# On the essential decreasing of the summation order in the Abel-Lidskii sense

Maksim V. Kukushkin

Russian Academy of Sciences, Kabardino-Balkarian Scientific Center,
Institute of applied Mathematics and Automation, 360051, Nalchik, Russia
<EMAIL_ADDRESS>
###### Abstract

In this paper we have a challenge - how to decrease the summation order in the
Abel-Lidskii sense. Firstly, we clarified the previously obtained results
related to decreasing the summation order to the value equal to the convergent
exponent. Secondly, produced some comprehensive reasonings on the artificial
construction of non-selfadjoint operators and finally represent a novel result
on essential decreasing of the summation order. In the conclusion we made some
remarks stressing the relevance and applicability of the obtained result
broadening the operator class involved in the concept of decomposition of the
evolution equation solution on the root vectors series.

Keywords: Strictly accretive operator; Abel-Lidskii basis property; Schatten-
von Neumann class; convergence exponent; counting function.

MSC 47B28; 47A10; 47B12; 47B10; 34K30; 58D25.

##  1 Introduction

Generally, the concept origins from the well-known fact that the eigenvectors
of the compact selfadjoint operator form a basis in the closure of its range.
The question what happens in the case when the operator is non-selfadjoint is
rather complicated and deserves to be considered as a separate part of the
spectral theory. Basically, the aim of the mentioned part of the spectral
theory are propositions on the convergence of the root vector series in one or
another sense to an element belonging to the closure of the operator range.
Here, we should note when we say a sense we mean Bari, Riesz, Abel (Abel-
Lidskii) senses of the series convergence [1],[8]. A reasonable question that
appears is about minimal conditions that guaranty the desired result, for
instance in the mentioned papers the authors considered a domain of the
parabolic type containing the spectrum of the operator. In the paper [1], non-
salfadjoint operators with the special condition imposed on the numerical
range of values are considered. The main advantage of this result is a weak
condition imposed upon the numerical range of values comparatively with the
sectorial condition (see definition of the sectorial operator). Thus, the
convergence in the Abel-Lidskii sense was established for an operator class
wider than the class of sectorial operators. Here, we make a comparison
between results devoted to operators with the discrete spectra and operators
with the compact resolvent, for they can be easily reformulated from one to
another realm.

The central idea of this paper is to formulate sufficient conditions of the
Abel-Lidskii basis property of the root functions system for a sectorial non-
selfadjoint operator of the special type. Considering such an operator class,
we strengthen a little the condition regarding the semi-angle of the sector,
but weaken a great deal conditions regarding the involved parameters.
Moreover, the central aim generates some prerequisites to consider technical
peculiarities such as a newly constructed sequence of contours of the power
type on the contrary to the Lidskii results [27], where a sequence of the
contours of the exponential type was considered. Thus, we clarify results [27]
devoted to the decomposition on the root vector system of the non-selfadjoint
operator. We use a technique of the entire function theory and introduce a so-
called Schatten-von Neumann class of the convergence exponent. Considering
strictly accretive operators satisfying special conditions formulated in terms
of the norm, using a sequence of contours of the power type, we invent a
peculiar method how to calculate a contour integral involved in the problem in
its general statement. Finally, we produce applications to differential
equations in the abstract Hilbert space. In particular, the existence and
uniqueness theorems for evolution equations with the right-hand side – a
differential operator with a fractional derivative in final terms are covered
by the invented abstract method. In this regard such operator as a Riemann-
Liouville fractional differential operator, Kipriyanov operator, Riesz
potential, difference operator are involved. Note that analysis of the
required conditions imposed upon the right-hand side of the evolution
equations that are in the scope leads us to relevance of the central idea of
the paper. Here, we should note a well-known fact [42],[15] that a particular
interest appears in the case when a senior term of the operator at least is
not selfadjoint, for in the contrary case there is a plenty of results devoted
to the topic wherein the following papers are well-known
[11],[24],[28],[29],[42]. The fact is that most of them deal with a
decomposition of the operator on a sum, where the senior term must be either a
selfadjoint or normal operator. In other cases, the methods of the papers
[16], [15] become relevant and allow us to study spectral properties of
operators whether we have the mentioned above representation or not. We should
remark that the results of the papers [1],[28], applicable to study non-
selfadjoin operators, are based on the sufficiently strong assumption
regarding the numerical range of values of the operator. At the same time, the
methods [15] can be used in the natural way, if we deal with more abstract
constructions formulated in terms of the semigroup theory [17]. The central
challenge of the latter paper is how to create a model representing a
composition of fractional differential operators in terms of the semigroup
theory. We should note that motivation arouse in connection with the fact that
a second order differential operator can be represented as a some kind of a
transform of the infinitesimal generator of a shift semigroup. Here, we should
stress that the eigenvalue problem for the operator was previously studied by
methods of theory of functions [36]. Thus, having been inspired by novelty of
the idea we generalize a differential operator with a fractional integro-
differential composition in the final terms to some transform of the
corresponding infinitesimal generator of the shift semigroup. By virtue of the
methods obtained in the paper [15], we managed to study spectral properties of
the infinitesimal generator transform and obtain an outstanding result –
asymptotic equivalence between the real component of the resolvent and the
resolvent of the real component of the operator. The relevance is based on the
fact that the asymptotic formula for the operator real component can be
established in most cases due to well-known asymptotic relations for the
regular differential operators as well as for the singular ones [39]. It is
remarkable that the results establishing spectral properties of non-
selfadjoint operators allow us to implement a novel approach regarding the
problem of the basis property of root vectors. In its own turn, the
application of results connected with the basis property covers many problems
in the framework of the theory of evolution equations. The abstract approach
to the Cauchy problem for the fractional evolution equation was previously
implemented in the papers [3],[4]. At the same time, the main advantage of
this paper is the obtained formula for the solution of the evolution equation
with the relatively wide conditions imposed upon the right-hand side, where
the derivative at the left-hand side is supposed to be of the fractional
order. This problem appeals to many ones that lie in the framework of the
theory of differential equations, for instance in the paper [34] the solution
of the evolution equation can be obtained in the analytical way if we impose
the conditions upon the right-hand side. We can also produce a number of
papers dealing with differential equations which can be studied by this paper
abstract methods [30], [32], [31], [38], [43]. The latter information gives us
an opportunity to claim that the offered approach is undoubtedly novel and
relevant.

##  2 Preliminaries

Notations and definitions

Let $C,C_{i},\;i\in\mathbb{N}_{0}$ be real constants. We assume that a value
of $C$ is positive and can be different in various formulas but values of
$C_{i}$ are certain. Denote by $\mathrm{int}\,M,\;\mathrm{Fr}\,M$ the interior
and the set of boundary points of the set $M$ respectively. Everywhere
further, if the contrary is not stated, we consider linear densely defined
operators acting on a separable complex Hilbert space $\mathfrak{H}$. Denote
by $\mathcal{B}(\mathfrak{H})$ the set of linear bounded operators on
$\mathfrak{H}.$ Denote by $\tilde{L}$ the closure of an operator $L.$ We
establish the following agreement on using symbols
$\tilde{L}^{i}:=(\tilde{L})^{i},$ where $i$ is an arbitrary symbol. Denote by
$\mathrm{D}(L),\,\mathrm{R}(L),\,\mathrm{N}(L)$ the domain of definition, the
range, and the kernel or null space of an operator $L$ respectively. The
deficiency (codimension) of $\mathrm{R}(L),$ dimension of $\mathrm{N}(L)$ are
denoted by $\mathrm{def}\,L,\;\mathrm{nul}\,L$ respectively. Let $L$ be a
closed operator, for any closable operator $S$ such that $\tilde{S}=L,$ its
domain $\mathrm{D}(S)$ will be called a core of $L.$ Denote by
$\mathrm{D}_{0}(L)$ a core of a closeable operator $L.$ Let $\mathrm{P}(L)$ be
the resolvent set of an operator $L$ and
$R_{L}(\zeta),\,\zeta\in\mathrm{P}(L),\,[R_{L}:=R_{L}(0)]$ denotes the
resolvent of an operator $L.$ Denote by $\lambda_{i}(L),\,i\in\mathbb{N}$ the
eigenvalues of an operator $L.$ Suppose $L$ is a compact operator and
$N:=(L^{\ast}L)^{1/2},\,r(N):={\rm dim}\,\mathrm{R}(N);$ then the eigenvalues
of the operator $N$ are called the singular numbers (s-numbers) of the
operator $L$ and are denoted by $s_{i}(L),\,i=1,\,2,...\,,r(N).$ If
$r(N)<\infty,$ then we put by definition $s_{i}=0,\,i=r(N)+1,2,...\,.$
According to the terminology of the monograph [8] the dimension of the root
vectors subspace corresponding to a certain eigenvalue $\lambda_{k}$ is called
the algebraic multiplicity of the eigenvalue $\lambda_{k}.$ Let $\nu(L)$
denotes the sum of all algebraic multiplicities of an operator $L.$ Denote by
$n(r)$ a function equals to a number of the elements of the sequence
$\\{a_{n}\\}_{1}^{\infty},\,|a_{n}|\uparrow\infty$ within the circle $|z|<r.$
Let $A$ be a compact operator, denote by $n_{A}(r)$ counting function a
function $n(r)$ corresponding to the sequence
$\\{s^{-1}_{i}(A)\\}_{1}^{\infty}.$ Let
$\mathfrak{S}_{p}(\mathfrak{H}),\,0<p<\infty$ be a Schatten-von Neumann class
and $\mathfrak{S}_{\infty}(\mathfrak{H})$ be the set of compact operators.
Suppose $L$ is an operator with a compact resolvent and $s_{n}(R_{L})\leq
C\,n^{-\mu},\,n\in\mathbb{N},\,0\leq\mu<\infty;$ then we denote by $\mu(L)$
order of the operator $L$ in accordance with the definition given in the paper
[42]. Denote by
$\mathfrak{Re}L:=\left(L+L^{*}\right)/2,\,\mathfrak{Im}L:=\left(L-L^{*}\right)/2i$
the real and imaginary components of an operator $L$ respectively. In
accordance with the terminology of the monograph [10] the set
$\Theta(L):=\\{z\in\mathbb{C}:z=(Lf,f)_{\mathfrak{H}},\,f\in\mathrm{D}(L),\,\|f\|_{\mathfrak{H}}=1\\}$
is called the numerical range of an operator $L.$ An operator $L$ is called
sectorial if its numerical range belongs to a closed sector
$\mathfrak{L}_{\iota}(\theta):=\\{\zeta:\,|\arg(\zeta-\iota)|\leq\theta<\pi/2\\},$
where $\iota$ is the vertex and $\theta$ is the semi-angle of the sector
$\mathfrak{L}_{\iota}(\theta).$ If we want to stress the correspondence
between $\iota$ and $\theta,$ then we will write $\theta_{\iota}.$ An operator
$L$ is called bounded from below if the following relation holds
$\mathrm{Re}(Lf,f)_{\mathfrak{H}}\geq\gamma_{L}\|f\|^{2}_{\mathfrak{H}},\,f\in\mathrm{D}(L),\,\gamma_{L}\in\mathbb{R},$
where $\gamma_{L}$ is called a lower bound of $L.$ An operator $L$ is called
accretive if $\gamma_{L}=0.$ An operator $L$ is called strictly accretive if
$\gamma_{L}>0.$ An operator $L$ is called m-accretive if the next relation
holds
$(A+\zeta)^{-1}\in\mathcal{B}(\mathfrak{H}),\,\|(A+\zeta)^{-1}\|\leq(\mathrm{Re}\zeta)^{-1},\,\mathrm{Re}\zeta>0.$
An operator $L$ is called m-sectorial if $L$ is sectorial and $L+\beta$ is
m-accretive for some constant $\beta.$ An operator $L$ is called symmetric if
one is densely defined and the following equality holds
$(Lf,g)_{\mathfrak{H}}=(f,Lg)_{\mathfrak{H}},\,f,g\in\mathrm{D}(L).$
Everywhere further, unless otherwise stated, we use notations of the papers
[8], [10], [12], [13], [41].

Series expansion

Consider a compact operator $B,$ we can choose a Jordan basis in the finite
dimensional root subspace $\mathfrak{N}_{q}$ corresponding to the eigenvalue
$\mu_{q}$ that consists of Jordan chains of eigenvectors and root vectors of
the operator $B.$ Each chain has a set of numbers

$e_{q_{\xi}},e_{q_{\xi}+1},...,e_{q_{\xi}+k},\,k\in\mathbb{N}_{0},$ (1)

where $e_{q_{\xi}},\,\xi=1,2,...,m$ are the eigenvectors corresponding to the
eigenvalue $\mu_{q}.$ Considering the sequence $\\{\mu_{q}\\}_{1}^{\infty}$ of
the eigenvalues of the operator $B$ and choosing a Jordan basis in each
corresponding space $\mathfrak{N}_{q},$ we can arrange a system of vectors
$\\{e_{i}\\}_{1}^{\infty}$ which we will call a system of the root vectors or
following Lidskii a system of the major vectors of the operator $B.$ Assume
that $e_{1},e_{2},...,e_{n_{q}}$ is the Jordan basis in the subspace
$\mathfrak{N}_{q}.$ We can prove easily (see [27, p.14]) that there exists a
corresponding biorthogonal basis $g_{1},g_{2},...,g_{n_{q}}$ in the subspace
$\mathfrak{M}_{q}^{\perp},$ where $\mathfrak{M}_{q},$ is a subspace wherein
the operator $B-\mu_{q}I$ is invertible. Using the reasonings [18], we
conclude that $\\{g_{i}\\}_{1}^{n_{q}}$ consists of the Jordan chains of the
operator $B^{\ast}$ which correspond to the Jordan chains (1), more detailed
information can be found in [18]. It is not hard to prove that the set
$\\{g_{\nu}\\}^{n_{j}}_{1},\,j\neq i$ is orthogonal to the set
$\\{e_{\nu}\\}_{1}^{n_{i}}$ (see [18]). Gathering the sets
$\\{g_{\nu}\\}^{n_{j}}_{1},\,j=1,2,...,$ we can obviously create a
biorthogonal system $\\{g_{n}\\}_{1}^{\infty}$ with respect to the system of
the major vectors of the operator $B.$ Consider a sum

$\mathcal{P}_{\nu}(\alpha,t)f:=\sum\limits_{q=N_{\nu}+1}^{N_{\nu+1}}\sum\limits_{\xi=1}^{m(q)}\sum\limits_{i=0}^{k(q_{\xi})}e_{q_{\xi}+i}c_{q_{\xi}+i}(t)$
(2)

where $k(q_{\xi})+1$ is a number of elements in the $q_{\xi}$-th Jourdan
chain, $m(q)$ is a geometrical multiplicity of the $q$-th eigenvalue,

$c_{q_{\xi}+i}(t)=e^{-\lambda^{\alpha}_{q}t}\sum\limits_{m=0}^{k(q_{\xi})-i}H_{m}(\varphi,\lambda_{q},t)c_{q_{\xi}+i+m},\,i=0,1,2,...,k(q_{\xi}),$
(3)

$c_{q_{\xi}+i}=(f,g_{q_{\xi}+k-i})/(e_{q_{\xi}+i},g_{q_{\xi}+k-i}),$
$\lambda_{q}=1/\mu_{q}$ is a characteristic number corresponding to
$e_{q_{\xi}},$

$H_{m}(\alpha,\lambda,t):=\frac{e^{\lambda^{\alpha}t}}{m!}\cdot\lim\limits_{\zeta\rightarrow
1/\lambda}\frac{d^{m}}{d\zeta^{\,m}}\left\\{e^{-\zeta^{-\alpha}t}\right\\},\;m=0,1,2,...\,,\,.$

More detailed information on the considered above Jordan chains can be found
in [18].

##  3 Main results

1\. Decreasing of the summation order to the convergent exponent

Recall that an arbitrary compact operator $B$ can be represented as a series
on the basis vectors due to the so-called polar decomposition $B=U|B|,$ where
$U$ is a concrete unitary operator, $|B|:=(B^{\ast}B)^{1/2},$ i.e. using the
system of eigenvectors $\\{e_{n}\\}_{1}^{\infty},$ we have

$Bf=\sum\limits_{n=1}^{\infty}s_{n}(B)(f,e_{n})g_{n},$ (4)

where $e_{n},s_{n}$ are the eigenvectors and eigenvalues of the operator $|B|$
respectively, $g_{n}=Ue_{n}.$ It is clear that the letter elements form a an
orthonormal system due to the major property of the unitary operator. The main
concept that lies in the framework of our consideration relates to the
following statement.

$(\mathrm{S}1)$ Under the assumptions
$B\in\mathfrak{S}^{\star}_{\rho},\,0<\rho<\infty,\,\Theta(B)\subset\mathfrak{L}_{0}(\theta),$
a sequence of natural numbers $\\{N_{\nu}\\}_{0}^{\infty}$ can be chosen so
that

$\frac{1}{2\pi i}\int\limits_{\vartheta(B)}e^{-\lambda^{\alpha}t}B(I-\lambda
B)^{-1}fd\lambda=\sum\limits_{\nu=0}^{\infty}\mathcal{P}_{\nu}(\alpha,t)f,$

where
$\vartheta(B):=\left\\{\lambda:\;|\lambda|=r>0,\,|\mathrm{arg}\lambda|\leq\theta+\varepsilon\right\\}\cup\left\\{\lambda:\;|\lambda|>r,\;|\mathrm{arg}\lambda|=\theta+\varepsilon\right\\},\,\alpha$
is a positive number. Moreover

$\sum\limits_{\nu=0}^{\infty}\left\|\mathcal{P}_{\nu}(\alpha,t)f\right\|<\infty.$
(5)

Note that in accordance with the Lemma 3 [18], we have

$\ln r\frac{n(r)}{r^{\rho}}\rightarrow 0,\Longrightarrow\ln
r\left(\int\limits_{0}^{r}\frac{n(t)}{t^{p+1}}dt+r\int\limits_{r}^{\infty}\frac{n(t)}{t^{p+2}}dt\right)r^{p-\rho}\rightarrow
0,\,r\rightarrow\infty,$ (6)

where $\rho$ is a convergence exponent (non integer) defined as follows
$\rho:=\inf\lambda,$

$\sum\limits_{n=1}^{\infty}s_{n}^{\,\lambda}(B)<\infty,$

$n(t)$ is a counting function corresponding to the singular numbers of the
operator $B,$ the number $\lambda=p+1$ is the smallest natural number for
which the latter series is convergent. To produce an operator class with more
subtle asymptotics, we can deal with representation (4) directly imposing
conditions upon the singular numbers instead of considering Hermitian real
component. Assume that the sequence of singular numbers has a non integer
convergent exponent $\rho$ and consider the following condition

$(\ln^{1+1/\rho}x)^{\prime}_{s^{-1}_{m}(B)}=o(m^{-1/\rho}).$

This gives us

$\frac{m\ln s^{-1}_{m}(B)}{s^{-\rho}_{m}(B)}\leq
C\cdot\alpha_{m},\;\alpha_{m}\rightarrow 0,\,m\rightarrow\infty.$

Taking into account the facts
$n(s^{-1}_{m})=m;\,\;n(r)=n(s^{-1}_{m}),\,s^{-1}_{m}<r<s^{-1}_{m+1},$ using
the monotonous property of the functions, we get

$\ln r\frac{n(r)}{r^{\rho}}<C\cdot\alpha_{m},\;s^{-1}_{m}<r<s^{-1}_{m+1},$ (7)

i.e. we obtain the following implication

$(\ln^{1+1/\rho}x)^{\prime}_{s^{-1}_{m}(B)}=o(m^{-1/\rho}),\Longrightarrow\left(\ln
r\frac{n(r)}{r^{\rho}}\rightarrow 0\right),$

therefore the premise in (6) holds and as a result the consequence holds.
Absolutely analogously, we can obtain the implication

$s^{-1}_{m}(B)=o(m^{-1/\rho}),\Longrightarrow\left(\frac{n(r)}{r^{\rho}}\rightarrow
0\right),$ (8)

here the corresponding example is given by $s_{m}=(m\ln m)^{-1/\rho}$ and the
reader can verify directly that $B\in\mathfrak{S}_{\rho}^{\star}.$ Using these
facts, we can reformulate Theorem 2, 4 [18] for an artificially constructed
compact operator. In this paper, we represent modified proofs since there are
some difficulties that require refinement, moreover, we produce the variant of
the proof of Theorem 4 [18] corresponding to the case
$\Theta(B)\subset\mathfrak{L}_{0}(\theta),\,\theta<\pi/2\alpha,$ that is not
given in [18].

######  Theorem 1.

Assume that $B$ is a compact operator,
$\Theta(B)\subset\mathfrak{L}_{0}(\theta),\,\theta<\pi/2\alpha,\;B\in\mathfrak{S}^{\star}_{\alpha},$

$s_{n}(B)=o(n^{-1/\alpha}),$

where $\alpha$ is positive non integer. Then the statement of Theorem 2 [18]
remains true, i.e. statement S1 holds.

###### Proof.

Applying corollary of the well-known Allakhverdiyev theorem, see Corollary 2.2
[8] (Chapter II, $\S$ 2.3), we get

$s_{(m+1)(k-1)+1}(B^{m+1})\leq s_{k}^{m+1}(B),\,m\in\mathbb{N}.$

Note that $n_{B}(r)=k,$ where $s_{k}^{-1}(B)<r<s_{k+1}^{-1}(B).$ Therefore

$s^{-1}_{(m+1)k+1}(B^{m+1})\geq s_{k+1}^{-m-1}(B)>r^{m+1},$

and we have

$n_{B^{m+1}}(r^{m+1})\leq(m+1)k=(m+1)n_{B}(r),$ (9)

hence using (8), we obtain

$\frac{n_{B^{m+1}}(r^{m+1})}{r^{\alpha}}\rightarrow
0,\,r\rightarrow\infty,\,m=[\alpha].$

The rest part of the proof is absolutely analogous to Theorem 2 [18]. ∎

######  Theorem 2.

Assume that $B$ is a compact operator,
$\Theta(B)\subset\mathfrak{L}_{0}(\theta),\,\theta<\pi/2\alpha,\;B\in\mathfrak{S}^{\star}_{\alpha},$

$(\ln^{1+1/\alpha}x)^{\prime}_{s^{-1}_{m}(B)}=o(m^{-1/\alpha}).$

where $\alpha$ is positive non integer. Then the statement of the Theorem 4
[18] remains true, i.e. statement S1 holds, moreover we have the following
gaps between the eigenvalues

$|\lambda_{N_{\nu}+1}|^{-1}-|\lambda_{N_{\nu}}|^{-1}\geq
K|\lambda_{N_{\nu}}|^{\alpha-1},\,K>0,$

and the eigenvalues corresponding to the partial sums become united to the
groups due to the distance of the power type

$|\lambda_{N_{\nu}+k}|^{-1}-|\lambda_{N_{\nu}+k-1}|^{-1}\leq
K|\lambda_{N_{\nu}+k}|^{\alpha-1},\;k=2,3,...,N_{\nu+1}-N_{\nu}.$

###### Proof.

Using the theorem conditions, we obtain easily

$s_{m}(B)=o(m^{-1/\alpha}),\;m\rightarrow\infty,$

In accordance with Corollary 3.2 [8] (Chapter II, $\S$ 3.3), it gives us

$|\lambda_{m}(B)|=o(m^{-1/\alpha}),\;m\rightarrow\infty.$

In consequence of the latter relation, we have $\mu_{m}/m^{1/\alpha}\geq C,$
where $\mu_{m}=\lambda^{-1}_{m}(B).$ Using the implication

$\lim\limits_{n\rightarrow\infty}(\mu_{n+1}-\mu_{n})/\mu^{(p-1)/p}_{n}=0,\;\Longrightarrow\lim\limits_{n\rightarrow\infty}\mu_{n}/n^{p}=0,\;p>0,$

the proof of which can be found in the proof of Lemma 2 [1], we can claim that
there exists a subsequence $\\{\lambda_{N_{\nu}}\\}_{\nu=0}^{\infty},$ such
that

$|\mu_{N_{\nu}+1}|-|\mu_{N_{\nu}}|\geq K|\mu_{N_{\nu}}|^{1-\alpha},\,K>0.$

Indeed, for the case $p\geq 1,$ making the auxiliary denotation
$\xi_{n}:=\mu^{1/p}_{n},$ we can rewrite the assumed implication as follows

$\lim\limits_{n\rightarrow\infty}(\xi^{p}_{n+1}-\xi^{p}_{n})/\xi^{p-1}_{n}=0,\;\Longrightarrow\lim\limits_{n\rightarrow\infty}\xi_{n}/n=0,$

denoting $\Delta\xi_{n}:=\xi_{n+1}-\xi_{n},$ we can easily obtain, due to the
application of the mean value theorem (or Lagrange theorem), the estimate
$\xi^{p}_{n+1}-\xi^{p}_{n}\geq p\,\xi^{p+1}_{n}\Delta\xi_{n}.$ It is clear
that it gives us the implication

$\left\\{\lim\limits_{n\rightarrow\infty}\Delta\xi_{n}=0,\Rightarrow\lim\limits_{n\rightarrow\infty}\xi_{n}/n=0\right\\}\Rightarrow\left\\{\lim\limits_{n\rightarrow\infty}(\mu_{n+1}-\mu_{n})/\mu^{(p-1)/p}_{n}=0,\;\Longrightarrow\lim\limits_{n\rightarrow\infty}\mu_{n}/n^{p}=0\right\\},$

however the fact that the implication in the first term holds can be
established with no difficulties. The proof corresponding to the case $p<1$ is
absolutely analogous, we just need notice

$\lim\limits_{n\rightarrow\infty}(\xi^{p}_{n+1}-\xi^{p}_{n})/\xi^{p-1}_{n}=0,\Rightarrow\lim\limits_{n\rightarrow\infty}\xi^{p}_{n}/\xi^{p}_{n+1}=1,$

and deal with the following implication

$\lim\limits_{n\rightarrow\infty}(\xi^{p}_{n+1}-\xi^{p}_{n})/\xi^{p-1}_{n+1}=0,\;\Longrightarrow\lim\limits_{n\rightarrow\infty}\xi_{n}/n=0$

in the analogous way.

Let us find $\delta_{\nu}$ from the condition
$R_{\nu}=K|\mu_{N_{\nu}}|^{1-\alpha}+|\mu_{N_{\nu}}|,\,R_{\nu}(1-\delta_{\nu})=|\mu_{N_{\nu}}|,$
then $\delta_{\nu}^{-1}=1+K^{-1}|\mu_{N_{\nu}}|^{\alpha}.$ Note that by virtue
of such a choice, we have $R_{\nu}<R_{\nu+1}(1-\delta_{\nu+1}).$

In accordance with the theorem condition, established above relation (7), we
have $n_{B}(r)=o(r^{\alpha}/\ln r).$ Consider a contour

$\vartheta(B):=\left\\{\lambda:\;|\lambda|=r_{0}>0,\,|\mathrm{arg}\lambda|\leq\theta+\varepsilon\right\\}\cup\left\\{\lambda:\;|\lambda|>r_{0},\;|\mathrm{arg}\lambda|=\theta+\varepsilon\right\\},$

here the number $r$ is chosen so that the circle with the corresponding radios
does not contain values $\mu_{n},\,n\in\mathbb{N}.$ Applying Lemma 5 [18], we
have that there exists such a sequence
$\\{\tilde{R}_{\nu}\\}_{0}^{\infty}:\;(1-\delta_{\nu})R_{\nu}<\tilde{R}_{\nu}<R_{\nu}$
that the following estimate holds

$\|(I-\lambda B)^{-1}\|_{\mathfrak{H}}\leq
e^{\gamma(|\lambda|)|\lambda|^{\alpha}}|\lambda|^{m},\,m=[\alpha],\,|\lambda|=\tilde{R}_{\nu},$

where

$\gamma(|\lambda|)=\beta(|\lambda|^{m+1})+(2+\ln\\{4e/\delta_{\nu}\\})\beta(\sigma|\lambda|^{m+1})\,\sigma^{\alpha/(m+1)},\,\sigma:=\left(\frac{2e}{1-\delta_{0}}\right)^{m+1},\,|\lambda|=\tilde{R}_{\nu},$
$\;\beta(r)=r^{-\frac{\alpha}{m+1}}\left(\int\limits_{0}^{r}\frac{n_{B^{m+1}}(t)dt}{t}+r\int\limits_{r}^{\infty}\frac{n_{B^{m+1}}(t)dt}{t^{2}}\right).$

Here we should explain that that the opportunity to obtain this estimate is
based upon the estimation of the Fredholm determinant
$\Delta_{B^{m+1}}(\lambda^{m+1})$ absolute value from bellow (see [27, p.8]).
In its own turn the latter is implemented via the general estimate from bellow
of the absolute value of a holomorphic function (Theorem 11 [26, p.33]).
However, to avoid any kind of misunderstanding let us implement a scheme of
reasonings of Lemma 5 consistently.

Using the theorem condition, we have
$B\in\mathfrak{S}_{\alpha+\varepsilon},\,\varepsilon>0.$ Obviously, we have

$(I-\lambda B)^{-1}=(I-\lambda^{m+1}B^{m+1})^{-1}(I+\lambda
B+\lambda^{2}B^{2}+...+\lambda^{m}B^{m}).$ (10)

In accordance with Lemma 3 [27], for sufficiently small $\varepsilon>0,$ we
have

$\sum\limits_{n=1}^{\infty}s^{\frac{\alpha+\varepsilon}{m+1}}_{n}(B^{m+1})\leq\sum\limits_{n=1}^{\infty}s^{\alpha+\varepsilon}_{n}(B)<\infty,$

Applying inequality (1.27) [27, p.10] (since $\rho/(m+1)<1$), we get

$\|\Delta_{B^{m+1}}(\lambda^{m+1})(I-\lambda^{m+1}B^{m+1})^{-1}\|\leq
C\prod\limits_{i=1}^{\infty}\\{1+|\lambda^{m+1}s_{i}(B^{m+1})|\\},$

where $\Delta_{B^{m+1}}(\lambda^{m+1})$ is a Fredholm determinant of the
operator $B^{m+1}$ (see [27, p.8]). Rewriting the formulas in accordance with
the terms of the entire functions theory, we get

$\prod\limits_{n=1}^{\infty}\\{1+|\lambda^{m+1}s_{n}(B^{m+1})|\\}=\prod\limits_{n=1}^{\infty}G\left(|\lambda|^{m+1}/a_{n},p\right),$

where $a_{n}:=-s^{-1}_{n}(B^{m+1}),\,p=[\alpha/(m+1)]=0,$ using Lemma 2 [18],
we get

$\prod\limits_{n=1}^{\infty}G\left(r^{m+1}/a_{n},p\right)\leq
e^{\beta(r^{m+1})r^{\alpha}},\,r>0.$

In accordance with Theorem 11 [26, p.33], and regarding to the case, the
following estimate holds

$|\Delta_{B^{m+1}}(\lambda^{m+1})|\geq
e^{-(2+\ln\\{2e/3\eta\\})\ln\xi_{m}},\;\xi_{m}=\\!\\!\\!\max\limits_{\psi\in[0,2\pi/(m+1)]}|\Delta_{B^{m+1}}([2eR_{\nu}e^{i\psi}]^{m+1})|,\,|\lambda|\leq
R_{\nu},$

except for the exceptional set of circles with the sum of radii less that
$4\eta R_{\nu},$ where $\eta$ is an arbitrary number less than $2e/3.$ Thus to
find the desired circle $\lambda=e^{i\psi}\tilde{R}_{\nu}$ belonging to the
ring, i.e. $R_{\nu}(1-\delta_{\nu})<\tilde{R}_{\nu}<R_{\nu},$ we have to
choose $\eta$ satisfying the inequality

$4\eta
R_{\nu}<R_{\nu}-R_{\nu}(1-\delta_{\nu})=\delta_{\nu}R_{\nu};\;\eta<\delta_{\nu}/4,$

for instance let us choose $\eta_{\nu}=\delta_{\nu}/6,$ here we should note
that $\delta_{\nu}$ tends to zero, this is why without loss of generality of
reasonings we are free in such a choice since the inequality $\eta_{\nu}<3e/2$
would be satisfied for a sufficiently large $\nu.$ Under such assumptions, we
can rewrite the above estimate in the form

$|\Delta_{B^{m+1}}(\lambda^{m+1})|\geq
e^{-(2+\ln\\{4e/\delta_{\nu}\\})\ln\xi_{m}},\;|\lambda|=\tilde{R}_{\nu}.$

Note that in accordance with the estimate (1.21) [27, p.10], Lemma 2 [18] we
have

$|\Delta_{B^{m+1}}(\lambda)|\leq\prod\limits_{i=1}^{\infty}\\{1+|\lambda
s_{i}(B^{m+1})|\\}=\prod\limits_{n=1}^{\infty}G\left(|\lambda|/a_{n},p\right)\leq
e^{\beta(|\lambda|)|\lambda|^{\alpha/(m+1)}}.$

Using this estimate, we obtain

$\xi_{m}\leq e^{\beta([2eR_{\nu}]^{m+1})(2eR_{\nu})^{\alpha}}.$

Substituting, we get

$|\Delta_{B^{m+1}}(\lambda^{m+1})|^{-1}\leq
e^{(2+\ln\\{4e/\delta_{\nu}\\})\beta([2eR_{\nu}]^{m+1})(2eR_{\nu})^{\rho}},\;|\lambda|=\tilde{R}_{\nu}.$

Having noticed the facts

$1-\delta_{\nu}=\frac{|\mu_{N_{\nu}}|^{\alpha}}{K+|\mu_{N_{\nu}}|^{\alpha}}=1-\frac{K}{K+|\mu_{N_{\nu}}|^{\alpha}}\geq
1-\frac{K}{K+|\mu_{N_{0}}|^{\alpha}}=1-\delta_{0};$
$\;R_{\nu}<\tilde{R}_{\nu}(1-\delta_{\nu})^{-1}\leq\tilde{R}_{\nu}(1-\delta_{0})^{-1},$

we obtain

$|\Delta_{B^{m+1}}(\lambda^{m+1})|^{-1}\leq
e^{(2+\ln\\{4e/\delta_{\nu}\\})\beta(\sigma\tilde{R}^{m+1}_{\nu})\tilde{R}^{\alpha}_{\nu}\sigma^{\alpha/(m+1)}},\;|\lambda|=\tilde{R}_{\nu}.$

Combining the above estimates, we get

$\|\Delta_{B^{m+1}}(\lambda^{m+1})(I-\lambda^{m+1}B^{m+1})^{-1}\|=|\Delta_{B^{m+1}}(\lambda^{m+1})|\cdot\|(I-\lambda^{m+1}B^{m+1})^{-1}\|\leq
e^{\beta(|\lambda|^{m+1})|\lambda|^{\alpha}};$
$\|(I-\lambda^{m+1}B^{m+1})^{-1}\|\leq
e^{\beta(|\lambda|^{m+1})|\lambda|^{\alpha}}|\Delta_{B^{m+1}}(\lambda^{m+1})|^{-1}=e^{\gamma(|\lambda|)|\lambda|^{\alpha}},\,|\lambda|=\tilde{R}_{\nu}.$

Consider relation (10), we have

$\|(I-\lambda B)^{-1}\|\leq\|(I-\lambda^{m+1}B^{m+1})^{-1}\|\cdot\|(I+\lambda
B+\lambda^{2}B^{2}+...+\lambda^{m}B^{m})\|\leq$
$\leq\|(I-\lambda^{m+1}B^{m+1})^{-1}\|\cdot\frac{|\lambda|^{m+1}\|B\|^{m+1}-1}{|\lambda|\cdot\|B\|-1}\leq
Ce^{\gamma(|\lambda|)|\lambda|^{\alpha}}|\lambda|^{m},\,|\lambda|=\tilde{R}_{\nu}.$

Applying the obtained estimate for the norm, we can claim that for a
sufficiently small $\delta>0,$ there exists an arch
$\tilde{\gamma}_{\nu}:=\\{\lambda:\;|\lambda|=\tilde{R}_{\nu},\,|\mathrm{arg}\lambda|<\theta+\varepsilon\\}$
in the ring $(1-\delta_{\nu})R_{\nu}<|\lambda|<R_{\nu},$ on which the
following estimate holds

$J_{\nu}:=\left\|\,\int\limits_{\tilde{\gamma}_{\nu}}e^{-\lambda^{\alpha}t}B(I-\lambda
B)^{-1}fd\lambda\,\right\|\leq\,\int\limits_{\tilde{\gamma}_{\nu}}e^{-t\mathrm{Re}\lambda^{\alpha}}\left\|B(I-\lambda
B)^{-1}f\right\||d\lambda|\leq$
$\leq\|f\|Ce^{\gamma(|\lambda|)|\lambda|^{\alpha}}|\lambda|^{m}\int\limits_{-\theta-\varepsilon}^{\theta+\varepsilon}e^{-t\mathrm{Re}\lambda^{\alpha}}d\,\mathrm{arg}\lambda,\,|\lambda|=\tilde{R}_{\nu}.$

Using the theorem conditions, we get
$\,|\mathrm{arg}\lambda|<\pi/2\alpha,\,\lambda\in\tilde{\gamma}_{\nu},\,\nu=0,1,2,...\,,$
we get

$\mathrm{Re}\lambda^{\alpha}\geq|\lambda|^{\alpha}\cos\left[(\pi/2\alpha-\delta)\alpha\right]=|\lambda|^{\alpha}\sin\alpha\delta,$

where $\delta$ is a sufficiently small number. Therefore

$J_{\nu}\leq
Ce^{|\lambda|^{\alpha}\\{\gamma(|\lambda|)-t\sin\alpha\delta\\}}|\lambda|^{m},\,m=[\alpha],\,|\lambda|=\tilde{R}_{\nu}.$
(11)

It is clear that within the contour $\vartheta(B),$ between the arches
$\tilde{\gamma}_{\nu},\tilde{\gamma}_{\nu+1}$ (we denote the boundary of this
domain by $\gamma_{\nu}$) there lie the eigenvalues only for which the
following relation holds

$|\mu_{N_{\nu}+k}|-|\mu_{N_{\nu}+k-1}|\leq
K|\mu_{N_{\nu}+k-1}|^{1-\alpha},\;k=2,3,...,N_{\nu+1}-N_{\nu}.$

Using Lemma 8, [18], we obtain a relation

$\frac{1}{2\pi i}\int\limits_{\gamma_{\nu}}e^{-\lambda^{\alpha}t}B(I-\lambda
B)^{-1}fd\lambda=\sum\limits_{q=N_{\nu}+1}^{N_{\nu+1}}\sum\limits_{\xi=1}^{m(q)}\sum\limits_{i=0}^{k(q_{\xi})}e_{q_{\xi}+i}c_{q_{\xi}+i}(t).$

It is clear that to obtain the desired result, we should prove that the series
composed of the above terms converges. Consider the following auxiliary
denotations originated from the reasonings
$\gamma_{\nu_{+}}:=\\{\lambda:\,\tilde{R}_{\nu}<|\lambda|<\tilde{R}_{\nu+1},\,\mathrm{arg}\lambda=\theta+\varepsilon\\},\,\gamma_{\nu_{-}}:=\\{\lambda:\,\tilde{R}_{\nu}<|\lambda|<\tilde{R}_{\nu+1},\,\mathrm{arg}\lambda=-\theta-\varepsilon\\},$

$J^{+}_{\nu}:=\left\|\,\int\limits_{\gamma_{\nu_{+}}}e^{-\lambda^{\alpha}t}B(I-\lambda
B)^{-1}fd\lambda\,\right\|,\;J^{-}_{\nu}:=\left\|\,\int\limits_{\gamma_{\nu_{-}}}e^{-\lambda^{\alpha}t}B(I-\lambda
B)^{-1}fd\lambda\,\right\|,$

we have

$\left\|\int\limits_{\gamma_{\nu}}e^{-\lambda^{\alpha}t}B(I-\lambda
B)^{-1}fd\lambda\right\|\leq J_{\nu}+J_{\nu+1}+J^{+}_{\nu}+J^{-}_{\nu}.$

Thus, it is clear that to establish S1, it suffices to establish the facts

$\sum\limits_{\nu=0}^{\infty}J_{\nu}<\infty,\;\sum\limits_{\nu=0}^{\infty}J^{+}_{\nu}<\infty,\;\sum\limits_{\nu=0}^{\infty}J^{-}_{\nu}<\infty.$
(12)

Consider the right-hand side of formula (11). Substituting
$\delta^{-1}_{\nu},$ we have

$\ln\\{4e/\delta_{\nu}\\}=1+\ln
4+\ln\\{1+K^{-1}|\mu_{N_{\nu}}|^{\alpha}\\}\leq C\ln|\mu_{N_{\nu}}|.$

Hence, to obtain the desired result we should prove that
$\beta(\sigma|\mu_{N_{\nu}}|^{m+1})\ln|\mu_{N_{\nu}}|\rightarrow
0,\,\nu\rightarrow\infty.$ Note that in accordance with relation (6), we can
prove the latter relation, if we show that

$\ln r\frac{n_{B^{m+1}}(r^{m+1})}{r^{\alpha}}\rightarrow
0,\;r\rightarrow\infty.$ (13)

In accordance with (7), we have

$(\ln^{1+1/\alpha}x)^{\prime}_{s^{-1}_{m}(B)}=o(m^{-1/\alpha}),\Longrightarrow\left(\ln
r\frac{n(r)}{r^{\rho}}\,\rightarrow 0,\,r\rightarrow\infty\right),$

Applying (9), we have

$n_{B^{m+1}}(r^{m+1})\leq(m+1)n_{B}(r),$

hence

$\ln r\frac{n_{B^{m+1}}(r^{m+1})}{r^{\alpha}}\leq(m+1)\ln
r\frac{n_{B}(r)}{r^{\alpha}},$

what gives us the desired result, i.e. we compleat the proof of the first
relation (12).

In accordance with Lemma 6 [18], we can claim that on each ray $\zeta$
containing the point zero and not belonging to the sector
$\mathfrak{L}_{0}(\theta)$ as well as the real axis, we have

$\|(I-\lambda B)^{-1}\|\leq\frac{1}{\sin\psi},\,\lambda\in\zeta,$

where
$\,\psi=\min\\{|\mathrm{arg}\zeta-\theta|,|\mathrm{arg}\zeta+\theta|\\}.$
Applying this estimate, the established above estimate
$\mathrm{Re}\lambda^{\alpha}\geq|\lambda|^{\alpha}\sin\alpha\delta,$ we get

$J^{+}_{\nu}\leq
C\|f\|\\!\\!\\!\int\limits_{\tilde{R}_{\nu}}^{\tilde{R}_{\nu+1}}e^{-t\mathrm{Re}\lambda^{\alpha}}|d\lambda|\leq
C\|f\|e^{-t\tilde{R}_{\nu}^{\alpha}\sin\alpha\delta}\\!\\!\\!\int\limits_{\tilde{R}_{\nu}}^{\tilde{R}_{\nu+1}}|d\lambda|=$
$=C\|f\|e^{-t\tilde{R}_{\nu}^{\alpha}\sin\alpha\delta}(\tilde{R}_{\nu+1}-\tilde{R}_{\nu}).$

It obvious, that the same estimate can be obtained for $J^{-}_{\nu}.$
Therefore, the second and the third relations (12) hold. The proof is
complete. ∎

######  Remark 1.

Application of the results established in paragraph 1, under the imposed
sectorial condition upon the compact operator $B,$ gives us

$\left\\{(\ln^{1+1/\rho}x)^{\prime}_{\lambda^{-1}_{m}(\mathfrak{Re}B)}=o(m^{-1/\rho})\right\\}\Longrightarrow\left\\{(\ln^{1+1/\rho}x)^{\prime}_{s^{-1}_{m}(B)}=o(m^{-1/\rho})\right\\},$

what becomes clear if we recall $s_{m}(B)\leq C\lambda_{m}(\mathfrak{Re}B).$
However to establish the equivalence

$B\in\mathfrak{S}^{\star}_{\rho}\Longleftrightarrow\mathfrak{Re}B\in\mathfrak{S}^{\star}_{\rho},$

we require the estimate from bellow $s_{m}(B)\geq
C\lambda_{m}(\mathfrak{Re}B),$ which in its own turn can be established due to
a more subtle technique and under stronger conditions regarding the operator
$B.$ In particular the second representation theorem is involved and
conditions upon the matrix coefficients of the operator $B$ are imposed in
accordance with which they should decrease sufficiently fast. In addition, the
application of Theorem 5 [15] (if $B$ satisfies its conditions) gives us the
relation
$\lambda_{n}(\mathfrak{Re}B)\asymp\lambda_{n}(H^{-1}),\,H=\mathfrak{Re}B^{-1},$
the latter allows one to deal with unbounded operators reformulating Theorem 2
in terms of the Hermitian real component. It should be noted that the made
remark remains true regarding the implication

$\left\\{\lambda_{m}(\mathfrak{Re}B)=o(m^{-1/\rho})\right\\}\Longrightarrow\left\\{s_{m}(B)=o(m^{-1/\rho})\right\\},$

and since the scheme of the reasonings is the same we left them to the reader.

######  Example 1.

Using decomposition (4), let us construct artificially the sectorial operator
$B$ satisfying the Theorem 1 condition. Observe the produced above example
$s_{m}(B)=(m\ln m)^{-1/\rho},$ as it was said above the latter condition
guarantees $B\in\mathfrak{S}_{\rho}^{\star}.$ However, the problem is how to
choose in the polar decomposition formula the unitary operator $U$ providing
the sectorial property of the operator $B.$ The following approach is rather
rough but can be used to supply the desired example, in terms of formula (4),
we get

$(Bf,f)=\sum\limits_{n=1}^{\infty}s_{n}(B)(f,e_{n})\overline{(f,g_{n})}.$

Therefore, if we impose the condition

$\left|\arg(f,e_{n})-\arg(f,g_{n})\right|<\theta,\,n\in\mathbb{N},\,f\in\mathrm{R}(B),$

then we obtain the desired sectorial condition, where $\theta$ is a semi-angle
of the sector. It is remarkable that that the selfadjoint case corresponds to
the zero difference since in this case $U=I.$ However, we should remark that
following to the classical mathematical style it is necessary to formulate
conditions of the sectorial property in terms of the eigenvectors of the
operator $|B|$ but it is not so easy matter requiring comprehensive analysis
of the operator $U$ properties provided with concrete cases.

Essential decreasing of the summation order

Another method allows us to decrease the summation order essentially!
Agranovich M. S. considered the condition

$\overline{\lim\limits_{n\rightarrow\infty}}\lambda_{n}(W)n^{-p}>0,\,p>0,$

which gives us the existence of the subsequence and $K>0$ so that

$|\lambda_{N_{\nu}+1}|-|\lambda_{N_{\nu}}|\geq
K|\lambda_{N_{\nu}}|^{1-1/p},\;\nu\in\mathbb{N}_{0}.$

Taking into account an opportunity to consider the corresponding projectors
$\mathcal{P}_{\nu}(\alpha,t)$ and the fact that if $\lambda_{n}=n^{p},\,p>1$
then we can suppose
$N_{\nu}=\nu^{q},\,q\in\mathbb{N};\,\lambda_{N_{\nu}}=\nu^{qp};\,\lambda_{N_{\nu}+1}=(\nu^{q}+1)^{p}$
preserving the previous estimate for eigenvalues, i.e.

$(\nu^{q}+1)^{p}-\nu^{qp}\geq p\nu^{qp(1-1/p)}.$

Thus, we conclude that it is rather reasonable to impose the restriction on
the counting function assuming

$N_{\nu}=O^{\ast}(\nu^{\beta+1/\eta}),\,\nu\rightarrow\infty,$

where $\beta,1/\eta\in\mathbb{N}.$ To make the writing more convenient we will
use a shortage $\gamma:=\beta+1/\eta.$ Consider a representation

$N_{\nu}=C_{\nu}(\nu+1)^{\gamma}\cdot\frac{\beta}{\gamma},\;\;C_{1}\leq
C_{\nu}\leq C_{2}.$

Then using simple reasonings, we obtain the following decomposition

$N_{\nu}=N^{\prime}_{\nu}+C_{\nu}\sum_{k=1}^{\nu^{1/\eta}}(\nu^{\beta}-k^{\eta\beta}).$

where $N^{\prime}_{\nu}$ is a newly constructed counting function having the
index in the asymptotic formula less than the given counting function. Indeed,
consider the estimate between the given sum and a corresponding definite
integral, i.e. calculating the integral, we have

$\frac{\nu^{\gamma}}{\eta\gamma}+\frac{\beta}{\gamma}\leq\sum_{k=1}^{\nu^{1/\eta}}k^{\eta\beta}\leq\frac{(\nu+1)^{\gamma}}{\eta\gamma}-\frac{1}{\eta\gamma}.$

It follows that

$C_{\nu}\nu^{\gamma}+\frac{C_{\nu}\beta}{\gamma}+\frac{C_{\nu}\beta}{\gamma}\left\\{(\nu+1)^{\gamma}-\nu^{\gamma}\right\\}\leq
N_{\nu}+C_{\nu}\sum_{k=1}^{\nu^{1/\eta}}k^{\eta\beta}\leq
C_{\nu}(\nu+1)^{\gamma}-\frac{C_{\nu}}{\eta\gamma},$

since $1/\gamma\eta+\beta/\gamma=1.$ Therefore

$\frac{C_{\nu}\beta}{\gamma}+C_{\nu}\beta\nu^{\gamma-1}\leq
N^{\prime}_{\nu}\leq
C_{\nu}\gamma(\nu+1)^{\gamma-1}-\frac{C_{\nu}}{\eta\gamma}.$

Hence

$N^{\prime}_{\nu}=O^{\ast}(\nu^{\gamma-1}),\,\nu\rightarrow\infty.$

Now, assume that

$N_{\nu}/\lambda^{\sigma}_{N_{\nu}}\leq C_{2}.$

In accordance with the asymptotic formula, we have the following relation

$C_{1}\leq N_{\nu+1}/N_{\nu}\leq C_{2}.$

Hence

$\frac{N_{\nu}+k}{\lambda^{\sigma}_{N_{\nu}+k}}<\frac{N_{\nu+1}}{\lambda^{\sigma}_{N_{\nu}}}<C_{2}\frac{N_{\nu}}{\lambda^{\sigma}_{N_{\nu}}}.$

Thus, we have the following relation for the counting function

$\lim\limits_{r\rightarrow\infty}\frac{n(r)}{r^{q}}=0,\,q>\sigma.$

Thus, using the obvious implication

$\frac{\nu}{\lambda^{\sigma/\gamma}_{N_{\nu}}}\leq
C_{2},\Longrightarrow\frac{N^{\prime}_{\nu}}{\lambda^{\sigma(1-1/\gamma)}_{N_{\nu}}}\leq
C_{2},$

in accordance with the above reasonings, we get

$\lim\limits_{r\rightarrow\infty}\frac{n_{1}(r)}{r^{s}}=0,\,s>\sigma(1-1/\gamma),\;\;\lim\limits_{r\rightarrow\infty}\frac{n_{k}(r)}{r^{q^{\prime}}}=0,\,q^{\prime}>\sigma\beta/\gamma,\,k=2,3,...\,,$

where $n_{k}(r)$ is a counting function corresponding to the sequence of
numbers $N^{\prime}_{\nu},\,k=1,$ and
$\nu^{\beta}-k^{\eta\beta},\,k=2,3,...,\nu^{1/\eta}$ respectively. Having
noticed the fact $(1-1/\gamma)\geq\beta/\gamma$ that holds for the acceptable
parameter values, we can clime that the order allows an essential decreasing
from $q$ to $s,$ where it is obviously supposed that $q>s.$ The latter
representation gives us the essential decreasing of the order since we can
split projectors and summarize them in the different way, considering the
counting functions for newly constructed operators $B_{k},\,k=1,2,...,$ which
are restrictions of the operator $B$ on the root vector subspaces
corresponding to the eigenvalues counted by the counting functions
$N^{\prime}_{\nu}$ and $\nu^{\beta}-v^{\beta}(k)$ respectively. The symbol $k$
indicates a position of the first eigenvalue of $B_{k}.$ Thus, due to the
connection with the asymptotics of the counting function and the estimate of
the norm of the resolvent studied above, we can claim that the following
double series is convergent, i.e.

$I:=\sum_{k=1}^{\infty}\sum_{\nu=v(k)}^{\infty}\left\|\,\oint\limits_{\vartheta_{\nu}}e^{-\lambda^{s}t}B_{k}\left(I-\lambda
B_{k}\right)^{-1}fd\lambda\right\|<\infty,$ (14)

where $\alpha$ less than the convergent exponent $\rho$ of the operator $B,$
the contours $\vartheta_{k}$ are defined so that the eigenvalues counted by a
corresponding counting function belong to the union of them. Let us fulfil the
scheme of reasonings represented in Theorem 2. For this purpose we will
estimate the integrals in the inner sum. Consider the restrictions $B_{k}$ of
the operator $B$ on the root vectors subspaces generated by the corresponding
sets of the eigenvalues. Consider an arbitrary eigenvalue of the operator
$B_{k},$ we denote it $\lambda_{k_{j}},$ it corresponds to the root vector
subspace $\mathfrak{M}_{k_{j}}.$ Thus we can put in correspondence to the
operator $B_{k}$ a sequence of subspaces, where the index $j$ is counted. Let
us show that the closures of the subspaces corresponding to the operators
$B_{k}$ and $B_{m},\,m\neq k$ do not not intersect. For this purpose, let us
notice the fact that the root vectors are linearly independent. It follows
easily if we consider a Riesz projector

$P_{\Gamma_{q}}e_{q_{\xi}+i}=\frac{1}{2\pi i}\int_{\Gamma_{q}}B(I-\lambda
B)^{-1}e_{q_{\xi}+i}d\lambda=\frac{1}{2\pi
i}\int_{\Gamma_{q}}\frac{1}{\lambda^{2}}\left\\{(\lambda^{-1}I-B)^{-1}-I\
\right\\}e_{q_{\xi}+i}d\lambda=$ $=\frac{1}{2\pi
i}\int_{\Gamma_{q}}\frac{1}{\lambda^{2}}(\lambda^{-1}I-B)^{-1}e_{q_{\xi}+i}d\lambda=\frac{1}{2\pi
i}\int_{\Gamma^{\prime}_{q}}(\lambda^{\prime}I-B)^{-1}e_{q_{\xi}+i}d\lambda^{\prime}=$
$=\frac{1}{2\pi
i}\sum\limits_{j=0}^{i}\int_{\Gamma^{\prime}_{q}}\frac{e_{q_{\xi}+j}}{(\lambda^{\prime}-\mu_{q})^{i-j+1}}d\lambda^{\prime}=e_{q_{\xi}+i},\,$

where, we have made the change of the variables
$\lambda^{-1}=\lambda^{\prime},$ and $B^{-1}e_{q}=\mu_{q}e_{q}.$ In accordance
with the Cauchy integral formula, we get easily
$P_{\Gamma_{q_{1}}}P_{\Gamma_{q_{2}}}=0,\,q_{1}\neq q_{2}.$ From what follows
linear independence between the root vectors corresponding to different
eigenvalues. Assume that the root vectors corresponding to an eigenvalue are
linearly dependent, then there exists a non-zero set $C_{j},\,j=0,1,...n$ so
that

$\sum\limits_{j=0}^{n}C_{j}e_{q_{\xi}+j}=0,$

where $n$ is a height of the eigenvalue. Applying the operators
$(B-\lambda_{q})^{k},\,k=n-1,n-2,...,1$ to the above sum consistently, we
obtain $C_{j}=0,\,j=0,1,...n$ what contradicts the the given assumptions.
Thus, we have proved that the root vectors are linearly independent. Define

$\mathfrak{M}_{k}=\mathfrak{M}_{k_{1}}\oplus\mathfrak{M}_{k_{2}}\oplus...\oplus\mathfrak{M}_{k_{n}}\oplus...\;.$

It is clear that this definition has an essential lack for we cannot define
limits in this way, thus we just consider finite sum of elements. Consider

$\mathfrak{M}_{k}\cup\mathfrak{M}_{m},\,k\neq m.$

Using the Gram-Schmidt orthogonalization procedure in a concrete way, we
obtain two orthogonal bases in the orthogonal Hilbert spaces
$\mathfrak{M}_{k},\mathfrak{M}_{m}.$ The way of choosing elements in the
orthogonalization procedure is in the following, we should take the elements
from $\mathfrak{M}_{k},\mathfrak{M}_{m}$ alternately. Note that the
constructed Hilbert spaces are invariant subspaces of the operator $B.$ This
fact follows easily from the fact

$x_{n}\stackrel{{\scriptstyle\mathfrak{H}}}{{\rightarrow}}x\Longrightarrow
Bx_{n}\rightarrow Bx,$

here we should note that $\mathfrak{M}_{k},\mathfrak{M}_{m}$ are closed in the
sense of the norm. It is clear that the restrictions $B_{k}$ are compact,
since $B$ is compact. Since $B_{k}$ is compact then in accordance with the
Hilbert theorem its spectrum consists of normal values. Thus, if a complex
value is not an eigenvalue then it is a regular value. Therefore, the
resolvent set of $B_{k}$ contains the rest part of eigenvalues of the operator
$B.$ This fact can be proved under more general assumptions. In accordance
with Theorem 6.17 Chapter III [10] (we can omit the compactness property ), we
have the decomposition

$\mathfrak{M}_{k}=\mathrm{M}^{\prime\prime}\oplus\mathrm{M}^{\prime},$

where $\mathrm{M}^{\prime},\mathrm{M}^{\prime\prime}$ correspond to finite and
infinite sets of eigenvalues respectively so that the restriction
$B_{\mathrm{M}^{\prime}}$ has a resolvent defined on the eigenvalues
corresponding to the space $\mathrm{M}^{\prime\prime}.$ Let us estimate the
inner sum (14), for this purpose let us fix the defined above constant
$q>\sigma$ and find $\delta_{\nu}$ from the condition
$R_{\nu}=K|\mu_{N_{\nu}}|^{1-1/q}+|\mu_{N_{\nu}}|,\,R_{\nu}(1-\delta_{\nu})=|\mu_{N_{\nu}}|,$
then $\delta_{\nu}^{-1}=1+K^{-1}|\mu_{N_{\nu}}|^{1/q}.$ Note that by virtue of
such a choice, we have $R_{\nu}<R_{\nu+1}(1-\delta_{\nu+1}).$ In accordance
with the above we have $n_{k}(r)=o(r^{s}).$ Consider a contour

$\vartheta(B):=\left\\{\lambda:\;|\lambda|=r_{0}>0,\,|\mathrm{arg}\lambda|\leq\theta+\varepsilon\right\\}\cup\left\\{\lambda:\;|\lambda|>r_{0},\;|\mathrm{arg}\lambda|=\theta+\varepsilon\right\\},$

here the number $r$ is chosen so that the circle with the corresponding radios
does not contain values $\mu_{n},\,n\in\mathbb{N}.$

Now, consider a formal representation for the resolvent

$(I-\lambda
B)^{-1}f=\sum\limits_{m=1}^{\infty}\left(\sum\limits_{l=1}^{\infty}(-1)^{l+m}\frac{\Delta^{lm}(\lambda)}{\Delta}f_{l}\right)e_{m},$

where

$\Delta^{lm}(\lambda)=\sum\limits_{p=0}^{\infty}(-1)^{p}\lambda^{p}\\!\\!\\!\\!\\!\sum\limits_{i_{1},i_{2},...,i_{p}=1}^{\infty}\\!\\!\\!B\begin{pmatrix}i_{1}&i_{2}&...&i_{p}\\\
i_{1}&i_{2}&...&i_{p}\end{pmatrix}_{lm},$

the used formula in brackets means a minor formed from the columns and rows
with $i_{1},i_{2},...,i_{p}$ numbers except for $l$ -th row and $m$ -th
column. Using analogous form of writing, we denote the Fredholm determinant of
the operator $B$ as follows

$\Delta(\lambda)=1-\lambda\sum\limits_{i=1}^{\infty}B\begin{pmatrix}i\\\
i\end{pmatrix}+\lambda^{2}\sum\limits_{i_{1},i_{2}=1}^{\infty}B\begin{pmatrix}i_{1}&i_{2}\\\
i_{1}&i_{2}\end{pmatrix}+...+(-1)^{p}\lambda^{p}\sum\limits_{i_{1},i_{2},...,i_{p}=1}^{\infty}B\begin{pmatrix}i_{1}&i_{2}&...&i_{p}\\\
i_{1}&i_{2}&...&i_{p}\end{pmatrix}+...\,,$

where the used formula in brackets means a minor formed from the columns and
rows with $i_{1},i_{2},...,i_{p}$ numbers. Note that if $B$ belongs to the
trace class then in accordance with the well-known theorems (see [8]), we have

$\sum\limits_{n=1}^{\infty}|b_{nn}|<\infty,\;\sum\limits_{n,k=1}^{\infty}|b_{nm}|^{2}<\infty,$
(15)

where $b_{nm}$ is the matrix coefficients of the operator $B.$ This follows
easily from the properties of the trace class operators and Hilbert-Schmidtt
class operators respectively. In accordance with the von Koch theorem the
conditions (15) guaranty the absolute convergence of the series

$\sum\limits_{i_{1},i_{2},...,i_{p}=1}^{\infty}B\begin{pmatrix}i_{1}&i_{2}&...&i_{p}\\\
i_{1}&i_{2}&...&i_{p}\end{pmatrix}.$

Moreover, the formal series $\Delta(\lambda)$ is convergent for arbitrary
$\lambda\in\mathbb{C},$ hence it represents an entire function. Note that in
accordance with Lemma 1 [26], we have

$s_{n}(B_{k})\leq s_{n}(B),\,k=1,2,...\,,$

since the operator $B=P_{k}BP_{k},$ where $P_{k}$ is orthogonal projector into
the corresponding invariant subspace of the operator $B_{k}.$ Thus, we get the
implication

$B\in\mathfrak{S}_{p},\Rightarrow B_{k}\in\mathfrak{S}_{p},\,0<p<1.$

Having denoted $\Delta_{k}(\lambda),\,\Delta^{lm}_{k}(\lambda)$ the considered
above constructions corresponding to the operator $B_{k},$ we come due to the
reasonings represented in Lemma 1 [26] to the formula

$|\Delta_{k}(\lambda)|\leq\prod\limits_{n=1}^{\infty}\left\\{1+|\lambda|s_{n}(B_{k})\right\\}.$

It implies that the entire function $\Delta_{k}(\lambda)$ is of the finite
order, hence in accordance with the Theorem 13 (Chapter I, §10) [26], it has a
representation by the canonical product, we have

$\Delta_{k}(\lambda)=\prod\limits_{n=1}^{\infty}\left\\{1-\lambda\lambda_{n}(B_{k})\right\\}.$

Now, let us chose an arbitrary element
$f\in\mathfrak{H}_{k},\;B_{k}:\mathfrak{H}_{k}\rightarrow\mathfrak{H}_{k}$ and
construct a new orthonormal basis having put $f$ as a first basis element.
Note that the relations (15) hold for the matrix coefficients of the operator
in new basis, this fact follows from the well-known theorem for the trace
class operator. Thus, using the given above representation for the resolvent,
we obtain the following relation

$\Delta_{k}(\lambda)\left((I-\lambda
B_{k})^{-1}f,f\right)_{\mathfrak{H}}=\Delta^{11}_{k}(\lambda,f).$

Let us observe the latter entire function more properly, we have

$\Delta^{11}_{k}(\lambda,f)=\sum\limits_{p=0}^{\infty}(-1)^{p}\lambda^{p}\\!\\!\\!\\!\\!\sum\limits_{i_{1},i_{2},...,i_{p}=1}^{\infty}\\!\\!\\!B\begin{pmatrix}i_{1}&i_{2}&...&i_{p}\\\
i_{1}&i_{2}&...&i_{p}\end{pmatrix}_{f11},$

where its construction reveals the fact that it is a Fredholm determinant of
the operator $Q_{1}B_{k}Q_{1},$ where $Q_{1}$ is the projector into orthogonal
complement of the element $f.$ Having applied the above reasonings, we obtain

$s_{n}(Q_{1}B_{k}Q_{1})\leq s_{n}(B_{k}),\,k=1,2,...\,.$

In the same way, we obtain that the entire function
$\Delta^{11}_{k}(\lambda,f)$ is of the finite order and obtain the
representation

$\Delta^{11}_{k}(\lambda,f)=\prod\limits_{n=1}^{\infty}\left\\{1-\lambda\lambda_{n}(Q_{1}B_{k}Q_{1})\right\\}.$

Note that the operator $Q_{1}B_{k}Q_{1}$ has no more eigenvectors than the
operator $B_{k}.$ This fact can be obtained easily due the the properties of
the inner product in the Hilbert space and the construction of the operator
$Q_{1}B_{k}Q_{1}.$ In accordance with the orthogonal projector properties, we
have for arbitrary element $h\in\mathfrak{H},$ on the one hand

$\left(Q_{1}B_{k}Q_{1}e,h\right)=\lambda\cdot(e,h),$

where $Q_{1}B_{k}Q_{1}e=\lambda e,\;\lambda\neq 0,$ on the other hand

$\left(Q_{1}B_{k}Q_{1}e,h\right)=\left(Q_{1}B_{k}Q_{1}e,Q_{1}h\right)=\lambda\cdot(e,Q_{1}h)=\lambda\cdot(Q_{1}e,h),$

hence

$\lambda\cdot(Q_{1}e-e,h)=0;\;Q_{1}e=e;\;B_{k}e=\lambda e.$

Thus, we have no more eigenvalues for the restricted operator, it implies that
we can preserve the obtained asymptotic for the operator $B_{k}.$ Evaluating,
we obtain

$|\Delta^{11}_{k}(\lambda,f)|\leq\prod\limits_{n=1}^{\infty}\left\\{1+|\lambda\lambda_{n}(Q_{1}B_{k}Q_{1})|\right\\}\leq\prod\limits_{n=1}^{\infty}\left\\{1+|\lambda\lambda_{n}(B_{k})|\right\\}.$

Since the right-hand side does not depend on $f,$ then using decomposition on
the Hermitian components, we obtain easily

$|\Delta_{k}(\lambda)|\cdot\|(I-\lambda B_{k})^{-1}\|\leq
2\prod\limits_{n=1}^{\infty}\left\\{1+|\lambda\lambda_{n}(B_{k})|\right\\}.$

To establish the letter relation Agaranovich M.S. used the polarization
formula, however we can prove it using the offered method. Consider the
operator function

$D_{B_{k}}(\lambda)=\Delta_{k}(\lambda)(I-\lambda B_{k})^{-1},$

having involved ordinary operator properties, we have

$\sup\limits_{\|f\|\leq 1}\|D_{B_{k}}(\lambda)f\|=\sup\limits_{\|f\|\leq
1}\|\mathfrak{Re}D_{B_{k}}(\lambda)f+i\,\mathfrak{Im}D_{B_{k}}(\lambda)f\|\leq\sup\limits_{\|f\|\leq
1}\|\mathfrak{Re}D_{B_{k}}(\lambda)f\|+\sup\limits_{\|f\|\leq
1}\|\mathfrak{Im}D_{B_{k}}(\lambda)f\|\leq$ $\leq\sup\limits_{\|f\|\leq
1}\left|\mathrm{Re}(B_{k}(\lambda)f,f)\right|+\sup\limits_{\|f\|\leq
1}\left|\mathrm{Im}(B_{k}(\lambda)f,f)\right|\leq 2\sup\limits_{\|f\|\leq
1}\left|(B_{k}(\lambda)f,f)\right|\leq
2\prod\limits_{n=1}^{\infty}\left\\{1+|\lambda\lambda_{n}(B_{k})|\right\\}.$

Using the properties of the canonical product, we obtain the estimate

$\|D_{B_{k}}(\lambda)\|\leq e^{\beta_{k}(|\lambda|)|\lambda|^{s}},$

where

$\;\beta_{1}(r)=r^{-s}\left(\int\limits_{0}^{r}\frac{n_{1}(t)dt}{t}+r\int\limits_{r}^{\infty}\frac{n_{1}(t)dt}{t^{2}}\right),$
$\;\beta_{k}(r)=r^{-s}\left(\int\limits_{k^{\eta}}^{r}\frac{n_{k}(t)dt}{t}+r\int\limits_{r}^{\infty}\frac{n_{k}(t)dt}{t^{2}}\right),\;r>k^{\eta},\;k=2,3,...\,,$
$\beta_{k}(r)=r^{1-s}\int\limits_{k^{\eta}}^{\infty}\frac{n_{k}(t)dt}{t^{2}},\;r<k^{\eta}.$

In accordance with the above, we have $\beta_{k}(r)\rightarrow
0,\,r\rightarrow\infty,$ uniformly with respect to $k,$ moreover
$\beta_{k}(r)\rightarrow 0,\,k\rightarrow\infty,$ uniformly with respect to
$r.$

Consider the entire function

$\Delta_{k}(\lambda)=\prod\limits_{n=1}^{\infty}\left\\{1-\lambda\lambda_{n}(B_{k})\right\\}.$

In accordance with the Joseph Cartan concept, we can obtain the following
estimate then holds on the complex plane except may be exceptional set of
circulus. The letter cannot be found but and we are compelled to make an
occlusion evaluating the measures. However, further we will produce the method
allowing to find exceptional set of the circulus. We have

$|\Delta_{k}(\lambda)|\geq
e^{-(2+\ln\\{4e/\delta_{\nu}\\})\beta_{k}(\varpi|\lambda|)\varpi^{1/s}},\,\varpi:=\frac{2e}{1-\delta_{0}},\,|\lambda|=\tilde{R}_{\nu},$

where $|\lambda_{N_{\nu}}|=(1-\delta_{\nu})R_{\nu}<\tilde{R}_{\nu}<R_{\nu}.$
Therefore, in accordance with the obtained above relations, we get

$\|(I-\lambda
B_{k})^{-1}\|=\frac{\|D_{B_{k}}(\lambda)\|}{|\Delta_{k}(\lambda)|}\leq
e^{\gamma_{k}(|\lambda|)|\lambda|^{s}}|,\,|\lambda|=\tilde{R}_{\nu},$

where

$\gamma_{k}(|\lambda|)=\beta_{k}(|\lambda|)+(2+\ln\\{4e/\delta_{\nu}\\})\beta_{k}(\varpi|\lambda|)\,\varpi^{1/s},\,s>\sigma(1-1/\gamma).$

Recalling the formula

$\ln\\{4e/\delta_{\nu}\\}=1+\ln
4+\ln\\{1+K^{-1}|\mu_{N_{\nu}}|^{\alpha}\\}\leq C\ln|\mu_{N_{\nu}}|\leq
C\ln|\mu_{N_{\nu}}|\leq C\ln\tilde{R}_{\nu},$

we get easily $\gamma_{k}(|\tilde{R}_{\nu}|)\rightarrow
0,\,\nu\rightarrow\infty,$ uniformly with respect to $k.$ To proceed to the
next step, we want to estimate termwise the following relation representing
unified reasonings for all values of $k\in\mathbb{N}$

$\left\|\,\int\limits_{\vartheta_{\nu}}e^{-\lambda^{s}t}B_{k}\left(I-\lambda
B_{k}\right)^{-1}fd\lambda\right\|=J^{-}_{\nu}+J^{+}_{\nu}+J_{\nu}+J_{\nu+1}.$

we have

$J_{\nu}:=\left\|\,\int\limits_{\tilde{\vartheta}_{\nu}}e^{-\lambda^{s}t}B_{k}\left(I-\lambda
B_{k}\right)fd\lambda\,\right\|\leq\,\int\limits_{\tilde{\vartheta}_{\nu}}e^{-t\mathrm{Re}\lambda^{s}}\left\|B_{k}\left(I-\lambda
B_{k}\right)f\right\||d\lambda|\leq$ $\leq
C\|f\|e^{\gamma_{k}(|\lambda|)|\lambda|^{s}}\int\limits_{-\theta-\varepsilon}^{\theta+\varepsilon}e^{-t\mathrm{Re}\lambda^{s}}d\,\mathrm{arg}\lambda,\,|\lambda|=\tilde{R}_{\nu}.$

The condition $\theta<\pi/2\sigma<\pi/2s$ guarantees the convergence of the
latter integral. To estimate other terms, we are rather satisfied with the
estimate represented above

$\|(I-\lambda B_{k})^{-1}\|\leq\frac{1}{\sin\psi},\,\lambda\in\zeta,$

where $\zeta$ is the ray containing the point zero and not belonging to the
sector $\mathfrak{L}_{0}(\theta)$ as well as the real axis,
$\,\psi=\min\\{|\mathrm{arg}\zeta-\theta|,|\mathrm{arg}\zeta+\theta|\\}.$
Absolutely analogously, we have

$J^{+}_{\nu}\leq
C\|f\|\\!\\!\\!\int\limits_{\tilde{R}_{\nu}}^{\tilde{R}_{\nu+1}}e^{-t\mathrm{Re}\lambda^{s}}|d\lambda|\leq
C\|f\|e^{-t\tilde{R}_{\nu}^{s}\sin
s\delta}\\!\\!\\!\int\limits_{\tilde{R}_{\nu}}^{\tilde{R}_{\nu+1}}|d\lambda|=$
$=C\|f\|e^{-t\tilde{R}_{\nu}^{s}\sin
s\delta}(\tilde{R}_{\nu+1}-\tilde{R}_{\nu}).$

Thus, we have come to the relation

$I_{k,\nu}:=\left\|\,\int\limits_{\vartheta_{\nu}}e^{-\lambda^{s}t}B_{k}\left(I-\lambda
B_{k}\right)^{-1}fd\lambda\right\|\leq
C\|f\|e^{\\{\gamma_{k}(\tilde{R}_{\nu})-t\\}\tilde{R}_{\nu}^{s}}.$

Note that

$n_{k}(\lambda_{N_{\nu}+m})\leq
n_{k}(\lambda_{N_{\nu+1}})=(\nu+1)^{\beta}-k^{\eta\beta},\,m=0,1,...,N_{\nu+1}-N_{\nu},\,k>1.$

Hence, we can estimate the set of the counting function by the newly
constructed step function
$\tilde{n}(r)=(\nu+1)^{\gamma-1},\,r=\lambda_{N_{\nu}},$ we have
$n_{k}(r)\leq\tilde{n}(r),\,k\in\mathbb{N}$ at the same time

$\tilde{\beta}(r):=r^{-s}\left(\int\limits_{0}^{r}\frac{\tilde{n}(t)dt}{t}+r\int\limits_{r}^{\infty}\frac{\tilde{n}(t)dt}{t^{2}}\right)\rightarrow
0,\,r\rightarrow\infty,\,s>\sigma(1-1/\gamma).$

This gives us a unified function $\tilde{\gamma}(r)\geq\gamma(r),$ constructed
due to the replacement in the formula for $\gamma(r)$ functions $\beta_{k}(r)$
with $\tilde{\beta}(r).$ Therefore

$I_{k,\nu}\leq
e^{\\{\tilde{\gamma}(\tilde{R}_{\nu})-t\\}\tilde{R}_{\nu}^{s}}.$

Therefore, for arbitrary $t>0$ there exists $m>0$ such that

$I\leq\sum_{k=1}^{\infty}\sum_{\nu=v(k)}^{\infty}e^{\\{\tilde{\gamma}(\tilde{R}_{\nu})-t\\}\tilde{R}_{\nu}^{s}}=\sum_{k=1}^{\infty}\sum_{\nu=v(k)}^{\infty}e^{\\{\tilde{\gamma}(\tilde{R}_{\nu})-t\\}\tilde{R}_{\nu}^{s}}=\sum_{k=1}^{m-1}\sum_{\nu=v(k)}^{\infty}e^{\\{\tilde{\gamma}(\tilde{R}_{\nu})-t\\}\tilde{R}_{\nu}^{s}}+\sum_{k=m}^{\infty}\sum_{\nu=v(k)}^{\infty}e^{\\{\tilde{\gamma}(\tilde{R}_{\nu})-t\\}\tilde{R}_{\nu}^{s}}\leq$
$\leq\sum_{k=1}^{m-1}\sum_{\nu=v(k)}^{\infty}e^{\\{\tilde{\gamma}(\tilde{R}_{\nu})-t\\}\tilde{R}_{\nu}^{s}}+\sum_{k=m}^{\infty}\sum_{\nu=v(k)}^{\infty}e^{-M\tilde{R}_{\nu}^{s}}=I_{1}+I_{2}.$
$I_{2}\leq\sum_{k=m}^{\infty}\sum_{\nu=v(k)}^{\infty}e^{-M\lambda_{N_{\nu}}^{s}}\leq\sum_{k=m}^{\infty}\sum_{\nu=v(k)}^{\infty}e^{-M\nu^{s\gamma/\sigma}}\leq\sum_{k=m}^{\infty}\sum_{\nu=v(k)}^{\infty}e^{-M\nu^{\gamma-1}}\leq\sum_{k=m}^{\infty}\sum_{\nu=v(k)}^{\infty}e^{-M\nu}.$

The latter relation holds since $\gamma\geq 2.$ Using simple formulas for
geometrical progression, we get for arbitrary $\eta>0$

$I_{2}\leq\frac{e^{M}}{e^{M}-1}\sum_{k=m}^{\infty}e^{-M(k^{\eta}+1)}<\infty.$

The fact that the series $I_{1}$ is convergent can be proved in the analogous
but rather simpler way. Thus, we have proved that the series (14) is
convergent. Now, note that

$\frac{1}{2\pi
i}\oint\limits_{\vartheta_{\nu}}e^{-\lambda^{s}t}B_{k}\left(I-\lambda
B_{k}\right)^{-1}fd\lambda=-\sum\limits_{q=\mathcal{K}_{\nu}+1}^{\mathcal{K}_{\nu+1}}\sum\limits_{\xi=1}^{m(q)}\sum\limits_{i=0}^{l(q_{\xi})}e_{q_{\xi}+i}c_{q_{\xi}+i}(t),\;h\in\mathfrak{H},$

where $m(q)$ is a geometrical multiplicity of the $q$-th eigenvalue,
$l(q_{\xi})+1$ is a number of elements in the $q_{\xi}$-th Jourdan chain of
the operator $B_{k},\;\mathcal{K}_{\nu+1}-\mathcal{K}_{\nu}$ is a subset of
the eigenvalues of the operator $B_{k}$ contained in the subset
$N_{\nu+1}-N_{\nu}$ of eigenvalues of the operator $B,$

$c_{q_{\xi}+j}(t):=e^{-\lambda^{s}_{q}t}\sum\limits_{m=0}^{l(q_{\xi})-j}c_{q_{\xi}+j+m}H_{m}(s,\lambda_{q},t),\,\lambda_{q}=\mu^{-1}_{q},$
$H_{m}(s,\lambda_{q},t):=\frac{e^{\lambda^{s}_{q}t}}{m!}\lim\limits_{\zeta\rightarrow\,\mu_{q}}\frac{d^{m}}{d\zeta^{m}}\left\\{e^{-\lambda^{-s}t}\right\\},\,Be_{q_{\xi}}=e_{q_{\xi}},\;Be_{q_{\xi}+i}=\mu_{q}e_{q_{\xi}+i}+e_{q_{\xi}+i-1},$
$\mathfrak{H}=\mathfrak{N}_{q}\oplus\mathfrak{M}_{q},\;f=f_{1}+f_{2},\;f_{1}\in\mathfrak{N}_{q},\,f_{2}\in\mathfrak{M}_{q},\;f_{1}=\sum\limits_{\xi=1}^{m(q)}\sum\limits_{i=0}^{l(q_{\xi})}e_{q_{\xi}+i}c_{q_{\xi}+i}.$

Thus, we can formulate the obtained result in the following form.

Theorem 2. Assume that $B$ is a compact operator,
$B\in\mathfrak{S}_{p}\,,\,p>0,\,\Theta(B)\subset\mathfrak{L}_{0}(\theta),\,\theta<\pi/2\rho,$
the following condition holds

$N_{\nu}=O^{\ast}(\nu^{\gamma}),\,\gamma=2,3,...\,,$

then the order of summation $\alpha$ can be essentially decreased to
$\rho(1-1/\gamma)<\alpha$, moreover there exists the splitting of the
projectors $P_{\nu}(\alpha,t)$ such that the series is still convergent, i.e.

$\sum_{k=1}^{\infty}\mathcal{P}_{k}(\alpha,t)\rightarrow I,\,t\rightarrow
0,\;\;\mathcal{P}_{\nu}(\alpha,t)=\sum_{k=k_{1}(\nu)}^{k_{2}(\nu)}\mathcal{P}_{k}(\alpha,t).$

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