V. B. LIDSKII

CONDITIONS FOR COMPLETE CONTINUITY OF THE RESOLVENT OF A NON-SELF-ADJOINT DIFFERENTIAL OPERATOR

(Presented by Academician A. N. Kolmogorov on 27 IX 1956)

A differential equation is considered

\[ -y''+p(x)y=\lambda y \tag{1} \]

with complex function \(p(x)\). It is assumed that \(p(x)\) has the form
\(p(x)=q(x)+ir(x)\), where \(q(x)\) and \(r(x)\) are real-valued functions, summable and bounded on each finite interval of the real axis. Along with equation (1) we consider the associated operator

\[ Ly=-y''+p(x)y, \tag{2} \]

acting in the Hilbert space \(L_2(-\infty,+\infty)\). The domain of definition \(D\) of the operator \(L\) consists of the functions \(y(x)\in L_2(-\infty,+\infty)\), absolutely continuous on each finite interval together with their derivative, and such that

\[ -y''+p(x)y\in L_2(-\infty,+\infty). \]

In the present paper conditions imposed on \(p(x)\) are given under which the operator \(L\) has a completely continuous resolvent and, consequently, a discrete spectrum.

In the case of a real-valued function \(p(x)\) bounded from below, a necessary and sufficient condition for the complete continuity of the resolvent of the self-adjoint operator (2) was given by A. M. Molchanov in his dissertation \((^1)\). We also note that, in the nature of the questions considered, the present work is close to the article of M. A. Naimark \((^2)\). We now proceed to the statement of the results.

Theorem 1. Let the function \(q(x)\), which is the real part of \(p(x)\), satisfy the condition

\[ \lim_{|x|\to\infty} q(x)=+\infty. \tag{3} \]

Then, whatever the function \(r(x)\), which is the imaginary part of \(p(x)\), may be, the operator \(Ly=-y''+p(x)y\) has a completely continuous resolvent, and hence a discrete spectrum.

Proof. Choose a real value of the parameter \(\lambda=\lambda_0\) for which \(q(x)-\lambda_0\geq 1\) \((-\infty<x<+\infty)\). With the aid of constructions close to those carried out by H. Weyl in the paper \((^3)\) in the case of a self-adjoint problem, it can be shown that equation (1) for \(\lambda=\lambda_0\) has two linearly independent solutions \(\psi_1(x)\) and \(\psi_2(x)\), belonging respectively to \(L_2(0,+\infty)\) and \(L_2(-\infty,0)\). Using these solutions, one can construct a bounded integral operator inverse to \(L-\lambda_0 E\). Denote it, as usual, by \(R_{\lambda_0}\), and show that \(R_{\lambda_0}\) is a completely continuous operator. For this we shall prove that the set of functions \(\{y\}: y=R_{\lambda_0}f\), where \(\|f\|\leq 1\), is compact in \(L_2(-\infty,+\infty)\). By the definition of \(R_{\lambda_0}\), we have:
\(-y''+(q(x)+ir(x))y-\lambda_0 y=f\). Multiplying

this relation by \(\overline{y}(x)\), take the integrals of both sides over the limits from \(-\infty\) to \(+\infty\), and equate the real parts. As a result we arrive at the equality

\[ \int_{-\infty}^{+\infty} |y'|^2\,dx+ \int_{-\infty}^{+\infty} (q(x)-\lambda_0)|y|^2\,dx = \operatorname{Re}\int_{-\infty}^{+\infty} f\overline{y}\,dx . \tag{4} \]

Putting \(R_{\lambda_0}f\) in place of \(y\) in the right-hand side of this formula and estimating it by the Cauchy–Bunyakovsky inequality, we obtain

\[ \int_{-\infty}^{+\infty} (q(x)-\lambda_0)|y|^2\,dx \leq \|R_{\lambda_0}\|. \tag{5} \]

This inequality is valid for all functions of the set \(\{y\}\). Let now \(\varepsilon>0\) be given; choose \(N\) so large that for \(|x|\geq N\) the inequality \(q(x)-\lambda_0 \geq 2\|R_{\lambda_0}\|\varepsilon^{-1}\) holds. From formula (5) it then follows that

\[ \int_{|x|\geq N} |y|^2\,dx \leq \frac{\varepsilon}{2}. \tag{6} \]

Let us now note that the functions \(\{y\}\), considered on the finite interval \([-N,N]\), form a compact family. This follows, for example, from the lemma proved in [4].* Having chosen on the interval \([-N,N]\) a finite \(\varepsilon/2\)-net for the functions \(\{y\}\), extend each function of the net by zero to the whole axis. In view of (6) we obtain a finite \(\varepsilon\)-net for the set \(\{y\}\) on the whole axis. Since \(\varepsilon\) is arbitrary, the set \(\{y\}\) is compact. Consequently, the operator \(R_{\lambda_0}\) is completely continuous. From the identity \(R_\lambda-R_{\lambda_0}=(\lambda-\lambda_0)R_\lambda R_{\lambda_0}\) it then follows that the resolvent is completely continuous for all \(\lambda\) for which it is defined. The theorem is proved.

Analogously to Theorem 1, with only some technical complications, Theorem 2 is proved.

Theorem 2. Let the function \(r(x)\), which is the imaginary part of \(p(x)\), satisfy the condition \(\lim_{|x|\to\infty} r(x)=+\infty\) (or \(-\infty\)).

Then, whatever the real part of \(p(x)\) may be, the operator

\[ Ly=-y''+p(x)y \]

has a completely continuous resolvent and, consequently, a discrete spectrum.

Theorem 2 needs explanation. The point is that if no restrictions are imposed on the real part of \(p(x)\), then for some \(q(x)\) all solutions of equation (1), for all \(\lambda\), may belong to \(L_2(-\infty,+\infty)\).** In this case the operator \(L\) is understood to be an operator acting not on the whole manifold \(D\), but on some submanifold of it consisting of functions satisfying certain boundary conditions at infinity (see in this connection [2]).

In conclusion, for a certain broad class of functions \(p(x)\) we shall give a necessary and sufficient condition for the complete continuity of the resolvent of operator (2):

* The compactness of the family \(\{y\}\) on the interval \([-N,N]\) under the conditions of the present theorem can be established quite simply on the basis of the inequality \(\int_{-\infty}^{+\infty}|y'|^2\,dx<C\), which easily follows from (4). We deliberately refer to a more general assertion, so that by means of the scheme presented it would be possible to reconstruct the proof of Theorem 2 formulated below.

** We emphasize that under the conditions of Theorem 1 (where \(q(x)\) is bounded below) this is impossible.

Theorem 3. Let the function \(q(x)\) be bounded below, and the function \(r(x)\) be semibounded\(^*\). Then, for the complete continuity of the resolvent of the operator

\[ Ly=-y''+(q(x)+ir(x))y \]

it is necessary and sufficient that, for any sequence of nonintersecting intervals \(\{D_n\}\) of the same length,

\[ \int_{D_n} (q(x)+|r(x)|)\,dx \to \infty, \]

when the interval goes to infinity\({}^{**}\).

For definiteness we shall assume \(r(x)\) to be bounded below. We shall also assume, without loss of generality, that \(q(x)\geqslant 1\) and \(r(x)\geqslant 1\). In this case, as we have already indicated, the operator \(L\) has a bounded inverse \(L^{-1}\). We shall first dwell on the proof of the sufficiency of the condition of the theorem. Denote by \(l\) the set of functions from \(D\) that are equal to zero outside a finite interval. It can be proved that, under the condition \(q(x)\geqslant 1\), the set \(Lu\), where \(u\in l\), is everywhere dense in \(L_2(-\infty,+\infty)\). Bearing in mind that we shall use this remark below, represent \(Lu\) in the form

\[ Lu=-u''+(q(x)+r(x))u+(i-1)r(x)u=Au+(i-1)ru. \tag{7} \]

By \(A\) we have denoted the self-adjoint operator \(-d^2/dx^2+q(x)+r(x)\), which, in view of the positivity of the function \(q(x)+r(x)\), is positive definite. Let us note here that, according to a theorem of A. M. Molchanov ((1), p. 175), the operator \(A\) has a completely continuous inverse \(A^{-1}\). We next transform formula (7) as follows:

\[ Lu=A^{1/2}\left(E+(i-1)A^{-1/2}rA^{-1/2}\right)A^{1/2}u. \tag{8} \]

It is not difficult to check that all operations on the right-hand side of this formula are successively executable. The operator applied to the function \(u\) on the right-hand side of formula (8) has a completely continuous inverse. Indeed, the operator \(A^{-1/2}rA^{-1/2}\) is a bounded self-adjoint operator\({}^{***}\). This fact can be proved analogously to Lemma 1 of our paper (5). Since, moreover, \(\operatorname{Im}(i-1)^{-1}\ne 0\), the operator standing in the parentheses has a bounded inverse. On the other hand, the operator \(A^{-1/2}\) is completely continuous together with \(A^{-1}\).

Therefore

\[ A^{-1/2}\left(E+(i-1)A^{-1/2}rA^{-1/2}\right)^{-1}A^{-1/2} \]

is also completely continuous. From formula (8) it follows that \(L^{-1}\), being bounded, coincides on the everywhere dense set \(\{Lu\}\) with a completely continuous operator. Hence \(L^{-1}\) is completely continuous. The first part of the theorem is proved.

We now outline in general terms the proof of the necessity of the condition of the theorem. Suppose that \(L^{-1}\) is completely continuous, but the condition of the theorem is not satisfied. Then there will be found a sequence of nonintersecting intervals \(\{D'_n\}\) of the same length \(2d\), for which

\[ \int_{D'_n} (q(x)+|r(x)|)\,dx < C. \tag{9} \]

Let us bring all \(D'_n\) into coincidence with the interval \([-d,+d]\) and consider the infinite system of equations
\(-y''+p_n(x)y=0,\ x\in[-d,d],\ n=1,2,\ldots;\ p_n(x)\) coincides with \(p(x)\) at the corresponding points of \(D'_n\). It can be proved that on

\(^*\) That is, bounded either below or above.
\({}^{**}\) Cf. (1), pp. 173–175.
\({}^{***}\) Strictly speaking, its closure has this property.

on \([-d,d]\) there exists a function \(\nu(x)\), the integral of the square of whose modulus is equal to one and which is orthogonal on \([-d,d]\) to all solutions of all the indicated equations. Consequently, each equation
\[ -y''+p(x)y=\bar{\nu}(x) \]
has on the interval \([-d,d]\) a nonzero solution \(\mu_n(x)\), which, together with its derivative, vanishes at the endpoints of the interval.

Consider two families of functions \(\{g_n(x)\}\) and \(\{f_n(x)\}\) possessing the following properties: each function with index \(n\) is nonzero only on \(D'_n\); the values of \(g_n(x)\) on \(D'_n\) coincide with the values of \(\mu_n(x)\) at the corresponding points of the interval \([-d,d]\), while the values of \(f_n(x)\) on \(D'_n\) coincide with the values of \(\bar{\nu}(x)\) on \([-d,d]\). Obviously, then \(\|f_n(x)\|=1\) and \(g_n=L^{-1}f_n\). In view of the complete continuity of the operator \(L^{-1}\), the family \(\{g_n\}\) is compact. Hence, owing to the orthogonality of the functions \(g_n\), it follows that \(\|g_n\|\to 0\) as \(n\to\infty\). It can be shown that this fact entails the uniform convergence to zero of the functions \(\mu_n(x)\) on the interval \([-d,d]\). Taking this into account, multiply each of the identities
\[ \bar{\nu}(x)=-\mu_n''+(q_n+i r_n)\mu_n \]
by \(e^{im\pi x/d}\) and integrate from \(-d\) to \(d\). Integrating the first of the integrals on the right twice by parts, we arrive, after simple estimates, at the inequality
\[ \left|\int_{-d}^{+d}\bar{\nu}(x)e^{\,i\frac{m\pi x}{d}}\,dx\right| \le \max|\mu_n| \left( \frac{2m^2\pi^2}{d} + \int_{-d}^{+d}q_n\,dx + \int_{-d}^{+d}|r_n|\,dx \right). \]

Since, as \(n\to\infty\), the expression in parentheses remains bounded by assumption, while \(\max|\mu_n|\to0\), it follows from this inequality that all Fourier coefficients of the function \(\nu(x)\) are equal to zero. This, however, contradicts the condition
\[ \int_{-d}^{+d}|\nu(x)|^2\,dx=1. \]

The contradiction obtained proves the assertion of the theorem.

Moscow Institute of Physics and Technology

Received
25 IX 1956

REFERENCES

¹ A. M. Molchanov, Tr. Mosk. matem. obshch., 2 (1953).
² M. A. Naimark, DAN, 85, No. 1 (1952).
³ G. Weyl, Math. Ann., 68, 222 (1910).
⁴ V. B. Lidskii, DAN, 112, No. 6 (1957).
⁵ V. B. Lidskii, DAN, 110, No. 2 (1956).