The lower bounds of non-real eigenvalues for singular indefinite Sturm–Liouville problems

Abstract

The present paper deals with the non-real eigenvalues for singular indefinite Sturm–Liouville problems. The lower bounds on non-real eigenvalues for this singular problem associated with a special separated boundary condition are obtained.

1. Introduction

In the present paper, we consider the eigenvalue problem

(1.1)\begin{equation} -(py')'+qy=\lambda w y,\quad [y,u_{{-}1}]({-}1)=0=[y,u_1](1), \end{equation}

where the functions $p,\, q,\,w$ are real-valued and $w$ changes sign on $(-1,\,1)$ and $u_{-1},\, u_{1}$ are the principal solutions of differential expression

(1.2)\begin{equation} -(py')'+qy=\lambda w y \end{equation}

at $-1,\, 1$ for $\lambda =0$, respectively. Such a problem is called indefinite and the indefinite nature, that non-real spectral points may appear, was noticed by Haupt [11] and Richardson [19] at the beginning of the last century and has attracted a lot of attention in the recent years, see [1, 7, 8, 10, 14].

A priori bounds on non-real eigenvalues for indefinite Sturm–Liouville problems were raised in [15] by Mingarelli and stressed by Kong et al. [13]. Recently, the regular indefinite case of this problem was solved by Qi et al. in [2, 17, 18, 21]. For the singular indefinite Sturm–Liouville problems with limit-point type endpoints

(1.3)\begin{equation} (Af)(x):=\mathrm{sgn}(x)({-}f''(x)+V(x)f(x))=\lambda f(x), \quad x\in \mathbb{R}, \end{equation}

the authors in [3] provided sufficient conditions for the existence of non-real eigenvalues. The explicit bounds on the non-real eigenvalues of (1.3) were obtained in [4]. In [5, 6], the authors not only estimated the absolute values of the non-real eigenvalues in terms of the $L^{1}\text {-norm}$ of the continuous potential, but also obtained the bounds on the imaginary parts and absolute values of these eigenvalues in terms of the $L^{1}\text {-norm}$ of the potential and its negative part. Recently in [20], the authors solved the estimates of absolute values on the non-real eigenvalues for the singular indefinite Sturm–Liouville eigenvalue problems with limit-circle type non-oscillation endpoints associated with a special self-adjoint boundary condition.

In the present paper, we will focus on the singular indefinite Sturm–Liouville eigenvalue problems with self-adjoint boundary conditions associated with principal solution at endpoints. The lower bounds of this eigenvalue problem are obtained under the some conditions. This paper is organized as follows: the preliminary knowledge and the main results, theorems 2.2 and 2.3, are stated in § 2 and their proofs are given in § 3.

2. Preliminary knowledge and main results

In this section, we give some basic knowledge for the singular differential equation (1.2) under the standard conditions that $p,\, q,\, w$ are real-valued functions satisfying

(2.1)\begin{equation} p>0, |w|>0\ \text{a.e. on}\ ({-}1,1),\ \frac{1}{p},w,q\in L^{1}_{\text{loc}}({-}1,1), \int^{1}_{{-}1}\left(\left|\frac{1}{p}\right|+|q|+|w|\right)=\infty \end{equation}

and $w$ changes its sign on $(-1,\,1)$ in the meaning that

\[ \mathrm{mes}\{x:w(x)>0\}>0, \quad \mathrm{mes}\{x:w(x)<0\}>0. \]

We first introduce some concepts. For fixed $\lambda \in \mathbb {R}$, a real solution $u(x)$ of (1.2) is called a principal solution at $1$ if there exists $c\in (-1,\,1)$ such that $u(x)\not =0, x\in (c,\,1),\, \ \int ^1_c 1/(pu^2)=\infty.$ A real solution $v(x)$ of (1.2) is called a non-principal solution at $1$ if there exists $c\in (-1,\,1)$ such that $v(x)\not =0$, $x\in (c,\,1),$ $\int ^1_c 1/(pv^2)<\infty.$ If $u$ and $v$ are principal and non-principal solutions at $1$, respectively, then $u(x) / v(x)\to 0$ as $x\to 1.$ (cf. [16] and [22, Theorem 6.2.2]). In order to give the asymptotic behaviours of eigenfunctions at the endpoints, we assume that

(2.2)\begin{equation} \Upsilon (t): =\sup\limits_{{-}1< x<1}\left|\frac{1}{p(t)}\int^x_tq(s)\text{d} s\right|\in L^{1}({-}1,1), \quad \int^x_c\frac{1}{p(t)}\text{d} t\in L^2_{|w|}({-}1,1) \end{equation}

for some (and hence for all) $c\in (-1,\,1)$. Throughout this paper, the functions $p,\,q,\,w$ always satisfy (2.1) and (2.2). Then

Lemma 2.1 [20, Lemma 2.3] Assume that (2.2) holds, $u_1(\cdot )$ is a principal solution of (1.2) at $1$ for $\lambda =0$. Let $y$ be an eigenfunction of (1.1) corresponding to the eigenvalue $\lambda$. If either $\int ^1_{c} 1/p(t)\text {d} t=\infty$ or $\int ^1_{c} 1/p(t)\text {d} t<\infty,$ $q\in L^1(-1,\,1),$ then $y$ is bounded and

(2.3)\begin{equation} [y,u_{1}](1)=0 \Leftrightarrow (py')(x)y(x)\to 0\ \text{as}\quad x\to 1. \end{equation}

The similar conclusion holds for $x\to -1$.

The operator $S$ associated with the right-definite problem

(2.4)\begin{equation} \left\{\begin{array}{l} -(py')'(x)+q(x)y(x)=\lambda |w(x)|y(x), \\ {}[y,u_{{-}1}]({-}1)=0,\ [y,u_1](1)=0 \end{array}\right. \end{equation}

is defined as $Sy=\frac {1}{|w|}\tau y$ for $y\in D(S)$, where $\tau y := -(py')'(x)+q(x)y(x)$,

\[ D(S)=\{y\in L^{2}_{|w|}({-}1,1):\ y,py'\in AC_{loc}({-}1,1), \tau y/ |w|\in L^{2}_{|w|}({-}1,1),\ \mathcal{B}y=0\} \]

and $\mathcal {B}y=0:=\ [y,\,u_{-1}](-1)=0=[y,\,u_1](1)$. It follows from [12] and [22, Theorem 10.6.2, p.195] that the operator $S$ is self-adjoint in the Hilbert space $(L^2_{|w|},\, (\cdot,\,\cdot )_{|w|})$ and it has discrete spectrum consisting of an infinite number of eigenvalues $\{\mu _{n}: n\in \mathbb {N}:=\{1,\,2,\,\cdots \}\}$, which are all real, unbounded from above and bounded from below, i.e. $-\infty <\mu _{1}<\mu _{2}<\mu _{3}<\cdots \rightarrow +\infty.$

Let $K=(L^{2}_{|w|}(-1,\,1),\,[\cdot,\,\cdot ]_{w})$ be the Krein space with the indefinite inner product ${[f,\,g]_{w}}=\int ^{1}_{-1}w(x)f(x)\overline {g(x)}\,{\rm d}x,$ $f,\, g\in L^{2}_{|w|}(-1,\,1)$ and $J=\mathrm {sgn}\ w$ the fundamental symmetry operator. The operator $T$ in $K$ is defined as

\[ Ty=\frac{1}{w}\tau y,\quad y\in D(T)=D(S). \]

Then $S=JT$, $[Tf,\,g]_{w}=(Sf,\,g)_{|w|}$, $f,\, g\in D(T)$ and $T$ is a self-adjoint operator in $K$ [3, 7, 9]. In the following, we denote the resolvent set of $S$ by $\rho (S)$.

Now, we state the lower bound result on $T$.

Theorem 2.2 Let $T$ and $S$ be defined as above. Suppose that $0\in \rho (S)$ and $S^{-1}$ is compact. Let $\mu ^+:=\min \ \sigma (S)\cap (0,\,\infty ),\, \mu ^{-}:= \min \ \left \{|\lambda |: \lambda \in \sigma (S)\cap (-\infty,\,0)\right \},$ where $\min \ \emptyset := \infty$. Then for each eigenvalue $\lambda$ of $T$ we have $|\lambda |\geq \min \ \{\mu ^+,\,\mu ^{-}\}.$

Moreover, if $\lambda$ corresponds to an eigenvector $\phi$ of $T$ with $[\phi,\,\phi ]_{w}=0$, then the following, in general stronger, estimate holds $|\lambda |^2\geq -\mu ^+\mu ^{-}.$

In order to give another result of the lower bound on $T$, we assume that $q_-(x)=\max \{0,\, -q(x)\}$ and for some $C,\,\ C_0,\,\ C_{1},\,\ C_2>0,\,\ x\in (-1,\,1)$

(2.5)\begin{align} & \left|\frac{1-x}{\sqrt{p(x)}}\int^x_{{-}1} q_-(t)\text{d} t\right|\leq C_{0},\quad \left|\frac{x+1}{\sqrt{p(x)}}\int^1_x q_-(t)\text{d} t\right|\leq C_{1}, \nonumber\\ & W(x)=\int^x_{{-}1} w(t)\text{d} t,\quad \frac{W(x)}{\sqrt{p(x)}}\leq C_2,\quad C=\sqrt{2}C_2. \end{align}

It is easy to verify that if $q\in L^{2}(-1,\,1),\,\ p(x)=1-x^2,\,\ w(x)=x$, then (2.5) holds. Let

(2.6)\begin{align} & \gamma_{t}:= \min \left\{\int^{t_0+t}_{t_0} |w(x)| \text{d} x: t\in ({-}1,1),\ t_0\in ({-}1,1) \right\},\nonumber\\ & \Delta_{w,1,n}=\|w\|_1 + C \left(\sqrt{ \Delta}+\sqrt{\Delta_{n}}\right),\quad \Delta= 2\int^1_{{-}1}q_{-}(t)\text{d} t + {8\alpha^2},\nonumber\\ & \Delta_n = 2\int^1_{{-}1}q_{-}(t)\text{d} t +{8\alpha^2}+ 2|\mu_{n}|\| w\|_1,\quad \alpha=\frac{C_{0}+C_{1}}{2}. \end{align}

With this notation, we give the following result.

Theorem 2.3 Assume that $\lambda$ and $\mu _n$ are the non-real eigenvalue of (1.1) and the $n$th eigenvalue of right-definite problem (2.4), respectively. Let $\mu _{h-1}<0<\mu _h$ for some positive integer $h\geq 2$, then the eigenvalue $\lambda$ satisfies

\[ |\lambda|^2 \geq \frac{ \mu_h \mu^2_{h-1} \gamma_\delta}{16 (\mu_h - \mu_{h-1})} \left(\sum^{h-1}_{n=1} \frac{ \Delta^2_{w,1,n} }{ \gamma_{\delta_n}}\right)^{{-}1}. \]

3. The proofs of theorems 2.2 and 2.3

In order to prove theorems 2.2 and 2.3, we prepare some lemmas. The following lemma is the estimates of $\|\sqrt {p}\phi '\|_2$, where $\phi$ is an eigenfunction of (1.1) corresponding to a non-real eigenvalue $\lambda$. That is $\mathcal {B}\phi =0$ and

(3.1)\begin{equation} -(p\phi')'+q \phi=\lambda w \phi. \end{equation}

Since problem (1.1) is a linear system and $\phi$ is continuous, we can choose $\phi$ satisfies $\max \{|\phi (x)|:x\in [-1,\,1]\}=1$ in the following discussion.

Lemma 3.1 Let $\lambda$ and $\phi$ be defined as above. Then

(3.2)\begin{equation} \int^1_{{-}1} w|\phi|^2=0,\quad \int^1_{{-}1} p|\phi'|^2 \leq \Delta, \end{equation}

where $\Delta$ is given by (2.6).

Proof. It follows from $\mathcal {B}\phi =0$ and lemma 2.1 that $\phi$ is bounded and satisfies $(p\phi '\phi )(x)\to 0$ as $x\to -1$ or $1.$ Multiplying both sides of (3.1) by $\overline {\phi }$ and integrating over the interval $[a,\,b]$, then

(3.3)\begin{equation} \int^1_{{-}1} p|\phi'|^2+\int^1_{{-}1} q|\phi|^2=\lambda\int^1_{{-}1} w|\phi|^2. \end{equation}

From $\operatorname {Im}\lambda \not =0$ and (3.3) one sees that $\int ^1_{-1} w|\phi |^2=0$, and hence

(3.4)\begin{equation} \int^1_{{-}1} p|\phi'|^2+ \int^1_{{-}1} q|\phi|^2 =0. \end{equation}

Set

\[ Q(x)=\int^x_{{-}1} q_-(t)\text{d} t-\frac{x+1}{2}\int^1_{{-}1} q_-(t)\text{d} t \]

Then one can verify that

\begin{align*} & Q({-}1)=0=Q(1),\ Q'(x)=q_-(x)-\frac{1}{2}\int^1_{{-}1}q_{-}(t)\text{d} t\ \text{a.e.} \ x\in ({-}1,1)\quad \text{and}\\ & |Q(x)|\leq\left|\frac{1-x}{2}\int^x_{{-}1} q_-(t)\text{d} t\right| +\left|\frac{x+1}{2}\int^1_x q_-(t)\text{d} t\right| \leq \frac{\sqrt{p(x)}}{2}(C_{0}+C_{1}) = \alpha\sqrt{p(x)}. \end{align*}

As a result, this together with (3.4) and Cauchy–Schwarz inequality yields that

(3.5)\begin{align} \int^1_{{-}1} q_{-}|\phi|^2 & =\int^1_{{-}1} \left(Q'+\frac{1}{2}\int^1_{{-}1}q_{-}\right)|\phi|^2\nonumber\\ & \leq \int^1_{{-}1}q_{-} -2\operatorname{Re}\left(\int^1_{{-}1} Q\phi'\overline{\phi}\right)\nonumber\\ & \leq \int^1_{{-}1}q_{-} + \frac{1}{2}\int^1_{{-}1}p|\phi'|^2 + {4\alpha^2}. \end{align}

It follows from (3.4), (3.5) and $q=q_+-q_-$, $q_{\pm } =\max \{0,\, \pm q \}$ that

\[ \int^1_{{-}1} p|\phi'|^2 ={-}\int^1_{{-}1} q|\phi|^2 \leq \int^1_{{-}1} q_{-}|\phi|^2 \leq \int^1_{{-}1}q_{-}+\frac{1}{2}\int^1_{{-}1} p|\phi'|^2 + {4\alpha^2}. \]

So the inequalities in (3.2) holds immediately.

Similarly with the argument of lemma 3.1, we give the estimates of $\|\sqrt {p}\psi '_n\|_2$, where $\psi _n$ is the eigenfunction that satisfies $\max \{|\psi _n(x)|: x\in [-1,\,1]\}=1$ corresponding to the $n$th eigenvalue $\mu _n$ of (2.4).

Lemma 3.2 Suppose that $\mu _n$ and $\psi _n$ are defined as above. Then $\int ^1_{-1} p|\psi _{n}'|^2 \leq \Delta _{n},$ where $\Delta _n$ is given by (2.6).

Proof. Let $\mu _n$ and $\psi _n$ be defined as above, then

(3.6)\begin{equation} -(p\psi'_{n})'+q\psi_{n}=\mu_{n}|w|\psi_{n},\quad \mathcal{B}\psi_{n}=0. \end{equation}

From $\mathcal {B}\psi _{n}=0$ and lemma 2.1 that $\psi _{n}$ is bounded and satisfies $(p\psi _{n}'\psi _{n})(x)\to 0$, $x\to -1$ or $1$. Multiplying both sides of (3.6) by $\psi _{n}$ and integrating over the interval $(-1,\,1)$, we have

(3.7)\begin{equation} \int^1_{{-}1} p|\psi_{n}'|^2+\int^1_{{-}1} q|\psi_{n}|^2 =\mu_{n} \int^1_{{-}1} |w| |\psi_{n}|^2. \end{equation}

With the similar argument in (3.5), one can prove that

\[ \int^1_{{-}1} q_-|\psi_{n}|^2 \leq \int^1_{{-}1}q_{-} +\frac{1}{2}\int^1_{{-}1}p|\psi'_{n}|^2 + {4\alpha^2}. \]

This together with (3.7) and $q=q_+-q_-$ yields that

\begin{align*} \int^1_{{-}1} p|\psi_{n}'|^2 & \leq \int^1_{{-}1}q_-|\psi_{n}|^2+\mu_{n}\int^1_{{-}1}|w||\psi_{n}|^2\\ & \leq \int^1_{{-}1}q_{-} +\frac{1}{2}\int^1_{{-}1}p|\psi'_{n}|^2 + {4\alpha^2}+ |\mu_{n}| \|w\|_1 . \end{align*}

And hence

\[ \int^1_{{-}1} p|\psi_{n}'|^2 \leq 2\int^1_{{-}1}q_{-} +{8\alpha^2}+2|\mu_{n}|\|w\|_1. \]

This completes the proof of lemma 3.2.

For any $\varepsilon >0$, let

(3.8)\begin{align} \delta& = \sup\left\{\min\left\{\tilde{\delta}, \frac{1}{2}\right\}: \int_{I} \frac{1}{p}\leq \frac{1}{4\Delta}, \text{for any}\ I\subset [{-}1/2,1/2]\ \text{with length}\ \tilde{\delta}\right\} \end{align}

(3.9)\begin{align} \delta_n& = \sup\left\{\min\left\{\tilde{\delta}, \frac{1}{2}\right\}: \int_{I} \frac{1}{p}\leq \frac{1}{4\Delta_{n}}, \text{for any}\ I\subset [{-}1/2,1/2]\ \text{with length}\ \tilde{\delta}\right\} \end{align}

From the definition of $\delta$ in (3.8), one sees that $\delta \in (0,\,1/2]$ and $\int _{I} \frac {1}{p}\leq \frac {1}{4\Delta }$ for any interval $I\subset [-1/2,\, 1/2]$ with length $\delta$. The conclusion holds for $\delta _n$.

Lemma 3.3 Let $\lambda$ and $\phi$ be defined as above. Then there exists an interval $\tilde {I}\subset (-1,\,1)$ with $\delta$ in length, such that $|\phi (\cdot )|\geq 1/2$ on $\tilde {I}$.

Proof. For any interval $I\subset [-1/2,\, 1/2]$ with length $\delta$, it follows from Cauchy–Schwarz inequality and lemma 3.1 that

(3.10)\begin{equation} \left(\int_{I}|\phi'|\right)^2\leq \int_{I}\frac{1}{p}\int^1_{{-}1}p|\phi'|^2\leq \frac{1}{4}. \end{equation}

Since $\max \{|\phi (x)|:x\in [-1,\,1]\}=1$, there exists $x_{0}\in [-1/2,\, 1/2]$ such that $|\phi (x_{0})|\leq 1.$ Hence, for $x\in (-1,\,1)$ and $|x-x_{0}|\leq \delta$,

\[ \Big| |\phi(x)|-1\Big| \leq \Big| |\phi(x)|-|\phi(x_{0})| \Big| \leq |\phi(x)-\phi(x_{0})| =\left|\int^x_{x_{0}}\phi'(t)\,{\rm d}t\right|\leq \frac{1}{2} \]

by (3.10), and hence

\[ |\phi(x)|\geq \frac{1}{2}\quad \text{on}\quad \tilde{I} = [-\delta+x_{0},x_{0}]\quad\text{or}\quad [x_{0},x_{0}+\delta ]. \]

From $\delta \in (0,\, 1/2]$ one sees that $(-1,\,1)$ contains at least one such interval $\tilde {I}$.

Similar with lemma 3.3 we have

Lemma 3.4 Assume that $\mu _n$ is an eigenvalue of (2.4) and $\psi _n$ is the corresponding eigenfunction. Then there exists an interval $\tilde {I}_{n}\subset (-1,\,1)$ with $\delta _n$ in length, such that $|\psi _n(\cdot )|\geq 1/ 2$ on $\tilde {I}_{n}$.

Applying the above lemmas we now prove the main results of theorems 2.2 and 2.3.

The proof of theorem 2.2 Let $\mu _{n}$ be the $n$th eigenvalue of right-definite problem (2.4) and $\psi _{n}$ the corresponding eigenfunction. From $\psi _{n}\in D(S)$ is linearly independent, one sees that $\{\psi _n: n\geq 1\}$ forms an orthonormal system. Let $\phi$ be an eigenfunction of $T$ associated with eigenvalue $\lambda$ such that $\int ^1_{-1} |w| |\phi |^2=1$. Since $S$ is a self-adjoint operator and $S^{-1}$ is compact, we can expand $\phi$ via the orthonormal system $\psi _{n}$, i.e. $\phi =\sum \limits ^{\infty }_{n=1}(\phi,\,\psi _{n})_{|w|} \psi _{n}.$ Then from $\int ^1_{-1} |w||\phi |^2=1$, we have

(3.11)\begin{equation} \sum^{\infty}_{n=1}|(\phi,\psi_{n})_{|w|}|^2=1,\quad \sum^{\infty}_{n=1}|(\phi,\psi_{n})_{|w|}|^2\mu^2_{n} =|\lambda|^2. \end{equation}

It follows from $\mu ^+ =\min \ \sigma (S)\cap (0,\,\infty )$ and $\mu ^{-} = \min \ \{|\lambda |: \lambda \in \sigma (S)\cap (-\infty,\,0)\}$ that $|\mu _{n}|\geq \min \ \{\mu ^+,\,\mu ^{-}\}$ for $n\geq 1$. This together with (3.11) yields that

\[ |\lambda| \geq \min\ \{\mu^+,\mu^{-}\}. \]

If $\lambda$ corresponds to an eigenvector $\phi$ of $T$ with $[\phi,\,\phi ]_{w}=0$, then it follows from (3.11) that

(3.12)\begin{equation} \sum^{\infty}_{n=1} \left|(\phi,\psi_{n})_{|w|}\right|^2 \left(\mu_{n}-\frac{1}{2}(\mu^+{+}\mu^{-})\right)^2 =|\lambda|^2+\frac{1}{4}(\mu^+{+}\mu^{-})^2. \end{equation}

From $|\mu _{n}| \geq \min \ \{\mu ^+,\,\mu ^{-}\}$, one sees that

(3.13)\begin{equation} \left(\mu_{n}-\frac{1}{2}(\mu^+{+}\mu^{-})\right)^2-\frac{1}{4}(\mu^+{-}\mu^{-})^2 =(\mu_{n}-\mu^+)(\mu_{n}-\mu^{-})\geq0. \end{equation}

From (3.11), (3.12) and (3.13), we have that

\begin{align*} |\lambda|^{2}& =\left(\mu_{n}-\frac{1}{2}(\mu^+{+}\mu^{-})\right)^2 -\frac{1}{4}(\mu^+{+}\mu^{-})^2\\ & =\left(\mu_{n}-\frac{1}{2}(\mu^+{+}\mu^{-})\right)^2 -\frac{1}{4}(\mu^+{-}\mu^{-})^2 -\mu^+\mu^{-} \geq{-}\mu^+\mu^{-}, \end{align*}

which completes the proof of theorem 2.2.

The proof of theorem 2.3 Let $\mu _n$ be the $n$th eigenvalue of (2.4) and $\psi _n$ the corresponding eigenfunction such that $\max \{|\psi _n(x)|:x\in [-1,\,1]\}=1,\,\ n\geq 1$. It follows from lemmas 3.3, 3.4 and the definition of $\gamma _{t}$ in (2.6) that

(3.14)\begin{align} & \|\phi\|_{|w|}^2=\int^1_{{-}1} |w| |\phi|^2 \geq \int_{I} |w| |\phi|^2 \geq \int_{I} \frac{|w|}{4} \geq \frac{\gamma_{\delta}}{4}, \end{align}

(3.15)\begin{align} & \|\psi_n\|_{|w|}^2=\int^1_{{-}1} |w| |\psi_n|^2 \geq \int_{I_n} |w| |\psi_n|^2 \geq \int_{I_n} \frac{|w|}{4} \geq \frac{\gamma_{\delta_n}}{4},\ n\geq 1. \end{align}

From the definition of $W(x)=\int ^x_{-1} w(t)\text {d} t$, one sees that

\[ [\phi, \psi_n]_w = \int^1_{{-}1} w\phi \overline{\psi_n} =\int^1_{{-}1} W' \phi \overline{\psi_n} = \phi(1) \overline{\psi_n}(1) \int^1_{{-}1} w- \int^1_{{-}1} W (\phi' \overline{\psi_n}+\phi \overline{\psi'_n}). \]

This together with (2.5) and lemmas 3.1 and 3.2 that

\begin{align*} \Big|[\phi, \psi_n]_w \Big| & \leq \int^1_{{-}1} |w|+ \left| \int^1_{{-}1} W (\phi' \overline{\psi_n} +\phi \overline{\psi'_n}) \right|\\ & \leq \|w\|_1 + \left|\int^1_{{-}1} \frac{W}{\sqrt{p}} \sqrt{p}\phi' \overline{\psi_n}+ \int^1_{{-}1} \frac{W}{\sqrt{p}} \phi \sqrt{p}\overline{\psi'_n}\right|\\ & \leq \|w\|_1 + C_2 \left(\int^1_{{-}1} |\sqrt{p}\phi' \overline{\psi_n}|+ \int^1_{{-}1} |\phi \sqrt{p}\overline{\psi'_n}|\right)\\ & \leq \|w\|_1 + C \left(\sqrt{ \Delta}+\sqrt{\Delta_{n}}\right)=\Delta_{w,1,n}. \end{align*}

Furthermore, for $n\geq 1$ we get

(3.16)\begin{equation} \lambda [\phi, \psi_n]_w=[T\phi, \psi_n]_w=(S\phi, \psi_n)_{|w|} =(\phi, S\psi_n)_{|w|}=\mu_n (\phi, \psi_n)_{|w|}. \end{equation}

If we set

\[ \Lambda_n=(\phi,\psi_{n})_{|w|} =\frac{(\phi, \psi_n)_{|w|}}{\|\phi\|_{|w|} \|\psi_n\|_{|w|}}, \]

then (3.14)–(3.16) give that

(3.17)\begin{equation} |\Lambda_n|\leq \frac{4 |\lambda| \Big|[\phi, \psi_n]_w \Big|} {|\mu_{n}| \sqrt{\gamma_\delta \gamma_{\delta_n}}} \leq \frac{4 |\lambda| \Delta_{w,1,n}} {|\mu_{n}| \sqrt{\gamma_\delta \gamma_{\delta_n}}}. \end{equation}

From $\sum \limits ^{\infty }_{n=1}|(\phi,\,\psi _{n})_{|w|}|^2\mu _{n}=\lambda [\phi,\,\phi ]_w$ and $[\phi,\,\phi ]_{w}=\int ^1_{-1} w|\phi |^2=0$ in lemma 3.1, we have

\[ 0=\lambda[\phi,\phi]_{w}= \sum^{\infty}_{n=1} \left|(\phi,\psi_{n})_{|w|}\right|^2 \mu_{n} =\sum^{\infty}_{n=1} |\Lambda_n|^2 \mu_{n} = \sum^{h-1}_{n=1} |\Lambda_n|^2 \mu_{n} +\sum^{\infty}_{n=h} |\Lambda_n|^2 \mu_{n}, \]

and hence $- \sum \limits ^{h-1}\limits _{n=1} |\Lambda _n|^2 \mu _{n}= \sum \limits ^{\infty }\limits _{n=h} |\Lambda _n|^2 \mu _{n}$. Thus, by (3.11) and $\mu _{h-1}<0<\mu _h$, $h\geq 2$,

(3.18)\begin{align} & \sum^{h-1}_{n=1} |\Lambda_n|^2 (\mu_{h}- \mu_n) = \sum^{h-1}_{n=1} |\Lambda_n|^2 \mu_{h} + \sum^{\infty}_{n=h} |\Lambda_n|^2 \mu_n \nonumber\\ & \quad = \sum^{h}_{n=1} |\Lambda_n|^2 \mu_{h} + \sum^{\infty}_{n=h+1} |\Lambda_n|^2 \mu_n \geq \sum^{h}_{n=1} |\Lambda_n|^2 \mu_{h} + \sum^{\infty}_{n=h+1} |\Lambda_n|^2 \mu_h \nonumber\\ & \quad = \sum^{\infty}_{n=1} |\Lambda_n|^2 \mu_h= \mu_h. \end{align}

By $\mu _h>0,\, \mu _n\leq \mu _{h-1}<0,\, 1\leq n \leq h-1,$ we have

\[ \frac{1}{\mu^2_n} \leq \frac{1}{\mu^2_{h-1}},\quad \frac{\mu_h}{\mu_n^2}\leq \frac{\mu_h}{\mu^2_{h-1}},\quad \frac{-1}{\mu_n}\leq \frac{-1}{\mu_{h-1}}= \frac{-\mu_{h-1}}{\mu^2_{h-1}}. \]

This together with the assumption that $\mu _1\leq \cdots \leq \mu _{h-1} <0< \mu _h,\, h\geq 2$ yields that

(3.19)\begin{equation} \frac{\mu_h- \mu_n}{\mu_n^2}= \frac{\mu_h}{\mu_n^2} - \frac{1}{\mu_n } \leq \frac{\mu_h}{\mu^2_{h-1}} - \frac{\mu_{h-1}}{\mu^2_{h-1}} = \frac{\mu_h - \mu_{h-1}}{\mu^2_{h-1}}, \quad 1\leq n\leq h-1. \end{equation}

Hence, by (3.17)–(3.19) we have

\begin{align*} \mu_h & = \sum^{h-1}_{n=1} |\Lambda_n|^2 (\mu_h\!- \!\mu_n) \leq \sum^{h-1}_{n=1} \frac{16 |\lambda|^2 \Delta^2_{w,1,n} (\mu_h\!-\! \mu_n) } {|\mu_{n}|^2 \gamma_\delta \gamma_{\delta_n}} =\frac{16 |\lambda|^2 }{\gamma_\delta} \sum^{h-1}_{n=1} \frac{ \mu_h- \mu_n } {|\mu_{n}|^2 \gamma_{\delta_n}}\Delta^2_{w,1,n} \\ & \leq \frac{16 |\lambda|^2 }{\gamma_\delta} \sum^{h-1}_{n=1} \frac{ \mu_h - \mu_{h-1} } {\mu^2_{h-1} \gamma_{\delta_n}}\Delta^2_{w,1,n} =\frac{16 |\lambda|^2 (\mu_h - \mu_{h-1}) }{\mu^2_{h-1} \gamma_\delta} \sum^{h-1}_{n=1} \frac{ \Delta^2_{w,1,n} }{ \gamma_{\delta_n}}. \end{align*}

Therefore,

\[ |\lambda|^2 \geq \frac{ \mu_h \mu^2_{h-1} \gamma_\delta}{16 (\mu_h - \mu_{h-1})} \left(\sum^{h-1}_{n=1} \frac{ \Delta^2_{w,1,n} }{ \gamma_{\delta_n}}\right)^{{-}1}. \]

This completes the proof of theorem 2.3.