## 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

_{n}

### |

^{2}

### ≥r

^{2}

## | 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

^{2}

## defined by h f, g i L

^{2}

## := R

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

^{2}

## = h f, f i ^{1/2} _{L}

2 ## . 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}

2## , k L _{α} f k ^{2} _{L}

2 ## ≤ 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}

^{2}2

## + 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

^{2}

## ≤ _{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

^{2}

## .

## 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

^{2}

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

^{2}

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

^{2}

## . 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

^{2}

## .

## 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

^{2}

## .

## (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

^{2}

## ( 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

^{2}

## = 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

^{2}

## 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

_{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}

^{−α|x−y|}

_{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}

^{2}

## =

## 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}

1
## 4α ^{2} . Hence k K _{α} k L

^{2}

## ≤ k U k L

^{1}

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

## |h f, K α g i L

^{2}

## | = Z

### R

## Z

### R

## f (x)K _{α} (x, y)g(y)dxdy

## ≤ Z

### R

2## | f (x) | ^{2} | g(y) | ^{2} dxdy

^{1}

2

## Z

### R

2## | K α (x, y) | ^{2} dxdy

^{1}

2

## = k K _{α} k L

^{2}

## k f k L

^{2}

## k g k L

^{2}

## .

## Hence kK α f k L

^{2}

## ≤ k K _{α} k L

^{2}

## k f k L

^{2}

## ≤ ( k U k L

^{1}

## /2α) k f k L

^{2}

## . 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

^{2}

## = h f, U ^{1/2} S _{α} U ^{1/2} f i L

^{2}

## = h U ^{1/2} f, S _{α} U ^{1/2} f i L

^{2}

## = h g, S _{α} g i L

^{2}

## hence, by using the definition of S _{α} ,

## h f, T _{α} f i L

^{2}

## = 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}

^{2}2

## − 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

^{2}

## .

## 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

^{2}

## = h U S _{α} √

## U f, χ i L

^{2}

## = h U ϕ, χ i L

^{2}

## . 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}

^{2}2

## − U belongs to [ −k U k ^{2} _{L}

1## /4, 0).

## Answer — Set λ = − α ^{2} with α > 0. Then λ is a eigenvalue of − _{dx} ^{d}

^{2}2

## − 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

^{1}

## /2α, which implies that α ≤ k U k L

^{1}

## /2. Hence the result.

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

^{∞}

### ( R , C ), any weak solution v ∈ H

^{1}

### ( R , C )

### of the equation − v

^{′′}

### + V v = 0 is smooth, i.e. v ∈ C

^{∞}