Open Access

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

Journal of Inequalities and Applications20152015:316

https://doi.org/10.1186/s13660-015-0823-0

Received: 20 April 2015

Accepted: 15 September 2015

Published: 6 October 2015

Abstract

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

Keywords

separable-Hilbert spaceself-adjoint operatorkernel operatorspectrumessential spectrumresolvent

MSC

47A1034L20

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 [813] 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:
  1. (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 \(x\in[0,\pi] \).

     
  2. (Q2)

    \(\|Q\|<\frac{5}{2}\).

     
  3. (Q3)
    There is an orthonormal basis \(\{\varphi_{n}\}_{n=1}^{\infty}\) of the space H such that
    $$\sum_{n=1}^{\infty}\bigl\| Q(x)\varphi_{n} \bigr\| < \infty. $$
     
  4. (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:
  1. (a)

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

     
  2. (b)
    Spectrum of the operator L is a subset of the union of pairwise disjoint intervals
    $$F_{m}=\biggl[\biggl(m+\frac{1}{2}\biggr)^{4}-\|Q\|, \biggl(m+\frac{1}{2}\biggr)^{4}+\|Q\| \biggr] \quad(m=0,1,2, \ldots), \sigma(L)\subset\bigcup_{m=0}^{\infty}F_{m}. $$
     
  3. (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.

     
  4. (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. □

Declarations

Acknowledgements

The authors would like to thank Professor Ehliman Adıgüzelov for his expert assistance and for contributions in detailed editing of the last version of this manuscript.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
Faculty of Arts and Science, Yildiz Technical University

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© Karayel and Sezer 2015