\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), \\ -\infty0, \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: \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} \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} $$\label{eq2.3} \begin{split} &\|\int_{\Delta_n} e^{-(S_n-\lambda E_n)(b_n-s)}f_n(s)ds\|^2\\ &\leq\frac{1}{2(\lambda_{r}-\lambda_{1}^{(n)})}[e^{2(\lambda_{r} -\lambda_{1}^{(n)})(b_n-a_n)}-1] \|f_n\|_{L_n^2}^2 \end{split}$$ Hence from \eqref{eq2.2} and \eqref{eq2.3}, we have \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} 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 $\lambda_{m}(A_n)\sim c_nm^{\alpha_n}$, $00$, 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 $N(A;\lambda)\sim c\lambda^{\frac{1}{\alpha}}$, $00$, 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