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This article is cited in 1 scientific paper (total in 1 paper) Molchanov's criterion for compactness of the resolvent for a nonselfadjoint Sturm–Liouville operator S. N. Tumanovab a Moscow Center of Fundamental and Applied Mathematics, Moscow, Russia b Faculty of Mechanics and Mathematics, Lomonosov Moscow State University, Moscow, Russia
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    § 1. IntroductionA significant result in the theory of qualitative spectral analysis for ordinary differential operators is Molchanov’s theorem, which provides a criterion for the complete continuity of the resolvent [1] for a singular Sturm–Liouville operator of the form The criterion, formulated originally for operators with real potentials $q(x)\geqslant\mathrm{const}$, $x\in\mathbb{R}$, was generalized in [2]–[8], mainly to relax the condition that $q$ is bounded below or to extend to the vector case when the role of $q$ is played by a Hermitian matrix, $q\geqslant0$. We do not consider partial differential operators. Generalizations concerning complex-valued $q$ are few, and the key work here is Lidskii’s paper [3], where Molchanov’s criterion was generalized to potentials taking values in the sector $0\leqslant\arg q(x)\leqslant\pi/2$ or $-\pi/2\leqslant\arg q(x)\leqslant0$ in the complex plane. Throughout what follows,
$$
\begin{equation}
q\in L_{1,\mathrm{loc}}(\mathbb{R}_+) \text{ is a complex-valued function}.
\end{equation}
\tag{1.1}
$$
We consider the case of operators on a half-axis (in the space $L_2(\mathbb{R}_+)$), whereas the results in the above papers concern operators in the space $L_2(\mathbb{R})$ on the whole axis. This difference is insignificant for our results: the theorems below can easily be extended to $L_2(\mathbb{R})$ with just one correction: the limit $x\to+\infty$ in Molchanov’s condition formulated below must be replaced by $x\to\infty$. We say that $q$ satisfies Molchanov’s condition if for each $a>0$.Even for real $q$, if we abandon boundedness below, Molchanov’s condition is not a criterion for the complete continuity of the resolvent any longer, for example, for ${q(x)=-x^2}$; see [9]. Nevertheless, it remains to be necessary even when no restrictions other than (1.1) are imposed on $q$. We prove this result in the more general case of ordinary differential operators of an arbitrary order $n\geqslant 2$. Supplementing it with sufficiency theorems and a counterexample, we draw a picture of possible generalizations of Molchanov’s theorem to the complex-valued case in terms of Molchanov’s condition itself. We consider the differential expression and the linear sets
$$
\begin{equation*}
\begin{gathered} \, \mathscr{D}=\bigl\{y\in L_2(\mathbb{R}_+)\mid y,y'\in \mathrm{AC}_{\mathrm{loc}}(\mathbb{R}_+),\ l(y)\in L_2(\mathbb{R}_+)\bigr\}, \\ \mathscr{D}_0=\bigl\{y\in\mathscr{D}\mid y(0)=y'(0)=0,\ \exists\, x_0>0\ \forall\, x\geqslant x_0\ y(x)=0\bigr\} \end{gathered}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\mathscr{D}_U=\bigl\{y\in\mathscr{D}\mid U(y)=0\bigr\},
\end{equation*}
\notag
$$
where $U$ is some form of a boundary condition at $x=0$:
$$
\begin{equation*}
U(y)=A y(0)+B y'(0), \qquad A,B\in\mathbb{C}, \quad |A|+|B|>0.
\end{equation*}
\notag
$$
Following [10], we define differential operators $L_0\subset L_U$ in $L_2(\mathbb{R}_+)$ on relevant domains $\mathscr D_0 \subset \mathscr D_U$ by the differential expression (1.2). Theorem 1. For $L_0$ to have an extension with compact resolvent, it is necessary that $q$ satisfy Molchanov’s condition. This is a consequence of our general result in Theorem 5, which we state and prove in § 3. Definition 1. A potential $q$ is said to satisfy the $\mathbb{R}^-$-condition if for all sufficiently large $x>x_0\geqslant0$ the values of $q(x)$ lie in the sector $\alpha\leqslant\arg (q(x)-q_0)\leqslant\beta$ for some $-\pi<\alpha\leqslant\beta<\pi$ and $q_0\in\mathbb{C}$. In other words, the $\mathbb{R}^-$-condition means such that for some $q_0\in\mathbb{C}$ the difference $q-q_0$ asymptotically takes values outside a small sector containing $\mathbb{R}^-$. Definition 2. A potential $q$ is said to be sectorial if it satisfies the $\mathbb{R}^-$-condition for $\beta-\alpha<\pi$. Theorem 2. Assume that a potential $q\in L_{1,\mathrm{loc}}(\mathbb{R}_+)$ is sectorial. Then the operator $L_U$ has a compact resolvent if and only if $q$ satisfies Molchanov’s condition. As the following theorem shows, the condition $\beta-\alpha<\pi$ cannot be relaxed. Theorem 3. There is a potential $q$ taking purely imaginary values $q(x)\in i\mathbb{R}$ for $x\in\mathbb{R}_+$ such that $|q|\to+\infty$ as $x\to+\infty$ but the operator $L_0$ has no extension with compact resolvent. For this potential we have $\beta-\alpha=\pi$, and it obviously satisfies Molchanov’s condition. The operator $L_U$ with the Dirichlet boundary condition $U(y)=y(0)$ has a bounded resolvent at least in the left-hand half-plane (see [3], Lemma 2, and [11]), but this resolvent is not a completely continuous operator. The following theorem gives a sufficient condition for the compactness of the resolvents of operators with potentials satisfying the $\mathbb{R}^-$-condition for $\beta - \alpha > \pi$. In this case the sectorial property of the operators under consideration is lost; in particular, the numerical image of $L_U$ can cover the whole complex plane [12]. Theorem 4. Assume that for some $x_0>0$ $|q(x)| \geqslant 1$ for $x \geqslant x_0$ and Under these conditions, for the resolvent of $L_U$ to be compact it is sufficient that for any $a>0$.The further presentation of is as follows: we devote § 2 to proving Theorems 2–4; in § 3 we return to Theorem 1, which is stated and proved there for general differential operators of order $n\geqslant 2$. § 2. Proof of Theorems 2–4Since neither Molchanov’s condition nor the complete continuity of the resolvent depend on the shift of the potential by a constant or its values on a finite interval $[0,x_0]$, we assume without loss of generality that $q_0=0$ and $x_0=0$. The proof of Molchanov’s criterion [1] is based on Rellich’s criterion of the compactness of the resolvent of a positive definite selfadjoint operator ([13], Ch. II, § 24, Theorem 11), which we generalize to $\mathrm m$-sectorial operators. For relevant definitions, see [14], Ch. V, § 3.10. Lemma 1. Let $A$ be an $\mathrm m$-sectorial operator in a Hilbert space $\mathfrak{H}$ with domain $\mathscr{D}_A\subset\mathfrak{H}$. Then $A$ has a compact resolvent if and only if the set of vectors $\varphi\in\mathscr{D}_A$ satisfying the condition is compact. Proof. We associate with $A$ a densely defined closed sectorial sesquilinear form $\mathfrak{a}$ ([14], Ch. VI, § 2, Theorem 2.7) such that its generating kernel is $\mathscr{D}_A$.
Recall that a linear set $\mathscr{D}_A$ is called a generating kernel of a closed form $\mathfrak{a}$ if the closure of the restriction of $\mathfrak{a}$ to $\mathscr{D}_A$ coincides with form $\mathfrak{a}$ itself. Given $\mathfrak{a}$, we construct the symmetric form $\mathfrak{t}=\operatorname{Re}\mathfrak{a}$: we set
$$
\begin{equation*}
\mathfrak{t}[u,v]=\frac{1}{2}(\mathfrak{a}[u,v]+\overline{\mathfrak{a}[v,u]})
\end{equation*}
\notag
$$
for $u,v\in\mathscr{D}_\mathfrak{a}$ (in the domain of $\mathfrak{a}$).
The form $\mathfrak{t}$ is closed on the domain $\mathscr{D}_\mathfrak{t}=\mathscr{D}_\mathfrak{a}$, densely defined, and nonnegative. According to the second representation theorem ([14], Ch. VI, § 2, Theorem 2.23), it is associated with a selfadjoint operator $T\geqslant0$, $\mathscr{D}_\mathfrak{t}=\mathscr{D}(T^{1/2})\supset\mathscr{D}_T$, where the domain $\mathscr{D}_T$ of $T$ is a generating kernel of $\mathfrak{t}$, and
$$
\begin{equation*}
\mathfrak{t}[u,v]=(T^{1/2}u,T^{1/2}v), \qquad u,v\in\mathscr{D}_\mathfrak{t}.
\end{equation*}
\notag
$$
The domain $\mathscr{D}_A$ is also a kernel of $\mathfrak{t}$. In view of the fact that $\mathscr{D}_A$ is a kernel of $\mathfrak{a}$, for any $u\in\mathscr{D}_\mathfrak{t}=\mathscr{D}_\mathfrak{a}$ there is a sequence $u_n\in\mathscr{D}_A$ such that $u_n\to u$ and $\mathfrak{a}[u_n,u_n]\to\mathfrak{a}[u,u]$ as $n\to\infty$. Clearly, the sequence $\mathfrak{t}[u_n,u_n]=\operatorname{Re}\mathfrak{a}[u_n,u_n]$ is convergent; owing to the closedness of $\mathfrak{t}$, the limit $\lim\mathfrak{t}[u_n,u_n]$ is equal to $\mathfrak{t}[u,u]$.
The resolvents of both $A$ and $T$ are compact or not simultaneously ([14], Ch. VI, § 3, Theorem 3.3). The subsequent reasoning uses Rellich’s criterion for $T$. We prove the necessity. Let $\Phi=\bigl\{\varphi\in\mathscr{D}_A \mid\operatorname{Re}(A\varphi,\varphi)\leqslant1\bigr\}$. Since $\mathscr{D}_T$ is a kernel of $\mathfrak{t}$, for any $\varepsilon>0$ and $\varphi\in\Phi$ there exists $u\in\mathscr{D}_T$ such that $|\mathfrak{t}[u,u]-\mathfrak{t}[\varphi,\varphi]|<\varepsilon$ and $|u-\varphi|<\varepsilon$. Since the resolvent of $A$ is compact, so is also the resolvent of $T$; since $(Tu,u)=\mathfrak{t}[u,u]\leqslant1+\varepsilon$, the whole of the set $\{u\}$ is compact and, as $\varepsilon$ is arbitrary, $\Phi$ is compact. We prove the sufficiency. Let $U=\bigl\{u\in\mathscr{D}_T\mid (Tu,u)\leqslant1\bigr\}$. Since $\mathscr{D}_A$ is the kernel of $\mathfrak{t}$, for any $\varepsilon>0$ and any $u\in U$ there exists $\varphi\in\mathscr{D}_A$ such that $|\mathfrak{t}[u,u]-\mathfrak{t}[\varphi,\varphi]|<\varepsilon$ and $|u-\varphi|<\varepsilon$. Since $\operatorname{Re}(A\varphi,\varphi)=\mathfrak{t}[\varphi,\varphi]\leqslant1+\varepsilon$, we conclude that the whole system $\{\varphi\}$ is compact and, as $\varepsilon>0$ is arbitrary, $U$ is compact. Therefore, $T$ has a compact resolvent; thus, $A$ also does. The lemma is proved. Proof of Theorem 2. The necessity follows from Theorem 1; we dwell only on the sufficiency.
Without loss of generality we assume that $-\pi<\alpha\leqslant0\leqslant\beta<\pi$ (otherwise we enlarge the sector in which the potential ranges so that the assumptions of the theorem still hold). We show that $L_U$ has a resolvent in some domain. Under the assumptions of the theorem the limit-point case ([11], Corollary to Theorem 4) takes place for the homogeneous equation $l(y)=\lambda y$, and the operators $L_{D}$ and $L_N$ with the Dirichlet and Neumann boundary conditions ($U(y)=y(0)$ and $U(y)=y'(0)$) have bounded resolvents $R_{D,\lambda}$ and $R_{N,\lambda}$, respectively, in some sector $\Lambda\subset\mathbb{C}$ ([15], Theorem 4.1). The explicit expressions for the corresponding resolvents in terms of Green’s functions imply that $L_U-\lambda$ for $\lambda\in\Lambda$ has at any rate a right inverse
$$
\begin{equation*}
R_\lambda=w_D(\lambda)R_{D,\lambda}+w_N(\lambda)R_{N,\lambda}, \qquad w_D(\lambda)+w_N(\lambda)=1,
\end{equation*}
\notag
$$
where $w_D$ and $w_N$ are meromorphic functions in $\Lambda$. This means that the operator $L_U-\lambda$ is surjective, at least in some subdomain $\lambda\in\Lambda_0\subset\Lambda$ not containing poles of $w_D$ and $w_N$. In view of the fact that the limit-point case occurs, the conjugate differential expression and the conjugate boundary condition specify the conjugate operator $(L_U-\lambda)^*$, which turns out to be surjective in a similar way. Hence $L_U-\lambda$ maps $\mathscr{D}_U$ bijectively onto the whole of $L_2(\mathbb{R}_+)$ for $\lambda\in\Lambda_0$, and the resolvent is well defined.
For each $\lambda\in\Lambda$ the operator $L_0-\lambda$ has defect 1 (the image of $L_0-\lambda$ is orthogonal to the only solution in $L_2(\mathbb{R}_+$) of the equation $y''=(\overline{q(x)}-\overline{\lambda})y$), and the resolvents of two arbitrary extensions of $L_0$ differ at most by a one-dimensional operator at each point $\lambda\in\Lambda$ where they exist simultaneously. Therefore, they are compact or noncompact simultaneously. So it suffices to carry out the rest of the proof only for the operator $L_D$ with the Dirichlet boundary condition at zero. As the resolvent exists for $\lambda\in\Lambda$, $L_D$ is closed. We denote the domain of $L_D$ by $\mathscr{D}_D$ and also consider the linear manifold $\mathscr{D}_{D0}$ of elements of $\mathscr{D}_D$ with compact support:
$$
\begin{equation}
\mathscr{D}_D=\bigl\{y\in\mathscr{D}\mid y(0)=0\bigr\}\quad\text{and} \quad \mathscr{D}_{D0}=\bigl\{y\in\mathscr{D}_D\mid \exists\, x_0>0\ \forall\, x\geqslant x_0\ y(x)=0\bigr\}.
\end{equation}
\tag{2.1}
$$
The domain $\mathscr{D}_{D0}$ is a generating kernel of $L_D$ in the following sense: the closure of the restriction of $L_D$ to $\mathscr{D}_{D0}$ coincides with $L_D$ ([3] Lemma 6).
We set $\theta=-(\alpha+\beta)/2$. The operator $M=e^{i\theta}L_D$ is $\mathrm m$-sectorial. Since the resolvent exists, this operator is maximal closed and
$$
\begin{equation*}
\begin{gathered} \, (My,y)=e^{i\theta}\int_0^{+\infty}|y'(x)|^2\,dx+\int_0^{+\infty}e^{i\theta}q(x)|y(x)|^2\,dx, \\ |{\arg(My,y)}|\leqslant\frac{\beta-\alpha}{2}<\frac{\pi}{2} \end{gathered}
\end{equation*}
\notag
$$
for any $y\in\mathscr{D}_{D0}$.
The further argument essentially repeats Molchanov’s reasoning in [1]; nevertheless, we present it here: it will be used in the proof of Theorem 4. Assume that Molchanov’s condition holds but the resolvent of $M$ is not compact. In view of Lemma 1 we can take a noncompact sequence $Y=\{y_n\}_{n=1}^{\infty}\subset\mathscr{D}_{D}$ such that $\operatorname{Re}(My_n,y_n)\leqslant1$, $n\in\mathbb{N}$. Since $\mathscr{D}_{D0}$ is a generating kernel of $M$, we assume without loss of generality that all the $y_n$ are in $\mathscr{D}_{D0}$. In fact, for every $\varepsilon>0$ it suffices to find for each $y_n\in Y$ an element $\widehat y_n\in\mathscr{D}_{D0}$ such that
$$
\begin{equation*}
\|y_n-\widehat y_n\|<\varepsilon\quad\text{and} \quad |{\operatorname{Re}(M\widehat y_n,\widehat y_n)-\operatorname{Re}(My_n,y_n)}|<\varepsilon.
\end{equation*}
\notag
$$
The system $\{\widehat y_n\}_{n=1}^{\infty}$ cannot be compact for all $\varepsilon>0$: there is $\varepsilon_0>0$ for which $\{\widehat y_n\}_{n=1}^{\infty}$ is not compact. In this case $\operatorname{Re}(M\widehat y_n,\widehat y_n)\leqslant1+ \varepsilon_0$. The suitable system $\{\widehat y_n/\sqrt{1+\varepsilon_0}\}_{n=1}^{\infty}$ consists of normalized elements.
There is $\varepsilon_0>0$ such that for any $T>0$ there exists $y_T\in Y$ such that
$$
\begin{equation}
\int_T^{+\infty} |y_T(x)|^2\,dx\geqslant\varepsilon_0.
\end{equation}
\tag{2.2}
$$
This follows from the compactness of the sequence of truncated functions $Y_T=\{y_{n,T}\}_{n=1}^{\infty}$, where $y_{n,T}(x)=y_n(x)$ for $x\in[0,T]$ and $y_{n,T}(x)=0$ for $x>T$ ([3], Lemma 8). If for any $\varepsilon>0$ there were $T_0=T_0(\varepsilon)$ such that for each $n\in\mathbb{N}$, then the sequence $Y$ itself would be compact (see the final part of the proof of Theorem 4 in [3]).
We take $d=(\varepsilon_0^{1/2}\cos^{1/2}\theta)/4>0$, an arbitrary $T>0$ and the corresponding $y_T$ and partition the ray $[T,+\infty)$ into intervals $D_n$ of equal length $d$, $n\in\mathbb{N}$; then on at least one of them, which we denote by $D_T$, it is true that
$$
\begin{equation}
\operatorname{Re}\biggl(e^{i\theta}\int_{D_T}|y_T'(x)|^2\,dx+\int_{D_T}e^{i\theta}q(x)|y_T(x)|^2\,dx\biggr) \leqslant\frac{1}{\varepsilon_0}\int_{D_T} |y_T(x)|^2\,dx
\end{equation}
\tag{2.3}
$$
and If the inequalities reverse to (2.3) were valid on all intervals $D_n$, $n\in\mathbb{N}$, such that (2.4) holds, then, summing them up and using (2.2), we would obtain the estimate $\operatorname{Re}(My_T,y_T)>1$, which is a contradiction.
We normalize $y_T$ by setting
$$
\begin{equation}
v_T=y_T \frac{d^{1/2}}{\bigl(\int_{D_T} |y_T(x)|^2\,dx\bigr)^{1/2}}.
\end{equation}
\tag{2.5}
$$
In view of the fact that the length $|D_T|$ is $d$, we have
$$
\begin{equation}
\begin{gathered} \, \operatorname{Re}\biggl( e^{i\theta}\int_{D_T}|v_T'(x)|^2\,dx+\int_{D_T}e^{i\theta}q(x)|v_T(x)|^2\,dx\biggr)\leqslant\frac{d}{\varepsilon_0}, \\ \frac{1}{|D_T|}\int_{D_T}|v_T(x)|^2\,dx=1. \end{gathered}
\end{equation}
\tag{2.6}
$$
In particular,
$$
\begin{equation*}
\operatorname{Re}\biggl(e^{i\theta}\int_{D_T}|v_T'(x)|^2\,dx\biggr)\leqslant\frac{d}{\varepsilon_0}, \quad\text{that is,}\quad \int_{D_T}|v_T'(x)|^2\,dx\leqslant\frac{d}{\varepsilon_0\cos\theta}.
\end{equation*}
\notag
$$
For any two points $x_1,x_2\in D_T$,
$$
\begin{equation*}
\begin{aligned} \, \bigl||v_T(x_1)|-|v_T(x_2)|\bigr| &\leqslant\bigl|v_T(x_1)-v_T(x_2)\bigr| \\ &\leqslant\int_{D_T}|v_T'(x)|\,dx\leqslant d^{1/2}\biggl(\int_{D_T}|v_T'(x)|^2\,dx\biggr)^{1/2}\leqslant \frac{1}{4}, \end{aligned}
\end{equation*}
\notag
$$
since $|v_T|$ is continuous on $D_T$ and the integral mean of $|v_T|^2$ is $1$ (see (2.6)). Therefore, there exists $x_0\in D_T$ such that $|v_T(x_0)|=1$. We conclude that ${|v_T(x)|\geqslant3/4}$ for all $x\in D_T$.
We turn to (2.6) again, which yields
$$
\begin{equation*}
\int_{D_T}\operatorname{Re}(e^{i\theta}q(x))|v_T(x)|^2\,dx \leqslant\frac{d}{\varepsilon_0}.
\end{equation*}
\notag
$$
However, $|\arg(e^{i\theta}q(x))|\leqslant(\beta-\alpha)/2$; hence
$$
\begin{equation*}
\operatorname{Re}(e^{i\theta}q(x))\geqslant|q|\cos\frac{\beta-\alpha}{2}
\end{equation*}
\notag
$$
and, consequently,
$$
\begin{equation*}
\int_{D_T}|q(x)|\,dx\leqslant\frac{d}{\varepsilon_0}\frac{4^2}{3^2}\sec\frac{\beta-\alpha}{2},
\end{equation*}
\notag
$$
which contradicts Molchanov’s condition for $q$ since $T>0$ can be arbitrary. This completes the proof.
Theorem 2 is proved. Proof of Theorem 3. We construct a counterexample, which will give us a proof.
We partition the half-axis into half-open intervals of length $\pi$ by setting $I_k=[\pi(k-1);\pi k)$, $k\in\mathbb{N}$, and we set $q_k(x)=ikr_{n_k}(x-\pi(k-1))$, $x\in I_k$, where
$$
\begin{equation*}
r_n(t)=(-1)^{j+1}, \qquad t\in\biggl[\pi\frac{j-1}{n},\pi\frac{j}{n}\biggr), \quad j=1,\dots,n.
\end{equation*}
\notag
$$
The procedure for choosing $\{n_k\}_{k=1}^{\infty}$ will become clear below.
Setting $q|_{I_k}=q_k$, we define $q$ for all $x\in\mathbb{R}_+$. It is obvious that $|q(x)|\to+\infty$ as ${x\to+\infty}$. As the boundary condition, we take the Dirichlet condition $U(y)=y(0)$. To simplify the notation, we denote the operator itself by $L$. The limit-point case takes place for the homogeneous equation $y''(x)=(q(x)- \lambda)y(x)$; as already noted above, the operator $L$ has a bounded resolvent in the left half-plane (see [3], Lemma 2, and [11]). We will show how, by choosing $\{n_k\}_{k=1}^{\infty}$ properly, we can ensure that the resolvent is not a compact operator. We show that for each $k\in\mathbb{N}$ there exists $n_k\in\mathbb{N}$ and a function $y_k\in\mathscr{D}_{U}$ vanishing everywhere outside $I_k$ such that $\|y_k\|=1$ and $\|Ly_k\|<2^{13}$. This will imply the noncompactness of the resolvent of the operator $L$ corresponding to the set $\{n_k\}_{k=1}^{\infty}$. This problem is local for each $I_k$. In this connection it suffices to show that for each $u>0$ there is $n\in\mathbb{N}$ such that for the operator in $L_2[0,\pi]$ defined in the domain
$$
\begin{equation*}
\mathscr{E}=\bigl\{y\in L_2[0,\pi]\mid y,y'\in \mathrm{AC}[0,\pi],\ y''\in L_2[0,\pi],\ y(0)=y(\pi)=0\bigr\}
\end{equation*}
\notag
$$
there exists $y\in\mathscr{E}$, $y(0)=y'(0)=y(\pi)=y'(\pi)=0$, $\|y\|=1$, such that ${\|L_n(u)y\|<2^{13}}$. Here and in what follows the norms in $L_2[0,\pi]$ are meant.
Along with the operators $L_n(u)$, we consider another operator $L_0$ with the same domain $\mathscr{E}$: we let $\mu_j=j^2$, $j\in\mathbb{N}$, denote the eigenvalues of $L_0$.The operator $B_ny=r_ny$ is bounded in $L_2[0,\pi]$, $\|B_n\|=1$, and $B_n\xrightarrow{w}0$ as ${n \to \infty}$ in the sense of weak operator convergence, which can easily be verified using the indicator functions $\chi_{[a,b]}$ of the intervals $[a,b] \subset [0,\pi]$, whose linear combinations are dense in $L_2[0,\pi]$. We show that, given $u>0$, the eigenvalues of the $L_n(u)$ converge pointwise to $\{\mu_j\}_{j=1}^{\infty}$ as $n\to\infty$: for any $\mu_j$ there is a sequence $\lambda_{j,n}\to \mu_j$ of eigenvalues of the operators $L_n(u)$, and for each compact set $\mathcal{C}$ not containing $\{\mu_j\}_{j=1}^{\infty}$ there is $n_0>0$ such that the compact set $\mathcal{C}$ does not contain eigenvalues of $L_n(u)$ for ${n>n_0}$. Fix $u>0$ and an integer $j_0>u-1/2$. We let $\Gamma_j$, $j\in\mathbb{N}$, denote the closed disc with centre $\mu_j$ and radius $u$. Set $\Gamma=\Gamma_1\cup\cdots\cup\Gamma_{j_0}$. The compact sets $\Gamma$, $\Gamma_{j}$, $j>j_0$, are pairwise disjoint; $\Gamma$ contains $j_0$ eigenvalues of the operator $L_0$, while each disc $\Gamma_{j}$ for $j>j_0$ contains one eigenvalue. For $n\in\mathbb{N}$ the eigenvalues of $L_n(u)=L_0+iuB_n$ lie in the union of $\Gamma$ and the $\Gamma_{j}$, $j>j_0$; $\Gamma$ contains $j_0$ eigenvalues of the perturbed operator $L_n(u)$ (taking account of multiplicities), while each disc $\Gamma_{j}$, $j>j_0$, contains one eigenvalue of it; see [14], Ch. V, § 4.3. We show that the limit points of the eigenvalues $\lambda_{j,n}$ of $L_n(u)$ as $n\to\infty$ coincide with the $\mu_j$, $j\in\mathbb{N}$ (the limit points are considered in the relevant compact sets $\Gamma$ and $\Gamma_{j}$, $j>j_0$). Assume the converse: as $l\to\infty$, a subsequence $\lambda_{n_l}$ converges to some ${\lambda_0\ne\mu_j}$, $j\in\mathbb{N}$. We let $f_{n_l}$ denote the eigenfunctions of $L_{n_l}(u)$ corresponding to the eigenvalues $\lambda_{n_l}$, $\|f_{n_l}\|=1$. Then
$$
\begin{equation*}
(L_0-\lambda_{n_l})f_{n_l}=-iuB_{n_l}f_{n_l}\quad\text{and} \quad \|{-iuB_{n_l}f_{n_l}}\|=u.
\end{equation*}
\notag
$$
Set $h_{n_l}=(L_0-\lambda_0)f_{n_l}$; the norms $\|h_{n_l}\|\leqslant u+|\lambda_{n_l}-\lambda_0|$ are bounded uniformly in $n_l$. Since $\lambda_0\ne\mu_j$, the operators $(L_0-\lambda_0)^{-1}$ and $(L_0-\overline{\lambda_0})^{-1}$ are defined on the whole of $L_2[0,\pi]$ and are compact. The sequence $f_{n_l}$ is also compact and, without loss of generality, can be assumed to be convergent: $f_{n_l}\to f$, $\|f\|=1$. Thus, $B_{n_l}f_{n_l}\xrightarrow{w}0$. We set $g=(L_0-\overline{\lambda_0})^{-1}f$; then
$$
\begin{equation}
\begin{aligned} \, \notag \|f\|^2 &=\lim_{l\to\infty}(f_{n_l},(L_0-\overline{\lambda_0})g)=\lim_{l\to\infty}((L_0-\lambda_0)f_{n_l},g) \\ &=\lim_{l\to\infty}\bigl\{((L_0-\lambda_{n_l})f_{n_l},g)+iu(B_{n_l}f_{n_l},g)\bigr\} \notag \\ &=\lim_{l\to\infty}((L_{n_l}(u)-\lambda_{n_l})f_{n_l},g)=0. \end{aligned}
\end{equation}
\tag{2.7}
$$
We have used the relations $(\lambda_{n_l}-\lambda_0)f_{n_l}\to0$ and $(B_{n_l}f_{n_l},g)\to0$.
However, $\|f\|\,{=}\,1$, and we arrive at a contradiction; hence $\lambda_n\,{\to}\,\mu_j$ for some ${j\,{\in}\,\mathbb{N}}$. For $j>j_0$ the disc $\Gamma_j$ contains a unique eigenvalue $\lambda_{j,n}$ for each $n\in\mathbb{N}$; thus, $\lambda_{j,n}\to\mu_j$ as $n\to\infty$. However, for $j=1,\dots,j_0$ the following inconvenient cases are potentially possible:
By eliminating these possibilities we will prove that for each $\mu_j$ there is a sequence of eigenvalues $\lambda_{j,n}$ converging to $\mu_j$. Assume that some sequence of eigenvalues $\lambda_n$ of $L_{n}(u)$ converges to $\mu_j$ as $n\to\infty$. As before, $f_n$, $\|f_n\|=1$, is the eigenfunction of $L_{n}(u)$ corresponding to $\lambda_n$. We show that we can extract a subsequence $f_{n_l}$ converging to the eigenfunction $f_0$, $\|f_0\|=1$, of $L_0$ that corresponds to $\mu_j$. We consider the orthogonal decomposition $L_2[0,\pi]=\mathfrak{M}\oplus\mathfrak{N}$, where $\mathfrak{M}$ is the one-dimensional eigenspace of $L_0$ corresponding to $\mu_j$ and $\mathfrak{N}$ is its orthogonal complement. Setting $f_n=f_{0n}+f_n^\perp$, where $f_{0n}\in\mathfrak{M}$ and $f_n^\perp\in\mathfrak{N}$, we write Again, we extract a convergent subsequence $f_{n_l}^\perp\to f_0^\perp\in\mathfrak{N}$. Using the boundedness of the $f_{0n}$ and the one-dimensionality of $\mathfrak{M}$, without loss of generality we can assume that $f_{0n_l}\to f_0\in\mathfrak{M}$. Repeating the reasoning in (2.7) for an arbitrary $g\in\mathfrak{N}$, we conclude that $f_0^\perp=0$. In other words, we have found a subsequence $f_{n_l}$ such that $f_{n_l}=f_{0n_l}+f_{n_l}^\perp\to f_0\in\mathfrak{M}$.The problem with inconvenient potential cases is dealt with as follows. $\bullet$ Assume that $\lambda_{j,n}\ne\lambda_{k,n}$ converge to the same $\mu_j$ as $n\to\infty$. Then we consider the normalized eigenfunctions $f_{n}$ and $g_n$ of $L_{n}(u)$ corresponding to $\lambda_{j,n}$ and $\lambda_{k,n}$. We extract subsequences $f_{n_l}\to f_0\in\mathfrak{M}$ and $g_{n_l}\to e^{i\gamma}f_0\in \mathfrak{M}$, $\gamma\in\mathbb{R}$ (because $\|f_0\|=\|g_0\|=1$). Since $\lambda_{j,n_l}\ne\lambda_{k,n_l}$, we have
$$
\begin{equation*}
0=\int_0^\pi f_{n_l}g_{n_l}\,dx\to e^{i\gamma}\int_0^\pi f^2_0\,dx\ne0,
\end{equation*}
\notag
$$
which is a contradiction.
$\bullet$ A multiple eigenvalue $\lambda_n$ (corresponding to a Jordan cell) converges to $\mu_j$. Let $f_n$, $\|f_n\|=1$, be eigenfunctions and $g_n\ne0$ be associated vectors (generally, not normalized): $g_n{\perp}\, f_n$ and $(L_n(u) - \lambda_n)g_n = f_n$, so that We show that the norms $\|g_n\|$ are above some $C>0$ for all $n\in \mathbb{N}$. Otherwise, there is a subsequence $g_{n_l}\to 0$. Without loss of generality we assume that $f_{n_l}\to f_0\in\mathfrak{M}$, $\|f_0\|=1$. We arrive at a contradiction: so
$$
\begin{equation*}
0=\lim_{l\to\infty}(g_{n_l}, (L_0-\mu_j)f_0)=\lim_{l\to\infty}((L_0-\mu_j)g_{n_l}, f_0)=(f_0,f_0)=1.
\end{equation*}
\notag
$$
We normalize $g_n$ in (2.8) by setting $\widehat g_n=g_n/\|g_n\|$ and obtain
$$
\begin{equation*}
(L_0-\mu_j)\widehat g_n=-iuB_n \widehat g_n+(\lambda_n-\mu_j)\widehat g_n+\frac{f_n}{\|g_n\|}.
\end{equation*}
\notag
$$
The right-hand side is bounded for all $n$; we introduce the representation $\widehat g_n=\widehat g_{0n}+\widehat g_n^\perp$, $\widehat g_{0n}\in \mathfrak{M}$, $\widehat g_n^\perp\in\mathfrak{N}$, and extract subsequences $\widehat g_{0n_l}\to \widehat g_0\in\mathfrak{M}$ and $\widehat g_{n_l}^\perp\to \widehat g_0^\perp \in\mathfrak{N}$ similarly to the construction for eigenfunctions. Without loss of generality we assume that $f_{n_l}\to f_0\in\mathfrak{M}$, $\|f_0\|=1$.
For any $h\in\mathfrak{N}$, $h=(L_0-\mu_j)v$, $v\perp f_0$, we have
$$
\begin{equation*}
(\widehat g_0^\perp, h)=\lim_{l\to\infty}(\widehat g_n^\perp,(L_0-\mu_j)v)=\lim_{l\to\infty}((L_0-\mu_j)\widehat g_n,v)=0,
\end{equation*}
\notag
$$
which yields the convergence $\widehat g_{n_l}\to \widehat g_0=e^{i\gamma}f_0\in\mathfrak{M}$, $\gamma\in\mathbb{R}$ (since $\|g_0\|=1$). The elimination of the inconvenient case under consideration follows from the contradiction
$$
\begin{equation*}
0=(\widehat g_{n_l},f_{n_l})\to e^{i\gamma}(f_0,f_0)\ne0.
\end{equation*}
\notag
$$
Thus, we have proved that for each $\mu_j$ there exists a sequence of eigenvalues such that $\lambda_{j,n}\to \mu_j$. This fact and the localization of eigenvalues in $\Gamma$ and $\Gamma_{j}$, $j>j_0$, imply, among other things, that for any compact set $\mathcal{C}$ not containing the points $\{\mu_j\}_{j=1}^{\infty}$ there is $n_0>0$ such that $\mathcal{C}$ does not contain eigenvalues of $L_n(u)$ for $n>n_0$, which completes the proof of pointwise convergence. Set $\mathscr{E}_0=\bigl\{y\in \mathscr{E}\mid y'(0)=y'(\pi)=0\bigr\}$. We show that for each $u>0$ there are $n_0>0$ and $y_{n_0} \in \mathscr{E}_0$, $\|y_{n_0}\| = 1$, such that $\|L_{n_0}(u)y_{n_0}\|<2^{13}$, which will complete the construction of the counterexample. We introduce the notation $H_n=L_n(u)\mathscr{E}_0=L_2[0,\pi]\ominus \langle y_{n,1},y_{n,2}\rangle$, where $y_{n,1}$ and $y_{n,2}$ form a fundamental system of solutions of the homogeneous equation Here $\langle y_{n,1},y_{n,2}\rangle$ is the 2-plane spanned by $y_{n,1}$ and $y_{n,2}$.We let $R_n(u)$ denote the inverse of $L_n(u)$. Its existence and the estimate $\|R_n(u)\|\leqslant 1$ follow from the straightforward calculation:
$$
\begin{equation*}
\|My\|\|y\|\geqslant\bigl|(My,y)\bigr|=|(Ly,y)\pm i(uB_ny,y)\bigr|\geqslant(Ly,y)\geqslant \|y\|^2,
\end{equation*}
\notag
$$
where $M=L_n(u)$ or $M=L_n(u)^*$.
We denote the restriction of $R_n(u)$ to $H_n$ by $\mathring{R}_n(u)$ and estimate its norm as an operator from $H_n$ to $L_2[0,\pi]$ as follows:
$$
\begin{equation*}
\|\mathring{R}_n(u)\|^2=\sup_{f\perp\langle y_{n,1},y_{n,2}\rangle}\frac{\|R_n(u)f\|^2}{\|f\|^2}\geqslant \min_{\mathscr{L},\,\dim\mathscr{L}=2}\ \max_{f\perp\mathscr{L}} \frac{\bigl((R_n(u))^*R_n(u)f,f\bigr)}{\|f\|^2}=s_3^2,
\end{equation*}
\notag
$$
where the minimum is taken over all possible two-dimensional subspaces of $L_2[0,\pi]$ and $s_3$ is the third singular number of the compact operator $R_n(u)$. The last equality follows from [16], Ch. II, § 1.
In view of the inequalities $1\geqslant \|R_n(u)\|=s_1\geqslant s_2$, using Weyl’s lemma we deduce that
$$
\begin{equation*}
s_3s_2s_1\geqslant|\lambda_1|\,|\lambda_2|\,|\lambda_3|, \quad \text{which yields } s_3\geqslant|\lambda_1|\,|\lambda_2|\,|\lambda_3|,
\end{equation*}
\notag
$$
where the $\lambda_j$, $j=1,2,3$, are the three largest (in modulus) eigenvalues of $R_n(u)$, which are the inverses of the three smallest (in modulus) eigenvalues of $L_n(u)$. The latter ones converge to the $\mu_j$ as $n\to\infty$, which implies that each $|\lambda_j|$ is above $1/\mu_4=1/2^4$ for some large $n_0$. Hence for $R_{n_0}(u)$ we have $s_3>1/2^{12}$ and
$$
\begin{equation*}
1\geqslant\|\mathring{R}_{n_0}(u)\|>\frac{1}{2^{12}}.
\end{equation*}
\notag
$$
Then there exists $f_{n_0}\in H_n$, $f_{n_0}\ne0$, such that $\|R_{n_0}(u)f_{n_0}\|\geqslant (1/2^{13})\|f_{n_0}\|$.
Setting $y_{n_0}=R_{n_0}(u)f_{n_0}/\|R_{n_0}(u)f_{n_0}\|$, we find the required element of $\mathscr{E}_0$. Theorem 3 is proved. Proof of Theorem 4. We extract a square root of $q$ by setting $p(x)=\sqrt{q(x)}$ for ${x\geqslant0}$ and choosing the branch so that $|{\arg p(x)}|\leqslant \pi/2-\varkappa/2$. Following [17], we set
Setting $C_0=(1-\delta)\sin\varkappa/2>0$, we derive the estimates
$$
\begin{equation}
\rho(x)=\operatorname{Re} p(x)-\frac{|p(x)|}{4}\biggl| \frac{q'(x)}{q^{3/2}(x)} \biggr|\geqslant\operatorname{Re} p(x)-|p(x)|\delta\sin\frac{\varkappa}{2}\geqslant C_0|p(x)|\geqslant C_0>0,
\end{equation}
\tag{2.9}
$$
since $\operatorname{Re} p(x)\geqslant |p(x)|\sin\varkappa/2$ in view of the inequality $|{\arg p(x)}|\leqslant \pi/2-\varkappa/2$ and since $|p(x)|\geqslant1$.
By [17], Theorem 1, $L_U$ has a bounded resolvent in some neighbourhood of zero $\lambda\in\Omega\subset\mathbb{C}$. The definite case (an analogue of the limit-point case) takes place for $L_U$. Reasoning like in the proof of Theorem 2, we assume that $L_U$ is specified by the Dirichlet boundary condition $U(y)=y(0)$ and defined on the domain $\mathscr{D}_D$ introduced in (2.1). We also derive from Theorem 1 in [17] that $\mathscr{D}_{D0}$ is a generating kernel for $L_U$. The operator $R_0=L_U^{-1}$ is bounded and can be extended to a bounded operator $\widetilde{R_0}\colon L_2(\mathbb{R}_+,1/|q|)\to L_2(\mathbb{R}_+)$ ([17], Theorem 4). This is equivalent to the boundedness of $Rg=\widetilde{R_0}(pg)$ as an ordinary operator in $L_2(\mathbb{R}_+)$. Since $R$ being compact implies that $R_0$ is too, we aim to prove the complete continuity of the operator $R$. The boundedness of $R$ means that has a closure as an operator in $L_2(\mathbb{R}_+)$. Since $L_2(\mathbb{R}_+)\subset L_2(\mathbb{R}_+,1/|q|)$ is dense in the metric of the weighted space, this closure $\overline M$ is generated by the kernel $\mathscr{D}_D$, and $\overline M^{-1}=R$.From the facts that $\mathscr{D}_{D0}$ is a generating kernel of $L_U$ and $|p|>1$ we conclude that $\mathscr{D}_{D0}$ is a generating kernel for $\overline M$. We turn to the proof of the compactness of the resolvent of $\overline M$. We show that $\overline M$ is an $\mathrm m$-sectorial operator. To do this, it suffices to show merely that for $y\in\mathscr{D}_{D0}$ the form $(My,y)$ is sectorial (see similar arguments in the proof of Theorem 2):
$$
\begin{equation}
\begin{aligned} \, \notag (My,y) &=\int_0^{+\infty}\biggl(y'(x)\biggl(-\frac{p'(x)}{p^2(x)}\overline{y(x)} +\frac{1}{p(x)}\overline{y'(x)}\biggr)+p(x)|y(x)|^2\biggr)\,dx \\ &=\int_0^{+\infty}\biggl(p(x)|y(x)|^2+\overline{p(x)}\biggl|\frac{y'(x)}{p(x)}\biggr|^2 -\frac{p'}{p}\biggl(\frac{y'(x)}{p(x)}\biggr)\overline{y(x)}\biggr)\,dx. \end{aligned}
\end{equation}
\tag{2.10}
$$
We set
$$
\begin{equation*}
\xi(x)=-\frac{p'}{p}\biggl(\frac{y'(x)}{p(x)}\biggr)\overline{y(x)}, \qquad |\xi(x)|\leqslant\frac{1}{2}\biggl|\frac{p'}{p}\biggr|\biggl( \biggl|\frac{y'(x)}{p(x)}\biggr|^2+|y(x)|^2\biggr);
\end{equation*}
\notag
$$
for convenience, we write
$$
\begin{equation*}
\xi(x)=m(x)\frac{1}{2}\biggl|\frac{p'}{p}\biggr|\biggl( \biggl|\frac{y'(x)}{p(x)}\biggr|^2+|y(x)|^2 \biggr)e^{i\varphi(x)},
\end{equation*}
\notag
$$
where $0\leqslant m(x)\leqslant1$ and $\varphi(x)=\arg\xi(x)$. As a result, (2.10) takes the following form:
$$
\begin{equation}
\begin{aligned} \, \notag (My,y) &=\int_0^{+\infty}\biggl(\biggl(p(x)+m(x)\frac{1}{2}\biggl|\frac{p'}{p}\biggr|e^{i\varphi(x)}\biggr)|y(x)|^2 \\ &\qquad +\biggl(\overline{p(x)}+m(x)\frac{1}{2}\biggl|\frac{p'}{p}\biggr|e^{i\varphi(x)} \biggr)\biggl|\frac{y'(x)}{p(x)}\biggr|^2\biggr)\,dx. \end{aligned}
\end{equation}
\tag{2.11}
$$
We have
$$
\begin{equation*}
\biggl|m(x)\frac{1}{2}\biggl|\frac{p'}{p}\biggr|e^{i\varphi(x)} \biggr|\leqslant\frac{1}{4}\biggl|\frac{q'(x)}{q^{3/2}(x)}\biggr|\,|p(x)|\leqslant \delta\sin\frac{\varkappa}{2}|p(x)|.
\end{equation*}
\notag
$$
We turn to the terms in (2.11). In view of the inequality $|{\arg p(x)}|\leqslant\pi/2-\varkappa/2$ and the above estimate it follows from geometric considerations that
$$
\begin{equation*}
\biggl|\arg\biggl( p(x)+m(x)\frac{1}{2}\biggl|\frac{p'}{p}\biggr|e^{i\varphi(x)} \biggr)\biggr|\leqslant\frac{\pi}2-\epsilon\quad\text{and} \quad \sin\biggl(\frac{\varkappa}{2}-\epsilon\biggr)\leqslant\delta\sin\frac{\varkappa}{2}
\end{equation*}
\notag
$$
for some $0<\epsilon\leqslant\varkappa/2$; hence $|\arg(My,y)|\leqslant\pi/2-\epsilon$, and the $\mathrm m$-sectoriality of $\overline M$ is proved.
In what follows we need an estimate from below for $y\in\mathscr{D}_{D0}$, which is immediately derived from (2.11) in view of (2.9):
$$
\begin{equation*}
\operatorname{Re} (My,y)\geqslant\int_0^{+\infty}C_0|p(x)|\biggl(|y(x)|^2+\biggl|\frac{y'(x)}{p(x)}\biggr|^2\biggr)\,dx.
\end{equation*}
\notag
$$
The reasoning that follows repeats essentially the proof of Theorem 2. We dwell only on the most important points. In a similar way, assuming that the resolvent of $\overline M$ is noncompact, we find a noncompact sequence $Y=\{y_n\}_{n=1}^{\infty}\subset\mathscr{D}_{D0}$ such that $\operatorname{Re} (My_n,y_n)<1$ for all ${n\in\mathbb{N}}$. We find $\varepsilon_0>0$ and, for each $T>0$, consider $y_T\in Y$ such that (2.2) holds. For $d=C_0\varepsilon_0/4$ we find an interval $D_T$ of length $d$ such that
$$
\begin{equation*}
\begin{gathered} \, C_0\int_{D_T}\biggl(|p(x)|\,|y_T(x)|^2+\frac{|y_T'(x)|^2}{|p(x)|}\biggr)\,dx\leqslant\frac{1}{\varepsilon_0}\int_{D_T} |y_T(x)|^2\,dx, \\ \int_{D_T} |y_T(x)|^2\,dx>0. \end{gathered}
\end{equation*}
\notag
$$
We perform the normalization (2.5) and finally arrive at the relations
$$
\begin{equation}
C_0\int_{D_T}\biggl(|p(x)|\,|v_T(x)|^2+\frac{|v_T'(x)|^2}{|p(x)|}\biggr)\,dx\leqslant \frac{d}{\varepsilon_0}\quad\text{and} \quad \frac{1}{|D_T|} \int_{D_T}|v_T(x)|^2\,dx=1.
\end{equation}
\tag{2.12}
$$
For any $x_1,x_2\in D_T$ we have
$$
\begin{equation*}
\begin{aligned} \, &\bigl||v_T(x_1)|^2-|v_T(x_2)|^2\bigr|\leqslant \bigl|v_T^2(x_1)-v_T^2(x_2)\bigr| \\ &\qquad\leqslant2\int_{D_T}|v_T(x)|\,|v_T'(x)|\,dx =2\int_{D_T}|p(x)|^{1/2}|v_T(x)|\frac{|v_T'(x)|}{|p(x)|^{1/2}}\,dx \\ &\qquad \leqslant\int_{D_T}\biggl(|p(x)|\,|v_T(x)|^2+\frac{|v_T'(x)|^2}{|p(x)|}\biggr)\,dx \leqslant \frac{d}{C_0\varepsilon_0}=\frac{1}{4}. \end{aligned}
\end{equation*}
\notag
$$
Taking account of (2.12) we conclude that $|v_T(x)|^2\geqslant3/4$ for all $x\in D_T$.
Turning to (2.12) again, we obtain the estimate
$$
\begin{equation*}
C_0\frac{3}{4}\int_{D_T}|p(x)|\,dx\leqslant C_0\int_{D_T}|p(x)|\,|v_T(x)|^2\,dx\leqslant\frac{d}{\varepsilon_0},
\end{equation*}
\notag
$$
or
$$
\begin{equation*}
\int_{D_T}|q(x)|^{1/2}\,dx=\int_{D_T}|p(x)|\,dx\leqslant\frac{1}{3},
\end{equation*}
\notag
$$
which leads to a contradiction and completes the proof.
Theorem 4 is proved. The conditions in Theorem 4 are also necessary for the compactness of the resolvent of the operator $\overline M$, which can be proved just as in [1]. This means that the method of the proof we propose does not make it possible to relax condition (1.3). § 3. Necessary condition for the compactness of the resolventWe establish a necessary condition for the compactness of the resolvent of an ordinary differential operator of an arbitrary order $n\geqslant2$. Since the results from the previous sections are not needed here, we use the same notation as there for similar objects. Assume we are given complex-valued functions $p_1\equiv C_0=\mathrm{const}$ and $p_j\in L_{1,\mathrm{loc}}(\mathbb{R}_+)$, $j=2,\dots,n$. We introduce the differential expression
$$
\begin{equation}
l(y)=\frac{d^n}{dx^n}y+\sum_{j=1}^n p_j\frac{d^{n-j}}{dx^{n-j}}y
\end{equation}
\tag{3.1}
$$
and the linear manifold
$$
\begin{equation*}
\begin{aligned} \, \mathscr{D}_0 &=\bigl\{y\in L_2(\mathbb{R}_+)\mid y,y',\dots,y^{(n-1)}\in \mathrm{AC}_{\mathrm{loc}}(\mathbb{R}_+),\ l(y)\in L_2(\mathbb{R}_+), \\ &\qquad y(0)=\dots=y^{(n-1)}(0)=0,\ \exists\, x_0>0\ \forall\, x\geqslant x_0\ y(x)=0\bigr\}. \end{aligned}
\end{equation*}
\notag
$$
We consider the differential operator $L_0$ in $L_2(\mathbb{R}_+)$ with domain $\mathscr{D}_0$ that is specified by the differential expression (3.1). Theorem 5. For $L_0$ to have an extension with compact resolvent, it is necessary that for each $a>0$. Before we prove the theorem, we state a lemma. Lemma 2. Assume that continuous complex-valued functions $\phi_\nu$ and $W_\nu$, $\nu=1,\dots,n$, $n\in\mathbb{N}$, defined on a interval $[0,d]$ are such that $W_n(0)=1$ and the system of functions $\{\phi_\nu\}_{\nu=1}^{n}$ is linearly independent on $[0,d_1]$ for each $d_1>0$, $0<d_1\leqslant d$. Then the operator $R$ in $L_2[0,d]$ defined by is infinite-dimensional (that is, its image is not a finite-dimensional linear set). Proof. We construct a sequence $g_l\in L_2[0,d]$, $l\in\mathbb{N}$, such that the sequence of images $f_l=Rg_l$ is linearly independent.
With this aim in view, first we construct disjoint intervals
$$
\begin{equation*}
I_l=(\alpha_l,\beta_l)\subset[0,d], \qquad l=0,1,\dots, \quad 0<\dots<\alpha_1<\beta_1<\alpha_0<\beta_0,
\end{equation*}
\notag
$$
such that the system $\{\phi_\nu\}_{\nu=1}^{n}$ is linearly independent on each $I_l$. It suffices to find one interval $I_0=(\alpha_0,\beta_0)$ of this kind and to construct further using recursion.
We choose an arbitrary $\beta_0$, $0<\beta_0<d$. If the system $\{\phi_\nu\}_{\nu=1}^{n}$ is linearly independent on $(\beta_0/2,\beta_0)$, then the construction is complete. Otherwise, let ${\mathscr{L}\subset\mathbb{C}^n}$ be a maximal linear subspace such that $\sum_\nu A_\nu\phi_\nu(x)\equiv0$ for all $A=(A_\nu)\in\mathscr{L}$ and all $x\in(\beta_0/2,\beta_0)$. Clearly, $1\leqslant\dim\mathscr{L}\leqslant n$. Let $\alpha'$ be the infimum of $\alpha>0$ such that for the interval $(\alpha,\beta_0)$ there is $A=(A_\nu)\in\mathscr{L}$ such that $\sum_\nu A_\nu\phi_\nu(x)\equiv0$ for $x\in(\alpha,\beta_0)$. It is obvious that ${0\leqslant\alpha'\leqslant\beta_0/2}$. We show that $\alpha'>0$. Otherwise, there exists $A_k=(A_{k\nu})\in\mathscr{L}$, $\|A_k\|=1$, $k\in \mathbb{N}$, such that $\sum_\nu A_{k\nu}\phi_\nu(x)\equiv0$ for $x\in(\beta_0/(k+1),\beta_0)$. We extract a convergent subsequence $A_{k_s}\to A_0=(A_{0\nu})\in\mathscr{L}$ as $s\to\infty$, $\|A_0\|=1$. For $s\geqslant s_0\geqslant1$ we have
$$
\begin{equation*}
\sum_\nu A_{0\nu}\phi_\nu(x)=\sum_\nu(A_{0\nu}-A_{k_s\nu})\phi_\nu(x) \quad\text{for all } x\in\biggl(\frac{\beta_0}{k_{s_0}+1},\beta_0\biggr).
\end{equation*}
\notag
$$
Letting $s\to\infty$, as $s_0$ is arbitrary and all the $\phi_\nu$ are continuous, we infer that $\sum_\nu A_{0\nu}\phi_\nu(x)=0$ for all $x\in[0,\beta_0]$, which contradicts the independence of the system $\{\phi_\nu\}_{\nu=1}^{n}$ on the interval $[0,\beta_0]$. Thus, $\alpha'>0$.
Set $\alpha_0\!=\!\alpha'/2$. Then $\{\mkern-1mu\phi_\nu\mkern-1mu\}_{\nu=1}^{n}$ turns out to be linearly independent on ${I_0\!=\!(\mkern-1mu\alpha_0,\beta_0)}$. Otherwise, for some $B=(B_\nu)\in\mathbb{C}^n$ we have
$$
\begin{equation*}
\sum_\nu B_\nu\phi_\nu(x)\equiv0 \quad\text{for } x\in(\alpha_0,\beta_0)
\end{equation*}
\notag
$$
and therefore for all $x\in(\beta_0/2,\beta_0)$, that is, $B\in\mathscr{L}$, which contradicts the choice of $\alpha'$.
Let $I_l=(\alpha_l,\beta_l)$, $l=0,1,\dots$, be a system of disjoint intervals of linear independence of $\{\phi_\nu\}_{\nu=1}^{n}$. Let $\beta_0$ be chosen so that $W_n(x)\ne0$ for $x\in[0,\beta_0]$. For $l\in\mathbb{N}$ we construct $g_l\in L_2[0,d]$ as follows: it is identically $0$ outside $I_l$, and the following procedure is used on $I_l$. On $I_1$ the system of functions $\{W_\nu\}_{\nu=1}^{n}$ can be expressed in terms of $1\leqslant k\leqslant n$ independent functions $W_{s_j}$ (at any rate, $W_n(x)\not\equiv0$ on $I_l$): We choose $g_l\in L_2(I_l)$ so that $\overline{g_l}$ is not orthogonal to $W_{s_1}$; if $k>1$, then we additionally require that $\overline{g_l}$ is orthogonal to all $\{W_{s_j}\}_{j=2}^{n}$ in the sense of the metric in $L_2(I_l)$.We set $f_l=Rg_l$, $l\in\mathbb{N}$. Then
$$
\begin{equation*}
f_l(x)= \begin{cases} 0, & x\in[0,\alpha_l], \\ \displaystyle \sum_{\nu=1}^nA_{\nu 1}\phi_\nu(x)\int_{\alpha_l}^{\beta_l} W_{s_1}(\xi)g_l(\xi)\,d\xi, & x\in[\beta_l,d]. \end{cases}
\end{equation*}
\notag
$$
It is significant for us that the integral in the above formula is not zero and
$$
\begin{equation*}
\sum_\nu A_{\nu 1}\phi_\nu(x)\not\equiv0 \quad\text{for } x\in I_j \quad\text{for all } 0\leqslant j<l.
\end{equation*}
\notag
$$
Thus, $f_l(x)\equiv0$ for $x\in I_j$ when $j>l$, but $f_l(x)\not\equiv0$ for $x\in I_j$ when $0\leqslant j<l$. The sequence $f_l$ is linearly independent. Lemma 2 is proved. Proof of Theorem 5. Assume the converse: there exists an extension $L\supset L_0$ with compact resolvent and for some system $D_k=[a_k,b_k]\subset\mathbb{R}_+$ of disjoint intervals of equal length $d=|D_k|$.
Since $p_1\equiv \mathrm{const}$, it is convenient to write (3.2) for all $p_j$. Without loss of generality we assume that $C_j>0$. If necessary, we reduce the lengths of the intervals so that We arrive at a contradiction if for each $k\in\mathbb{N}$ there exists a function $y_k\in\mathscr{D}_0$ vanishing outside $D_k$ and such that $\|y_k\|=1$ and $\|L_0y_k\|<C$. To find such functions we consider the operators $L_{k,0}$ in $L_2[0,d]$ specified by the differential expressions $l_k(y(x))=l(y(x+a_k))$, $x\in[0,d]$, on the domains
$$
\begin{equation*}
\begin{aligned} \, \widetilde{\mathscr{D}_{0k}} &=\bigl\{y\in L_2[0,d]\mid y,y',\dots,y^{(n-1)}\in \mathrm{AC}[0,d],\ l_k(y)\in L_2[0,d], \\ &\qquad y(0)=\dots=y^{(n-1)}(0)=y(d)=\dots=y^{(n-1)}(d)=0\bigr\}. \end{aligned}
\end{equation*}
\notag
$$
Once we find a sequence $f_k\in\widetilde{\mathscr{D}_{0k}}$ such that $\|f_k\|=1$ and $\|L_{k,0}f_k\|<C$ for ${k\in\mathbb{N}}$, we solve the problem by setting $y_k(x)=f_k(x-a_k)$ for $x\in D_k$ and extending $y_k$ by zero outside $D_k$. For each homogeneous equation $l_k(y)=0$, $k\in\mathbb{N}$, we consider the special solution $\psi_{k,\nu}$ of the Cauchy problem that is specified by the initial conditions
$$
\begin{equation*}
\psi_{k,\nu}^{(j-1)}(0)=\delta_\nu^j, \qquad j,\nu=1,\dots,n, \quad k\in\mathbb{N},
\end{equation*}
\notag
$$
where $\delta_\nu^j$ is the Kronecker symbol.
The existence of these solutions follows from [13], Ch. V, § 16, Theorem 1. All the $\psi_{k,\nu}$ and their derivatives up to the $(n-1)$st order inclusive are absolutely continuous on $[0,d]$. We fix $k\in\mathbb{N}$. For any $g\in L_2[0,d]$ the equation $l_k(f)=g$ has a unique solution such that $f_k^{(j)}(0)=0$, $j=0,\dots,n-1$ ([13], Ch. V, § 16, Theorem 1). It can be represented as $f=R_kg$, where $R_k$ (the Cauchy operator) is a bounded Volterra operator in $L_2[0,d]$, namely,
$$
\begin{equation}
f=(R_kg) (x)= \int_0^x\frac{1}{W(\xi)} \begin{vmatrix} \psi_{k,1}(\xi) & \psi_{k,2}(\xi) & \cdots & \psi_{k,n}(\xi)\\ \psi_{k,1}'(\xi) & \psi_{k,2}'(\xi) & \cdots & \psi_{k,n}'(\xi)\\ \vdots & \vdots & \ddots & \cdots \\ \psi_{k,1}^{(n-2)}(\xi) & \psi_{k,2}^{(n-2)}(\xi) & \cdots & \psi_{k,n}^{(n-2)}(\xi)\\ \psi_{k,1}(x) & \psi_{k,2}(x) & \cdots & \psi_{k,n}(x) \end{vmatrix} g(\xi)\,d\xi,
\end{equation}
\tag{3.4}
$$
where $W(\xi)$ is the Wronskian:
$$
\begin{equation*}
\begin{aligned} \, W(\xi)&= \begin{vmatrix} \psi_{k,1}(\xi) & \psi_{k,2}(\xi) & \cdots & \psi_{k,n}(\xi)\\ \psi_{k,1}'(\xi) & \psi_{k,2}'(\xi) & \cdots & \psi_{k,n}'(\xi)\\ \vdots & \vdots & \ddots & \cdots \\ \psi_{k,1}^{(n-2)}(\xi) & \psi_{k,2}^{(n-2)}(\xi) & \cdots & \psi_{k,n}^{(n-2)}(\xi)\\ \psi_{k,1}^{(n-1)}(\xi) & \psi_{k,2}^{(n-1)}(\xi) & \cdots & \psi_{k,n}^{(n-1)}(\xi)\\ \end{vmatrix} \\ &=\exp\biggl(-\int_0^\xi p_1(a_k{+}\,t)\,dt\biggr)=e^{-\xi C_0}. \end{aligned}
\end{equation*}
\notag
$$
Differentiating (3.4) $n-1$ times, we see that $f\in\widetilde{\mathscr{D}_{0k}}$ if and only if
$$
\begin{equation*}
\int_0^d e^{\xi C_0} \begin{vmatrix} \psi_{k,1}(\xi) & \psi_{k,2}(\xi) & \cdots & \psi_{k,n}(\xi)\\ \psi_{k,1}'(\xi) & \psi_{k,2}'(\xi) & \cdots & \psi_{k,n}'(\xi)\\ \vdots & \vdots & \ddots & \cdots \\ \psi_{k,1}^{(n-2)}(\xi) & \psi_{k,2}^{(n-2)}(\xi) & \cdots & \psi_{k,n}^{(n-2)}(\xi)\\ \psi_{k,1}^{(s)}(d) & \psi_{k,2}^{(s)}(d) & \cdots & \psi_{k,n}^{(s)}(d) \end{vmatrix} g(\xi)\,d\xi=0, \qquad s=0,\dots,n-1,
\end{equation*}
\notag
$$
or, in other words, if $g\in\mathring{\mathfrak{H}}_k=L_2[0,d]\ominus\mathfrak{K}_k$, where $\mathfrak{K}_k$ is the finite-dimensional subspace of $L_2[0,d]$ spanned by the complex conjugates of the functions
$$
\begin{equation*}
W_{ks}(x)= e^{x C_0} \begin{vmatrix} \psi_{k,1}(x) & \psi_{k,2}(x) & \cdots & \psi_{k,n}(x)\\ \psi_{k,1}'(x) & \psi_{k,2}'(x) & \cdots & \psi_{k,n}'(x)\\ \vdots & \vdots & \ddots & \cdots \\ \psi_{k,1}^{(n-2)}(x) & \psi_{k,2}^{(n-2)}(x) & \cdots & \psi_{k,n}^{(n-2)}(x)\\ \psi_{k,1}^{(s)}(d) & \psi_{k,2}^{(s)}(d) & \cdots & \psi_{k,n}^{(s)}(d) \end{vmatrix}, \qquad s=0,\dots,n-1;
\end{equation*}
\notag
$$
thus, $\dim\mathfrak{K}_k\leqslant n$.
The operator $L_{k,0}$ maps $\widetilde{\mathscr{D}_{0k}}$ bijectively onto $\mathring{\mathfrak{H}}_k$. We let $\mathring{R}_k\subset R_k$ denote its inverse, which acts bijectively from $\mathring{\mathfrak{H}}_k$ onto $\widetilde{\mathscr{D}_{0k}}$. To find the required sequence $f_k$, it suffices to show that the norms $\|\mathring{R}_k$ are uniformly bounded below for all $k\in\mathbb{N}$: $\|\mathring{R}_k\|\geqslant C>0$. Then for each $k$ there exists $u_k\in\mathring{\mathfrak{H}}_k$, $u_k\not\equiv 0$, such that $\|v_k\|\geqslant (C/2)\|u_k\|$ for $v_k=\mathring{R}_ku_k$. The required $f_k$ is $v_k/\|v_k\|$; we have $\|L_{k,0}f_k\|\leqslant 2/C$, and the following estimate holds:
$$
\begin{equation*}
\|\mathring{R}_k\|^2=\sup_{g\in\mathring{\mathfrak{H}}_k}\frac{\|R_kg\|^2}{\|g\|^2} \geqslant\min_{\substack{\mathfrak{L}\subset L_2[0,d]\\ \dim\mathfrak{L}\leqslant n}}\ \max_{\substack{g\in L_2[0,d]\\ g\perp\mathfrak{L}}} \frac{(R_k^*R_k g,g)}{(g,g)}=s_{n+1}^2(R_k).
\end{equation*}
\notag
$$
The minimum is taken over all possible linear sets $\mathfrak{L}\subset L_2[0,d]$ of dimension at most $n$, and $s_{n+1}(R_k)$ is the $(n+1)$st singular number of the operator $R_k$ The last equality follows from [16], Ch. II, § 1.
Next, we show that from the sequence of operators $R_k$ we can extract a subsequence $R_{k_m}\to R$, $m\to\infty$, converging uniformly to an infinite-dimensional operator $R$, all of whose $s$-numbers are thus positive. The singular numbers $s_l(R_{k_m})$ converge to $s_l(R)>0$, $l\in\mathbb{N}$ (see [16], Ch. II, § 2, Corollary 2.3). In particular, for ${m>m_0}$ $\|\mathring{R}_{k_m}\|\geqslant s_{n+1}(R_{k_m})\geqslant s_{n+1}(R)/2=C>0$. Considering the $D_k$ with indices $k_m$, we can assume without loss of generality that $\|\mathring{R}_k\|\geqslant C>0$ for all $k\in\mathbb{N}$ . This completes the proof. We show how to extract a uniformly convergent subsequence from $R_k$. We integrate the identity $l_k(\psi_{k,\nu})=0$ $s$ times (for $s=1,\dots,n$):
$$
\begin{equation}
\begin{aligned} \, \notag \psi_{k,\nu}^{(n-s)}(x) &=\sum_{r=0}^{s-1}\frac{x^r}{r!}\psi_{k,\nu}^{(n-s+r)}(0) \\ &\qquad-\int_0^x d\xi_1\int_0^{\xi_1}\cdots\,d\xi_{s-1}\int_0^{\xi_{s-1}} \sum_{j=1}^n \widetilde{p_{jk}}(\xi_s)\psi_{k,\nu}^{(n-j)}(\xi_s)\,d\xi_s, \end{aligned}
\end{equation}
\tag{3.5}
$$
where $\widetilde{p_{jk}}(x)=p_j(x+a_k)$. There are no inner integrals for $s=1$.
We introduce the notation $M_{k,\nu,l}=\max|\psi_{k,\nu}^{(l)}(x)|$, $x\in[0,d]$, $l=0,\dots,n-1$. Using (3.5) and (3.2) we conclude that
$$
\begin{equation*}
M_{k,\nu,n-s}\leqslant e^d+d^{s-1}\sum_{j=1}^nC_j M_{k,\nu,n-j}.
\end{equation*}
\notag
$$
Multiplying both sides by $C_s$ and summing over $s\,{=}\,1,\dots,n$ we obtain the inequality
$$
\begin{equation*}
\sum_{s=1}^{n}C_sM_{k,\nu,n-s}\leqslant e^d\sum_{s=1}^{n}C_s+\eta\sum_{j=1}^nC_j M_{k,\nu,n-j},
\end{equation*}
\notag
$$
which, in view of (3.3), yields the estimate
$$
\begin{equation*}
\sum_{s=1}^{n}C_sM_{k,\nu,n-s}\leqslant\frac{e^d}{1-\eta}\sum_{s=1}^{n}C_s.
\end{equation*}
\notag
$$
Since $C_s>0$, this estimate means that all quantities $M_{k,\nu,l}$ are bounded uniformly in $k\in\mathbb{N}$, $\nu=1,\dots,n$ and $l=0,\dots,n-1$. For fixed $\nu=1,\dots,n$ and $l=0,\dots,n-2$, we consider the sequence $\{\psi_{k,\nu}^{(l)}\}_{k\in\mathbb{N}}$. It is precompact in the uniform metric on $[0,d]$ by the Arzelà–Ascoli theorem. We choose a sequence of indices $k_m$ so that the subsequences $\{\psi_{k_m,\nu}^{(l)}\}_{m\in\mathbb{N}}$ are uniformly convergent on $[0,d]$ for all $\nu=1,\dots,n$ and $l=0,\dots,n-2$:
$$
\begin{equation*}
\psi_{k_m,\nu}^{(l)}\Rightarrow\phi_\nu^l, \qquad m\to\infty,
\end{equation*}
\notag
$$
where the $\phi_\nu^l$ are continuous functions with indices $\nu=1,\dots,n$ and $l=0,\dots,n-2$.
For convenience we set $\phi_\nu=\phi_\nu^0$. In view of uniform convergence it follows from the well-known theorem on limit transition ([18], Ch. XVI, Theorem 4) that each function $\phi_\nu$ has $ n-2 $ derivatives and $\phi_\nu^l=\phi_\nu^{(l)}$, $l=0,\dots,n-2$. Noting that in the definition of $R_k$ the determinant depends only on the derivatives $\psi_{k,\nu}^{(l)}$ of order at most $n-2$, we conclude that $R_{k_m}\Rightarrow R$ as $m\to\infty$ in the sense of uniform operator convergence, where
$$
\begin{equation}
(Rg) (x)= \int_0^xe^{\xi C_0} \begin{vmatrix} \phi_1(\xi) & \phi_2(\xi) & \cdots & \phi_n(\xi)\\ \phi_1'(\xi) & \phi_2'(\xi) & \cdots & \phi_n'(\xi)\\ \vdots & \vdots & \ddots & \cdots \\ \phi_1^{(n-2)}(\xi) & \phi_2^{(n-2)}(\xi) & \cdots & \phi_n^{(n-2)}(\xi)\\ \phi_1(x) & \phi_2(x) & \cdots & \phi_n(x) \end{vmatrix} g(\xi)\,d\xi, \qquad g\,{\in}\, L_2[0,d].
\end{equation}
\tag{3.6}
$$
The operator $R$ is compact as a uniform limit of compact operators. It remains to prove that $R$ is infinite dimensional. Note that the system of functions $\{\phi_\nu\}_{\nu=1}^{n}$ is linearly independent on each interval $[0,d_1]\subset[0,d]$. Assume the converse: there are $A_\nu\in\mathbb{C}$ such that $\sum_\nu A_\nu\phi_\nu(x) \equiv 0$ for all $x\in[0,d_1]$. We turn to (3.5) to express the linear combinations $\sum_\nu A_\nu\psi_{k,\nu}^{(n-s)}$ for ${s=n}$. We obtain
$$
\begin{equation*}
\begin{aligned} \, \sum_{\nu=1}^n A_\nu\psi_{k,\nu}(x) &=\sum_{r=0}^{n-1}\frac{x^r}{r!}\sum_{\nu=1}^nA_\nu\psi_{k,\nu}^{(r)}(0) \\ &\qquad-C_0\int_0^x d\xi_1\int_0^{\xi_1}\cdots\,d\xi_{n-1}\int_0^{\xi_{n-1}} \sum_{\nu=1}^nA_\nu\psi_{k,\nu}^{(n-1)}(\xi_n)\,d\xi_n \\ &\qquad-\int_0^x d\xi_1\int_0^{\xi_1}\cdots\,d\xi_{n-1}\int_0^{\xi_{n-1}} \sum_{j=2}^n \widetilde{p_{jk}}(\xi_n)\sum_{\nu=1}^nA_\nu\psi_{k,\nu}^{(n-j)}(\xi_n)\,d\xi_n. \end{aligned}
\end{equation*}
\notag
$$
Setting $k=k_m$ and taking the limit as $m\to\infty$, uniformly with respect to $x\in[0,d_1]$ we arrive at the identity
$$
\begin{equation*}
0\equiv\sum_{r=0}^{n-1}\frac{x^r}{r!}\sum_{\nu=1}^nA_\nu\delta_\nu^{r+1}(0) +C_0 \sum_{r=1}^{n-1}\frac{x^r}{r!}\sum_{\nu=1}^nA_\nu\delta_\nu^r,
\end{equation*}
\notag
$$
or
$$
\begin{equation*}
0\equiv A_1+\sum_{r=1}^{n-1}\bigl(A_{r+1}+C_0A_r\bigr)\frac{x^r}{r!},
\end{equation*}
\notag
$$
which shows that $A_\nu=0$, $\nu=1,\dots,n$.
We represent $R$ in the form
$$
\begin{equation*}
(Rg) (x)=\int_0^x\sum_{\nu=1}^n\phi_\nu(x)W_\nu(\xi)g(\xi)\,d\xi, \qquad g\in L_2[0,d],
\end{equation*}
\notag
$$
where the $W_\nu$ are continuous functions. These functions are algebraic complements, multiplied by $e^{\xi C_0}$, in the expansion of the determinant in (3.6) with respect to the last row, and $W_n(0)=1$. Lemma 2 shows that $R$ is infinite dimensional.
Theorem 5 is proved. AcknowledgmentsThe author is grateful to the corresponding member of the RAS, professor Andrei Andreevich Shkalikov for his attention, support and keen interest. 
Citation in format AMSBIB
\Bibitem{Tum24} |
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