\documentclass[reqno]{amsart}
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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2011 (2011), No. 89, pp. 1--11.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2011 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2011/89\hfil Compact inverses]
{Compact inverses of multipoint normal differential
 operators for first order}

\author[Z. I. Ismailov, E. Otkun \c{C}evik, E. Unluyol\hfil EJDE-2011/??\hfilneg]
{Zameddin I. Ismailov, Elif Otkun \c{C}evik, Erdal Unluyol}  % in alphabetical order

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

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

\address{Erdal Unluyol \newline
Department of Mathematics, Faculty of Sciences,
Karadeniz Technical University, 61080, Trabzon, Turkey}
\email{eunluyol@yahoo.com}


\thanks{Submitted May 7, 2011. Published July 5, 2011.}
\subjclass[2000]{47A05, 47A20}
\keywords{Direct sum of Hilbert spaces and operators;
 extension; \hfill\break\indent
 multipoint selfadjoint operator; formally normal operator;
 normal operator}

\begin{abstract}
 In this work, we describe all normal extensions of a multipoint
 minimal operators generated by linear multipoint
 differential-operator expressions for first order in the
 Hilbert space of vector functions, in terms of boundary values
 at the endpoints of infinitely many separated subintervals.
 Also we investigate compactness properties of the inverses
 of such extensions.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{definition}[theorem]{Definition}
\allowdisplaybreaks

\section{Introduction}

It is known that the traditional infinite direct sum of Hilbert
spaces $H_n$, $n\ge 1$ and infinite direct sum of operators
$A_n $ in $H_n$, $n\ge 1$ are defined as
\begin{gather*}
 H=\oplus_{n=1}^{\infty } H_n =\{u=(u_n ):u_n \in H_n ,\;
n\ge 1,\;\sum_{n=1}^{\infty }\| u_n \|_{H_n }^2 <+\infty \},
\\
A=\oplus_{n=1}^{\infty } A_n, \\
D(A)=\{ u=(u_n)\in H:u_n \in D(A_n),\; n\ge 1,\; Au=(A_n u_n )\in H
\}.
\end{gather*}
Note that $H$ is a Hilbert space with norm induced by the
inner product
\[
(u,v)_{H}=\sum_{n=1}^{\infty}(u_n,v_n)_{H_n},\quad
u, v\in H\,;
\]
see \cite{Dun,Lin,NaFo}.
The general theory of linear closed operators in Hilbert spaces
and its applications to physical problems have been investigated
by many  researches (see for example \cite{Dun,RoKh}). Furthermore,
many physical problems require studying  the theory of linear
operators in direct sums in Hilbert spaces. This is the case in
\cite{GaKr,Ism,Koc,So,Tim,Ze} and their references, which is
the motivation for this work. We note
that a detail analysis of normal subspaces and operators in
Hilbert spaces have been studied in \cite{Cod} and the references
there in.

 Besides the introduction, this study contains three sections. In
section 2, the multipoint minimal and maximal operators for the
first order differential-operator expression are determined.
In section 3, all normal extensions of multipoint formally normal
operators are described in terms of boundary values in the
endpoints of the infinitely many separated subintervals. Finally
in section 4, compactness properties of the inverses of such
extensions have been established.

\section{Minimal and maximal operators}

 Throughout this work $(a_n )$ and $(b_n )$ will be sequences
of real numbers such that
\[
-\infty <a_n <b_n < a_{n+1} <\dots <+\infty ,
\]
 $H_n $ is a Hilbert space, $\Delta_n =(a_n ,b_n )$,
$L_n^2 =L^2 (H_n,\Delta_n )$,
$L^2 =\oplus_{n=1}^{\infty }L^2(H_n ,\Delta_n )$,  $n\ge 1$,
$\sup_{n\geq1}(b_n-a_n)<+\infty$,
$W_{2}^{1} =\oplus_{n=1}^{\infty } W_{2}^{1} (H_n ,\Delta_n )$,
$\mathaccent"7017{W_2^1} =\oplus_{n=1}^{\infty }
 \mathaccent"7017{W_2^1} (H_n ,\Delta_n )$,
$H=\oplus_{n=1}^{\infty } H_n$,
$\operatorname{cl}(T)$ is the closure of the operator $T$.
$l(\cdot )$ is a linear multipoint differential-operator
expression for first order in $L^2 $ in the following form:
\begin{equation} \label{eq2.1}
l(u)=(l_n (u_n )),
\end{equation}
and for each $n\ge 1$,
\begin{equation} \label{eq2.2}
l_n (u_n )=u'_n + A_n u_n ,
\end{equation}
where $A_n :D(A_n )\subset H_n \to H_n $ is a linear positive
defined selfadjoint operator in $H_n$.
 It is clear that formally adjoint expression to
\eqref{eq2.2} in the Hilbert space $L_n^2 $ is in the form
\begin{equation} \label{eq2.3}
l_n^{+} (v_n )=-v'_n + A_n v_n , n\ge 1.
\end{equation}
We define an operator $L'_{n0} $ on the dense manifold of vector
functions $D'_{n0} $ in $L_n^2 $ as
\begin{align*}
D'_{n0} :=\Big\{&u_n \in L_n^2 :u_n =\sum_{k=1}^{m}\phi_k
f_k , \phi_k \in C_0^{\infty } ({\Delta_n}),\\
& f_k \in D(A_n),k=1,2,\dots ,m; m\in {\mathbb N} \Big\}
\end{align*}
with $L'_{n0} u_n :=l_n (u_n ), n\ge 1$.
Since the operator $A_n>0$, $n\geq1$, then  the relation
\[
\operatorname{Re}(L'_{n0} u_n ,u_n )_{L_n^2 }
=2(A_n u_n ,u_n )_{L_n^2 } \ge 0,
\quad u_n \in D'_{n0}
\]
implies that $L'_{n0} $ is an accretive in $L_n^2$, $n\ge 1$.
Hence the operator $L'_{n0} $ has a closure in $L_n^2$, $n\ge 1$.
The closure $\operatorname{cl}(L'_{n0} )$ of the operator
$L'_{n0} $ is called the minimal operator generated by
differential-operator expression \eqref{eq2.2} and is denoted
by $L_{n0} $ in $L_n^2$, $n\ge 1$. The operator $L_0$ defined by
\[
D(L_0 ):=\{u=(u_n):u_n \in D(L_{n0} ),\; n\ge 1,\;
 \sum_{n=1}^{\infty }\| L_{n0} u_n \|_{L_n^2 }^2 <+\infty  \}
\]
with
$L_0 u:=(L_{n0}u_n)$, $u\in D(L_0 )$,
$ L_0 :D(L_0 )\subset L^2 \to L^2$
is called a minimal operator (multipoint) generated by
differential-operator expression \eqref{eq2.1} in Hilbert space
$L^2 $ and denoted by $L_0 =\oplus_{n=1}^{\infty } L_{n0} $.
In a similar way the minimal operator for two points  denoted
by $L_{n0}^{+} $ in $L_n^2 , n\ge 1$ for the formally adjoint
linear differential-operator expression \eqref{eq2.3} can be
constructed.
 In this case the operator  $L_0^{+} $ defined by
\[
D(L_0^{+} ):=\{v:=(v_n ):v_n \in D(L_{n0}^{+} ),\; n\ge 1,\;
 \sum_{n=1}^{\infty }\| L_{n0}^{+} v_n \|_{L_n^2 }^2 <+\infty  \}
\]
with $L_0^{+} v:=(L_{n0}^{+} v_n )$, $v\in D(L_0^{+} )$,
$L_0^{+} :D(L_0^{+} )\subset L^2 \to L^2$
is called a minimal operator (multipoint) generated by
$l^{+} (v)=(l_n^{+}(v_n))$ in the Hilbert space $L^2 $
and denoted by $L_0^{+} =\oplus_{n=1}^{\infty } L_{n0}^{+}$.

 Now we state the following relevant result.

\begin{theorem} \label{thm2.1}
The minimal operators $L_0 $ and $L_0^{+} $ are densely defined
closed operators in $L^2 $.
\end{theorem}

\begin{proof}
Let $w=(w_n)$ be any element in $L^2=\oplus_{n=1}^{\infty } L_n^2$
and $\varepsilon$ be an arbitrary positive number. In this case
$w_n \in L_n^2$, $n\geq1$ and
$\sum_{n=1}^{\infty }\| w_n\|_{L_n^2 }^2 <+\infty$.
Since the linear manifold $D(L_{n0})$, $n\geq 1$ is densely defined
in $L_n^2$, $n\geq1$ , then there exist $w_n(\varepsilon)\in D(L_{n0})$
such that
\[
\|w_n-w_n(\varepsilon)\|_{L_n^2}<\frac{\varepsilon}{n}, \quad n\geq1.
\]
Then for the element $w(\varepsilon)=(w_n(\varepsilon))$ we have
\begin{align*}
\|w(\varepsilon)\|_{L^2}^2
&=\sum_{n=1}^{\infty }\|w_n(\varepsilon)\|_{L_n^2}^2\\
&\leq \sum_{n=1}^{\infty } (\|w_n-w_n(\varepsilon)\|_{L_n^2}
  +\|w_n(\varepsilon)\|_{L_n^2})^2\\
&\leq 2\sum_{n=1}^{\infty }\|w_n-w_n(\varepsilon)\|_{L_n^2}^2
  +2\sum_{n=1}^{\infty }\|w_n(\varepsilon)\|_{L_n^2}^2 \\
&\leq 2{\varepsilon}^2\sum_{n=1}^{\infty }\frac{1}{n^2}
  +2 \sum_{n=1}^{\infty }\|w_n(\varepsilon)\|_{L_n^2}^2<+\infty;
\end{align*}
that is, $w(\varepsilon)\in L^2$. On the other hand since
$$
\|w-w(\varepsilon)\|_{L^2}^2=\sum_{n=1}^{\infty }
\|w_n-w_n(\varepsilon)\|_{L_n^2}^2\leq {\varepsilon}^2
\sum_{n=1}^{\infty }\frac{1}{n^2}<2{\varepsilon}^2,
$$
 the linear manifold $D(L_0)$ is dense in $L^2$.

 Now we show that the minimal operator $L_0$ is closed in $L^2$.
Let $w^{(m)}\subset D(L_0)$ be any sequence such that for
$w,z \in L^2$, $w^{(m)}\to w$ as $m\to \infty$ and
$L_0w^{(m)}\to z$ as $m\to \infty$ in $L^2$.
In this situation, in the space $L_n^2$,
 $w_n^{(m)}\to w_n$ as $m\to \infty$ and
$L_{n0}w_n^{(m)}\to z_n$ as  $m\to \infty$, $n\geq 1$,
where $w=(w_n)$ and $z=(z_n)$. Since the operator $L_{n0}$
is closed in $L_n^2$, then $w_n \in D(L_{n0})$ and
$z_n=L_{n0}w_n, n\geq1$.

The above relations and $(w_n),(z_n)\in L^2$  imply that
$w=(w_n)\in D(L_0)$ and $z=L_0w$.
In similar way it could be shown that the minimal operator
$L_0^+$ is densely defined and closed operator in $\in L^2$.

The operators in $L^2$ defined by
$L:=(L_0^{+} )^{*} =\oplus_{n=1}^{\infty } L_n $ and
$L^{+} :=(L_0 )^{*} =\oplus_{n=1}^{\infty } L_n^{+}$
are called maximal operators (multipoint) for the
differential-operator expression $l(\cdot)$ and $l^{+} (\cdot )$
respectively.It is clear that $Lu=(l_n (u_n )), u\in D(L)$,
\begin{gather*}
D(L):=\{u=(u_n )\in L^2 :u_n \in D(L_n ),\; n\ge 1,\;
\sum_{n=1}^{\infty }\| L_n u_n \|_{L_n^2 }^2 <\infty  \},\\
L^{+} v=(l_n^{+} (v_n )), v\in D(L^{+} ), \\
D(L^{+} ):=\{v=(v_n )\in L^2 :v_n \in D(L_n^{+} ),\; n\ge 1 ,\;
\sum_{n=1}^{\infty }\| L_n^{+} v_n \|_{L_n^2 }^2 <\infty  \}
\end{gather*}
and $L_0 \subset L$, $L_0^{+} \subset L^{+} $.
\end{proof}

From \cite{Krall} and the definition of direct sum
of operators, the validity of following theorem is clear.

\begin{theorem} \label{thm2.2}
The domain of the operator $L$  consists of
$u=(u_n )\in L^2$ such that:
(1) for  each $n\ge 1$  vector function $u_n \in L_n^2$,
$u_n$ is absolutely continuous in the interval $\Delta_n$;
 (2) $l_n (u_n )\in L_n^2$, $n\ge 1$;
 (3) $l(u)=(l_n (u_n ))\in L^2 \}$;
i.e.,
\[
D(L)=\{u=(u_n )\in L^2 :u_n \in D(L_n ),\; n\ge 1,\;
l(u)=(l_n (u_n ))\in L^2 \}.
\]
The domain of  $L_0$ is
$ D(L_0)=\{u=(u_n )\in D(L):u_n (a_n )=u_n (b_n )=0,\; n\ge 1\}$.
\end{theorem}

\begin{remark} \label{rmk2.3}\rm
 If $A_n \in B(H)$, $n\ge 1$ and
$\sup_{n\ge 1} \| A_n \| \le c<+\infty $, then for any
$u=(u_n )\in L^2 $ we have $(Au)=(A_n u_n )\in L^2 $.
\end{remark}

\begin{theorem} \label{thm2.4}
If a minimal operator $L_0 $ is formally normal in $L^2 $,
then $D(L_0 )\subset \mathaccent"7017{W_2^1} $ and
$AD(L_0 )\subset L^2 $.
\end{theorem}

\begin{proof}
In this case for every $u=(u_n)\in D(L_0)\subset D(L^+)$ we have
 $u'+Au \in L^2$ and $-u'+Au \in L^2$.
From this it is obtained that
 $u' \in L^2, Au \in L^2$.
This means that
 $D(L_0)\subset \mathaccent"7017{W_2^1}$ and $AD(L_0)\subset L^2$.
\end{proof}

\begin{theorem} \label{thm2.5}
If $A^{1/2} W_{2}^{1} \subset W_{2}^{1} $, then minimal operator
$L_0 $ is formally normal in $L^2 $.
\end{theorem}

\begin{proof}
In this case from the  relations
\[
L_0^{+} u=L_0 u-2Au, u\in D(L_0 ),\quad
L_0 u=L_0^{+} u+2Au, u\in D(L_0^{+} )
\]
imply that $D(L_0 )=D(L_0^{+} )$.
Since $D(L_0^{+} )\subset D(L_0^{*} )=D(L^{+} )$,
it is obtained that $D(L_0 )\subset D(L^{+} )$.

On the other hand for any $u\in D(L_0 )$,
\begin{align*}
\| L_0 u\|_{L^2 }^2 &=(u'+Au,u'+Au)_{L^2 } \\
&=\| u'\|_{L^2 }^2 +[(u',Au)_{L^2 } +(Au,u')_{L^2 }]
+\| Au\|_{L^2 }^2  \\
&=\| u'\|_{L^2 }^2 +\| Au\|_{L^2 }^2
\end{align*}
and
\begin{align*}
\| L^{+} u\|_{L^2 }^2
&=(-u'+Au,-u'+Au)_{L^2 } \\
&=\| u'\|_{L^2 }^2 -[(u',Au)_{L^2 } +(Au,u')_{L^2 }]
+\| Au\|_{L^2 }^2  \\
&=\| u'\|_{L^2 }^2 +\| Au\|_{L^2 }^2 .
 \end{align*}
Thus, it is established that operator $L_0 $ is formally normal
in $L^2 $.
\end{proof}

\begin{remark} \label{rmk2.6}\rm
 If  $A_n \in B(H), n\ge 1$ and
 $\sup_{n\ge 1} \| A_n \| \le c<+\infty $, then
$D(L_0 )=D(L_0^{+} )$ and $D(L)=D(L^{+} )$.

 If $AW_{2}^{1} \subset L^2 $, then $D(L_0 )=D(L_0^{+} )$
and $D(L)=D(L^{+} )$.
\end{remark}

\section{Description of normal extensions of the minimal operator}

 In this section the main purpose is to describe all normal
extensions of the minimal operator  $L_0 $ in  $L^2 $
in terms in the boundary values of the endpoints of the
subintervals.
Firstly, we will show that there exists normal extension
of the minimal operator $L_0 $. Consider the following
extension of the minimal operator $L_0$,
\begin{gather*}
\widetilde{L}u:=u'+Au,\quad  AW_{2}^{1} \subset W_{2}^{1} , \\
D(\widetilde{L})=\{u=(u_n )\in W_{2}^{1} :u_n (a_n )=u_n (b_n ),
n\ge1\}.
\end{gather*}
 Under the condition on the coefficient $A$, we have
\begin{align*}
(\widetilde{L}u,v)_{L^2}
&=(u',v)_{L^2 } +(Au,v)_{L^2 }\\
&=(u,v)_{L^2} ' +(u,-v'+Av)_{L^2}\\
&=\sum_{n=1}^{\infty }[(u_n(b_n),v_n(b_n ))_{H_n}
- (u_n(a_n),v_n(a_n ))_{H_n}]+(u,-v'+Av)_{L^2}
\end{align*}
From this it is obtained that
\begin{gather*}
\widetilde{L}^{*} v:=-v'+Av,\\
D(\widetilde{L}^{*})=\{v=(v_n )\in W_{2}^{1} :v_n (a_n )=v_n (b_n ),
n\ge 1\}.
\end{gather*}
In this case it is clear that $D(\widetilde{L})=D(\widetilde{L}^{*} )$.
On the other hand, since for each $u\in D(\widetilde{L})$,
\begin{gather*}
\| \widetilde{L}u\|_{L^2 }^2 =\| u'\|_{L^2 }^2
+[(u',Au)_{L^2 } +(Au,u')_{L^2 } ]+\| Au\|_{L^2 }^2 , \\
\| \widetilde{L}^{*} u\|_{L^2 }^2 =\| u'\|_{L^2 }^2
  -[(u',Au)_{L^2 } +(Au,u')_{L^2 }]+\| Au\|_{L^2 }^2
\end{gather*}
and
\begin{align*}
(u',Au)_{L^2 } +(Au,u')_{L^2 }
&=(u,Au)_{L^2 } '  \\
&=\sum_{n=1}^{\infty}[(u_n(b_n),A_nu_n(b_n))_{H_n}-(u_n(a_n),
A_nu_n(a_n))_{H_n} ]=0.
\end{align*}
Then $\| \widetilde{L}u\|_{L^2 } =\| \widetilde{L}^{*} u\|_{L^2 } $
for every $u\in D(\widetilde{L})$. Consequently, $\widetilde{L}$
is a normal extension of the  minimal operator $L_0 $.

 The following result establishes the relationship between normal
extensions of $L_0 $ and normal extensions of  $L_{n0}$, $n\ge 1$.

\begin{theorem} \label{thm3.1}
The extension $\widetilde{L}=\oplus_{n=1}^{\infty }\widetilde{L_n}$
of the minimal operator $L_0$ in $ L^2$ is a normal if and only if
for any $n\geq 1$, $\widetilde{L_n}$ is so in $L_n^2$.
\end{theorem}

\begin{proof}
Let $\widetilde{L}=\oplus_{n=1}^{\infty }\widetilde{L_n}$
 be a normal extension of the minimal operator
$L_0=\oplus_{n=1}^{\infty }L_{n0}$. In this case it is clear
that for each $n\geq1$ an operator
$\widetilde{L_n}:D(\widetilde{L_n})\subset L_n^2 \to L_n^2$
defined by
\[
\widetilde{L_n}u_n=L_nu_n,\quad
D(\widetilde{L_n})=P_nD(\widetilde{L}),
\]
where $P_n$ is an orthogonal projection operator from $L^2$
to $L_n^2$, is a normal extension of the minimal operator $L_{n0}$.
On the contrary, assume that
$\widetilde{L_n}:D(\widetilde{L_n})\subset L_n^2 \to L_n^2$
is any normal extension of the minimal operator for each $n\geq1$,
and $\widetilde{L}:=\oplus_{n=1}^{\infty }\widetilde{L_n}$.
In this case for any $u=(u_n)\in D(\widetilde{L})$ we have
$\|\widetilde{L}u\|_{L^2}=\sum_{n=1}^{\infty }\|
\widetilde{L_n}u_n\|_{L_n^2}^2< +\infty$.
Since for any $n\geq1$, $\widetilde{L_n}$ is a normal extension,
it follows that $u_n\in D(\widetilde{L_n})$, $n\geq1$ and
$\sum_{n=1}^{\infty }\| \widetilde{L_n}^*u_n\|_{L_n^2}^2
=\sum_{n=1}^{\infty }\| \widetilde{L_n}u_n\|^2< +\infty$.
From this for every $u=(u_n)\in D(\widetilde{L})$ it is obtained that
 $(u_n)\in D(\widetilde{L}^*)$; i.e.,
$D(\widetilde{L}) \subset D(\widetilde{L}^*)$.

  In the similar way it can be shown that
$D(\widetilde{L}^*)\subset D(\widetilde{L})$.
Hence $D(\widetilde{L})=D(\widetilde{L}^*)$.
 On the other hand for any $n\geq1$ since $\widetilde{L_n}$ is
a normal extension, then for each $u=(u_n)\in D(\widetilde{L})$,
\[
\| \widetilde{L}u\|_{L^2}^2
=\sum_{n=1}^{\infty }\| \widetilde{L_n}u_n\|_{L_n^2}^2
=\sum_{n=1}^{\infty }\| \widetilde{L_n}^*u_n\|_{L_n^2}^2
=\| \widetilde{L}^*u\|_{L^2}^2 <+ \infty.
\]
This complete the proof.
\end{proof}

 Now using Theorem \ref{thm3.1} and \cite{Krall} we can formulate
the following main result of this section,
where it is given a description of all normal extension
of the minimal operator $L_0 $ in $L^2 $ in terms
of boundary values of vector functions at the endpoints
of subintervals.

\begin{theorem} \label{thm3.2}
Let $A^{1/2} W_{2}^{1} \subset W_{2}^{1} $.
If $\widetilde{L}=\oplus_{n=1}^{\infty }\widetilde{L_n }$
is a normal extension of the minimal operator $L_0 $ in $L^2 $,
then it is generated by differential-operator expression \eqref{eq2.1}
with boundary conditions
\begin{equation} \label{eq3.0}
u_n(b_n)=W_nu_n(a_n), u_n\in D(L_n),
\end{equation}
where $W_n$ is a unitary operator in $H_n$ and
$W_nA_n^{-1}=A_n^{-1}W_n, n\geq 1$. The unitary operator
$W=\oplus_{n=1}^{\infty }W_n$ in $H=\oplus_{n=1}^{\infty }H_n$
is determined uniquely by the extension $\widetilde{L}$; that is,
$\widetilde{L}=L_{W} $.
\end{theorem}

 On the contrary, the restriction of the maximal operator
$L$ to the linear manifold $u\in D(L)$ satisfying the condition
\eqref{eq3.0} with any unitary operator
$W=\oplus_{n=1}^{\infty }W_n$ in $H$  with property
$WA^{-1}=A^{-1}W$ is a normal extension of the minimal
operator $L_0$ in $L^2$.

\section{Some compactness properties of the normal extensions}

In this work  $\mathscr{H}_n, \mathfrak{H_n}, H_n, n\geq1$ will
denoted Hilbert spaces. First, we prove the following result.

\begin{theorem} \label{thm4.1}
 For the point spectrum of
${\mathscr{A}}=\oplus_{n=1}^{\infty }\mathscr{A}_n$ in the direct
sum ${\mathscr{H}}=\oplus_{n=1}^{\infty }\mathscr{H}_n$
of Hilbert spaces $\mathscr{H}_n, n\geq1$, it is true that
\[
\sigma_{p}(\mathscr{A})=\cup_{n=1}^{\infty}\sigma_{p}(\mathscr{A}_n)
\]
\end{theorem}

\begin{proof} Let $\lambda \in \sigma_p(\mathscr{A})$. In this case
there exist non zero element $u=(u_n)\in D(\mathscr{A})$ such that
$Au=\lambda u, u\ne 0$; i.e., $A_nu_n=\lambda u_n, n\geq1$.
Since $u\ne0$, then there exist some $m \in \mathbb{N}$ such that
$u_m \ne0$ and $A_mu_m=\lambda u_m$. This means that
$\lambda \in \sigma_p(\mathscr{A}_{m})$. From this it is obtained
that  $\sigma_p(\mathscr{A})\subset \cup_{n=1}^{\infty }
\sigma_p(\mathscr{A}_n)$.
 On the contrary, if for any $m \in \mathbb{N}$,
$\lambda \in \sigma_p(A_m)$, then there exist $u_m\in D(A_m)$
such that $u_m\ne0$ and $A_mu_m=\lambda u_m$.
In this case, if we choose the element
$u_* :=\{0,0,\dots,u_m,0,\dots\}$,
then $u_* \in D(\mathscr{A}), u_* \ne 0$ and
$Au_*=\lambda u_*$. Hence $\lambda \in \sigma_{p}(\mathscr{A})$.
This implies that
$\cup_{n=1}^{\infty }\sigma_p(\mathscr{A}_n)\subset
\sigma_p(\mathscr{A})$. Therefore, the claim of the theorem is valid.
\end{proof}

\begin{theorem}[\cite{NaFo}] \label{thm4.2}
Let $\mathscr{A}_n\in B(\mathscr{H}_n)$, $n\geq 1$,
$\mathscr{A}=\oplus_{n=1}^{\infty }\mathscr{A}_n$ and
$\mathscr{H}=\oplus_{n=1}^{\infty }\mathscr{H}_n$.
In order to $\mathscr{A}\in B(\mathscr{H})$ the necessary and
sufficient condition is $\sup_{n\geq1}\|\mathscr{A}_n\|<+\infty$.
In this case $\|\mathscr{A}\|= \sup_{n\geq1}\|\mathscr{A}_n\|$.
\end{theorem}

Let $C_{\infty}(\cdot)$ and $C_{p}(\cdot), 1\leq p<\infty$
denote the class of compact operators and the Schatten-von Neumann
subclasses of compact operators in corresponding spaces respectively.

\begin{definition}[\cite{Gor}] \label{def4.3}
Let $T$ be a linear closed and densely defined operator in any
 Hilbert space $\mathscr{H}$. If $\rho(T)\neq\emptyset$ and for
$\lambda\in\rho(T)$ the resolvent operator
$R_{\lambda}(T)\in C_{\infty}(\mathscr{H})$,
then operator $T:D(T)\subset{\mathscr{H}}\to{\mathscr{H}}$
is called an operator with discrete spectrum.
\end{definition}

First we note  the following result.

\begin{theorem} \label{thm4.4}
 If the operator $\mathscr{A}=\oplus_{n=1}^{\infty }\mathscr{A}_n$
as an operator with discrete spectrum in
$\mathscr{H}=\oplus_{n=1}^{\infty }\mathscr{H}_n$, then for every
$n\geq1$ the operator $\mathscr{A}_n$ is so in $\mathscr{H}_n$.
\end{theorem}

\begin{remark} \label{rmk4.5}\rm
 Unfortunately, the converse of the Theorem \ref{thm4.4} is not true
in general case.
Indeed, consider the sequence of operators
$\mathscr{A}_nu_n=u_n, 0<\dim\mathscr{H}_n=d_n<\infty$, $n\geq1$.
In this case for every $n\geq1$ operator $\mathscr{A}_n$ is an
operator with discrete spectrum. But an inverse of the direct
 sum operator $\mathscr{A}=\oplus_{n=1}^{\infty }\mathscr{A}_n$
is not compact operator in
$\mathscr{H}=\oplus_{n=1}^{\infty }\mathscr{H}_n$, because
$\dim\mathscr{H}=\infty$ and $\mathscr{A}$ is an identity
operator in $\mathscr{H}$.
\end{remark}

\begin{theorem} \label{thm4.6}
If $\mathscr{A}=\oplus_{n=1}^{\infty }\mathscr{A}_n, \mathscr{A}_n$
is an operator with discrete spectrum in $\mathscr{H}_n$, $n\geq1$,
$\cap_{n=1}^{\infty}{\rho(\mathscr{A}_n)}\neq \emptyset$ and for any
$\lambda \in \cap_{n=1}^{\infty}\rho(\mathscr{A}_n)$,
 $\lim_{n\to\infty}\|R_{\lambda}(\mathscr{A}_n)\|=0$,
then $\mathscr{A}$ is an operator with discrete spectrum
in $\mathscr{H}$.
\end{theorem}

\begin{proof}
For each $\lambda \in \cap_{n=1}^{\infty}{\rho(\mathscr{A}_n)}$
we have $R_{\lambda}(\mathscr{A}_n)\in C_{\infty}(\mathscr{H}_n)$,
$n\geq1$.
 Now define the  operators
$\mathscr{K}_{m}:\mathscr{H}\to\mathscr{H}$, $m\geq1$, as
\[
\mathscr{K}_{m}:=\{R_{\lambda}(\mathscr{A}_1)u_1,
 R_{\lambda}(\mathscr{A}_{2})u_{2},\dots,
R_{\lambda}(\mathscr{A}_{m})u_{m}, 0, 0, \dots \},\quad
 u=(u_n)\in \mathscr{H}.
\]
 The convergence of the operators $\mathscr{K}_{m}$ to the operator
$\mathscr{K}$ in operator norm will be investigated.
For  $u=(u_n)\in \mathscr{H}$, we have
\begin{align*}
\|\mathscr{K}_{m}u-\mathscr{K}u\|_{\mathscr{H}}^2
& =  \sum_{n=m+1}^{\infty}\|R_{\lambda}(\mathscr{A}_n)
 u_n\|_{\mathscr{H}_n}^2 \\
&\leq\sum_{n=m+1}^{\infty}\|R_{\lambda}(\mathscr{A}_n)\|^2\|
\|u_n\|_{\mathscr{H}_n}^2\\
& \leq  \Big(\sup_{n\geq m+1}{\|R_{\lambda}(\mathscr{A}_n)\|}\Big)^2
\sum_{n=1}^{\infty}\|u_n\|_{\mathscr{H}_n}^2\\
&=\Big(\sup_{n\geq m+1}{\|R_{\lambda}(\mathscr{A}_n)\|}
\Big)^2\|u\|_{\mathscr{H}}^2
\end{align*}
thus we get $\|\mathscr{K}_{m}u-\mathscr{K}u\|
\leq \sup_{n\geq m+1}{\|R_{\lambda}(\mathscr{A}_n)\|}$, $m\geq1$.
 This implies that sequence of operators $(\mathscr{K}_{m})$
converges in operator norm to the operator $\mathscr{K}$.
Then by the important theorem of the theory of compact operators
it implies that $\mathscr{K}\in C_{\infty}(\mathscr{H})$ \cite{Dun},
because for any $m\geq1$, $\mathscr{K}_{m}\in C_{\infty}(\mathscr{H})$.
\end{proof}

Using the Theorem \ref{thm3.2} and Theorem \ref{thm4.6} can be proved the following
result.

\begin{theorem} \label{thm4.7}
If ${A}_n^{-1}\in C_{\infty}(H_n)$, $n\geq1$,
$\sup_{n\geq1}(b_n-a_n)<\infty$ and the sequence of first
minimal eigenvalues $\lambda_1({A}_n)$ of the operators
${A}_n$, $n\geq1$  satisfy the condition
\[
\lambda_1({A}_n)\to\infty \quad\text{as }  n \to \infty,
\]
then the normal extension $\widetilde{L}$ of the minimal operator
$L_0$ is an operator with discrete spectrum in $L^2$.
\end{theorem}

From the definition of the characteristic numbers $\mu(.)$
of any compact operator in any Hilbert space \cite{Dun}
and Theorem \ref{thm4.1} it is easy to prove that the validity
of the following result.

\begin{theorem} \label{thm4.8}
\begin{itemize}
\item[(i)]
 If $\mathscr{A}=\oplus_{n=1}^{\infty }{\mathscr{A}_n}$,
$\mathscr{H}=\oplus_{n=1}^{\infty }{\mathscr{H}_n}$ and
$\mathscr{A}\in C_\infty(\mathscr{H})$, then for every $n\geq1$,
$\mathscr{A}_n \in C_\infty(\mathscr{H}_n)$ and
$\{\mu_m(\mathscr{A}):m\geq1\}=\cup_{n=1}^{\infty }
\{\mu_k(\mathscr{A}_n):k\geq1\}$;

\item[(ii)]
If $\mathscr{A}=\oplus_{n=1}^{\infty }{\mathscr{A}_n}$,
$\mathscr{H}=\oplus_{n=1}^{\infty }{\mathscr{H}_n}$ and
 $\mathscr{A}\in C_p(\mathscr{H})$, $1\leq p\leq+\infty$,
 then for every $n\geq1$,  $\mathscr{A}_n\in C_p(\mathscr{H}_n)$.
\end{itemize}
\end{theorem}

\begin{theorem} \label{thm4.9}
Let $\mathscr{H}=\oplus_{n=1}^{\infty }{\mathscr{H}_n}$,
$\mathscr{A}=\oplus_{n=1}^{\infty }{\mathscr{A}_n}$ and
$\mathscr{A}_n\in C_{p}(\mathscr{H}_n)$, $n\geq1$, $1\leq p<\infty$.
In this case $\mathscr{A}\in C_{p}(\mathscr{H})$ if and only if
the series $\sum_{n=1}^{\infty }{\sum_{k=1}^{\infty }}\mu_k^{p}
(\mathscr{A}_n)$ is convergent.
\end{theorem}

\begin{proof} Let us $\mathscr{A}\in C_p(\mathscr{H})$.
Then the series $\sum_{m=1}^{\infty }\mu_m^p(\mathscr{A})$
is convergent. In this case by the Theorem \ref{thm4.8} (i) and important
theorem on the convergence of rearrangement series it is obtained
that the series $\sum_{n=1}^{\infty }{\sum_{k=1}^{\infty }}
\mu_k^{p}(\mathscr{A}_n)$ is convergent.

 On the contrary, if the series $\sum_{n=1}^{\infty }
{\sum_{k=1}^{\infty }}\mu_k^{p}(\mathscr{A}_n)$ is convergent,
then the series $\sum_{m=1}^{\infty }\mu_m^p(\mathscr{A})$
 being a rearrangement of the above series,  is also convergent.
 So $\mathscr{A}\in C_p(\mathscr{H})$.
\end{proof}

 Now we will present an application the last theorem.
For all $n\geq1$, Let $\mathfrak{H}_n$ be a Hilbert space,
$\Delta_n=(a_n,b_n)$, $-\infty<a_n<b_n<a_{n+1}<\dots<\infty$,
$A_n:D(A_n)\subset \mathfrak{H}_n\to \mathfrak{H}_n$,
$A_n=A_n^{*}\geq E, U_n:\mathfrak{H}_n\to \mathfrak{H}_n$ is
unitary operator, $A_n^{-1}U_n=U_nA_n^{-1}$,
$L_{U_n}u_n=u'_n+A_nu_n, A_nW_{2}^{1}(\mathfrak{H}_n,\Delta_n)
\subset W_{2}^{1}(\mathfrak{H}_n,\Delta_n)$,
\[
H_n=L^2(\mathfrak{H}_n,\Delta_n), D(L_{U_n})
=\{u_n\in W_{2}^{1}(\mathfrak{H}_n,\Delta_n): u_n(b_n)=U_nu_n(a_n)\},
\]
$L_{U_n}:H_n\to H_n$, $U=\oplus_{n=1}^{\infty }{U_n}$,
$L_U=\oplus_{n=1}^{\infty }{L_{U_n}}$,
$H=\oplus_{n=1}^{\infty }{H_n}$ and
$h=\sup_{n\geq 1}(b_n-a_n)<\infty$.

 Since for all $n\geq1$, $U_n$ is a unitary operator in
$\mathfrak{H}_n$, then $L_{U_n}$ is normal operator
in $H_n$ \cite{Ism2}. Also for
$L_U:D(L_U)\subset H\to H$, the relation
$L_UL_{U^{*}}=L_{U^{*}}L_U$ is true; i.e.,
$L_U$ is a normal operator in $H$.
It is known that if $A_n^{-1}\in C_{p}(\mathfrak{H}_n)$ for $p>1$,
then $L_{U_n}^{-1}\in C_{2p}(H_n), p>1$ for all $n\geq1$\cite{Ism2}.
On the other hand, if $A_n^{-1}\in C_{\infty}(\mathfrak{H}_n), n\geq1$,
 then eigenvalues $\lambda_q(L_{U_n})$, $q\geq1$ of operator
 $L_{U_n}$ is in the form
\[
\lambda_q(L_{U_n})=\lambda_{m}(A_n)+\frac{i}{a_n-b_n}
(\arg\lambda_{m}(U_n^{*}e^{(-A_n(b_n-a_n))})+2k\pi ),
\]
where
$m\geq1$, $k\in{\mathbb Z}$, $n\geq1$,
$q=q(m,k)\in {\mathbb N}$.
Therefore, we have the following result.

\begin{theorem} \label{thm4.10}
If $A=\oplus_{n=1}^{\infty}A_n, \mathfrak{H}
=\oplus_{n=1}^{\infty}\mathfrak{H}_n$ and
$A^{-1}\in C_{p/2}(\mathfrak{H}), 2<p<\infty$, then
$L_U^{-1}\in C_{p}(H)$.
\end{theorem}

\begin{proof}
The operator $L_U$ is a normal in $H$. Consequently, for the
characteristic numbers of normal operator $L_U^{-1}$ an
equality $\mu_q(L_U^{-1})=|\lambda_q(L_U^{-1})|$, $q\geq1$
holds \cite{Dun}. Now we search for convergence of the series
$\sum_{q=1}^{\infty}\mu_q^{p}(L_U^{-1})$, $2<p<\infty$.
\begin{align*}
&\sum_{n=1}^{\infty }\sum_{q=1}^{\infty }\mu_q^{p}(L_{U_n}^{-1})\\
& =  \sum_{n=1}^{\infty }\sum_{k=-\infty}^{\infty}\sum_{m=1}^{\infty }
\Big( \lambda_{m}^2(A_n)+\frac{1}{(b_n-a_n)^2}(\delta(m,n)+2k\pi)^2
 \Big)^{-p/2} \\
& \leq  \sum_{n=1}^{\infty }\sum_{k=-\infty}^{\infty}\sum_{m=1}^{\infty }
\Big( \lambda_{m}^2(A_n)+\frac{4k^2\pi^2}{(b_n-a_n)^2}\Big)^{-p/2} \\
& \leq  \sum_{n=1}^{\infty }\sum_{m=1}^{\infty}
( \lambda_{m}^2(A_n))^{-p/2}
+2\sum_{n=1}^{\infty }\sum_{k=1}^{\infty}\sum_{m=1}^{\infty}
\Big( \lambda_{m}^2(A_n)+\frac{4k^2\pi^2}{(b_n-a_n)^2} \Big)^{-p/2}
\end{align*}
where $\delta(m,n)=arg\lambda_{m}(U_n^{*}e^{(-A_n(b_n-a_n))})$,
$n\geq1$, $m\geq1$. Then from the inequality
$\frac{\mid ts \mid}{t^2+s^2}\leq \frac{1}{2}$ for all
$t,s\in \mathbb{R}\setminus \{0\}$ and last equation we have
the inequality
\begin{align*}
&\sum_{n=1}^{\infty }\sum_{k=1}^{\infty}\sum_{m=1}^{\infty}
\big( \lambda_{m}^2(A_n)+\frac{4k^2\pi^2}{(b_n-a_n)^2} \big)^{-p/2}\\
&\leq 2^{-p}\pi^{-p/2}h^{p/2}
\Big(\sum_{n=1}^{\infty }\sum_{m=1}^{\infty}
|\frac{1}{\lambda_{m}(A_n)}|^{p/2}
\sum_{k=1}^{\infty}| \frac{1}{k} |^{p/2}\Big)
\end{align*}
  Since $A^{-1}\in C_{p/2}(\mathfrak{H})$, then the series
$\sum_{n=1}^{\infty }\sum_{m=1}^{\infty}| \lambda_{m}(A_n) |^{-p/2}$
is convergent. Thus the series
\[
\sum_{n=1}^{\infty }\sum_{k=1}^{\infty}
\sum_{m=1}^{\infty}\Big( \lambda_{m}^2(A_n)
+\frac{4k^2\pi^2}{(b_n-a_n)^2}\Big)^{-p/2}
\]
is also convergent. Then from the relation
\[
\sum_{n=1}^{\infty }\sum_{m=1}^{\infty}|
\lambda_{m}(A_n)|^{-p}\leq\sum_{n=1}^{\infty }\sum_{m=1}^{\infty}|
\lambda_{m}(A_n)|^{-p/2}
\]
  and the convergence of the series
$\sum_{n=1}^{\infty }\sum_{m=1}^{\infty}| \lambda_{m}(A_n)|^{-p/2}$
we get that the series $\sum_{n=1}^{\infty }
\sum_{m=1}^{\infty}| \lambda_{m}(A_n)|^{-p}$ is convergent too.
Consequently the series $\sum_{q=1}^{\infty}\mu_q^{p}(L_U^{-1})$,
$2<p<\infty$ is convergent and thus $L_U^{-1}\in C_{p}(H), 2<p<\infty$.
\end{proof}

Theorems \ref{thm4.8} and \ref{thm4.9}. can be can generalized as follows.

\begin{corollary} \label{coro4.11}
For $n\geq1$, let  $\mathscr{A}_n\in C_{p_n}(\mathscr{H}_n)$,
$1\leq p_n<\infty$ and $p=\sup_{n\geq1} p_n<\infty$.
Then  $\mathscr{A}=\oplus_{n=1}^{\infty}\mathscr{A}_n
\in C_{p}(\mathscr{H})$ if and only if the series
$\sum_{n=1}^{\infty }\sum_{k=1}^{\infty}\mu_k^{p}(\mathscr{A}_n)$
 converges.
\end{corollary}

\begin{proof}
In this case for each $n\geq1$, $\mathscr{A}_n\in C_p(\mathscr{H}_n)$.
So, by using Theorem \ref{thm4.8}(i), the validity of the proposition is clear.
\end{proof}

The following result it is obtained from the above claim.

\begin{corollary} \label{coro4.12}
If for any $n\geq1$, $A_n^{-1}\in C_{{p_n}/2}(\mathfrak{H_n})$,
$2<p=\sup_{n\geq1}p_n<\infty$, and
 $\sum_{n=1}^{\infty }\sum_{k=1}^{\infty}\mu_k^{p/2}(A_n^{-1})<+\infty$, \\
then $L_U^{-1}\in C_p({H})$.
\end{corollary}

\begin{proof}
Indeed,  by Corollary \ref{coro4.11} the operator
$A^{-1}\in C_{p/2}(\mathfrak{H})$ and
$A^{-1}=\oplus_{n=1}^{\infty}A_n^{-1}$,
$\mathfrak{H}=\oplus_{n=1}^{\infty}\mathfrak{H}_n$.
 Consequently, from  Theorem \ref{thm4.10}  it follows that
$L_U^{-1}\in C_p({H})$.
\end{proof}

Corollary \ref{coro4.11} can be generalized the following sense.

\begin{theorem} \label{thm4.13}
Let  $\mathscr{H}=\oplus_{n=1}^{\infty}\mathscr{H}_n$,
$\mathscr{A}=\oplus_{n=1}^{\infty}\mathscr{A}_n \in L(\mathscr{H})$,
for each $n\geq1, 1\leq p_n<+\infty$,
$\mathscr{A}_n\in C_{p_n}(\mathscr{H}_n)$ and
$p=sup_{n\geq1}p_n<+\infty$. If the series
$\sum_{n=1}^{\infty }\sum_{k=1}^{\infty}\mu_k^{p_n}(\mathscr{A}_n)$
is convergent, then $\mathscr{A}\in C_p(\mathscr{H})$.
\end{theorem}

\begin{proof}
If for every $n\geq1$, $\|\mathscr{A}_n\|\leq1$, then the
validity of the above theorem is clear
from Corollary \ref{coro4.11} and the inequality
\[
\sum_{n=1}^{\infty }\sum_{k=1}^{\infty}\mu_k^{p}(\mathscr{A}_n)
\leq \sum_{n=1}^{\infty }\sum_{k=1}^{\infty}
\mu_k^{p_n}(\mathscr{A}_n)<+\infty\,.
\]
Now consider the general case. In this case the operator
$\mathscr{A}$ can be written in form
\[
\mathscr{A}=CB, \quad
C=\oplus_{n=1}^{\infty}((1+\|\mathscr{A}_n\|)E_n), \quad
B=\oplus_{n=1}^{\infty}(\frac{\mathscr{A}_n}{1+\|\mathscr{A}_n\|}),
\]
 here $\sup_{n\geq1}\|(1+\|\mathscr{A}_n\|)E_n\|
=1+\sup_{n\geq1}\|\mathscr{A}_n\|<+\infty$. Then from
Theorem \ref{thm4.2} it is obtained that $C\in L(\mathscr{H})$.

On the other hand, since
 $\|B_n\|=\frac{\|\mathscr{A}_n\|}{1+\|\mathscr{A}_n\|}\leq1$,
$n\geq1$,   and
\[
\sum_{n=1}^{\infty }\sum_{k=1}^{\infty}\mu_k^{p_n}(B_n)
=\sum_{n=1}^{\infty }\sum_{k=1}^{\infty}
\frac{\mu_k^{p_n}(\mathscr{A}_n)}{(1+\|\mathscr{A}_n\|)^{p_n}}
\leq\sum_{n=1}^{\infty }\sum_{k=1}^{\infty}\mu_k^{p_n}(\mathscr{A}_n)
<+\infty,
\]
then from Theorem \ref{thm4.13},  $B\in C_p(\mathscr{H})$ with
$p=\sup_{n\geq1}p_n$. So, on the important result of the operator
theory $\mathscr{A}=CB \in C_p(\mathscr{H})$ \cite{Dun}.
\end{proof}

 The following results give some information in case when
$\sup_{n\geq1}p_n=+\infty$.

\begin{theorem} \label{thm4.14}
 Let $\mathscr{H}=\oplus_{n=1}^{\infty}\mathscr{H}_n$,
$\mathscr{A}=\oplus_{n=1}^{\infty}\mathscr{A}_n \in L(\mathscr{H})$,
for each $n\geq1$, $\mathscr{A}_n\in C_{p_n}(\mathscr{H}_n)$,
$1\leq p_n<+\infty$,
$p_n=\inf\{\alpha\in [1,+\infty): \mathscr{A}_n
\in C_\alpha(\mathscr{H}_n) \}$ and
$\sup_{n\geq1}p_n=+\infty$. Then for every
$p, 1\leq p<+\infty, \mathscr{A}\notin C_p(\mathscr{H})$.
\end{theorem}

\begin{proof}
Assume that for some $p, 1\leq p<+\infty$,
$\mathscr{A}\in C_p(\mathscr{H})$. Then by the Theorem \ref{thm4.8} (ii),
for every $n\geq1$, $\mathscr{A}_n\in C_p(\mathscr{H}_n)$,
$1\leq p<+\infty$. Since
$\sup_{n\geq1}p_n=+\infty$, then there exist
$p_m, 1\leq p_m<+\infty$ such that $p<p_m$ and
$\mathscr{A}_m\in C_{p_m}(\mathscr{H}_m)$. On the other hand
since $C_p\subset C_{p_m}$ and $C_p \ne C_{p_m}$,
$p_n=\inf\{\alpha\in [1,+\infty): \mathscr{A}_n\in
C_\alpha(\mathscr{H}_n) \}$, then
$\mathscr{A}_m \notin C_p(\mathscr{H}_m)$. This contradiction shows
that the claim of the theorem is true.
\end{proof}

\begin{corollary} \label{coro4.15}
Let $\mathscr{H}=\oplus_{n=1}^{\infty}\mathscr{H}_n$,
$\mathscr{A}=\oplus_{n=1}^{\infty}\mathscr{A}_n \in L(\mathscr{H})$,
for every $n\geq1$  $\mathscr{A}_n\in C_{p_n}(\mathscr{H}_n)$,
$1\leq p_n\leq+\infty$,
$p_n=inf\{\alpha\in [1,+\infty): \mathscr{A}_n\in
C_\alpha(\mathscr{H}_n)\}$ and $sup_{n\geq1}p_n=+\infty$.
If for some $m \in \mathbb{N}$,
 $\mathscr{A}_m\in C_{\infty}(\mathscr{H}_m)$, then for every
$p, 1\leq p<+\infty$, $\mathscr{A}\notin C_p(\mathscr{H})$.
\end{corollary}

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\end{document}