For the following we recall that kernel linear operators with a kernel in L 2 are compact.

Universit´ e Denis Diderot 2014-2015

Math´ ematiques M1 Spectral Theory

Corrig´ e de l’examen du 5 janvier 2015 Dur´ee 3 heures

For the following we recall that kernel linear operators with a kernel in L ² are compact.

1. 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 ^k 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 }

2. 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−→ h y, T a x i

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. h y, T _a x i = h z, x i , ∀ x ∈ D (T _a ). But

h y, T _a x i = 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 1 – (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.

3. In the following α ∈ (0, + ∞ ). The space L ² ( R , C ) is endowed with the Hermitian product h· , ·i L

defined by h f, g i L

:= R

R f(x)g(x)dx, ∀ f, g ∈ L ² ( R , C ) and we set k f k L

= h f, f i ^1/2 _L

. The space H ¹ ( R , C ) is endowed with the Hermitian product h· , ·i α defined by : h f, g i α :=

R

R (f ^′ (x)g ^′ (x)) + α ² f (x)g(x))dx, ∀ f, g ∈ H ¹ ( R , C ) and we set k f k α = h f, f i ^1/2 α . (a) For any f ∈ C c ^∞ ( R , C ) and we define

(R α f )(x) = e ^αx Z +∞

x

e ^−αy f (y)dy, (L α f)(x) = e ^−αx Z x

−∞

e ^αy f (y)dy.

Compute (R _α f ) ^′ in function of R _α f and of f ; compute (L _α f ) ^′ in function of L _α f and of f .

Answer — (R _α f ) ^′ = αR _α f − f , (L _α f ) ^′ = − αL _α f + f.

(b) Let f ∈ C c ^∞ ( R , C ), R > 0 such that supp(f) ⊂ [ − R, R] and k f k ∞ = sup _x | f (x) | . Show

that | R _α f(x) | ≤ Ce ^αx and | L _α f(x) | ≤ Ce ^−αx , where you can express C in terms of

R and k f k ∞ . Study the support of R _α f and L _α f . Deduce that R _α f, L _α f ∈ L ² ( R , C ).

Answer — | R _α f(x) | ≤ 2 k f k ^{∞ sinhαR} _α e ^αx and | L _α f(x) | ≤ 2 k f k ^{∞ sinhαR} _α e ^−αx . Moreo- ver supp(R _α f) ⊂ ( −∞ , R] and supp(L _α f ) ⊂ [ − R, + ∞ ). Hence

k R _α f k ² _L

, k L _α f k ² _L

≤ k f k ² ∞

2 sinh ² αR α ^3/2 e ^2αR .

(c) For any f ∈ C ^∞ c ( R , C ), we define (Σ _α f )(x) = _2α ¹ (R _α + L _α ). Show that ( − _dx ^d

+ α ² )Σ α f = f and that Σ α f ∈ H ¹ ( R , C ).

Answer — Compute, observe that (Σ _α f ) ^′ = ¹ ₂ (R _α f − L _α f ) and use the previous question.

(d) Show that the injection map j : H ¹ ( R , C ) −→ L ² ( R , C ), f 7−→ f is continuous.

Answer — k f k L

≤ _α ¹ k f k α .

(e) Show that, for any f ∈ L ² ( R , C ), there exists a unique u ∈ H ¹ ( R , C ) such that :

∀ ϕ ∈ H ¹ ( R , C ), h u, ϕ i α = h f, ϕ i L

.

Prove that this defines a linear bounded operator S α : L ² ( R , C ) −→ H ¹ ( R , C ) such that S _α f = u.

Answer — Let L L

: L ² ( R , C ) −→ (L ² ( R , C )) ^′ be the Riesz anti-isomorphism, i.e. such that ∀ g ∈ L ² ( R , C ), ( L L

f )(g) = h f, g i L

. Similarly define the Riesz anti- isomorphism L α : H ¹ ( R , C ) −→ (H ¹ ( R , C )) ^′ by ( L α f )(g) = h f, g i α , ∀ f, g ∈ H ¹ ( R , C ).

Then S _α = L ^∗ α ◦ j ^∗ ◦ L L

.

In the following we make a frequent use of the property : ∀ ϕ ∈ H ¹ ( R , C ), h S α f, ϕ i α = h f, ϕ i L

.

(f) Abusing notations, we also denote by S α the operator L ² ( R , C ) −→ L ² ( R , C ), f 7−→

S _α f . Show that S _α is self-adjoint. What can we say about the spectrum of S _α ? Answer — Actually S _α is nonnegative because of h S _α f, ϕ i α = h f, ϕ i L

( R , C ), ∀ f ∈ L ² , ∀ ϕ ∈ H ¹ ( R , C ) : indeed, by letting ϕ = S α f, this implies h f, S α f i L

= h S α f, S α f i α = k S _α f k ² α ≥ 0. Hence S _α is self-adjoint and its spectrum is contained in [0, + ∞ ) and is composed only of eigenvalues and of continuous spectral values (i.e. does not contain residual spectral values).

(g) Show that, ∀ f ∈ C c ^∞ ( R , C ),

∀ ϕ ∈ H ¹ ( R , C ), h Σ _α f, ϕ i α = h f, ϕ i L

and deduce that the restriction of S _α on C c ^∞ ( R , C ) coincides with Σ _α .

Answer — For any f ∈ C c ^∞ ( R , C ), set u = Σ _α f . Then, by the results of question (c), u ∈ H ¹ ( R , C ) and − u ^′′ + α ² u = f . Hence

∀ ϕ ∈ H ¹ ( R , C ), Z

R

ϕ(x)( − u ^′′ (x) + α ² u(x))dx = Z

R

ϕ(x)f (x)dx which is equivalent to :

∀ ϕ ∈ H ¹ ( R , C ), Z

R

(ϕ ^′ (x)u ^′ (x) + α ² ϕ(x)u(x))dx = Z

R

ϕ(x)f(x)dx,

i.e. the required result. The fact that S _α | C

( R , C ) = Σ _α follows from the uniqueness

of the solution of this problem shown in question (e).

(h) We let G _α ∈ C ⁰ ( R ² , R ) be defined by G _α (x, y) = 1

2α e ^−α|x−y| .

Find an explicit expression for Σ _α in terms of G _α . Does G _α belong to L ² ( R ² , R ) ? Answer — For any f ∈ C c ^∞ ( R , C ),

(Σ α f )(x) = Z

R

G α (x, y)f (y)dy

because of the result of Question (c). We observe that G _α does not belong to L ² ( R ² , R ). Indeed

Z

R

Z

R | G _α (x, y) | ² dxdy = 1 4α ²

Z

R

dx Z

R

dye ^{−2α|x−y|}

which gives by the change of variable z = x − y

= 1 4α ²

Z

R

dx Z

R

dze ^−2α|z| = 1 4α ²

Z

R

dx dz

α = + ∞ .

(i) We consider a smooth function U ∈ C c ^∞ ( R , R ) with compact support suppU ⊂ [ − 1, 1]

and nonnegative values (U (x) ≥ 0, ∀ x ∈ R ). We define T α := U ^1/2 S α U ^1/2

where U ^1/2 : f 7−→ U ^1/2 f is the operator of multiplication by U ^1/2 . Show that

∀ f ∈ C c ^∞ ( R , C ),

(T _α f)(x) = Z

R

K _α (x, y)f (y)dy, (1)

where K _α ∈ C ⁰ ( R ² , R ).

In the following we will denote by K α the operator acting on C c ^∞ ( R , C ) defined by ( K α f)(x) = R

R K _α (x, y)f (y)dy.

Answer — K α (x, y) = U (x) ^1/2 (x) ^e

_2α U (x) ^1/2 (y) = U (x) ^1/2 (x)G α (x, y)U (x) ^1/2 (y).

(j) Show that K _α ∈ L ² ( R ² , R ). Deduce that the operator K α has an unique continuous extension from L ² ( R , C ) to itself. Show that Relation (1) is actually true for any f ∈ L ² ( R , C ).

Answer — We compute k K _α k ² _L

=

Z

R

Z

R

U (x)U(y) e ^{−2α|x−y|}

4α ² dxdy ≤ 1 4α ²

Z

R

Z

R

U (x)U (y)dxdy = k U k ² _L

4α ² . Hence k K _α k L

≤ k U k L

/2α. This implies in particular that, ∀ f, g ∈ C c ^∞ ( R , C ),

|h f, K α g i L

| = Z

R

Z

R

f (x)K _α (x, y)g(y)dxdy

≤ Z

R

| f (x) | ² | g(y) | ² dxdy

2

Z

R

| K α (x, y) | ² dxdy

2

= k K _α k L

k f k L

k g k L

.

Hence kK α f k L

≤ k K _α k L

k f k L

≤ ( k U k L

/2α) k f k L

. This implies that K α admits an unique continuous extension from L ² ( R , C ) to itself.

But T _α is also bounded. Since both operators are continuous and coincide on a dense subspace (i.e. C c ^∞ ( R , C )), they coincide : T α = K α .

(k) Show that T α is compact.

Answer — T _α is a kernel operator, hence is Hilbert–Schmidt. Hence it is compact.

(l) Show that T _α is nonnegative and self-adjoint. What can we say about the spectrum of T _α ?

Answer — For any f ∈ L ² ( R , C ), by setting g = U ^1/2 f in the following, h f, T _α f i L

= h f, U ^1/2 S _α U ^1/2 f i L

= h U ^1/2 f, S _α U ^1/2 f i L

= h g, S _α g i L

hence, by using the definition of S _α ,

h f, T _α f i L

= h S _α g, S _α g i α = k S _α g k ² α ≥ 0.

This implies that T α is nonnegative and hence in particular self-adjoint. Hence, since T _α is also compact, its spectrum is equal to { 0 } ∪ { λ _n ; n ∈ N } , where N ⊂ N and (λ _n ) _n∈N is a sequence of positive real numbers which, either is finite, or tends to 0 as n → + ∞ , if N is infinite.

(m) Show the Birman–Schwinger principle : Ker(1 − T α ) 6 = { 0 } if and only if − α ² is an eigenvalue of the operator − _dx ^d

− U , i.e. ∃ ϕ ∈ H ¹ ( R , C ) ∩ C ^∞ ( R , C ) such that

− ϕ ^′′ − U ϕ = − α ² ϕ in the weak sense ¹ .

Answer — First of all observe that, ∀ ϕ ∈ H ¹ ( R , C ), − ϕ ^′′ − U ϕ = − α ² ϕ in a weak sense iff ∀ χ ∈ L ² ( R , C ), h ϕ, χ i α = h U ϕ, χ i L

.

Now let f ∈ L ² ( R , C ) such that f 6 = 0 and T α f = f, i.e. f = √ U S α

√ U f. Then

√ U f = U S _α √

U f. Hence, by setting ϕ := S _α √

U f ∈ H ¹ ( R , C ),

∀ χ ∈ L ² ( R , C ), h ϕ, χ i α = h S _α √

U f, χ i α = h √

U f, χ i L

= h U S _α √

U f, χ i L

= h U ϕ, χ i L

. Thus ϕ is a weak solution of − ϕ ^′′ − U ϕ = − α ² ϕ. Conversely let ϕ 6 = 0 be a solution in C ^∞ ( R , C ) ∩ H ¹ ( R , C ) of − ϕ ^′′ − U ϕ = − α ² ϕ. Then f := √

U ϕ ∈ C ^∞ ( R , C ) ∩ H ¹ ( R , C ) is a solution of T _α f = f.

(n) Show that any negative eigenvalue λ of − _dx ^d

− U belongs to [ −k U k ² _L

/4, 0).

Answer — Set λ = − α ² with α > 0. Then λ is a eigenvalue of − _dx ^d

− U iff 1 is an eigenvalue of T α . If so, by using the result of Question (j),

1 ≤ r(T _α ) ≤ k T _α k ≤ k U k L

/2α, which implies that α ≤ k U k L

/2. Hence the result.

1. We admit the following regularity result : for any function V ∈ C

( R , C ), any weak solution v ∈ H

( R , C )

of the equation − v

+ V v = 0 is smooth, i.e. v ∈ C

( R , C ).