\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 31, pp. 1--12.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2016/31\hfil 
 Trace formula for an even-order differential operator]
{Regularized trace formula for higher order differential operators with
unbounded coefficients}

\author[E. \c{S}en, A. Bayramov, K. Oru\c{c}o\u{g}lu \hfil EJDE-2016/31\hfilneg]
{Erdo\u{g}an \c{S}en, Azad Bayramov, Kamil Oru\c{c}o\u{g}lu}

\address{Erdo\u{g}an \c{S}en \newline
Department of Mathematics,
Faculty of Arts and Science,
Nam\i kKemal University,
59030, Tekirda\u{g}, Turkey}
\email{erdogan.math@gmail.com}

\address{Azad Bayramov \newline
Department of Mathematics Education,
Faculty of Education,
Recep Tayyip Erdogan University,
 Rize, Turkey}
\email{azadbay@gmail.com}


\address{Kamil Oru\c{c}o\u{g}lu \newline
Department of Mathematics Engineering,
Istanbul Technical University,
 Maslak, 34469, Istanbul, Turkey}
\email{koruc@itu.edu.tr}

\thanks{Submitted April 1, 2015. Published January 20, 2016.}
\subjclass[2010]{47A10, 47A55}
\keywords{Hilbert space; self-adjoint operator; spectrum; trace class operator; 
\hfill\break\indent regularized trace}

\begin{abstract}
 In this work we obtain the regularized trace formula for an
 even-order differential operator with unbounded operator coefficient.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\allowdisplaybreaks

\section{Introduction}

The first work about the theory of regularized traces of differential
operators belongs to Gelfand and Levitan \cite{bib1}. They considered the
Sturm-Liouville operator
\[
-y''+[ q(x)-\lambda ] y=0,
\]
with boundary conditions
\[
y'(0)=y'(\pi) =0,
\]
where $q(x)\in C^{1}[ 0,\pi ] $. Under the condition
$\int_0^{\pi }q(x) dx=0$ they obtained the formula
\[
\sum_{n=0}^{\infty }(\mu _n-\lambda _n) =\frac{1}{4}(
q(0) +q(\pi)) .
\]

Gul \cite{bib2} obtained the formula
\[
\lim_{m\to \infty }\sum_{k=1}^{n_{m}}(\lambda _{k}-\mu
_{k}) =\frac{1}{4}[ \operatorname{tr}Q(\pi )-\operatorname{tr}Q(0) ]
\]
for the regularized trace of the second order differential operator
\[
l[ y] =-y''(x) +Ay(x)+Q(x)y(x)
\]
with unbounded operator coefficient and with the boundary conditions
$y(0) =y'(\pi) =0$. Here $\lambda_{k}$ and $\mu_{k}$ are the eigen-elements
of the operators
\begin{gather*}
l_0[ y]  =-y''(x) +Ay(x) \\
l[ y]  =-y''(x) +Ay(x)+Q(x)y(x)
\end{gather*}
with the same boundary conditions $y(0) =y'(\pi) =0$ respectively.

 Ad\i g\"{u}zelov and Sezer \cite{bib3} obtained a regularized
trace formula for a self-adjoint differential operator of higher order with
unbounded operator coefficient. Articles \cite{bib4, bib5, bib6, bib7,
bib8} are devoted to study of regularized trace formulas of differential
operators with bounded operator coefficient. 
Bayramov et al \cite{bib9} obtained the second regularized trace formula for the
differential operator equation with the semi-periodic boundary conditions.
Makin \cite{bib10} established a formula for the first regularized trace of
the Sturm-Liouville equation with a complex-valued potential and with
irregular boundary conditions.

Let $H$ be a separable Hilbert space and let $H_1=L_2(H;[0,\pi]) $
denote the set of all strongly measurable functions $
f$ with values in $H$ and such that
\[
\int_0^{\pi}\| f(t) \| _H^2dt<\infty.
\]
and the scalar function $(f(t),g) $ is Lebesgue measurable for
every $g\in H$ in the interval $[ 0,\pi] $. Here $(\cdot,\cdot) $ denotes
the inner product in $H$ and $\|\cdot\| $ denotes the norm in $H$.

If the inner product of two arbitrary elements $f$ and $g$ of the space
 $H_1$ is defined by
\[
(f,g) _{H_1}=\int_0^{\pi }(f(t),g(t)) _Hdt,\quad f,g\in H_1
\]
then $H_1$ becomes a separable Hilbert space \cite{bib11}.
$\sigma_{\infty }(H) $ denotes the set of all compact operators from
$H$ into $H$. If $A\in \sigma _{\infty }(H) $, then $AA^{\ast }$ is
a nonnegative self-adjoint operator and
$(A^{\ast }A) ^{1/2}\in \sigma _{\infty }(H)$. Let the non-zero eigen-elements
 of the operator $(A^{\ast }A) ^{1/2}$ be $s_1\geq s_2\geq \dots\geq s_q$
$(0\leq q\leq \infty)$. Here, each eigen-element is counted
according to its own multiplicity. The numbers $s_1,s_2,\dots,s_q$ are
called $s$-numbers of the the operator $A$. $\sigma _1(H) $ is
the set of all the operators $A\in \sigma _{\infty }(H) $ such
that the $s$-numbers of which satisfy the condition
$\sum_{q=1}^{\infty}s_q<\infty $. An operator is called a trace class
operator if it belongs to $\sigma _1(H) $.

Let us consider the operators $l_0$ and $l$ in $H_1$ defined by
\begin{gather}
l_0[ u]  =(-1) ^{m}u^{(2m)}(t) +Au(t), \label{e1} \\
l[ u]  =(-1) ^{m}u^{(2m)}(t)
+Au(t)+Q(t)u(t)  \label{e2}
\end{gather}
with the same boundary conditions 
$y^{(2i-2) }(0)=y^{(2i-1) }(\pi) =0$ $(i=1,2,\dots,m) $ respectively.
Here $A:\Omega (A) \to H$ is a densely defined
self-adjoint operator in $H$ with $A=A^{\ast }\geq E$ where 
$E:H\to H$ is identity operator and $A^{-1}\in \sigma _{\infty }(H) $. We
also should note that our problem's boundary conditions are different from
the considered problem's boundary conditions in \cite{bib3} which arise new
difficulties.

Let $\eta_1\leq\eta_2\leq\dots\leq\eta_n\leq\dots$ be the eigen-elements
of the operator $A$ and $\varphi_1,\varphi_2,\dots,\varphi_n,\dots$ be the
orthonormal eigenvectors corresponding to these eigen-elements. Here, each
eigenvalue is counted according to its own multiplicity number. Let $
\Omega(L_0') $ denote the set of the functions $u(t)$
of the space $H_1$ satisfying the following conditions:
\begin{itemize}
\item[(a)] $u(t)$ has continuous derivative of the $2m$ order with respect to the
norm in the space $H$ in the interval $[ 0,\pi] $;

\item[(b)] $u(t)\in\Omega(A) $ for every $t\in[ 0,\pi] $ and
$Au(t)$ is continuous with respect to the norm in the space $H$.

\item[(c)] $y^{(2i-2) }(0)=y^{(2i-1) }(\pi) =0$
$(i=1,2,\dots,m) $.
\end{itemize}
Here $\overline{\Omega (L_0') }=H_1$. Let us
consider the linear operator $L_0'u=l_0u$ from $D(L_0)$ to $H_1$. 
$L_0'$ is a symmetric operator. The eigen-elements of $L_0'$ are
$(\frac{1}{2}+k) ^{2m}+\eta _j$ $(k=0,1,2,\dots;j=1,2,\dots) $.
and the orthonormal eigenvectors corresponding to these eigen-elements are
\[
\sqrt{\frac{2}{\pi }}\sin \big((\frac{1}{2}+k) t\big)
\varphi _j(k=0,1,2,\dots ;j=1,2,\dots) .
\]

\section{Some relations about the eigen-elements and resolvents}

Let the eigenvalues of the operators $L_0$ and $L$ be 
$\mu _1\leq \mu_2\leq \dots\leq \mu _n\leq \dots$
 and $\lambda _1\leq \lambda _2\leq \dots\leq \lambda _n\leq \dots$ respectively.
Let $N(\mu )$ be the number of
eigen-elements of operator $L_0$ which is not greater than a positive
number $\mu $. If $\eta _j\sim aj^{\alpha }$ as $j\to \infty $
$(a>0,\alpha >\frac{2m}{2m-1}) $ that is, if
\[
\lim_{j\to \infty }\frac{\eta _j}{aj^{\alpha }}=1
\]
then using the same method in \cite{bib12} it can be found that
$N(\mu) \sim d\mu ^{\frac{2m+\alpha }{2m\alpha }}$, where
\[
d=\frac{2}{\alpha a^{1/a}}\int_0^{\pi /2}(\sin \tau) ^{\frac{2
}{\alpha }-1}(\cos \tau) ^{1+\frac{1}{m}}d\tau
\]
and hence
\begin{equation}
\mu _n\sim d_0n^{\frac{2m\alpha }{2m+\alpha }}\quad \text{as }
j\to \infty \quad (d_0=d^{\frac{2m\alpha }{2m+\alpha }
}) .  \label{e3}
\end{equation}
Let $Q(t)$ be an operator function satisfying the following conditions:
\begin{enumerate}
\item $Q(t):H\to H$ is a self-adjoint operator for every $t\in[
0,\pi] $;

\item $Q(t)$ is weakly measurable in the interval $[ 0,\pi] $;

\item The norm function $\| Q(t)\| $ is bounded in the
interval $[ 0,\pi] $;

\item $Q(t)$ has weak derivative of the second order in the interval $[
0,\pi] $;

\item The function $(Q'(t)f,g) $ is continuous for every $
f,g\in H$;

\item $Q^{(i)}(t):H\to H$ $(i=0,1,2)$ are self-adjoint trace class
operators and the functions $\| Q^{(i)}(t)\| _{\sigma
_1(H)}$ $(i=0,1,2) $ are bounded and measurable in the
interval $[ 0,\pi ] $.
\end{enumerate}
Since $Q$ is a self-adjoint operator from $H_1$ to $H_1$ for every 
$y\in H_1$ we have
\[
| (Qy,y) _{H_1}|
\leq \|Qy\| _{H_1}\| y\| _{H_1}
\leq \|Q\| _{H_1}\| y\| _{H_1}^2
\]
or
\[
(-\| Q\| y,y) _{H_1}\leq (Qy,y)_{H_1}\leq (\| Q\| y,y) _{H_1}.
\]
This means that
\[
-\| Q\| _{H_1}E\leq Q\leq \| Q\|_{H_1}E.
\]
And so
\[
L_0-\| Q\| _{H_1}E\leq L=L_0+Q\leq L_0+\|Q\| _{H_1}E.
\]
In this situation, it is well-known that (Smirnov, \cite{bib13})
\[
\mu _n-\| Q\| _{H_1}\leq \lambda _n\leq \mu_n+\| Q\| _{H_1}\,.
\]
According to this, we can write
\[
1-\frac{\| Q\| _{H_1}}{\mu _n}\leq \frac{\lambda _n}{
\mu _n}\leq 1+\frac{\| Q\| _{H_1}}{\mu _n}.
\]
By applying limit to each side of this inequality and by considering the
equality
\[
\lim_{n\to \infty }\frac{\mu _n}{d_0n^{\frac{2m\alpha }{
2m+\alpha }}},
\]
we get $\lim_{n\to \infty }\lambda _n/ \mu _n=1$. Thus we
have
\[
\lim_{n\to \infty }\frac{\lambda _n}{d_0n^{\frac{2m\alpha }{
2m+\alpha }}}=\lim_{n\to \infty }\frac{\lambda _n}{\mu _n}\frac{
\mu _n}{d_0n^{\frac{2m\alpha }{2m+\alpha }}}=\lim_{n\to \infty }
\frac{\lambda _n}{\mu _n}\lim_{n\to \infty }\frac{\mu _n}{
d_0n^{\frac{2m\alpha }{2m+\alpha }}}=1
\]
or as $n\to \infty $,
\begin{equation}
\lambda _n\sim d_0n^{\frac{2m\alpha }{2m+\alpha }}.  \label{e4}
\end{equation}

By using the formula \eqref{e3}, it is easily seen that the sequence 
$\{ \mu_n\} $ has a subsequence $\mu _{n_1}<\mu _{n_2}<\dots<\mu_{n_p}<\dots$
such that
\begin{equation}
\mu _q-\mu _{n_p}>d_0\big(q^{\frac{2m\alpha }{2m+\alpha }}-n_p^{
\frac{2m\alpha }{2m+\alpha }}\big) \quad \big(
q=n_p+1,n_p+2,\dots;d_0=d^{\frac{2m\alpha }{2m+\alpha }}\big) .
\label{e5}
\end{equation}

Let $R_{\lambda }^0=(L_0-\lambda E) ^{-1}$,
$R_{\lambda}=(L-\lambda E) ^{-1}$ be the resolvents of the operators $L_0$
and $L$ respectively. If $\alpha >\frac{2m}{2m-1}$ then, by the formulas 
\eqref{e3} and \eqref{e4}, $R_{\lambda }^0$ and\ $R_{\lambda }$
are trace class operators for $\lambda \neq \lambda _q,\mu _q$ $(q=1,2,\dots) $.
 In this situation
\begin{equation}
\operatorname{tr}(R_{\lambda }-R_{\lambda }^0) =trR_{\lambda }-trR_{\lambda
}^0=\sum_{q=1}^{\infty }(\frac{1}{\lambda _q-\lambda }-\frac{1}{
\mu _q-\lambda }) ,  \label{e6}
\end{equation}
see Cohberg and Krein \cite{bib14}.

Let  $| \lambda | =d_p=2^{-1}(\mu _{n_p+1}+\mu_{n_p}) $.
It is easy to see that for large values of $p$ the
inequalities $\mu _{n_p}<d_p<\mu _{n_p+1}$ and
$\lambda_{n_p}<d_p<\lambda _{n_p+1}$ are satisfied. The series 
$\sum_{q=1}^{\infty }\frac{\lambda }{\lambda _q-\lambda }$ and
$\sum_{q=1}^{\infty }\frac{\lambda }{\mu _q-\lambda }$ are uniform
convergent on the circle $| \lambda | =d_p$. Hence with
the help of inequality \eqref{e5}, we obtain
\begin{equation}
\sum_{q=1}^{n_p}(\lambda _q-\mu _q) =-\frac{1}{2\pi i}
\int_{| \lambda | =d_p}\lambda \operatorname{tr}(R_{\lambda
}-R_{\lambda }^0) d\lambda ,  \label{e7}
\end{equation}
where $i^2=-1$.

\begin{lemma} \label{lem2.1}
If $\ \eta_j\sim aj^{\alpha}$ as $j\to\infty$
$(a>0,\alpha>\frac{2m}{2m-1}) $ then 
$\| R_{\lambda}^0\|_{\sigma_1(H_1)}<2d_0^{-1}\frac{(2\delta +1) }{\delta
n_p^{\delta-1}}$ $(\delta=\frac{2m\alpha }{2m+\alpha}-1) $ on
the circle $| \lambda| =d_p$.
\end{lemma}

\begin{proof}
For $\lambda \notin \{ \mu _q\} _{q=1}^{\infty }$, since
$R_{\lambda }^0$ is a normal operator we have
$\| R_{\lambda}^0\| _{\sigma _1(H_1)}=\sum_{q=1}^{\infty }\frac{1}{
| \mu _q-\lambda | }$ \cite{bib14}. On the circle
$| \lambda | =d_p$ we have
\begin{equation} \label{e8}
\begin{aligned}
&\| R_{\lambda }^0\| _{\sigma _1(H_1)}\\
& \leq \sum_{q=1}^{\infty }\frac{1}{| | \lambda | -\mu
_q| }=\sum_{q=1}^{n_p}\frac{2}{\mu _{n_p}+\mu
_{n_p+1}-2\mu _q}+\sum_{q=n_p+1}^{\infty }\frac{2}{2\mu _q-\mu
_{n_p}-\mu _{n_p+1}}   \\
& \leq \sum_{q=1}^{n_p}\frac{2}{\mu _{n_p+1}-\mu _q}
+\sum_{q=n_p+1}^{\infty }\frac{2}{2\mu _q-\mu _{n_p}}
=\sum_{q=1}^{n_p}\frac{2}{\mu _{n_p+1}-\mu _q}+2D_p,
\end{aligned}
\end{equation}
where $D_p=\sum_{k=n_p+1}^{\infty }(\mu _{k}-\mu _{n_p})^{-1}$ 
$(p=1,2,\dots) $. By using the inequality \eqref{e5} we obtain
\begin{equation} \label{e9}
\begin{aligned}
\sum_{q=1}^{n_p}\frac{2}{\mu _{n_p+1}-\mu _q}
&<\frac{n_p}{\mu _{n_p+1}-\mu _{n_p}}
 <\frac{n_p}{d_0((n_p+1)^{1+\delta }-n_p^{1+\delta }) }\\
&<\frac{n_p}{d_0(n_p+1) ^{\delta }}<\frac{n_p^{1-\delta }}{d_0},
\end{aligned}
\end{equation}
\begin{equation}
\begin{aligned}
D_p& =\sum_{k=n_p+1}^{\infty }(\mu _{k}-\mu _{n_p}) ^{-1}<
\frac{1}{d_0}\sum_{k=n_p+1}^{\infty }\frac{1}{k^{1+\delta
}-n_p^{1+\delta }}   \\
& =d_0^{-1}[ \frac{1}{((n_p+1) ^{1+\delta
}-n_p^{1+\delta }) }+\sum_{k=n_p+2}^{\infty }\frac{1}{k^{1+\delta
}-n_p^{1+\delta }}] . 
\end{aligned} \label{e10}
\end{equation}
It is easy to see that
\begin{gather*}
\sum_{k=n_p+2}^{\infty }\frac{1}{k^{1+\delta }-n_p^{1+\delta }}
\leq\int_{n_p+1}^{\infty }\frac{dt}{t^{1+\delta }-n_p^{1+\delta }},
\\
\int_{n_p+1}^{\infty }\frac{dt}{t^{1+\delta }-n_p^{1+\delta }}
<\frac{1}{\delta [ ((n_p+1) ^{1+\delta }-n_p^{1+\delta
}) ] ^{\frac{\delta }{1+\delta }}}.
\end{gather*}
Taking into account the inequality \eqref{e10} we obtain 
\begin{equation}
D_p<d_0^{-1}\frac{\delta +1}{\delta [ ((n_p+1)
^{1+\delta }-n_p^{1+\delta }) ] ^{\frac{\delta }{1+\delta }}}
<d_0^{-1}\frac{\delta +1}{\delta n_p^{\frac{\delta ^2}{1+\delta }}}.
\label{e11}
\end{equation}
With the help of \eqref{e8}, \eqref{e9} and \eqref{e11}, it follows that
on the circle  $| \lambda | =d_p$,
\[
\| R_{\lambda }^0\| _{\sigma _1(H_1)}
<2d_0^{-1} \frac{(2\delta +1) }{\delta n_p^{\delta -1}}.
\]
\end{proof}

\begin{lemma} \label{lem2.2}
If the operator function $Q(t)$ satisfies the conditions 
{\rm(1)--(3)}, and $\eta _j\sim aj^{\alpha}$ as $j\to\infty$
$(a>0,\alpha>\frac {2m}{2m-1}) $, then for $| \lambda| =d_p$ and large
values of $p$,
\[
\| R_{\lambda}\| _{H_1}<4d_0^{-1}n_p^{-\delta}.
\]
\end{lemma}

\begin{proof}
Since the $s$-numbers of the trace class operator $R_{\lambda }$ are
 $\{\frac{1}{\lambda _1-\lambda },\frac{1}{\lambda _2-\lambda },\dots,\\
\frac{1}{ \lambda _q-\lambda },\dots\} $, it follows that
\begin{equation}
\| R_{\lambda }\| _{H_1}=\max \{ \frac{1}{\lambda
_1-\lambda },\frac{1}{\lambda _2-\lambda },\dots,\frac{1}{\lambda
_q-\lambda },\dots\} .  \label{e12}
\end{equation}
On the circle $| \lambda | =d_p$,
\begin{equation}
\big| | \lambda _q| -| \lambda| \big|
=\big| | \lambda _q|-2^{-1}(\mu _{n_p}+\mu _{n_p+1}) \big|
=2^{-1}\big| \mu _{n_p}+\mu _{n_p+1}-2| \lambda
_q| \big| .  \label{e13}
\end{equation}
Using the inequality $q\leq n_p$ and for the large values of $p$, 
since $| \lambda _q| \leq \lambda _{n_p}$, we have
\begin{equation} \label{e14}
\begin{aligned}
&\mu _{n_p}+\mu _{n_p+1}-2| \lambda _q| \\
& \geq \mu _{n_p}+\mu _{n_p+1}-2\lambda _{n_p}=\mu _{n_p+1}-\mu
_{n_p}+2(\mu _{n_p}-\lambda _{n_p})   \\
& \geq \mu _{n_p+1}-\mu _{n_p}-2| \mu _{n_p}-\lambda_{n_p}| .  
\end{aligned}
\end{equation}
Considering $| \mu _q-\lambda _q| \leq \|Q\| _{H_1}$ $(q=1,2,\dots) $
 by \eqref{e14} we obtain
\begin{equation}
\mu _{n_p}+\mu _{n_p+1}-2| \lambda _q| \geq \mu
_{n_p+1}-\mu _{n_p}-2\| Q\| _{H_1}\text{ \ }(
q\leq n_p) .  \label{e15}
\end{equation}
With the help of inequality $q\geq n_p+1$ and for the large values of $p$,
since $| \lambda _q| =\lambda _q\geq \lambda
_{n_p+1}$ then
\begin{align*}
2| \lambda _q| -\mu _{n_p}-\mu _{n_p+1}
& \geq 2\lambda _{n_p+1}-\mu _{n_p}-\mu _{n_p+1}\\
&=2(\lambda_{n_p+1}-\mu _{n_p+1}) +\mu _{n_p+1}-\mu _{n_p} \\
& \geq \mu _{n_p+1}-\mu _{n_p}-2| \lambda _{n_p+1}-\mu
_{n_p+1}| .
\end{align*}
Using the above inequality,
\begin{equation}
2| \lambda _q| -\mu _{n_p}-\mu _{n_p+1}
\geq \mu _{n_p+1}-\mu _{n_p}-2\| Q\| _{H_1}\quad (q\geq n_p+1) .  \label{e16}
\end{equation}
Taking into account that $\lim_{p\to \infty }(\mu
_{n_p+1}-\mu _{n_p}) =\infty $ and by \eqref{e13}, \eqref{e15} and \eqref{e16}, 
on the circle $| \lambda | =d_p$ we have
\begin{equation}
\big| | \lambda _q| -| \lambda | \big| >4^{-1}(\mu _{n_p+1}-\mu _{n_p}) .
\label{e17}
\end{equation}
By \eqref{e5} and \eqref{e17} we obtain
\[
\big| | \lambda _q| -| \lambda| \big|
>4^{-1}d_0((n_p+1) ^{1+\delta
}-n_p^{1+\delta }) >4^{-1}d_0(n_p+1) ^{\delta }.
\]
From the above inequality and $| \lambda | =d_p$ for the
sufficiently large values of $p$, we have
\[
| \lambda _q-\lambda | >4^{-1}d_0n_p^{\delta }.
\]
From \eqref{e12} and the above inequality we have
$4d_0^{-1}n_p^{-\delta }$.
\end{proof}


\section{Regularized trace formula}

We know from operator theory that for the resolvents of the operators 
$L_0$ and $L$ the following formula holds:
\[
R_{\lambda }=R_{\lambda }^0-R_{\lambda }QR_{\lambda }^0\quad
(\lambda \in \rho (L_0) \cap \rho (L)) .
\]
Using the above formula and \eqref{e7}, it can be easily shown that
\begin{equation}
\sum_{q=1}^{n_p}(\lambda _q-\mu _q)
=\sum_{j=1}^{s}U_{pj}+U_p^{(s) },  \label{e18}
\end{equation}
where
\begin{gather}
U_{pj}=\frac{(-1) ^{j}}{2\pi ij}\int_{| \lambda
| =d_p}tr[ (QR_{\lambda }^0) ^{j}]
d\lambda \text{ \ }(i^2=-1;j=1,2,\dots),  \label{e19}
\\
U_p^{(s) }=\frac{(-1) ^{s}}{2\pi i}
\int_{| \lambda | =d_p}\lambda tr[ R_{\lambda
}(QR_{\lambda }^0) ^{s+1}] d\lambda \text{ \ }(
i^2=-1) .  \label{e20}
\end{gather}

\begin{theorem} \label{thm3.1}
If the operator function $Q(t)$ satisfies the the conditions 
{\rm(1)--(3)} and $\eta_j\sim aj^{\alpha}$ as
$j\to\infty$ $(a>0,\alpha >\frac{2m(1+\sqrt{2}) }{2\sqrt{2}m-\sqrt{2}-1}) $ then
\[
\lim_{p\to\infty}U_{pj}=0\quad (j=2,3,4,\dots) .
\]
\end{theorem}

\begin{proof}
According to \eqref{e19} for $U_{p2}$ we have the equality
\begin{equation}
U_{p2}=\frac{1}{2\pi i}\sum_{j=1}^{n_p}\sum_{k=n_p+1}^{\infty }\Big[
\int_{| \lambda | =d_p}\frac{d\lambda }{(\lambda
-\mu _j) (\lambda -\mu _{k}) }\Big] (Q\psi
_j,\psi _{k}) _{H_1}(Q\psi _{k},\psi _j) _{H_1}.
\label{e21}
\end{equation}
Therefore, 
\begin{equation}
| U_{p2}| \leq \| Q\| _{H_1}^2D_p.\label{e22}
\end{equation}
By \eqref{e11} and \eqref{e22} we obtain
\begin{equation}
\lim_{p\to \infty }U_{p2}=0\quad (\alpha >\frac{2m}{2m-1}) .  \label{e23}
\end{equation}
Let us show that
\begin{equation}
\lim_{p\to \infty }U_{p3}=0.  \label{e24}
\end{equation}
By using  \eqref{e19} it follows that
\begin{equation}  \label{e25}
\begin{aligned}
U_{p3}& =\sum_{J=1}^{n_p}\sum_{k=1}^{n_p}\sum_{s=n_p+1}^{\infty }[
G(j,k,s) +G(s,k,j)+G(j,s,k]   \\
&\quad +\sum_{J=1}^{n_p}\sum_{k=n_p+1}^{\infty }\sum_{s=n_p+1}^{\infty }
[ G(j,k,s) +G(s,k,j)+G(k,j,s] ,
\end{aligned}
\end{equation}
where
\begin{gather*}
G(j,k,s) =g(j,k,s)(Q\psi _j,\psi _{k})
_{H_1}(Q\psi _{k},\psi _{s}) _{H_1}(Q\psi _{s},\psi
_j) _{H_1}, \\
g(j,k,s)  =\frac{1}{6\pi i}\int_{| \lambda |
=d_p}\frac{d\lambda }{(\lambda -\mu _j) (\lambda -\mu
_{k}) (\lambda -\mu _{s}) }.
\end{gather*}
Taking into account $g(j,k,s) =\overline{g(j,k,s) }$
and $Q=Q^{\ast }$ we obtain
\begin{equation}
G(s,k,j) =\overline{G(j,k,s) },\quad
G(k,j,s) =\overline{G(j,k,s) },\quad
G(j,s,k) =\overline{G(j,k,s) }.  \label{e26}
\end{equation}
With the help of \eqref{e25} and \eqref{e26} we obtain
\[
U_{p3}=E_1+E_2
\]
with
\begin{gather*}
E_1 =\sum_{J=1}^{n_p}\sum_{k=1}^{n_p}\sum_{s=n_p+1}^{\infty }(
G(j,k,s) +2\overline{G(j,k,s) }) , \\
E_2 =\sum_{J=1}^{n_p}\sum_{k=n_p+1}^{\infty }\sum_{s=n_p+1}^{\infty
}(G(j,k,s) +2\overline{G(j,k,s) })
\end{gather*}
and
\begin{equation}
E_1=E_{11}+\overline{E_{11}},\quad E_2=E_{21}+\overline{E_{21}},
\label{e27}
\end{equation}
with
\begin{gather*}
E_{11}=\sum_{J=1}^{n_p}\sum_{k=1}^{n_p}\sum_{s=n_p+1}^{\infty }G(j,k,s), \\
E_{21}=\sum_{J=1}^{n_p}\sum_{k=n_p+1}^{\infty }\sum_{s=n_p+1}^{\infty}G(j,k,s) .
\end{gather*}
It is not hard to see that the following inequalities hold:
\begin{gather}
| E_{11}|  \leq \frac{1+\delta }{d_0^2\delta }
\| Q\| _{H_1}^{3}n_p^{\frac{1-2\delta ^2}{1+\delta }},
\label{e28} \\
| E_{21}|  \leq (\frac{1+\delta }{d_0\delta }
) ^2\| Q\| _{H_1}^{3}n_p^{\frac{-2\delta ^2}{
1+\delta }}.  \label{e29}
\end{gather}
It follows that
\[
\lim_{p\to \infty }U_{p3}=0.
\]
Now, let us show that the equality $\lim_{p\to \infty }U_{pj}=0$
 $(j=4,5,\dots) $ holds. According to \eqref{e19},
\begin{equation}  \label{e30}
\begin{aligned}
| U_{pj}|
& \leq \frac{1}{2\pi j}\int_{| \lambda
| =d_p}| \operatorname{tr}(QR_{\lambda }^0)^{j}| | d\lambda | \\
&\leq \int_{|\lambda | =d_p}\| (QR_{\lambda }^0)
^{j}\| _{\sigma _1(H_1)}| d\lambda |
\\
& \leq \int_{| \lambda | =d_p}\| (
QR_{\lambda }^0) \| _{\sigma _1(H_1)}\| (
QR_{\lambda }^0) ^{j-1}\| _{H_1}| d\lambda|   \\
& \leq \int_{| \lambda | =d_p}\| Q\|
_{H_1}\| R_{\lambda }^0\| _{\sigma
_1(H_1)}\| (QR_{\lambda }^0) ^{j-1}\|
_{H_1}| d\lambda |   \\
& \leq \| Q\| _{H_1}\int_{| \lambda |
=d_p}\| R_{\lambda }^0\| _{\sigma
_1(H_1)}\| (QR_{\lambda }^0) \|
_{H_1}^{j-1}| d\lambda |   \\
& \leq const.\int_{| \lambda | =d_p}\|
R_{\lambda }^0\| _{\sigma _1(H_1)}\| R_{\lambda
}^0\| _{H_1}^{j-1}| d\lambda | .
\end{aligned}
\end{equation}
Since $R_{\lambda }=R_{\lambda }^0$ for $Q(t) \equiv 0$
according to Lemma \ref{lem2.2}, on the circle $| \lambda | =d_p$,
\begin{equation}
\| R_{\lambda }^0\| _{H_1}<4d_0^{-1}n_p^{-\delta }
\quad (\delta =\frac{2m\alpha }{2m+\alpha }-1) .  \label{e31}
\end{equation}
Using Lemma \ref{lem2.1} and the inequalities \eqref{e30} and \eqref{e31} one obtains
\[
| U_{pj}| <const.n_p^{1-\delta j}\int_{|
\lambda | =d_p}| d\lambda |
<\mathrm{const.}n_p^{1-\delta j}d_p.
\]
For the sufficiently large values of $p$, since $d_p=2^{-1}(\mu
_{n_p}+\mu _{n_p+1}) \leq \mathrm{const.}n_p^{1+\delta }$, then we obtain
\[
| U_{pj}| <\mathrm{const.}n_p^{2-\delta (j-1) }.
\]
It is easy to see that if $\delta >\frac{2}{3}$ or $\alpha >\frac{10m}{6m-5}$,
 then
\begin{equation}
\lim_{p\to \infty }U_{pj}=0\quad (j=4,5,\dots) .  \label{e32}
\end{equation}
However, if
\[
\frac{2m(1+\sqrt{2}) }{2\sqrt{2}m-\sqrt{2}-1}>\frac{10m}{6m-5},
\]
considering  \eqref{e23} and \eqref{e24} as
$\alpha >\frac{2m(1+ \sqrt{2}) }{2\sqrt{2}m-\sqrt{2}-1}$ one obtains
$\lim_{p\to \infty }U_{pj}=0$ $(j=2,3,\dots)$.
\end{proof}

Since the eigen-elements of the operator $L_0$ are
\[
\big(k+\frac{1}{2}\big) ^{2m}+\eta _j\quad (k=0,1,2,\dots;j=1,2,\dots) ,
\]
then for $q=1,2,\dots$,
\begin{equation}
\mu _q=\big(k_q+\frac{1}{2}\big) ^{2m}+\eta _{j_q}.  \label{e34}
\end{equation}

\begin{theorem} \label{thm3.2}
If the operator function $Q(t)$ satisfies the conditions {\rm(4)--(6)}and 
 $\eta_j\sim aj^{\alpha}$ as $j\to\infty$ $(a>0,\alpha >\frac{
2m(1+\sqrt{2}) }{2\sqrt{2}m-\sqrt{2}-1}) $ then 
\[
\lim_{p\to\infty}\sum_{q=1}^{n_p}\Big[ \lambda_q-\mu_q-
\pi^{-1}\int_0^{\pi}(Q(t)\varphi_{j_q},\varphi_{j_q}) dt
\Big] =4^{-1}[ \operatorname{tr}Q(\pi) -\operatorname{tr}Q(0) ]
\]
where $\{ j_q\} _{q=1}^{\infty}$ is a set of natural numbers
satisfying \eqref{e34}.
\end{theorem}

\begin{proof}
From \eqref{e19},
\begin{equation}
U_{p1}=-\frac{1}{2\pi i}\int_{| \lambda | =d_p}\operatorname{tr}(
QR_{\lambda }^0) d\lambda .  \label{e35}
\end{equation}
Since $QR_{\lambda }^0$ is a trace class operator for each 
$\lambda \in \rho (L_0) $ and 
$\{ \Psi _1(t) ,\Psi_2(t) ,\dots\} $ is an orthonormal basis of the space 
$H_1$, then
\[
\operatorname{tr}(QR_{\lambda }^0) =\sum_{q=1}^{\infty }(QR_{\lambda
}^0\Psi _q,\Psi _q) _{H_1}.
\]
By putting $\operatorname{tr}(QR_{\lambda }^0) $ into \eqref{e35} and
considering
\[
R_{\lambda }^0\Psi _q=(L_0-\lambda E) ^{-1}\Psi
_q=(\mu _q-\lambda) ^{-1}\Psi _q,
\]
one obtains
\begin{equation} \label{e36}
\begin{aligned}
U_{p1}& =-\frac{1}{2\pi i}\int_{| \lambda | =d_p}[
\sum_{q=1}^{\infty }(QR_{\lambda }^0\Psi _q,\Psi _q)
_{H_1}] d\lambda   \\
& =-\frac{1}{2\pi i}\int_{| \lambda | =d_p}[
\sum_{q=1}^{\infty }(\mu _q-\lambda) ^{-1}(Q\Psi
_q,\Psi _q) _{H_1}] d\lambda   \\
& =[ \sum_{q=1}^{\infty }(Q\Psi _q,\Psi _q) _{H_1}
] \frac{1}{2\pi i}\int_{| \lambda | =d_p}(
\lambda -\mu _q) ^{-1}d\lambda .
\end{aligned}
\end{equation}
Since the orthonormal eigenvectors according to the eigen-elements
$(k+\frac{1}{2}) ^{2m}+\eta _j$ $(k=0,1,2,\dots;j=1,2,\dots) $
of the operator $L_0$ are $\sqrt{\frac{2}{\pi }}\sin ((k+\frac{
1}{2}) t) \varphi _j$ ($k=0,1,2,\dots$; $j=1,2,\dots$), it follows that
\begin{equation}
\Psi _q(t) =\sqrt{\frac{2}{\pi }}\sin \big((k+\frac{1}{
2}) t\big) \varphi _{j_q}\quad q=1,2,\dots.  \label{e37}
\end{equation}
Further,
\begin{equation}
\frac{1}{2\pi i}\int_{| \lambda | =d_p}(\lambda-\mu _q) ^{-1}d\lambda
=\begin{cases}
1, & q\leq n_p, \\
0, & q>n_p
\end{cases} \label{e38}
\end{equation}
and by using \eqref{e36}--\eqref{e38} we find that
\begin{align*}
U_{p1}
& =\sum_{q=1}^{n_p}(Q\Psi _q,\Psi _q)_{H_1}
 =\sum_{q=1}^{n_p}\int_0^{\pi }(Q(t) \Psi_q(t) ,\Psi _q(t)) \\
& =\sum_{q=1}^{n_p}\int_0^{\pi }\Big(Q(t)
\sqrt{\frac{2}{ \pi }} \sin \big((k_q+\frac{1}{2}) t\big) \varphi _{j_q},
\sqrt{\frac{2}{\pi }} \sin \big((k_q+\frac{1}{2}) t\big)
\varphi _{j_q}\Big) dt \\
& =\frac{2}{\pi }\sum_{q=1}^{n_p}\int_0^{\pi }\sin ^2\big((
k_q+\frac{1}{2}) t\big) (Q(t)\varphi _{j_q},\varphi_{j_q}) dt \\
& =\frac{1}{\pi }\sum_{q=1}^{n_p}\int_0^{\pi }\big(1-\cos \big(
(2k_q+1) t\big)\big) \big(Q(t)\varphi _{j_q},\varphi
_{j_q}\big) dt
\\
&=\frac{1}{\pi }\sum_{q=1}^{n_p}\int_0^{\pi }(Q(t)\varphi
_{j_q},\varphi _{j_q}) dt-\frac{1}{\pi }\sum_{q=1}^{n_p}
\int_0^{\pi }\cos ((2k_q+1) t) (
Q(t)\varphi _{j_q},\varphi _{j_q}) dt.
\end{align*}
By subtracting and adding the expression
$(Q(t)\varphi_{j_q},\varphi _{j_q}) \cos (2k_qt) $ into the
second integral on the right side of last equality one obtains
\begin{align*}
U_{p1}
& =\frac{1}{\pi }\sum_{q=1}^{n_p}\int_0^{\pi }(Q(t)\varphi _{j_q},
\varphi _{j_q}) dt+\frac{1}{\pi }\sum_{q=1}^{n_p}
\int_0^{\pi }\cos (2k_qt) (Q(t)\varphi _{j_q},\varphi _{j_q}) dt \\
& -\frac{1}{\pi }\sum_{q=1}^{n_p}\int_0^{\pi }[ \cos ((
2k_q+1) t) +\cos (2k_qt) ] (
Q(t)\varphi _{j_q},\varphi _{j_q}) dt
\end{align*}
We  can write the expresssion
\[
-\frac{1}{\pi }\sum_{q=1}^{n_p}\int_0^{\pi }\cos (r_qt)
(Q(t)\varphi _{j_q},\varphi _{j_q}) dt,
\]
instead of first term in the right-hand side of the above  equality.
 Thus, we have
\begin{align*}
U_{p1}
& =\frac{1}{\pi }\sum_{q=1}^{n_p}\int_0^{\pi }(Q(t)\varphi
_{j_q},\varphi _{j_q}) dt+\frac{1}{\pi }\sum_{q=1}^{n_p}
\int_0^{\pi }\cos (2k_qt) (Q(t)\varphi
_{j_q},\varphi _{j_q}) dt \\
&\quad  -\frac{1}{\pi }\sum_{q=1}^{n_p}\int_0^{\pi }\cos (r_qt)
(Q(t)\varphi _{j_q},\varphi _{j_q}) dt.
\end{align*}
We can write this equation in the form
\begin{align*}
\lim_{p\to \infty }U_{p1}
& =\frac{1}{\pi }\sum_{j=1}^{\infty
}\int_0^{\pi }(Q(t)\varphi _j,\varphi _j) dt-\frac{1}{\pi }
\sum_{j=1}^{\infty }\sum_{r=1}^{\infty }\int_0^{\pi }\cos (rt)
(Q(t)\varphi _j,\varphi _j) dt \\
&\quad +\frac{1}{2\pi }\sum_{j=1}^{\infty }\sum_{k=1}^{\infty }\Big[
\int_0^{\pi }\cos (kt) (Q(t)\varphi _j,\varphi _j) dt\\
&\quad +(-1) ^{k}\int_0^{\pi }\cos (kt)
(Q(t)\varphi _j,\varphi _j) dt\Big]
\end{align*}
and so we have
\begin{equation} \label{e39}
\begin{aligned}
\lim_{p\to \infty }U_{p1}
& =\frac{1}{\pi }\sum_{j=1}^{\infty
}\int_0^{\pi }(Q(t)\varphi _j,\varphi _j) dt\\
&\quad -\frac{1}{2}
\sum_{j=1}^{\infty }\Big\{ \sum_{r=1}^{\infty }\Big[ \frac{2}{\pi }
\int_0^{\pi }(Q(t)\varphi _j,\varphi _j) \cos rt\,dt\Big]
\cos r0\Big\}   \\
&\quad +\frac{1}{4}\sum_{j=1}^{\infty }\Big\{ \sum_{k=1}^{\infty }\Big[ \frac{2
}{\pi }\int_0^{\pi }(Q(t)\varphi _j,\varphi _j) \cos ktdt
\Big] \cos k0   \\
& \quad  +\sum_{k=1}^{\infty }\Big[ \frac{2}{\pi }\int_0^{\pi }(
Q(t)\varphi _j,\varphi _j) \cos kt\,dt\Big] \cos k\pi \Big\} .
\end{aligned}
\end{equation}
The sum with respect to $r$ in the first term on the right side of this
expression in the value at $0$ of Fourier series according to functions
$\{ \cos rx\} _{r=0}^{\infty }$ in the interval
$[ 0,\pi] $ of the function $\int_0^{\pi }(Q(t)\varphi _j,\varphi_j) _H$
having the derivative of second order. Analogically, the
sums in the second term with respect to $k$ are the values at the points $0$
and $\pi $ respectively of Fourier series with respect to the functions
$\{ \cos kx\}_{k=0}^{\infty }$ in the same interval of that
function.

Also
\begin{equation}
\big| \sum_{j=1}^{p}(Q(t)\varphi_j,\varphi_j)\big| 
\leq\sum_{j=1}^{p}| (Q(t)\varphi_j,\varphi_j) | 
\leq\| Q(t) \|_{\sigma_1(H) },\quad (p=1,2,\dots) .  \label{e40}
\end{equation}
Since the norm function $\| Q(t) \|_{\sigma_1(H) }$ is bounded and measurable 
in the interval $[ 0,\pi] $, we have
\begin{equation}
\int_0^{\pi}\| Q(t) \| _{\sigma_1(H) }dt<\infty.  \label{e41}
\end{equation}

By using \eqref{e40}, \eqref{e41} and Lebesgue theorem we obtain
\begin{equation}
\sum_{j=1}^{\infty }\int_0^{\pi }(Q(t)\varphi _j,\varphi
_j) dt=\int_0^{\pi }\Big[ \sum_{j=1}^{\infty }(Q(t)\varphi
_j,\varphi _j) \Big] dt=\int_0^{\pi }\operatorname{tr}Q(t) dt.
\label{e42}
\end{equation}
Furthermore, as in the proof of \eqref{e32} by Lemma \ref{lem2.1} and
Lemma \ref{lem2.2}, we can show that 
\begin{equation}
\lim_{p\to \infty }U_p^{(s) }=0\quad (s>\frac{3}{\delta })\,.  \label{e43}
\end{equation}
Therefore by \eqref{e18}, \eqref{e39}, \eqref{e42} and \eqref{e43}, we obtain
\begin{equation}
\lim_{p\to \infty }\sum_{q=1}^{n_p}\Big[ \lambda _q-\mu
_q-\pi ^{-1}\int_0^{\pi }(Q(t)\varphi _{j_q},\varphi
_{j_q}) dt\Big] 
=4^{-1}[ \operatorname{tr}Q(\pi) -\operatorname{tr}Q(0) ] .  \label{e44}
\end{equation}
\end{proof}

The limit on the left hand side of the equality (44) is said to be
regularized trace of the operator $L$ (see \cite{bib1}).


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\end{document}