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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 210, pp. 1--8.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2012 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2012/210\hfil Spectrum of the direct sum of operators]
{Spectrum of the direct sum of operators}

\author[E. Otkun \c{C}evik, Z. I. Ismailov \hfil EJDE-2012/210\hfilneg]
{Elif Otkun \c{C}evik, Zameddin I. Ismailov}  % in alphabetical order

\address{Elif Otkun \c{C}evik \newline
Institute of Natural Sciences, 
Karadeniz Technical University, 
61080, Trabzon, Turkey}
\email{e\_otkuncevik@hotmail.com}

\address{Zameddin I. Ismailov \newline
Department of Mathematics, Faculty of Sciences,
Karadeniz Technical University, 
61080, Trabzon, Turkey}
\email{zameddin@yahoo.com}

\thanks{Submitted November 30, 2011. Published November 27, 2012.}
\subjclass[2000]{47A10}
\keywords{Direct sum of Hilbert spaces; spectrum; resolvent set;
\hfill\break\indent
compact operators; discrete spectrum; eigenvalues}

\begin{abstract}
 We study the connection between spectral properties
 of direct the sum of operators in the direct sum of Hilbert spaces
 and its coordinate operators.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks

\section{Introduction}

 It is known that infinite direct sum of Hilbert spaces $H_n$, $n\geq1$ and
infinite direct sum of operators $A_n$ in $H_n$, $n\geq1$ are defined as
 \[ 
 H=\oplus_{n=1}^{\infty } H_n
 =\Big\{u=(u_n):
 u_n \in H_n , {\rm \; }n\ge 1, \| u \| _{H}^2= \sum _{n=1}^{\infty }
 \| u_n \| _{H_n }^2 <+\infty \Big\},
\]
 and
\begin{gather*}
A=\oplus_{n=1}^{\infty } A_n, D(A)
=\{ u=(u_n)\in H:
 u_n \in D(A_n),{\rm \; }n\ge 1, {\rm \;} Au=(A_n u_n )
 \in H \}, \\
A:D(A)\subset H\to H
\end{gather*}
(see \cite{Dun}).


The general theory of linear closed operators in Hilbert spaces and its 
applications to physical problems has been investigated by many 
mathematicians (see for example \cite{Dun}).

 However, many physical problems of today arising in the modeling of processes of
multi-particle quantum mechanics, quantum field theory and in the physics of rigid bodies
support to study a theory of linear direct sum of operators in the direct sum of Hilbert spaces
(see \cite{Gan,Ism1,Koc,Tim,Ze} and references in it).

 In this paper, a connection between spectrum, resolvent sets, discreteness of the spectrum
(sec. 2) and asymptotical behavior of the eigenvalues (sec. 3) of direct sum of operators
defined in the direct sum of Hilbert spaces and suitable properties of coordinate operators
has been established. The obtained results has been supported with applications.

 These and related problems in the case continuous direct sum of the Hilbert space operators
have been investigated in works (\cite{Azo,Cho,Fia,Nai}).But in these works has not been considered a connection
between parts of the spectrum of direct sum operator and suitable parts of the spectrum their
coordinate operators.In this paper given sharp formulaes in the this sense.


\section{On the spectrum of direct sum of operators}

 In this section, the relationship between the spectrum and resolvent sets
of the direct sum of operators and its coordinate operators will be investigated.

	 First of all it will be investigated the continuity and compactness properties
of the operator $A=\oplus_{n=1}^{\infty } A_n$ in
$H=\oplus_{n=1}^{\infty } H_n$ in case when $ A_n\in L(H_n)$
for each $n\geq1$.

 It is easy to see that the following propositions are true in general.

\begin{theorem} \label{thm1} 
Let $A=\oplus_{n=1}^{\infty } A_n$,
 $H=\oplus_{n=1}^{\infty } H_n$ and for any $n\geq1$, $ A_n\in L(H_n)$.
 In order for $A\in L(H)$ the necessary and sufficient condition is
 $\underset{{n\geq1}}{sup}\|A_n\|<+\infty$.
\end{theorem}

In addition, in this case when $A\in L(H)$ it is true 
$\|A\|=\underset{{n\geq1}}{sup}\|A_n\|$ (see \cite{Nai}).

\begin{theorem} \label{thm2}
Let $A_n\in C_{\infty}(H_n)$ for each $n\geq1$. In this case
 $A=\oplus_{n=1}^{\infty } A_n\in C_{\infty}(H)$ if and only if
 $\lim_{n\to\infty}\|A_n\|=0$.
\end{theorem}

 Furthermore, the following main result can be proved.

\begin{theorem} \label{thm3} 
For the parts of spectrum and resolvent sets of the operator
 $A=\oplus_{n=1}^{\infty } A_n$ in Hilbert space
 $H=\oplus_{n=1}^{\infty } H_n$ the following statements are true
\begin{gather*}
 \sigma_{p}(A)=\cup_{n=1}^{\infty}\sigma_{p}(A_n),
 \\
\begin{aligned}
\sigma_{c}(A)&=\Big\{(\cup_{n=1}^{\infty}\sigma_{p}(A_n))^{c}
\cap (\cup_{n=1}^{\infty}\sigma_{r}(A_n))^{c}\cap
 (\cup_{n=1}^{\infty}\sigma_{c}(A_n))\Big\}\\
&\quad \cup
 \Big\{\lambda\in \cap_{n=1}^{\infty}\rho(A_n):\sup\|R_{\lambda}(A_n)\|
 =\infty\Big\},
\end{aligned}\\
\sigma_{r}(A)=\big(\cup_{n=1}^{\infty}\sigma_{p}(A_n)\big)^{c}\cap
\big(\cup_{n=1}^{\infty}\sigma_{r}(A_n)\big),
\\
\rho(A)=\big\{\lambda\in \cap_{n=1}^{\infty}\rho(A_n):
 \sup\|R_{\lambda}(A_n)\|<\infty\big\}.
\end{gather*}
\end{theorem}

\begin{proof} 
The validity of first claim of given relations is clear.
Moreover, it is easy to prove the fourth equality using the 
Theorem \ref{thm1}.

	 Now we prove the second relation on the continuous spectrum.
Let $\lambda\in \sigma_{c}(A)$. In this case by the definition of continuous 
spectrum $A-\lambda E$  is a one-to-one operator, $R(A-\lambda E)\neq H$ 
and $R(A-\lambda E)$  is dense in $H$.
 Consequently, for any $n\geq1$ an operator $A_n-\lambda E_n$ is a one-to-one operator in $H_n$,
 there exists $m\in \mathbb{N}$ such that $R(A_{m}-\lambda E_{m})\neq H_m$ and for any $n\geq1$
 linear manifold $R(A_n-\lambda E_n)$ is dense in $H_n$ or $\lambda\in \rho(A_m)$ for each
 $m\geq1$ but $\sup \{\|R_{\lambda}(A_m)\|: m\geq1 \}=\infty $. This means that
\begin{align*}
\lambda &\in \Big\{(\cap_{n=1}^{\infty} [\sigma_{c}(A_n)\cup \rho(A_n)])
\cap (\cup_{n=1}^{\infty}\sigma_{c}(A_n)) \Big\} \\
&\quad \cup
\Big\{\lambda\in \cap_{n=1}^{\infty}\rho(A_n):\sup\|R_{\lambda}(A_n)\|
=\infty\Big\}
\end{align*}
On the contrary, now suppose that for the point $\lambda\in \mathbb{C}$ 
the above relation is satisfied.
 Consequently, either for any $n\geq1$,
\[
\lambda\in \sigma_{c}(A_n)\cup \rho(A_n),
\]
or
 \[
\lambda\in \cap_{n=1}^{\infty} \rho(A_n): \sup\|R_{\lambda}(A_n)\|=\infty ,
\]
and there exist $m\in \mathbb{N}$ such that
\[
\lambda\in \sigma_{c}(A_m).
\]
That is, for any $n\geq1$, $A_n$ is a one-to-one operator, 
$\overline{R(A_n-\lambda E_n)}=H_n$ and
$R(A_{m}-\lambda E_{m})\neq H_m$. And from this it implies that the operator
$A=\oplus_{n=1}^{\infty } A_n$ is a one-to-one operator,
$R(A-\lambda E)\neq H$ and
$\overline{R(A-\lambda E)}=H$. Hence $\lambda\in \sigma_{c}(A_m)$.

 	On the other hand the simple calculations show that
\begin{align*}
&[\cap_{n=1}^{\infty}(\sigma_{c}(A_n)\cup\rho(A_n))]\cap
[\cup_{n=1}^{\infty}\sigma_{c}(A_n)]\\
&=[\cup_{n=1}^{\infty}\sigma_{p}(A_n)]^{c}\cap
[\cup_{n=1}^{\infty}\sigma_{r}(A_n)]^{c}
 \cap[\cup_{n=1}^{\infty}\sigma_{c}(A_n)].
\end{align*}
By a similarly technique, we can proved the validity of the third 
equality of the theorem.
\end{proof}

\begin{example} \label{exa4}\rm
 Consider the  multi-point differential operator for first order,
\begin{gather*}
A_nu_n=u_n^{'}(t), \quad
H_n=L^2(\Delta_n),\quad
\Delta_n=(a_n,b_n), \\
-\infty<a_n<b_n<a_{n+1}<\dots <+\infty;
\\
A_n:D(A_n)\subset H_n\to H_n,\quad
D(A_n)=\{u_n\in W_{2}^{1}(\Delta_n):
 u_n(a_n)=u_n(b_n)\},\ n\geq1;
\\
 A=\oplus_{n=1}^{\infty } A_n, \quad
 H=\oplus_{n=1}^{\infty } H_n.
\end{gather*}
For any $n\geq1$ operator $A_n$ and $A$ are normal,
$\sigma(A_n)=\sigma_{p}(A_n)=\{\frac{2k\pi i}{b_n-a_n}:k\in \mathbb{Z}\}$
and eigenvectors according to the eigenvalue $\lambda_{nk}$, $n\geq1$,
 $k\in \mathbb{Z}$ are in the form
\[
u_{nk}(t)=c_{nk}exp(\lambda_{nk}(t-a_n)), \quad
 t\in \Delta_n, \;  c_{nk}\in \mathbb{C}-\{0\}
\] 
(see \cite{Ism}). In this case
 \[ 
\sum_{n=1}^{\infty}\|u_{nk}\|_{H_n}^2=
 \sum_{n=1}^{\infty}\int_{\Delta_n} |c_{nk}|^2 |
\exp(\lambda_{nk}(t-a_n))|^2dt=
 \sum_{n=1}^{\infty}|c_{nk}|^2(b_n-a_n).
\]
 The coefficients $c_{nk}$ may be chosen such that the last series to be 
convergent.
 This means that $\lambda_{nk}\in \sigma_{p}(A)$. 
From this and Theorem \ref{thm1} it is obtained that
 $\sigma_{p}(A)= \cup_{n=1}^{\infty}\sigma_{p}(A_n)$.
\end{example}

\begin{definition}[\cite{Gor}] \label{def5} \rm
Let $T$ be a linear closed and densely defined operator
in any Hilbert space $H$. If $\rho(T)\neq \emptyset$ and for 
$\lambda\in \rho(T)$ the  resolvent operator 
$R_{\lambda}(T)\in C_{\infty}(H)$, then operator $T:D(T)\subset H \to H$
is called a operator with discrete spectrum .
\end{definition}

 Note that if the operator $A=\oplus_{n=1}^{\infty } A_n$
is an operator with discrete spectrum in $H=\oplus_{n=1}^{\infty } H_n$,
then for every $n\geq1$ the operator $A_n$ is also in $H_n$.

 The following proposition is proved by using the Theorem \ref{thm2} .


\begin{theorem} \label{thm6}
 If $A=\oplus_{n=1}^{\infty } A_n$, $A_n$ is an operator
 with discrete spectrum in $H_n$, $n\geq1$, 
$\cap_{n=1}^{\infty}\rho(A_n)\neq \emptyset$
 and $\lim_{n\to\infty}\|R_{\lambda}(A_n)\|=0$, then $A$ is an 
operator with discrete spectrum in $H$.
\end{theorem}

\begin{proof} 
In this case for each $\lambda\in \cap_{n=1}^{\infty}\rho(A_n)$ we have
 $R_{\lambda}(A_n)\in C_{\infty}(H_n)$, $n\geq1$.
 Now we define the operator $K:=\oplus_{n=1}^{\infty }R_{\lambda} (A_n)$ in $H$.
 In this case for every $u=(u_n)\in D(A)$, we have
 \begin{align*}
 K(A-\lambda E)(u_n)
&=\oplus_{n=1}^{\infty }R_{\lambda} (A_n)
 (\oplus_{n=1}^{\infty }(A_n-\lambda E_n))(u_n)\\
&= \oplus_{n=1}^{\infty }R_{\lambda} (A_n)((A_n-\lambda E_n)u_n)\\
&=(R_{\lambda} (A_n)(A_n-\lambda E_n)u_n)=(u_n)
\end{align*}
 and
\begin{align*}
(A-\lambda E)K(u_n)
&=(A-\lambda E)(\oplus_{n=1}^{\infty }R_{\lambda} (A_n))(u_n)\\
&=(A-\lambda E)(R_{\lambda} (A_n)u_n) \\
&=(\oplus_{n=1}^{\infty }(A_n-\lambda E_n))(R_{\lambda} (A_n)u_n)\\
&=((A_n-\lambda E_n)R_{\lambda} (A_n)u_n)=(u_n)
\end{align*}
These relations show that 
$R_{\lambda}(A)=\oplus_{n=1}^{\infty }R_{\lambda}(A_n)$.
Furthermore, we define the following operators
 $K_{m}:H \to H$, $m\geq1$ in the form
\[ 
K_{m}u:=\{R_{\lambda}(A_1)u_{1},R_{\lambda}(A_2)u_{2},\dots ,
R_{\lambda}(A_m)u_{m},0,0,0,\dots \},
 \quad u=(u_n)\in H.
 \]
Now the convergence in operator norm of the operators $K_m$ to the 
operator $K$  will be investigated. For the $u=(u_n)\in H$ we have
\begin{align*}
 \|K_{m}u-Ku\|_{H}^2
&=\sum_{n=m+1}^{\infty }\|R_{\lambda}(A_n)u_n\|_{H_n}^2\\
&\leq \sum_{n=m+1}^{\infty }\|R_{\lambda}(A_n)\|^2\|u_n\|_{H_n}^2
\\
&\leq \Big(\sup_{n\geq m+1} \|R_{\lambda}(A_n)\|\Big)^2
 \sum_{n=1}^{\infty }\|u_n\|_{H_n}^2\\
&=\Big(\sup_{n\geq m+1} \|R_{\lambda}(A_n)\|\Big)^2\|u\|_{H}^2
 \end{align*}
From this,  
$\|K_{m}-K\|\leq \sup_{n\geq m+1} \|R_{\lambda}(A_n)\|$ for
 $m\geq1$.
This means that sequence of the operators $(K_m)$ converges in operator 
norm to the operator $K$.
 Then by the important theorem of the theory of compact operators 
$K\in C_{\infty}(H)$ (\cite{Dun}),
 because for any $m\geq1$, $K_{m}\in C_{\infty}(H)$.
\end{proof}

\begin{example} \label{exa7} \rm
Consider that the  family of the operators in the form
\begin{gather*}
 A_n := \frac{d}{dt}+S_n, \quad S_n^{*}=S{n}>0, \quad
  S_n^{-1}\in C_{\infty}(H),\\
 A_n : D(A_n)\subset L_n^2\to L_n^2, \quad
 \Delta_n=(a_n,b_n),\quad
 \sup_{n\geq 1} (a_n,b_n)<\infty,\\
 D(A_n) = \{u_n\in W_{2}^{1}(H,\Delta_n):u_n(b_n)=W_nu_n(a_n),\;
 A_n^{-1}W_n=W_nA_n^{-1}\},
 \end{gather*}
where $L_n^2=L^2(H,\Delta_n)$, $n\geq1$, $H$ is any Hilbert space and
$W_n$ is a unitary operator in $H$, $n\geq1$ (for this see \cite{Ism}). 
For any $n\geq1$  an operator $A_n$ is normal with discrete spectrum and 
$\cap_{n=1}^{\infty }\rho(A_n)\neq\emptyset$.
 For the $\lambda\in \cap_{n=1}^{\infty }\rho(A_n)$ and sufficiently large
 $n\geq1$ a simple calculation shows that
\begin{align*}
R_{\lambda}(A_n)f_n(t)
&=e^{-(S_n-\lambda E_n)(t-a_n)}
\Big(E-W_n^{*}e^{-(S_n-\lambda E_n)(b_n-a_n)}\Big)^{-1}\\
&\quad\times W_n^{*} \int_{\Delta_n} e^{-(S_n-\lambda E_n)(b_n-s)}f_n(s)ds \\
&\quad + \int_{a_n}^{t}e^{-(S_n-\lambda E_n)(t-s)}f_n(s)ds, \quad
 f_n\in L_n^2, \quad n\geq1.
\end{align*}
On the other hand the following estimates hold:
\begin{equation} \label{eq2.1}
\begin{aligned}
&\big\|\int_{a_n}^{t}e^{-(S_n-\lambda E_n)(t-s)}f_n(s)ds\big\|_{L_n^2}^2\\
&\leq  \int_{\Delta_n} \Big(\int_{a_n}^{t}\|e^{-(S_n-\lambda E_n)(t-s)}\|\,
 \|f_n(s)\|_{H}ds\Big)^2dt
\\
 &\leq \int_{\Delta_n} \Big(\int_{a_n}^{t}\|e^{-(S_n-\lambda E_n)(t-s)}\|^2ds\Big)dt
 \int_{\Delta_n} \|f_n(s)\|_{H}^2ds \\
 &= \int_{\Delta_n} \Big(\int_{a_n}^{t}\|e^{-(S_n-\lambda_{r} E_n)(t-s)}\|^2
   ds\Big)dt\|f_n\|_{L_n^2}^2 \\
 &= \int_{\Delta_n} \Big(\int_{a_n}^{t}e^{2\lambda_{r}(t-s)}\|e^{-S_n(t-s)}\|^2ds
\Big)dt  \|f_n\|_{L_n^2}^2 \\
 &= \int_{\Delta_n} \Big(\int_{a_n}^{t}e^{2(\lambda_{r}-\lambda_{1}^{(n)})(t-s)}ds
\Big)dt  \|f_n\|_{L_n^2}^2
\\
&= \frac{1}{4(\lambda_{r}-\lambda_{1}^{(n)})^2}[2(\lambda_{r}-\lambda_{1}^{(n)})(a_n-b{n})-1+
 e^{2(\lambda_{r}-\lambda_{1}^{(n)})(b_n-a_n)}]\|f_n\|_{L_n^2}^2 ,
\end{aligned}
\end{equation}
\begin{equation}  \label{eq2.2}
 \begin{aligned}
\|(E-W_n^{*}e^{-(S_n-\lambda E_n)(b_n-a_n)})^{-1}W_n^{*}\|
&=\|\sum_{m=0}^{\infty}(W_n^{*}e^{-(S_n-\lambda E_n)(b_n-a_n)})^{m}\| \\
&\leq \sum_{m=0}^{\infty}\|e^{-(S_n-\lambda_{r} E_n)(b_n-a_n)}\|^{m} \\
& =(1-\|e^{-(S_n-\lambda_{r} E_n)(b_n-a_n)}\|)^{-1}\\
&=(1-e^{(\lambda_{r}-\lambda_{1}^{(n)})(b_n-a_n)})^{-1},
\end{aligned}
\end{equation}
 \begin{equation} \label{eq2.3}
\begin{split}
&\|\int_{\Delta_n} e^{-(S_n-\lambda E_n)(b_n-s)}f_n(s)ds\|^2\\
&\leq\frac{1}{2(\lambda_{r}-\lambda_{1}^{(n)})}[e^{2(\lambda_{r}
-\lambda_{1}^{(n)})(b_n-a_n)}-1]
 \|f_n\|_{L_n^2}^2
\end{split}
 \end{equation}
Hence from \eqref{eq2.2} and \eqref{eq2.3}, we have
\begin{equation} \label{eq2.4}
\begin{aligned}
&\|e^{-(S_n-\lambda E_n)(t-a_n)}
\Big(E-W_n^{*}e^{-(S_n-\lambda E_n)(b_n-a_n)}\Big)^{-1}\\
&\times W_n^{*}
 \int_{\Delta_n} e^{-(S_n-\lambda E_n)(b_n-s)}f_n(s)ds\|_{L_n^2}^2
\\
&\leq \int_{\Delta_n} e^{2\lambda_{r}(t-a_n)}\|e^{-S_n(t-a_n)}\|^2dt
 \|(E-W_n^{*}e^{-(S_n-\lambda E_n)(b_n-a_n)})^{-1}W_n^{*}\|^2\\
&\quad\times \|\int_{\Delta_n} e^{-(S_n-\lambda E_n)(b_n-s)}f_n(s)ds\|_{L_n^2}^2
\\
&\leq\frac{1}{4\lambda_{r}(\lambda_{r}-\lambda_{1}^{(n)})}
\Big(e^{2\lambda_{r}(b_n-a_n)}-1\Big)
\Big(1-e^{(\lambda_{r}-\lambda_{1}^{(n)})(b_n-a_n)}\Big)^{-1}\\
&\quad\times \Big(e^{2(\lambda_{r}-\lambda_{1}^{(n)})(b_n-a_n)}-1\Big)
\|f_n\|_{L_n^2}^2,
\end{aligned}
\end{equation}
where  $\lambda_r$ is the real part of $\lambda$ and $\lambda_{1}^{(n)}$
is the first  eigenvalue of the operator $S_n$, $n\geq1$.
\end{example}

From estimates \eqref{eq2.1} and \eqref{eq2.4} the following result is obtained.

\begin{proposition} \label{pro8} 
If $\lambda\in \cap_{n=1}^{\infty}\rho(A_n)$,
 $\underset{{n\geq 1}}{sup}(b_n-a_n)<\infty$ and 
$\lambda_{1}^{(n)}(S_n)\to \infty$
 as $n\to \infty$, then $\|R_{\lambda}(A_n)\|\to 0$ as $n\to \infty$.
 Consequently , the operator $A=\oplus_{n=1}^{\infty } A_n$ is
 an operator with discrete spectrum in $L^2=\oplus_{n=1}^{\infty } L_n^2$.
\end{proposition}

\section{Asymptotical behavior of the eigenvalues}

 In this section, asymptotical behavior for the eigenvalues of the operator
 $A=\oplus_{n=1}^{\infty } A_n$ in $H=\oplus_{n=1}^{\infty } H_n$
 is investigated, in a special case.

\begin{theorem} \label{thm3.1} 
Assume that the operators $A$ in $H$ and $A_n$ in $H_n$, $n\geq1$
 are operators with discrete spectrum and for $i,j\geq1$, $i\neq j$, 
$\sigma(A_i)\cap\sigma(A_j)=\emptyset$.
If $\lambda_{m}(A_n)\sim c_nm^{\alpha_n}$, $0<c_n, \alpha_n<\infty$,
 $m\to \infty$, $\sum_{n=1}^{\infty}c_n^{\frac{-1}{\alpha_n}}<\infty$, $n\geq1$
 and there exists $q\in \mathbb{N}$ such that 
$\alpha_{q}=\inf_{n\geq 1}\alpha_n>0$,
 then $\lambda_n(A)\sim \gamma n^{\alpha}$, 
$0<\gamma, \alpha=\alpha_{q}<\infty$ as $n\to \infty$.
\end{theorem}

\begin{proof} 
First of all note that by the Theorem \ref{thm3}
 $\sigma_{p}(A)=\cup_{n=1}^{\infty}\sigma_{p}(A_n)$.
Here it is denoted by 
$N(T;\lambda):=\underset{{|\lambda(T)|\leq \lambda}}{\sum}1$, $\lambda\geq 0$, 
that is, a number of eigenvalues of the some linear closed operator 
$T$ in any Hilbert space with modules of these eigenvalues 
less than or equal to $\lambda$, $\lambda\geq0$. 
This function takes values in the set of non-negative integer numbers 
and in case of unbounded operator $T$ it is nondecreasing and tends 
to $\infty$ as $\lambda\to \infty$.

 Since for every $i,j\geq1$, $i\neq j$, $\sigma(A_i)\cap \sigma(A_j)=\emptyset$, 
then $N(A;\lambda)=\sum_{n=1}^{\infty}N(A_n;\lambda)$. In this case it is 
clear that
\[ 
\frac{N(A;\lambda)}{\lambda^{\frac{1}{\alpha}}}\sim \sum_{n=1}^{\infty}c_n^{\frac{-1}{\alpha_n}}\lambda^{\frac{1}{\alpha_n}-\frac{1}{\alpha}}
 =\sum_{n=1}^{\infty}c_n^{\frac{-1}{\alpha_n}}\lambda^{\frac{\alpha-\alpha_n}{\alpha \alpha_n}}
 =\sum_{n=1}^{\infty}c_n^{\frac{-1}{\alpha_n}}
(\frac{1}{\lambda})^{\frac{\alpha_n-\alpha}{\alpha \alpha_n}}, \quad
\lambda\geq0. 
\]
 The last series is uniformly convergent in $(1,\infty)$ on $\lambda$. 
Then
\[
\lim_{\lambda\to\infty}\sum_{n=1}^{\infty}c_n^{\frac{-1}{\alpha_n}}
 (\frac{1}{\lambda})^{\frac{\alpha_n-\alpha}{\alpha \alpha_n}}
=c_{q}^{\frac{-1}{\alpha_q}}.
\]
 Therefore $N(A;\lambda)\sim c\lambda^{\frac{1}{\alpha}}$, 
$0<c=c_{q}^{-1/\alpha_q}$,
 $\alpha<\infty$ as $\lambda\to \infty$.
\end{proof}

We have the following asymptotic behavior of eigenvalues of the operator 
$A$ in $H$  $\lambda_n(A)\sim\gamma n^{\alpha}$, $0<\gamma,\alpha<\infty$ 
as $n\to \infty$.

\begin{remark} \label{rem3.2}  \rm
If in the above theorem the coefficients $\alpha_n$, $n\geq1$ satisfy
 the  condition $\inf_{n\geq 1}\alpha_n>0$, then for every
 $0<\alpha<\inf_{n\geq 1}\alpha_n$,
$N(A;\lambda)=o(\lambda^{\frac{1}{\alpha}})$  as $\lambda\to \infty$.
\end{remark}

\begin{remark} \label{rem3.3} \rm
If the every finitely many sets of the family $\sigma(A_n)$, $n\geq1$ in
complex plane intersect in the finitely many points, then it can be proved 
that claim of  Theorem \ref{thm3.1} is also valid in this case.
\end{remark}

\begin{example} \label{exa3.4} \rm
Let $H=\oplus_{n=1}^{\infty } H_n$, $H_n=l^2(\mathbb{N})$,
 $n\geq1$, $A_n:D(A_n)\subset H_n\to H_n$, $A_n(u_m):=(c_{nm}u_{m})$,
 $u=(u_m)\in D(A_n)$, $c_{nm}\in \mathbb{C}$, $c_{nm}\neq c_{km}$, 
$n\neq k$, $n,k,m\geq1$,
 $c_{nm}\sim k_nm^{\alpha_n}$, $0<k_n<\infty$, $1\leq \alpha_n<\infty$ 
as $m\to \infty$,
 $ \sum_{n=1}^{\infty}k_n^{\frac{-1}{\alpha_n}}$ is convergent and there exists
 $q\in \mathbb{N}$ such that $\alpha_{q}=\inf_{n\geq 1}\alpha_n$.
In this case, for any $n\geq1$, $A_n$ is a linear normal operator and 
 $\sigma(A_n)=\sigma_{p}(A_n)=\overline{\cup_{m=1}^{\infty}\{c_{nm}\}}$.
\end{example}

 Now we obtain the resolvent operator of $A_n$. 
Let $\lambda\in \cap_{n=1}^{\infty}\rho(A_n)$.
 Then from the relation $(A_n-\lambda E_n)(u_m)=(v_m)$, $n\geq1$, 
$(v_m)\in H_n$, i.e
 $c_{nm}u_{m}-\lambda u_{m}=v_{m}$, $m\geq1$.
It is established that $u_{m}=\frac{v_m}{c_{nm}-\lambda}$, $m\geq1$; i.e.,
 $ R_{\lambda}(A_n)(v_m)=(\frac{v_m}{c_{nm}-\lambda})$, $n\geq1$.

On the other hand since $c_{nm}\sim k_nm^{\alpha_n}$, $\alpha_n\geq1$ 
as $m\to \infty$,  then for any $v=(v_m)\in H_n$ we have
 \begin{align*}
\|R_{\lambda}(A_n)(v_m)\|_{H_n}^2
&=\sum_{m=1}^{\infty}|\frac{v_m}{c_{nm}-\lambda}|^2
 \leq \sum_{m=1}^{\infty}|\frac{1}{c_{nm}-\lambda}|^2\sum_{m=1}^{\infty}|v_m|^2\\
&=\sum_{m=1}^{\infty}|\frac{1}{c_{nm}-\lambda}|^2\|v\|_{H_n}^2.
\end{align*}
Consequently, for any $n\geq1$,
\begin{equation} \label{eq3.1}
 \|R_{\lambda}(A_n)\|\leq
\Big(\sum_{m=1}^{\infty}|\frac{1}{c_{nm}-\lambda}|^2\Big)^{1/2}
\end{equation}
Moreover, it is known that a resolvent operator $R_{\lambda}(A_n)$, $n\geq1$
is compact  if and only if
$\frac{1}{c_{nm}-\lambda}\underset{{m\to \infty}}{\to }0$ (\cite{Hun}).
Since $\lambda\neq c_{nm}$, $n, m\geq1$ and conditions on $c_{nm}$,
then the last condition is satisfied.
 Hence for any $n\geq1$, $R_{\lambda}(A_n)\in C_{\infty}(H_n)$.

 On the other hand since the series 
$\sum_{n=1}^{\infty}k_n^{-\frac{1}{\alpha_n}}$
 is convergent, then from the inequality \eqref{eq3.1}
for the $R_{\lambda}(A_n)$, $n\geq1$ it is easy to
 see that $\underset{{n\to \infty}}{lim}\|R_{\lambda}(A_n)\|=0$,
 $\lambda\in \cap_{n=1}^{\infty}\rho(A_n)$.
 Hence by the Theorem \ref{thm6} for the 
$\lambda\in \cap_{n=1}^{\infty}\rho(A_n)$ it is
 established that $R_{\lambda}(A)\in C_{\infty}(H)$. 
Then by the Theorem \ref{thm3} it is true that
 $\sigma_{p}(A)=\cup_{n=1}^{\infty}\sigma_{p}(A_n)$.
 Furthermore, the validity of the relation 
$\sigma(A_i)\cap \sigma(A_j)=\emptyset$, $i,j\geq1$,
 $i\neq j$ is clear. Therefore by the Theorem \ref{thm3.1} 
$\lambda_n(A)\sim \gamma n^{\alpha}$,
 $0<\gamma,\alpha<\infty$ as $n\to \infty$.

\subsection*{Acknowledgments}
 The authors are grateful to Research Asst. R. \"Ozt\"urk Mert 
(Department of Mathematics, Hitit University) for the interesting and 
helpful comments.

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\end{document}