of L1 ([−π, π]) with 2π periodic functions on R.
• Then for any f ∈ L1 ([−π, π]),
�
� π
dt f (t) =
−π
π
−π
dt f (t − x)
• For any k ∈ Z+ let Pn be the projections
Pk = |ϕk � �ϕk |
such that
(Pk f )(x) = (ϕk , f )ϕk (x) =
where
Pk (x) =
�
π
−π
dt Pk (x − t) f (t)
1 ikx
e
2π
is the integral kernel of Pk .
• Let Gk be the projections (not orthogonal)
Gj =
j
�
Pk
k=− j
with the integral kernel
j
1 � ikx
G j (x) =
e
2π k=− j
• Also define the sequence of operators
Kn =
n
n
1 �
1 �
Gj =
(n + 1 − |k|)Pk
n + 1 j=0
n + 1 k=−n
with the integral kernel
n
�
1
(n + 1 − |k|) eikx
Kn (x) =
2π(n + 1) k=−n
called the Fejer’s kernel.
mathphys15.tex; October 28, 2015; 15:35; p. 78
4.5. TRIGONOMETRIC FOURIER SERIES
81
• Lemma 4.5.1 We have
�
�
sin2 (n + 1) 2x
1
� �
Kn (x) =
2π(n + 1)
sin2 x
2
Proof: Direct calculation.
• A sequence Kn of 2π-periodic continuous functions is a summability kernel if it satisfies the conditions:
1.
�
π
−π
dt Kn (t) = 1,
∀n ∈ Z+
2. There is M ∈ R such that ∀n ∈ Z+
� π
dt |Kn (t)| ≤ M,
−π
3. For any δ ∈ (0, π)
2π−δ
�
lim
dt |Kn (t)| = 0
n→∞
δ
• A summability kernel converges formally to
lim Kn (x) =
n→∞
∞
�
k=−∞
δ(x − 2πk)
• Theorem 4.5.1 Let (Kn ) be a summability kernel. Then for any f ∈ L1 ([−π, π])
the sequence Fn = Kn f strongly converges to f in L1 norm, that is,
lim ||Fn − f || = lim ||(Kn − I) f || = 0
n→∞
n→∞
Proof: Use the properties of the summability kernel. Basically because
Kn → I (delta function).
• Lemma 4.5.2 The Fejer’s kernel is a summability kernel.
Proof: Direct calculation.
mathphys15.tex; October 28, 2015; 15:35; p. 79
82
CHAPTER 4. HILBERT SPACES
• Theorem 4.5.2 Let f ∈ L1 ([−π, π]). If all Fourier coefficients fn = (ϕn , f )
vanish, then f = 0 almost everywhere.
Proof: Let Fn = Kn f. Suppose fn = 0. Then Fn = 0. Since Fn → f , then
a.e.
f = 0.
• Theorem 4.5.3 The sequence
1
ϕn (x) = √ einx ,
2π
n∈Z
is a complete orthonormal sequence, (an orthonormal basis), in L2 ([−π, π]).
a.e.
Proof: Let f ∈ L2 . Then f ∈ L1 . Suppose (ϕn , f ) = 0. Then f = 0 . That
is, f = 0 in L2 .
• Let f ∈ L2 ([−π, π]). The series
f (x) =
∞
�
fn ϕn (x),
n=−∞
where
fn =
�
π
−π
dt ϕ̄n (t) f (t),
is the Fourier series. The scalars fn are the Fourier coefficients.
• Fourier series does not converge pointwise!
• Fourier series of a function f ∈ L2 ([−π, π]) converges almost everywhere.
4.6
Orthonormal Complements and Projection Theorem
• A subspace of a Hilbert space is an inner product space.
• A closed subspace of a Hilbert space is a Hilbert space.
• Orthogonal Complement. Let H be a Hilbert space and S ⊂ H be a
nonempty subset of H. We say that x ∈ H is orthogonal to S , denoted
by x ⊥ S , if ∀y ∈ S , (x, y) = 0.
mathphys15.tex; October 28, 2015; 15:35; p. 80
4.6. ORTHONORMAL COMPLEMENTS AND PROJECTION THEOREM 83
The set
S ⊥ = {x ∈ H | x ⊥ S }
of all vectors orthogonal to S is called the orthogonal complement of S .
Two subsets A and B of H are orthogonal, denoted by A ⊥ B, if every
vector of A is orthogonal to every vector of B.
• If x ⊥ H, then x = 0, that is
H ⊥ = {0},
{0}⊥ = H .
• If A ⊥ B, then A ∩ B = {0} or ∅.
• Orthogonal decomposition. If every element of H can be uniquely represented as the sum of an element of S and an element of S ⊥ , then H is the
direct sum of S and S ⊥ , which is denoted by
H = S ⊕ S⊥
• The union of a basis of S and a basis of S ⊥ gives a basis of H.
• Orthogonal projection. An orthogonal decomposition H = S ⊕S ⊥ induces
a projection map P : H → S defined by
P(y + z) = y,
where y ∈ S and z ∈ S ⊥ .
• Examples.
• Theorem 4.6.1 The orthogonal complement of any subset of a Hilbert space
is a Hilbert subspace.
Proof:
1. Let H be a Hilbert space and S ⊂ H.
2. Check directly that S ⊥ is a vector subspace.
3. Claim: S ⊥ is closed.
4. Let (xn ) be a sequence in S ⊥ such that xn → x ∈ H.
mathphys15.tex; October 28, 2015; 15:35; p. 81
84
CHAPTER 4. HILBERT SPACES
5. Then ∀y ∈ S
(x, y) = lim (xn , y) = 0.
n→∞
⊥
6. Thus x ∈ S .
�
• Remarks.
• S does not have to be a vector subspace.
• Theorem 4.6.2 Orthogonal Projection. Let H be a Hilbert space and S
be a closed subspace of H. Then
H = S ⊕ S ⊥.
Proof: We need to show that ∀x ∈ H there exist unique y ∈ S and z ∈ S ⊥
such that
x = y + z.
1. Let (ei , f j ) be an orthonormal basis in H such that ei ∈ S and f j ∈ S ⊥ .
2. Let x ∈ H. Then
x=
∞
�
yi ei +
i=1
3. Let
y=
∞
�
z j f j.
j=1
∞
�
yi ei
i=1
z=
∞
�
zj fj
j=1
4. Then
x=y+z
�
• Theorem 4.6.3 Let H be a Hilbert space and S be a closed subspace of H.
Then
(S ⊥ )⊥ = S .
Proof:
mathphys15.tex; October 28, 2015; 15:35; p. 82
4.7. LINEAR FUNCTIONALS AND THE RIESZ REPRESENTATION THEOREM85
1. Let x ∈ S .
2. Then x ⊥ S ⊥ , or x ∈ S ⊥⊥ .
3. So, S ⊆ S ⊥⊥ .
4. Let x ∈ S ⊥ .
5. Since S is closed, there exist y ∈ S and z ∈ S ⊥ such that x = y + z.
6. Then y ∈ S ⊥⊥ .
7. Since S ⊥⊥ is a vector space, z = x − y ∈ S ⊥⊥ .
8. Since z ∈ S ⊥⊥ and z ∈ S ⊥ .
9. Thus, z = 0, and x = y ∈ S .
10. Therefore, S ⊥⊥ ⊆ S .
4.7
�
Linear Functionals and the Riesz Representation Theorem
• A linear functional on a Hilbert space H is a linear map
F : H → C.
• Examples.
• C0∞ (R)
F( f ) = f (x0 )
• Note that there is no g ∈ C0∞ such that F( f ) = (g, f )
• L2 ([a, b])
• l02 .
∞
�
xn
F(x) =
n
n=1
There is no y = (yn ) such that F(x) = (y, x).
mathphys15.tex; October 28, 2015; 15:35; p. 83
86
CHAPTER 4. HILBERT SPACES
• �g| is a linear bounded functional defined by
ϕg ( f ) = (g, f )
with the norm
||ϕg || = ||g||
• Let x0 ∈ (a, b). Then ϕ( f ) = f (x0 ) is a linear but unbounded functional.
• Remarks. The set H � of all bounded linear functionals on a Hilbert space
is a Banach space, called the dual space.
• The dual space H � of a Hilbert space H is isomorphic to H.
• If f = 0, then the null space N( f ) = E, and, therefore, (N( f ))⊥ = {0} and
dim(N( f ))⊥ = 0.
• Lemma 4.7.1 Let E be an inner product space and f : E → C be a nonzero bounded linear functional on E. Then the orthogonal complement of
the null space
dim(N( f ))⊥ = 1.
Proof:
1. Since f is bounded and linear it is continuous.
2. Therefore, N( f ) is a closed subspace of E.
3. Thus, (N( f ))⊥ is not empty.
4. Let x, y ∈ (N( f ))⊥ be two nonzero vectors.
5. Then f (x) � 0 and f (y) � 0.
6. Therefore, there exists α � 0 ∈ C such that f (x + αy) = 0.
7. Hence, x + αy ∈ N( f ).
8. Since x, y ∈ (N( f ))⊥ , we also have x + αy ∈ (N( f ))⊥ .
9. Thus x + αy = 0.
10. Therefore, x and y are linearly dependent, and, therefore,
dim(N( f ))⊥ = 1.
�
mathphys15.tex; October 28, 2015; 15:35; p. 84
4.7. LINEAR FUNCTIONALS AND THE RIESZ REPRESENTATION THEOREM87
• There holds
E = N( f ) ⊕ (N( f ))⊥
• Theorem 4.7.1 Riesz Representation Theorem. Let H be a Hilbert space
and f : H → C be a bounded linear functional on H. There exists a unique
x0 ∈ H such that
f (x) = (x0 , x)
for all x ∈ H. Moreover,
� f �=� x0 � .
Proof:
1. (I). Existence.
If f = 0, then x0 = 0.
2. Suppose f � 0.
3. Then dim(N( f ))⊥ = 1.
4. Let u ∈ (N( f ))⊥ .
5. Then ∀x ∈ H,
x=y+z
where y = x − (x, u)u ∈ N( f ) and z = (x, u)u ∈ (N( f ))⊥ .
6. Therefore, f (y) = 0.
7. Further,
f (x) = f (z) = (x, u) f (u) = (x, x0 ),
where
x0 = ( f (u))∗ u.
8. (II). Uniqueness. Suppose there exists x0 and x1 such that ∀x ∈ H
f (x) = (x, x0 ) = (x, x1 ).
9. Then ∀x ∈ H
(x, x0 − x1 ) = 0.
10. Thus, (x0 − x1 ) ∈ H ⊥ = {0}.
11. So, x0 = x1 .
mathphys15.tex; October 28, 2015; 15:35; p. 85
88
CHAPTER 4. HILBERT SPACES
12. Finally, we have
� f �= sup
x�0
| f (x)|
|(x, x0 )|
= sup
≤� x0 � .
�x�
x�0 � x �
13. On another hand
| f (x0 )| |(x0 , x0 )|
=
=� x0 � .
� x0 �
� x0 �
14. Thus, � f �=� x0 �.
�
• Theorem 4.7.2 Let H be a Hilbert space and S be a subspace of H. Let
f : S → C be a bounded linear functional on S . There exists a unique
linear functional g : H → C such that
g(x) = f (x)
for all x ∈ S and
� g �=� f � .
Proof: Use Riesz theorem.
mathphys15.tex; October 28, 2015; 15:35; p. 86