ME(EE) 550 Foundations of Engineering Systems Analysis
Chapter 06: Analysis of Bounded Linear Transformations
While some authors have used the terms ”Operator” and ”Transformation”
synonymously, others specifically restrict operators as those transformations that
map a vector space into itself. In this chapter, linear transformations, that map
a vector space V (over a field F) into (possibly) another vector space W (over
the same field F) is denoted as L(V, W ) and bounded linear transformations as
BL(V, W ). After introducing the basic concepts of bounded linear transformations,
this chapter focuses on bounded linear operators on Hilbert spaces that are denoted
as BL(H, H), where H is a Hilbert space. In Chapter 5, we have shown that the
dual space ℓ⋆2 is isometrically isomorphic to the ℓ2 space itself. By Reisz-Frechét
Theorem, introduced and proved in Chapter 5, it follows that any Hilbert space
H is isometrically isomorphic to its dual space H ⋆ . In this chapter, we show that
there exists an isometric isomorphism between any two Hilbert spaces, defined over
the same field, provided that they have the same Hilbert dimension. This chapter
should be read along with Chapter 5 and Chapter 6 of Naylor & Sell, where some
of the solved examples and exercises are very useful.
1
Basic Concepts of Bounded Linear Transformations
This section first presents background materials on bounded linear transformations
in Banach spaces beyond what have been addressed in Chapter 5. In particular,
the following four theorems, which are useful for analysis of a variety of advanced
engineering problems, are presented below.
1. Baire Category Theorem
2. Banach-Steinhaus Uniform Boundedness Theorem
3. Open Mapping Theorem
4. Closed Graph Theorem
1.1
Background for Baire Category Theorem
Let (X, d) be a metric space. Among all subsets of X let us define a class of large
sets and a class of small sets with the following properties:
(i) A large set is nonempty.
(ii) A set S ⊆ X is large if and only if its complement X \ S is a small set.
(iii) A countable intersection of large sets is a large set.
1
For example, let a probability measure P be defined on a measurable space (X, E).
Then, a measurable set L ⊆ X with probability P (L) = 1 is a large set and a
measurable S with P (S) = 0 is a small set.
Theorem 1.1. (Baire Category Theorem)
Let (X, d) be a complete metric space and let {Vk } be a sequence of (nonempty)
open, dense subsets of X. Then, V , ∩∞
k=1 Vk is a (nonempty) dense subset of X.
Proof. Let U ⊆ X be nonempty and open; and the choice of U is arbitrary. It
(
)
suffices to show that ∩∞
k=1 Vk ∩ U is nonempty.
Let x0 ∈ U and r0 ∈ (0, 1) such that B3r0 (x0 ) ⊆ U . Then, by induction, for
(
)
every k ≥ 1, choose xk ∈ U and rk > 0 such that B3rk (xk ) ⊆ Vk ∩ Brk−1 (xk−1 ) ;
this is possible because Vk is open and dense. For each k ≥ 1, it implies that
rk+1 ≤
rk
1
and d(xk+1 , xk ) ≤ rk < k ⇒ {xk } is a Cauchy sequence
3
3
Since the metric space (X, d) is complete, the Cauchy sequence {xk } converges to
some x⋆ ∈ X. Then, it follows, by triangular inequality, that
d(xk , x⋆ ) ≤
∞
∑
d(xj , xj+1 ) ≤
j=k
∞
∑
rj ≤
j=k
∞
∑
3k−j rk =
j=k
3
rk
2
Therefore, for every k ≥ 1, x⋆ ∈ B3rk (xk ) ⊆ Vk . For k = 0, the same argument
(
)
yields B3r0 (x0 ) ⊆ U . Hence, x⋆ ∈ ∩∞
k=1 Vk ∩ U .
1.2
Background for Uniform Boundedness Theorem
Let (V, ∥ • ∥V ) and (W, ∥ • ∥W ) be Banach spaces over the same field. Let Λ : V →
W be a linear bounded transformation (i.e., Λ ∈ BL(V, W )), where the induced
norm of Λ is defined as: ∥ Λ ∥ , sup∥x∥V =1 ∥ Λx ∥W . Based on some results of
Baire category theorem, Banach-Steinhaus uniform boundedness theorem is stated
and proved below.
Theorem 1.2. (Banach-Steinhaus Uniform Boundedness Theorem)
Let L ⊆ BL(V, W ) ba a family of bounded linear transformations and let Λ ∈ L.
Then, either L is uniformly bounded, i.e.,
sup ∥ Λ ∥< ∞
(1)
Λ∈L
or else there exists a dense set S ⊆ V such that
sup ∥ Λx ∥= ∞ ∀x ∈ S
(2)
Λ∈L
Proof. For every integer n ∈ N, let us consider an open set
Sn , {x ∈ V : ∥ Λx ∥> n for some Λ ∈ L}
(3)
If one of these sets, say Sk , is not dense in V , then there exists x0 ∈ V and radius
r > 0 such that the closed ball B̄r (x0 ) does not intersect Sk , i.e., ∥ Λx ∥≤ k for
some x ∈ B̄r (x0 ) and every Λ ∈ L. Now if ∥ x ∥≤ r, then
∀Λ ∈ L, ∥ Λx ∥ = ∥ Λ(x0 + x) − Λx0 ∥ ≤ ∥ Λ(x0 + x) ∥ + ∥ Λx0 ∥ ≤ 2k
2
Therefore,
∀Λ ∈ L, ∥ Λ ∥ , sup ∥ Λx ∥ =
∥x∥=1
1
r
sup ∥ Λx ∥ ≤
∥x∥=r
2k
r
In this case, the family of linear transformations L is uniformly bounded.
The other possibility is that the open sets Sn in Eq. (3) are all dense in V .
Then, by Baire category theorem, the intersection S , ∪n≥1 Sn is dense in V . By
construction, each x ∈ S and n ≥ 1, there exists a linear transformation Λ ∈ L
such that Λx ∥< n. Hence, Eq. (1) holds.
Remark 1.1. It follows from Banach-Steinhaus Uniform Boundedness Theorem
(Theorem 1.2) that if a family of linear transformation Λ ∈ BL(V, W ) is bounded
on the unit ball, then it is uniformly bounded, i.e., the following condition holds.
sup Λx ∥< ∞ ∀x with ∥ x ∥≤ 1
Λ∈L
This justifies the name “uniform boundedness principle.”
Corollary 1.1. (Continuity of the Pointwise Limit)
Let (V, ∥ • ∥V ) and (W, ∥ • ∥W ) be Banach spaces over the same field, and let
{Λk : k ∈ N} be a sequence of linear transformations with the pointwise limit
Λx = lim Λn x ∀x ∈ V
(4)
n→∞
Then, the map Λ in Eq. (4) is a bounded linear transformation.
Proof. The sequence {Λk : k ∈ N} is bounded for every x ∈ V . Hence, by
Theorem 1.2, the sequence {Λk } is uniformly bounded, which implies that
(
)
∥ Λ ∥ , sup ∥ Λx ∥ = sup
lim ∥ Λn x ∥ ≤ sup ∥ Λn ∥ < ∞
∥x∥≤1
∥x∥≤1
n→∞
n≥1
Hence, the linear transformation Λ is bounded.
1.3
Background for Open Mapping Theorem
First let us introduce the notion of open mapping from one metric space to another.
Definition 1.1. (Open Mapping) Let (X, dX ) and (Y, dY ) be metric spaces. Then,
a map f : X → Y is said to be (dX -dY ) open if, for every dX -open set A ⊆ X, the
image f (A) is dY -open.
The implication of Definition 1.1 is that, for every x ∈ X and r ∈ (0, ∞), the
(
)
image f Br (x) of the dX -open ball Br (x) at center x and radius r is a dY -open
set that contains a dY -open ball centered at f (x).
Theorem 1.3. (Open Mapping Theorem) Let (V, ∥ • ∥V ) and (W, ∥ • ∥W ) be
Banach spaces over the same field. If Λ : U → V is a linear, surjective and
bounded transformation, then Λ is open, i.e., for every ∥ • ∥V -open X ⊆ U , the
image Λ(X) is ∥ • ∥W -open.
Proof. The proof of open mapping theorem is presented in three steps.
3
Step 1 By linearity, the image of an open ball Br (x), where the ball radius r > 0
and x ∈ V , is expressed as:
(
)
(
)
(
)
Λ Br (x) = Λ(x) + Λ Br (0V ) = Λ(x) + rΛ B1 (0V )
(
)
To prove the theorem, it suffices to show that the image Λ B1 (0V ) contains
an open ball around the origin 0W in (W, ∥ • ∥W ).
(
)
∪∞
Step 2 Since Λ is surjective, W = n=1 Λ Bn (0V ) . Recalling that (W, ∥ • ∥W ) is
a Banach space, it follows from Baire Category Theorem (Theorem 1.1) that
(
)
at least one of the closures Λ Bn (0V ) has nonempty interior. By rescaling,
it follows that
(
)
)
1 (
Λ B1 (0V ) = Λ Bn (0V )
n
will also have non-empty interior, i.e., there exists w ∈ W and r > 0 such
(
)
that the ball Br (w) ⊂ Λ B1 (0V ) . Furthermore, since the unit ball B1 (0V )
(
)
is convex and symmetric, the same is true of its image Λ Bn (0V ) and of its
(
)
(
)
closure Λ Bn (0V ) . By symmetry Br (−w) ⊆ Λ B1 (0V ) , while convexity
implies that the ball Br (0W ) ⊂ W satisfies
(1
(
)
)
1
Br (0W ) =
Br (w) + Br (−w) ⊆ Λ B1 (0V )
2
2
Using the linearity of Λ and by rescaling, it follows that, for all n ≥ 1,
(
)
B2−n r (0W ) ⊆ Λ Bn (0V )
(
)
Step 3 The proof is concluded by showing that Br/2 (0W ) ⊆ Λ B1 (0V ) for all
n ≥ 1. Letting w ∈ Br/2 (0W ), w.e proceed by induction as follows.
• By density, we find v1 ∈ B2−1 (0V ) such that ∥ w − Λv1 ∥< 2−2 r.
• Next, we find v2 ∈ B2−2 (0V ) such that ∥ (w − Λv1 ) − Λv2 ∥< 2−3 r.
• Continuing by induction, for each n ∈ N, we have
w−
n−1
∑
(
)
Λvj ∈ B2−n (0V ) ⊆ Λ Bn (0V )
j=1
Therefore, we can select a point vn ∈ B2−n (0V ) such that
(
n−1
)
∑
−n−1
w −
Λv
−
Λv
r
j
n < 2
j=1
∑
∑
Since V is a Banach space and n ∥ vn ∥< ∞, the series n vn converges,
∑∞
say n=1 vn = v. Then, we observe that
∥ v ∥≤
∞
∑
n=1
∥ vn ∥<
∞
∑
2−n = 1 and Λv = lim
n→∞
n=1
n
∑
Λvj = w
j=1
Hence, the image Λ(B1 (0V ) contains all points w ∈ W with ∥ w ∥< 2r .
Remark 1.2. If (V, ∥ • ∥V and (W, ∥ • ∥W are Banach spaces over the same field,
and if Λ : U → V is a linear, continuous bijection, then the inverse Λ−1 : W → V is
continuous as well. This is so because of the following fact – since Λ is a bijection,
Λ is open if and only if Λ−1 is continuous.
4
1.4
Background for Closed Graph Theorem
Let (V, ∥ • ∥V ) and (W, ∥ • ∥W ) be Banach spaces over the same field. The product
space V × W is the set of all ordered couples (v, w) with v ∈ V and w ∈ W . The
product space V × W is a Banach space with the following norm:
∥ (v, w) ∥V ×W , ∥ v ∥V + ∥ w ∥W .
Definition 1.2. (Closed Graph) Let Λ be a (possibly unbounded) linear transformation with domain Dom(Λ) ⊆ V and values in W . Then, the linear transformation Λ is said to be closed if its graph
{
}
Graph(Λ) , (v, w) : v ∈ Dom(Λ), w = Λv
is a closed subset of the product space V × W .
In other words, Definition 1.2 states that the linear transformation Λ : V → W
is closed provided that the following condition holds:
Given two sequences of points vk ∈ Dom(Λ) and wk = Λvk ∈ W , if
vk → v and wk → w, then v ∈ Dom(Λ) and Λv = w.
It is noted that every continuous linear transformation Λ : V → W is closed.
The next result shows that the converse is also true provided that the domain
Dom(Λ) is the entire space V .
Theorem 1.4. (Closed Graph Theorem)
Let (V, ∥ • ∥V ) and (W, ∥ • ∥W ) be Banach spaces over the same field. If Λ : U → V
is a closed linear transformation defined on the entire space V , i.e., Dom(Λ) = V .
Then, Λ is continuous.
Proof. Let us denote Γ ≡ Graph(Λ). By Definition 1.2, Γ is a closed subspace of
the product space V × W ; hence, Γ is a Banach space as well. Now, let us consider
two projections, π1 : Γ → V and π2 : Γ → W , respectively defined as:
x , π1 (x, Λx), and Λx , π2 (x, Λx).
Since the projection map π1 is a (linear) bijection between Γ and V , the inverse π1−1
is continuous. Hence, the composition of two continuous functions Λ = π2 ◦ π1−1 is
continuous.
Example 1.1. Let the space V = C 0 (R), which is the space of all bounded continuous functions f : R → R, with norm ∥ f ∥= sup∥x∥=1 |f (x)|. Let Λ be the
differentiation operator defined as: Λf = f ′ . The domain of Λ is the subspace of
V defined as:
{
}
Dom(Λ) , f ∈ C 0 (R) : f ′ ∈ C 0 (R) = C 1 (R)
consisting of all continuously differentiable functions with bounded derivatives.
Next we observe that the linear transformation Λ is not bounded and hence
not continuous. For example, the functions fk (x) = sin kx are uniformly bounded,
i.e., ∥ fn ∥C 0 = 1 for all n ≥ 1. However, the sequence of derivatives Λfk = fk′ is
unbounded, because fk′ (x) = k cos kx and hence ∥ fk′ ∥C 0 = k.
5
On the other hand, the linear transformation Λ : f → f ′ has closed graph. To
see this, let us consider a sequence fk ∈ Dom(Λ) such that, for some functions
f, g ∈ C 0 (R), one may have
∥ fk − f ∥→ 0 and ∥ fk′ − g ∥→ 0.
If the above two conditions hold, then f is continuously differentiable and f ′ = g.
Hence, the point (f, g) ∈ V × V lies in the graph of Λ. Notice that this example
does not contradict the closed theorem (Theorem 1.4), because Λ is not defined on
the entire space V = C 0 (R), i.e., Dom(Λ) is a proper subspace of V .
2
Operators on Hilbert Spaces
Let H be a Hilbert space over a complete field F (i.e., R or C). It follows from
Reisz-Frechét Theorem that ∀y ∈ H there exists a unique functional y ⋆ ∈ H ⋆ such
that y ⋆ (x) = ⟨x, y⟩ ∀x ∈ H. Now, let T ∈ BL(H, H) be a bounded linear operator
mapping a Hilbert space H into itself such that
⟨T x, y⟩ = ⟨x, y ⋆ ⟩
In this context, we introduce the notion of adjoint operator T ⋆ : H → H defined by
T ⋆ y = y ⋆ , where y ⋆ ∈ H ⋆ is unquely defined as y ⋆ (x) , ⟨x, y⟩∀x ∈ H. The above
statement on adjoint operator is summarized below.
If H is a Hilbert space and H ⋆ be its dual space, then every functional
in H ⋆ uniquely identifies a vector in H, i.e., ∀y ⋆ ∈ H ⋆ ∃ a unique y ∈
H such that y ⋆ (x) = ⟨x, y⟩H ∀x ∈ H, and ∥y ⋆ ∥H ⋆ = ∥y∥H .
Definition 2.1. (Adjoint operator) Let H be a Hilbert space and let T ∈ BL(H, H).
Then, T ⋆ : H → H, called the (Hilbert-) adjoint operator of T , is defined as
⟨T x, y⟩ = ⟨x, T ⋆ y⟩ ∀x, y ∈ H
We denote y ⋆ = T ⋆ y ∀y ∈ H, where y ⋆ is a bounded linear functional on H, i.e.,
y⋆ ∈ H ⋆ .
Proposition 2.1. (Uniqueness of the adjoint operator) Let H be a Hilbert space.
Then, the adjoint operator T ⋆ is unique for every operator T ∈ BL(H, H),
Proof. The proof follows from Riesz-Frechèt Theorem (see Chapter 5).
Remark 2.1. The following identities are derived directly from Proposition 2.1.
(i) I ⋆ = I, where I is the identity operator, i.e., Ix = x ∀x ∈ H.
(ii) 0⋆L(H,H) = 0L(H,H) , where 0L(H,H) is the zero operator on H,
i.e., 0L(H,H) x = 0H ∀x ∈ H.
(iii) (αT )⋆ = α T ⋆ ∀T ∈ BL(H, H) and every scalar α.
(iv) (S + T )⋆ = S ⋆ + T ⋆ and (ST )⋆ = T ⋆ S ⋆ ∀S, T ∈ BL(H, H).
Theorem 2.1. Let H be a Hilbert space and let K ∈ BL(H, H). Then,
6
(i) K ⋆ ∈ BL(H, H).
(ii) K ⋆⋆ = K.
(iii) ∥K ⋆ ∥ = ∥K∥.
Proof. (i) Linearity: Let H be a Hilbert space over a field F. Let x, y, z ∈ H and
α, β ∈ F. Then, ⟨x, K ⋆ (αy+βz)⟩ = α⟨x, K ⋆ y⟩+β⟨x, K ⋆ z⟩ = ⟨x, K ⋆ αy⟩+⟨x, K ⋆ βz⟩.
Boundedness: ∥K ⋆ y∥2 = |⟨K ⋆ y, K ⋆ y⟩| = |⟨KK ⋆ y, y⟩| ≤ ∥KK ⋆ y∥∥y∥ ≤ ∥K∥∥K ⋆ y∥∥y∥
⇒ ∥K ⋆ y∥ ≤ ∥K∥ ∥y∥ ⇒ ∥K ⋆ | ≤ ∥K∥ < ∞.
(ii) ⟨x, Ky⟩ = ⟨Ky, x⟩ = ⟨y, K ⋆ x⟩ = ⟨K ⋆ x, y⟩ ⇒ ⟨x, Ky⟩ = ⟨K ⋆ x, y⟩ ∀x, y ∈ H.
Replacing K by K ⋆ in Definition 2.1, we have ⟨K ⋆ x, y⟩ = ⟨x, K ⋆⋆ y⟩ ∀x, y ∈ H
⇒ ⟨x, Ky⟩ = ⟨x, K ⋆⋆ y⟩ ∀x, y ∈ H ⇒ K ⋆⋆ = K.
(iii) It is shown in part (i) that ∥K ⋆ | ≤ ∥K∥ < ∞. To prove ∥K ⋆ ∥ = ∥K∥, it
suffices to show that ∥K∥ ≤ ∥K ⋆ |, which is obtained by replacing K with K ⋆ in
part (ii).
Definition 2.2. (Invariance) A subspace V of a vector space H is called invariant
under an operator T ∈ BL(H, H) if the following condition holds:
Since T (V ) ⊆ V, i.e., ∀x ∈ V, it follows that T x ∈ V
Theorem 2.2. Let H be a Hilbert space and let T ∈ BL(H, H). A closed subspace
V of H is invariant under the operator T if and only if V ⊥ is T ⋆ -invariant.
Proof. (only if part): Let the closed subspace V(of H be T -invariant,
)
( i.e., T V ⊆ V).
Then, ∀x ∈ V and ∀y ∈ V ⊥ , it follows that ⟨y, T x⟩ = 0 ⇒ ⟨T ⋆ y, x⟩ = 0 .
Hence, V ⊥ is T ⋆ -invariant.
⋆ ⊥
⊥
(if part) Let the (closed subspace) V ⊥ of H(be T ⋆ -invariant,
) i.e.,
( T V ⊆ V ).
Then, ∀x ∈ V and ∀y ∈ V ⊥ , it follows that ⟨T ⋆ y, x⟩ = 0 ⇒ ⟨y, T x⟩ = 0 .
Hence, V is T -invariant.
2.1
Convergence in the Space of Linear Bounded Operators
{
}
Let H be a Hilbert space over a field F. Let T k ∈ BL(H, H) : k ∈ N be a
sequence of bounded linear operators. The following definitions are introduced for
different notions of convergence of {T k }.
Definition 2.3. (Convergence in operator norm or uniform operator convergence)
{
}
Let T k ∈ BL(H, H) : k ∈ N be a sequence of bounded linear transformations
from H into H. Then, {T k } converges to some T ∈ BL(H, H) in the operator
norm, also called uniform convergence, if
lim
u
sup ∥(T − T k )x∥H = 0, which is denoted as T k → T
k→∞ ∥x∥H =1
{
}
Definition 2.4. (Strong convergence) Let T k ∈ BL(H, H) : k ∈ N be a sequence
of bounded linear transformations from H into H. Then, {T k } converges strongly
to some T ∈ BL(H, H) if
s
lim ∥(T − T k )x∥H = 0 ∀x ∈ H, which is denoted as T k → T
k→∞
7
{
}
Definition 2.5. (Weak convergence) Let T k ∈ BL(H, H) : k ∈ N be a sequence
of bounded linear transformations from H into H. Then, {T k } converges weakly
to some T ∈ BL(H, H) if
(
)
w
lim | f (T x) − f (T k x) | = 0 ∀x ∈ H ∀f ∈ H ⋆ which is denoted as T k → T
k→∞
Proposition 2.2. (Convergence in operator norm) ⇒ (Strong convergence) ⇒
(Weak Convergence). The converse is not true, in general.
Proof. (Convergence in operator norm) ⇒ (Strong convergence): Since ∀x ∈ H,
|(T − T k ) x∥H ≤ ∥(T − T k )∥ind ∥x∥H ,limk→∞ ∥(T − T k )∥ind = 0
s
⇒ limk→∞ ∥(T − T k )x∥H = 0 ∀x ∈ H, i.e., T k → T .
To show (Strong convergence) ⇒ (Weak convergence), we proceed as follows:
s
T k = T ⇒ limk→∞ ∥(T − T k )x∥H = 0 ∀x ∈ H. If f ∈ H ⋆ ; then, it follows from
linearity and boundedness of the functional f that f (x) ≤∥ f ∥ ∥ x ∥. Then,
(
)
w
lim f (T x) − f (T k x) = 0 ∀x ∈ H ∀f ∈ H ⋆ ⇒ T k → T
k→∞
We prove falsity of the converse by two counterexamples, one for each case.
(Strong convergence) ; (Convergence in operator norm): Let us define a sequence
of bounded linear operators T k : ℓ2 → ℓ2 ∀k ∈ N as:
T k x = {0, 0, 0, · · · , 0, ξk+1 , ξk+2 , · · · }, where x , {ξn : n ∈ N}
|
{z
}
f irst k terms
Clearly, T is a bounded linear operator, i.e., T k ∈ BL(ℓ2 , ℓ2 ). Since x ∈ ℓ2 , it
follows that
lim ∥T k x∥ℓ2 = 0 ⇒ lim ∥T k ∥ =s 0BL(ℓ2 ,ℓ2 )
k
k→∞
k→∞
However, the induced norm limk→∞ sup∥x∥ℓ =1 ∥T k x∥ℓ2 = 1 as seen by choosing
2
x = {0, 0, 0, · · · , 0, ξk+1 , ξk+2 , · · · } with ∥x∥ℓ2 = 1 ⇒ limk→∞ ∥T k ∥ ̸=u 0BL(ℓ2 ,ℓ2 ) .
|
{z
}
f irst k terms
Therefore, (Strong convergence) ; (Convergence in operator norm).
(Weak convergence) ; (Strong convergence): Let us define a sequence of
bounded linear operators T k : ℓ2 → ℓ2 ∀k ∈ N as:
T k x = {0, 0, 0, · · · , 0, ξ1 , ξ2 , · · · }, where x , {ξn : n ∈ N}
|
{z
}
f irst k terms
It is given that T is a bounded linear operator, i.e., T k ∈ BL(ℓ2 , ℓ2 ). Furthermore,
in this Hilbert space setting, it follows from the Riesz-Frechèt Theorem that every
f ∈ ℓ⋆2 can be represented as:
k
f (x) = ⟨x, y⟩ℓ2 =
∞
∑
ξn ηn , where y = {ηk : k ∈ N}
n=1
It follows by Cauch-Schwarz inequality that
|f (T k x)|2 = |⟨T k x, y⟩|2 ≤
∞
∑
|ξn |2
n=1
∞
∑
|ηm |2 → 0 as k → ∞
m=k+1
However,
∥T k x∥ℓ2 = ∥x∥ℓ2 ∀k ∈ N ⇒ lim ∥T k x∥ℓ2 ̸= 0 ∀x ̸= 0ℓ2 ⇒ lim T k ̸=s 0BL(ℓ2 ,ℓ2 )
k→∞
k→∞
Therefore, (Weak convergence) ; (Strong convergence).
8
2.2
Convolution in Time-varying Systems
Let H = L2 [α, β], where [α, β] ⊂ R. Let h : [α, β] × [α, β] → R be bounded in
L2 [α, β] × L2 [α, β] in the following sense, i.e.,
∫β
• ∥h(t, •)∥2 , α dτ |h(t, τ )|2 < ∞
∫β
• ∥h(•, τ )∥2 , α dt |h(t, τ )|2 < ∞
∫β ∫β
• ∥h∥2 , α dt α dτ |h(t, τ )|2 < ∞
( )
∫t
Let T ∈ BL(H, H) be defined as: T x (t) , α dτ h(t, τ )x(τ ) ∀x ∈ H ∀t ∈ [α, β].
Then, boundedness of T follows from the fact that, ∀x ∈ H,
2 ∫ β
∫ β
∫ β ∫ t
2
2
dτ h(t, τ )x(τ ) =
dt
∥T x∥ =
dt |(T x)| =
dt α
α
α
∫
β
≤
˙ 2=
dt ∥h(t, •)∥2 ∥x∥
α
∫
∫
β
β
dτ |h(t, τ )|2 < ∞
dt
α
α
⟨
⟩
h(t, •), x(•) α
Next we find the adjoint operator T ⋆ .
Proposition 2.3. (Adjoint operator T ⋆ ) The adjoint operator T ⋆ is given as:
∫
(T ⋆ y)(t)
β
dτ h(τ, t) y(τ ) ∀y ∈ H ∀t ∈ [α, β]
=
|{z}
t
L2 −sense
Proof. Following Fubuni’s theorem, the order of integration is exchanged as:
(∫ t
)
∫ β
∫ β
∫ β
∀x, y ∈ H, ⟨T x, y⟩ =
dt
dτ h(t, τ ) x(τ ) y(t) =
dτ x(τ )
dt h(t, τ ) y(t)
α
α
α
τ
By exchanging τ and t in the last double integral, it follows that
∫
∫
β
⟨T x, y⟩ =
α
∫
β
dτ h(τ, t) y(τ ) = ⟨x,
dt x(t)
t
β
dτ h(τ, t) y(τ )⟩
t
⟨ (
)⟩
∫β
Since ⟨T x, y⟩ = ⟨x, T ⋆ y⟩, we have x, T ⋆ y − t dτ h(τ, t) y(τ ) = 0 ∀x, y ∈ H.
Hence,
∫ β
⋆
(T y)(t) |{z}
=
dτ h(τ, t) y(τ ) ∀y ∈ H ∀t ∈ [α, β]
L2 −sense
t
Remark 2.2. It follows from from Proposition 2.3 that the adjoint of the impulse
response h(t, τ ) is h(τ, t) in the setting of the (real) time-varying impulse response.
The adjoint of the complex-valued time-varying impulse response h(t, τ ) is h(τ, t).
Notice that the arguments are interchanged in the adjoint operator. This is interpreted as the time t is running backwards for the adjoint operator. This becomes
very obvious if we consider the time-invariant case, where the adjoint of the impulse
response h(t − τ ) is h(τ − t).
Proposition 2.4. (Finite-dimensional Linear Systems) Let a finite-dimensional
linear time-varying dynamical system be modeled as:
dx(t)
= A(t)x(t) + B(t)u(t); x(to ) = xo and to ∈ R
dt
9
where the state vector x ∈ Cn and the input vector u ∈ Cm ; and the time-dependent
complex matrices A(t) and B(t) are of compatible dimensions and whose elements
are continuous functions of time t ∈ R. Since A(•) are B(•) are continuous,
there exists a unique solution of the above differential equation for a given initial
condition x(to ) = xo and an input profile {u(t) : t ∈ [to , ∞)}. The solution of the
above differential equation is:
∫
x(t) =
Φ(t, to )xo
| {z }
t
+
zero−input response
|
to
dτ Φ(t, τ )B(τ )u(τ ) ∀t ∈ [to , ∞)
{z
}
zero−state response
where the state transition matrix Φ(t, τ ) is the unique solution of the matrix differential equation
∂
Φ(t, τ ) = A(t) Φ(t, τ ); Φ(τ, τ ) = I, where I is the identity matrx.
∂t
with the following properties:
(i) Φ(t, t) = I ∀t ∈ [to , ∞), where I is the identity matrix.
(ii) Φ(t, τ ) = Φ(t, θ) Φ(θ, τ ) ∀t, τ ∈ R and the choice of θ ∈ R is arbitrary.
(iii) The matrix Φ(t, τ ) is invertible for all t, τ ∈ R).
Proof. The proof follows by substituting the proposed solution in the differential
equation as follows.
∫ t
dx(t)
∂Φ(t, to )
∂
=
xo +
dτ Φ(t, τ )B(τ )u(τ )
dt
∂t
∂t to
∫ t
∂
Φ(t, τ )B(τ )u(τ )
= A(t)Φ(t, to )xo + Φ(t, t)B(t)u(t) +
dτ
∂t
to
∫ t
= A(t)Φ(t, to )xo + B(t)u(t) +
dτ A(t)Φ(t, τ )B(τ )u(τ )
to
∫ t
[
]
= A(t) Φ(t, to )xo +
dτ Φ(t, τ )B(τ )u(τ ) + B(t)u(t) = A(t)x(t) + B(t)u(t
to
and also x(to ) = Φ(t0 , to )xo +
∫ to
to
dτ Φ(t, τ )B(τ )u(τ ) = xo .
Proposition 2.5. (Adjoint System) The adjoint of the finite-dimensional linear
dz(t)
H
system dx(t)
dt = A(t)x(t) is dt = −A (t)z(t), whose state transition matrices are
Φ(t, to ) and ΦH (to , t), respectively.
Proof. It follows from Proposition 2.4 that Φ(t, to )Φ(to , t) = Φ(t, t) = I. It follows,
by taking the time derivatives on both sides, that
)
∂(
∂Φ(t, to )
∂Φ(to , t)
Φ(t, to )Φ(to , t) = 0 ⇒
Φ(to , t) = −Φ(t, to )
∂t
∂t
∂t
⇒ A(t)Φ(t, to ) Φ(to , t) = −Φ(t, to )
∂Φ(to , t)
∂Φ(to , t)
⇒ A(t) = −Φ(t, to )
∂t
∂t
Taking Hermitian on both sides yields:
)
∂( H
Φ (to , t) ΦH (t, to ) = −AH (t)
∂t
Therefore, the state transition matrix of the system
10
dz(t)
dt
= −AH (t)z(t) is ΦH (to , t).
H
Remark 2.3. Note that, in the adjoint system dz(t)
dt = −A (t)z(t), the negative
sign on the right hand side implies that time is running backwards, i.e., t could
be replaced by −t and the adjoint of the operator A is A⋆ = AH for linear finitedimensional systems.
2.3
Unitary Operators and Equivalent Inner Product Spaces
Analogous to the concepts of homeomorphism in topological spaces and isomorphism in vector spaces, we introduce an analog of unitary equivalence in inner
product spaces.
Definition 2.6. (Unitary operators) Let (V, ⟨·, ·⟩V ) and (W, ⟨·, ·⟩W ) be inner product spaces over the same field. Then, V and W are said to be unitarily equivalent
if there is an isomorphism T : V → W of V onto W that preserves inner products,
i.e., ⟨T x, T y⟩W = ⟨x, y⟩V ∀x, y ∈ V . Then, it follows that T T ⋆ = T ⋆ T = I and
the operator T is called unitary.
Proposition 2.6. Let V and W be inner product spaces over the same field. A
mapping T is an isometric isomorphism of V onto W if and only if T is unitary.
Proof. The only issue here is to show that if ∥T x∥ = ∥x∥ ∀x ∈ V , then
⟨T x, T y⟩W = ⟨x, y⟩V ∀x, y ∈ V . The proof follows by use of the polarization
identity.
Remark 2.4. Let T : Fn → Fn be a unitary operator. Then, the n rows of T form
an orthonormal basis for Fn , and n columns of T form an orthonormal basis for Fn .
2.4
Normal and Self-adjoint Operators
This subsection is devoted to normal and self-adjoint operators.
Definition 2.7. (Normal and self-adjoint operators) Let H be a Hilbert space, and
let T ∈ BL(H, H). Then, T is said to be normal if T T ⋆ = T ⋆ T , i.e., if T commutes
with its adjoint T ⋆ ; and T is said to be self-adjoint if T = T ⋆ .
Remark 2.5. If T is self-adjoint, then T is normal and if T is unitary, then T is
normal. However, self-adjoint operators may or may not be unitary and vice versa.
While all orthogonal projections are self-adjoint, they are not unitary except for
the trivial cases of the identity operator I and the zero operator 0.
Proposition 2.7. The space of all self-adjoint operators on a Hilbert space H is
closed in BL(H, H).
Proof. Let {Lk } be a sequence of self-adjoint operators in BL(H, H). Let ∥L −
Lk ∥ → 0 as k → ∞, where L ∈ BL(H, H). It suffices to show that L is self-adjoint,
i.e., ⟨Lx, y⟩ = ⟨x, Ly⟩ ∀x, y ∈ H. Since Lk is self-adjoint for all k ∈ N, it follows
that |⟨Lx, y⟩ − ⟨x, Ly⟩| = |⟨Lx, y⟩ − ⟨Lk x, y⟩ + ⟨Lk x, y⟩ − ⟨x, Ly⟩|
= |⟨(L − Lk )x, y⟩ + ⟨x, (L − Lk )y⟩| ≤ 2∥L − Lk ∥∥x∥∥y∥ → 0 as n → ∞.
Remark 2.6. It is not true, in general, that if A and B are self-adjoint, then AB
is self-adjoint. However, if A and B are self-adjoint, then AB is self-adjoint if and
only if AB = BA, i.e., if and only if A and B commute.
11
Remark 2.7. Let H be a Hilbert space, and let T ∈ BL(H, H). The following
important results are presented below.
(i) If T is self-adjoint, then the operator norm of T is given by ∥T ∥ = sup∥x∥H =1 ⟨T x, x⟩
and ∥T ∥ = sup∥x∥H =1,∥y∥H =1 ⟨T x, y⟩.
(ii) A continuous projection P on H is orthogonal if and only if T is self-adjoint.
(iii) T is normal if and only if ∥T ⋆ x∥ = ∥T x∥ ∀x ∈ H.
(iv) The set of all normal operators on H is closed in BL(H, H) and is closed
under scalar multiplication.
(v) Let A and B be normal operators on H. Then, A+B, AB, and BA are normal
if AB = BA, i.e., if A and B commute with each other.
(vi) Let A and B be normal operators on H. Then, A + B, AB, and BA are
normal if AB = BA, or if AB ⋆ = B ⋆ A, i.e., if A commutes with the adjoint
of B, or if BA⋆ = A⋆ B, i.e., if B commutes with the adjoint of A.
3
Compact Linear Transformations
Infinite-dimensional compact linear operators are close cousins of finite-dimensional
linear operators that are inherently compact. However, compact operators neither
include normal operators, or are included in normal operators. Since the properties
of compact operators do not require to be defined on an inner product space, we
abandon such a restriction. Hence, we consider the general case of Banach spaces
that, of course, include Hilbert spaces. We also revert back to linear transformations
instead of being restricted to linear operators.
Definition 3.1. (Compact linear transformations) Let V and W be Banach spaces
over the same field. Let D , {x ∈ V : ∥x∥V ≤ 1} be a closed disk in V . Let
T ∈ L(V, W ); then, T is said to be compact if the image T (D) of D under T is
compact in W .
Remark 3.1. All finite-dimensional linear transformations are compact. However,
for example, the identity operator on an infinite-dimensional Banach space is not
compact. As an example, consider the ℓ2 space and the identity operator I : ℓ2 →
ℓ2 . The unit closed disk D , {x ∈ ℓ2 ; ∥x∥ℓ2 ≤ 1} is not compact in ℓ2 .
Proposition 3.1. Let T ∈ L(V, W ) where V and W are Banach spaces over the
same field. If the range space R(T ) is finite-dimensional, then T is compact.
Proof. Since R(T ) is finite-dimensional, the image T (D) is compact in R(T ).
Now we very briefly introduce the concepts of ε-net and total boundedness. For
more details, see Naylor & Sell, pp. 134-151.
Definition 3.2. (ε-net) Let S be a set in a metric space (X, d). Given ε > 0, a
subset Sε ⊆ S is called an ε-net in S if the following two conditions are satisfied.
1. ∀y ∈ S there exists x ∈ Sε such that d(x, y) < ε.
2. Sε is finite.
12
Definition 3.3. (Totally bounded) A set S in a metric space (X, d) is called totally
bounded if ∀ε > 0 there exists an ε-net Sε in S.
Remark 3.2. For all spaces, totally bounded ⇒ bounded. Only for finite-dimensional
spaces, bounded ⇒ totally bounded.
Theorem 3.1. (Compactness and Totally Boundedness) Let (S, d) be a metric
space. Then the following statements are equivalent.
1. The metric space (S, d) is compact.
2. The metric space (S, d) is sequentially compact.
3. The metric space (S, d) is closed and totally bounded.
4. Every infinite subset of S has an accumulation point.
Proof. See Naylor & Sell (pp. 142-145).
Remark 3.3. Let T ∈ L(V, W ) where V and W are Banach spaces over the same
field. The following important results are presented below.
(i) ∀ε > 0 there exists a finite-dimensional subspace M of the range space R(T )
such that inf m∈M ∥T x − m∥W ≤ ε. In other words, the finite-dimensional
space M is within the ε-radius of R(T ). Therefore, if R(T ) is infinitedimensional, then smaller ε is, larger would be the dimension of M .
(ii) The following statements are equivalent.
1. T is compact.
2. If B is any bounded set in V , then the image T (B) lies in a compact
subset of W .
3. If B is any bounded set in V , then the image T (B) lies in a sequentially
compact subset of W .
4. If {xk } is a bounded sequence of vectors in V , then {T xk } contains a
convergent subsequence in W .
5. If B is any bounded set in V , then the image T (B) is a totally bounded
set in W .
(iii) Let T ∈ L(V, W ) and s ∈ L(V, W ) be compact, where V and W are Banach
spaces over the same field. Then, S + T is compact.
(iv) Let T ∈ L(V, V ) be compact, where V is a Banach space. Then, both ST and
T S are compact.
(v) let {T k : V → W } be a sequence of compact linear transformations converging
to T ∈ BL(V, W ), i.e., ∥T k − T ∥ → 0 as k → ∞. Then, T is compact. Hence
the space of compact linear transformations is a closed subspace of bounded
linear transformations.
Theorem 3.2. (Convergence of Sequences Compact Transformation) Let V and
W be two normed vector spaces (defined over the same field F) and let T : V → W
be a compact linear transformation. Let {xk } be a weakly convergent sequence in V ,
w
s
i.e., xk → x. Then, {T xk } is a strongly convergent sequence in W , i.e., T xk → T x.
13
w
Proof. Let y k = T xk and y = T x. We first show that y k → y and then show that
s
y k → y. Let a bounded linear functional g ∈ W ⋆ be arbitrarily selected and let us
construct a functional f : V → F with f (z) 7→ g(T z). Therefore, for all z ∈ V ,
|f (z)| = |g((T z)| ≤ ∥ g ∥ ∥ T z ∥ ≤ ∥ g ∥ ∥ T ∥∥ z ∥
It is noted from the above inequalities that f is linear because both g and T are
linear; and f is bounded because T is compact and hence bounded.
w
Given that xk → x, which implies that f (xk ) → f (x), it follows that
w
g(T xk ) → g(T x), i.e., g(y k ) → g(y). Since the choice of g is arbitrary, y k → y.
s
Next we use the method of contradiction to prove that y k → y. Suppose it
is false; then, {y k } has a subsequence {y kn } such that ∥ y kn − y ∥≥ η for some
η ≥ 0. Now {xk } is bounded because it is weakly convergent. Compactness of
T implies existence of a convergent subsequence {z k } and let z k → z. Hence, it
follows that z = y Consequently, z k → y, which contradicts the assertion that
∥ z k − y ∥≥ η > 0.
4
Spectra of Bounded Linear Operators
We are familiar with eigenvalues and eigenvectors of square matrices that are representatives of bounded linear operators in finite-dimensional vector spaces. The
eigenvalues that belong to the complex field C form the spectrum σ(A) of a given
square matrix A and its complement ρ(A) , C \ σ(A) is called the resolvent set of
A. What can we say about the spectrum and resolvent set of a linear operator T
(that may not be bounded) in an infinite-dimensional vector space V ? It is possible
that an infinite-dimensional operator T ∈ L(V, V ) may not have an eigenvalue and
yet its spectrum is non-empty. Then, what are the contents of the spectrum σ(T )?
These issues are addressed in this section.
Let V be a Banach space and T ∈ L(V, V ), where V could be finite-dimensional
or infinite-dimensional, and let I : V → V be the identity operator on V , i.e.,
Ix = x ∀x ∈ V . In addition, let λ ∈ C.
Definition 4.1. If there exists x ∈ V \ {0V } such that T x = λx, then x is called an
eigenvector of T corresponding to the eigenvalue λ, and the null space N (λI − T )
of the operator λI − T is called the eigenmanifold of eigenspace corresponding to
the eigenvalue λ.
Definition 4.2. The resolvent set, denoted as ρ(T ), of T ∈ BL(V, V ), is a set
of complex numbers (i.e., ρ(T ) ⊂ C) that satisfies each of the following three
conditions for all λ ∈ ρ(T ):
• (λI − T ) : V → V is injective, i.e., N (λI − T ) = {0V }.
• The range R(λI − T ) is dense in V , i.e., R(λI − T ) = V .
• The inverse operator (λI − T )−1 on the range R(λI − T ) is bounded, i.e.,
(λI − T ) has a continuous inverse on on its range.
The spectrum of the operator is defined as: σ(T ) = C \ ρ(T ), i.e., σ(T ) = {λ ∈
C:λ∈
/ ρ(T ).
14
It follows from Definition 4.2 that the resolvent set and the spectrum of a linear
bounded operator T form a partitioning of the complex field C. Next we introduce
a further partitioning of the spectrum.
Definition 4.3. The spectrum σ(T ) of the linear operator T : V → V is further
partitioned as follows.
• The point spectrum σp (T ) = {λ ∈ σ(T ) : (λI − T ) is not injective}. In
other words, the point spectrum σ(T ) is the set of eigenvalues of T .
• The residual spectrum σr (T ) = {λ ∈ σ(T ) \ σp (T ) : R(λI − T ) V }. In
other words, (λI − T ) is injective and R(λI − T ) is not dense in V .
(
)
• The continuous spectrum σc (T ) = {λ ∈ σ(T ) \ σp (T ) ∪ σr (T ) : (λI −
T )−1 unbounded on R(λI − T )}. In other words, (λI − T ) is injective,
R(λI − T ) is dense in V , and the operator (λI − T )−1 does not have a
continuous inverse on its domain R(λI − T ).
Proposition 4.1. (Finite-dimensional linear operators) Let V be a finite-dimensional
vector space (of dimension n ∈ N) over the complex field C. Let T ∈ L(V, V ) and
σp (T ) = {λ ∈ C : det(λI − T ) = 0} and σc (T ) = ∅ and σr (T ) = ∅.
Proof. Because of the isomorphism between the spaces L(V, V ) and Cn×n , i.e., T
is isomorphic to an (n × n) matrix, it follows that
dim N (λI − T ) + dim R(λI − T ) = n
Then, it follows from the definition of the point spectrum that
σp (T ) = {λ ∈ C : det(λI − T ) = 0}
i.e., σp (T ) is the set of all eigenvalues of the matrix T .
If (λI − T ) is injective, i.e., dim N (λI − T ) = 0, then dim R(λI − T ) = n, i.e.,
R(λI − T ) = V . Therefore, λ ∈
/ σr (T ) ∀λ ∈ C. Hence, σr (T ) = ∅.
If (λI − T ) is injective and surjective (because R(λI − T ) = V ), then the
inverse (λI − T )−1 exists and is bounded for all λ ∈ C, because the space V is
finite-dimensional. Hence, σc (T ) = ∅.
Example 4.1. (Finite-degree polynomials) Let V = P n−1 (R) be the space of all
real polynomials for an arbitrary n ∈ N. Let T ∈ L(V, V ) be defined as: (T x)(t) ,
d
dt x(t). Then, σp (T ) = {0} and σc (T ) = ∅ and σr (T ) = ∅ and ρ(T ) = C \ {0}.
Proof. Since V ∼ Rn is finite-dimensional, it follows from Proposition 4.1 that
σr (T ) = ∅ and σc (T ) = ∅. The point spectrum
σp (T ) = {λ ∈ σ(T ) :
d
x(t) = λx(t) for some non-zero x ∈ P n−1 (R)}
dt
By equating the coefficients of polynomials, it follows that λ = 0 and the constant
polynomial x(t) = c is the corresponding eigenvector, where c is a non-zero real.
Hence, σp (T ) = {0} and ρ(T ) = C \ {0}.
15
Example 4.2. (Right shift operator Sr on ℓ2 (−∞, ∞)) Let H be the Hilbert
space ℓ2 (−∞, ∞) with the usual inner product. The right shift operator Sr :
ℓ2 (−∞, ∞) → ℓ2 (−∞, ∞) is defined as y = Sr x, where yk = xk−1 ∀k ∈ Z. Now we
show that σp (Sr ) = ∅, i.e., (λI − Sr ) is injective for all λ ∈ C. In other words, the
operator Sr : ℓ2 (−∞, ∞) → ℓ2 (−∞, ∞) has no eigenvalues.
Proof. Let x ∈ H, λ ∈ C such that (λI − Sr )x = 0H , i.e., λxk − xk−1 = 0 ∀k ∈ Z. If
λ = 0, then the only possible solution is x = 0H , which implies that Sr is injective.
For λ ̸= 0, it follows that x = {· · · , cλ2 , cλ, c, cλ−1 , cλ−2 , · · · }, where c is a scalar
constant. For an arbitrary nonzero λ, the sequence x ∈ H only if c = 0. Therefore,
λI − Sr is injective for all λ ∈ C. Therefore, σp (Sr ) = ∅.
It is proven in Naylor & Sell (pp. 416-420) that σr (Sr ) = ∅ and σc (Sr ) = {λ ∈
C : |λ| = 1}, i.e., the circumference of the unit circle with the center at the origin
in the complex plane.
Example 4.3. (Left shift operator Sℓ on ℓ2 (−∞, ∞)) Let H be the Hilbert space
ℓ2 (−∞, ∞) with the usual inner product. The left shift operator Sℓ : ℓ2 (−∞, ∞) →
ℓ2 (−∞, ∞) is defined as y = Sℓ x, where yk = xk+1 ∀k ∈ Z. In a similar way, it
can be shown that σp (Sℓ ) = ∅, i.e., (λI − Sℓ ) is injective for all λ ∈ C. In other
words, the operator Sℓ on ℓ2 (−∞, ∞) has no eigenvalues and that σr (Sℓ ) = ∅ and
σc (Sℓ ) = {λ ∈ C : |λ| = 1}. Note that Sℓ (ℓ2 (−∞, ∞)) is the adjoint as well as the
inverse of Sr (ℓ2 (−∞, ∞)).
Example 4.4. (Right shift operator Sr0 on ℓ2 [0, ∞)) Let H be the Hilbert space
ℓ2 [0, ∞) with the usual inner product. The right shift operator Sr0 : ℓ2 [0, ∞) →
ℓ2 [0, ∞) is defined as y = Sr0 x = {0, x0 , x1 , x2 , · · · }. It is proven in Naylor & Sell
(pp. 416-420) that σp (Sr0 ) = ∅ and σr (Sr0 ) = {λ ∈ C : |λ| < 1} and σc (Sr0 ) = {λ ∈
C : |λ| = 1}. Consequently, ρ(Sr0 ) = {λ ∈ C : |λ| > 1}.
Example 4.5. (Left shift operator Sℓ0 on ℓ2 [0, ∞)) Let H be the Hilbert space
ℓ2 [0, ∞) with the usual inner product. The left shift operator Sℓ0 : ℓ2 [0, ∞) →
ℓ2 [0, ∞) is defined as y = Sℓ0 x = {x1 , x2 , x3 , · · · }. It is proven in Naylor & Sell (pp.
422-423) that σp (Sr0 ) = {λ ∈ C : |λ| < 1} and σr (Sr0 ) = ∅ and σc (Sr0 ) = {λ ∈ C :
|λ| = 1}. Consequently, ρ(Sr0 ) = {λ ∈ C : |λ| > 1}. Note that Sℓ (ℓ2 [0, ∞)) is the
adjoint as well as the left inverse of Sr (ℓ2 [0, ∞)).
Next we establish an important property of bounded self-adjoint operators on
a complex Hilbert space.
Proposition 4.2. (Resolvent Set) Let H be a complex Hilbert space and let T ∈
BL(H, H) be a self-adjoint operator, i.e., T ⋆ = T . Then, λ ∈ ρ(T ) if and only if
there exists a real positive c, i.e., c ∈ (0, ∞), such that
∥ (λI − T ) ∥≥ c ∥ x ∥ ∀x ∈ H
Proof. See Kreyszig (pp. 461-463).
Proposition 4.3. (Real spectrum of a self-adjoint Hilbert operator)) Let H be a
complex Hilbert space and let T ∈ BL(H, H) be a self-adjoint operator, i.e., T ⋆ = T .
Then, σ(T ) ⊂ R.
16
Proof. Let λ ∈ σ(T ) ⊂ C. We will show that λ ∈ R, i.e., λ̄ = λ. Let λ = a + ib,
where a, b ∈ R and we will prove that b = 0.
Since T is self-adjoint, it follows from the property of an inner product that
⟨T x, x⟩ = ⟨x, T x⟩ and ⟨T x, x⟩ = ⟨x, T x⟩, which imply that ⟨T x, x⟩ is real. Therefore, ⟨(λI − T )x, x⟩ = λ̄⟨x, x⟩ − ⟨T x, x⟩. Since both ⟨T x, x⟩ and ⟨x, x⟩ are real,
⟨(λI − T )x, x⟩ − ⟨(λI − T )x, x⟩ = (λ − λ̄)⟨x, x⟩ = i2b ∥ x ∥2
Since the left side of the above equation is twice the imaginary part of ⟨(λI −
T )x, x⟩, it follows that
|b| ∥ x ∥2 ≤ |⟨(λI − T )x, x⟩| ≤∥ ⟨(λI − T )x ∥∥ x ∥
Then, for any x ̸= 0H , it follows that |b| ∥ x ∥≤∥ (λI −T ) ∥. By Proposition 4.2,
λ ∈ ρ(T ) if |b| > 0. Therefore, b = 0 if and only if λ ∈
/ ρ(T ), i.e. λ ∈ σ(T ).
Proposition 4.4. (Empty residual spectrum) Let H be a complex Hilbert space and
let T ∈ BL(H, H) be a self-adjoint operator, i.e., T ⋆ = T . Then, σr (T ) = ∅.
Proof. For a proof by contradiction, let us assume that σr (T ) ̸= ∅, i.e., there exists
a complex number λ ∈ σr (T ). Then, it follows from Definition 4.3 that (λI − T ) :
H → H is injective and its range R(λI − T ) is not dense in H. Then, by Riesz
representation Theorem, it follows that there exists a nonzero vector y ∈ N (λI −T ),
i.e., ⟨(λI − T )x, y⟩ = 0 ∀x ∈ H.
Since T is self-adjoint, it follows from Proposition 4.3 that λ is real. Therefore,
⟨(λI − T )x, y⟩ = ⟨x, (λI − T )y⟩ = 0 ∀x ∈ H. Hence, (λI − T )y = 0H and y ̸= 0H ,
which implies that (λI − T ) is non-injective, i.e., λ ∈
/ σr (T ). So, σr (T ) = ∅.
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