1. Let A, B, T ∈ B(H) with T = T ∗ . We suppose that AT = T B and T A = BT . (a) Let g ∈ C( R ) be an even function. Show that g(T ) commutes with A and B.

Université Denis Diderot 2010-2011

Mathématiques M1 Spectral Theory

Examen Durée 3 heures

1. Let A, B, T ∈ B(H) with T = T ^∗ . We suppose that AT = T B and T A = BT . (a) Let g ∈ C( R ) be an even function. Show that g(T ) commutes with A and B.

(b) Let f ∈ C( R ) be an odd function. Show that Af (T ) = f (T )B.

Answer — We first show by recursion on n the property :

P n : T ²ⁿ A = AT ²ⁿ , T ²ⁿ B = BT ²ⁿ , T ²ⁿ⁺¹ A = T ²ⁿ⁺¹ B, T ²ⁿ⁺¹ B = AT ²ⁿ⁺¹ . For n = 0, P 0 is a straightforward consequence of the hypotheses AT = T B and T A = BT . Now assume that P n has been proved. Then we write :

T ²ⁿ⁺² A = T ²ⁿ⁺¹ T A = T ²ⁿ⁺¹ BT = AT ²ⁿ⁺¹ T = AT ²ⁿ⁺²

and we obtain by a similar computation that T ²ⁿ⁺² B = AT ²ⁿ⁺² . Similarly we obtain T ²ⁿ⁺³ A = T ²ⁿ⁺² T A = T ²ⁿ⁺² BT = BT ²ⁿ⁺² T = BT ²ⁿ⁺³

and also T ²ⁿ⁺³ B = AT ²ⁿ⁺³ . Hence P n+1 holds. It follows that, for any odd polynomial P , we have AP (T) = P(T )B and, for any even polynomial Q, [A, Q(T )] = [B, Q(T )] = 0. Now we use the fact that T is self-adjoint, so that we can define a continuous functional calculus C ⁰ (SpT, C ) ∋ f 7−→ f(T ) ∈ B(H), which satisfies in particular ||f (T )|| _{B(H )} ≤ ||f || _C

₍ Sp T ) . Moreover by the Stone–Weierstrass theorem the set of restrictions of polynomial functions on SpT is dense in C ⁰ (SpT, C ). Hence for any, say, even function g ∈ C ⁰ (SpT, C ), there exists a sequence of polynomials (Q _n ) _n s.t. Q _n | Sp T converges to g. Then one proves easily that if we set : Q b _n (x) = ¹ ₂ (Q _n (x) + Q _n (−x)), then Q b _n is obviously even and Q b _n | Sp T

converges to g also in C ⁰ (SpT, C ) and hence Q b _n (T) converges to g(T ) in B(H). Since [ Q b _n (T ), A] = [ Q b _n (T), B] = 0, (a) follows. The proof of (b) is similar.

2. Let U be the bilateral shift on ℓ ² ( Z ) ie, U e _n = e _n+1 where (e _n ) _n∈ Z is the canonical basis.

We recall that the spectrum of U is the unit circle. Let k ∈ Z . Show that e _k is a cyclic vector for U and compute the associated spectral measure.

Answer — The vector e _k is cyclic iff the vector subspace spanned by (U ⁿ (U ^∗ ) ^m e _k ) _n,m∈ N

is dense in ℓ ² ( Z ). Since U ^∗ = U ⁻¹ (this can be seen from the very definition of U or as

a consequence of the fact that U is unitary because its spectrum is a subset of the unit

circle), we have U ⁿ (U ^∗ ) ^m = U ^n−m , so that this amounts to show that the vector subspace

spanned by (U ⁿ e _k ) _n∈ Z is dense in ℓ ² ( Z ). But this is clearly the fact, for (e _n+k ) _n∈ Z being

a Hilbert basis of ℓ ² ( Z ).

The spectral measure dµ _e

on the circle S ¹ ⊂ C is defined by the relation

∀f ∈ C ⁰ (S ¹ , C ), Z

S

f (e ^iθ )dµ _e

(θ) = he k , f (U )e _k i.

By using a Fourier series decomposition f (e ^iθ ) = P

n∈ Z f b _n e ^inθ and the fact that U ^∗ = U ⁻¹ , we obtain f (U ) = P

n∈ Z f b _n U ⁿ . Hence he _k , f (U )e _k i = X

n∈ Z

he _k , f b _n U ⁿ e _k i = X

n∈ Z

f b _n he _k , e _n+k i = f b ₀ = 1 2π

Z

S

f (e ^iθ )dθ, so that

dµ _e

(θ) = 1 2π dθ.

3. Let a = (a _n ) _n∈ N be a sequence of complex numbers and M _a be the following operator on ℓ ² ( N )

D(M _a ) = {x = (x _n ) _n ∈ ℓ ² ( N ) : X

n

|a _n | ² |x _n | ² < ∞}, M _a x = (a _n x _n ) _n . (a) Show that M _a is densely defined and closed.

Answer — The fact that D(M _a ) is dense in ℓ ² ( N ) is a consequence of the fact that D(M a ) contains all finite sequences and that finite sequences are dense in ℓ ² ( N ). We now need to show that GrM _a is a closed subspace of ℓ ² ( N )×ℓ ² ( N ). Consider a sequence (x ^k , M _a x ^k ) _k∈ N with values in GrM _a and assume that (x ^k , M _a x ^k ) converges to some (x, y) ∈ ℓ ² ( N )×ℓ ² ( N ), when k → +∞. Setting x ^k = (x ¹ ₀ , x ^k ₁ , · · · ) and x = (x ₀ , x ₁ , · · · ), this implies immediately that, for any fixed n ∈ N , lim _k→∞ x ^k _n = x _n in C and hence that lim _k→∞ a n x ^k _n = a n x n in C . Hence since we also have lim _k→∞ a n x ^k _n = y n , this implies that a _n x _n = y _n , ∀n ∈ N . Hence, since y ∈ ℓ ² ( N ), P

n∈ N |a n | ² |x n | ² < +∞ and we deduce that x ∈ D(M a ) and y = M _a x, i.e. (x, y) ∈ GrM _a . Hence GrM _a is closed.

(b) Show that sp(M _a ) = {a n : n ∈ N }.

Answer — We first observe that any value a _n is an eigenvalue of M _a for at least the eigenvector e n . Hence {a n : n ∈ N } ⊂ Sp _p M a ⊂ SpM a . Since we know from the course that the spectrum of any operator is closed, this implies that {a n : n ∈ N } ⊂ SpM _a . Now let b ∈ C \ {a n : n ∈ N }. Then, since {a n : n ∈ N } is closed, there exists some ε > 0 s.t. B(b, ε) ∩ {a n : n ∈ N } = ∅. We will then first show that M a − b is a bijection between D(M a ) and ℓ ² ( N ) : given some y ∈ ℓ ² ( N ) we need to prove that there exists an unique x ∈ D(M a ) s.t. (M _a − b)x = y. If such an x would exist, it would be the unique solution of the equation (a n − b)x n = y n ⇐⇒ x n = y n /(a n − b),

∀n ∈ N . Lastly one needs to prove that (b − M _a ) ⁻¹ : ℓ ² ( N ) −→ ℓ ² ( N ) is bounded, a consequence of :

||(b − M _a ) ⁻¹ y|| ² = X

n∈ N

|y n | ²

|b − a _n | ² ≤ X

n∈ N

|y n | ²

ε ² = ||y|| ² ε ² .

Hence the resolvent set of M _a contains C \ {a _n : n ∈ N }, which is equivalent to say

that SpM _a ⊂ {a n : n ∈ N }.

4. Let a = (a _n ) _n∈ Z be a sequence of complex numbers indexed by Z and T _a be the operator defined on ℓ ² ( Z ) by

D(T a ) = {x = (x n ) _n∈ Z ∈ ℓ ² ( Z ) : X

n∈ Z

|a n | ² |x −n | ² < ∞}, T a x = (a n x −n ) _n∈ Z . (a) Show that T _a is densely defined and closed.

Answer — Use the same method as in question 3 – (a).

(b) Compute T _a ^∗ .

Answer — We first compute its domain D(T _a ^∗ ) : this is the set of y ∈ ℓ ² ( Z ) s.t. the linear form

D(T a ) −→ C x 7−→ hy, T a xi

admits a continuous extension on ℓ ² ( Z ) (which is then unique since D(T a ) is dense in ℓ ² ( Z )). By Riesz’ theorem this property is equivalent to say that there exists some z ∈ ℓ ² ( Z ) s.t. hy, T _a xi = hz, xi, ∀x ∈ D(T _a ). But

hy, T a xi = X

n∈ Z

y _n a _n x _−n = X

n∈ Z

y _−n a _−n x _n . Hence such a z exists, its satisfies z _n = a _−n y _−n . Thus y ∈ D(T _a ^∗ ) iff P

n∈ Z |a −n y _−n | ² <

+∞, i.e.

D(T _a ^∗ ) = {y ∈ ℓ ² ( Z ) : X

n∈ Z

|a n y _n | ² < +∞} and ∀y ∈ D(T _a ^∗ ), T _a ^∗ y = (a _−n y _−n ) _n∈ Z . (c) Find a necessary and sufficient condition on a for T _a to be normal.

Answer — For any x ∈ D(T a T _a ^∗ ) := {x ∈ D(T _a ^∗ ) : T _a ^∗ x ∈ D(T a )}, T a T _a ^∗ x = T a ((a −n x −n ) _n ) = (a n (a n x n ) _n ) = |a n | ² x n

n , wheras for any x ∈ D(T _a ^∗ T a ),

T _a ^∗ T a x = T _a ^∗ ((a n x −n ) _n ) = (a −n (a −n x n ) _n ) = |a −n | ² x n

n ,

Whatever D(T a T _a ^∗ ) and D(T _a ^∗ T a ) are, they contain the space of finite sequences. Hence a necessary condition for T _a to be normal is that, for any finite sequence x = (x _n ) _n ,

|a n | ² x _n = |a −n | ² x _n , ∀n ∈ Z , which implies that :

|a n | = |a −n |, ∀n ∈ Z .

Conversely it is clear from the preceding computation that, if the above condition holds, then D(T a T _a ^∗ ) = D(T _a ^∗ T _a ) and T _a T _a ^∗ = T _a ^∗ T _a , i.e. T _a is normal.

(d) Find a necessary and sufficient condition on a for T a to be bounded.

Answer — The operator T _a is bounded iff D(T _a ) = ℓ ² ( Z ) and there exists a constant

C ∈ [0, +∞) s.t. ∀x ∈ ℓ ² ( Z ), ||T a x|| ≤ C||x||. Testing this condition with x = e _n ,

for any n ∈ Z implies that |a n | ≤ C. Conversely, it is easy to check that the latter

condition implies that T _a is bounded. Hence the necessary and sufficient condition

is : the sequence (a _n ) _n∈ Z is bounded.

(e) Compute sp(T _a ^∗ T _a ) and sp(T _a T _a ^∗ ).

Answer — We have seen in question c) that T _a ^∗ T _a has the domain {x ∈ ℓ ² ( Z ) : P

n∈ Z |a −n | ⁴ |x n | ² < +∞} and is defined by T _a ^∗ T a (x) = (|a −n | ² x n ) n . Hence by a reasoning similar to the question 3 – (b), we deduce that the spectrum of T _a ^∗ T _a is {|a −n | ² : n ∈ Z } = {|a n | ² : n ∈ Z }. Similarly T _a T _a ^∗ has the domain {x ∈ ℓ ² ( Z ) : P

n∈ Z |a n | ⁴ |x n | ² < +∞} and is defined by T _a ^∗ T a (x) = (|a n | ² x n ) n . Hence the spectrum of T _a T _a ^∗ is also {|a n | ² : n ∈ Z }.

(f) Find a necessary and sufficient condition on a for T _a ^∗ T _a (resp. T _a T _a ^∗ ) to be compact.

Answer — Assume that T _a ^∗ T a is compact. Note also that this operator is self-adjoint.

Then it follows from the course that the spectrum of T _a ^∗ T _a is equal to {0} ∪ Λ, where Λ is a subset of C \ {0} which is at most countable and has no accumulation point, excepted may be 0. In particular, for any r > 0, SpT _a ^∗ T a ∩ ( C \ B(0, r ² )) is finite. Moreover Λ is composed of eigenvalues λ associated with finite dimensional vector eigenspaces. However we have seen in the preceding question that SpT _a ^∗ T _a = {|a n | ² : n ∈ Z } and the dimension of the eigenspace corresponding to any value λ ∈ Λ is the cardinal of {n ∈ Z : |a n | ² = λ}. We hence deduce that the number of values n ∈ Z s.t. |a n | ² ≥ r ² is finite. In particular, if we set N (r) := sup{|n| ∈ Z : |a n | ² ≥ r}, we have ∀n ∈ Z s.t. |n| > N (r), |a n | < r. Hence lim _|n|→∞ a n = 0.

Conversely if lim _|n|→∞ a _n = 0, then we define for any r > 0 the operator K _r on ℓ ² ( Z ) by

K _r x = X

n∈ Z ;|a

|

≥r

|a −n | ² x _n e _n .

Then ||T _a ^∗ T _a − K _r || ≤ r ² , so that lim _r→0 ||T _a ^∗ T _a − K _r || = 0 and each K _r is a finite rank operator. Hence T _a ^∗ T _a is compact.

A similar reasonning shows that T _a T _a ^∗ is compact iff the same condition holds, i.e.

|n|→∞ lim a _n = 0.

5. Let H be a complex separable Hilbert space. Let A be a non bounded self-adjoint operator such that its domain D(A) is dense in H. We assume that A is positive, i.e. that ∀ϕ ∈ D(A), hϕ, Aϕi ≥ 0.

(a) recall the spectral decomposition theorem for A ;

Answer — There exists a topological space X endowed with a Borelian measure µ, an unitary operator U : H −→ L ² (X, µ; C ) and a real-valued Borelian ¹ function f : X −→ C s.t., for any function ϕ ∈ U (D(A)),

(U AU ⁻¹ ϕ)(x) = f (x)ϕ(x), for µ − a.e. x ∈ X.

In the following we will write L ² (X) = L ² (X, µ; C ) for shortness. Note that the positivity condition translates here as ∀ϕ ∈ U (D(A)), R

X f (x)|ϕ(x)| ² dµ(x) ≥ 0.

Moreover since D(A) is dense in H, U (D(A)) is dense in L ² (X) and hence the latter condition implies that f is real and f ≥ 0, µ−a.e.

1. not necessarily bounded

(b) Prove that, for any t > 0, one can define the operator e ^−tA and that this operator is bounded on H ;

Answer — Since the function x 7−→ e ^−tf(x) is Borelian and bounded, it suffices to define e ^−tA by : ∀ϕ ∈ L ² (X),

(U e ^−tA U ⁻¹ ϕ)(x) = e ^−tf(x) ϕ(x), for µ − a.e. x ∈ X.

This operator is bounded and its norm is equal to the supremum of x 7−→ e ^−tf(x) over X.

Beware : it is not possible to use the series P

n≥0 (−tA) ⁿ /n!, which does not converge since A is not bounded.

(c) Prove that ∀t, s > 0, e ^−tA e ^−sA = e ^−(t+s)A ;

Answer — This follows from the definition of e ^−tA and the fact that e ^−tf(x) e ^−sf(x) = e ^−(t+s)f(x) .

(d) Prove that ∀t 0 > 0, lim _t>0,t→0 e ^−tA = e ^−t

^A ;

Answer — We need to evaluate the limit of ||e ^−tA − e ^−t

^A || H , when t → t ₀ . Without loss of generality we can assume that t 0 /2 < t < 3t 0 /2. For any u ∈ H, let ϕ := U u ∈ L ² (X), so that u = U ⁻¹ ϕ. Then

||(e ^−tA − e ^−t

^A )u|| ² _H = ||(e ^−tf − e ^−t

^f )ϕ|| ² _L

= Z

X

|e ^−tf(x) − e ^−t

^{f (x)} | ² |ϕ(x)| ² dµ(x).

By the mean value theorem, for any x ∈ X, there exists some τ between t ₀ and t s.t. |e ^−tf(x) − e ^−t

^f(x) | = | − f(x)e ^{−τ f(x)} (t − t 0 )| = |f(x)|e ^−t

^{f (x)/2} |t − t 0 | ≤ C|t − t 0 |, where C := sup _z>0 ze ^−t

^z/2 (actually C = 2/et ₀ ). Hence

||(e ^−tA −e ^−t

^A )u|| ² _H ≤ C ² (t−t ₀ ) ² Z

X

|ϕ(x)| ² dµ(x) = C ² (t−t ₀ ) ² ||ϕ|| ² _L

= C ² (t−t ₀ ) ² ||u|| ² _H and ||e ^−tA − e ^−t

^A || _H ≤ C|t − t 0 |, which implies the result.

(e) Prove that ∀u ∈ D(A),

t>0,t→0 lim

e ^−tA u − u

t = −Au

Answer — For any u ∈ D(A) and ϕ := U u ∈ L ² (X), so that u = U ⁻¹ ϕ, we have, for t > 0 :

e ^−tA − 1 t + A

u

2 H

=

e ^−tf − 1 t + f

ϕ

2

L

= Z

X

e ^−tf(x) − 1

t + f (x)

2

|ϕ(x)| ² dµ(x)

Using the mean value theorem, we know that, for any x ∈ X, there exists some τ ∈ (0, t) s.t.

e ^−tf(x) − 1

t + f(x) = −f (x)e ^{−τ f (x)} + f(x) = f (x)(1 − e ^{−τ f (x)} ),

and hence

e ^−tA − 1 t + A

u

2 H

= Z

X

|1 − e ^{−τ f(x)} | ² f(x)|ϕ(x)| ² dµ(x).

The function |1 − e ^{−τ f(x)} | ² f (x)|ϕ(x)| ² is dominated by f (x)|ϕ(x)| ² , which is µ- integrable and it converges pointwise to 0 when t (and hence τ ) converges to 0.

Hence we can apply Lebesgue’s dominated convergence theorem to conclude that

e

u−u

t − Au converges to 0.

(f) We define D(A ⁿ ) for all n ∈ N recursively by : D(A ⁰ ) := H and ∀n ∈ N , D(A ⁿ⁺¹ ) :=

{u ∈ D(A ⁿ )| A ⁿ u ∈ D(A)}. Prove that ∀t > 0, ∀n ∈ N , ∀u ∈ H, e ^−tA u ∈ D(A ⁿ ).

Answer — This is just a consequence of the fact that, for any n ∈ N , x 7−→

f (x) ⁿ e ^−tf(x) is bounded on X, since f is a nonnegative function.

Supplementary questions (an application) ²

(g) Let Ω ⊂ R ⁿ be a bounded domain with smooth boundary. Let A be the non bounded operator on L ² (Ω, C ) with domain D(A) = H ² (Ω, C ) ∩ H ₀ ¹ (Ω, C ) and such that ∀u ∈ D(A), Au = −∆u. Show that A is symmetric.

Answer — We must show that D(A) ⊂ D(A ^∗ ) and that A and A ^∗ coincides on D(A). In other words it suffices to check that, for any u ∈ D(A), the linear form

D(A) −→ C

v 7−→ hu, Avi

admits a continuous extension on L ² (Ω), which coincides with v 7−→ hAu, vi. But, since u and v belongs to H ² (Ω, C ) ∩ H ₀ ¹ (Ω, C ), we have

hu, Avi = − Z

Ω

u∆v = − Z

Ω

v∆u + Z

∂Ω

v ∂u

∂n − u ∂v

∂n = − Z

Ω

v∆u = hAu, vi, where we have used the fact that the boundary integral cancels because u and v vanish on ∂Ω (u, v ∈ H ₀ ¹ (Ω, C )).

In the following we will admit that A is self-adjoint ; (h) Prove that, for any f ∈ L ² (Ω, C ), there exists ³ a C ¹ map

u : (0, +∞) −→ D(A) t 7−→ u(t, ·) such that :

2. One could use the following formula, valid for f, g ∈ H

(Ω, C ) Z

∂Ω

f ∂g

∂n − g ∂f

∂n = Z

Ω

f ∆g − g∆f,

where

denotes the derivatives with respect to the exterior normal vector to the boundary ∂Ω .

3. Sorry, there were a slight mistake in the subject : the equation was

(t, x) = ∆u(t, x) (and not =

− ∆u(t, x)).

– u(0, ·) = f ;

– ∀t > 0, u(t, ·) ∈ C ^∞ (Ω) and u(t, ·) vanishes on ∂Ω ; – ∀t > 0, ∀x ∈ Ω,

∂u

∂t (t, x) = ∆u(t, x).

Answer — We just take u(t, x) = e ^t∆ f

(x) = e ^−tA f

(x), ∀t ≥ 0, ∀x ∈ X, with the convention that, for t = 0, e ^−0·A = 1 is the identity map.

We first note that, ∀t > 0, e ^t∆ maps L ² (Ω) to ∩ n∈ N H ₀ ⁿ (Ω, C ), because of the result of question (f) (where H ₀ ⁿ (Ω, C ) is the space of functions f in L ² (Ω, C ) such that all derivatives of f of order less than or equal to n are in L ² (Ω, C ) and such that f vanishes on ∂Ω). But this space is contained in the space of C ^∞ functions on Ω which vanish on ∂Ω.

The map u is clearly continuous on (0, +∞), as a consequence of question (d). It is moreover continuous at t = 0 : this can be proved by starting the same computation as in question (d), which gives

||e ^−tA u − u|| ² _H = Z

X

|e ^−tf(x) − 1| ² |ϕ(x)| ² dµ(x)

and by using Lebesgue’s dominated convergence theorem as in question (e). However u is not C ¹ at 0 : we can only prove that, for any f ∈ D(A), the map t 7−→ e ^t∆ f is derivable at 0, with a derivative equal to ∆f , as a straightforward consequence of question (e). Lastly u is continuously derivable on (0, +∞), since, we can write, for any f ∈ L ² (Ω, C ) and for any t > 0 :

u(t + s, ·) − u(t, ·)

s = e ^(t+s)∆ f − e ^t∆ f

s = e ^s∆ e ^t∆ f

− e ^t∆

s f

and we remark that e ^t∆ f ∈ D(A) because of question (f). Hence we can apply question (e) with the function e ^t∆ f to prove that this quotient has a limit when s goes to 0, which is equal to ∆ e ^t∆ f

= ∆u(t, ·). Hence ^∂u _∂t (t, x) = ∆u(t, x), ∀t > 0,

∀x ∈ X. The fact that t 7−→ ^∂u _∂t (t, ·) = ∆u(t, ·) is continuous can be seen by writing

∆u(t, ·) = ∆ e ^t∆ f

= e ^(t−τ)∆ ∆e ^τ∆ f ,

for some fixed τ such that 0 < τ < t : question (f) says us again that, for any

f ∈ L ² (Ω, C ), ∆e ^τ∆ f belongs to L ² (Ω, C ) and question (d) says us that its image by

e ^(t−τ)∆ depends continuously on t.