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.

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.

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 $\tilde {I}$.

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.