Mathematics

Z. L. Leibenzon

On the Estimation of Eigenvalues of Self-Adjoint Operators*

(Presented by Academician S. L. Sobolev on 31 V 1957)

1°. Let \(A\) be a self-adjoint (not necessarily bounded) operator in a Hilbert space \(H\). Suppose that the operator \(A\) has a discrete spectrum.

Take an element \(x \in H\); then

\[ x=\sum_i x_i e_i, \]

where all \(x_i \ne 0\), and the normalized eigenfunctions \(e_1, e_2,\ldots\) of the operator \(A\) correspond to distinct eigenvalues \(\lambda_1,\lambda_2,\ldots\).

In the totality \(M\) of functions \(F(\lambda)\) for which the series

\[ \sum_i |F(\lambda_i)|^2 |x_i|^2=\|F\|^2 \]

converges, introduce the scalar product

\[ (F,G)=\sum_i F(\lambda_i)\overline{G(\lambda_i)}|x_i|^2 . \tag{1} \]

Lemma. Let \(F(\lambda)\) and \(\Phi(\lambda)\) be functions from \(M\) and \(\|\Phi\|>0\). Then there exists an eigenvalue \(\lambda\) of the operator \(A\) such that \(\Phi(\lambda)\ne 0\) and

\[ \frac{|F(\lambda)|}{|\Phi(\lambda)|}\le \frac{\|F\|}{\|\Phi\|}. \]

Proof. Since

\[ \sum_i |\Phi(\lambda_i)|^2 |x_i|^2=\|\Phi\|^2>0, \]

we have \(\Phi(\lambda_i)\ne 0\) for some \(\lambda_i\). As is easy to see, either

\[ \frac{\|F\|^2}{\|\Phi\|^2} = \frac{\sum_i |F(\lambda_i)|^2 |x_i|^2} {\sum_i |\Phi(\lambda_i)|^2 |x_i|^2} > \inf_{i,\Phi(\lambda_i)\ne 0} \frac{|F(\lambda_i)|^2}{|\Phi(\lambda_i)|^2}, \]

or

\[ \frac{|F(\lambda_i)|}{|\Phi(\lambda_i)|}=\frac{\|F\|}{\|\Phi\|}, \]

if \(\Phi(\lambda_i)\ne 0\). In both cases the lemma is true.

Suppose it is known that \(A^n x \in H\). Then \(M\) contains the subspace \(M_n\) of polynomials in \(\lambda\) of degree not exceeding \(n\). The scalar products of polynomials in \(\lambda\) of degree not exceeding \(n\), and in particular their norms, are easily compu-

* Reported at the seminar on differential equations at the Institute of Mathematics of the Academy of Sciences of the Ukrainian SSR in 1950.

compute, knowing the numbers

\[ p_{\nu+\mu}=(\lambda^\nu,\lambda^\mu)=\sum_i \lambda_i^{\nu+\mu}|x_i|^2=(A^\nu x,A^\mu x),\qquad 0\leq \nu,\mu\leq n. \tag{2} \]

Applying the lemma, taking in it as \(F(\lambda)\) and \(\Phi(\lambda)\) polynomials of degree not exceeding \(n\), differing by a linear factor, we shall obtain intervals containing some eigenvalues of the operator \(A\).

Theorem 1. Let

\[ \varepsilon= \frac{\left\|(a_0+a_1\lambda+\cdots+a_{n-1}\lambda^{n-1})(\lambda-\alpha)\right\|} {\left\|a_0+a_1\lambda+\cdots+a_{n-1}\lambda^{n-1}\right\|} = \frac{ \left[ \displaystyle\sum_{\nu,\mu=0}^{n-1} \bigl(p_{\nu+\mu+2}-2\alpha p_{\nu+\mu+1}+\alpha^2p_{\nu+\mu}\bigr)a_\nu\overline{a_\mu} \right]^{1/2} }{ \left[ \displaystyle\sum_{\nu,\mu=0}^{n-1} p_{\nu+\mu}a_\nu\overline{a_\mu} \right]^{1/2} }, \]

where \(a_0,a_1,\ldots,a_{n-1}\) are arbitrary complex numbers; \(\alpha\) is an arbitrary real number.

Then the interval \([\alpha-\varepsilon,\alpha+\varepsilon]\) contains one of the eigenvalues of the operator \(A\).

If the elements \(A^\nu x\in H\) are known for \(0\leq \nu\leq n\), then one can compute the numbers \(p_{\nu+\mu}\), equal to their scalar products \((A^\nu x,A^\mu x)\) in the Hilbert space \(H\); then it is easy to compute the interval \([\alpha-\varepsilon,\alpha+\varepsilon]\). Putting \(n=1\) in Theorem 1, we obtain that for any real value \(\alpha\) the interval

\[ \left[ \alpha-\sqrt{\frac{p_2-2\alpha p_1+\alpha^2p_0}{p_0}}, \quad \alpha+\sqrt{\frac{p_2-2\alpha p_1+\alpha^2p_0}{p_0}} \right] \]

contains some eigenvalue of the operator \(A\).

For a self-adjoint differential operator \(A\), this assertion for \(\alpha=p_1/p_0\) coincides with the result of N. N. Bogolyubov and N. M. Krylov (¹).

2°. In Theorem 1 the length \(2\varepsilon\) of the interval containing one of the eigenvalues is the ratio of the norms of polynomials whose degrees differ by 1. Therefore, in order to make \(\varepsilon\) as small as possible, we use polynomials of least norm. Namely, from all polynomials of some degree \(m\) (\(m\leq n\)) with leading coefficient 1, choose the polynomial \(D_m(\lambda)\) with the least norm \(\|D_m\|\).

Let \(D_m(r)=0\), so that \(D_m(\lambda)=(\lambda-r)\varphi(\lambda)\). Here \(\varphi(\lambda)\) is a polynomial of degree \(m-1\) with leading coefficient 1, and therefore \(\|\varphi\|\geq \|D_{m-1}\|\). Let in the lemma \(F(\lambda)=D_m(\lambda)\) and \(\Phi(\lambda)=\varphi(\lambda)\). On the basis of the lemma there exists an eigenvalue \(\lambda\) of the operator \(A\) such that

\[ |\lambda-r|\leq \frac{\|D_m\|}{\|\varphi\|}\leq \frac{\|D_m\|}{\|D_{m-1}\|}. \]

For the scalar product (1), the orthogonal polynomials of the theory of moments (²) are the polynomials \(D_0(\lambda),D_1(\lambda),\ldots,D_n(\lambda)\); in the theory of moments (²) they are studied.

We have the formulas

\[ D_m(\lambda)= \frac{ \begin{vmatrix} p_0&\cdots&p_{m-1}&1\\ \cdot&&\cdot&\cdot\\ \cdot&&\cdot&\cdot\\ \cdot&&\cdot&\cdot\\ p_m&\cdots&p_{2m-1}&\lambda^m \end{vmatrix} }{ \begin{vmatrix} p_0&\cdots&p_{m-1}\\ \cdot&&\cdot\\ \cdot&&\cdot\\ p_{m-1}&\cdots&p_{2m-2} \end{vmatrix} }; \qquad \|D_m\|^2= \frac{ \begin{vmatrix} p_0&\cdots&p_m\\ \cdot&&\cdot\\ \cdot&&\cdot\\ \cdot&&\cdot\\ p_m&\cdots&p_{2m} \end{vmatrix} }{ \begin{vmatrix} p_0&\cdots&p_{m-1}\\ \cdot&&\cdot\\ \cdot&&\cdot\\ p_{m-1}&\cdots&p_{2m-2} \end{vmatrix} }. \tag{3} \]

The computation of the polynomials \(D_m(\lambda)\) and the numbers \(\|D_m\|\) is much easier to carry out not by formulas (3), but by using the orthogonality property. For example, we have the recurrence formulas (3)

\[ \|D_{m-1}\|^2=(\lambda^{m-1},D_{m-1}),\qquad a_m=\frac{(\lambda^m,D_{m-1})}{\|D_{m-1}\|^2}, \]

\[ D_m(\lambda)=\lambda D_{m-1}(\lambda)+(a_{m-1}-a_m)D_{m-1}(\lambda) -\frac{\|D_{m-1}\|^2}{\|D_{m-2}\|^2}D_{m-2}(\lambda). \]

The polynomial \(D_m(\lambda)\) has \(m\) distinct real roots \({}^{(2)}\). From what has been said it follows:

Theorem 2. Let \(r_1,\ldots,r_m\) be the \(m\) distinct real roots of the polynomial \(D_m(\lambda)\), defined by formula (3).

There exist eigenvalues \(\lambda^{(1)},\ldots,\lambda^{(m)}\) of the operator \(A\) such that

\[ |\lambda^{(k)}-r_k| \le \frac{\|D_m\|}{\|D_m(\lambda)/(\lambda-r_k)\|} \le \frac{\|D_m\|}{\|D_{m-1}\|}, \qquad k=1,\ldots,m. \]

\(3^\circ\). It is useful to be able to choose the initial element \(x\) close to an eigenfunction of the operator \(A\) or to a linear combination of eigenfunctions. Then the intervals obtained in \(1^\circ\) and \(2^\circ\) will give narrower bounds for the corresponding eigenvalues.

We can always choose the initial element close to a linear combination of eigenfunctions corresponding to the largest eigenvalues of the operator \(A\), taking the iterated element \(A^t x\) instead of \(x\).

Let, as the initial element, instead of \(x\), the element \(y=A^t x\) be taken, where \(t\ge 0\) is an integer. Then the norm \(\|F\|\), the scalar product \((F,G)\), the numbers \(p_k\), and the polynomials \(D_m(\lambda)\) are replaced, respectively, by the norm \(\|F\|_t\), the scalar product \((F,G)_t\), the numbers \(q_k\), and the polynomials \(D_{m,t}(\lambda)\). It is easy to see that \(q_k=p_{k+2t}\).

If \(A\) is a bounded operator, then for any \(n\ge 0\), \(A^n x\in H\), and the numbers \(q_m=p_{m+2t}\) and the polynomials \(D_{m,t}(\lambda)\) are defined by us for all \(t,m\ge 0\).

Theorem 3. Suppose that for some \(m\)

\[ \gamma=\sup_{j>m}|\lambda_j|<\min_{i=1,\ldots,m}|\lambda_i|. \]

Then the roots \(r_{1,t},\ldots,r_{m,t}\) of the polynomial \(D_{m,t}(\lambda)\) converge, as \(t\to\infty\), to the eigenvalues \(\lambda_1,\ldots,\lambda_m\) of the operator \(A\). Moreover,

\[ |r_{i,t}-\lambda_i|\le C\left(\frac{\gamma}{\lambda_i}\right)^{2t} \qquad (i=1,\ldots,m), \]

where \(C\) does not depend on \(t\).

Received
5 VII 1956

REFERENCES

1. N. M. Krylov, N. N. Bogolyubov, Izv. AN SSSR, OMEN, No. 5, 471 (1929).
2. N. I. Akhiezer, M. G. Krein, On Certain Questions in the Theory of Moments, Kharkov, 1938.
3. J. Jackson, Fourier Series and Orthogonal Polynomials, IL, 1948, p. 174.