Proof. Since $E$ has more than one point, we choose $(\theta _1, K_1), (\theta _2, K_2)\in \Pi$ (with $\theta _1\neq \theta _2({\rm mod}\,\pi )$) and $\lambda _0\in \Lambda _{\theta _1, K_1}\cap \Lambda _{\theta _2, K_2}$. Then by the definition of $\Lambda _{\theta, K}$, we obtain

(3.6)\begin{equation} {\rm Re}\left\{{\rm e}^{i\theta_j}(q-\lambda_0\,w)\right\}\ge\delta_j w, \ {\rm Re}\left\{{\rm e}^{i\theta_j}p\right\}\ge 0,\ j=1, 2 \end{equation}

for some $\delta _1, \delta _2>0$. Set

(3.7)\begin{equation} \begin{aligned} & r(x)=|q(x)-\lambda_0\,w(x)|, \ q(x)-\lambda_0\,w(x)=r(x)\,{\rm e}^{i\alpha(x)}, \ \alpha_j(x)=\theta_j+\alpha(x), \\ & \widetilde{r}(x)=|p(x)|, \ p(x)=\widetilde{r}(x)\,{\rm e}^{i\beta(x)}, \ \beta_j(x)=\theta_j+\beta(x),\ j=1, 2. \end{aligned} \end{equation}

For $j=1$, Let $y$ be a solution of (1.1) with $\lambda =\lambda _0$. Set $u=y$, $v=-i\,{\rm e}^{i\theta _1}py'$, then (1.1) is transformed into the Hamiltonian differential system

(3.8)\begin{equation} u'=B\,v+\xi w_2\,v,\quad v'=C\,u-\xi w_1\,u \end{equation}

with the new spectral parameter $\xi =i$, where

(3.9)\begin{equation} \begin{aligned} & C(x)=r(x)\sin\alpha_1(x), \ w_1(x)=r(x)\cos\alpha_1(x), \\ & B(x)=\sin\beta_1(x)/\widetilde{r}(x), \ w_2(x)={\cos\beta_1}(x)/{\widetilde{r}(x)}. \end{aligned} \end{equation}

This is the Hamiltonian differential system (3.1) with $n=1$, $A(x)\equiv 0$ and $\xi =i$. Clearly, $w_1={\rm Re}\left \{{\rm e}^{i\theta _1}(q-\lambda _0\,w)\right \}\ge \delta _0\,w>0$, $w_2=\frac {{\rm Re}\left \{{\rm e}^{i\theta _1}p(t)\right \}}{r^2_2}\ge 0$ by (3.6). We note further that the coefficients of the Hamiltonian system (3.8) are real functions. It is easy to verify that the definiteness condition holds for the system (3.8). Therefore, (1.1) is in case I or II w.r.t. $(\theta _1, K_1)\in \Pi$ if and only if (3.8) is in the limit point case at $b$.

Let $D(\theta _1)$ be the maximal domain associated to (3.8), i.e., $(u, v)^T\in D(\theta _1)$ if and only if $(u, v)^T\in AC_{\text {loc}}\cap L^2_W$ and there exists an element $(f, g)^T\in L^2_W$ such that

(3.10)\begin{equation} u'=B\,v+w_2\,g, \quad v'=C\,u-w_1\,f. \end{equation}

For $y_0\in D(\tau )$, set

(3.11)\begin{equation} f_0=\tau y_0,\ g_0=f_0-\lambda_0 y_0. \end{equation}

Then $y_0$ satisfies

(3.12)\begin{equation} (\tau-\lambda_0)y={w}^{{-}1}[-(py')'+(q-\lambda_0\,w)y]=g_0. \end{equation}

Set $u_0=y_0$, $v_0=-i\,{\rm e}^{i\theta _1}py'_0$. Then $(u_0, v_0)$ satisfies

(3.13)\begin{equation} u'=B\,v+i w_2\,v, \ v'=C\,u-i w_1\,u-w_1f_1, \ f_1=\frac{w}{w_1}\left({-}i\,{\rm e}^{i\theta_1}g_0\right). \end{equation}

Conversely, if $(u, v)$ satisfies (3.13), then $y=u$ solves (3.12).

Let $g_0$ be given in (3.11). Consider the equation (3.13), we get from (3.4) that (3.13) has a solution $(u_1, v_1)^T$ such that $u_1\in L^2_{w_1}, v_1\in L^2_{w_2}$ and $v_1=-i\,{\rm e}^{i\theta _1}p\,u'_1$. Set $y_1=u_1$. Then $y_1$ satisfies (3.12), hence $(\tau -\lambda _0)(y_0-y_1)=0$. Note that $y_1=u_1\in L^2_{w_1}$ and $w_1\ge \delta w$ implies that $y_1\in L^2_w$. Thus, $y_1-y_0$ is an $L^2_w$-solution of $\tau y=\lambda _0 y$. Since $\tau$ is in case I w.r.t. $(\theta _1, K_1)$, it follows from (2.8) that $y_1-y_0\in L^2_{w_1}$ and $v_1-v_0\in L^2_{w_2}$. This together with $y_1\in L^2_{w_1}$ and $v_1\in L^2_{w_2}$ gives that $y_0\in L^2_{w_1}$ and $v_0\in L^2_{w_2}$. In fact, we have proved that for $y\in D(\tau )$,

(3.14)\begin{equation} \int^b_a |q-\lambda_0 w| \cos\alpha_1 |y|^2<\infty,\ \int_a^b |p| \cos\beta_1 |y'|^2<\infty, \end{equation}

where $\alpha _1$ and $\beta _1$ are defined in (3.7). Recall that $g_0\in L^2_w$, or $-i\,{\rm e}^{i\theta _1}g_0\in L^2_w$ and $w_1\ge \delta w$ implies $f_1\in L^2_{w_1}$. It follows (3.13) that $(y_0, v_0)$ satisfies (3.10) with $f=i y_0+f_1$ and $g=i v_0$. This yields that

(3.15)\begin{equation} y\in D(\tau)\quad \Longrightarrow (y, v)^T\in D(\theta_1)\ {\rm with}\ v={-}i\,{\rm e}^{i\theta_1}p y'. \end{equation}

Note that $Y\in D(\theta _1)$ if and only if $\overline {Y}\in D(\theta _1)$. Then for $\overline {y}\in D(\tau )$, we have from (3.15) that $(\overline {y}, \overline {v})^T\in D(\theta _1)$ with $\overline {v}=-i\,{\rm e}^{i\theta _1}p \overline {y}'$, hence

(3.16)\begin{equation} \overline{y}\in D(\tau) \Longrightarrow (y, v)^T\in D(\theta_1)\ {\rm with}\ v=i\,{\rm e}^{{-}i\theta_1}\overline{p} y'. \end{equation}

Let $y_1, \overline {y}_2\in D(\tau )$. Since (3.8) is in the limit point case at $b$ and $(y_j, v_j)^T\in D(\theta _1)$ for $j=1, 2$ with $v_1=-i\,{\rm e}^{i\theta _1}p y'_1$ and $v_2=i\,{\rm e}^{-i\theta _1}\overline {p} y'_2$ by (3.15) and (3.16), respectively, we get from (3.3) that

(3.17)\begin{equation} (\overline{y}_2, \overline{v}_2)\begin{pmatrix} 0 & -1\\ 1 & 0\end{pmatrix}\begin{pmatrix}y_1\\v_1\end{pmatrix} =i\,{\rm e}^{i\theta_1}\,p (y_1'\,\overline{y}_2-y_1\,\overline{y}'_2)\to 0 \end{equation}

as $x\to b$. Furthermore, for $y_1, y_2\in D(\tau )$, since $(y_j, v_j)^T\in D(\theta _1)$ by (3.15) with $v_j=-ie^{i\theta _1}p y'_j$, $j=1, 2$, (3.3) also gives that

\[ (\overline{y}_2, \overline{v}_2)\begin{pmatrix} 0 & -1\\ 1 & 0\end{pmatrix}\begin{pmatrix}y_1\\v_1\end{pmatrix} =i\,{\rm e}^{{-}i\theta_1}(\overline{p}\,\overline{y}_2'\,y_1+{\rm e}^{2i\theta_1} p\,y'_1\,\overline{y}_2)\to 0 \]

as $x\to b$, or

(3.18)\begin{equation} \overline{p}\,\overline{y}_2'\,y_1+{\rm e}^{2i\theta_1} p\,y'_1\,\overline{y}_2\to 0, \ y_1, y_2\in D(\tau) \end{equation}

as $x\to b$. Similarly, for $j=2$, the above methods also give that (3.18) holds for $y_1, y_2\in D(\tau )$ with $\theta _1$ replaced by $\theta _2$. Therefore, we have

(3.19)\begin{equation} \left({\rm e}^{2i\theta_1}-{\rm e}^{2i\theta_0}\right) p\,y_1'\,\overline{y}_2 \to 0 \end{equation}

as $x\to b$. This clearly gives (3.5) for $y_1, y_2\in D(\tau )$ since $\theta _1\ne \theta _2$ (mod $\pi$).

Finally, we prove $y\in L^2_{|q|}$ and $y'\in L^2_{|p|}$ for $y\in D(\tau )$. In fact, the similar proof as in (3.14) gives that for $y\in D(\tau )$,

(3.20)\begin{equation} \int_a^b |q-\lambda_0 w| \cos\alpha_2 |y|^2<\infty, \ \int_a^b |p| \cos\beta_2 |y'|^2<\infty. \end{equation}

Therefore, (3.14), (3.20) and (2.17) together yield that $\int _a^b |q-\lambda _0 w||y|^2<\infty$ and $\int _a^b |p||y'|^2<\infty$, or $y\in L^2_{|q|}$ and $y'\in L^2_{|p|}$ since $y\in L^2_w$. This completes the proof.

Proof. Since $E$ has more than one point, we choose $(\theta _1, K_1), (\theta _2, K_2)\in \Pi$ (with $\theta _1\neq \theta _2({\rm mod}\,\pi )$) and $\lambda _0\in \Lambda _{\theta _1, K_1}\cap \Lambda _{\theta _2, K_2}$. Since (1.1) is in case I, we know that $T$ is a J-self-adjoint extension of $T_0$ by theorem 4.4 in paper [Reference Brown, McCormack, Evans and Plum3]. Next, we define

(3.22)\begin{equation} \textsf{t}[u, v]=\int_a^b[pu'\overline{v'}+qu\overline{v}], u, v\in D(\textsf{t}), \end{equation}

(3.23)\begin{align} D(\textsf{t})=\left\{u\in AC_{\rm loc}[a, b): u(a)=0, u\in L_w^2[a, b)\cap L_{|q-\lambda_0w|}^2[a, b), u'\in L_{|p|}^2[a, b) \right\}. \end{align}

Since (1.1) is in case I, we see by theorem 3.1 that $D(T_0)\subseteq D(\textsf {t})$, so $\textsf {t}$ is densely defined. Similar method as in lemma 2.6 can deduce $\textsf {t}$ as a sectorial operator. Now we turn to prove that $\textsf {t}$ is closed.

Suppose that $y_n\in D(\textsf {t})$, $y_n\to y$ in $L_w^2[a, b)$ and

\[ \textsf{t}[y_n-y_m]=\int_a^b[p|y_n'-y_m'|^2+q|y_n-y_m|^2]\to 0 \]

for $n, m \to \infty$. Let $y_{nm}=y_n-y_m$ for convenience. Since $\textsf {t}[y_{nm}]\to 0$, we obtain $({\rm e}^{i\theta _1}+{\rm e}^{i\theta _2})\textsf {t}[y_{nm}]\to 0$, and

\[ \lambda_0\int_a^bw|y_{nm}|^2\leq 2|\lambda_0|\int_a^bw\left(|y_{n}-y|^2+|y_{m}-y|^2\right)\to 0. \]

So we obtain that

(3.24)\begin{equation} \int_a^b[p_{\theta}|y_{nm}'|^2+(q-\lambda_0w)_{\theta}|y_{nm}|^2]\to 0. \end{equation}

Since ${\rm Re}\left \{(q-\lambda _0 w)_{\theta }\right \} \ge \delta _0|q-\lambda _0 w|$ and ${\rm Re}\left \{p_{\theta }\right \} \ge \epsilon _0|p|$ for some constants $\delta _0, \epsilon _0>0$ by (2.19) and (2.20), and noticing $\textsf {t}$ is a sectorial operator, we obtain

\[ \epsilon_0\int_a^b|p|y_{nm}'|^2+\delta_0\int_a^b|q-\lambda_0w||y_{nm}|^2\to 0. \]

So

\[ \int_a^b|p|y_{nm}'|^2\to 0, \int_a^b|q-\lambda_0w||y_{nm}|^2\to 0 \ {\rm as}\ n, m\to\infty. \]

This shows that $\{y_n'\}$ and $\{y_n\}$ are Cauchy sequences in $L^2_{|p|}[a, b)$ and $L^2_{|q-\lambda _0w|}[a, b)$, respectively. We may assume $y_n'\to z$ in $L^2_{|p|}[a, b)$. For each fixed $x\geq a$,

\begin{align*} |y_{nm}(x)|& =\left|\int_a^xy_{nm}'(s)\,{\rm d}s\right|\\ & \leq \left(\int_a^x\frac{1}{|p(s)|}\,{\rm d}s\right)^{1/2}\left(\int_a^x|p(s)||y_{nm}'(s)|\,{\rm d}s\right)^{1/2}\\ & \leq \left(\int_a^x\frac{1}{|p(s)|}\,{\rm d}s\right)^{1/2}\left(\int_a^b|p(s)||y_{nm}'(s)|\,{\rm d}s\right)^{1/2}\to 0 \end{align*}

as $n, m\to \infty$, since $y_n\to y$ in $L^2_{w}[a, b)$, we obtain $y_n(x)\to y(x)$ point-wise for $x\geq a$ as $n\to \infty$, hence $y(0)=\lim _{n\to \infty }y_n(0)=0$, and $y'(x)=z(x)$, so $y\in D(\textsf {t})$, and $\textsf {t}$ is a closed sesquilinear form.

By the first representation theorem [Reference Kato14, P322, theorem 2.1], there exists an m-sectorial operator $T=T_{\textsf {t}}$ such that $D(T)\subseteq D(\textsf {t})$, $\textsf {t}[u, v]=\langle Tu, v\rangle$ for $u, v\in D(T)$, and $D(T)$ is a core of $\textsf {t}$. Since $T=T_{\textsf {t}}$, so $T_F=T_{\textsf {t}}$ is the Friedrichs extension of $T_0$. We prove $D(T_F)=\{y\in D_{\max }: y(a)=0\}:=\tilde {D}$ to complete the proof. For each $y\in \tilde {D}$, then $y\in D_{\max }$, by theorem 3.1, $y'\in L^2_{|p|}[a, b)$ and $y\in L^2_{|q|}[a, b)$, this together with $y\in L^2_{w}[a, b)$ implies

\[ \int_a^b|q-\lambda_0w|||y|^2\leq \int_a^b|q|||y|^2+|\lambda_0|\int_a^bw|y|^2<\infty, \]

so $y\in L^2_{|q-\lambda _0w|}[a, b)$, thus we obtain $y\in T_F$. Conversely, for each $y\in D(T_F)$, set $T_Fy=f$. By the definition of $T_F$, we obtain $\textsf {t}[y, v]=\langle T_Fy, v\rangle =\langle f, v\rangle$ for all $v\in D(\textsf {t})$, or

(3.25)\begin{equation} \int_a^bwf\overline{v}=\int_a^b\left[py'\overline{v'}+qy\overline{v}\right]. \end{equation}

Set $z'(x)=w(x)f(x)-q(x)y(x)$ and for arbitrary $d< b$,

\[ A_d=\{v:\ v'\in L^2[a, d], \ v(a)=v(d)=0 \}. \]

Then for $v\in A_d$

(3.26)\begin{equation} \int^d_a z'\,\overline{v}=z\,\overline{v}\big{|}^d_a-\int^d_a z\,\overline{v'}={-}\int^d_a z\,\overline{v}'. \end{equation}

If we set

\[ v_0(x)=\left\{\begin{array}{@{}ll} v(x), & x\in[a, d], \\ 0, & x\in [d, b), \end{array}\right. \]

then $v_0\in D(\textsf {t})$, hence (3.25) and (3.26) imply that

(3.27)\begin{equation} \int^d_a (py'+z)\overline{v'}=0. \end{equation}

Note that

\[ \{v':\ v\in A_d \}=\{f:\ f\in L^2[a, d], \ \int^d_a f=0 \}, \]

which is the set of orthogonal to $1$ in $L^2[a, d]$. Therefore, $py'+z=C_d$, where $C_d$ is a constant, hence $z'(x)=-(py')'(x)$ a.e. $x\in [a, d]$, or $-(py')'+qy=wf$ on $[a, d]$. By the arbitrary of $d$ we know that

(3.28)\begin{equation} -(py')'+qy=wf\ {\rm on}\ [a, b). \end{equation}

This means $y\in \tilde {D}$. So $D(T_F)= \tilde {D}$, this completes the proof.

Proof. Similar to the proof of theorem 3.1, we choose $(\theta _1, K_1), (\theta _2, K_2)\in \Pi$ (with $\theta _1\neq \theta _2({\rm mod}\,\pi )$) and $\lambda _0\in \Lambda _{\theta _1, K_1}\cap \Lambda _{\theta _2, K_2}$. Since (1.1) is in case II (with $\lambda$ replaced by $\lambda _0$), we obtain the Hamiltonian system (3.8) is in the limit point case at $b$.

Let $D(\theta _1)$ denote the maximal domain associated to (3.8), we obtain that $y\in D_{\max }$ with $y'\in L^2_{|p|}[a, b)$ yields $(y, v)^T\in D(\theta _1)$, where$v=-ie^{i\theta _1}p y'$, and $\overline {y}\in D_{\max }$ with $\overline {y}'\in L^2_{|p|}[a, b)$ yields $(y, v)^T\in D(\theta _1)$, where$v=ie^{-i\theta _1}\overline {p} y'$. Then as $x\to b$, we obtain

\[ \overline{p}\,\overline{y}_2'\,y_1+{\rm e}^{2i\theta_1} p\,y'_1\,\overline{y}_2\to 0\quad {\rm for\ all}\ y_1, y_2\in D_{\max}\quad{\rm with}\ y_1', y_2'\in L^2_{|p|}[a, b). \]

Similar method implies

\[ \overline{p}\,\overline{y}_2'\,y_1+{\rm e}^{2i\theta_2} p\,y'_1\,\overline{y}_2\to 0\quad {\rm for\ all}\ y_1, y_2\in D_{\max}\quad{\rm with}\ y_1', y_2'\in L^2_{|p|}[a, b). \]

The above two formula imply (4.1) holds, so we complete the first part of theorem 4.1. Similar argument as theorem 3.1 deduce $y\in L^2_{|q|}[a, b)$ for all $y\in D_{\max }$ with $y'\in L^2_{|p|}$. This completes the proof.

Proof. Since $E$ has more than one point, we choose $(\theta _1, K_1), (\theta _2, K_2)\in \Pi$ (with $\theta _1\neq \theta _2({\rm mod}\,\pi )$) and $\lambda _0\in \Lambda _{\theta _1, K_1}\cap \Lambda _{\theta _2, K_2}$. We define the sesquliniear form $\textsf {t}$ as (3.22), then the similar method implies $\textsf {t}$ is a densely defined closed sectorial operator. By the first representation theorem [Reference Kato14, P322, theorem 2.1], the m-sectorial operator $T=T_{\textsf {t}}$ defined by $\textsf {t}[u, v]=\langle Tu, v\rangle$ for $u, v\in D(T)$ is the Friedrichs extension of $T_0$. Now, using the similar method as theorem 3.2, we obtain

\[ D(T_F)=\{y\in D_{\max}: y(a)=0, y'\in L_{|p|}^2[a, b)\}:=\tilde{D}. \]

Finally, we turn to prove $T_F$ is a J-self-adjoint operator by three steps.

1. (1) $JT_{F}J\subset T^*_{F}$. It is equivalent to prove that

(4.3)\begin{equation} \overline{y}_0\in \tilde{D} \Longrightarrow y_0\in D(T^*_{F}), J T_{F} J y_0=T^*_{F} y_0. \end{equation}

Note that for $Jy_0=\overline {y}_0\in \tilde {D}$ and all $y\in \tilde {D}$

\begin{align*} \langle T_{F} y, y_0\rangle- \langle y, J T_{F} J y_0\rangle& =\int_a^{b} \left(\overline{y_0}w\tau y-y w\overline{\tau^+ y_0}\right)\\ & =\int_a^{b} \left(\overline{y_0}\left[-(py')'+qy\right]-y\left[ \overline{-(\overline{p}y'_0)'+\overline{q}y_0}\right]\right)\\ & =\int_a^{b}\left(y\overline{(\overline{p}y'_0)'}-\overline{y_0}(py')'\right) =\int_a^{b}\left(y(p\overline{y'_0})'-\overline{y_0}(py')'\right)\\ & =p\left(y\overline{y'_0}-\overline{y_0}y'\right)\big|^b_a. \end{align*}

This equality together with the boundary condition $y(a)=\overline {y_0}(a)=0$ gives

(4.4)\begin{equation} \langle T_{F} y, y_0\rangle- \langle y, J T_{F} J y_0\rangle=p\left(y\overline{y'_0}-\overline{y_0}y'\right)(b), \end{equation}

(here we denote $\lim _{x\to b}p(y\overline {y'_0}-\overline {y_0}y')(x)$ by $p(y\overline {y'_0}-\overline {y_0}y')(b)$ since this limit always exists). Since $E$ has more than one point and (1.1) is in case II, we know from theorem 4.1 that $p(y\overline {y'_0}-\overline {y_0}y')(b)=0$, and hence we have $y_0\in D(T^*_{F})$ and $J T_{F} J y_0=T^*_{F} y_0$ by (4.4). This proves that $T_{F}\subset J T^*_{F} J$.

1. (2) $T_{F}$ is closed operator. Since for $u, v\in D(\alpha )$, we find that

\[ \langle (T_{F}-\lambda_0)\,u, v\rangle=\int_a^{b}\left[ p\,u'\,\overline{v}'+(q-\lambda_0 w)\,u\,\overline{v}\right]. \]

Suppose $y_n\in \tilde {D}$ such that $y_n\to y_0$ and $T_{F} y_n\to f_0$, or

\[ y_n\to y_0, \ (T_{F}-\lambda_0 I)y_n\to g_0=f_0-\lambda_0 y_0 \quad{\rm in}\ L^2_w[a, b). \]

Let $y_{nm}=y_n-y_m$, as (3.24), we have that $\{y_n\}$ and $\{y'_n\}$ are Cauchy sequences in $L^2_{|q|}[a, b)$ and $L^2_{|p|}[a, b)$, respectively. Since $y_n\to y$ in $L^2_w[a, b)$ as $n\to \infty$ and $\{y_n'\}$ is convergent in $L^2_{|p|}[a, b)$, we see that,

\[ y_n(x)= \int^x_a\frac{1}{p} py_n'\to y_0(x) \]

pointwise in $x$ as $n\to \infty$ on $[a, b)$. It follows from $\tau y_n= T_{F}y_n=f_n$ that

\[ y_n(x)=\int^x_a\frac{1}{p}\left(py_n'(a)+\int^s_a[w\,f_n-q\,y_n]\right). \]

Note that the convergence of each sequence in the above equality except for $\{py_n'(a)\}$ implies $\{py_n'(a)\}$ is convergent, and hence by letting $n\to \infty$ we have that

\[ y(x)=\int^x_a\frac{1}{p}\left(\xi+\int^s_a[w\,f-q\,y]\right), \]

where $py'_n(a)\to \xi$ as $n\to \infty$. This means that $\tau y=f$ and $y(a)=0$, and hence $y\in \tilde {D}$ and $T_{F}y=f$. This proves the closedness of $T_{F}$.

1. (3) $T_{F}$ is a $J$-self-adjoint operator. By the definition of an $m$-sectorial operator we know def $T_F=0$. So

\[ {\text{def}}(T_{F}-\lambda_0\,I)={\text{def}}\,T_F=0, \]

where $\lambda _0\in \Lambda _{\theta _1, K_1}\cap \Lambda _{\theta _2, K_2}$. Then it follows from [Reference Edmunds and Evans6, p115, theorem 5.5] that $T_{F}$ is a $J$-self-adjoint extension of $T_0$.

Proof. Since each spectral point of given operator lies in the numerical range of the operator, we turn to prove the numerical range of $T_F$

\[ \Theta(T_F)=\{\langle T_Fu, u\rangle: u\in D(T_F), ||u||=1\} \]

is contained in $\Omega$. For all $\lambda \in \Theta (T_F)$ there exists $u\in D(T_F), ||u||=1$ such that

\[ \lambda=\langle T_Fu, u \rangle=\textsf{t}[u, u]=\int_a^b[p|u'|^2+q|u|^2]. \]

First, we assume $u\neq 0$ in any subinterval of $[a, b)$. Then

\begin{align*} \lambda& =\int_a^b\left(p(x)\frac{|u'(x)|^2}{w(x)|u(x)|^2}+\frac{q(x)}{w(x)}\right)w(x)|u(x)|^2\,{\rm d}x\\ & :=\int_a^b\left(p(x)r_1(x)+\frac{q(x)}{w(x)}\right)w(x)|u(x)|^2\,{\rm d}x, \end{align*}

where $\displaystyle {r_1(x)=\frac {|u'(x)|^2}{w(x)|u(x)|^2}\in (0, \infty )}$. For arbitrary $\epsilon >0$, there exists $T'\in (a, b)$ such that

\begin{align*} & \left|\int_a^{T'}\left(p(x)r_1(x)+\frac{q(x)}{w(x)}\right)w(x)|u(x)|^2\,{\rm d}x-\lambda\right|<\frac{\epsilon}{4}, \\ & \left|\int_{T'}^b\left(p(x)r_1(x)+\frac{q(x)}{w(x)}\right)w(x)|u(x)|^2\,{\rm d}x\right|<\frac{\epsilon}{4}. \end{align*}

Since $||u||=1$ implies $\int _a^bw(x)|u(x)|^2\,{\rm d}x=1$, there exists $T^{''}\in (a, b)$ such that

\[ \int_{T^{''}}^bw(x)|u(x)|^2\,{\rm d}x<\frac{\epsilon}{4}\left|p(T')r_1(T')+\frac{q(T')}{w(T')}\right|^{{-}1}. \]

Let $T=\max \{T', T^{''}\}$ and $\displaystyle {\lambda _T=\int _a^{T}(p(x)r_1(x)+\frac {q(x)}{w(x)})w(x)|u(x)|^2\,{\rm d}x}$. Then

\begin{align*} & |\lambda_{T}-\lambda|<\frac{\epsilon}{2}, \\ & \int_{T}^bw(x)|u(x)|^2\,{\rm d}x<\frac{\epsilon}{4}\left[p(T')r_1(T')+\frac{q(T')}{w(T')}\right]^{{-}1}. \end{align*}

Now we consider the constant $\lambda _{T}$, by the definition of integrand, for $\epsilon >0$, there exists $\delta >0$, and a partition $a=x_0< x_1< x_2<\cdots < x_n=T$,

\[ \left| \sum_{i=1}^n\left[p(\xi_i)r_1(\xi_i)+\frac{q(\xi_i)}{w(\xi_i)}\right] \Delta\left(\int_{x_{i-1}}^{x_i}w(x)|u(x)|^2\,{\rm d}x\right)-\lambda_{T}\right|<\frac{\epsilon}{4} \]

provided $\displaystyle {\Delta (T)=\max _{1\leq i\leq n}\{x_i-x_{i-1}\}<\delta }$, where $\xi _i\in (x_{i-1}, x_i)$. Since $\int _a^bw(x)|u(x)|^2 {\rm d}x=1$, we obtain

\begin{align*} \rho(\Delta)& :=\sum_{i=1}^n\left[p(\xi_i)r_1(\xi_i)+\frac{q(\xi_i)}{w(\xi_i)}\right] \Delta\left(\int_{x_{i-1}}^{x_i}w(x)|u(x)|^2\,{\rm d}x\right)\\ & \quad+\left[p(T')r_1(T')+\frac{q(T')}{w(T')}\right]\int_{T}^bw(x)|u(x)|^2\,{\rm d}x\\ & \in {\rm co}\left\{\frac{q(x)}{w(x)}+rp(x):\,r>0, x\in(a, b)\right\}. \end{align*}

So we obtain

\begin{align*} |\lambda-\rho(\Delta)|& \leq |\lambda-\lambda_T|+\left|\lambda_{T}- \sum_{i=1}^n\left[p(\xi_i)r_1(\xi_i)+\frac{q(\xi_i)}{w(\xi_i)}\right] \Delta\left(\int_{x_{i-1}}^{x_i}w(x)|u(x)|^2\,{\rm d}x\right)\right|\\ & \quad+\left| \sum_{i=1}^n\left[p(\xi_i)r_1(\xi_i)+\frac{q(\xi_i)}{w(\xi_i)}\right] \Delta\left(\int_{x_{i-1}}^{x_i}w(x)|u(x)|^2\,{\rm d}x\right)-\rho(\Delta)\right|\\ & < \frac{\epsilon}{2}+\frac{\epsilon}{4}+\left[p(T')r_1(T')+\frac{q(T')}{w(T')}\right]\int_{T}^bw(x)|u(x)|^2\,{\rm d}x\\ & = \frac{3\epsilon}{4}+\left|p(T')r_1(T')+\frac{q(T')}{w(T')}\right|\int_{T}^bw(x)|u(x)|^2\,{\rm d}x<\epsilon, \end{align*}

which implies

\[ \lambda\in \overline{\rm co}\left\{\frac{q(x)}{w(x)}+rp(x):\,r>0, x\in(a, b)\right\}=\Omega. \]

For the case where $u(t)\equiv 0$ in some sub-interval of $[a, b)$, we define $V=\bigcup _{i=1}^n(a_i, b_i)$, where $n$ is finite or $n=\infty$, $(a_i, b_i)$ are disjoint open intervals such that $u(x)\equiv 0$ for $x\in (a_i, b_i)$, $1\leq i\leq n$. We assume further that $a< a_i< b_i< a_{i+1}$ and the set $\{a_i\}$ has no finite accumulation point if $b=\infty$, and the only possible finite accumulation point is $b$ if $b<\infty$.

Let $\tau =\tau (x)=m([a, x]-V)$, where $m$ denotes the usual linear Lebesgue measure, and let $B=\tau (b)$, $A_i=\tau (a_i)$, $i=1, 2, \cdots, n$. $[a, B)$ is obtained from $[a, b)$ by shrinking each interval $(a_i, b_i)$ to its left endpoint.

Let $D_I$ denote the set of piecewise continuous functions on $I$. We construct a class of transformation $L_V: D_{[a, b)}\to D_{[a, B)}$ as follows: Let $f\in D_{[a, b)}$. Then $F=L_V(f)$ is defined by

\[ F(\tau)=f(x)\ {\rm if}\ \tau =\tau(x), \tau \neq A_i,\ {\rm and}\ F(A_i)=f(a_i). \]

The function $F$ is obtained from $f$ by collapsing each interval $(a_i, b_i)$ to a point. For all $\lambda \in \Theta (T_F)$, we obtain

\begin{align*} \lambda& =\int_a^b[p(x)|u'(x)|^2+q(x)|u(x)|^2]\,{\rm d}x\\ & = \int_a^B\left[(L_Vp)(\tau)|(L_Vu)'(\tau)|^2+(L_Vq)(\tau)|(L_Vu)(\tau)|^2\right]{\rm d}\tau. \end{align*}

The remainder is similar as above, so we omit the details. In both cases, we have proved $\Theta (T_F)\subseteq \Omega$. This completes the proof.