5. Existence of an eigenvalue of a completely continuous operator in a Hilbert space.

The largest and smallest eigenvalues.

Spectral decomposition of self-adjoint operators.

The Hilbert-Schmidt theorem on expansion in eigenvectors

We now turn to the fundamental theorem on symmetric completely continuous operators.

Theorem 8. (Hilbert-Schmidt). In a separable Hilbert space, every symmetric completely continuous operator possesses a complete orthogonal system of eigenvectors.

We shall carry out the proof of this theorem in several stages.

Lemma 7. If and A – is a symmetric operator, then

moreover, equality is possible only in the case when e is an eigenvector of the operator with eigenvalue

Proof. By virtue of the symmetry of the operator and the Cauchy – Bunyakovsky inequality we have:

(12)

The Cauchy – Bunyakovsky inequality becomes an equality only when the vectors involved in it are collinear, hence in the case of equality we have i.e., it is an eigenvector of the operator A2. Substituting the resulting expression into (12), we find :

The lemma is proved.

We call a maximal vector of a bounded operator A such a unit vector at which the quantity attains its greatest value Generally speaking, not every bounded operator has a maximal vector.

Lemma 8. A symmetric completely continuous operator possesses a maximal vector.

Proof. Choose a sequence , where such that From the sequence we can extract, by virtue of the complete continuity of A, a convergent subsequence; discarding the extra vectors and renumbering, we may assume that the sequence itself converges as ; let By continuity of the norm Let us show that the vector is the sought maximal vector. First of all, by continuity of the operator A we have:

The vectors belong to the unit ball, and therefore the vectors do not exceed in length . Applying Lemma 7, we obtain:

.

Whence it follows that

i.e., is the maximal vector of the operator A. The lemma is proved.

Lemma 9. If is a maximal vector for a symmetric operator , then is an eigenvector for the operator with eigenvalue

Proof. By Lemma 7 and by the definition of the operator norm we have:

whence it follows that

By virtue of Lemma 7 the vector is an eigenvector of the operator with eigenvalue The lemma is proved.

Lemma 10. If the operator possesses an eigenvector with eigenvalue , then the operator A has an eigenvector with eigenvalue or

Proof. The equality can be written in the form

Suppose that . Then from the condition or, equivalently, it follows that is an eigenvector of the operator with eigenvalue If, however, , then and then the vector is an eigenvector of the operator A with eigenvalue The lemma is proved.

Lemmas 7-10 show that every symmetric completely continuous operator A possesses an eigenvector with eigenvalue We shall now show that from the eigenvectors of the operator A we can construct an orthogonal system in the space H.

Lemma 6 allows us to draw certain conclusions concerning the collection of all eigenvectors and eigenvalues of the operator A. Consider on the real axis the set of all eigenvalues of the operator A. By virtue of Lemma 6, there exists only a finite number of eigenvalues exceeding in absolute value a given positive number , therefore, if the set of eigenvalues is infinite (obviously countable), then they form a sequence converging to zero. Consequently, we can number all the eigenvalues with natural numbers in order of decreasing absolute value. Let us agree that in doing so we shall assign to each eigenvalue as many consecutive numbers as the dimension of the corresponding eigenspace (this dimension is called the multiplicity of this eigenvalue). In this case the sequence of nonzero eigenvalues of the operator A

we can associate a sequence of eigenvectors moreover We may assume that the vectors are mutually orthogonal and normalized. Indeed, if then orthogonality holds by virtue of Lemma 5; if, on the other hand, then within the finite eigenspace corresponding to the eigenvalue we can always perform orthogonalization. Normalization of all the vectors obtained completes the construction.

Let us now show that every vector , orthogonal to all the constructed vectors is carried by the operator A into zero.

Lemma 11. Let – be a subspace in a Hilbert space H, invariant with respect to a symmetric operator A (i.e., every vector of the subspace is carried by the operator A into a vector of this same space). Then the orthogonal complement of the subspace is also invariant with respect to the operator A.

Proof. Let – be any vector from the subspace , – any vector from the subspace . By assumption Then by virtue of the symmetry of the operator A it follows that This means that the vector is orthogonal to any vector and, consequently, The lemma is proved.

Now let us consider the collection P of all vectors orthogonal to all the constructed vectors This collection P is a closed subspace as the orthogonal complement of the subspace L, generated by the orthogonal system Since L, obviously, is invariant with respect to the operator A, its orthogonal complement P (by Lemma 11) is also invariant with respect to the operator A. Denote by M(P) the exact upper bound of the values on the unit sphere of the subspace P. By virtue of Lemmas 9 and 10, in the subspace P there is an eigenvector with eigenvalue But by the very construction of the subspace P it cannot contain a single eigenvector with a nonzero eigenvalue. Hence ; but this means that for any vector which was to be proved.

Every vector can be represented in the form of a sum

The vector y can further be expanded in a Fourier series with respect to the system complete in the space L; the vector z, by what has been proved, is carried by the operator A into the zero vector. We have obtained the following basic theorem:

Theorem 9. In a Hilbert space , in which a symmetric completely continuous operator A is given, every vector can be represented in the form of an orthogonal sum where (a finite or infinite) system of eigenvectors of the operator with nonzero eigenvalues and

From this theorem also follows the Hilbert theorem. Indeed, in the separable Hilbert space H the subspace P is also separable, and in it one can choose a complete orthogonal system together with the already constructed vectors one obtains a complete orthogonal system in the whole space H. Each of the vectors of this system is an eigenvector of the operator A: the vectors with eigenvalues and the vectors with eigenvalue 0. Thereby the Hilbert theorem is proved.

From the Hilbert theorem it follows that i.e., any vector , where , admits an expansion in the eigenvectors of the operator with nonzero eigenvalues.

Problems

1. Prove the following statements:

A) any linear operator A: Rn→Rm is completely continuous;

B) any linear operator A: E1→E2 is completely continuous, if E1 – is a finite-dimensional space;

C) any bounded linear operator A: E1→E2 is completely continuous, if E2 – is a finite-dimensional space;

D) a bounded linear operator, whose image lies in a finite-dimensional space, is completely continuous.

2. Are the following operators completely continuous in the space C[0, 1]? In the space L2[0, 1]?

1) ;

2) ;

3) ;

4) ;

5) ;

6) ;

7) ;

8) ;

9) ;

10) Ax(t)=x(t2).

3. Does the operator have eigenvalues in the space ?

4. Show that for the equation , where – is a Volterra operator, and is continuous for , all values of the parameter are regular.

Show that if the value of the parameter l is regular for the operator, then it will also be regular for the operator A + B, when is sufficiently small.

5. Show that every completely continuous (compact) operator in a separable Hilbert space H is the limit of operators that map the entire space onto a finite-dimensional subspace.

Hint: one may assume that . If , then let .

6. Show that the operator A in a separable Hilbert space, given in an orthonormal basis by the matrix by the formulas is completely continuous, if

Hint: see Problem 5.

7. Let, for , , . Which of these operators are completely continuous?

8. For , let , show that A – is a completely continuous operator.

9. What are the eigenfunctions of the Fredholm integral operator with kernel on the intervals a) , b) ?

10. Solve the equation .

11. In the space C[0, 1] consider the operator

.

Find the spectrum and resolvent of the operator A.

12. In the real linear space C[-p, p] find the eigenvalues and eigenvectors of the operators

a) (Ax)(t) = x(-t);

b) .

Do these operators have a continuous spectrum in this space? Construct the resolvents on the set of regular values of each operator.

13. In the complex space C[0, 1] consider the operator (Ax)(t) = x(0) + tx(1). Find the point and continuous spectra of the operator A and construct the resolvent on the set of regular values.

14. In the space C[0, 2p] consider the operator (Ax)(t) = eittx(t). Prove that the spectrum of the operator A is the set {l ∈C: |l| = 1}, and no point of the spectrum is an eigenvalue.

15. Find the spectrum and resolvent of the operator A in the space L2[-1, 1]

.

16. What should the function j∈C[a, b], be so that the multiplication operator A: C[a, b] → C[a, b], defined by the equality (Ax)(t) = j(t)×x(t) was completely continuous.

17. Find the spectrum and eigenvalues of the multiplication operator by a fixed continuous function in the space C[a, b].

18. Find the spectrum of the operator A in the space L2(R):

.

19. Let the number p > 1 and q – be conjugate to it, i.e. 1/p + 1/q = 1. Consider the operator A: lp → lq, which is defined by the formula

,

where the numerical matrix is such that the double series converges. Prove that the operator A is completely continuous.

20. Consider the operator A: lp → lq, which is defined by the formula

Ax = (l1x1, l2x2, …), x = (x1, x2, …)

where lk – is a given sequence of numbers, k =1, 2, … What should this sequence of numbers be so that the operator A was completely continuous.

21. Let a bounded linear operator A be given in a Hilbert space H such that A*A is a completely continuous operator in H. Prove that the operator A is completely continuous?