(a) Recall the relation seen in the course which exists between Sp(P (A)) and SpA for a self-adjoint operator A. Show that the same result holds for an arbitrary bounded operator.

Université Denis Diderot 2012-2013

Mathématiques M1 Spectral Theory

Corrigé de l’examen duu 16 janvier 2013 Durée 3 heures. Documents interdits.

1. Let H be complex Hilbert space, A ∈ L(H) and P ∈ C[X].

(a) Recall the relation seen in the course which exists between Sp(P (A)) and SpA for a self-adjoint operator A. Show that the same result holds for an arbitrary bounded operator.

Answer — The relation is

Sp(P (A)) = {P (λ); λ ∈ SpA}.

It is proved by showing the inclusions Sp(P (A)) ⊂ {P (λ); λ ∈ SpA} and Sp(P(A)) ⊃ {P (λ); λ ∈ SpA}. For the first inclusion one takes any µ ∈ Sp(P (A)), decomposes P (X) − µ = a(X − λ

) · · · (X − λ

) and apply this identity by replacing X by A to conclude that µ = P (λ

), for some λ

∈ SpA. For the second inclusion we take any λ ∈ SpA and decompose the polynomial P(X) − P (λ) = (X − λ)Q(X) and use the same reasoning to conclude that P (λ) ∈ Sp(P(A)).

In both reasoning the same kinds of arguments are used, which do not rely on the fact that A is self-adjoint. Hence the result actually extend to any bounded operator.

(b) Prove that if P (A) = 0, then SpA is contained in the set of roots of P .

Answer — If P(A) = 0 we deduce from the previous question that {P (λ); λ ∈ SpA} = Sp(P(A)) = {0}. Hence any λ in SpA is contained in the set of roots of P . 2. Let H = L

( R , λ) be the Hilbert space of complex-valued square integrable functions

with respect to the Lebesgue measure. Let f : R → C be a measurable and λ-locally square integrable function on R (i.e. R

K

|f|

dλ < ∞ for all K ⊂ R compact). Define the unbounded operator M on H :

D(M ) = {ξ ∈ H : Z

|f ξ|

dλ < ∞} (M ξ)(x) = f (x)ξ(x) ξ ∈ D(M), x ∈ R . (a) Show that M is densely defined.

Answer — It suffices to show that D(M ) contains a dense subset of H, for instance the space L

(R, C) of essentially bounded measurable functions with compact sup- port. Indeed if ξ ∈ L

( R , C ), then supp ξ ⊂ K, a compact subset of R and there exists some M > 0 s.t. |ξ(x)| ≤ M , for a.e. x ∈ R . Hence R

|f ξ|

dλ ≤ M

R

K

|f |

dλ < +∞,

hence ξ ∈ D(M). Thus L

(R, C) ⊂ D(M ). The fact that L

(R, C) is dense in

H is a consequence of the fact that it contains the space C

( R , C ) of continuous

functions with compact support, which is known to be dense in H. Alternatively

one may prove directly that L

(R, C) is dense in H by proving that any function

ξ ∈ H is approximated by the sequence (ξ

)

with values in L

( R , C ) defined by

ξ

(x) := 1

(x) inf(|ξ(x)|, n)

and by using Lebesgue’s dominated convergence

theorem.

(b) Let T be the unbounded operator on H defined by

D(T ) = D(M) (T ξ)(x) = f (x)ξ(x) ξ ∈ D(T ), x ∈ R . Show that T is densely defined and M = T

. Deduce that M is closed.

Answer — First D(T) = D(M ) is dense by the result of the previous question. The domain of T

is the subspace of ζ ∈ H such that the linear form

D(T ) −→ C ξ 7−→ R

R

ζ(x)f(x)ξ(x)dλ(x)

admits a continuous linear extension to H. This linear form is uniquely defined be- cause D(T ) is dense in H. By the Riesz–Fréchet theorem, this property is equiva- lent to the fact that [x 7−→ f(x)ζ (x)] belongs to L

( R , C ). This latter condition is equivalent to ζ ∈ D(M ) and, if so, this linear map coincides with T . Hence D(T

) = D(T ) = D(M ) and T

= M . It follows that M is closed since it coin- cides with the adjoint operator of an operator defined on a dense domain.

(c) Compute M

.

Answer — By changing f 7−→ f , we change T to M . Hence by applying the previous results, we deduce that M

= T .

(d) Show that sp(M ) = EssIm(f ) where

EssIm(f ) = {λ ∈ C : ∀ > 0 λ(f

(B (λ, ))) > 0}.

Answer — We first show that EssIm(f ) ⊂ Sp(M ). Let λ ∈ EssIm(f ). Then for any n ∈ N

, λ[f

(B(λ, 1/n))] > 0. Let A

be a compact subset of f

(B(λ, 1/n)) such that λ(A

) > 0. Set ξ

:= 1

/ p

λ(A

), then R

|ξ

|

dλ = λ(A

)

R

An

dλ = 1.

Moreover, ∀x ∈ A

, |f(x) − λ| <

, so f |

is bounded by |λ| + 1/n. Hence T ξ

, defined by (T ξ

)(x) = f(x)ξ

(x), is in L

(R, C), so ξ

∈ D(T ). Now

k(T − λ)ξ

k

= Z

R

|f (x) − λ|

|ξ

(x)|

dλ ≤ 1 n

Z

An

|ξ

(x)|

dλ = 1 n

.

Since kξ

k

= 1, this shows that T − λ does not have a bounded inverse. Hence λ ∈ Sp(T ).

Let us now show that ( C \ EssIm(f)) ⊂ ( C \ Sp(T )), i.e. that, for any λ which does not belong to EssIm(f), T − λ is invertible. Indeed if λ 6∈ EssIm(f ), then ∃ε > 0 such that λ[f

(B(λ, ε))] = 0, i.e. λ({x; |f (x) − λ| < ε}) = 0. Hence, for a.e. x ∈ R ,

1

|f(x)−λ|

≤

. This proves that the operator ξ 7−→

is bounded, but this operator is the inverse of T − λ. Hence λ 6∈ Sp(T ).

(e) Show that if f is continuous then sp(M ) = f ( R ).

Answer — Because of the previous question, it suffices to show that, if f is conti- nuous, EssIm(f )) = f (R). On the one hand, if λ ∈ EssIm(f )), then, ∀ε > 0, f

(B(λ, ε)) has a non vanishing measure, thus is non empty, i.e. f ( R ) ∩ B(λ, ε) 6= ∅.

Hence λ ∈ f ( R ) (note that, here, we do not need f to be continuous). On the other

hand, if λ ∈ f (R), then ∀ε > 0, B (λ, ε) ∩ f (R) 6= ∅, i.e. f

(B(λ, ε)) 6= ∅. Since f is

continuous f

(B (λ, ε)) is also open, hence it has a non vanishing measure.

(f) Let A ⊂ R be a Borel subset such that λ(A) < ∞, ν be the finite measure on R defined by ν(B) = λ(A ∩ B) for all B Borel subset of R . Observe that the integral with respect to ν is

Z

gdν = Z

A

gdλ.

Let f

(ν) the measure image of ν by f i.e., f

(ν)(B) = ν (f

(B)) for all Borel subset B ⊂ C . Let ξ = 1

∈ H. Suppose that f is bounded and continuous. Show that M is bounded and the spectral measure µ

of M associated to ξ is µ

= f

(ν).

Answer — The fact that M is bounded follows from the fact that f is bounded, i.e.

there exists M > 0 such that |f (x)| ≤ M , ∀x ∈ R . The spectral measure µ

is defined by : ∀ϕ ∈ C( R , C ) (or for any Borelian function)

Z

ϕdµ

= hξ, ϕ(M )ξi = Z

R

ξ(x)ϕ(f (x))ξ(x)dλ = Z

A

ϕ ◦ f dλ = Z

ϕ ◦ f dν.

In the particular case where ϕ = 1

(where B is a Borelian subset of C ), Z

1

dµ

= Z

1

◦ f dν = Z

R

1

dν = ν(f

(B )) = f

(ν)(B) and we deduce that µ

(B) = R

1

dµ

= f

(ν )(B), for all Borelian subset B. This implies µ

= f

(ν).

3. Let H be a complex Hilbert space and T ∈ L(H) be self-adjoint.

(a) Show that : ∀n ∈ N , ||T

|| = ||T ||

.

Answer — We first show this result for n = 1, i.e. ||T

|| = ||T ||

. By using the fact that T

is self-adjoint and a result from the course, we have

||T

|| = sup

|hx, T

xi| = sup

|hT

x, T xi| = sup

|hT x, T xi|

= sup

||T x||

=

sup

||T x||

= ||T ||

.

Similarly, by using the fact that T

is self-adjoint, we show that ||T

|| = ||T

||

. We conclude by recursion.

(b) Show by recursion that the following property P(n) is true for any n ∈ N : P (n) : ∀k ∈ N such that 0 ≤ k ≤ 2

, ||T

|| = ||T ||

. [Hint : for 2

< k ≤ 2

, consider ` := 2

− k and T

T

.]

Answer — The property P(0) is straightforward (||T

|| = ||T||

et ||T || = ||T ||).

Assume that we have shown P (n) let us prove P (n + 1) : it is enough to show that

||T

|| = ||T||

for 2

< k ≤ 2

. Let ` := 2

− k, then 0 ≤ ` < 2

. Hence by using the result of the previous question, we get

||T ||

= ||T

|| = ||T

T

|| ≤ ||T

|| ||T

||.

We can then use P (n) with `, i.e. ||T

|| = ||T||

. We hence obtain

||T ||

≤ ||T ||

||T

||

which gives by simplification ||T ||

≤ ||T

||. The reverse inequality ||T ||

≥ ||T

|| is

actually true for any operator. We deduce ||T ||

= ||T

||.

Conclusion : we have shown that, for any self-adjoint operator T ∈ L(H),

∀n ∈ N, ||T

|| = ||T ||

.

4. Let H be a complex Hilbert space. The goal of this exercise is to show that there are no bounded self-adjoint

operators P, Q ∈ L(H) such that [P, Q] = λ1

, where 1

∈ L(H) is the identity operator, λ is a complex constant

and [P, Q] := P Q − QP .

(a) Assume that H has a finite dimension n. Show, by using a simple argument, that there are no operators P, Q ∈ L(H) such that [P, Q] = 1

.

Answer — On the one hand the trace of [P, Q] vanishes, on the other hand the trace of 1

is n.

(b) In the following we assume that the dimension of H is infinite. Show that, for any pair of operators P, Q ∈ L(H), we have :

∀n ∈ N

, [P

, Q] =

n

X

j=1

P

[P, Q]P

.

Answer — We write that [P

, Q] = P

Q − QP

= (P

Q − P

QP ) + (P

QP − P

QP

) + · · · + (P QP

− QP

)

= P

j=1

P

(P Q − QP )M

.

(c) We argue by contradiction and we assume that there exists operators P, Q ∈ L(H) such that [P, Q] = 1

. By using the result of the previous question and of the previous exercise, show that, for any n ∈ N

,

2||P || ||Q|| ≥ n and conclude to a contradiction.

Answer — First method (assuming that P and Q are self-adjoint) : Because of the previous question and the hypothesis [P, Q] = 1

, we have [P

, Q] = nP

. Hence, by using the result of Exercise 1,

n||P ||

= ||nP

|| = ||[P

, Q]|| ≤ 2||P

|| ||Q|| = 2||P||

||Q||,

which gives us, by dividing by ||P||

, that n ≤ 2||P|| ||Q||. Since n can be chosen arbitrarily, this is impossible, since P and Q are bounded.

Second method (without assuming that P and Q are selfadjoint) :

nkP

k = k[P, Q]k = kP

P Q − QP P

k ≤ kP

k(kPk kQk + kQk kP k)

implies n ≤ 2kP k kQk.

1. Sorry, this hypothesis was missing : the aim of this question was to use the result of the previous exercise on self-adjoint operators. We shall see, however, that the hypothesis thatP andQbe self-adjoint is not essential.

2. The reason for this other modification of the hypothesis (we set λ= 1in the original exam) is that, ifP andQ are self-adjoint, then[P, Q]is necessarily skew adjoint, i.e. satisfies[P, Q]^∗ =−[P, Q], so that condition [P, Q] =λ1H makes sense iffλis complex imaginary.

(d) Give an example of a complex Hilbert space H and two non bounded operators P and Q such that there exists a dense vector subspace V ⊂ H such that ∀ϕ ∈ V , [P, Q]ϕ = ϕ

Answer — We may choose H := L

(R, C), D(P ) = H

(R, C), P = i

, D(Q) = {f ∈ L

( R , C ); R

R

|xf (x)|

dx < +∞} and Q defined by (Qf )(x) = xf (x).

5. Let H := `

(Z, C) ' `

(Z). We denote by (

)

the canonical Hilbertian Hermitian orthogonal basis of H (i.e.

is the sequence which vanishes for all relative integer, excepted for n, for which it takes the value 1). We note L, R ∈ L(H) the operators defined by :

L

=

et R

=

, ∀n ∈ Z and A := L + R ∈ L(H).

(a) Compute L

and R

. Deduce that A is self-adjoint.

Answer — For all x = P

n∈Z

x

and y = P

n∈Z

y

, we have Ly = P

n∈Z

y

= P

n∈Z

y

and thus :

hx, Lyi = X

n∈Z

x

y

= X

n∈Z

x

y

= hRx, yi,

since Rx = P

n∈Z

x

= P

n∈Z

x

. Hence L

= R. Similarly we obtain that R

= L. We deduce then that A is self-adjoint.

(b) We note U : `

( Z ) −→ L

( R / Z ) the unitary operator defined by : (U

) (θ) = e

,

∀n ∈ Z (Fourier series isomorphism). For all function m ∈ L

( R / Z ) we note m b ∈ L L

( R / Z )

the multiplication operator defined by :

∀f ∈ L

(R/Z), ( mf b ) (θ) = m(θ)f (θ), p.p.

Find U LU

and U RU

and show that they coincide with multiplication operators by functions to be precised. Deduce U AU

.

Answer — For any function f ∈ L

( R / Z ) the Fourier series of which is f (θ) = P

n∈Z

a

e

, we have U

f = P

n∈Z

a

and thus LU

f = P

n∈Z

a

, which implies :

U LU

f (θ) = X

n∈Z

a

e

= e

X

n∈Z

a

e

= e

f(θ).

Hence U LU

coincides with the multiplication the multiplication operator by the function θ 7−→ e

. Similarly U RU

coincides with the multiplication operator by the function θ 7−→ e

. We deduce that U AU

= U LU

+ U RU

coincides with the multiplication operator by the function θ 7−→ e

+ e

= 2 cos(2πθ), i.e. U AU

= 2 cos(2πθ). \

(c) Let ψ ∈ H be different of 0 and g := U ψ. We note F

:= {P (A)ψ| P ∈ C [X]} =

Vect

{A

ψ| n ∈ N }.

Show that the U F

, the image of F

by U , is equal to : P

g := {f g| f ∈ P

}, where P

is the subspace of L

( R / Z ) of polynomials in e

and e

which are even function of θ.

Answer — A consequence of the previous question is that, ∀n ∈ N , U (A

ψ) = U A

U

U ψ = (U AU

)

g = (2 cos(2πθ))

g,

hence, more generally, for any polynomial P ∈ C [X], U(P (A)ψ) = P (2 cos(2πθ))g.

We deduce that U F

coincides with {P (2 cos(2πθ))g| P ∈ C [X]}. Hence it suffices to remark

that the set Q := {P (2 cos(2πθ))| P ∈ C [X]} coincides with P

Lastly the closure of P

dans L

( R / Z ) is the set of even functions of the variable θ.

(d) Let ψ and g be as in the preceding question. We define h ∈ L

( R / Z ) by : h(θ) = 2i sin(2πθ)g(−θ).

Show that, ∀f

, f

∈ P

, hf

g, f

hi

= 0.

Answer — For any functions f

, f

∈ P

, we have : hf

g, f

hi

=

Z

R/Z

f

(θ)g(θ)f

(θ)2i sin(2πθ)g(−θ)dθ,

which vanishes since the functions θ 7−→ f

(θ)f

(θ) and θ 7−→ g(θ)g(−θ) are even and the function θ 7−→ 2i sin(2πθ) is odd, so that hf

g, f

hi

is equal to the value of the integral of an odd function.

(e) Deduce from the preceding questions that A does not admit a cyclic vector.

Answer — For any ψ ∈ H \ {0}, let g := U ψ and χ := U

h ∈ H, where h is defined in terms of g as in the previous question. Then, because of Parseval’s inequality, for any polynomials P

, P

∈ C [X],

hP

(A)ψ, P

(A)χi = hU P

(A)ψ, U P

(A)χi

= hf

g, f

hi

,

where f

, f

∈ P

are such that f b

= U P

(A)U

and f b

= U P

(A)U

(recall that we denote by f b the multiplication operator by the function f). Hence, by the result of the previous question, hP

(A)ψ, P

(A)χi = 0. This implies that F

⊥ F

. Since F

6= {0}, we deduce that F

is not dense in H, i.e. that ψ is not cyclic for A.

3. Let us show it by recursion on N : the set PN :=n P

−N≤n≤Nane^i2πnθ|an∈C, a−n=an

o

is equal to QN :={P(2 cos(2πθ))|degP ≤N}. ForN = 0, the result is straightforward. Assume thatPN−1=QN−1 let us show thatPN=QN. We use the identity :

(e^i2πθ+e^−i2πθ)^N =

N

X

j=0

N!

(N−j)!j!e^{i2π(2j−N)θ}=e^{i2πN θ}+e^{−i2πN θ}+RN(θ), (1) whereRN(θ) =PN−1

j=1 N!

(N−j)!j!e^{i2π(2j−N)θ}. It is clear thatRN∈ PN−1=QN−1(note in particular thatRN(θ) = RN(−θ)). We deducee^{i2πN θ}+e^{−i2πN θ}∈ QN and(e^i2πθ+e^−i2πθ)^N ∈ PN. Any element ofPN readsa(e^{i2πN θ}+ e^{−i2πN θ}) +α(θ), whereα∈ PN−1, and is thus contained in QN. Similarly any element ofQN readsb(e^i2πθ+ e^−i2πθ)^N+β(θ), whereβ∈ QN−1, and is thus also contained inPN.