Lecture 3 Spectrum of operators -...

Lecture 3

Spectrum of operators

1 Resolvent set, spectrum, point spectrum

As usual, we denote by H a Hilbert space. Let T : D(T ) → H be a closedlinear operator with D(T ) dense in H.

The resolvent set of T , denoted ρ(T ), is the set of all λ ∈ C such thatλI − T is a bijection (from D(T ) to H) and (λI − T )−1 : H → H is abounded operator.

The spectrum of T , denoted σ(T ), is the set C \ ρ(T ).By definition, we have ρ(T )∪σ(T ) = C and ρ(T )∩σ(T ) = ∅. In addition,

ρ(T ) is an open subset of C, and consequently σ(T ) is a closed subset of C.In order to study the spectrum of T , it is useful to classify subsets of

σ(T ) according to the reason why its elements fail to be in the resolventset. There are different ways of doing such classification (and we will nottalk much about this). In this course, we are particularly interested in thesubset of σ(T ) (called point spectrum of T ) that consists of eigenvalues ofT (in the classical sense of linear algebra). Here is the definition of pointspectrum:

A number λ ∈ C is called an eigenvalue of T if λI − T is not injective,that is, Ker(λI −T ) 6= 0. The point spectrum of T , denoted σp(T ), is theset of all eigenvalues of T .

Clearly σp(T ) ⊂ σ(T ).Let λ ∈ σ(T ). A vector x ∈ D(T ) such that Tx = λx and x 6= 0 is called

an eigenvector of T associated to λ.

Example 1. Suppose that M is an n×n matrix that is diagonalizable (thisis the case if M is symmetric or hermitian). Then there exists a matrix Vsuch that V −1MV = D, where D is a diagonal matrix. The diagonal entriesof D are the eigenvalues of M . The columns of V are the correspondingeigenvectors of M . In this example σ(T ) = σp(T ).

More generally, if T : H → H is a bounded operator and dim(H) <∞,then σ(T ) = σp(T ).

Example 2 (Multiplication operator). Let (X,Σ, µ) be a σ-finite measurespace, and let F : X → C be a measurable function that is finite almost

September 10, 2016 1 http://bit.ly/specgeo2016

everywhere on X. Set

MF :

D(MF )→ L2(X,µ)

u 7→ Fu

with D(MF ) = u ∈ L2(X,µ) | Fu ∈ L2(X,µ). We have the followingproperties:

(i) D(MF ) is dense in L2(X,µ).

(ii) If F is real-valued, then MF is self-adjoint.

(iii) If F is bounded, then MF is bounded on L2(X,µ) and

‖MF ‖ = ‖F‖L∞(X,µ).

Furthermore

ρ(MF ) = λ ∈ C | ∃ ε > 0 such that |λ−F (x)| ≥ ε almost everywhere

andσp(MF ) = λ ∈ C | µ(x ∈ X | F (x) = λ) > 0.

For instance, if (X,Σ, µ) = (R,B, dx) and F (x) = x, we obtain σp(MF ) = ∅.

2 Resolvent operator

Let T : D(T )→ H be an operator. The resolvent of T at λ ∈ ρ(T ), denotedRλ(T ), is the operator defined by

Rλ(T ) = (λI − T )−1.

It follows from the definition that Rλ(T ) is a bounded operator on H.The resolvent of T satisfies the following identities for all λ, µ ∈ ρ(T ):

• Rλ(T )−Rµ(T ) = (µ− λ)Rλ(T )Rµ(T ).

• Rλ(T )Rµ(T ) = Rµ(T )Rλ(T ).

In addition, the mapping ρ(T )→ B(H)

λ 7→ Rλ(T )

is analytic. (Recall that B(H) denotes the set of all bounded operatorson H.) The notion of analytic function from complex analysis generalizesto functions with values in Banach spaces. Many features of the theory ofanalytic functions generalizes as well. Here, we just mention one definition


that characterizes analytic functions in a general setting: Let Ω be an opensubset of C and let X be a Banach space. A mapping f : Ω→ X is said tobe analytic at z0 ∈ Ω if

limz→z0

f(z)− f(z0)

z − z0exists. If f : Ω→ X is analytic at all z0 ∈ Ω, then f is said to be analytic.

Theorem 1. Let T be a bounded operator on H. For all z ∈ C such that|z| > ‖T‖, the operator (zI − T ) is bijective, the operator (zI − T )−1 isbounded, and

(zI − T )−1 =∞∑j=0

1

zj+1T k.

Since the mapping λ 7→ Rλ(T ) is analytic, we can use the theory ofanalytic functions to study this mapping and obtain information about thespectrum of T . The proof of the next proposition illustrates this idea. Ituses the above theorem and properties of analytic functions.

Proposition 1. Let T be a bounded operator on H. Then σ(T ) 6= ∅.

Idea of proof. By contradiction. Suppose that σ(T ) = ∅. Then Rλ(T ) is anentire analytic function. In addition, for |z| > ‖T‖, we have

Rλ(T ) =1

z

∑j=0

(T

z

)j.

Thus ‖Rλ(T )‖ → 0 as |z| → ∞. Hence, by Liouville’s Theorem, we wouldhave Rλ(T ) = 0 for all z ∈ C. This is a contradiction because Rλ(T ) 6= 0.Therefore σ(T ) 6= ∅.

Theorem 2. Let T : D(T ) → H be a closed operator such that ρ(T ) 6= ∅.For any λ0 ∈ ρ(T ), we have

σp(Rλ0) \ 0 =

1

λ0 − µ

∣∣∣∣ µ ∈ σp(T )

and

σ(Rλ0) \ 0 =

1

λ0 − µ

∣∣∣∣ µ ∈ σ(T )

.

Proof of the first equality. Suppose that λ ∈ σp(Rλ0(T )) with λ 6= 0. Thenthere exists ϕ ∈ H such that (λ0I−T )−1ϕ = λϕ and ϕ 6= 0. Since (λ0I−T )is a bijection from D(T ) to H, we conclude that ϕ ∈ D(T ). Calculating, we


obtain

Tϕ = Tλ−1(λ0I − T )−1ϕ

= −λ−1(λ0I − T − λ0I)(λ0I − T )−1ϕ

= λ−1(λ0Rλ0(T )− I)ϕ

= λ−1(λ0λ− I)ϕ

= µϕ

with µ = λ0 − λ−1. Thus µ ∈ σp(T ) and λ = (λ0 − µ)−1. Therefore

σp(Rλ0(T )) \ 0 ⊂ (λ0 − µ)−1 | µ ∈ σp(T ).

Similarly we prove that

(λ0 − µ)−1 | µ ∈ σp(T ) ⊂ σp(Rλ0(T )) \ 0 .

This proves the first equality of the theorem.

3 The self-adjoint case

Theorem 3. Let T : D(T ) → H be a self-adjoint operator. We have thefollowing properties:

(i) σ(T ) ⊂ R.

(ii) Eigenvectors associated to different eigenvalues are orthogonal.

To prove the Part (i) of the above theorem, we will use the followingproposition:

Proposition 2. Let T : D(T )→ H be a self-adjoint operator. If there existsC > 0 such that ‖(λI − T )u‖ ≥ C‖u‖ for all u ∈ D(T ), then λ ∈ ρ(T ).

Proof of Theorem 3. (i) For z = λ+ iµ with µ 6= 0, we have

‖(zI − T )u‖2 = ‖(λI − T )u‖2 + |µ|2‖u‖2

≥ |µ|2‖u‖2

for all u ∈ D(T ). By the above proposition, we conclude that z ∈ ρ(T ).Since ρ(T ) and σ(T ) are disjoint, we have z ∈ σ(T ) only if <(z) = 0.Therefore σ(T ) ⊂ R.

(ii) Let u and v be eigenvectors of T associated to different eigenvaluesλ and η, respectively. Then

λ〈u, v〉 = 〈Tu, v〉 = 〈u, Tv〉 = 〈u, v〉η.

Hence 〈u, v〉 = (λ− η)−10 = 0. Thus u and v are orthogonal.


Theorem 4. Let T be a bounded self-adjoint operator on H. Define

m = infu∈H‖u‖=1

〈Tu, u〉 and M = supu∈H‖u‖=1

〈Tu, u〉.

We have σ(T ) ⊂ [m,M ] and m,M ∈ σ(T ).

Proposition 3. Let T be a bounded self-adjoint operator on H. We have

‖T‖ = supu∈H‖u‖=1

|〈Tu, u〉|.

Corollary 1. Let T be a bounded self-adjoint operator on H. If σ(T ) = 0,then T = 0.

Proof. Apply the theorem and the proposition above.

4 The spectral theorem

Theorem 5 (Spectral theorem—Multiplication operator form). Supposethat T : D(T )→ H is a self-adjoint operator. There exist

1. a measure space (X,µ),

2. a measurable function F : X → R,

3. a unitary operator U : H → L2(X,µ)

such thatUTU−1 = MF ,

where MF is the multiplication operator

MF :

D(MF )→ L2(X,µ)

u 7→ Fu

with D(MF ) = u ∈ L2(X,µ) | Fu ∈ L2(X,µ). If H is separable, then µcan be chosen to be a finite measure.

References

[1] O. Lablee, Spectral Theory in Riemannian Geometry, EMS Textbooksin Mathematics, European Mathematical Society, 2015.

[2] P. D. Hislop and I. M. Sigal, Introduction to Spectral Theory WithApplication to Schrodinger Operator, Applied Mathematical Sciences113, Springer, 1996.


Lecture 3 Spectrum of operators -...

Documents

Transcript of Lecture 3 Spectrum of operators -...