The regularized trace formula for a fourth order differential operator given in a finite interval

Abstract

In this work, a regularized trace formula for a differential operator of fourth order with bounded operator coefficient is found.

1 Introduction

Investigations into the regularized trace formulas of scalar differential operators started with the work [1] firstly. After that work, regularized trace formulas for several differential operators have been studied in some works as [2, 3] and [4]. In [5] a formula for the second regularized trace of the problem generated by a Sturm-Liouville operator equation with a spectral parameter dependent boundary condition is found. The list of the works on this subject is given in [6] and [7]. The trace formulas for differential operators with operator coefficient are investigated in the works [8–13] and [14]. The boundary conditions in our work are completely different from those in [9].

In this work, we find the following regularized trace formula for a self-adjoint differential operator L of fourth order with bounded operator coefficient:

$$\sum_{m=0}^{\infty} \Biggl[\sum _{n=1}^{\infty} \biggl(\lambda _{mn}-\biggl(m+ \frac{1}{2}\biggr)^{4} \biggr)-\frac{1}{\pi} \int _{0}^{\pi }\operatorname{tr}Q(x)\,dx \Biggr]= \frac{1}{4} \bigl[\operatorname{tr}Q(\pi)-\operatorname{tr}Q(0) \bigr]. $$

Here $\{\lambda_{mn}\}_{n=1}^{\infty}$ are the eigenvalues of the operator L which belong to the interval $[(m+\frac{1}{2})^{4}-\|Q\|,(m+\frac{1}{2})^{4}+\|Q\|]$.

1.1 Notation and preliminaries

Let H be a separable Hilbert space with infinite dimension. Let us consider the operators $L_{0}$ and L in the Hilbert space $H_{1}=L_{2}(0,\pi; H)$ which are formed by the following differential expressions:

$$\begin{aligned}[b] &\ell_{0}(y)=y^{{\i}v}(x),\\ &\ell(y)=y^{{\i}v}(x)+Q(x)y(x) \end{aligned} $$

with the same boundary conditions $y(0)=y^{\prime\prime}(0)=y^{\prime}(\pi)=y^{\prime\prime\prime}(\pi)=0$. Suppose that the operator function $Q(x)$ in the expression $\ell(y)$ satisfies the following conditions:

• (Q1) $Q(x):H \rightarrow H$ is a self-adjoint kernel operator for every $x\in[0,\pi]$. Moreover, $Q(x)$ has second order weak derivative in this interval and $Q^{(i)}(x):H \rightarrow H$ ($i=1,2$) are self-adjoint kernel operators for every $x\in[0,\pi]$.

• (Q2) $\|Q\|<\frac{5}{2}$.

• (Q3) There is an orthonormal basis $\{\varphi_{n}\}_{n=1}^{\infty}$ of the space H such that

  $$\sum_{n=1}^{\infty}\bigl\| Q(x)\varphi_{n} \bigr\| < \infty. $$

• (Q4) The functions $\|Q^{(i)}(x)\|_{\sigma_{1}(H)}$ are bounded and measurable in the interval $[0,\pi]$ ($i=0,1,2$).

Here $\sigma_{1}(H)$ is the space of kernel operators from H to H as in [15]. Moreover, we denote the norms by $\|\cdot\|_{H}$ and $\|\cdot\|$ and inner products by $(\cdot,\cdot)_{H}$ and $(\cdot,\cdot)$ in H and $H_{1}$, respectively, and we also denote the sum of eigenvalues of a kernel operator A by $\operatorname{tr}A=\operatorname{trace}A$.

Spectrum of the operator $L_{0}$ is the set

$$\sigma(L_{0})=\biggl\{ \biggl(\frac{1}{2}\biggr)^{4}, \biggl(\frac{3}{2}\biggr)^{4},\ldots,\biggl(m+\frac {1}{2} \biggr)^{4},\ldots\biggr\} . $$

Every point of this set is an eigenvalue of $L_{0}$ which has infinite multiplicity. The orthonormal eigenfunctions corresponding to eigenvalue $(m+\frac{1}{2})^{4}$ are in the form

$$ \psi_{mn}(x)=\sqrt{\frac{2}{\pi}}\sin\biggl(m+\frac{1}{2} \biggr)x\cdot\varphi_{n}\quad (n=1,2,\ldots). $$

(1.1)

2 Some relations between spectrums of operators $L_{0}$ and L

Let $R_{\lambda}^{0}$, $R_{\lambda}$ be resolvents of the operators $L_{0}$ and L, respectively. If the operator $Q:H_{1} \rightarrow H _{1}$ satisfies conditions (Q2) and (Q3), the following can be proved:

• (a) $QR_{\lambda}^{0} \in\sigma_{1}(H_{1})$ for every $\lambda\notin\sigma(L_{0})$.

• (b) Spectrum of the operator L is a subset of the union of pairwise disjoint intervals

  $$F_{m}=\biggl[\biggl(m+\frac{1}{2}\biggr)^{4}-\|Q\|, \biggl(m+\frac{1}{2}\biggr)^{4}+\|Q\| \biggr] \quad(m=0,1,2, \ldots), \sigma(L)\subset\bigcup_{m=0}^{\infty}F_{m}. $$

• (c) Each point of the spectrum of L, different from $(m+\frac{1}{2})^{4}$ in $F_{m}$ is an isolated eigenvalue which has finite multiplicity.

• (d) The series $\sum_{n=1}^{\infty} [\lambda_{mn}-(m+\frac {1}{2})^{4} ]$ ($m=0,1,2,\ldots$) are absolutely convergent where $\{\lambda_{mn}\}_{n=1}^{\infty}$ are eigenvalues of the operator L in the interval $F_{m}$.

Let $\rho(L)$ be resolvent set of the operator L; $\rho(L)={ \mathbf{C}\backslash\sigma(L)}$. Since $QR_{\lambda}^{0} \in\sigma_{1}(H_{1})$ for every $\lambda\in\rho(L)$, from the equation $R_{\lambda}=R_{\lambda}^{0}-R_{\lambda}QR_{\lambda}^{0}$ we obtain $R_{\lambda}-R_{\lambda}^{0}\in\sigma_{1}(H_{1})$.

On the other hand, if we consider the series

$$\sum_{n=1}^{\infty} \biggl[\lambda_{mn}- \biggl(m+\frac{1}{2}\biggr)^{4} \biggr]\quad (m=0,1,2,\ldots) $$

are absolutely convergent then we have

$$\operatorname{tr}\bigl(R_{\lambda}-R_{\lambda}^{0}\bigr)= \sum_{m=0}^{\infty}\sum _{n=1}^{\infty} \biggl[\frac{1}{\lambda_{mn}-\lambda}-\frac{1}{(m+\frac {1}{2})^{4}-\lambda} \biggr] $$

for every $\lambda\in\rho(L)$ [10]. If we multiply both sides of this equality with $\frac{\lambda}{2\pi i}$ and integrate this equality over the circle $|\lambda|=b_{p}=(p+1)^{4}$ ($p\in\mathbf{N}$, $p\geq1$), then we find

$$\begin{aligned} &\frac{1}{2\pi i}\int_{|\lambda|=b_{p}}\lambda \operatorname{tr}\bigl(R_{\lambda }-R_{\lambda}^{0}\bigr)\,d\lambda \\ &\quad=\frac{1}{2\pi i}\int_{|\lambda|=b_{p}}\lambda\sum _{m=0}^{p}\sum_{n=1}^{\infty} \biggl[\frac{1}{\lambda_{mn}-\lambda}- \frac{1}{(m+\frac {1}{2})^{4}-\lambda} \biggr]\,d\lambda \\ &\qquad{}+\frac{1}{2\pi i}\int_{|\lambda|=b_{p}}\lambda\sum _{m=p+1}^{\infty}\sum_{n=1}^{\infty} \biggl[\frac{1}{\lambda_{mn}-\lambda}-\frac {1}{(m+\frac{1}{2})^{4}-\lambda} \biggr]\,d\lambda. \end{aligned}$$

(2.1)

If we consider the relations $|\lambda_{mn}|< b_{p}$ ($m=0,1,2,\ldots,p$) and $|\lambda_{mn}|> b_{p}$ ($m=p+1, p+2,\ldots$) for $n=1,2,3,\ldots$ , then from (2.1) we get

$$\begin{aligned} &\frac{1}{2\pi i}\int_{|\lambda|=b_{p}}\lambda \operatorname{tr}\bigl(R_{\lambda }-R_{\lambda}^{0}\bigr)\,d\lambda \\ &\quad=\sum_{m=0}^{p}\sum _{n=1}^{\infty} \biggl[\frac{1}{2\pi i}\int _{|\lambda|=b_{p}}\frac{\lambda \,d\lambda}{\lambda -(m+\frac{1}{2})^{4}}-\frac{1}{2\pi i}\int _{|\lambda|=b_{p}}\frac {\lambda \,d\lambda}{\lambda-\lambda_{mn}} \biggr] \\ &\qquad{}+\sum_{m=p+1}^{\infty}\sum _{n=1}^{\infty} \biggl[\frac {1}{2\pi i}\int _{|\lambda|=b_{p}}\frac{\lambda \,d\lambda}{\lambda-(m+\frac{1}{2})^{4}}-\frac{1}{2\pi i}\int _{|\lambda|=b_{p}}\frac{\lambda \,d\lambda}{\lambda-\lambda_{mn}} \biggr] \\ &\quad=\sum_{m=0}^{p}\sum _{n=1}^{\infty} \biggl[\biggl(m+\frac {1}{2} \biggr)^{4}-\lambda_{mn} \biggr]. \end{aligned}$$

(2.2)

Moreover, from the formula $R_{\lambda}=R_{\lambda }^{0}-R_{\lambda}QR_{\lambda}^{0}$, we obtain the following equality:

$$ R_{\lambda}-R_{\lambda}^{0}=-R_{\lambda}^{0}QR_{\lambda}^{0}+R_{\lambda }^{0} \bigl(QR_{\lambda}^{0}\bigr)^{2}-R_{\lambda} \bigl(QR_{\lambda}^{0}\bigr)^{3}. $$

(2.3)

If we put this equality into (2.2), we have

$$ \sum_{m=0}^{p}\sum _{n=1}^{\infty} \biggl[\lambda _{mn}-\biggl(m+ \frac{1}{2}\biggr)^{4} \biggr]=M_{p1}+M_{p2}+M_{p}. $$

(2.4)

Here

$$\begin{aligned}& M_{pj}=\frac{(-1)^{j}}{2\pi i}\int_{|\lambda|=b_{p}}\lambda \operatorname{tr}\bigl[R_{\lambda}^{0}\bigl(QR_{\lambda}^{0} \bigr)^{j}\bigr]\,d\lambda\quad (j=1,2), \end{aligned}$$

(2.5)

$$\begin{aligned}& M_{p}=\frac{-1}{2\pi i}\int_{|\lambda|=b_{p}}\lambda \operatorname{tr}\bigl[R_{\lambda}\bigl(QR_{\lambda}^{0} \bigr)^{3}\bigr]\,d\lambda. \end{aligned}$$

(2.6)

Theorem 2.1

If the operator function $Q(x)$ satisfies condition (Q3), then we have

$$M_{pj}=\frac{(-1)^{j}}{2\pi ij}\int_{|\lambda |=b_{p}} \operatorname{tr}\bigl[\bigl(QR_{\lambda}^{0}\bigr)^{j} \bigr]\,d\lambda. $$

Proof

It can be shown that the operator function $QR_{\lambda}^{0}$ is analytic with respect to the norm in the space $\sigma_{1}(H_{1})$ in domain $\rho(L_{0})=\mathbf{C}\setminus\sigma(L_{0})$ and

$$ \operatorname{tr}\bigl\{ \bigl[\bigl(QR_{\lambda}^{0} \bigr)^{j}\bigr]'\bigr\} =j\operatorname{tr}\bigl[\bigl(QR_{\lambda}^{0} \bigr)'\bigl(QR_{\lambda }^{0}\bigr)^{j-1} \bigr]. $$

(2.7)

Considering $(QR_{\lambda}^{0})'=(QR_{\lambda}^{0})^{2}$, we can write the formula (2.7) as

$$ \operatorname{tr}\bigl\{ \bigl[\bigl(QR_{\lambda}^{0} \bigr)^{j}\bigr]'\bigr\} =j\operatorname{tr}\bigl[R_{\lambda}^{0} \bigl(QR_{\lambda }^{0}\bigr)^{j}\bigr]. $$

(2.8)

From (2.5) and (2.8), we obtain

$$M_{pj}=\frac{(-1)^{j+1}}{2\pi ij}\int_{|\lambda|=b_{p}}\lambda \operatorname{tr}\bigl\{ \bigl[\bigl(QR_{\lambda}^{0} \bigr)^{j}\bigr]'\bigr\} \,d\lambda. $$

From here, we find

$$\begin{aligned} M_{pj}&=\frac{(-1)^{j+1}}{2\pi ij}\int_{|\lambda|=b_{p}} \operatorname{tr}\bigl\{ \bigl[\lambda \bigl(QR_{\lambda}^{0} \bigr)^{j}\bigr]'-\bigl(QR_{\lambda}^{0} \bigr)^{j}\bigr\} \,d\lambda \\ &=\frac{(-1)^{j}}{2\pi ij}\int_{|\lambda|=b_{p}}\operatorname{tr}\bigl[ \bigl(QR_{\lambda}^{0}\bigr)^{j}\bigr]\,d\lambda+ \frac {(-1)^{j+1}}{2\pi ij}\int_{|\lambda|=b_{p}}\operatorname{tr}\bigl\{ \bigl[ \lambda\bigl(QR_{\lambda}^{0}\bigr)^{j} \bigr]'\bigr\} \,d\lambda. \end{aligned}$$

(2.9)

It can be easily shown that

$$\operatorname{tr}\bigl\{ \bigl[\lambda\bigl(QR_{\lambda}^{0} \bigr)^{j}\bigr]'\bigr\} =\bigl\{ \operatorname{tr}\bigl[ \lambda\bigl(QR_{\lambda}^{0}\bigr)^{j}\bigr]\bigr\} '. $$

Therefore, we have

$$ \int_{|\lambda|=b_{p}}\operatorname{tr}\bigl\{ \bigl[\lambda \bigl(QR_{\lambda}^{0}\bigr)^{j}\bigr]' \bigr\} \,d\lambda=\int_{|\lambda|=b_{p}} \bigl\{ \operatorname{tr}\bigl[ \lambda\bigl(QR_{\lambda }^{0}\bigr)^{j}\bigr]\bigr\} '\,d\lambda. $$

(2.10)

The integral on the right-hand side of the last equality can be written as

$$ \int_{|\lambda|=b_{p}}\bigl\{ \operatorname{tr}\bigl[\lambda \bigl(QR_{\lambda}^{0}\bigr)^{j}\bigr]\bigr\} '\,d\lambda= \mathop{\int_{|\lambda|=b_{p}}}_{\hphantom{a}\operatorname{Im}\lambda\geq 0}\bigl\{ \operatorname{tr} \bigl[\lambda\bigl(QR_{\lambda}^{0}\bigr)^{j}\bigr]\bigr\} '\,d\lambda +\mathop{\int_{|\lambda|=b_{p}}}_{\hphantom{a}\operatorname{Im}\lambda\leq 0}\bigl\{ \operatorname{tr} \bigl[\lambda\bigl(QR_{\lambda}^{0}\bigr)^{j}\bigr]\bigr\} '\,d\lambda. $$

(2.11)

Let $\varepsilon_{0}$ be a constant satisfying the condition $0<\varepsilon_{0}<b_{p}-(p+\frac{1}{2})^{4}$. Consider the function $\operatorname{tr}[\lambda(QR_{\lambda}^{0})^{j}]$ is analytic in simple connected domains

$$\begin{aligned}& G_{1}=\{\lambda\in\mathbf{C}:b_{p}-\varepsilon_{0}< \lambda< b_{p}+\varepsilon _{0}, \operatorname{Im}\lambda>- \varepsilon_{0}\}, \\& G_{2}=\{\lambda\in\mathbf{C}:b_{p}-\varepsilon_{0}< \lambda< b_{p}+\varepsilon _{0}, \operatorname{Im}\lambda< \varepsilon_{0}\} \end{aligned}$$

and

$$\begin{aligned}& \bigl\{ \lambda\in\mathbf{C}:|\lambda|=b_{p}, \operatorname{Im}\lambda\geq0 \bigr\} \subset G_{1}, \qquad \bigl\{ \lambda\in\mathbf{C}:|\lambda|=b_{p}, \operatorname{Im}\lambda\leq0 \bigr\} \subset G_{2} . \end{aligned}$$

From (2.11) we obtain

$$\begin{aligned} \int_{|\lambda|=b_{p}}\bigl\{ \operatorname{tr}\bigl[\lambda \bigl(QR_{\lambda}^{0}\bigr)^{j}\bigr]\bigr\} '\,d\lambda={}&\operatorname{tr}\bigl[-b_{p} \bigl(QR_{-b_{p}}^{0}\bigr)^{j}\bigr] - \operatorname{tr}\bigl[b_{p}(QR_{b_{p}})^{j}\bigr] \\ &{}+\operatorname{tr}\bigl[b_{p}\bigl(QR_{b_{p}}^{0} \bigr)^{j}\bigr] -\operatorname{tr}\bigl[-b_{p} \bigl(QR_{-b_{p}}^{0}\bigr)^{j}\bigr]=0. \end{aligned}$$

(2.12)

From (2.9), (2.10) and (2.12) we find

$$M_{pj}=\frac{(-1)^{j}}{2\pi ij}\int_{|\lambda|=b_{p}} \operatorname{tr}\bigl[\bigl(QR_{\lambda}^{0}\bigr)^{j} \bigr]\,d\lambda. $$

 □

3 The formula of the regularized trace of the operator L

In this section, we find a formula for the regularized trace of the operator L. According to Theorem 2.1,

$$ M_{p1}=-\frac{1}{2\pi i}\int_{|\lambda|=b_{p}} \operatorname{tr}\bigl[\bigl(QR_{\lambda }^{0}\bigr)\bigr]\,d\lambda. $$

(3.1)

Since $\{\psi_{mn}\}_{m=0,n=1}^{\infty \infty}$ is an orthonormal basis of the space $H_{1}$, from (3.1) we obtain

$$\begin{aligned} M_{p1}&=-\frac{1}{2\pi i}\int_{|\lambda|=b_{p}} \sum_{m=0}^{\infty}\sum _{n=1}^{\infty}\bigl(QR_{\lambda }^{0} \psi_{mn},\psi_{mn}\bigr)\,d\lambda \\ &=-\frac{1}{2\pi i}\int_{|\lambda|=b_{p}}\sum _{m=0}^{\infty}\sum_{n=1}^{\infty} \frac{(Q\psi_{mn},\psi_{mn})}{(m+\frac{1}{2})^{4}-\lambda }\,d\lambda \\ & =\sum_{m=0}^{\infty}\sum _{n=1}^{\infty}(Q\psi _{mn},\psi_{mn})\cdot \frac{1}{2\pi i }\int_{|\lambda|=b_{p}}\frac{1}{\lambda- (m+\frac{1}{2} )^{4}}\,d\lambda. \end{aligned}$$

(3.2)

Considering $(m+\frac{1}{2})^{4}< b_{p}=(p+1)^{4}$ for $m\leq p$ and $(m+\frac{1}{2})^{4}>b_{p}=(p+1)^{4}$ for $m>p$, from formula (3.2)

$$\begin{aligned} M_{p1}&=\sum_{m=0}^{p} \sum_{n=1}^{\infty}(Q\psi_{mn},\psi _{mn})\frac{1}{2\pi i }\int_{|\lambda|=b_{p}} \frac{1}{\lambda- (m+\frac{1}{2} )^{4}}\,d\lambda =\sum_{m=0}^{p}\sum _{n=1}^{\infty}(Q\psi_{mn},\psi _{mn}) \end{aligned}$$

(3.3)

is obtained.

From (1.1) and (3.3), we have

$$\begin{aligned} M_{p1}&=\sum_{m=0}^{p} \sum_{n=1}^{\infty}\int_{0}^{\pi} \biggl(Q(x)\sqrt{\frac{2}{\pi}} \sin\biggl(m+\frac{1}{2}\biggr)x\cdot \varphi _{n},\sqrt{\frac{2}{\pi}}\sin\biggl(m+\frac{1}{2} \biggr)x\cdot\varphi_{n} \biggr)_{H}\,dx \\ &=\frac{2}{\pi}\sum_{m=0}^{p}\sum _{n=1}^{\infty}\int_{0}^{\pi} \bigl(Q(x)\varphi_{n}, \varphi_{n}\bigr)_{H} \sin^{2}\biggl(m+\frac {1}{2}\biggr)x\,dx \\ &=\frac{1}{\pi}\sum_{m=0}^{p}\int _{0}^{\pi} \Biggl[\sum_{n=1}^{\infty} \bigl(Q(x)\varphi_{n},\varphi_{n}\bigr)_{H} \Biggr]\bigl(1-\cos (2m+1)x\bigr)\,dx. \end{aligned}$$

(3.4)

If we consider the formula $\sum_{n=1}^{\infty }(Q(x)\varphi_{n},\varphi_{n})_{H}=\operatorname{tr}Q(x)$, then we get

$$ M_{p1}=\frac{p+1}{\pi}\int_{0}^{\pi} \operatorname{tr}Q(x)\,dx-\frac{1}{\pi}\sum_{m=0}^{p} \int_{0}^{\pi}\operatorname{tr}Q(x)\cos(2m+1)x\,dx. $$

(3.5)

Lemma 3.1

If the operator function $Q(x)$ satisfies conditions (Q2) and (Q3), then we have

$$\|R_{\lambda}\|< p^{-3} $$

over the circle $|\lambda|=b_{p}$.

Proof

Since the operator function $Q(x)$ satisfies conditions (Q2) and (Q3), we have

$$\begin{aligned}& \{\lambda_{mn}\}_{n=1}^{\infty}\subset \biggl(\biggl(m+ \frac{1}{2}\biggr)^{4}-\|Q\| ,\biggl(m+\frac{1}{2} \biggr)^{4}+\|Q\| \biggr)\quad (m=0,1,2,\ldots), \\& \biggl|\lambda_{mn}-\biggl(m+\frac{1}{2}\biggr)^{4}\biggr|< \|Q\|< \frac{5}{2}\quad (m=0,1,2,\ldots;n=1,2,\ldots). \end{aligned}$$

If we consider this relation, we get

$$\begin{aligned} |\lambda_{mn}-\lambda|&= \biggl|\lambda-\biggl(m+ \frac{1}{2}\biggr)^{4}- \biggl(\lambda _{mn}- \biggl(m+\frac{1}{2}\biggr)^{4}\biggr) \biggr| \\ &\geq \biggl|\lambda-\biggl(m+\frac{1}{2}\biggr)^{4} \biggr|- \biggl| \lambda_{mn}-\biggl(m+\frac {1}{2}\biggr)^{4} \biggr| \\ &>|\lambda|-\biggl(m+\frac{1}{2}\biggr)^{4}- \frac{5}{2}=(p+1)^{4}-\biggl(m+\frac {1}{2} \biggr)^{4}-\frac{5}{2} \\ &\geq(p+1)^{4}-\biggl(p+\frac{1}{2}\biggr)^{4}- \frac{5}{2}>2\biggl(p+\frac{1}{2}\biggr)^{3}- \frac {5}{2}>p^{3} \end{aligned}$$

(3.6)

for $m\leq p$ and

$$\begin{aligned} |\lambda_{mn}-\lambda|&= \biggl|\biggl(m+\frac{1}{2} \biggr)^{4}-\lambda- \biggl(\biggl(m+\frac {1}{2} \biggr)^{4}-\lambda_{mn} \biggr) \biggr| \\ &\geq \biggl|\biggl(m+\frac{1}{2}\biggr)^{4}-\lambda \biggr|- \biggl|\biggl(m+ \frac {1}{2}\biggr)^{4}-\lambda_{mn} \biggr|\geq\biggl(m+ \frac{1}{2}\biggr)^{4}-|\lambda|-\frac{5}{2} \\ &\geq\biggl(p+\frac{3}{2}\biggr)^{4}-(p+1)^{4}- \frac{5}{2}>2(p+1)^{3}-\frac {5}{2}>p^{3} \end{aligned}$$

(3.7)

for $m\geq p+1$.

On the other hand, we can write

$$ \| R_{\lambda} \|=\mathop{\max_{m=0,1,\ldots}}_{n=1,2,\ldots}\bigl\{ |\lambda_{mn}- \lambda|^{-1}\bigr\} . $$

(3.8)

From (3.6), (3.7) and (3.8) we get

$$\| R_{\lambda} \|< p^{-3}\quad \bigl(|\lambda|=b_{p}\bigr). $$

 □

Lemma 3.2

If the operator function $Q(x)$ satisfies condition (Q3), then we have

$$\bigl\| QR_{\lambda}^{0}\bigr\| _{\sigma_{1}(H_{1})}< 5p^{-2}\sum _{n=1}^{\infty}\bigl\| Q(x)\varphi_{n}\bigr\| $$

over the circle $|\lambda|=b_{p}$.

Proof

Let us show that the series $\sum_{m=0}^{\infty}\sum_{n=1}^{\infty} \|QR_{\lambda}^{0}\psi_{mn} \|$ is convergent.

For $\lambda\notin\sigma(L_{0})$, we get

$$\begin{aligned} &\sum_{m=0}^{\infty}\sum _{n=1}^{\infty} \bigl\| QR_{\lambda }^{0} \psi_{mn} \bigr\| \\ &\quad=\sum_{m=0}^{\infty} \sum _{n=1}^{\infty} \biggl|\biggl(m+\frac{1}{2} \biggr)^{4}-\lambda \biggr|^{-1} \|Q\psi _{mn} \| \\ &\quad=\sum_{m=0}^{\infty}\sum _{n=1}^{\infty} \biggl|\biggl(m+\frac {1}{2} \biggr)^{4}-\lambda \biggr|^{-1} \biggl[\int_{0}^{\pi} \biggl\| Q(x)\sqrt{\frac{2}{\pi}}\sin\biggl(m+\frac{1}{2}\biggr)x\cdot \varphi_{n} \biggr\| _{H}^{2}\,dx \biggr]^{\frac{1}{2}} \\ &\quad=\sqrt{\frac{2}{\pi}}\sum_{m=0}^{\infty} \sum_{n=1}^{\infty } \biggl|\biggl(m+\frac{1}{2} \biggr)^{4}-\lambda \biggr|^{-1} \int_{0}^{\pi} \bigl\| Q(x)\varphi_{n} \bigr\| _{H}^{2}\sin^{2} \biggl(m+\frac{1}{2}\biggr)x\,dx \\ &\quad< \sum_{m=0}^{\infty}\sum _{n=1}^{\infty} \biggl|\biggl(m+\frac {1}{2} \biggr)^{4}-\lambda \biggr|^{-1} \biggl[\int_{0}^{\pi} \bigl\| Q(x)\varphi_{n} \bigr\| _{H}^{2}\,dx \biggr]^{\frac{1}{2}} \\ &\quad=\sum_{m=0}^{\infty} \biggl|\biggl(m+ \frac{1}{2}\biggr)^{4}-\lambda \biggr|^{-1}\sum _{n=1}^{\infty} \bigl\| Q(x)\varphi_{n} \bigr\| . \end{aligned}$$

(3.9)

From this relation we obtain

$$\sum_{m=0}^{\infty}\sum _{n=1}^{\infty} \bigl\| QR_{\lambda }^{0} \psi_{mn} \bigr\| < \infty \quad\bigl(\lambda\notin\sigma(L_{0})\bigr). $$

On the other hand, since the sequence $\{\psi_{mn}\} _{m=0,n=1}^{\infty \infty}$ is an orthonormal basis of the space $H_{1}$, we get

$$ \bigl\| QR_{\lambda}^{0} \bigr\| _{\sigma_{1}(H_{1})}\leq\sum _{m=0}^{\infty}\sum_{n=1}^{\infty} \bigl\| QR_{\lambda}^{0}\psi _{mn} \bigr\| $$

(3.10)

[15]. From (3.9) and (3.10) we obtain

$$ \bigl\| QR_{\lambda}^{0} \bigr\| _{\sigma_{1}(H_{1})}\leq\sum _{n=1}^{\infty} \bigl\| Q(x)\varphi_{n} \bigr\| \sum _{m=0}^{\infty} \biggl|\biggl(m+\frac{1}{2} \biggr)^{4}-\lambda \biggr|^{-1}. $$

(3.11)

Furthermore, over the circle $|\lambda|=b_{p}$ we get

$$\begin{aligned} &\sum_{m=0}^{\infty} \biggl|\biggl(m+ \frac{1}{2}\biggr)^{4}-\lambda \biggr|^{-1} \\ &\quad=\sum_{m=0}^{p} \biggl|\biggl(m+ \frac{1}{2}\biggr)^{4}-\lambda \biggr|^{-1}+\sum _{m=p+1}^{\infty} \biggl|\biggl(m+\frac{1}{2} \biggr)^{4}-\lambda \biggr|^{-1} \\ &\quad< \sum_{m=0}^{p} \biggl(|\lambda|- \biggl(m+\frac{1}{2}\biggr)^{4} \biggr)^{-1}+\sum _{m=p+1}^{\infty} \biggl(\biggl(m+\frac{1}{2} \biggr)^{4}-|\lambda | \biggr)^{-1} \\ &\quad=\sum_{m=0}^{p} \biggl((p+1)^{4}-\biggl(m+\frac{1}{2}\biggr)^{4} \biggr)^{-1}+\sum_{m=p+1}^{\infty} \biggl( \biggl(m+\frac {1}{2}\biggr)^{4}-(p+1)^{4} \biggr)^{-1} \\ &\quad< \sum_{m=0}^{p} \biggl((p+1)^{4}-\biggl(p+\frac{1}{2}\biggr)^{4} \biggr)^{-1} \\ &\qquad{}+\sum_{m=p+1}^{\infty} \biggl( \biggl(m+\frac {1}{2}\biggr)^{2}-(p+1)^{2} \biggr)^{-1} \biggl(\biggl(m+\frac {1}{2}\biggr)^{2}+(p+1)^{2} \biggr)^{-1} \\ &\quad< \frac{1}{2}(p+1) \biggl(p+\frac{1}{2}\biggr)^{-3}+ \frac{1}{2}p^{-2}\sum_{m=p+1}^{\infty} \biggl(\biggl(m+\frac{1}{2}\biggr)^{2}-(p+1)^{2} \biggr)^{-1}. \end{aligned}$$

(3.12)

It can be easily shown that

$$ \biggl(m+\frac{1}{2}\biggr)^{2}-(p+1)^{2}> \frac{1}{4}\bigl(m^{2}-p^{2}\bigr)\quad (m\geq p+1) $$

(3.13)

and

$$ \sum_{m=p+1}^{\infty}\bigl(m^{2}-p^{2} \bigr)^{-1}< 2p^{-\frac{1}{2}}. $$

(3.14)

From (3.12), (3.13) and (3.14) we get

$$ \sum_{m=o}^{\infty} \biggl|\biggl(m+\frac{1}{2} \biggr)^{4}-\lambda \biggr|^{-1}< p^{-2}+4p^{-\frac{5}{2}}< 5p^{-2}. $$

(3.15)

From (3.11) and (3.15) we obtain

$$\bigl\| QR_{\lambda}^{0}\bigr\| _{\sigma_{1}(H_{1})}< 5p^{-2}\sum _{n=1}^{\infty}\bigl\| Q(x)\varphi_{n}\bigr\| \quad \bigl(|\lambda|=b_{p}\bigr). $$

 □

Theorem 3.3

If the operator function $Q(x)$ satisfies conditions (Q1), (Q2), (Q3), and (Q4), then we have the formula

$$\begin{aligned} &\sum_{m=0}^{\infty} \Biggl\{ \sum _{n=1}^{\infty} \biggl[\lambda_{mn}-\biggl(m+ \frac{1}{2}\biggr)^{4} \biggr]-\frac{1}{\pi}\int _{0}^{\pi}\operatorname{tr}Q(x)\,dx \Biggr\} \\ &\quad=\frac{1}{4}\bigl[\operatorname{tr}Q(\pi)-\operatorname{tr}Q(0) \bigr]. \end{aligned}$$

Proof

By using Theorem 2.1, Lemma 3.1 and Lemma 3.2, we find

$$\begin{aligned} |{M_{p2}}| = & {\frac{1}{4\pi}} \biggl|\int _{|\lambda|=b_{p}} \operatorname{tr} \bigl[\bigl(QR_{\lambda}^{0} \bigr)^{2} \bigr]\,d\lambda \biggr| \\ < & {\frac{1}{4\pi}}\int_{|\lambda|=b_{p}} \bigl| \operatorname{tr} \bigl[\bigl(QR_{\lambda}^{0}\bigr)^{2} \bigr] \bigr||\,d\lambda| \\ \leq&{\frac{1}{4\pi}}\int_{|\lambda|=b_{p}}\bigl\| \bigl(QR_{\lambda}^{0}\bigr)^{2}\bigr\| _{\sigma_{1}(H_{1})}|\,d\lambda| \\ \leq& {\frac{1}{4\pi}}\int_{|\lambda|=b_{p}}\bigl\| QR_{\lambda}^{0} \bigr\| \bigl\| QR_{\lambda}^{0} \bigr\| _{\sigma_{1}(H_{1})}|\,d\lambda| \\ \leq& {\frac{\|Q\|}{4\pi}}\int_{|\lambda|=b_{p}}\bigl\| R_{\lambda}^{0}\bigr\| \bigl\| QR_{\lambda}^{0} \bigr\| _{\sigma_{1}(H_{1})}|\,d\lambda| \\ < & c_{1}\int_{|\lambda|=b_{p}}p^{-5}\,d\lambda \\ = & 2\pi\cdot b_{p}\cdot c_{1}\cdot p^{-5}= 2 \pi c_{1}(p+1)^{4}p^{-5}< c_{2}p^{-1}. \end{aligned}$$

(3.16)

Here c is a positive constant.

By using formula (2.6), Lemma 3.1 and Lemma 3.2, we find

$$\begin{aligned} |{M_{p}}| = & {\frac{1}{2\pi}} \biggl|\int _{|\lambda|=b_{p}} \lambda \operatorname{tr} \bigl[R_{\lambda} \bigl(QR_{\lambda}^{0}\bigr)^{3} \bigr]\,d\lambda \biggr| \\ \leq& {\frac{b_{p}}{2\pi}}\int_{|\lambda|=b_{p}} \bigl\| R_{\lambda} \bigl(QR_{\lambda}^{0}\bigr)^{3} \bigr\| |\,d\lambda| \\ \leq&{\frac{b_{p}}{2\pi}}\int_{|\lambda|=b_{p}}\| R_{\lambda}\| \bigl\| \bigl(QR_{\lambda}^{0}\bigr)^{3}\bigr\| _{\sigma_{1}(H_{1})}|\,d\lambda| \\ \leq& {\frac{b_{p}}{2\pi}}\int_{|\lambda|=b_{p}}\| R_{\lambda}\| \bigl\| \bigl(QR_{\lambda}^{0}\bigr)^{2}\bigr\| \bigl\| \bigl(QR_{\lambda}^{0}\bigr)\bigr\| _{\sigma_{1}(H_{1})}|\,d\lambda| \\ \leq& {\frac{b_{p}}{2\pi}}\int_{|\lambda|=b_{p}}\| R_{\lambda}\| \| Q\|^{2}\bigl\| R_{\lambda}^{0}\bigr\| ^{2}\bigl\| QR_{\lambda}^{0}\bigr\| _{\sigma_{1}(H_{1})}|\,d\lambda| \\ < & c_{3}\cdot b_{p}^{2}p^{-11} \\ = & c_{3}(p+1)^{8}p^{-11}< c_{4}p^{-3}. \end{aligned}$$

(3.17)

From (3.16) and (3.17) we get

$$ \lim_{p\rightarrow\infty} {M}_{p2}=\lim_{p\rightarrow \infty} {M}_{p}=0. $$

(3.18)

From (2.4) and (3.5) we obtain

$$\begin{aligned} &\sum_{m=0}^{p} \Biggl\{ \sum _{n=1}^{\infty} \biggl[\lambda _{mn}- \biggl(m+\frac{1}{2}\biggr)^{4} \biggr]-\frac{1}{\pi}\int _{0}^{\pi} \operatorname{tr} Q(x)\,dx \Biggr\} \\ &\quad=-\frac{1}{\pi}\sum_{m=0}^{p} \int_{0}^{\pi}\operatorname{tr} Q(x)\cos (2m+1)x\,dx+M_{p2}+M_{p}. \end{aligned}$$

(3.19)

From (3.18) and (3.19) we find

$$\begin{aligned} &\sum_{m=0}^{\infty} \Biggl\{ \sum _{n=1}^{\infty} \biggl[\lambda_{mn}- \biggl(m+\frac{1}{2}\biggr)^{4} \biggr]-\frac{1}{\pi}\int _{0}^{\pi} \operatorname{tr} Q(x)\,dx \Biggr\} \\ &\quad=-\frac{1}{\pi}\sum_{m=0}^{\infty}\int _{0}^{\pi}\operatorname{tr} Q(x) \cos(2m+1)x\,dx. \end{aligned}$$

(3.20)

Moreover, using conditions (Q1) and (Q4), we get

$$\begin{aligned} &{-}\frac{1}{\pi}\sum_{m=0}^{\infty} \int_{0}^{\pi}\operatorname{tr} Q(x)\cos(2m+1)x\,dx \\ &\quad=-\frac{1}{2\pi}\sum_{m=1}^{\infty} \biggl[\int_{0}^{\pi}\operatorname{tr} Q(x)\cos mx \,dx-(-1)^{m}\int_{0}^{\pi} \operatorname{tr} Q(x)\cos mx \,dx \biggr] \\ &\quad=-\frac{1}{4} \Biggl\{ \sum_{m=1}^{\infty} \biggl[\frac{2}{\pi}\int_{0}^{\pi} \operatorname{tr} Q(x)\cos mx\,dx \biggr]\cos m0+ \biggl[\frac{1}{\pi}\int _{0}^{\pi }\operatorname{tr}Q(x)\,dx \biggr]\cos0 \Biggr\} \\ &\qquad{}+\frac{1}{4} \Biggl\{ \sum_{m=1}^{\infty} \biggl[\frac{2}{\pi}\int_{0}^{\pi} \operatorname{tr} Q(x)\cos mx\,dx \biggr]\cos m \pi+ \biggl[\frac{1}{\pi}\int _{0}^{\pi}\operatorname{tr} Q(x)\,dx \biggr]\cos0\pi \Biggr\} \\ &\quad=\frac{1}{4}\bigl[\operatorname{tr} Q(\pi)-\operatorname{tr} Q(0) \bigr]. \end{aligned}$$

(3.21)

From (3.20) and (3.21) we find

$$\sum_{m=0}^{\infty} \Biggl\{ \sum _{n=1}^{\infty} \biggl[\lambda_{mn}-\biggl(m+ \frac{1}{2}\biggr)^{4} \biggr]-\frac{1}{\pi}\int _{0}^{\pi}\operatorname{tr}Q(x)\,dx \Biggr\} = \frac{1}{4}\bigl[\operatorname{tr} Q(\pi)-\operatorname{tr} Q(0)\bigr]. $$

The theorem is proved. □