UDC 517.948.35

MATHEMATICS

A. S. LIBIN

ON TRACE FORMULAS FOR SELF-ADJOINT OPERATORS

(Presented by Academician L. S. Pontryagin, 13 XII 1967)

Let \(A\) be a self-adjoint operator with discrete spectrum; let \(\lambda_1,\ldots,\lambda_n,\ldots\) be the eigenvalues of \(A\), each eigenvalue being counted as many times as its multiplicity; denote by \(e_1,\ldots,e_n,\ldots\) the corresponding eigenvectors. The operator \(A\) is such that

\[ \sum_k \frac{1}{\lambda_k}<\infty \qquad (\lambda_k\ne 0). \tag{1} \]

Consider the operator \(C=A+B\), where \(B\) is a bounded self-adjoint operator. As is known, \(C\) also has a discrete spectrum \(\mu_1,\ldots,\mu_n,\ldots\). In the work of Halberg and Kramer \(\left({}^{2}\right)\) the following theorem is proved.

If

\[ \sum_{k=1}^{\infty}(\mu_k-\lambda_k)<\infty \quad\text{and}\quad \sum_{k=1}^{\infty}(Be_k,e_k)<\infty \]

then

\[ \sum_{k=1}^{\infty}(\mu_k-\lambda_k) = \sum_{k=1}^{\infty}(Be_k,e_k). \tag{2} \]

In our paper a somewhat stronger condition than (1) is imposed on \(A\), but it is proved that, under this condition, for the existence of the right-hand side of equality (2) it is necessary and sufficient that its left-hand side exist. In addition, the question of the relation between

\[ \sum_{k=1}^{\infty}(Be_k,e_k) \quad\text{and}\quad \sum_{k=1}^{\infty}(Bf_k,f_k) \]

for orthonormal bases \(\{e_k\}_1^\infty\) and \(\{f_k\}_1^\infty\) that are close in a certain sense is considered.

For simplicity of exposition we shall assume the operator \(A\) to be positive definite, with the \(\lambda_n\) arranged in increasing order. By condition (1), there exists a sequence of neighboring pairs \(\lambda_{n_i}, \lambda_{n_i+1}\) such that

\[ \frac{1}{\lambda_{n_i+1}-\lambda_{n_i}} \longrightarrow 0. \]

Introduce the notation \(\omega_n=\lambda_{n+1}-\lambda_n\); \(\nu_n=(\lambda_n+\lambda_{n+1})/2\).

Theorem 1. For any \(n\ge 2\) there exists an \(M\) such that, for sufficiently large \(N_i\), the inequality

\[ \left| \sum_{k=1}^{N_i}(\mu_k-\lambda_k) - \sum_{k=1}^{N_i}(Be_k,e_k) \right| \le M\left( 1+\frac{\nu_{N_i}}{\omega_{N_i}} \right) \sum_{k=1}^{\infty} \frac{1}{|\nu_{N_i}-\lambda_k|}. \]

In what follows we shall omit the index \(i\).

Theorem 1 follows from several preliminary lemmas, which are proved by computing and estimating contour integrals.

Lemma 1.

\[ \int_{C_R}\lambda\,\operatorname{Sp} R_\lambda^A (BR_\lambda^A)^n (E+BR_\lambda^A)^{-1}\,d\lambda \to 0 \]

as \(R\to\infty\), where \(R_\lambda^A=(A-\lambda E)^{-1}\), and \(C_R\) is an arc of the circle \(|\lambda|=R\); \(\operatorname{Re}\lambda\le v_N\).

Lemma 2.

\[ \left| \int_{v_N-i\infty}^{v_N+i\infty} \lambda\,\operatorname{Sp} R_\lambda^A (BR_\lambda^A)^{n+1}(E+BR_\lambda^A)^{-1}\,d\lambda \right| \le A_1\frac{v_N}{\omega_N^n}\sum_{k=1}^{\infty}\frac{1}{|v_N-\lambda_k|}. \]

Lemma 3.

\[ \left| \int_{C_N}\operatorname{Sp}(BR_\lambda^A)^n\,d\lambda \right| \le A_2\sum_{k=1}^{N}\frac{1}{v_N-\lambda_k}, \]

where \(C_N\) is a contour containing the points \(\lambda_1,\ldots,\lambda_N\); \(n\ge 2\).

Lemma 4.

\[ \int_{C_N}\operatorname{Sp}(BR_\lambda^A)\,d\lambda = -2\pi i\sum_{k=1}^{N}(Be_k,e_k). \]

The lemma is obvious.

Lemma 5.

\[ \int_{C_N}\lambda\,\operatorname{Sp}R_\lambda^A(BR_\lambda^A)^n\,d\lambda = -\frac{1}{n}\int_{C_N}\operatorname{Sp}(BR_\lambda^A)^n\,d\lambda . \]

Proof of the theorem. Choose the contour \(C_{R,N}\) in the complex plane as follows: the arc of the circle \(|\lambda|=R,\ \operatorname{Re}\lambda\le v_N\), and the segment cut out by this circle on the line \(\operatorname{Re}\lambda=v_N\). For sufficiently large \(N\) this contour contains the points \(\lambda_1,\ldots,\lambda_N\) and \(\mu_1,\ldots,\mu_N\), since \(|\mu_n-\lambda_n|\le \|B\|\). We shall next use the identity

\[ \sum_{k=1}^{N}(\mu_k-\lambda_k) = \frac{1}{2\pi i}\int_{C_{R,N}}\lambda\,\operatorname{Sp}(R_\lambda^A-R_\lambda^C)\,d\lambda, \tag{3} \]

where \(R_\lambda^C=(C-\lambda E)^{-1}\). But

\[ R_\lambda^C=R_\lambda^A(E+BR_\lambda^A)^{-1}, \]

\[ (E+BR_\lambda^A)^{-1} = \sum_{k=1}^{n}(-1)^k(BR_\lambda^A)^k + (-1)^{n+1}(BR_\lambda^A)^{n+1}(E+BR_\lambda^A)^{-1}. \tag{4} \]

Substituting (4) into (3) and using Lemmas 1–5, we obtain our theorem.

Corollary. If

\[ \left(1+\frac{v_{N_i}}{\omega_{N_i}^{\,n}}\right) \sum_{k=1}^{\infty}\frac{1}{|v_{N_i}-\lambda_k|} \to 0 \quad \text{as } N_i\to\infty, \]

then for the existence of the sum

\[ \sum_{k=1}^{\infty}(\mu_k-\lambda_k) \]

it is necessary and sufficient that the sum

\[ \sum_{k=1}^{\infty}(Be_k,e_k) \]

exist, and in that case they are equal.

Example. If \(\lambda_n=O(n^\alpha)\) and the multiplicity of \(\lambda_n\) is \(O(n^\beta)\), then it can be shown that in this case our theorem gives

\[ \left| \sum_{k=1}^{N}(\mu_k-\lambda_k) - \sum_{k=1}^{N}(Be_k,e_k) \right| \le M\frac{\ln N}{N^{\,n[(\alpha-\beta)-1]-1}}. \]

if, however, \(\alpha-\beta>1\), then, by virtue of the arbitrariness of \(n\), the assertion of the corollary holds.

Thus, in order to compute
\[ \sum_{k=1}^{\infty}(\mu_k-\lambda_k), \]
in many cases it is sufficient to be able to compute
\[ \sum_{k=1}^{\infty}(Be_k,e_k), \]
where \(\{e_k\}_1^\infty\) is a proper basis of the operator \(A\). Therefore the question of the connection between the matrix traces of the operator \(B\) in various orthonormal bases of Hilbert space is of interest.

Let \(\{e_k\}_1^\infty\) and \(\{f_k\}_1^\infty\) be two orthonormal bases such that
\[ \sum_{k=1}^{\infty}\|f_k-e_k\|^2<\infty. \]
Let \(Ue_k=f_k,\ U=E+S\), where \(S\) is a Hilbert–Schmidt operator.

Theorem 2.
\[ \sum_{k=1}^{n}(Be_k,e_k)-\sum_{k=1}^{n}(Bf_k,f_k) = \sum_{k=1}^{n}([SB]e_k,e_k)+o(1). \]

Proof.
\[ \begin{aligned} \sum_{k=1}^{n}(Be_k,e_k)-\sum_{k=1}^{n}(Bf_k,f_k) &=\sum_{k=1}^{n}(Be_k,e_k)-\sum (U^{-1}BUe_k,e_k)\\ &=-\sum_{k=1}^{n}((S^*B+BS+S^*BS)e_k,e_k), \end{aligned} \]
since \(U^{-1}=E+S^*\). The operator \(S^*BS\) is nuclear; therefore
\[ \operatorname{Sp}(S^*BS)=\operatorname{Sp}(SS^*B); \]
\[ UU^{-1}=E=E+S+S^*+SS^*; \]
whence, in view of the fact that
\[ \sum_{k=1}^{n}(S^*BSe_k,e_k) = \sum_{k=1}^{n}(SS^*Be_k,e_k)+o(1) = -\sum_{k=1}^{n}((SB+S^*B)e_k,e_k)+o(1), \]
our theorem follows.

Corollary 1. If the matrix trace \([SB]\) exists in the basis \(\{e_k\}_1^\infty\), then for the convergence of
\[ \sum_{1}^{\infty}(Bf_k,f_k) \]
it is necessary and sufficient that the series
\[ \sum_{k=1}^{\infty}(Be_k,e_k) \]
converge, and in this case
\[ \sum_{k=1}^{\infty}(Bf_k,f_k) = \sum_{k=1}^{\infty}(Be_k,e_k) + \sum_{k=1}^{\infty}([BS]e_k,e_k). \]

Corollary 2. If \(f_n=e_n+g_n+\varepsilon_n\), where
\[ \sum_{k=1}^{\infty}\|g_k\|^2<\infty \quad\text{and}\quad \sum_{k=1}^{\infty}\|\varepsilon_k\|<\infty, \]
i.e., if \(S=T_1+T_2\), where \(T_2\) is a nuclear operator, then in the assertion of Corollary 1 one may replace \(S\) by \(T_1\).

Indeed,
\[ [BS]=[BT_1]+[BT_2], \]
but
\[ \operatorname{Sp}[BT_2]=\sum_{k=1}^{\infty}([BT_2]e_k,e_k)=0. \]

Example. Consider the Sturm–Liouville problem
\[ -y''+\lambda y=0,\qquad y'(0)-hy(0)=0,\qquad y'(\pi)+Hy(\pi)=0. \]

The eigenvalues of this problem have the asymptotics

\[ \lambda_n=n^2+\frac{H+h}{n\pi}+O\left(\frac{1}{n^2}\right). \]

The eigenfunctions have the asymptotics:

\[ f_n(x)=\sqrt{\frac{2}{\pi}}\left\{\cos nx+\frac{h}{n}\sin nx-\frac{H+h}{n\pi}x\sin nx\right\} +O\left(\frac{1}{n^2}\right). \]

Let \(\mu_n\) be the eigenvalues of the problem

\[ -y''+q(x)y+\mu y=0 \]

under the same boundary conditions.

Here the operator \(B\) is multiplication by \(q(x)\), where \(q(x)\) is continuously differentiable and its mean value is equal to 0. In order to compute

\[ \sum_{k=1}^{\infty}(Bf_n,f_n), \]

we use Corollary 2.

Put

\[ e_n(x)=\sqrt{\frac{2}{\pi}}\cos nx,\qquad T_1e_n(x)=\sqrt{\frac{2}{\pi}}\frac{\sin nx}{n}\left[h-\frac{H+h}{\pi}x\right], \]

\[ \sum_{k=1}^{\infty}(Bf_n,f_n) = \sum_{k=1}^{\infty}(Be_k,e_k) + \sum_{k=1}^{\infty}([BT_1]e_k,e_k), \]

but it can be shown that

\[ \sum_{k=1}^{\infty}([BT_1]e_k,e_k)=0, \]

whence we obtain

\[ \sum_{k=1}^{\infty}(\mu_k-\lambda_k) = \frac{2}{\pi}\sum_{k=1}^{\infty}\int_0^\pi q(x)\cos^2 nx\,dx = \frac{2}{\pi}\sum_{k=1}^{\infty}\int_0^\pi q(x)\cos 2nx\,dx = \]

\[ =\frac{q(0)+q(\pi)}{4}, \]

which is well known \((^1)\).

Moscow State University
named after M. V. Lomonosov

Received
31 X 1967

CITED LITERATURE

\(^1\) I. M. Gel'fand, B. M. Levitan, DAN, 88, No. 4 (1953). \(^2\) C. J. A. Halberg, V. A. Kramer, Duke Math. J., 27, No. 4 (1960).