J. E. ALLAKHVERDIEV

ON THE COMPLETENESS OF A SYSTEM OF EIGEN- AND ASSOCIATED ELEMENTS OF NON-SELF-ADJOINT OPERATORS

(Presented by Academician M. V. Keldysh on 15 VI 1964)

Let \(B\) be a completely continuous operator possessing a complete system of eigen- and associated (e.a.) elements forming a basis in the Hilbert space \(\mathcal H\). Consider the operator

\[ L(\lambda)=A_0+\lambda A_1B+\cdots+\lambda^{n-1}A_{n-1}B^{n-1}+\lambda^nB^n. \]

In the present note one sufficient condition is established for the completeness of the system of e.a. elements of the operator \(L(\lambda)\). Along the way, some properties of the resolvents \((E-\lambda B)^{-1}\) and \((E-L(\lambda))^{-1}\) are studied.

In what follows \(y_{i,k}\) \((k=0,1,\ldots,n_i)\) will denote an eigen-element corresponding to the eigenvalue \(\lambda_i\) for \(k=0\), and the corresponding associated elements for \(k=1,2,\ldots,n_i\), \(a_{i,j}(f)=(f,z_{i,j})\), where \(z_{i,k}\) is the system biorthogonal to \(y_{i,k}\). In the work \((^1)\) of M. V. Keldysh it is shown that \(z_{i,n}\) is a certain system of e.a. elements of the adjoint operator \(B^*\).

Lemma 1. The resolvent of the operator \(B\) has the form

\[ (E-\lambda B)^{-1}f = \sum_{i=1}^{\infty}\sum_{j=1}^{n_i} \left(\sum_{k=j}^{n_i} a_{i,k} A_{i,j}^{k}\right)y_{i,j}, \]

where \(A_{i,j}^{k}\) are found from the recurrence relations

\[ A_{i,k}^{k}=\frac{\lambda_i}{\lambda_i-\lambda}\equiv \varphi_i(\lambda) \quad \text{for all } k \text{ and } i, \]

\[ A_{i,j}^{k} = \sum_{l=j}^{k-1}(-1)^{k-l-1}A_{i,j}^{l}\varphi_i(\lambda) \frac{\lambda}{\lambda_i^{\,k-l+1}}. \tag{1} \]

Proof. Let

\[ f=\sum_{i=1}^{\infty}\sum_{k=0}^{n_i} a_{i,k}y_{i,k}; \]

then

\[ (E-\lambda B)^{-1}f = \sum\sum a_{i,k}(E-\lambda B)^{-1}y_{i,k}. \]

Since

\[ y_{i,k}=\lambda_i By_{i,k}+By_{i,k-1}, \quad k=1,2,\ldots,n_i, \]

and \(y_{i,0}=\lambda_i By_{i,0}\), it follows that

\[ By_{i,k}=\frac{1}{\lambda_i}y_{i,k}-\frac{1}{\lambda_i}By_{i,k-1}; \qquad By_{i,0}=\frac{1}{\lambda_i}y_{i,0}. \]

Hence we easily obtain that

\[ By_{i,k} = \sum_{j=1}^{k+1}(-1)^{j+1}\frac{1}{\lambda_i^{\,j}}\,y_{i,k-j+1} \]

and, consequently,

\[ (E-\lambda B)^{-1}y_{i,k} = y_{i,k} - \lambda\sum_{j=1}^{k+1}(-1)^{j+1}\frac{1}{\lambda_i^{\,j}}\,y_{i,k-j+1}. \]

Applying \((E-\lambda B)^{-1}\) and solving with respect to \((E-\lambda B)^{-1}y_{i,k}\), we obtain

\[ (E-\lambda B)^{-1}y_{i,k} = \varphi_i(\lambda)y_{i,k} + \varphi_i(\lambda)\lambda \sum_{j=2}^{k+1}(-1)^{j+1}\frac{1}{\lambda_i^{\,j}} (E-\lambda B)^{-1}y_{i,k-j+1}. \tag{2} \]

Since \((E-\lambda B)^{-1}y_{i,0}=\dfrac{\lambda_i}{\lambda_i-\lambda}y_{i,0}\equiv\varphi_i(\lambda)y_{i,0}\), it follows from (2) that \((E-\lambda B)^{-1}y_{i,k}\) must have the form
\[ (E-\lambda B)^{-1}y_{i,k}=\sum_{j=0}^{k}A_{i,j}^{k}y_{i,j}, \]
where \(A_{i,j}^{k}\) depends on \(\lambda\). To find \(A_{i,j}^{k}\), let us turn to formula (2) and replace \((E-\lambda B)^{-1}y_{i,l}\) on the left- and right-hand sides by \(\sum_{j=0}^{l}A_{i,j}^{l}y_{i,j}\); we obtain
\[ \sum_{j=0}^{k}A_{i,j}^{k}y_{i,j} =\varphi_i(\lambda)y_{i,k} +\lambda\varphi_i(\lambda)\sum(-1)^{j+1}\frac{1}{\lambda_i^{j}} \left(\sum_{r=0}^{k-j+1}A_{i,r}^{k-j+1}y_{i,r}\right). \]
Comparing the coefficients of \(y_{ij}\) on the left- and right-hand sides, we obtain:
\[ A_{i,j}^{k}=\lambda\varphi_i(\lambda)\sum_{l=j}^{k-1}(-1)^{k-l-1}A_{i,j}^{l}\frac{1}{\lambda_i^{k-l+1}}, \qquad j<k;\qquad A_{ij}^{j}=\varphi_i(\lambda). \]
This proves the lemma.

For what follows, let us note that \(A_{i,j}^{k}\) are determined not by the indices \(k\) and \(j\) themselves, but by the difference \(k-j\). Indeed:
\[ A_{i,j}^{k} =\lambda\varphi_i(\lambda)\sum_{l=j}^{k-1}(-1)^{k-l-1}A_{i,j}^{l}\frac{1}{\lambda_i^{k-l+1}} =\lambda\varphi_i(\lambda)\sum_{s=1}^{k-j}(-1)^{s+1}A_{i,j}^{k-s}\frac{1}{\lambda_i^{s+1}}. \]
It is seen from this that if \(A_{i,j}^{k-s}\), \(s=1,2,\ldots,k-j\), depend only on the difference \(k-s-j\) (i.e. for \(l=j,j+i,\ldots,k-1\) depend on \(l-j\)), then \(A_{i,j}^{k}\) depends only on \(k-j\), and not on \(k\) and \(j\). But since \(A_{i,k}^{k}\) depends only on \(i\), by induction \(A_{i,j}^{k}\) will depend on \(i\) and \(k-j\); therefore one may introduce the notation \(A_{i,j}^{k}=B_{i,k-j}=B_{i,r}\).

Definition. We shall say that a ray belongs to the class \(\mathcal K_\beta\) if it is the bisector of some angle of aperture not smaller than \(2\beta\), inside which there are only finitely many eigenvalues of the operator \(B\).

Lemma 2. If the orders of the eigenelements of the operator \(B\) are bounded in the aggregate (i.e. \(n_i\le m\) for all \(i\)) and the system of root elements of the operator \(B\) forms a basis of Riesz type (a \(P\)-basis), then the resolvent of the operator \(A+\lambda B\) on each ray from \(\mathcal K_\beta\) has the form
\[ (E-A-\lambda B)^{-1}=(E+M(\lambda))(E-\lambda B)^{-1}, \]
where \(M(\lambda)\) on rays from \(\mathcal K_\beta\to 0\) as \(|\lambda|\to\infty\).

Proof.
\[ Y=(A+\lambda B)y+f,\qquad Y=(E-\lambda B)^{-1}Ay+(E-\lambda B)^{-1}f. \]
Using the expression for \((E-\lambda B)^{-1}\), let us estimate \(\|(E-\lambda B)^{-1}A\|\):
\[ \|(E-\lambda B)^{-1}Af\| =\left\|\sum_{i=1}^{\infty}\sum_{j=0}^{n_i}\left(\sum_{k=j}^{n_i}a_{i,k}(Af)A_{i,j}^{k}\right)y_{i,j}\right\|\le \]
\[ \le \left\|\sum_{i=1}^{N}\sum_{j=0}^{n_i}\left(\sum_{k=j}^{n_i}a_{ik}(Af)A_{i,j}^{k}\right)y_{i,j}\right\| + \left\|\sum_{i=N+1}^{\infty}\sum_{j=0}^{n_i}\left(\sum_{k=j}^{n_i}a_{i,k}(Af)A_{i,j}^{k}\right)y_{i,j}\right\|\le \]
\[ \le M\|A\|m\left(\sup_{\substack{i\le N\\ j<n_i}}\sum |A_{i,j}^{k}|^2\right)^{1/2} + Mm\left(\sup_{j,i}\sum_{k=j}^{n_i}|A_{i,j}^{k}|^2\right)\|Af\|_{P_N\mathcal H}, \]
where \(M\) is the constant appearing in the definition of a Riesz basis; \(m=\max n_i\); \(P_N\) is the projection operator of the subspace generated by the elements \(y_{i,k}\), \(i\ge N+1\) (in general not orthogonal, but equivalent to an orthogonal one).

Since \(A\) is a completely continuous operator, \(\|Af\|_{P_N\mathcal H}\) can be made arbitrarily small by choosing \(N\) sufficiently large (\(\|f\|=1\));

therefore for any \(\varepsilon>0\) one can find such an \(N_0(\varepsilon)\) that, for \(N\geq N_0(\varepsilon)\),

\[ \|Af\|_{P_N\mathcal H}<\varepsilon,\qquad \|Af\|_{P_N\mathcal H}=\|P_NAf\|. \]

We shall prove that if \(\lambda\in\mathcal K_\beta\), then

\[ \sup_{j,i}\sum_{i=j}^{n_i}|A^k_{i,j}|^2<P, \]

where \(P\) is some number independent of \(\lambda\), and

\[ \lim_{|\lambda|\to\infty}\sup_{i\leq N,\; j\leq m}\sum |A^k_{i,j}|^2=0. \]

First we show that if \(B_{i,r}=\varphi_i(\lambda)O(1)\), \(r\leq t\), then \(B_{i,t+1}=\varphi_i(\lambda)O(1)\). Indeed:

\[ B_{i,r+1}=\lambda\varphi_i(\lambda)\sum_{s=1}^{r+1}(-1)^{s+1}B_{i,r-s+1}\frac{1}{\lambda_i^{s+1}}, \]

\[ |B_{i,r+1}|\leq |\varphi_i|\left(\sum\left|\frac{\lambda}{\lambda_i}\varphi_i(\lambda)\right||O(1)|\left|\frac{1}{\lambda_i^s}\right|\right)\leq |\varphi_i|O(1), \]

since

\[ \frac{\lambda}{\lambda_i}\varphi_i(\lambda)=\frac{\lambda}{\lambda_i-\lambda}=\psi_i(\lambda);\qquad |\psi_i|\leq \frac{1}{\sin\beta}. \]

It follows that

\[ |A^k_{i,j}|^2\leq |\varphi_i(\lambda)|^2O(1); \]

but since

\[ \varphi_i(\lambda)=\frac{\lambda_i}{\lambda_i-\lambda}, \]

we have

\[ |\varphi_i|\leq \frac{1}{\sin\beta}, \]

and if \(i\leq N\), then

\[ \lim_{|\lambda|\to\infty}|\varphi_i|=0. \]

Taking into account the assumptions of the theorem, namely that \(k\) and \(j\) are bounded by the number \(m\), and using the formula

\[ (E-A-\lambda B)^{-1}=(E-(E-\lambda B)^{-1}A)^{-1}(E-\lambda B)^{-1}, \]

we arrive at the assertion of Lemma 2. Indeed, if one sets

\[ (E-(E-\lambda B)^{-1}A)^{-1}=E+M(\lambda), \]

then we have

\[ \|(E-\lambda B)^{-1}A\|\leq Mm\left\{\left(\sup\sum_{k=j}^{n_i}|A^k_{i,j}|^2\right)^{1/2}\|A\|+\sup_{j,i}\sum_{k=j}^{n_i}|A^k_{ij}|^2\|P_NAf\|\right\}. \]

Since for any \(\varepsilon>0\) one can find such an \(R\) that, for \(|\lambda|>R\), the expression in braces does not exceed \(\varepsilon\), we have

\[ (E-(E-\lambda B)^{-1}A)^{-1}=\sum_{i=0}^{\infty}C_i(\lambda),\qquad \text{where } C(\lambda)=(E-\lambda B)^{-1}A. \]

Thus Lemma 2 is proved.

An analogous lemma is also true for the operator \(L(\lambda)\).

We now prove the main theorem.

Theorem. If \(B\) is a completely continuous operator, \(B\in\gamma_\rho\), whose eigen- and associated elements form a basis of Riesz type, and the class of rays \(\mathcal K_\beta^n\) is \(\varepsilon\)-dense in \(G\) for \(\varepsilon\leq \pi/\rho n\), \(\beta\geq\beta_0>0\), then the system of eigen- and associated elements of the operator is \(n\)-fold complete in the space \(\mathcal H\).

Proof. Consider the equation

\[ y=L^*(\lambda)y+f, \]

where

\[ f(\lambda)=\sum_{i=0}^{n-1}\lambda^i f_i \]

is chosen so that the element \(f=\{f_0,\ldots,f_{n-1}\}\) is orthogonal to the elements of the operator \(L(\lambda)\); the solution \(y(\lambda)\) must be an entire function of order not exceeding \(n\rho\).

Since on the rays of \(\mathcal K_\beta^n\) (a ray belongs to \(\mathcal K_\beta^n\) if, upon rotation by \((n-1)\varphi\), where \(\varphi\) is the argument of the ray, it coincides with some ray from \(\mathcal K_\beta\))

this function grows no faster than a polynomial of degree \(n-1\), then on the basis of the Phragmén—Lindelöf theorem we conclude that \(y(\lambda)\) is a polynomial of degree not higher than \(n-1\):

\[ y(\lambda)=\sum_{i=0}^{n-1}\lambda^i y_i . \]

We shall show that \(y(\lambda)\equiv 0\). Suppose this is not so and \(y_{i_0}\) is the nonzero coefficient of the highest power of \(\lambda\). We have:

\[ y(\lambda)=\sum_{i=0}^{n-1}\lambda^i(A_iB^i)^*y(\lambda)+\lambda^n(B^i)^*y(\lambda)+f(\lambda). \]

Comparing the coefficients of equal powers of \(\lambda\) in the left- and right-hand sides, we obtain \((B^n)^*y_{i_0}=0\).

\((B^{n*}y_{i_0},g)=0\) for any \(g\). \((y_{i_0},B^n g)=0\) shows that \(y\) is orthogonal to all associated elements of the operator \(B^n\), and these elements are complete in \(\mathcal H\); hence it follows that \(y_{i_0}=0\); consequently, all \(f_i=0\). The theorem is proved.

Let now \(H\) be a complete self-adjoint operator having finite order \(\rho\), and let \(A_i\) be completely continuous operators.

Theorem 2. For any \(0\le \alpha\le 1\), the system of eigen and associated elements of the operator

\[ A_0+\lambda H^{\alpha/n}A_1H^{(1-\alpha)/n}+\cdots+\lambda^{n-1}H^{(n-1)\alpha/n}A_{n-1}H^{(n-1)(1-\alpha)/n}+\lambda^nH \]

is \(n\)-fold complete in \(\mathcal H\).

The proof of Theorem 2 is, in essence, close to the proof of Theorem 1.

We note that, analogously to how this was done in (2) for \(A_0\), the conditions on \(A_i\) can be weakened by requiring that the norm of the purely bounded part be sufficiently small.

We shall give one more theorem concerning operators of the form

\[ A_0+\lambda H^{1/n}A_1+\cdots+\lambda^{n-1}H^{(n-1)/n}A_{n-1}+\lambda^nH . \]

For simplicity of formulation of the theorem, we shall assume that \(H\) is a positive operator, although in fact similar results are valid even if \(H\) is a normal operator whose eigenvalues lie inside certain angles.

Consider the equation \(y=\lambda^nHy\). If \(H\) is a complete self-adjoint operator, then the eigenvalues \(\lambda_{i,k}\) and \(\mu_i\) of the equations \(y=\lambda^nHy\) and \(y=\mu Hy\) are connected by the relation \(\lambda_{i,k}^n=\mu_i\) \((k=0,1,\ldots,n-1)\), i.e., the eigenvalues of the first equation lie on rays passing through the \(n\)th roots of unity. Take \(n\) nonintersecting angles with vertex at the origin such that the indicated rays lie strictly inside these angles, and for each number \(r\) form the domains \(\Gamma_{\psi_i,r}\) \((i=0,1,\ldots,n-1)\), consisting of the angle \(\psi_i\) and the circle of radius \(r\) with center at the origin.

Theorem 3. Suppose that for some \(0<\rho<1\) the following conditions are satisfied: either the operators \(H^{-p}A_i\) \((i=0,\ldots,n-1)\) are bounded and \(\lim_{k\to\infty}\dfrac{k}{\mu_k^p}=0\), or else the operators \(H^{-p}A_i\) \((i=0,\ldots,n-1)\) are completely continuous and \(\lim_{k\to\infty}\dfrac{k}{\mu_k^p}<\infty\). Then for each set of numbers \(\alpha_1,\ldots,\alpha_j^{(j)}\), \(\alpha_j\le n\), \(\alpha_i\ne\alpha_k\), one can find an \(r\) such that the system of eigen and associated elements corresponding to eigenvalues lying in the domain

\[ \sum_{i=1}^j \Gamma_{\psi_i,r} \]

is \(j\)-fold complete in \(\mathcal H\), and some subsequence of the partial sums of the corresponding expansion (see (1)) converges.

I express my gratitude to Academician M. V. Keldysh and Professor M. A. Naimark for their attention to this work and for valuable comments during discussion.

Received
28 V 1964

References

\({}^{1}\) M. V. Keldysh, DAN, 77, No. 1 (1951). \({}^{2}\) Dzh. E. Allakhverdiev, DAN, 115, No. 2 (1957).