Spectral analysis of some non-normal operators arising  in  Reggeon field theory
Abdelkader INTISSAR

Equipe d'Analyse spectrale, Faculté des Sciences et Techniques
Université de Corté, 20250 Corté, France

Le Prador

129,  rue commandant Rolland

13008 Marseille France

In this work, we present a complete spectral study of a family of non-normal operators arising in Reggeon field theory. This family of operators is an original example who permit us to discover the recent theory of physical requirement of space-time reflection symmetry (PT symmetry) without losing any of the essential physical features of quantum mechanics  [Bender].

Early studies of Reggeon field theory, in the late 1970's led a number of investigators to observe that model cubic quantum-mechanical Hamiltonians might have real eigenvalues  [Bower et al].

The study of this family of operators, permit us to discover some fine results of Spectral Theory and Functional Analysis in particular the results connected with completeness of elementary solutions of mathematical physics problems.

We use knowledge  basics of holomorphic functions of one complex variable and the properties of Hilbert spaces. In this work the  knowledge of spectral theory and the functional analysis of standard level is required (see [Kato] , [Ghohberg2 et al] and [Markus]). It is also requested to know the basic properties of semigroup theory see for example ([Pazy]).

Keywords : Spectral theory ; Gribov operators ; semigroups ; Non-self-adjoint operators ; Bargmann space ; Reggeon field theory.

§  INTRODUCTION

For self-adjoint and more generally normal operators $T$ on some complex Hilbert space $\mathcal{H}$,  we have a nice theory, including the spectral theorem and the wellknown and important resolvent estimate,

\begin{equation}
\displaystyle{\mid\mid (H - \lambda I)^{-1} \mid\mid \leq \frac{1}{(dist (\lambda, \sigma(T))}}
\end{equation}

where $\sigma(H)$ denotes the spectrum of $T$. The spectral theorem also gives very nice control over functions of self-adjoint operators, so for instance if $H$ is
self-adjoint with spectrum contained in the half interval $\displaystyle{[\lambda_{0}, +\infty[}$, then

\begin{equation}
\displaystyle{\mid\mid e^{-tH} \mid\mid \leq e^{-\lambda_{0}t} , t \geq 0 }
\end{equation}

However, non-normal operators appear frequently in different problems;

Convection-diffusion problems, Theory of elasticity, transport equation, damped wave equations, linearized operators in fluid dynamics, Reggeon field theory etc.

Then typically,  $\displaystyle{\mid\mid (H - \lambda I)^{-1} \mid\mid}$ may not exist or may be very large even when $\lambda$ is far from the spectrum and this implies mathematical difficulties:

- When studying the existence of eigenvalues and their distribution,

- When studying functions of the operator, like $\displaystyle{e^{-tH}}$,

- When studying the operators $H$ in the form $\displaystyle{H = H_{0} + i H_{1} }$; $i^{2} = -1$ where  $H_{0}$ is self adjoint and  $H_{1}$  is a strong perturbation of $H_{0}$  with $\sigma(H_{1}) = \mathbb{C}$,

- When the domains of the adjoint and anti-adjoint parts of $H$ are not included in one another.

When choosing the ambient Hilbert space, like in problems for scattering poles or  in problems for Reggeon field  theory.

This is a source of mathematical and numerical difficulties.

This text is devoted to some old and recent general results for non-self-adjoint operators

\begin{equation}
\displaystyle{H_{\mu, \lambda} = \mu \mathcal {N} + i\lambda (\mathcal{N} A + A^{*}\mathcal{N})}
\end{equation}

\begin{equation}
\displaystyle{H_{\lambda', \mu, \lambda} =  \lambda'A^{*^{2}}A^{2} + \mu \mathcal {N} + i\lambda (\mathcal{N} A + A^{*}\mathcal{N})}
\end{equation}

\begin{equation}
\displaystyle{H_{ \lambda'', \mu, \lambda} =   \lambda''A^{*^{3}}A^{3} + \mu \mathcal {N} + i\lambda (\mathcal{N} A + A^{*}\mathcal{N})}
\end{equation}

arising in Reggeon field theory  where

- $\mu$, $\lambda$; $\lambda'$ and $\lambda''$ are real parameters and $i^{2} = -1$.

- $A^{*}$ and $A$, are the creation and annihilation operators. These operators satisfy the commutation relation:

\begin{equation}
[A, A^{*}] = I
\end{equation}

and  $\mathcal{N} = A^{*}A$ is Harmonic oscillator.

$H_{\mu, \lambda}$  can be defined  in Bargmann space as the closure of its restriction on the polynomials. Then the first mathematical difficulty when studying the minimal domain and the maximal domain of  our operator is to show their equality,

$H_{\mu, \lambda}$  acts as Hamiltonian of Reggeon field theory. In this theory, it is the simplest Hamiltonian that can describe a field with interaction and it is plausible that it describes the dominant interaction between two hadrons.

In the above, the word <<simple>> only applies to the algebraic expression of the operator. If we look at the family a little more generally,

\begin{equation}
\displaystyle{H_{\lambda', \mu, \lambda} = \lambda' A^{*^{2}}A^{2} +  H_{\mu, \lambda}}
\end{equation}

For $\lambda' \neq 0$,  the study of $H_{\lambda', \mu, \lambda}$ is easier than the study of $H_{\mu, \lambda}$ and permit us  by taking the limit $\lambda' = 0$ to confirm  that model cubic quantum-mechanical Hamiltonian  has the eigenvalues and which are real.

For $\lambda'' \neq 0$,  the study of $H_{\lambda'', \mu, \lambda}$ is easier than the study of $H_{\lambda', \mu, \lambda}$ and permit us to give a regularized  trace formula of $H_{\lambda'', \mu, \lambda}$ and  a regularized  trace formula of its semigroup.

Here is the plan of this work.

In Section 2  we give an overview of Reggeon Field Theory (RFT) and the operators which caracterize this theory for one site. This Reggeon field theory (RFT) is an attempt to predict the high-energy behaviour of soft pro- cesses; the RFT can be derived from the assumed softness of hadronic interactions at low transverse momenta, which seems to be well established experimentally in hadron-hadron and hadron-nucleus interaction.

The Reggeon Field Theory (RFT) was invented by Vladimir Naumovich Gribov who was  one of the creators of modern theoretical physics. The concepts and methods that Gribov has developed in the second half of the 20th century became cornerstones of the physics of high energy hadron interactions (relativistic theory of complex angular momenta, a notion of the vacuum pole-Pomeron, effective Reggeon field theory), condensed matter physics (critical phenomena), neutrino oscillations, and nuclear physics.

Originally, RFT was formulated by Gribov in 1969 as a field theory (or as quantum mechanics in zero transverse dimensions) of pomerons. The basic degrees of freedom in this formulation are the Gribov fields  $\psi$ and $\psi^{+}$ that create and annihilate the pomeron. The action defining the theory with triple pomeron
couplings only (MRFT) is defined in the following way:

\begin{equation}
\displaystyle{\mathcal{S} = \int dy\{\psi^{+}\partial_{y}\psi  - \mu \psi^{+}\psi  + i \lambda \psi^{+}( \psi + \psi^{+})\psi\}}
\end{equation}

where $\mu$ is the bare intercept of the pomeron and $\lambda$ is the coupling of the triple pomeron interaction and $i^{2} = -1$.

After redefinition of the Gribov fields  by simple harmonic oscillator annihilation and creation operators :

$\psi \longrightarrow A$    $\overline{\psi} \longrightarrow A^{*}$

Then, the Hamiltonian of the problem is given by:

\begin{equation}
\displaystyle{ H_{\mu, \lambda} = - \mu A^{*}A + i\lambda A^{*}(A + A^{*})A}
\end{equation}

The commutation relation of the pomeron annihilation and creation operators, $[\psi^{+}, \psi] = 1 $ implies $[A^{*}, A]= I$

We choose like ambient Hilbert space the Bargmann space

\begin{equation}
\displaystyle{  \mathbb{B }= \{\varphi : \mathbb{C} \longrightarrow \mathbb{C}  \, entire \, ; \int_{\mathbb{C} }\mid \varphi(z)\mid^{2}e^{-\mid z \mid^{2}}dx dy < \infty \}}
\end{equation}

to  give a complete spectral  analysis of $H_{\mu, \lambda}$ called Gribov-Intissar operators.

This naming of these operators was suggested by a referee of Advance in Mathematics  (CHINA) who reviewed an article of  author titled “On Spectral Approximation of Unbounded Gribov-Intissar Operators in Bargmann Space”, Advances in mathematics (China), Vol. 46. N0. 1, Jan., 2017  13-33, doi: 10.11845/sxjz.2015026b.

In Bargmann representation $\displaystyle{A = \frac{\partial}{\partial z}}$  and  $A^{*} = z$  where $z = x + iy ; (x, y) \in \mathbb{R}^{2}$ and the Hamiltonian takes the form:

\begin{equation}
\displaystyle{H_{\mu, \lambda}  = i \lambda z \frac{\partial^{2}}{\partial z^{2}} + (i\lambda z^{2} - \mu z)\frac{\partial}{\partial z}}
\end{equation}

The interacting Pomeron system whose state is given by  $u(t, z)$ evolves according to a Schrodinger equation

\begin{equation}
\displaystyle{\frac{\partial u(t, z)}{\partial t} = H_{\mu, \lambda}u(t,z)}
\end{equation}

The complete definition of the problem requires that the domain of $H_{\mu; \lambda}$ and the boundary conditions are specified.

The asymptotic behaviour of reggeon quantum mechanics (i.e., reggeon field theory without transverse dimensions) has been analyzed in a series of papers [Alessandrini et al], [Jengo],   [Bronzan et al] and [Ciafaloni et al].

However, White [White] has tried to cast doubt on the conclusion of refs. [Amati et al] and [Ciafaloni et al], as the mathematics used in ref. [Ciafaloni et al] is not beyond criticism (although it is quite respectable according to 'physicists' standards).

It was during this period that the professor Michel Lebellac of University of Nice submitted the suject to Professor Martin Zerner and that the latter in turn brought the subject to the author for a mathematical study. We present in the next sections the evolution of significant mathematical results from $1980$ to the present.

Also, we note that  Grassberger has given an interesting interpretation of RFT; following Feynman, he assumes that hadron cross sections at high energy are governed by the interactions of wee partons. If a hadron is boosted, the wee partons can split and recombine, they can be converted into hard partons, and thus are lost from the point of view of the collision, or they can recombine to give hard partons.

And with  Sundermeyer  they show that  Reggeon field theory with a quartic coupling in addition to the standard cubic one is mathematically equivalent to a chemical process where a radical can undergo diffusion, absorption, recombination, and autocatalytic production. Physically, these “radicals” are wee partons.

In Section 3,   we do not assume any prior knowledge of holomorphic function spaces. But we will have an idea of their usefulness as soon as we discover the properties of the Bargmann space which will be our ambient space for our operators throughout the work.

The Bargmann-Segal space has originally been introduced by V. Bargmann, see  [Bargmann 1], [Bargmann 2] and [Bargmann 3] and E.
Segal, see  [Segal 1],  [Segal 2] and  [Segal 3].

Despite the bountiful results on this space, we present some original results on it that we will applied to the study of our operator in particular to give an explicit inversion of $H_{\mu, \lambda}, \mu > 0$  on the axis of negative imaginaries ($z = - iy ; y \geq 0$) and to study an extension of integral operator associated to $H_{\mu, \lambda} ^ {- 1} $ at a Hilbet-Shmidt operator on $\displaystyle{L_{2}(]-\infty, 0], \theta(y)dy)}$ with a suitable choice of the function  $\theta$ to prove the existence of the eigenvalues of $H_{\mu, \lambda} $.

Notice that if $\mu = 0$, $H_{0, \lambda}$ is chaotic. A complete spectral analysis  of this operator is given in [Intissar 11](1998) and the study of its chaoticity is is given in [Intissar 2 et al] (2017).

In similar way, we show that the integral operator $H_{\lambda', \mu, \lambda} ^ {- 1}, \lambda' \neq 0 $ can be extended toi Hilbet-Shmidt operator on $\displaystyle{L_{2}(]-i\rho', 0], r(y)dy)}$; $\rho' = \frac{\lambda}{\lambda'}, \lambda' \neq 0$ with a suitable choice of the function  $r(y)$. these techniques allow us to show that :

(i)  The spectral radius of $\displaystyle{ H_{\lambda', \mu, \lambda} ^ {- 1} }$ converges to spectral radius of $\displaystyle{ H_{\mu, \lambda} ^ {- 1} }$ as $\lambda' \longrightarrow 0$.

(ii) Each of the above spectral radius can be extended analytically over $\mathbb{R}$ with respect to the parameter $\mu$.

Notice also that it is shown in this section that $\displaystyle{ H_{\mu, \lambda} ^ {- 1} }$ is the class of Carlamann $\displaystyle{\mathcal{C}_{1+ \epsilon}}$ of order $ 1 + \epsilon $   $\forall \,\, \epsilon > 0$.

where  we define  the class of Carleman $\mathcal{C}_{p}$ of order $p$ as follow :

Let $K$ be a compact operator in the Hilbert space $\mathcal{H}$. $K$ is said to belong to the Carleman-class $\mathcal{C}_{p}, p > 0$, with order $p$, if the
series $\displaystyle{\sum_{n=1}^{+\infty}s_{n}^{p}(K)}$. converges, where $s_{n}(K)$. are the eigenvalues of the operator $\displaystyle{\sqrt{K^{*}K}}$.

In the particular case $p = 2$, $\mathcal{C}_{2}$ is exactly the Hilbert-Schmidt class and for $p = 1$, $\mathcal{C}_{1}$ is said class of nuclear operators or trace operators.
For a systematic treatment of the operators of Carleman-class, we refer to the Gohberg and Krein's book [Gohberg et al].

From a mathematical point of view, one of the results which greatly contributed to meeting the desires of physicists is the work of professors T.Ando and M.Zerner (1984) on the extension of the small eigenvalue of $H_{\mu, \lambda} $ over any $\mathbb{R}$ with respect to the parameter $ \mu $  [Ando et al].

We have regularized $H_{\mu, \lambda}$  by adding a perturbation of the form $\displaystyle {\lambda' A ^ {* ^ {2}} A ^ {2} }$ which greatly facilitates the study of the operator:

\begin{equation}
\displaystyle{H _ {\lambda ', \mu, \lambda} = \lambda' A ^ {* ^ {2}} A ^ {2} + \mu A ^ {*} A + i \lambda A ^ {*} ( A + A ^ {*}) A }
\end{equation}

\begin{equation}
\displaystyle{H _ {\lambda ', \mu, \lambda} = (\lambda'z^{2} + i\lambda z)\frac{\partial^{2}}{\partial z^{2}} + (i \lambda z^{2} + \mu z) \frac{\partial}{\partial z}}
\end{equation}

For $\lambda' \neq 0$, $ H_{\lambda', \mu, \lambda}$ has a rich set of spectral properties desired by physicists  and which allows us to deduce the results of Ando-Zerner when $\lambda'$ approaches zero.

In Section 4, for the study of eigenvalue problem associated to $ H_{\lambda', \mu, \lambda}$, $\lambda' \neq 0$, we begin by to study  the singularity  at $\infty$  of

\begin{equation}
\displaystyle{ H_{\lambda',\mu, \lambda}\varphi(z) = \sigma(\lambda', \mu, \lambda) \varphi(z)}
\end{equation}

i.e. We consider the equation

\begin{equation}
\displaystyle{(\lambda'z^{2} + i\lambda z)\varphi^{''}(z) + (i \lambda z^{2} + \mu z)\varphi^{'}(z) = \sigma(\lambda', \mu, \lambda) \varphi(z)  \, at \,\quad \infty}
\end{equation}

This equation  belongs to family of the bi-confluent  Heun equations of class $(0, 1, 1_{4})$  which has three regular singular points $ z_{0} = 0$ and $z_{1} = -i \rho'$ where $\displaystyle{\rho' = \frac{\lambda}{\lambda'}}$; $\lambda' \neq 0$ and one irregular singular point $z_{\infty} = \infty$.

If $ \lambda' = 0$,  it has one regular singular points $ z_{0} = 0$ and one irregular singular point $z_{\infty} = \infty$.

As  $\displaystyle{ \mathbb{B}_{0} = \{ \varphi \in \mathbb{B}; \varphi (0) = 0 \}}$  is invariant under the  actions of $ H_{\lambda', \mu, \lambda}$ and $H_{\mu, \lambda}$ respectively then it is naturl to take $\varphi(0) = 0$ and at $\infty $  we  will prove that a solution $\varphi$ of (0.10)  is in Bargmann space.

Let $\psi \in \mathbb{B}_{0}$, we consider the equation  $H_{\lambda',\mu,\lambda}\varphi = \psi$  (respectively  $H_{\mu,\lambda}\varphi = \psi$) which can be written under following form :

\begin{equation}
\displaystyle{(\lambda'z^{2} + i\lambda z)\varphi''(z) + (i\lambda z^{2}  + \mu z)\varphi'(z) = \psi(z)}
\end{equation}

respectively :

\begin{equation}
\displaystyle{ i\lambda z\varphi''(z) + (i\lambda z^{2}  + \mu z)\varphi'(z) = \psi(z)}
\end{equation}

Now if $\psi \in \mathbb{B}_{0}$ and choosing the straight line connecting $-i\rho' , z \in \mathbb{C}$ parametrized  by

\begin{equation}
\displaystyle{\gamma : [0, 1] \rightarrow \mathbb{C}, \gamma(t) = -i\rho' + t(z + i\rho') \quad \quad (\gamma(0) = -i\rho' , \gamma(1) = z)}
\end{equation}

Then we define $\displaystyle{\int_{-i\rho'}^{z}\psi(\xi)d\xi:=\int_{\gamma}\psi(\xi)d\xi:= \int_{0}^{1}}\psi(\gamma(t))\gamma'(t)dt$ to deduce that our  equation can be transformed to following integral equation :

\begin{equation}
\displaystyle{\varphi(z) = \frac{1}{\lambda'}\int_{-i\rho'}^{z}e^{-i\rho'\eta}(\eta + i\rho')^{-(\delta +1)} [\int_{-i\rho'}^{\eta}e^{i\rho'\xi}(\xi + i\rho')^{\delta}\frac{\psi(\xi)}{\xi}d\xi]d\eta}
\end{equation}

In the same way we get on $]-i\infty, 0]$ an integral  equation with $\lambda' = 0$

However, the integral representation of $\varphi(z)$ in  the equations is hard to study in $\mathbb{C}$ for some existing results on  eigenvalues and eigenfunctions of our operators.

To overcome this difficulty, the study of integral equations associated to our equations are restricted on negative imaginary axis by applying an original result on Bargmann space  which say :

If $\varphi \in \mathbb{B}$ then its restriction on $x + i\mathbb{R}$ is square integrable function with measure $\displaystyle{e^{-|y|^{2}} dy, y \in  \mathbb{R}}$ for all fixed $x \in  \mathbb{R}$.

Observe the difference of above result  between the version for almost $x \in \mathbb{R}$ which  follows easily from Fubini theorem and the version  all fixed $x \in  \mathbb{R}$  which was showed  for the first time by the author  in 1987 in [Intissar2] which has been reproduced at the end of the section above.

Therefore led to put $u(y) = \varphi(-iy)$ ; $y \in [0, + \infty[$ to study the following problems :

\begin{equation}
\displaystyle{-\lambda y u_{\mu, \lambda}^{''}(y) + (\lambda y^{2} + \mu y)u_{\mu, \lambda}^{'}(y) = \sigma(\mu, \lambda)u_{\mu, \lambda}(y)}
\end{equation}

\begin{equation}
\displaystyle{(\lambda' y^{2} - \lambda y)u_{\lambda', \mu, \lambda}^{"}(y) + (\lambda y^{2} + \mu y)u_{\mu, \lambda}^{'}(y) = \sigma(\lambda', \mu, \lambda)u_{\lambda', \mu, \lambda}(y)}
\end{equation}

Let  $\displaystyle{\rho'  = \frac{\lambda}{\lambda'}; \lambda' \neq 0 \, > 0,  y \in [0, \rho']}$ and $\displaystyle{L_{2}((0, \rho'), r(y)dy)}$ be the Hilbert space of square integrable functions on $[0, \rho']$ with inner product

\begin{equation}
\displaystyle{< u, v > = \int_{0}^{\rho'}u(y)\overline{v(y)}r(y)dy}
\end{equation}

with $r(y)$ called weight function.

In this section we restrict (0.10) on the axis of negative imaginaries ($z = - iy ; y \geq 0$) that  gives the following equation :

\begin{equation}
\displaystyle{-y (\lambda - \lambda' y)u^{"}(y) + y(\lambda y +\mu y)u'(y) = \sigma(\lambda', \mu, \lambda)u(y)}
\end{equation}

with $u(y) = \varphi(-iy)$ ; $y \geq 0$ and we construct the fonction $r(y)$ such that $H_{\lambda', \mu, \lambda}$ is symmetric on $\displaystyle{L_{2}((0, \rho'), r(y)dy)}$.

Precisely, we get :

If $\displaystyle{\rho'(\rho + \rho')  \geq 1}$  where  $\displaystyle{\rho = \frac{\mu}{\lambda}; \lambda \neq 0 }$ then $H_{\lambda', \mu, \lambda}$ is symmetric on $\displaystyle{L_{2}((0, \rho'), r(y)dy)}$ and takes the following form

\begin{equation}
\displaystyle{H_{\lambda', \mu, \lambda}u(y) = \frac{1}{r(y)}\frac{\partial}{\partial y}(p(y)u'(y))},  where \displaystyle{p(y) = \lambda' y(\rho' - y) r(y)}
\end{equation}

In Section 5, we recall a list of fundamental spectral properties of $H_{\lambda',\mu, \lambda}$ that we have presented in above sections and we give an explicit inverse of $H_{\lambda', \mu, \lambda.}$ and  $H_{\mu, \lambda.}$ respectively.

$\bullet$ Precisely, for $ y \in [0, \rho^{'}]$, $\displaystyle{\rho' = \frac{\lambda}{\lambda'}}$ and $\displaystyle{\rho = \frac{\mu}{\lambda}}$, we get:

\begin{equation}
\displaystyle{H_{\lambda', \mu, \lambda}^{-1} \varphi(-iy) = \int_{0}^{\rho'}\mathcal{N}_{\lambda', \mu,\lambda}(y, s)\varphi(-is)ds}
\end{equation}

where $\displaystyle{\mathcal{N}_{\mu,\lambda}(y, s) = \frac{1}{\lambda s}e^{\rho' s}(1 -  \frac{s}{\rho'})^{\delta}\int_{0}^{min(y, s)}e^{-\rho' u }(1 - \frac{u}{\rho'})^{-(\rho + 1)} du}$

$\bullet$ For $ y \in [0, +\infty[$, we get:

\begin{equation}
\displaystyle{H_{\mu, \lambda}^{-1} \varphi(-iy) = \int_{0}^{\infty}\mathcal{N}_{\mu,\lambda}(y, s)\varphi(-is)ds}
\end{equation}

where  $\displaystyle{\mathcal{N}_{\mu, \lambda}(y, s) = \frac{1}{s}e^{-\frac{s^{2}}{2} - \frac{\mu}{\lambda}s}\int_{0}^{min(y, s)}e^{\frac{u^{2}}{2} +\frac{\mu}{\lambda}u}du}$

In Section 6, we shall consider a solution $u$ of the operational differential equation

\begin{equation}
\displaystyle{u'(t) =  H_{ \lambda'', \lambda', \mu , \lambda}u(t) }  for t > 0
\end{equation}

$\displaystyle{(\lambda'', \lambda', \mu, \lambda) \in \mathbb{R}^{4}}$

\begin{equation}
\displaystyle{H_{ \lambda'', \, \lambda', \, \mu, \, \lambda} = \lambda'' G + \lambda' S + \mu \mathcal{N} + i\lambda H_{I} ; \quad  i^{2} = -1}
\end{equation}

\begin{equation}
\displaystyle{G = A^{*^{3}}A^{3}}
\end{equation}

\begin{equation}
\displaystyle{S = A^{*^{2}}A^{2}}
\end{equation}

\begin{equation}
\displaystyle{\mathcal{N} = A^{*}A}
\end{equation}

\begin{equation}
\displaystyle{H_{I} = A^{*}(A + A^{*})A }
\end{equation}

to establish the following results

$\bullet$ For $\lambda'' > 0$ , we prove that the system of generalized eigenvectors of the operator  $H_{ \lambda'', \lambda', \mu , \lambda}$ is an unconditional basis in Bargmann space $\mathbb{B}$.

$\bullet$   For $\lambda'' > 0$, we give a generalized diagonalization in the Abel sense of the semigroups associated to $H_{ \lambda'', \lambda', \mu , \lambda}$ and to its square root, respectively.

$\bullet$   For $\lambda'' = 0$ , $\lambda' > 0$ and $\displaystyle{\lambda'^{2} \leq \mu \lambda' + \lambda^{2}}$, we prove  that the solution of Cauchy's problem :

\begin{equation}
\displaystyle{u'(t) = - H_{ 0, \lambda', \mu , \lambda}u(t) } ; t > 0 \quad and \quad u(0) = \phi \in \mathbb{B}
\end{equation}

can be expanded  in the following form :

$\displaystyle{u(t) = \sum_{k=1}^{+\infty}\frac{< \phi, \phi_{k}^{*} >}{\phi_{k}, \phi_{k}^{*} >}e^{-\sigma_{k}t}\phi_{k}}$.

$\displaystyle{H_{ 0, \lambda', \mu , \lambda}\phi_{k} = \sigma_{k} \phi_{k}}$ and $\displaystyle{H_{ 0, \lambda', \mu , \lambda}^{*}\phi_{k}^{*} = \sigma_{k} \phi_{k}^{*}}$.

$\bullet$   For $\lambda'' = 0$ and $\lambda' = 0$ , we consider the Cauchy problem :

\begin{equation}
\displaystyle{ u'(t) = - H_{ 0, 0, \mu , \lambda}u(t) , t > 0} \quad ; \quad  u(0) = \phi \in \mathbb{B}
\end{equation}

For $\mu > 0$, the semigroup $e^{-tH_{ 0, 0, \mu , \lambda}}$ associated to $H_{ 0, 0, \mu , \lambda}$ is compact in particular we have

\begin{equation}
\sigma (e^{-tH_{ 0, 0, \mu , \lambda}}) = e^{-t\sigma(H_{ 0, 0, \mu , \lambda})} \cup \{0\}
\end{equation}

Let $\sigma_{0}$ et $\sigma_{1}$ are respectively the smallest and the second eigenvalue of the operator $H_{ 0, 0, \mu , \lambda}$.

\begin{equation}
(i) \mid\mid e^{-tH_{ 0, 0, \mu , \lambda}} \mid\mid = e^{-\sigma_{0}t} + O(e^{-\epsilon t})\, \forall \epsilon < \sigma_{1}
\end{equation}

\begin{equation}
(ii) < e^{-tH_{ 0, 0, \mu , \lambda}}\phi, \psi > = <\phi, \phi_{0} > <\phi_{0} , \psi >e^{-\sigma_{0}t} + O(e^{-\sigma_{0}t})\, \forall \phi \in \mathbb{B}, \forall \psi \in \mathbb{B}
\end{equation}

where $\phi_{0}$  is the eigenfunction of  $H_{ 0, 0, \mu , \lambda}$ associated to $\sigma_{0}$.

$\bullet$ Notice that $H_{I}$ does not generates a semigroup in Bargmann space, which implies that we cannot apply the usual Trotter formula to the operator $H_{ 0, 0, \mu , \lambda}$.

Nevertheless  in 2005,  we have given in [Intissar 13] a “generalized Trotter product formula” for this operator $H_{ 0, 0, \mu , \lambda}$.

In Section 7,   we consider the operator :

\begin{equation}
\displaystyle{H_{\lambda'', \mu, \lambda} = \lambda''G + H_{\mu, \lambda}}
\end{equation}

where $\displaystyle{G = A^{*^{3}}A^{3}}$ and $\lambda''$ is called  the magic coupling of Pomeron by the specialists of high energies.

AS $\displaystyle{H_{\lambda'', \mu, \lambda}}$ is a  ordinary differential operator, we present a regularized trace formula for this family operators.

Notice that the theory of regularized traces of ordinary differential operators has a long history. First, the trace formulas for the Sturm-Liouville operator with the Dirichlet boundary conditions and sufficiently smooth potential $q(x)$ were established in [[Dikii 1],  [Gelfand et al]].

Afterwards these investigations were continued in many directions, for instance, the trace formulas for the Sturm-Liouville operator with periodic or antiperiodic
boundary conditions were obtained in [[Lax], [Sansuc et al]

and for regular but not strongly regular ones [Naimark]

similar formulas were found in [Makin].

A method for calculating trace formulas for general problems involving ordinary differential equations on a finite interval was proposed in [Lidskii et al].

The bibliography on the subject is very extensive and we refer to the list of the works in [[Levitan et al], [Sadovnichi et al] (2006)].

The trace formulas can be used for approximate calculation of the first eigenvalues of an operator.

we use the results of  [Sadovnichi et al] (2002)] to give the number of corrections sufficient for the existence of finite formula of the trace of  $\displaystyle{H_{\lambda'', \mu, \lambda}}$.

Precisely we prove the following statement.

Let $ \mathbb{B}$ be the Bargmann space, $\displaystyle{ H_{\lambda'', \mu, \lambda} = \lambda{''}G + H_{\mu,\lambda}}$ acting on $\mathbb{B}$ where $ G = A^{*^{3}}A^{3}$ and $ H_{\mu,\lambda} = \mu A^{*}A + i\lambda A^{*}( A + A^{*})A$ , $A$ and $A^{*}$ are the standard Bose annihilation and creation operators.

Then there exists an increasing sequence of radius $r_{m}$ such that $r_{m} \rightarrow \infty$ as $ m \rightarrow \infty$  and

\begin{equation}
\displaystyle{\lim\limits_{m \rightarrow \infty} (\sum_{n=0}^{m}(\sigma_{n} - \lambda{''}\lambda_{n}) + \frac{1}{2i\pi}\int_{\gamma_{m}} Tr[\sum_{k=1}^{4}\frac{(-1)^{k-1}}{k}[H_{\mu,\lambda}(\lambda{''}G - \sigma I)^{-1}]^{k}]d\sigma) = 0}
\end{equation}

- $\sigma_{n}$ are the eigenvalues of the operator $ H_{\lambda'', \mu, \lambda}  = \lambda{''}G + H_{\mu,\lambda}$

- $\lambda_{n} = n(n-1)(n-2)$ are the eigenvalues of the operator $ G $

- $ (\lambda{''}G - \sigma I)^{-1}$ is the resolvent of the operator $\lambda{''}G$

- $\gamma_{m}$ is the circle of radius $r_{m}$ centered at zero in complex plane.

In Section 8, We show that  $\displaystyle{- H_{\lambda'', \mu, \lambda}}$  generates a semigroup $e^{-t\mathbb{H}_{\lambda''}}$ of Carleman class $\mathcal{C}_{p}$ for all $ p > 0$ and all $ t > 0$. In particular, we go to derive an asymptotic expansion of the trace of $e^{-t\mathbb{H}_{\lambda''}}$ as $t\rightarrow 0^{+}$.

Precisely we prove the following statement.

\begin{equation}
\mid\mid e^{-tH_{\lambda''}} - e^{-t\lambda''G} \mid\mid_{1} = t\mid\mid e^{-t\lambda''G} H_{\mu,\lambda}\mid\mid_{1} + \mid\mid (\lambda''G)^{\delta} e^{-\frac{t}{3}\lambda''G }\mid\mid_{1}O(t^{2})
\end{equation}

$\displaystyle{\forall \, \delta \geq \frac{1}{2}}$.

§ PRESENTATION OF NON-NORMAL OPERATORS ARISING IN REGGEON FIELD THEORY

§.§ A short introduction of Reggeon Field Theory (RFT)

In this section we present a short introduction of Reggeon Field Theory (RFT) without transverse dimensions and the operators which characterize this theory.

The Reggeon field theory (see Poghosyan in  [Poghosyan] for an elementary introduction to this theory) is one of the attempts to understand strong interactions that is, the interactions between, among other less stable particles, protons and neutrons. For Gribov's reggeon calculus, we can see   [Gribov1], [Gribov2],   [Intissar1 et al] and  [Baker et al.]

Many experiments in physics (atomic and nuclear physics, solid state physics and high energy physics) consist of directing a beam of particles (1) onto a target of particles (2), and studying the resulting collisions. This is possible since the various particles constituting the final state of the system (state after the collision) are
detected and their characteristics (direction of emission, energy, etc.) are measured.

Aim of such experiments: Determine the interactions that occur between the various particles entering the collision. The collisions give rise to reactions:

$$\displaystyle{{\color{red}(1) + (2) \quad {\color{blue}\longrightarrow} \quad  (3) + (4) + (5) + . . . .}}$$

In quantum mechanics, we can only speak of probabilities of the possible states coming out of the collision. Among the reactions possible under given conditions, scattering
reactions are defined as those in which the final state and the initial state are composed of the same particles (1) and (2). In an elastic scattering reaction, none of the particles? internal states change during the collision.

The experimental quantity (see [Roy] and  [Levin1, Levin2]) that is wanted to be computed is the total cross-section that is, the proportion of particles that are destroyed to create new ones.

§.§ The representation of Hamiltonian of Reggeon Field Theory in Bargmann space

In Bargmann representation [Barg1] and [Barg2], the Reggeon's field theory in zero transverse dimensions is characterized by Gribov's operator [Gribov1]:

\begin{equation}
H_{\mu, \lambda} = \mu A^{*}A + i\lambda A^{*}(A + A^{*})A
\end{equation}

where $\mu$ and $\lambda$ are real parameters , ($\mu$ is Pemeron's intercept, $\lambda$ is  triple coupling of Pemeron and $i^{2} = -1$.

$A^{*}$ and $A$, are the creation and annihilation operators represented in Bargmann representation by $\displaystyle{A = \frac{\partial}{\partial z}}$  and  $A^{*} = z$  where $z = x + iy ; (x, y) \in \mathbb{R}^{2}$. These operators satisfy the commutation relation:

\begin{equation}
[A, A^{*}] = I
\end{equation}

The operator $H_{\mu, \lambda}$ can be regularized by  adding the term $\displaystyle{\lambda'A^{*^{2}}A^{2}}$ to (2.1)  to get:

\begin{equation}
H_{\lambda', \mu, \lambda} = \lambda' A^{*^{2}}A^{2}  + \mu A^{*}A + i\lambda A^{*}(A + A^{*})A
\end{equation}

where $\lambda'$ is quartic Reggeon coupling

The operator $H_{\lambda', \mu, \lambda}$ can be regularized by adding the term $\displaystyle{\lambda'' A^{*^{3}}A^{3}}$ to (2.3) to get:

\begin{equation}
H_{\lambda'', \lambda', \mu, \lambda} =\lambda'' A^{*^{3}}A^{3} +  \lambda' A^{*^{2}}A^{2}  + \mu A^{*}A + i\lambda A^{*}(A + A^{*})A
\end{equation}

where $\lambda''$ is magic Reggeon coupling

In Bargmann representation :

\begin{equation}
\displaystyle{  \mathbb{B }= \{\varphi : \mathbb{C} \longrightarrow \mathbb{C}  \, entire \, ; \int_{\mathbb{C} }\mid \varphi(z)\mid^{2}e^{-\mid z \mid^{2}}dx dy < \infty \}}
\end{equation}

the operators $H_{\mu, \lambda}$, $H_{\lambda',\mu, \lambda}$ and $H_{\lambda'', \lambda',\mu, \lambda}$ takes respectively  the  following form:

\begin{equation}
\displaystyle{H_{\mu, \lambda} = i\lambda z\frac{\partial^{2}}{\partial z^{2}} +  (i\lambda z^{2} + \mu z)\frac{\partial}{\partial z}}
\end{equation}

acting on its maximal domain $\displaystyle{D(H_{\mu, \lambda}) = \{\varphi \in  \mathbb{B} ; H_{\mu, \lambda} \varphi \in  \mathbb{B}\}}$.

\begin{equation}
\displaystyle{H_{\lambda', \mu, \lambda} = (\lambda' z^{2} + i\lambda z)\frac{\partial^{2}}{\partial z^{2}} +  (i\lambda z^{2} + \mu z)\frac{\partial}{\partial z}}
\end{equation}

acting on its maximal domain $\displaystyle{D(H_{\lambda', \mu, \lambda}) = \{\varphi \in  \mathbb{B} ; H_{\lambda' , \mu, \lambda} \varphi \in  \mathbb{B}\}}$.

\begin{equation}
\displaystyle{H_{\lambda'', \lambda', \mu, \lambda} = \lambda'' z^{3}\frac{\partial^{3}}{\partial z^{3}} +  (\lambda' z^{2} + i\lambda z)\frac{\partial^{2}}{\partial z^{2}} +  (i\lambda z^{2} + \mu z)\frac{\partial}{\partial z}}
\end{equation}

acting on its maximal domain $\displaystyle{D(H_{\lambda'', \lambda', \mu, \lambda}) = \{\varphi \in  \mathbb{B} ; H_{\lambda'', \lambda' , \mu, \lambda} \varphi \in  \mathbb{B}\}}$.

Now, we suppose that $ \lambda'' > 0$, $\lambda' > 0$, $\mu > 0$ and $\lambda \neq 0$, hence  we consider the evolution problems associated respectively to $H_{\mu, \lambda}$, $H_{\lambda', \mu, \lambda}$ and $H_{\lambda'', \lambda', \mu, \lambda}$ with their eigenvalue problems:

\begin{equation}
\left \{ \begin{array} [c] {l}\displaystyle{ \frac{\partial u(t, z)}{\partial t} = - H_{\mu, \lambda} u(t, z); \quad t > 0, z \in \mathbb{C}} \\ \quad\\
\displaystyle{ u(0, z) =\phi(z) ; \,  \phi \in \mathbb{B} } \\
\end{array} \right.
\end{equation}

with associated eigenvalue problem.

\begin{equation}
\left \{ \begin{array} [c] {l}\displaystyle{H_{\mu, \lambda}\varphi_{\mu, \lambda} = \sigma(\mu, \lambda)\varphi_{\mu, \lambda}}\\ \quad \\
\varphi_{\mu, \lambda} \in D(H_{\mu, \lambda}), \sigma(\mu, \lambda) \in \mathbb{C}\\
\end{array} \right .
\end{equation}

In (2.9) the physical literature does not put the minus sign. But it very different from a real Schrodinger equation for two reasons at least :

(i) $t$ is not time but rapidity (the logarithm of collision energy).

(ii) The $i$ factor is missing and anyway $H_{\mu, \lambda}$ (called three pomerons coupling) is neither self-adjoint nor skew-adjoint. As will be seen, this is the major source of mathematical difficulties.

(iii) At first glance it is not obvious that imaginary time can be used instead of real time. To motivate the use of imaginary time we can consult  [Eppens] (section 4.3, pp: 32-34) entitled “Real time vs. Imaginary time ” where the imaginary time Schrodinger equation is used to simplify the numerical simulation of quantum adiabatic computation.

The perturbation of $H_{\mu, \lambda}$ by $\lambda'A^{*^{2}}A^{2}$ (four Pomeron) with the parameter $\lambda' $ (called four coupling of Pomeron) allows us to regularize   $H_{\mu, \lambda}$, then we also consider the following problem:

\begin{equation}
\left \{ \begin{array} [c] {l}\displaystyle{ \frac{\partial u(t, z)}{\partial t} = - H_{\lambda' , \mu, \lambda} u(t, z); \quad t > 0, z \in \mathbb{C}} \\ \quad\\
\displaystyle{ u(0, z) =\phi(z) ; \,  \phi \in \mathbb{B} } \\
\end{array} \right.
\end{equation}

with associated eigenvalue problem.

\begin{equation}
\left \{ \begin{array} [c] {l}\displaystyle{H_{\lambda',\mu, \lambda}\varphi_{\lambda', \mu, \lambda} = \sigma(\lambda', \mu, \lambda)\varphi_{\lambda', \mu, \lambda}}\\ \quad \\
\varphi_{\lambda', \mu, \lambda} \in D(H_{\lambda', \mu, \lambda}), \, \sigma(\lambda', \mu, \lambda) \in \mathbb{C}\\
\end{array} \right .
\end{equation}

The perturbation of $H_{\lambda', \mu, \lambda}$ by $\lambda''A^{*^{3}}A^{3}$ where the parameter $\lambda'' $ is called by the specialists of high energies “magic coupling of Pomeron” allows us to regularize   $H_{\lambda', \mu, \lambda}$, then we also consider the following problem:

\begin{equation}
\left \{ \begin{array} [c] {l}\displaystyle{ \frac{\partial u(t, z)}{\partial t} = - H_{\lambda'', \lambda' , \mu, \lambda} u(t, z); \quad t > 0, z \in \mathbb{C}} \\ \quad\\
\displaystyle{ u(0, z) =\phi(z) ; \,  \phi \in \mathbb{B} } \\
\end{array} \right.
\end{equation}

with associated eigenvalue problem.

\begin{equation}
\left \{ \begin{array} [c] {l}\displaystyle{H_{\lambda'', \lambda',\mu, \lambda}\varphi_{\lambda'', \lambda', \mu, \lambda} = \sigma(\lambda'', \lambda', \mu, \lambda)\varphi_{\lambda'', \lambda', \mu, \lambda}}\\ \quad \\
\varphi_{\lambda', \mu, \lambda} \in D(H_{\lambda', \mu, \lambda}), \, \sigma(\lambda'', \lambda', \mu, \lambda) \in \mathbb{C}\\
\end{array} \right .
\end{equation}

And the abstract elliptic problem on $\mathbb{B}$ of Dirichlet type,

\begin{equation}
\left \{ \begin{array} [c] {l}\displaystyle{ \frac{\partial^{2} u(t, z)}{\partial t^{2}} = - H_{\lambda'', \lambda' , \mu, \lambda} u(t, z); \quad t > 0, z \in \mathbb{C}} \\ \quad\\
\displaystyle{ u(0, z) =\phi(z) ; \,  \phi \in D(H_{\lambda'', \lambda' , \mu, \lambda}) } \\
\end{array} \right.
\end{equation}

i.e. to consider the evolution problem associated to $\displaystyle{H_{\lambda'', \lambda' , \mu, \lambda}^{\frac{1}{2}}}$ and to show that Then the solution $u(t)$ can  be expanded in $\mathbb{B}$ into the series :

$\displaystyle{u(t) = \sum_{k=1}^{+\infty}P_{k}e^{-tH_{\lambda'', \lambda' , \mu, \lambda}^{\frac{1}{2}}} = \sum_{k=1}^{+\infty}e^{-tH_{\lambda'', \lambda' , \mu, \lambda}^{\frac{1}{2}}}P_{k}}$ fot $t >0$.

where the $P_{k}$'s are the projector operators given by $\displaystyle{P_{k} = \frac{1}{2i\pi}\int_{C_{k}}^{ } (H_{\lambda'', \lambda' , \mu, \lambda} - \sigma I)^{-1}d\sigma}$  with the $C_{k}$'s as small circles centered at the eigenvalues $\{\sigma_{k}\}$ of the operator $H_{\lambda'', \lambda' , \mu, \lambda}$ in the Bargmann space $\mathbb{B}$.

The complete definition of the above problems requires that the domains respectively  of $H_{\mu, \lambda}$, $H_{\lambda',\mu, \lambda}$ and $H_{\lambda'', \lambda',\mu, \lambda}$ and the boundary conditions be specified.

The asymptotic behaviour of Reggeon quantum mechanics (i.e., Reggeon field theory without transverse dimensions) has been analyzed by certain specialists of high energies in a series of papers see [Alessandrini et al], [Jengo], [Bronzan et al] and [Ciafaloni et al].

Around the 1980s a controversy raised by Professor R; White with his physic colleagues on certain results established in certain articles by Professors Amati, Ciafaloni and Le bellac without being demonstrated mathematically.

We recall that the Reggeon field theory (RFT) is an attempt to predict the high-energy behaviour of soft processes; the RFT can be derived from the assumed softness of hadronic interactions at low transverse momenta, which seems to be well established experimentally in hadron-hadron and hadron-nucleus interactions. Unfortunately, nobody has been able up to now to derive the RFT from the fashionable theory of strong interactions, quantum chromodynamics. For this reason, the RFT is at present somewhat out of fashion, but it must be remembered that it is the most elaborate model of soft processes, and may well prove to be a valuable tool for the analysis of forthcoming data from new colliders. We think
that it is still an important task to understand its spectral properties in particular the the Reggeon field theory for $n$ sites which have been began by the author in $1987$ in [Intissar 2].

The most important question is that of the asymptotic behaviour of total cross sections. Since it seems that the intercept of the bare Pomeron $\alpha := 1 -\mu$ is above one, we have to deal with the problem of a "supercritical" RFT. A solution to the problem has been given in ref.  [Amati et al], and one of its ingredients is the solution of the RFT in zero transverse dimension. In that case, the hamiltonian is given by (2.1),

The properties of $H_{\mu, \lambda}$ have been studied in ref.  [Amati et al], and this study has been refined in ref.  [Ciafaloni et al].

However around the 1980s, White  [White] has tried to cast doubt on the conclusion of refs.  [Amati et al]  and [Ciafaloni et al], as the mathematics used in ref.  [Ciafaloni et al ] is not beyond criticism (although it is quite respectable according to physicists' standards).

It was during this period that the professor Michel Lebellac of University of Nice submitted the subject to Professor Martin Zerner and that the latter in turn brought the subject to the author for a mathematical study.

We present in the next sections the evolution of significant mathematical results from 1980 to present

Ciafaloni-Onofri in [Ciafaloni-Onofri] have confirmed the results of ref.  [Ciafaloni et al] by adding a term $\displaystyle{ \lambda'A^{*^{2}}A^{2}}$ to (2.1) and taking the limit $\lambda' = 0$. However, it is worthwhile to have a completely rigorous study of the hamiltonian (2.1), and this work sums up such a mathematic study which confirms the results of ref. [Ciafaloni et al]. Then there cannot be any doubt about the high-energy behaviour of a supercritical RFT.

The mathematical difficulties of this problem come of course from the non-self-adjointness of $H_{\lambda'', \lambda' , \mu, \lambda}$; $\lambda \neq 0$. Notice that this non-self-adjointness is a rather wild one; the
word “wild” meaning here that the domains of the adjoint and anti-adjoint parts are not included in one another, nor is the domain of their commutator. To make
this more precise we want to point out two unpleasant properties of non-self-adjoint operators. First recall that a complex number $z$ is said to be in the spectrum of an operator $V$ if $V - zI$ has no inverse and it is an eigenvalue of V if there is some non-zero vector $u$ such that $Vu = zu$. Now if $V$ is compact, zero is in the spectrum and all other points of the spectrum are eigenvalues. This is a rather powerful result which we will use; but it should not fool us into thinking that all compact operators have eigenvalues (not to speak of a basis of eigenvectors!).

For instance look at the operator $V$ defined by $\displaystyle{V(u(y) = \int_{0}^{y}u(s)ds}$

It is a compact operator in $L^{2}(0,1)$ (and in fact in any reasonable Banach space of functions); still it has no eigenvalue whatsoever. The second property has to do with the relationship between the spectrum and the asymptotic behaviour of $\displaystyle{e^{ -tH_{\mu, \lambda}}}$, $\displaystyle{e^{ -tH_{ \lambda', \mu, \lambda}}} $ and $\displaystyle{e^{ -tH_{ \lambda'',  \lambda', \mu, \lambda}}}$.

It is possible to construct an operator $V$ with the two following properties:

- All points of the spectrum of $V$ are pure imaginary eigenvalues.

- There are vectors $u$ such that the norm of $\displaystyle{e^{t V}u}$ tends to $\infty $ with $t$.

§ FIRST SPECTRAL PROPERTIES OF THE HAMILTONIAN $\DISPLAYSTYLE{ H_{\LAMBDA',\MU, \LAMBDA}}$ OF  REGGEON FIELD THEORY

§.§ Some original results on Bargmann space

Let be $O(\mathbb{C})$ the set of entire functions, $\mathbb{M}$ is the operator of multiplication by $z = x + iy ; (x, y) \in \mathbb{R}^{2}$ and $i^{2} = -1$ (i.e $\mathbb{M}\varphi = z\varphi, \varphi \in O(\mathbb{C})$) and  $\mathbb{D}$ is the operator of derivative in order $z$ (i.e   $\mathbb{D}\varphi = \frac{\partial}{\partial z}\varphi, \varphi \in O(\mathbb{C})$). then we have:

If we request that $\mathbb{M}$ is adjoint to $\mathbb{D}$ in $O(\mathbb{C})$ with inner product :

$\displaystyle{  < \varphi, \psi > = \int_{\mathbb{C}} \varphi(z) \overline{\psi(z)}\rho(z, \bar{z})dxdy}$ where  $\rho(z, \bar{z})$ is a  measure.

Then $\displaystyle{ \rho(z, \bar{z}) = e^{-\mid z\mid^{2}}}$ (Gaussian measure).

The requirement that $\mathbb{M} = \mathbb{D}^{*}$ gives :

$\displaystyle{\int_{\mathbb{C}}\frac{\partial}{\partial z }[\varphi(z)]\overline{\psi(z)}\rho((z, \bar{z})dxdy =  \int_{\mathbb{C}}\varphi(z)\overline{z\psi(z)}\rho((z, \bar{z})dxdy}$

$\displaystyle{\frac{\partial}{\partial z }[\varphi(z)]\overline{\psi(z)}\rho((z, \bar{z}) = \frac{\partial}{\partial z }[\varphi(z)\overline{\psi(z)}\rho((z, \bar{z})] - \varphi(z)\frac{\partial}{\partial z }[\overline{\psi(z)}]\rho((z, \bar{z}) - \varphi(z)\overline{\psi(z)}\frac{\partial}{\partial z }[\rho((z, \bar{z})]}$

$\displaystyle{\int_{\mathbb{C}}\frac{\partial}{\partial z }[\varphi(z)]\overline{\psi(z)}\rho((z, \bar{z})dxdy = \int_{\mathbb{C}}\frac{\partial}{\partial z }[\varphi(z)\overline{\psi(z)}\rho((z, \bar{z})]dxdy - \int_{\mathbb{C}}\varphi(z)\frac{\partial}{\partial z }[\overline{\psi(z)}]\rho((z, \bar{z})dxdy }$

$\displaystyle{- \int_{\mathbb{C}}\varphi(z)\overline{\psi(z)}\frac{\partial}{\partial z }[\rho((z, \bar{z})]dxdy}$

In the right hand side, the first term of the integrand vanishes if we assume that the inner product between $\varphi$ and $\psi$  is finite, so that $\varphi \overline{\psi}\rho \longrightarrow 0$  sufficiently fast  as $\mid z \mid \longrightarrow \infty$. The second term also vanishes,because  $\psi$ is holomorphic, so that  $\overline{\psi}$ is anti-holomorphic and hence $\displaystyle{\frac{\partial}{\partial z}\overline{\psi} = 0}$.This gives:

\begin{equation}
\displaystyle{ \int_{\mathbb{C}}\varphi(z)\overline{z\psi(z)}\rho((z, \bar{z})dxdy + \int_{\mathbb{C}}\varphi(z)\frac{\partial}{\partial z }[\overline{\psi(z)}]\rho((z, \bar{z})dxdy = 0}
\end{equation}

which is solved for arbitrary $\varphi$ and $\psi$ if

\begin{equation}
\displaystyle{\bar{z}\rho((z, \bar{z}) + \frac{\partial}{\partial z }[\rho((z, \bar{z})] = 0}
\end{equation}

\begin{equation}
\displaystyle{ \rho((z, \bar{z}) = Ce^{-\mid z\mid^{2}}}
\end{equation}

.  $\square$

The constant $C$ is chosen to be $\displaystyle{\frac{1}{\pi}}$, so that the norm of the constant function $\varphi(z) = 1$ is one. This explains why the space of holomorphic functions equipped with a Gaussian measure gives Bargmann space:

\begin{equation}
\displaystyle{ \mathbb{B }= \{\varphi : \mathbb{C} \longrightarrow \mathbb{C}  \, entire \, ; \int_{\mathbb{C} }\mid \varphi(z)\mid^{2}e^{-\mid z \mid^{2}}dx dy < \infty \}}
\end{equation}

\begin{equation}
\displaystyle{ \mathbb{B}_{s} = \{(a_{n})_{n=0}^{\infty} \in \mathbb{C} ; \sum_{n=0}^{\infty}n!\mid a_{n}\mid^{2} < \infty \}}
\end{equation}

$\mathbb{B}_{s}$ is also a Hilbert space. The scalar product on $\mathbb{B}_{s}$ is defined by

\begin{equation}
\displaystyle{ < (a_{n}) , (b_{n} >_{s} = \sum_{n=0}^{\infty}n!a_{n}\overline{b_{n}}}
\end{equation}

and the associated norm is denoted by $\mid\mid . \mid\mid_{s}$.

It is well known that  $\mathbb{B}$ is related to $\mathbb{B}_{s} $ by unitary transform of  $\mathbb{B}$ onto $\mathbb{B}_{s} $, given by the following transform :

\begin{equation}
\displaystyle{ I: \mathbb{B} \longrightarrow \mathbb{B}_{s} , \varphi(z) = \sum_{n=0}^{\infty}a_{n}z^{n} \longrightarrow I(\varphi) = (\frac{1}{n!}\varphi^{(n)}(0))_{n=0}^{\infty} = (a_{n})_{n=0}^{\infty}}
\end{equation}

As the operator of the derivative on Bargmann space $\mathbb{B}$ is unbounded then we have the following result :

$\displaystyle{\varphi \in \mathbb{B}\, \Longleftrightarrow \,   z \longrightarrow \frac{\varphi'(z) - \varphi'(0)}{z} \in \mathbb{B}}$

For an entire function  $\displaystyle{\phi(z) = }\displaystyle{\sum_{k=0}^{+\infty}a_{k}z^{k}}$  on $\mathbb{C}$ we have:

$\displaystyle{\phi'(z) = \displaystyle{\sum_{k=0}^{+\infty}(k+1)a_{k+1}z^{k}}}$, $\displaystyle{\phi'(0) = a_{1}}$  and  $\displaystyle{\frac{\phi'(z) - \phi'(0)}{z} =  \sum_{k=0}^{+\infty}(k+2)a_{k+2}z^{k}}$

By using the isometry between Bargmann space $\mathbb{B}$ and $\mathbb{B}_{s}$, we deduce that:

$\displaystyle{\mid\mid \frac{\varphi'(z) - \varphi'(0)}{z} \mid\mid^{2} =  \displaystyle{\sum_{k=0}^{+\infty}k!(k+2)^{2}\mid a_{k+2}\mid^{2}}}$

\begin{equation}
\displaystyle{\mid\mid \frac{\varphi'(z) - \varphi'(0)}{z} \mid\mid^{2} =  \displaystyle{\sum_{k=0}^{+\infty}\frac{k+2}{k+1}(k+2)!\mid a_{k+2}\mid^{2}}}
\end{equation}

As $\displaystyle{\frac{k+2}{k+1} \leq 2}$, then   $\displaystyle{\mid\mid \frac{\varphi'(z) - \varphi'(0)}{z} \mid\mid^{2}  \leq 2\sum_{k=0}^{+\infty}(k+2)!\mid a_{k+2}\mid^{2}}$,
it follows that if
$\displaystyle{\varphi \in \mathbb{B} \,\, then \,   z \longrightarrow \frac{\varphi'(z) - \varphi'(0)}{z} \in \mathbb{B}}$

For the reciprocity, we remark that  $ \displaystyle{\frac{k+2}{k+1} > 1 }$, and from (3.8) we deduce that $\displaystyle{ \mid\mid \varphi \mid\mid \leq \mid\mid \frac{\varphi'(z) - \varphi'(0)}{z} \mid\mid}$, it follows that if $\displaystyle{ z \longrightarrow \frac{\varphi'(z) - \varphi'(0)}{z} \in \mathbb{B} \,\, then \,  \varphi \in \mathbb{B} }$.   $\square$

The Segal-Bargmann transform $\displaystyle{B : L_{2}(\mathbb{R}) \longrightarrow \mathbb{B} : f  \longrightarrow \varphi}$ defined by

$\displaystyle{B[f](z) = \varphi(z) = \int_{\mathbb{R}}\mathcal{A}(z, u)f(u)du}$   ($\star$)

$\displaystyle{\mathcal{A}(z, u) = e^{-\frac{u^{2}}{2} + \sqrt{2}uz -\frac{z^{2}}{2}}}$    ($\star\star$)

is surjective isometry.

It is well Known that the proof of this theorem consists of three steps. First by showing that for any $f \in L_{2}(\mathbb{R})$ the image $Bf$ is a holomorphic function. Secondly by introducing the complete orthogonal sequences $\{f_{k}\} \in L_{2}(\mathbb{R})$ and $\{z^{k}\} \in \mathbb{B}$ and showing that $\mathcal{B}f_{k} = z^{k}$ . Lastly by gathering all the data obtained from above steps, we can conclude that $B$ maps into $\mathbb{B}$ isometrically and surjectively.   $\square$

The Segal-Bargmann transform was introduced (depently) by I. Segal and V. Bargmann near 1960 and also by F.A Berezin in the same time give a generalization of this transformation).

Let $\mathbb{B}$ be Bargmann space,  if $\varphi \in \mathbb{B}$ then its restriction on  $x + i\mathbb{R}$ is square integrable function with measure $\displaystyle{e^{-\mid y \mid^{2}}dy , y \in \mathbb{R}}$ for all  fixed $x \in \mathbb{R}$.

Bargmann has built an isometry between the space $\mathbb{B}$ and $L_{2}(\mathbb{R})$ so that,  $\forall \,\varphi \in \mathbb{B}$ is uniquely represented by $f \in  L_{2}(\mathbb{R})$ by means of the following integral:

\begin{equation}
\displaystyle{\varphi (z) = c\int_{\mathbb{R}}e^{- \frac{z^{2}}{2} - \frac{q^{2}}{2} +\sqrt{2}zq } f(q)dq  }
\end{equation}

\begin{equation}
\displaystyle{\mid\mid \varphi \mid\mid = \mid\mid f \mid\mid_{L_{2}( \mathbb{R})}}
\end{equation}

Now, we put $\displaystyle{g_{z}(q) = e^{- \frac{q^{2}}{2} +\sqrt{2}zq }}$, which is function belonging to $L_{2}(\mathbb{R})$. Then we can write (1.2.9) in the following form :

\begin{equation}
\displaystyle{e^{\frac{z^{2}}{2}}\phi (z) = }c\displaystyle{\int_{\mathbb{R}}g_{z}(q)f(q)dq  }
\end{equation}

As $g_{z}(q)$ and $f$ are in  $L_{2}(\mathbb{R})$, we can apply the Parseval's identity at (3.11) to get :

\begin{equation}
\displaystyle{e^{\frac{z^{2}}{2}}\varphi (z) = c_{1}}\displaystyle{\int_{\mathbb{R}}\hat{g}_{z}(p)\hat{f}(p)dp}
\end{equation}

As $\displaystyle{g_{z}(q)}$ is a Gaussian's function,  one knows how to calculate its Fourier transform:

$\displaystyle{\hat{g}_{z}(p) = \int_{\mathbb{R}}e^{ipq}g_{z}(q)dq =\int_{\mathbb{R}}e^{- \frac{q^{2}}{2} + (\sqrt{2}z - ip)q} dq = (2\pi)^{\frac{1}{2}}e^{\frac{1}{2}(\sqrt{2}z - ip)^{2}}}$

\begin{equation}
\displaystyle{\hat{g}_{z}(p) = (2\pi)^{\frac{1}{2}}e^{\frac{1}{2}(\sqrt{2}z - ip)^{2}}}
\end{equation}

Let $z = x + iy$, then for all fixed $x \in \mathbb{R}$, the function $\hat{g}_{z}(p)$ can be written under following form:

\begin{equation}
\displaystyle{\hat{g}_{z}(p) = (2\pi)^{\frac{1}{2}}e^{\frac{1}{2}(\sqrt{2}x - i(p - \sqrt{2}y)^{2}}}
\end{equation}

If we put $\displaystyle{h_{x}(p) = e^{x^{2}}e^{-\frac{1}{2}(p^{2} + i\frac{\sqrt{2}}{2}px)}}$ then $\displaystyle{e^{\frac{z^{2}}{2}}\varphi (z)}$ is the convolution product of $h_{x}$ with $\hat{f}$ evaluated in $\sqrt{2}y$, i.e.,

\begin{equation}
\displaystyle{e^{\frac{(x + iy)^{2}}{2}}\varphi (x + iy) = C_{2} h_{x}*\hat{f}(\sqrt{2}y) }
\end{equation}

It follows that $\displaystyle{y \rightarrow e^{\frac{(x + iy)^{2}}{2}}\varphi (x + iy)}$ is in  $L_{2}(\mathbb{R})$ for all $x \in \mathbb{R}$

By applying the Young's inequality we deduce that :

\begin{equation}
\mid\mid e^{\frac{(x + iy)^{2}}{2}}\varphi (x + iy)\mid\mid^{2}_{L_{2}(\mathbb{R})} \leq \mid\mid h_{x}\mid\mid^{2}_{L_{1}((\mathbb{R}^{N})}.\mid\mid \hat{f}\mid\mid^{2}_{L_{2}(\mathbb{R})}
\end{equation}

where $C_{1}, C_{2}$ and $C$ are constants.

As $\mid\mid h_{x}\mid\mid^{2}_{L_{1}(\mathbb{R})} = C_{3}e^{2\mid x \mid^{2}}$ et $\mid\mid \hat{f}\mid\mid^{2}_{L_{2}(\mathbb{R})} = C_{4}\mid\mid f \mid\mid^{2}_{L_{2}(\mathbb{R})}$, then we deduce that :

\begin{equation}
\displaystyle{\int_{\mathbb{R}}e^{-\mid y \mid^{2}}\mid \varphi (x + iy)\mid^{2}dy \leq C_{5}e^{-\mid x \mid^{2}}\mid\mid \varphi \mid\mid^{2}}
\end{equation}

where $C_{5}$ depends of  all the previous constants.  $\square$

(3.17) is valid on $\mathbb{C}^{n}$ by taking :

\begin{equation}
\displaystyle{ \mathbb{B }= \{\varphi : \mathbb{C}^{n} \longrightarrow \mathbb{C}  \, entire \, ; \int_{\mathbb{C}^{n} }\mid \varphi(z)\mid^{2}e^{-\mid z \mid^{2}}dx dy < \infty \}}
\end{equation}

$\displaystyle{x = (x_{1}, ..., x_{j}, ...,x_{n })\in \mathbb{R}^{n},  y = (y_{1}, ..., y_{j}, ...,y_{n })\in \mathbb{R}^{n}}$ and

$\displaystyle{ z = (z_{1}, ..., z_{j}, ...,z_{n })\in \mathbb{C}^{n}; z_{j} = x_{j} + iy_{j}}$

§.§ Representation of $H_{\lambda', \mu, \lambda}$ in $\displaystyle{\mathbb{B}_{s} = \{ (a_{n})_{n=0}^{\infty} \in \mathbb{C} ; \sum_{n=0}^{\infty} n! \mid a_{n} \mid^{2} < \infty\}}$

The Segal-Bargmann transform is a very well-known operator acting from the space $L_{2}(\mathbb{R})$ and the Segal-Bargmann-Fock space which is unitary and allows identifying these two spaces see theorem 3.3.

In this under-section, we  built an isometry $I$ between the space $\mathbb{B}$ and

$\displaystyle{\mathbb{B}_{s} = \{ (a_{n})_{n=0}^{\infty} \in \mathbb{C} ; \sum_{n=0}^{\infty} n! \mid a_{n} \mid^{2} < \infty\}}$ to represent the annihilation and creation operators in this representation $\displaystyle{\mathbb{B}_{s}}$.

Now  we recall that $\mathbb{B}$ is defined as follow

\begin{equation}
\displaystyle{ \mathbb{B }= \{\varphi : \mathbb{C} \longrightarrow \mathbb{C}  \, entire \, ; \int_{\mathbb{C} }\mid \varphi(z)\mid^{2}e^{-\mid z \mid^{2}}dx dy < \infty \}}
\end{equation}

The scalar product on $\mathbb{B}$ is defined by

\begin{equation}
<\phi,\psi> = \displaystyle {\int_{\mathbb{C}}}\displaystyle{\phi(z)\overline{\psi(z)}e^{-\mid z\mid^{2}}dxdy}
\end{equation}

and the associated norm is denoted by $\mid\mid . \mid\mid $.

$\mathbb{B}$ is closed in $L_{2}(\mathbb{C}, d\mu(z))$ where the measure $d\mu(z) = e^{-\mid z\mid^{2}}dxdy$ and it is closed related to $L_{2}(\mathbb{R} )$ by an unitary transform of $L_{2}(\mathbb{R} )$ onto $\mathbb{B}$ given in [Bargmann1] by the following integral transform

\begin{equation}
\phi(z) = \displaystyle{\int_{\mathbb{R} }\,e^{-\frac{1}{2}(z^{2} + u^{2}) +\sqrt{2}uz}f(u)du }
\end{equation}

if $f$ $\in L_{2}(\mathbb{R} )$ the integral converges absolutely.

Let $\phi $ be in $\mathbb{B}$ with $\phi(z) = \displaystyle{\sum_{n=0}^{\infty}a_{n}z^{n}; (a_{n})_{n=0}^{\infty} \in \mathbb{C} }$ then
$\displaystyle{\mid\mid \phi \mid\mid^{2} = \pi \sum_{n=0}^{\infty}n!\mid a_{n}\mid^{2}}$

Let $\phi$ be in $B$ and we put $z = re^{i\theta}$ with $r \in [0, +\infty[$ and $\theta \in [0, 2\pi]$ then
$\phi(z) = \phi(re^{i\theta}) = \displaystyle{\sum_{n=0}^{+\infty}a_{n}r^{n}e^{in\theta}}$

We write $\mid\mid \phi \mid\mid^{2}$ and $\mid \phi(re^{i\theta})\mid^{2}$ as following:

$\mid\mid \phi \mid\mid^{2} =\displaystyle{\int_{\mathbb{C}}e^{-\mid z\mid^{2}}\mid\phi(z)\mid^{2}dzd\bar{z}} =\displaystyle{ _{_{_{\sigma \rightarrow +\infty}}}\!\!\!\!\!\!\!\!\!\!Lim \int_{0}^{\sigma}\int_{0}^{2\pi}e^{-r^{2}}\mid \phi(re^{i\theta})\mid^{2}rdrd\theta}$

$\mid \phi(re^{i\theta})\mid^{2} = \displaystyle{\sum_{n=0}^{+\infty}\displaystyle{\sum_{p+q =n}a_{p}\bar{a}_{q}r^{p+q}e^{i(p-q)\theta}}}$

By Beppo-Levi and Fubini theorems, we obtain that

$\displaystyle{\int_{\mathbb{C}}e^{-\mid z\mid^{2}}\mid\phi(z)\mid^{2}dzd\bar{z}}$ =
$\displaystyle{ _{_{_{\sigma \rightarrow +\infty}}}\!\!\!\!\!\!\!\!\!\!lim \int_{0}^{\sigma}e^{-r^{2}}}$
$\displaystyle{\sum_{n=0}^{+\infty}}$ $\displaystyle{\sum_{p+q=n}a_{p}\bar{a}_{q}r^{p+q+1}}$

As $\displaystyle{\int_{0}^{2\pi}e^{i(p-q)\theta}d\theta} = 2\pi\delta_{pq}$ where $\delta_{pq}$ is Kronoecker's symbol then we get

$\displaystyle{\int_{\mathbb{C}}e^{-\mid z\mid^{2}}\mid\phi(z)\mid^{2}dzd\bar{z}} = \displaystyle{ _{_{_{\sigma \rightarrow +\infty}}}\!\!\!\!\!\!\!\!\!\!lim \int_{0}^{\sigma}e^{-r^{2}}\displaystyle{\sum_{n=0}^{+\infty}\displaystyle{\sum_{2p=n}a_{p}\bar{a}_{p}r^{2p+1}}}}2\pi dr$

$= $
$2\pi\displaystyle{_{_{_{\sigma \rightarrow +\infty}}}\!\!\!\!\!\!\!\!\!\!lim \int_{0}^{\sigma}e^{-r^{2}}}$

Again by applying Beppo-Levi's theorem, we obtain

$\displaystyle{\int_{\mathbb{C}}e^{-\mid z\mid^{2}}\mid\phi(z)\mid^{2}dzd\bar{z}}$ = $\displaystyle{\sum_{n=0}^{+\infty}\mid a_{n}\mid^{2}\displaystyle{ _{_{_{\sigma \rightarrow +\infty}}}\!\!\!\!\!\!\!\!\!\!lim \int_{0}^{\sigma}e^{-r^{2}}r^{2n+1}2\pi dr}}$

$\displaystyle{_{_{_{\sigma \rightarrow +\infty}}}\!\!\!\!\!\!\!\!\!\!lim \int_{0}^{\sigma}e^{-r^{2}}r^{2n+1}2\pi dr}$
= $\pi n!$

$\displaystyle{\int_{\mathbb{C}}e^{-\mid z\mid^{2}}\mid\phi(z)\mid^{2}dzd\bar{z}} = \pi\displaystyle{\sum_{n=0}^{+\infty}n! \mid a_{n}\mid^{2} < +\infty}$.  $\square$

Now let

\begin{equation}
\displaystyle{ B_{s} = \{(a_{n})_{n=0}^{\infty} \in \mathbb{C} \, such\, that\, \sum_{n=0}^{\infty}n!\mid a_{n}\mid^{2} < \infty \}}
\end{equation}

$\mathbb{B}_{s}$ is also a Hilbert space. The scalar product on $\mathbb{B}_{s}$ is defined by

\begin{equation}
< (a_{n})_{n=0}^{\infty}, (b_{n})_{n=0}^{\infty}> = \displaystyle{\sum_{n=0}^{\infty}n!a_{n}\overline{b_{n}}}
\end{equation}

and the associated norm is denoted by $\mid\mid . \mid\mid_{s} $.

$\mathbb{B}$ related to $\mathbb{B}_{s}$ by an unitary transform of $\mathbb{B}$ onto $\mathbb{B}_{s}$, given  by the following transform I

\begin{equation}
{\bold I} : B \mapsto \mathbb{B}_{s} ; I(\phi) = (\frac{1}{n!}\phi^{(n)}(0))_{0}^{\infty} = (a_{n})_{n=0}^{\infty}
\end{equation}

\begin{equation}
B_{s}^{0} = \{(a_{n})_{n=0}^{\infty} \in B_{s} \, such \, that \, a_{0} = 0\}
\end{equation}

This space is associated to Bargmann space

\begin{equation}
\mathbb{B}_{0} = \{\phi \in \mathbb{B}\, such\,\, that \, \phi(0) = 0\}
\end{equation}

The realization in $\mathbb{B}_{s}$ of operators $\displaystyle{A\phi(z) = \frac{d}{dz}\phi(z)}$ with domain $\displaystyle{D(A) = \{\phi \in \mathbb{B}; \frac{d}{dz}\phi \in \mathbb{B}\}}$

and $\displaystyle{A^{*}\phi(z) = z\phi(z)}$ with domain $\displaystyle{D(A^{*}) = \{\phi \in \mathbb{B}; z\phi \in \mathbb{B}\}}$

is given by

\begin{equation}
\left\{\begin{array}[c]{l}A(a_{n})_{n=0}^{\infty}=((n+1)a_{n+1})_{n=0}^{\infty}\\ \quad \\ with \quad domain:\\

\n D(A) = \{(a_{n})_{n=0}^{\infty} \in \mathbb{B}_{s}; \displaystyle{\sum_{n=0}^{\infty}n!n\mid a_{n}
\mid^{2}} < \infty\} \\ \end{array}\right.
\end{equation}

\begin{equation}
\left\{\begin{array}[c]{l}A^{*}(a_{n})_{n=0}^{\infty}=(a_{n-1})_{n=0}^{\infty}\quad;\quad a_{-1} = 0 \\ \quad \\ with \quad domain: \\
D(A^{*}) = \{(a_{n})_{n=0}^{\infty} \in \mathbb{B}_{s};\displaystyle{\sum_{n=0}^{\infty}n!\mid a_{n-1}\mid^{2}} < \infty\} \\ \end{array}\right.
\end{equation}

In $\mathbb{B}_{s}$ realization, the first result on the domains of $A$ and $A^{*}$ is

i) $D(A)$ is dense in $\mathbb{B}_{s}$.

ii) $D(A) = D(A^{*})$

iii) $D(A) \hookrightarrow \mathbb{B}_{s}$ is compact.

i) and ii) is obvious.

iii) Let $C_{s}$ be the unit ball of $D(A)$ equipped with graph norm:

$C_{s} =\{(a_{n})_{n=0}^{\infty} \in D(A)$ such that $\mid\mid(a_{n})_{n=0}^{\infty}\mid\mid_{s}^{2} + \mid\mid A((a_{n})_{n=0}^{\infty})\mid\mid_{s}^{2} \leq 1\}$ i.e.

$C_{s} =\{(a_{n})_{n=0}^{\infty} \in D(A)$ such that $\displaystyle {\mid a_{0}\mid^{2} +\sum_{n=1}^{\infty}(n+1)!\mid a_{n}\mid^{2} \leq 1} \}$

To prove that $C_{s}$ is relatively compact for the norm associated to $\mathbb{B}_{s}$, we observe that $\mathbb{B}_{s}$ is complete space then
it suffices to show that $C_{s}$ can be covering by finite number of unit balls of radius $\frac{1}{p}$ with arbitrary $p\in \mathbb{N}-\{0\}$

Writing $\displaystyle {\sum_{n=1}^{\infty}(n+1)!\mid a_{n}\mid^{2}}$ as follows

$\displaystyle {\sum_{n=1}^{\infty}(n+1)!\mid a_{n}\mid^{2} = 2!\mid a_{1}\mid^{2} + 3!\mid a_{2}\mid^{2}+4!\mid a_{3}\mid^{2}+ ............} $

$ = $ $1!\mid a_{1}\mid^{2} + 2!\mid a_{2}\mid^{2}+3!\mid a_{3}\mid^{2}+ ............................+$

$\quad$ $1!\mid a_{1}\mid^{2} + 2!\mid a_{2}\mid^{2}+3!\mid a_{3}\mid^{2}+ ............................+$

$\quad$ $0 \quad \quad \quad + 2!\mid a_{2}\mid^{2}+3!\mid a_{3}\mid^{2}+ .............................+$

$\quad$ $0 \quad \quad \quad +  0 \quad \quad \quad +  3!\mid a_{3}\mid^{2}+ ............................+$

$\quad$ $............................................................................... +$

$\quad$ $................................................................................+$

$ = 2 \displaystyle {\sum_{n=1}^{\infty}n!\mid a_{n}\mid^{2}}$ + $\displaystyle {\sum_{n=2}^{\infty}n!\mid a_{n}\mid^{2}}$ + ........ + $\displaystyle{\sum_{n=p}^{\infty}n!\mid a_{n}\mid^{2}}$ + .......

then we deduce that $(p+1)\displaystyle{\sum_{n=p}^{\infty}n!\mid a_{n}\mid^{2}} \leq 1$
i.e. $(0,0, ......, 0,a_{p}, a_{p+1}, ......)$ is in $C_{s}^{0}(0,\frac{1}{p})$ the ball
with radius $\frac{1}{p}$ around the origin of $\mathbb{B}_{s}$.

As the set $\{(a_{1}, a_{2}, .....,a_{p-1}) \in \mathbb{C}^{p-1}$ ; $\displaystyle {\sum_{n=1}^{p-1}n!\mid a_{n}\mid^{2}}\leq 1\}$ is compact then they exist $m$ balls $K(x_{i},\frac{1}{p})$ with center $x_{i}\in \mathbb{C}^{p-1}$ of radius $\frac{1}{p}$ such that
$\{(a_{1}, a_{2}, .....,a_{p-1} \in \mathbb{C}^{p-1}$;$\displaystyle {\sum_{n=1}^{p-1}n!\mid a_{n}\mid^{2}}\leq 1\} = \bigcup_{i=1}^{m} K(x_{i},\frac{1}{p})$

then we get

$C_{s}\subset \displaystyle {\bigcup_{i=1}^{m} }K(x_{i},\frac{1}{p})\bigcup C_{s}^{0}(0,\frac{1}{p})$
and we deduce the property iii)..  $\square$

We can also prove iii) by applying a following classic proposition

Let $K \subset \mathbb{B}_{s}$ such that

i) $K$ is bounded and closed.

ii) $\forall \varepsilon > 0$, $\exists N_{\varepsilon} > 0$ such that $\displaystyle{\sum_{n= N_{\varepsilon}}^{\infty}n!\mid a_{n}\mid^{2} \leq \varepsilon \quad \forall \{a_{n}\}_{n=1}^{\infty}} \in K$

then $ K $ is compact.  $\square$

In $\mathbb{B}$ representation or $\mathbb{B}_{s}$  representation we have the following fundamental lemma

Let $e_{1}=^{t}(0,1,0,.........)$ and $S(a_{n})_{n=0}^{\infty}=(a_{n+1})_{n=0}^{\infty}$ then we have

i) In $\mathbb{B}$, $\displaystyle{\phi \in \mathbb{B}  \Leftrightarrow z \rightarrow \frac{\phi^{'}(z) - \phi^{'}(0)}{z} \in \mathbb{B}}$

ii) In $\mathbb{B}_{s}$, $\displaystyle{(a_{n})_{n=0}^{\infty}\in \mathbb{B}_{s} \Leftrightarrow S(a(a_{n})_{n=0}^{\infty} -a_{1}e_{1}) \in \mathbb{B}_{s} \Leftrightarrow  \sum_{n=0}^{\infty}n!(n +2)^{2}\mid a_{n+2}\mid^{2}< \infty}$

i) For $\phi(z) = \displaystyle {\sum_{n=0}^{\infty}a_{n}z^{n}}$ then $\phi^{'}(z) = \displaystyle {\sum_{n=0}^{\infty}(n+1)a_{n+1}z^{n}}$ and $\frac{\phi^{'}(z) - \phi^{'}(0)}{z} = \displaystyle {\sum_{n=0}^{\infty}(n + 2)a_{n+2}z^{n}}$

By using the isometry  between Bargmann space $\mathbb{B}$ and $\mathbb{B}_{s}$  we deduce that

$\mid\mid \frac{\phi^{'}(z) - \phi^{'}(0)}{z}\mid\mid^{2} = \displaystyle {\sum_{n=0}^{\infty}n!(n+2)^{2}\mid a_{n+2}\mid^{2}}$, i.e.

$\mid\mid \frac{\phi^{'}(z) - \phi^{'}(0)}{z}\mid\mid^{2} = \displaystyle {\sum_{n=0}^{\infty}\frac{n+2}{n+1}(n+2)!\mid a_{n+2}\mid^{2}}$ $\hfill { }  (*)$

Now, as $\frac{n+2}{n+1} < 2$ then $\mid\mid \frac{\phi^{'}(z) - \phi^{'}(0)}{z}\mid\mid^{2} \leq \displaystyle {2\sum_{n=0}^{\infty}(n+2)!\mid a_{n+2}\mid^{2}}$

if $\phi \in \mathbb{B}$ then $z\rightarrow \frac{\phi^{'}(z) - \phi^{'}(0)}{z} \in \mathbb{B}$

For the reciprocity, we remak that $\frac{n+2}{n+1} > 1$ and from (*) we deduce that
$\mid\mid \phi \mid\mid \leq  \mid\mid \frac{\phi^{'}(z) - \phi^{'}(0)}{z}\mid\mid$

ii) is a simple version of i) in $\mathbb{B}_{s}$.  $\square$

Now starting by a list of notations and remarks

In $\mathbb{B}$ representation the operator  $ H := H_{\mu,\lambda}$ is defined by

\begin{equation}
\left\{\begin{array}[c]{l}H\phi(z) = \quad i\lambda z\phi^{''}(z)  + (i\lambda z^{2} + \mu z)\phi^{'}(z)\\ \quad \\ with \quad maximal \quad domain: \\\quad \\
D(H_{max}) = \{\phi \in \mathbb{B} \quad such \quad that \quad  H\phi \in \mathbb{B}\}\\ \end{array}\right.
\end{equation}

i) Let $\mathcal{P}$ be space of polynomials then it is dense in $\mathbb{B}$.

ii) Let  $H_{\mid_{\mathcal{P}}}$ be the restriction of $H_{\mu, \lambda}$ to polynomials space,  we can define $H_{min}$ as the closure of operator $H_{\mid_{\mathcal{P}}}$ in Bargmann space:

\begin{equation}
\left\{\begin{array}[c]{l}H_{min}\phi= i\lambda z\phi^{''}  + (i\lambda z^{2} + \mu z)\phi^{'} \\ \quad \\
with  \quad minimal \quad domain: \\ \quad \\
D(H_{min}) = \{\phi \in \mathbb{B}; \, \exists p_{n} \in \mathcal{P}\, and \, \psi \in \mathbb{B} ; p_{n}\rightarrow \phi \, and\, Hp_{n}\rightarrow \psi\}\\ \end{array}\right.
\end{equation}

\begin{equation}
H_{min}^{*} = H_{\mu,-\lambda} \,  \,  with \, \, domain  \, \, D(H_{min}^{*}) = D(H_{max})
\end{equation}

iv) An orthonormal basis of $B$ is given by :

\begin{equation}
\displaystyle{ e_{n}(z) = \frac{z^n}{\sqrt{n!}}; n = 0, 1, ....}
\end{equation}

v)  The action of the operators $A$, $A^{*}$ and $H$ on the basis $\displaystyle{e_{n}(z)= \frac{z^{n}}{\sqrt{n!}} n = 0, 1, .....}$ is:

$\bullet$ $A(e_{n}) = \sqrt{n}e_{n-1}$ with the convention $e_{-1}=0$

$\bullet$ $A^{*}(e_{n}) = \sqrt{n+1}e_{n+1}$

$\bullet$ $H_{\mu, \lambda}(e_{n}) =  i\lambda(n-1)\sqrt{n}e_{n-1} + \mu ne_{n} + i\lambda n\sqrt{n+1}e_{n+1}$

vi) let $l^{2}(\mathbb{N})$ = $\displaystyle{\{ (\phi_{n})_{n=0}^{\infty}\in \mathbb{C}\quad such \quad that\quad \sum_{n=0}^{\infty}\mid\phi_{n}\mid^{2} < +\infty\}}$ with the inner product:

\begin{equation}
\displaystyle{< (\phi_{n})_{n=0}^{\infty}, (\psi_{n})_{n=0}^{\infty} > = \sum_{n=0}^{\infty}\phi_{n}\bar{\psi}_{n}}
\end{equation}

In the representation $l^{2}(\mathbb{N})$ where the coefficients $\phi_{n}$ define an entire function

$\phi(z) = \displaystyle{\sum_{n=0}^{\infty}\phi_{n}e_{n}(z)}$ in Bargmann space we have

\begin{equation}
\left\{\begin{array}[c]{l}(A\phi)_{n}=\sqrt{n}\phi_{n-1}, \phi_{-1}= 0\\ \quad \\ with \quad domain: \\ \quad \\ D(A) = \{(\phi)_{n} \in l^{2}(\mathbb{N}); \displaystyle{\sum_{n=0}^{\infty}n\mid \phi_{n}\mid^{2} < \infty} \}\\ \end{array}\right.
\end{equation}

\begin{equation}
\left\{\begin{array}[c]{l}(A^{*}\phi)_{n} =\sqrt{n+1}\phi_{n+1}\\ \quad \\ with \quad domain: \\ \quad \\ D(A^{*})=\{(\phi)_{n} \in l^{2}(\mathbb{N});\displaystyle {\sum_{n=0}^{\infty}n\mid \phi_{n}\mid^{2} < \infty }\} \\ \end{array}\right.
\end{equation}

\begin{equation}
\left\{\begin{array}[c]{l} (H_{\mu, \lambda}\phi)_{n} = i\lambda (n-1)\sqrt{n}\phi_{n-1} + \mu n\phi_{n} + i\lambda n\sqrt{n+1}\phi_{n+1}\\ \quad \\ with \quad domain: \\ \quad \\ D(H_{\mu, \lambda}) = \{\phi \in l^{2}(\mathbb{N}); H\phi \in l^{2}(I\!\!N)\}\end{array}\right.
\end{equation}

vi) $\mathbb{B} = \mathbb{B}_{0}\bigoplus\{constants\}$  and zero is eigenvalue of  $H_{\mu, \lambda}$ without interest.

vii) We define $l_{0}^{2}(\mathbb{N}) = \{\phi =(\phi_{n})_{n=0}^{\infty} \in l^{2}(\mathbb{N}); \phi_{0} = 0\}$ and the operators $A$ and $A^{*}$ by (3.34) and (3.35) and

$H_{\mu, \lambda} = \mu A^{*}A + i\lambda A^{*}(A + A^{*})A$ with domain $D(H_{\mu, \lambda}) = \{\phi \in l_{0}^{2}(\mathbb{N}); H\phi \in l_{0}^{2}(\mathbb{N})\}$

In the representation $l_{0}^{2}(\mathbb{N})$ where the coefficients $\phi_{n}$ define an entire function
$\phi(z) = \displaystyle{\sum_{n=1}^{\infty}\phi_{n}e_{n}(z)}$ in Bargmann space $B_{0}$, we study, in the present paper, a class of Jacobi-Gribov matrices with unbounded entries:

\begin{equation}
\left \{ \begin{array}{c} (H\phi)_{n} = \alpha_{n-1}\phi_{n-1} + q_{n}\phi_{n} +\alpha_{n}\phi_{n+1}, n\geq 2 \\
\quad\\
with\,\, the \,\, initial \,\, condition\\
\quad\\(H\phi)_{1} = q_{1}\phi_{1} + \alpha_{1}\phi_{2},\\
\end{array} \right.
\end{equation}

$q_{n} = \mu n$, and $\alpha_{n} = i\lambda n\sqrt{n+1}$, ( $\mu$ and $\lambda$ are real numbers and $i^{2} = -1$).

We will write from now the tridiagonal Jacobi-Gribov matrix as

$H = $ $\left(\begin{array}{c c c c c c c c} \mu & i\lambda \sqrt{2} & 0 & \cdots\\i\lambda \sqrt{2} & 2\mu & i\lambda 2\sqrt{3} &0 & .\\0 &i\lambda 2\sqrt{3} & 3\mu & i\lambda 3\sqrt{4}& 0 &.\\\vdots & 0 & i\lambda 3\sqrt{4} & 4\mu & * & 0 & .\\\vdots & \ddots & 0 & * & * & * & \ddots & .\\\vdots & \ddots & \ddots & 0 & * & * & * & \ddots\\
\end{array}\right).$

\begin{equation}
\left\{\begin{array}[c]{l}H = (h_{m,n})_{m,n=1}^{\infty}\quad with \quad the \quad elements:\\\quad \\h_{nn} = \mu n,h_{n,n+1}=h_{n+1,n}= i\lambda n\sqrt{n+1}; n = 1,2,...\\ \quad \\and \\ \quad \\h_{mn} = 0 \quad for \quad \mid m - n \mid > 1 \\ \end{array}\right.
\end{equation}

The above “Jacobi-Gribov” matrix determines two linear operators in $l_{0}^{2}(\mathbb{N})$ by the formal matrix product $H\phi$. the first operator is defined in the linear manifold of vectors in $l_{0}^{2}(\mathbb{N})$ with finite support related to the set $\mathcal{P}$ of polynomials in Bargmann space, this operator is densely defined and closable. Let $H_{min}$ be its closure.

The second operator $H_{max}$ has the domain $D(H_{max}) =\{ \phi = (\phi_{n})_{n=0}^{\infty} \in l_{0}^{2}(\mathbb{N}); H\phi \in l_{0}^{2}(\mathbb{N})\}$

The  family of infinite “Jacobi-Gribov” matrices is tridiagonal of the form $J + i\lambda G$, where the matrix $ J$ is diagonal with entries $J_{nn} := q_{n}= \mu n$ , and the matrix $G$ is off-diagonal, with nonzero entries $G_{n,n+1} = G_{n+1,n} := G_{n} = n\sqrt{n+1}$.

The Jacobi operators with zero main diagonal and $G_{n}\sim n^{\alpha}$ with $\frac{1}{2} \leq \alpha\leq 1$ has been studied in  [Janas3 et al],
[Janas4 et al] and   [Janas5 et al].

The spectral analysis of Jacobi operators corresponding to matrices with rapidly growing weights has been treated in few occasions see  [Janas2 et al],   [Moszynski] ] and [Silva].

In contrast, literature on Jacobi matrices with slowly growing weights is more abundant (see for example [Tur] and references therein) :

In  [Janas1 et al] Janas and Malejkib have studied  the asymptotic behaviour of the point spectrum for special classes of Jacobi matrices with analytic models where $J_{n} = \delta n^{\alpha}$ and $G_{n} = \gamma n^{\beta}(1 + \Delta_{n}) , \Delta_{n}$ goes to zero as $n$ goes to infinity, with $\alpha - \beta$ equals, respectively, to $1, \frac{1}{2}, 0 $ . In particular  their analytic model of example 4 with $\alpha - \beta = \frac{1}{2}$ uses the Bargmann space.

The spectral analysis of “Jacobi-Gribov” operator as analytical model is based on the above representations.

Let the infinite matrix $ ^{t\!}H $ be obtained from $H$ by transposing of the elements and the infinite matrix $ H^{\bot}$ be obtained from $H$ by transposing and by taking complex conjugates of the elements. Then

i) $ H $ is symmetric complex matrix i.e. $ H = ^{t\!}H$.

ii) $ H \neq  H^{\bot}$

iii) $ H_{min}^{*} = H_{max}$ and $ H_{max}^{*} = H_{min}$, in particular the maximal operator $H_{max}$ is a closed extension of $H_{min}$.

iv) As $G_{n} = O(n^{\alpha})$ with $\alpha = \frac{3}{2} > 1$ then the standard perturbation theory is not applicable.

For a systematic treatment of complex Jacobi matrices, we refer to  [Beckermann] and the references therein.

§.§ Some properties of domains of annihilation and creation operators and of domains of $H_{\lambda', \mu, \lambda}$ in $\mathbb{B}_{s}$

For an operator $T$ acting on Hilbert space $\mathcal{H}$, we denote by $D(T)$; $R(T)$ or $\Im m(T)$ (French notation); $N(T)$, $\sigma (T)$ and $\rho(T)$, its domain of definition, its range, its kernel, its spectrum, and its resolvent set.

In $l_{0}^{2}(\mathbb{N})$ we have:

1) $\mid\mid A\phi\mid\mid \geq \mid\mid \phi\mid\mid \quad \forall \phi \in D(A)$

2) $ < G\phi , \phi > = \displaystyle{\sum_{n=1}^{\infty}n\sqrt{n+1}[\phi_{n}\overline{\phi_{n+1}} + \phi_{n+1}\overline{\phi_{n}}]  \quad \forall \phi \in D(G_{max})}$

3) For $ \mu \neq 0$, $\mid\mid H_{min}\phi\mid\mid \geq \mid \mu \mid\mid\mid \phi\mid\mid \quad \forall \phi \in D(H_{min})$

4) $D(H_{min}) \hookrightarrow D(A)$ is continuous.

5) For $ \mu \neq 0$,   $ 0 \in \rho(H_{min}) $

6) The spectrum of $H_{min}$ is discrete.

1) We see that $\mid\mid A\phi\mid\mid^{2} = \displaystyle{\sum_{n=1}^{\infty}(n+1)\mid \phi_{n}\mid^{2} \, \forall \phi =\{\phi_{n}\}_{n=1}^{\infty} \, \in D(A)}$

$\mid\mid A\phi\mid\mid \geq \mid\mid \phi\mid\mid \, \forall \phi \in D(A)$

2) This property is obvious

3) For $\lambda \in \mathbb{R}$ we observe that
$\mid \Re e < H_{min}\phi, \phi>\mid = \mid \mu \mid\mid\mid A\phi\mid\mid^{2}$

and by using the property 1) we deduce that

$\mid \Re e < H_{min}\phi, \phi>\mid$ $\geq$ $\mid\mu\mid.\mid\mid \phi\mid\mid^{2} \quad \forall \phi \in D(H_{min})$

$\mid\mid H_{min}\phi\mid\mid$ $\geq$ $\mid\mu\mid.\mid\mid \phi\mid\mid \quad \forall \phi \in D(H_{min})$.

From this calculations, we deduce that  $N(H_{min}) = \{0\}$  $\Im m (H_{min})$ is closed and the property 4) i.e. $D(H_{min}) \hookrightarrow D(A)$ is continuous.

Now by using the property 4) and above lemma, we will deduce that

$D(H_{min}) \hookrightarrow l_{0}^{2}(\mathbb{N})$ is compact.    (**)

5) We begin by showing that $Im(H_{min})$ is dense in $l_{0}^{2}(\mathbb{N})$

In fact $ < H_{min}\phi,\psi> = 0 \Leftrightarrow H_{max}^{\perp}\psi = 0$ and in Bargmann representation, we have

\begin{equation}
-i\lambda z\psi^{''}(z) + (\mu z - i\lambda z^{2}) \psi^{'}(z) = 0
\end{equation}

For $\lambda \neq 0$ then

\begin{equation}
\displaystyle{\psi^{'}(z) = e^{-\frac{1}{2}z^{2} + i\frac{\mu}{\lambda}z}}
\end{equation}

Now we show that the norma-lizability requirement for $ \psi(z) $ is not verified in some direction of the $z$-plane.

In fact we will use the fundamental lemma 3.8 for $ z\rightarrow -i\infty$

Let $ z = \Re e z + i\Im m z$, then

$e^{-\mid z \mid^{2}}\mid \frac{\psi^{'}(z)}{z}\mid^{2}$ = $e^{-\mid z \mid^{2}}\mid \frac{e^{-\frac{1}{2}z^{2} + i\frac{\mu}{\lambda}z}}{z}\mid^{2}$ = $\frac{1}{\mid z \mid^{2}}e^{-2(\Re z)^{2}}e^{-2\frac{\mu}{\lambda}\Im z}$

For $ \mu $ and $\lambda $ given, there exists an direction of the z-plane where the function

$\psi (z) = \displaystyle{\int_{0}^{z} e^{-\frac{1}{2}\xi^{2} + i\frac{\mu}{\lambda}\xi}d\xi}$ cannot be considered in Bargmann space ( for example $\displaystyle{\frac{\mu}{\lambda} > 0}$ and $\Im m z < 0$).

By using the property 3), we deduce that $H_{min}^{-1}$ exists, i.e. $ 0 \in \rho(H_{min}) $.

6)  As $H_{min}^{-1}$ exists, we can write it in the form

$H_{min}^{-1}$ = $i_{2}oi_{1}oH_{min}^{-1}$ where $i_{1}:D(H_{min}) \hookrightarrow D(A)$ is continuous(see the property 4) and

$\displaystyle{ i_{2}:D(A) \hookrightarrow l_{0}^{2}(\mathbb{N})}$ is compact , i.e. $H_{min}^{-1}$ is compact.

By Chapter III, Theorem 6.29  [Kato], we deduce that the spectrum of $H_{min}$ consists entirely

of isolated eigenvalues with finite multiplicities, and $(H_{min} - \sigma I)^{-1}$ is compact $\forall \sigma \in \rho(H_{min})$.  $\square$

The first interesting result on this “Jacobi-Gribov” matrix is the following theorem

For $\mu \neq 0 $, the “Jacobi-Gribov” matrix is proper, i.e $H_{min}$  = $H_{max}$ and its spectrum $\sigma(H_{max})$ is discrete

i) $ D(H_{min})\subset D(H_{max})$ is obvious.

ii) $ D(H_{max})\subset D(H_{min})$.

In fact let $\psi = (\psi_{n})_{n=1}^{\infty} \in D(H_{max})$, according that $H_{min}$ is invertible by using the last proposition then there exists $ \phi = (\phi_{n})_{n=1}^{\infty} \in D(H_{min})$ such that $ H_{min}\phi = \psi$.

In particular there exists a sequence of polynomials $q_{m}$ such that  $q_{m} \rightarrow \psi ; m \rightarrow \infty$ then $H_{min}^{-1}q_{m} \rightarrow H_{min}^{-1}\psi = \phi$ but $H_{min}^{-1}q_{m}$ can be not  polynomials.

We proceed as follows to resolve this difficulty

We consider the polynomials $p_{m}$ such that $\mid\mid\mid p_{m} - H_{min}^{-1}q_{m} \mid\mid\mid \leq \frac{1}{m}$ where $\mid\mid\mid . \mid\mid\mid$ is graph norm.

\begin{equation}
\displaystyle{\mid\mid p_{m} - \phi \mid\mid \leq \mid\mid p_{m} - H_{min}^{-1}q_{m} \mid\mid + \mid\mid H_{min}^{-1}q_{m} - \phi \mid\mid}
\end{equation}

In particular we obtain

$\mid\mid p_{m} - \phi \mid\mid \leq \frac{1}{m} + \mid\mid H_{min}^{-1}q_{m} - \phi \mid\mid$, i.e. $p_{m}\rightarrow \phi$ as $ m \rightarrow \infty$

Now to show that $H_{max}p_{m}\rightarrow H_{max}\phi$ as $ m \rightarrow \infty$, we observe that

$\mid\mid H_{max} p_{m} - H_{max}\phi \mid\mid \leq \mid\mid H_{max}p_{m} - q_{m}\mid\mid  + \mid\mid q_{m} - H_{max}\phi \mid\mid \leq \mid\mid\mid H_{max}\mid\mid\mid(\frac{1}{m} + \mid\mid H_{min}^{-1}q_{m} - \phi \mid\mid)$.

where $\mid\mid\mid H_{max}\mid\mid\mid $ denote the graph norm of operator $H_{max}$.

$\displaystyle{H_{max} p_{m} \longrightarrow H_{max} \phi}$  as $m \longrightarrow +\infty$.  $\square$

(i) Let $\mathcal{P}_{0}$ = $\{p \in \mathcal{P}; p(0) = 0\}$ Then $H[\mathcal{P}]$ is dense in $B_{0}$.

(ii)  For $\mu = 0 $, the Jacobi-Gribov matrix is not proper, i.e $H_{min}$  $\neq $ $H_{max}$ and  $\sigma(H_{max}) =  \mathbb{C} $.

(i) Let $\psi \in B_{0}$, as $ 0 \in \rho(H_{min})$ and $H_{min} = H_{max}: = H$ then there exists $\phi \in D(H):= D(H_{max}) = D(H_{min})$ such that
$ \psi = H\phi$, hence there exists $p_{m} \in \mathcal{P}_{0}$ ;$ p_{m} \rightarrow \phi$ and $\psi_{1} \in B_{0}$ ;$Hp_{m} \rightarrow \psi_{1}$ when $ m \rightarrow \infty$.
As $Hp_{m} \rightarrow H\phi$ when $ m \rightarrow \infty$ of distributions sense, we deduce that $\psi = \psi_{1}$.

(ii) For $\mu = 0$ the boundary conditions at infinity had used in a description of all maximal dissipative extensions of $H_{min}$ and the characteristic functions of these dissipative extensions had computed. Completeness theorems had obtained for the system of generalized eigenvectors.
.  $\square$

To end this section, we observe that  our operator can be  generalized  to :

$\displaystyle{ \mathbb{H}_{\lambda',  \mu, \lambda} = \lambda' A^{*^{p+1}} A^{p+1} + \mu A^{*^{p}}A^{p} + i\lambda A^{*^{p}}(A + A^{*})A^{p}\, (p = 0, 1, 2 ...)}$   ((i))

and  If $p = 0$, $\mu= 0$ and $\lambda \in i\mathbb{R}$ then $\displaystyle{ \mathbb{H}_{\lambda',  \mu, \lambda}}$ is the displaced harmonic oscillator of the following form:

$\displaystyle{\mathbb{H}_{\omega} = \omega A^{*}A + \lambda (A + A^{*}) = \omega (A^{*} + \frac{\lambda}{\omega})(A +  \frac{\lambda}{\omega}) - \frac{\lambda^{2}}{\omega}}.$    ((ii))

As it is well known that this Hamiltonian is trivially solved by introducing the operators :

$\displaystyle{B = A + \frac{\lambda}{\omega}}$ and $\displaystyle{B^{*} = A^{*} + \frac{\lambda}{\omega}}$ which are unitarily equivalent to the $A$ and $A^{*}$.

As $\displaystyle{[B, B^{*}] = \mathbb{I}}$ and $\displaystyle{\mathbb{H}_{\omega} = \omega B^{*}B  - \frac{\lambda^{2}}{\omega}}$ then the energy spectrum is given by :

$\displaystyle{\sigma_{n} = n\omega -   \frac{\lambda^{2}}{\omega}}$    ((iii))

In terms of Bargmann realization $\displaystyle{A \longrightarrow \frac{d}{dz}}$, $\displaystyle{A^{*} \longrightarrow z}$ the eigenvalue equation

$\displaystyle{\mathbb{H}_{\omega}\phi = \sigma \phi}$    ((iv))

becomes :

$\displaystyle{[z\frac{d}{dz} + \frac{\lambda}{\omega}(z + \frac{d}{dz}) - \frac{\sigma}{\omega}]\phi(z) = 0}$     ((v))

By using the power series method, we choose $\displaystyle{\phi(z) = \sum_{n=0}^{\infty}a_{n}(z +  \frac{\lambda}{\omega})^{n + s}}$ to get the indicial equation :

$\displaystyle{s = \frac{\sigma}{\omega} + \frac{\lambda^{2}}{\omega^{2}}}$  ((vi))

and the recurrence relation for the $a_{n}$'s would read

$\displaystyle{a_{n} = -\frac{\lambda}{n\omega}a_{n-1}  = \frac{(-\lambda)^{n}}{(\omega)^{n}\Gamma(n +1)}a_{0}} $   ((vii))

The series $\displaystyle{\sum_{n}a_{n}z^{n}}$ thus converges for all $z$, and

$\displaystyle{\phi(z) = a_{0}(z + \frac{\lambda}{\omega})^{\frac{\sigma}{\omega} + \frac{\lambda^{2}}{\omega^{2}}}\sum_{n=0}^{\infty}\frac{(-\lambda)^{n}}{\omega^{n}\Gamma(n + 1)}(z + \frac{\lambda}{\omega})^{n}}$    ((viii))

From the requirement that  $\phi(z)$ be entire we obtain $\displaystyle{\frac{\sigma}{\omega} + \frac{\lambda^{2}}{\omega^{2}} = n \in \mathbb{N}}$

The  $\phi_{n}(z)$ associated to $\displaystyle{\sigma_{n} = n \omega - \frac{\lambda^{2}}{\omega}}$ are in domain of $\mathbb{H}_{\omega}$.

§.§ On existence of eigenvalues of $\displaystyle{H_{\mu, \lambda} \, and \, H_{\lambda', \mu, \lambda} ; \lambda' \neq 0}$

(1) Explicit inversion of $H_{\mu,\lambda}$ on $]-\infty, \, 0]$

We consider the eigenvalue problem:

$ \left \{ \begin{array} {c}  \displaystyle{ i\, \lambda z \frac{\partial^{2}\varphi_{\mu, \lambda}}{\partial z^{2}}(z) + (i\lambda z^{2} + \mu z)\frac{\partial \varphi_{\mu, \lambda}}{\partial z}(z) = \sigma(\mu, \lambda)\varphi_{\mu, \lambda} (z)}\\
\quad\\
\varphi \in D(H_{\mu, \lambda})\quad \quad \quad \quad \quad \quad  \quad \quad \quad \quad \quad  \quad \quad \quad \quad \quad  \quad \quad \quad \quad   \\
\quad\\
\sigma(\mu, \lambda) \in \mathbb{C}  \quad \quad \quad \quad \quad  \quad \quad \quad \quad \quad  \quad \quad \quad \quad \quad  \quad \quad \quad \quad \quad \\
\end{array} \right. $   (*)

Denoting $\displaystyle{\varphi_{\mu, \lambda}(z) = \varphi(z),  \sigma(\mu, \lambda) = \lambda\sigma, \frac{\partial}{\partial z}\varphi (z) = \varphi^{'}(z)\, and \, \frac{\partial^{2}}{\partial z^{2}}\varphi (z) = \varphi^{"}(z) }$ then the eigenvalue problem associated to $H_{\mu, \lambda}; \lambda \neq 0$ can be written as follows

\begin{equation}
\displaystyle{\varphi^{"}(z) + p(z) \varphi^{'}(z) + q(z) \varphi(z) = 0}; \displaystyle{ p(z) = (z - i\rho)}\, and\, \displaystyle{ q(z) =  \frac{i\sigma}{z}}
\end{equation}

We restrict the study of the equation (3.42) on the lines parallel to the imaginary axis :

Let $z = a + iy$ and $\varphi(a + iy) = \varphi_{a}(y)$ where $a$ is fixed. Then we get

\begin{equation}
\left \{ \begin{array} [c] {l} \displaystyle{\varphi_{a}^{"}(y) - (y - \gamma) \varphi_{a}^{'}(y) - \frac{i\sigma}{a + iy}\varphi_{a}(y) = 0}\\
\quad\\
\gamma = \rho + ia\\
\end{array} \right .
\end{equation}

We put

\begin{equation}
\displaystyle{\varphi_{a}^{'}(y) = K_{a}(y)e^{\frac{1}{2}(y - \gamma)^{2}}}
\end{equation}

and substitute it in (3.43) then we get

\begin{equation}
\displaystyle{K_{a}^{'}(y) = i\sigma e^{-\frac{1}{2}(y - \gamma)^{2}}\frac{\varphi_{a}(y)}{a + iy}}
\end{equation}

We choose a primitive representation of the function $\displaystyle{ y \longrightarrow K_{a}^{'}(y)}$ as follows

\begin{equation}
\displaystyle{K_{a}(y) = i\sigma \int_{-\infty}^{y}e^{-\frac{1}{2}(u - \gamma)^{2}}\frac{\varphi_{a}(u)}{a + iu}du + c_{a} \, ; y \in ]-\infty, 0]}
\end{equation}

where $c_{a}$ is a constant.

Consequently, equation (3.44) can be written as follows

\begin{equation}
\displaystyle{\varphi_{a}^{'}(y) = i\sigma e^{\frac{1}{2}(y - \gamma)^{2}} \int_{-\infty}^{y}e^{-\frac{1}{2}(u - \gamma)^{2}}\frac{\varphi_{a}(u)}{a + iu}du  + c_{a}e^{\frac{1}{2}(y - \gamma)^{2}} }
\end{equation}

For $\rho > 0$ the representation (3.45) has a sense.

For $\varphi \in \mathbb{B}$ we have $\displaystyle{\int_{\mathbb{R}^{2}}e^{-(x^{2} + y^{2})}\mid \varphi(x + iy) \mid^{2} dxdy < + \infty}$ then we apply the Fubini theorem to deduce that

\begin{equation}
\displaystyle{\int_{\mathbb{R}}e^{- y^{2}}\mid \varphi_{x}(y) \mid^{2} dy < + \infty}
\end{equation}

Now, as  $\displaystyle{ \int_{-\infty}^{y}e^{-\frac{1}{2}(u - \gamma)^{2}}\frac{\varphi_{a}(u)}{a + iu}du = e^{-\frac{1}{2}\gamma^{2}}\int_{-\infty}^{y}e^{-\frac{1}{2}u^{2}}\varphi_{a}(u)\frac{e^{\gamma u}}{a + iu}du }$ and we look for $\varphi$ in Bargmann space $\mathbb{B}$, we have on the one hand $\displaystyle{u \longrightarrow  e^{-\frac{1}{2}u^{2}}\varphi_{a}(u)}$ is square integrable and on the other hand for $\rho > 0$ the function $\displaystyle{u \longrightarrow  \frac{e^{\gamma u}}{ a + iu}}$ is also square integrable on $]-\infty, 0]$. Consequently, $\displaystyle{ \int_{-\infty}^{y}e^{-\frac{1}{2}(u - \gamma)^{2}}\frac{\varphi_{a}(u)}{a + iu}du}$ has a sense.   $\square$

(i) If $\rho < 0$, we choose a primitive representation of the function $\displaystyle{ y \longrightarrow K_{a}^{'}(y)}$ as follows

\begin{equation}
\displaystyle{K_{a}(y) = i\sigma \int_{y}^{+\infty}e^{-\frac{1}{2}(u - \gamma)^{2}}\frac{\varphi_{a}(u)}{a + iu}du + c_{a} }, c_{a} \, is \, constant
\end{equation}

This representation  has  a sense.

(ii) From (3.43) and (3.45), we deduce that

\begin{equation}
\displaystyle{\varphi_{a}^{'}(y) = i\sigma e^{\frac{1}{2}(y - \gamma)^{2}} \int_{y}^{+\infty}e^{-\frac{1}{2}(u - \gamma)^{2}}\frac{\varphi_{a}(u)}{a + iu}du + c_{a}e^{\frac{1}{2}(u - \gamma)^{2}}}
\end{equation}

(i) In the representation (3.45) we have $c_{a} = 0$

\begin{equation}
\displaystyle{Lim \, e^{-\frac{1}{2}(y - \gamma)^{2}}\varphi_{a}^{'}(y) = 0}\, as \, y \longrightarrow -\infty\n (boundary condition)
\end{equation}

(i) By multiplying the two members of the equation (3.49) by $\displaystyle{e^{-\frac{1}{2}(u - \gamma)^{2}}}$, we get

\begin{equation}
\displaystyle{e^{-\frac{1}{2}(u - \gamma)^{2}}\varphi_{a}^{'}(y) = i\sigma  \int_{y}^{+\infty}e^{-\frac{1}{2}(u - \gamma)^{2}}\frac{\varphi_{a}(u)}{a + iu}du + c_{a}}
\end{equation}

As the function $\displaystyle{u \longrightarrow e^{-\frac{1}{2}(u - \gamma)^{2}}\frac{\varphi_{a}(u)}{a + iu}}$ is integrable we deduce that

\begin{equation}
\displaystyle{Lim \, e^{-\frac{1}{2}(y - \gamma)^{2}}\varphi_{a}^{'}(y) = c_{a}}\, as \,y \longrightarrow - \infty
\end{equation}

Now, if $c_{a} \neq 0$ then we get that

\begin{equation}
\displaystyle{\varphi_{a}^{'}(y) \thicksim c_{a}e^{\frac{1}{2}(y - \gamma)^{2}}}\, and   \,\displaystyle{\frac{e^{-\frac{1}{2}y^{2}}}{y}\varphi_{a}^{'}(y) \thicksim c_{a}\frac{e^{- \rho y}}{y}e^{\frac{\gamma^{2}}{2} - ia}}
\end{equation}

We consider the space

\begin{equation}
\displaystyle{\mathbb{B}_{a} = \{ \varphi_{a} : \mathbb{R} \longrightarrow \mathbb{C} entire ; \int_{\mathbb{R}}e^{-\frac{1}{2}y^{2}}\mid \varphi_{a}(y)\mid^{2}dy < +\infty\}}
\end{equation}

($\alpha$) From theorem 3.4, we deduce  that $\forall \, \varphi \in \mathbb{B}$ then $\varphi_{a} \in \mathbb{B}_{a}$

($\beta$) From lemma 3.2, we deduce  that $\forall \, \varphi_{a} \in \mathbb{B}_{a}$ then the function
$\displaystyle{y \longrightarrow  \frac{\varphi_{a}^{'}(y) - \varphi_{a}^{'}(0)}{y} \in \mathbb{B}_{a}}$

As in (3.53) the function $\displaystyle{y \longrightarrow \frac{e^{- \rho y}}{y}e^{\frac{\gamma^{2}}{2} - ia}}$ is not square integrable then $\varphi_{a} \notin \mathbb{B}_{a}$

which is contradictory with the property ($\beta$) and therefore $c_{a} = 0$.

(ii) we apply (i) to (3.53).  $\square$

By applying the above theorem, the last equation  can be written as follows

\begin{equation}
\displaystyle{\varphi_{a}^{'}(y) = i\sigma e^{\frac{1}{2}(y - \gamma)^{2}} \int_{-\infty}^{y}e^{-\frac{1}{2}(u - \gamma)^{2}}\frac{\varphi_{a}(u)}{a + iu}du; \, y \in \,]-\infty, 0] }
\end{equation}

(2) Justification for crossing to the limit in the equation (3.56) as $a \longrightarrow 0$

If $a \longrightarrow 0$ in (3.56) then we get

\begin{equation}
\displaystyle{\varphi_{0}^{'}(y) = \sigma e^{\frac{1}{2}(y - \rho)^{2}} \int_{-\infty}^{y}e^{-\frac{1}{2}(u - \rho)^{2}}\frac{\varphi_{0}(u)}{u}du; \, y \in \,]-\infty, 0] }
\end{equation}

As the functions $\displaystyle{a \longrightarrow \varphi_{a}^{'}(y)}$ and $\displaystyle{a \longrightarrow \frac{e^{-\frac{1}{2}(u - \rho - ia)^{2}}}{a + iu}\varphi_{a}(u)}$ are continuous  then

\begin{equation}
\displaystyle{\lim\limits_{a \rightarrow 0} \varphi_{a}^{'}(y) = \varphi_{0}^{'}(y)} \, and \, \displaystyle{\lim\limits_{a \rightarrow 0}\frac{e^{-\frac{1}{2}(u - \rho - ia)^{2}}}{a + iu}\varphi_{a}(u) =  \frac{e^{-\frac{1}{2}(u - \rho)^{2}}}{iu}\varphi_{0}(u)}
\end{equation}

We have to show that $\displaystyle{u \longrightarrow \frac{e^{-\frac{1}{2}(u - \rho)^{2}}}{iu}\varphi_{0}(u)}$ is integrable. For this purpose we consider the following function

\begin{equation}
\displaystyle{f_{a}(u) = \frac{e^{-\frac{1}{2}(u - \rho - ia)^{2}}}{a + iu}\varphi_{a}(u)}
\end{equation}

As for $\epsilon > 0$ enough small  we have $\varphi_{a}(a + iu) \thicksim a +iu ; u \in ]-\epsilon, 0]$, we deduce that

\begin{equation}
\displaystyle{\mid f_{a}(u)\mid \leq e^{\rho u}}
\end{equation}

Now, for $u \in ]-\infty , -\epsilon[$, we observe that

\begin{equation}
\displaystyle{\mid f_{a}(u)\mid \leq e^{-\frac{\rho^{2}}{2}}\frac{e^{\rho u}}{u}}
\end{equation}

As the function

\begin{equation}
g(u) = \left \{ \begin{array}[c] {l} \displaystyle{e^{-\frac{\rho^{2}}{2}}\frac{e^{\rho u}}{u}}; u \in ]-\infty , -\epsilon[\\
\quad\\
\displaystyle{e^{\rho u}}; u \in ]-\epsilon, 0]\\
\end{array} \right .
\end{equation}

is integrable and the the sequence $f_{a}(u)$ converges pointwise to  $\displaystyle{\frac{e^{-\frac{1}{2}(u - \rho^{2}}}{iu}\varphi_{0}(u)}$ as $a \longrightarrow 0$ then by applying the Lebesgue's Dominated Convergence Theorem, we deduce that

\begin{equation}
\displaystyle{\lim\limits_{a \rightarrow 0} \int_{-\infty}^{y}\frac{e^{-\frac{1}{2}(u - \rho - ia)^{2}}}{a + iu}\varphi_{a}(u)du = \int_{-\infty}^{y}\frac{e^{-\frac{1}{2}(u - \rho^{2}}}{iu}\varphi_{0}(u)}
\end{equation}

Consequently, we get

\begin{equation}
\displaystyle{\varphi_{0}^{'}(y) = \sigma e^{\frac{1}{2}(y - \rho)^{2}} \int_{-\infty}^{y}e^{-\frac{1}{2}(u - \rho)^{2}}\frac{\varphi_{0}(u)}{u}du; \, y \in \,]-\infty, 0] }
\end{equation}

.  $\square$

(i) An operator integral associated to $H_{\mu, \lambda}$ is given as follows

\begin{equation}
\displaystyle{K\psi(y) = \int_{-\infty}^{0}\mathcal{N}(y, s)\psi(s)ds}
\end{equation}

($\alpha$) $\rhd$

\begin{equation}
\displaystyle{\psi(y) = \frac{\varphi_{0}^{'}(y)}{y}e^{-\frac{y^{2}}{2}}\theta(y)}
\end{equation}

$\theta(y) = \left \{ \begin{array}[c] {l}\,  y ; \quad y \in [-1, 0]\\
\quad\\
- 1; \quad y \in ]-\infty, -1]\\
\end{array} \right .$

($\beta$) $\rhd$

\begin{equation}
\displaystyle{\mathcal{N}(y, s) = e^{-\rho y}\frac{\theta(y)}{y}.\frac{s}{\theta(s)}e^{\frac{s^{2}}{2}}\int_{-\infty}^{min(y, s)}e^{-\frac{1}{2}(u - \rho)^{2}}{u}du}
\end{equation}

$min(y, s) = \left \{ \begin{array}[c] {l} y ; \quad y \leq s\\
\quad\\
s; \quad  s \leq y \\
\end{array} \right .$

($\gamma$) $\rhd$

\begin{equation}
\displaystyle{\psi \in L_{2}(]-\infty, 0], \theta(y)dy)}
\end{equation}

(ii) For $\psi \in \mathbb{B}_{0}$, an explicit inverse of $H_{\mu, \lambda}$ restricted on imaginary axis ; $y \in [0,  +\infty[$ is given by

\begin{equation}
\displaystyle{H_{\mu, \lambda}^{-1} \psi(-iy) = \int_{0}^{\infty}\mathcal{N}_{\mu,\lambda}(y, s)\psi(-is)ds}
\end{equation}

where  $\displaystyle{\mathcal{N}_{\mu,\lambda}(y, s) = \frac{1}{\lambda s}e^{-\frac{s^{2}}{2} - \frac{\mu}{\lambda}s}\int_{0}^{min(y, s)}e^{\frac{u^{2}}{2} +\frac{\mu}{\lambda}u}du}$

(i) We substitue $\displaystyle{\varphi_{0}(u) = -\int_{u}^{0} \varphi_{0}^{'}(s)ds}$ in (3.64) to obtain

\begin{equation}
\displaystyle{\varphi_{0}^{'}(y) = -\sigma e^{\frac{1}{2}(y - \rho)^{2}} \int_{-\infty}^{y}\frac{e^{-\frac{1}{2}(u - \rho)^{2}}}{u}\int_{u}^{0}\varphi_{0}^{'}(s)ds; \, y \in \,]-\infty, 0] }
\end{equation}

By applying the Fubini theorem we get

\begin{equation}
\displaystyle{\int_{-\infty}^{y}\frac{e^{-\frac{1}{2}(u - \rho)^{2}}}{u}\int_{u}^{0}\varphi_{0}^{'}(s)ds = \int_{-\infty}^{0}[\int_{-\infty}^{min(y, s)}\frac{e^{-\frac{1}{2}(u - \rho)^{2}}}{u}du] \varphi_{0}^{'}(s)ds}
\end{equation}

\begin{equation}
\displaystyle{\varphi_{0}^{'}(y) = -\sigma e^{\frac{1}{2}(y - \rho)^{2}}\int_{-\infty}^{0}[\int_{-\infty}^{min(y, s)}} \displaystyle{\frac{e^{-\frac{1}{2}(u - \rho)^2}}{u}du] \varphi_{0}^{'}(s)ds}
\end{equation}

\begin{equation}
\displaystyle{e^{-\frac{y^{2}}{2}}\varphi_{0}^{'}(y)} \displaystyle{ = -\sigma e^{\frac{\rho^{2}}{2}}\int_{-\infty}^{0}[ e^{\rho y} \int_{-\infty}^{min(y, s)}\frac{e^{-\frac{1}{2}(u - \rho)^{2}}}{u}du] \varphi_{0}^{'}(s)ds}
\end{equation}

Now, if we put :

\begin{equation}
\displaystyle{\psi(y) = \frac{\varphi_{0}^{'}(y)}{y}e^{-\frac{y^{2}}{2}}\theta(y)}
\end{equation}

with $\theta(y) = \left \{ \begin{array}[c] {l}\,  y ; \quad y \in [-1, 0]\\
\quad\\
- 1; \quad y \in ]-\infty, -1]\\
\end{array} \right .$

and if we substitute $\psi$ in (3.73) then we get

\begin{equation}
\displaystyle{\psi(y) = -\sigma e^{\frac{\rho^{2}}{2}}\int_{-\infty}^{0}\mathcal{N}(y, s) \psi(s)ds }
\end{equation}

Where $\mathcal{N}(y, s)$ is given by (3.67).

(ii) Let $\varphi \in \mathbb{B}_{0}$ and  $\psi \in \mathbb{B}_{0}$, we consider the equation $H_{\mu, \lambda}\varphi = \psi (z)$ i.e.

\begin{equation}
\displaystyle{i\lambda z \varphi^{"}(z)  + (i\lambda z^{2} + \mu z)\varphi'(z) = \psi(z)}
\end{equation}

Let $ z = -iy$, $u(y) = \varphi(-iy)$ and $ f(y) = \psi(-iy)$ with $y \in [0, \infty[$ then (3.76) can be written in the following form:

\begin{equation}
\displaystyle{ - \lambda y u''(y) + (\lambda y^{2} + \mu y) u'(y) = f(y)}
\end{equation}

\begin{equation}
\displaystyle{ - u''(y) + (y + \rho ) u'(y) = \frac{f(y)}{\lambda y}}
\end{equation}

with help theorem 3.4, Lemma 3.2 and a similar technique used in proof of (i) of this theorem we get (3.69).  $\square$

(3) The extension of integral operator  associated to the restriction of $H_{\mu, \lambda}$  on negative  imaginary axis to Hilbert-Scmidt operator on  $\displaystyle{L_{2}(]-\infty, 0], \theta(y)dy)}$

Le $ \mathcal{N}(y, s)$ = $ e^{-\rho y} \frac{\theta(y)}{y} \frac{s}{\theta(s)}e^{\frac{1}{2}s^{2}}$ $\displaystyle{\int_{-\infty}^{min(y,s)}}$ $\frac{e^{-\frac{1}{2}(u - \rho)^{2}}}{u}du$ be the kernel of the equation:

\begin{equation}
\displaystyle{-\lambda y\varphi''(iy) + (\lambda y^{2} + \mu y)\varphi'(iy)  = \psi(iy)  , y < 0}
\end{equation}

where  $\displaystyle{\rho = \frac{\mu}{\lambda}}$ then:

(i) $ \mathcal{N}(y, s)$ is Hilbert-Shmidt i.e. $\displaystyle{\mathcal{N}(y, s)}$ belongs to

$\displaystyle{L^{2}(]-\infty, 0]\times]-\infty, 0], \theta(y)\theta(s)dyds)}$.

(ii) $\displaystyle{\mathcal{N}(y, s)}$ belongs to $\displaystyle{L^{2}(]-\infty, 0]\times]-\infty, 0], \theta(y)\theta(s)dyds)}$.

(i) To apply the theorem of dominated convergence, we will increase $K(y, s)$ by a function $\tilde{K}(y, s) \in L^{2}(]-\infty, 0]\times]-\infty, 0], \theta(y)\theta(s)dyds)$

For this we pose $min(y, s) = m$  then we have :

\begin{equation}
\displaystyle{\mid \frac{e^{-\frac{1}{2}(u - \rho)^{2}}}{u}\mid \leq e^{-\frac{1}{2}\rho^{2}}\frac{1}{m(m - \rho)}\mid (\rho - u) e^{-\frac{1}{2}u^{2} + \rho u}\mid}
\end{equation}

The function $ u \longrightarrow \displaystyle{\frac{1}{u(u - \rho)}} $ is bounded for $-\infty < u \leq m < 0$

Hence a first increase:

\begin{equation}
\mid \mathcal{N}(y, s) \mid \leq \displaystyle{e^{-\frac{1}{2}\rho^{2}}\mid \frac{e^{-\rho y}}{y}\frac{\theta(y)}{\theta(s)}se^{\frac{1}{2}s^{2}}\frac{e^{-\frac{1}{2}m^{2} + \rho^{m}}}{m(m - \rho)}\mid}
\end{equation}

If $y \leq s$, we have:

$ m = y$ et $\displaystyle{e^{\frac{1}{2}s^{2} - \frac{1}{2}y^{2}} < 1}.$

Then :

\begin{equation}
\mid \mathcal{N}(y, s) \mid \leq \displaystyle{e^{-\frac{1}{2}\rho^{2}}\mid \frac{\theta(y)}{y}\frac{s}{\theta(s)}\frac{1}{y(y - \rho)}}\mid
\end{equation}

If $s \leq y$, we have:

$ m = s$ et $e^{\rho (s - y)} < 1.$

\begin{equation}
\mid \mathcal{N}(y, s) \mid \leq e^{-\frac{1}{2}\rho^{2}}\mid \frac{\theta(y)}{y}\frac{s}{\theta(s)}\frac{1}{s(s - \rho)}\mid
\end{equation}

Finally we put :

\begin{equation}
\tilde{\mathcal{N}}(y, s) = \left\{
\begin{array}{ c }
e^{-\frac{1}{2}\rho^{2}}\mid \frac{\theta(y)}{y}\frac{s}{\theta(s)}\frac{1}{y(y - \rho)} \quad pour \quad y \leq s \\
e^{-\frac{1}{2}\rho^{2}}\mid \frac{\theta(y)}{y}\frac{s}{\theta(s)}\frac{1}{s(s - \rho)} \quad pour \quad s \leq y \\
\end{array} \right .
\end{equation}

to get :

\begin{equation}
\mid \mathcal{N}(y, s) \mid \leq \mid \tilde{N}(y, s) \mid
\end{equation}

it remains to show that $ \tilde{N}(y, s) \in L^{2}(]-\infty, 0]\times]-\infty, 0], \theta(y)\theta(s)dyds)$

Let $\Delta = ]-\infty, 0]\times]-\infty, 0] -]-1, 0]\times]-1, 0], \Delta_{1} = \{(y,s) \in \Delta ; y \leq s\}$ et

$\Delta_{2} = \{(y,s) \in \Delta ; s \leq y\}$. Then

$\displaystyle{\int_{-\infty}^{0}\theta(y)dy}\displaystyle{\int_{-\infty}^{0}\theta(s)\mid \tilde{N}(y, s) \mid^{2} ds} = $

$\displaystyle{\int_{\Delta_{1}}\theta(y)\theta(s)\mid \frac{\theta(y)}{y}\frac{s}{\theta(s)}\frac{1}{y(y - \rho)}\mid^{2}dyds} + \displaystyle{\int_{\Delta_{2}}\theta(y)\theta(s)\mid \frac{\theta(y)}{y}\frac{s}{\theta(s)}\frac{1}{s(s - \rho)}\mid^{2}dyds}$

As $\rho > 0$  and the function  $\displaystyle{ u \rightarrow \frac{1}{(u - \rho)^{2}}}$  is integrable at the origin, we deduce that:

\begin{equation}
\displaystyle{\int_{-\infty}^{0}\theta(y)dy}\displaystyle{\int_{-\infty}^{0}\theta(s)\mid \tilde{N}(y, s) \mid^{2} ds < +\infty}
\end{equation}

then $\mathcal{N}(y, s)$ is a kernel of Hilbert-Shmidt.

We note that the functions :

$ y \rightarrow \displaystyle{\int_{-\infty}^{y}\frac{e^{-\frac{1}{2}(u - \rho)^{2}}}{u}du} ;\quad y \leq s$

$s \rightarrow \displaystyle{\int_{-\infty}^{s}\frac{e^{-\frac{1}{2}(u - \rho)^{2}}}{u}du} ;\quad s \leq y$

are decreasing and tend to zero respectively when$ y \rightarrow -\infty $ and $ s \rightarrow -\infty $,we deduce that:

$-\mathcal{N}(y, s)$ non negative.  $\square$

The operator $\mathbb{K}\psi(iy) = \displaystyle{\int_{-\infty}^{0} \mathcal{N}(y, s)\psi(is)ds}$ is:

(i)  Hilbert-Shmidt operator.

(ii) The spectrum of $\mathbb{K}$ is not empty and discrete.

It follows that we can give an explicit form of inverse of $H_{0, \mu,\lambda : H_{\mu, \lambda}}$ on negative imaginary axis; $z = -iy $ with $y > 0$.

$H_{0,\mu,\lambda}^{-1}$ can be written as integral operator in following form :

\begin{equation}
\displaystyle{H_{0,\mu,\lambda}^{-1}\psi(-iy) = \int_{0}^{+\infty}N_{0,\mu,\lambda}(y, s)\psi(-is)ds}
\end{equation}

\begin{equation}
\displaystyle{N_{0,\mu,\lambda}(y, s) = \frac{1}{\lambda s}e^{\frac{-s^{2}}{2} - \frac{\mu}{\lambda}s}\int_{0}^{min(y, s)}e^{\frac{u^{2}}{2} + \frac{\mu}{\lambda}u}du}
\end{equation}

Let $\displaystyle{L_{2}([0, +\infty[, e^{-x^{2} -2\omega x}dx)}$ be the space of square integrable  functions with respect to the measure $e^{-x^{2} -2\omega x}dx$  then  we have the following result :

(i) $\forall \, \mu \in \mathbb{C}; \mathcal{R}e \mu \geq \omega$  then $H_{0,\mu,\lambda}^{-1}$ can be extended to Hilbert-Schmidt operator of $\displaystyle{L_{2}([0, +\infty[, e^{-y^{2} -2\omega y}dy)}$ to $\displaystyle{L_{2}([0, +\infty[, e^{-y^{2} -2\omega y}dy)}$ .

(ii) For $\mu > 0$, let $\sigma(0,\mu)$ be the smallest eigenvalue of $H_{0,\mu,\lambda}$ then $\sigma (0,\mu)$
is positive, increasing and analytic function on the whole real line with respect to $\mu$.

(i) From the above proposition, we deduce (i).

(ii) is the main result of Ando-Zerner (see proposition 1 of  [Ando et al] in  Commun. Math. Phys. 93, (1984), p:123-139).  $\square$

(continuity of $H_{\mu, \lambda}^{-1}$ with respect to $\lambda$)

Let $\mu > 0$ and $\displaystyle{f: \mathbb{R} \longrightarrow \mathcal{B}_{0} ; \lambda \mapsto f(\lambda) = H_{\mu, \lambda}^{-1}}$, then $f$ is continuous.

As  for any $\lambda \in \mathbb{R}$ and $\mu > 0$ we have $\displaystyle{\mu \mid\mid \varphi \mid\mid \leq \mid\mid H_{\mu, \lambda}\varphi \mid\mid}$  on $\mathcal{B}_{0}$ then

$\displaystyle{\mid\mid H_{\mu, \lambda}^{-1}\psi \mid\mid \leq \frac{1}{\mu}\mid\mid \psi \mid\mid}$ $\forall\, \psi \in \mathcal{B}_{0}$.   (*)

Now, for $\lambda_{0} \in \mathbb{R}$ we have :

$\displaystyle{H_{\mu, \lambda}^{-1} - H_{\mu, \lambda_{0}}^{-1} = i(\lambda_{0} - \lambda)H_{\mu, \lambda}^{-1} A^{*}(A + A^{*})H_{\mu, \lambda_{0}}^{-1} }$.

By applying (*) we get

$\displaystyle{\mid\mid H_{\mu, \lambda}^{-1} - H_{\mu, \lambda_{0}}^{-1}\mid\mid \leq  \frac{1}{\mu}\mid\mid A^{*}(A + A^{*})H_{\mu, \lambda_{0}}^{-1}\mid\mid. \mid \lambda_{0} - \lambda \mid}$.

and as

$\displaystyle{\mid\mid A^{*}(A + A^{*})H_{\mu, \lambda_{0}}^{-1}\mid\mid = \frac{1}{\mid \lambda_{0}\mid }( 1 + \mu \mid\mid A^{*}AH_{\mu, \lambda_{0}}^{-1}\mid\mid}$.

From the density of the polynomial set $\mathcal{P}_{0}$ in $\mathbb{B}_{0}$, the equality of minimal domain with maximal domain of $\displaystyle{H_{\mu, \lambda}}$  and $\displaystyle{[A^{*}A]^{-1}(\mathcal{P}_{0}) = \mathcal{P}_{0}}$, we deduce the following fundamental inequality :

$\displaystyle{ \forall\, \epsilon > 0 , (\mu - \epsilon)\mid\mid A^{*}A\varphi \mid\mid \leq \mid\mid H_{\mu, \lambda}\varphi \mid\mid  + \frac{\mid \lambda\mid^{2}}{4\epsilon}\mid\mid \varphi \mid\mid  \, \forall\, \varphi \in D(H_{\mu, \lambda})}$   (**)

We choose $\displaystyle{\epsilon = \frac{\mu}{2}}$ to get :

$\displaystyle{\frac{1}{\mid \lambda_{0}\mid }( 1 + \mu \mid\mid A^{*}AH_{\mu, \lambda_{0}}^{-1}\mid\mid) \leq \frac{1}{\mid\lambda_{0}\mid}(3 + \frac{\mid\lambda_{0}\mid^{2}}{\mu})}$.

Consequently, we get:

$\displaystyle{\mid\mid H_{\mu, \lambda}^{-1} - H_{\mu, \lambda_{0}}^{-1}\mid\mid  \leq \frac{1}{\mu} \frac{1}{\mid\lambda_{0}\mid}(3 + \frac{\mid\lambda_{0}\mid^{2}}{\mu})\mid \lambda_{0} - \lambda\mid}$

i.e the continuity of  $H_{\mu, \lambda}^{-1}$ in $\lambda_{0} \neq 0$.

Now for $\lambda_{0} = 0$, we have:

$\displaystyle{\mid\mid H_{\mu, \lambda}^{-1}\psi - \frac{1}{\mu}[A^{*}A]^{-1}\psi \mid\mid \leq \mid\mid H_{\mu, \lambda_{0}}^{-1}\mid\mid. \mid\mid \psi - \frac{1}{\mu}H_{\mu, \lambda}[A^{*}A]^{-1}\psi \mid\mid \forall\, [A^{*}A]^{-1}\psi \in D(H_{\mu, \lambda})}$
i.e. $\forall\, \psi \in A^{*}A[D(H_{\mu, \lambda})]$.

Hence we use the fundamental inequality (**) to get:

$\displaystyle{\mid\mid H_{\mu, \lambda}^{-1}\psi - \frac{1}{\mu}[A^{*}A]^{-1}\psi \mid\mid \leq \frac{\mid \lambda \mid}{\mu^{2}}\mid\mid A^{*}(A + A^{*})A[A^{*}A]^{-1}\psi\mid\mid}$ which converges to zero as $\lambda \longrightarrow 0$ $\forall\, \psi \in A^{*}A[D(H_{\mu, \lambda})]$.

Now, as $\displaystyle{A^{*}A[D(H_{\mu, \lambda})]}$ is dense in $\mathbb{B}_{0}$, we deduce the continuity of $f$ at zero.  $\square$

§.§ what class Carleman operators belong the inverse of $H_{\mu, \lambda}$ ?

$\bigstar$ On the factorization of $H_{\mu, \lambda}^{-1}$

Which bounded linear operator on a Hilbert space can be factored as the product of finitely many normal operators ?  What is the answer if is normal operators is replaced by involutions ? partial isometrics ? or other classes of familiar operators ? Just as in the case of the factorization of integers, polynomials, or other objects in mathematics, such operator factorization problems seem to arise very naturally in the course of study of operators.

In the cases when they are solved, the solutions are usually neat and elegant; otherwise, they pose interesting and challenging questions whose solutions may lead us to a deeper understanding of the nature of the operators under consideration.

For the products of two Hermitian operators on finite-dimensional spaces, we have the following classical characterization

The following statements are equivalent for a finite matrix $T$:

(1) $T$ is the product of two Hermitian matrices;

(2) $T$ is the product of two Hermitian matrices, one of which is nonsingular;

(3) $T$ is similar to a matrix with real entries;

(4) $T$ is similar to $T^{*}$.

The proof is an easy exercise in linear algebra. In the context of infinite- dimensional spaces, the implication $(1) \Longrightarrow (4)$ is no longer true.  As a consolation, it was conjectured that $T$ is similar to $T^{*}$ if $T$ is the product of two Hermitian Fredholm operators. This seems still to be open.

As for the implication $(4)  \Longrightarrow  (1)$ despite the abundance of supporting special cases, its validity is still unconfirmed even under the extra assumption that $T$ is invertible.  $\square$

The following conditions are equivalent for an operator $T$ on an infinite-dimensional space:

(1) $T$ is a commutator;

(2) $T$ is not the sum of nonzero scalar and a compact operator.

In this context, the products of commutators seem not to have been considered. It would also be interesting to characterize products of a special kind of commutators: self-commutators, operators of the form $T^{*}T - TT^{*}$.

Any compact operator on an infinite-dimensional space is the product of finitely many nonnegative operators.

For our operator $H_{\mu, \lambda}$, we will answered to following question: Its inverse $H_{\mu, \lambda}^{-1}$ can be factorized as the product of two operators  of Carleman's classes ?

Let $(\mu, \lambda) \in \mathbb{R}^{2}$ and  $H_{\mu, \lambda} = \mu H_{0} + i\lambda H_{1}$ with $H_{0} = A^{*}A$ and $H_{1} =  A^{*}( A + A^{*})A$ then  we can  to write $H_{0}^{-1}H_{1}$ but not $H_{1}H_{0}^{-1}$ because the domain of $H_{0}$  and the domain of $H_{1}$ are not included  in each other, nor is the domain of their commutator.

Observing that $H_{0}^{-1}(\mathcal{P}) = \mathcal{P} = H_{0}(\mathcal{P})$ then $H_{1}H_{0}^{-1}$ is well defined on the polynomial space $\mathcal{P}$ and we have:

$\displaystyle{\forall\, p \in \mathcal{P}, H_{\mu, \lambda} H_{0}^{-1}p = \mu p + i\lambda H_{1} H_{0}^{-1}p = \mu p + i\lambda A^{*}A^{2}H_{0}^{-1}p + i\lambda A^{*^{2}}AH_{0}^{-1}p}$

$\displaystyle{ = \mu p + i\lambda (AA^{*} - I)AH_{0}^{-1}p + i\lambda A^{*}p = \mu p + i\lambda(A + A^{*})p - i\lambda AH_{0}^{-1}p}$.

In particular, we get:

$\displaystyle{\mid \Re e < H_{\mu, \lambda} H_{0}^{-1}p, p > \mid \geq \mu \mid\mid p  \mid\mid^{2} -  \mid \lambda \mid \mid \Im m < AH_{0}^{-1}p, p > \mid  \forall \, p \in \mathcal{P}}$    ($\bigstar_{1}$)

$\displaystyle{\mid\mid H_{\mu, \lambda} H_{0}^{-1}p\mid\mid.\mid\mid p \mid\mid \geq \mu \mid\mid p  \mid\mid^{2} -  \mid \lambda \mid \mid\mid AH_{0}^{-1}p \mid\mid. \mid \mid p \mid \mid \forall \, p \in \mathcal{P}}$     ($\bigstar_{2}$)

Let $\displaystyle{p(z) = \sum_{k= 1}^{n}a_{k}\frac{z^{k}}{\sqrt{k!}}}$ then from  ($\bigstar_{1}$),  ($\bigstar_{2}$) and :

(i) $\displaystyle{AH_{0}^{-1}p(z) = \sum_{k= 1}^{n}a_{k}AH_{0}^{-1}\frac{z^{k}}{\sqrt{k!}} = \sum_{k= 1}^{n}\frac{a_{k}}{\sqrt{k}}\frac{z^{k-1}}{\sqrt{(k-1)!}};}$

(ii) $\displaystyle{\mid\mid A\varphi\mid\mid^{2} = < A^{*}A\varphi, \varphi > = < H_{0}\varphi, \varphi >; \varphi \in D(H_{0})}$,

we deduce that

$\displaystyle{\mid\mid AH_{0}^{-1}p\mid\mid^{2} \leq \mid\mid p \mid\mid^{2} \forall \, p \in \mathcal{P};}$    ($\bigstar_{3}$)

$\displaystyle{\mid\mid A\varphi\mid\mid \leq \mid\mid H_{0}\varphi\mid\mid^{\frac{1}{2}}\mid\mid \varphi\mid\mid^{\frac{1}{2}}, \forall\, \varphi \in D(H_{0});}$    ($\bigstar_{4}$)

$\displaystyle{\mid\mid AH_{0}^{-1}p \mid\mid \leq \mid\mid \varphi \mid\mid^{\frac{1}{2}}\mid\mid H_{0}^{-1}p \mid\mid^{\frac{1}{2}}\leq \frac{\epsilon}{\mid \lambda \mid} \mid\mid p \mid\mid  + \frac{\mid \lambda \mid }{\mid 4\epsilon\mid} \mid\mid H_{0}^{-1}p \mid\mid.}$    ($\bigstar_{5}$)

$\displaystyle{(\mu - \epsilon)\mid\mid p \mid\mid \leq \mid\mid H_{\mu, \lambda}H_{0}^{-1}p \mid\mid +  \frac{\mid \lambda \mid^{2}}{4\epsilon} \mid\mid H_{0}^{-1}p \mid\mid  \forall \, p \in \mathcal{P} }$    ($\bigstar_{6}$)

Now, as $\mathcal{P}$ is dense in $\mathcal{B}_{0}$ and the maximal domain of $H_{\mu, \lambda}$ coincides with its minimal domain, we deduce the following fundamental inequality:

$\displaystyle{\forall \epsilon > 0, (\mu - \epsilon)\mid\mid H_{0}H_{\mu, \lambda}^{-1}\psi \mid\mid\, \, \leq \, \, (1 +  \frac{\mid \lambda \mid^{2}}{4\mid \mu\epsilon\mid}) \mid\mid \psi \mid\mid  \forall \, \psi \in \mathcal{B}_{0} }$    ($\bigstar_{7}$)

We summarize the results of above inequalities  as follows :

(i) $A$ is $\frac{1}{2}$-subordinated to $H_{0}$,

(ii) $D(H_{\mu, \lambda}) \subset D(H_{0})$,

(iii) $H_{0}$ is subordinated to $H_{\mu, \lambda}$,

(iv) the operator $T = H_{0}H_{\mu, \lambda}^{-1}$ is bounded on Bargmann space $\mathcal{B}_{0}$ and $H_{\mu, \lambda}^{-1}$ is factored as product of $H_{0}^{-1}$ with $T$,

(v) the operators $H_{0}^{-1}$ and $H_{\mu, \lambda}^{-1}$ are of Carleman class of order $1 + \epsilon$, $\forall \, \epsilon > 0$.

Open question

Is there exist a strictly positive bounded operator $\mathbb{T}$ acting on Bargmann space such that  $\mathbb{T}H_{\mu, \lambda}^{-1} = H_{\mu, \lambda}^{-1^{*}}\mathbb{T}$ to apply the followiong Krein's theorem:

(see [Krein])

Let $\mathcal{H}$ be  an  infinite  dimensional,  separable,  complex  Hilbert  space  and  let $\mathcal{C}_{p} = \mathcal{C}_{p}(\mathcal{H})$ be the Schatten-von Neumann class of compact operators acting on $\mathcal{H}$ and let $\mathbb{M} \in \mathcal{C}_{p}(\mathcal{H})$ be a linear bounded operator with the property that there exists a strictly positive bounded operator $\mathbb{A}$ such that $\mathbb{A}\mathbb{M} = \mathbb{M}^{*}\mathbb{A}$. Then the spectrum of $\mathbb{M}$ is real and for every non-zero eigenvalue $\lambda$, if $(\mathbb{M} -\lambda I)^{m}f = 0$ for some $m > 1$, then $(\mathbb{M} -\lambda I)f = 0$.

Moreover, the eigenvectors of $\mathbb{M}^{*}$, including the null vectors, span $\mathcal{H}$.

Although complex symmetric matrices (complex matrices coinciding with their transposes) are as ubiquitous as Hermitian matrices (those coinciding with their
complex adjoint), the first class is less well-known than the second. The matrix associated to $H_{\mu, \lambda}$ belongs to first class.Takagi , was  the first who had remarked that the antilinear eigenvalue problem $T\varphi = \sigma\bar{\varphi}$, where $T$ is a complex symmetric matrix and $\bar{\varphi}$ denotes complex conjugation, entry by entry, of the vector $\varphi$, solves a fundamental interpolation problem for bounded analytic functions in the disk. He had remarked there that the largest such positive skew-eigenvalue $\sigma$ coincides with the operator norm of $T$. Later on, this observation was extended to bounded linear operators with certain symmetries. About half a century ago, Glazman laid the foundations of the theory of unbounded complex symmetric operators. We will tested the Glazman's theory on our operator $H_{\mu, \lambda}$.

We begin by recalling a few definitions and facts about complex symmetric operators.

We consider a separable Hilbert space $\mathcal{H}$ which carries a conjugation operator $\displaystyle{\mathcal{C} : \mathcal{H} \longrightarrow \mathcal{H}}$ (an antilinear operator satisfying the conditions $\displaystyle{\mathcal{C}^{2} = I}$ and $\displaystyle{ < \mathcal{C}\varphi, \mathcal{C}\psi > = < \varphi, \psi >  \, \forall \, (\varphi, \psi) \in \mathcal{H}\times \mathcal{H} }$.

For a fixed $\mathcal{C}$, we define the transpose $^{t}T$ of a bounded linear operator $T$ to be  $^{t}T =  \mathcal{C}T^{*} \mathcal{C}$ where $ T^{*}$ is adjoint of $T$.

We say that $T$ is $\mathcal{C}$-symmetric if $T = ^{t}T$ (equivalently, if $\displaystyle{\mathcal{C}T = T\mathcal{C}}$). More generally, we say that T is complex symmetric if there exists a $\mathcal{C}$ such that $T$ is $\mathcal{C}$-symmetric.

We say that a densely defined, closed graph operator is $\mathcal{C}$-symmetric if $\displaystyle{T \subset  \mathcal{C}T^{*} \mathcal{C}}$ and $\mathcal{C}$-selfadjoint if $\displaystyle{T =  \mathcal{C}T^{*} \mathcal{C}}$.

By applying the Glazman's theory on our operator $H_{\mu, \lambda}$ then we obtain the following theorem:

Let  $\mu > 0$ and $\displaystyle{H_{\mu, \lambda} : D(H_{\mu, \lambda}) \longrightarrow \mathcal{B}_{0}}$ then

(i) $H_{\mu, \lambda}$ is an unbounded $\mathcal{C}$-selfadjoint operator.

(ii) There exists an orthonormal basis $(\varphi_{n})_{n=0}$ of $\mathcal{B}_{0}$ consisting of solutions of the antilinear eigenvalue problem $\displaystyle{H_{\mu, \lambda} \varphi_{n} = \sigma{_n}\mathcal{C}\varphi_{n}}$ where $(\sigma{_n})_{n=0}$ is an increasing sequence of positive numbers tending to $\infty$.

(iii)  $\displaystyle{\mid\mid H_{\mu, \lambda}^{-1}\mid\mid = \frac{1}{inf_{n}\sigma_{n}}}$

(i) As $ H_{\mu, \lambda} =  H_{\mu, -\lambda}^{*}$ then $ H_{\mu, \lambda}$ is $\mathcal{C}$-symmetric and as $ H_{\mu, \lambda}^{-1}$ exists then by applying  a criterion which goes back to Zhikhar : << if a $\mathcal{C}$-symmetric operator $T$ with dense domain $D(T)$ $\displaystyle{ T: D(T) \subset \mathcal{H} \longrightarrow  \mathcal{H}}$  satisfies $\displaystyle{ (T - z I)D(T) =   \mathcal{H}}$ for some complex number $z$, then $T$ is  $\mathcal{C}$-selfadjoint >>, we deduce that $H_{\mu, \lambda}$ is an unbounded $\mathcal{C}$-selfadjoint operator.

(i) and (ii) is a direct consequence of results established by Garcia and Putinar in [Garcia et al].   $\square$

§.§ Explicit inversion of $H_{\lambda',\mu,\lambda}$ on $[0, -i\frac{\lambda}{\lambda'}]$ and convergence of spectral radius of $H_{\lambda',\mu, \lambda}^{-1}$ to spectral radius of $H_{\mu, \lambda}^{-1}$ as  $\lambda' \longrightarrow 0$

Using the same above technique used to get the kernel of $H_{\mu,\lambda}^{-1}$ on $]-\infty, 0]$, we  explicit the inverse of  $H_{\lambda',\mu,\lambda}$ on $[0, -i\frac{\lambda}{\lambda'}]$.

Let $\psi \in \mathbb{B}_{0}$ and $\phi \in D(H_{\lambda',\mu,\lambda})$, we consider the equation  $H_{\lambda',\mu,\lambda}\phi = \psi$ which can be written under following form :

\begin{equation}
\displaystyle{(\lambda'z^{2} + i\lambda z)\phi''(z) + (i\lambda z^{2}  + \mu z)\phi'(z) = \psi(z)}
\end{equation}

Choosing the straight line connecting $-i\rho' , z \in \mathbb{C}$ parametrized  by

\begin{equation}
\displaystyle{\gamma : [0, 1] \rightarrow \mathbb{C}, \gamma(t) = -i\rho' + t(z + i\rho') \quad \quad (\gamma(0) = -i\rho' , \gamma(1) = z)}
\end{equation}

If we define $\displaystyle{\int_{-i\rho'}^{z}\psi(\xi)d\xi:=\int_{\gamma}\psi(\xi)d\xi:= \int_{0}^{1}}\psi(\gamma(t))\gamma'(t)dt$ then our equation can be transformed to following integral equation :

\begin{equation}
\displaystyle{\phi(z) = \frac{1}{\lambda'}\int_{-i\rho'}^{z}e^{-i\rho'\eta}(\eta + i\rho')^{-(\delta +1)} [\int_{-i\rho'}^{\eta}e^{i\rho'\xi}(\xi + i\rho')^{\delta}\frac{\psi(\xi)}{\xi}d\xi]d\eta}
\end{equation}

However, the integral representation of $\phi(z)$ in this last equation  is hard to study in $\mathbb{C}$ for some existing results on eigenvalues and eigenfunctions of our operators. To overcome this difficulty, the study of integral operator is restricted on negative imaginary axis by setting :

$ z = -iy$, $u(y) = \phi(-iy)$ and $ f(y) = \psi(-iy)$ with $y \in [0, \rho']$ then (3.89) can be written in the following form:

\begin{equation}
\displaystyle{(\lambda'y^{2} - \lambda y)u''(y) + (\lambda y^{2} + \mu y) u'(y) = f(y)}
\end{equation}

As $\psi \in \mathbb{B}_{0}$ then $f(0) = 0$ and if we put $\delta = \rho'(\rho + \rho') -1$, $\eta = -is$ and $\xi = y_{1}$ with $s \in [0, y]$ and $y_{1} \in [s, \rho']$ then (3.91) can be transformed to  following integral equation

\begin{equation}
\displaystyle{\phi(-iy) = \frac{1}{\lambda'}\int_{0}^{y}e^{-\rho's}(\rho' - s)^{-(\delta +1)}\int_{s}^{\rho'}e^{\rho'y_{1}}(\rho' - y_{1})^{\delta}\frac{\psi(-iy_{1})}{y_{1}}dy_{1}ds}
\end{equation}

then we get

\begin{equation}
\displaystyle{H_{\lambda',\mu,\lambda}^{-1}\psi(-iy) = \int_{0}^{\rho'}N_{\lambda',\mu,\lambda}(y, y_{1})\psi(-iy_{1})dy_{1}}
\end{equation}

\begin{equation}
\displaystyle{N_{\lambda',\mu,\lambda}(y, y_{1}) = \frac{1}{\lambda'y_{1}} e^{\rho'y_{1}}(\rho' - y_{1})^{\delta}\int_{0}^{min(y,y_{1})}e^{-\rho's}( \rho' - s)^{-(\delta +1)}ds}
\end{equation}

\begin{equation}
\displaystyle{N_{\lambda',\mu,\lambda}(y, y_{1}) = \frac{1}{\lambda y_{1}} e^{\rho'y_{1}}(1 - \frac{y_{1}}{\rho'})^{\delta}\int_{0}^{min(y,y_{1})}e^{-\rho's}(1 - \frac{s}{\rho'})^{-(\delta +1)}ds}
\end{equation}

Let $ \Upsilon = min(y,y_{1})$ and $ \displaystyle{\Theta(\Upsilon) = \int_{0}^{\Upsilon}e^{-\rho's}(1 - \frac{s}{\rho'})^{-(\delta +1)}ds}$

then we get the following elementary properties on the function  $\Theta(\Upsilon)$  near to points zero and $\rho'$ respectively:

(i) $\displaystyle{\Theta(\Upsilon) \equiv \Upsilon}$ near zero.

(ii) $\displaystyle{\Theta(\Upsilon) \equiv (\rho' -\Upsilon)^{-\delta}}$ near $\rho'$.

Let $L_{2}([0, \rho'], r(y)dy)$ be  the space of square integrable  functions with respect to the weight  $r(y) = e^{2\rho'y}(1 - \frac{y}{\rho'})^{2\delta}; \rho' = \frac{\lambda}{\lambda'}$ and $\delta = \rho'(\rho + \rho') - 1$.

Then for $\lambda' > 0$, $\mu > 0$ and $\delta \geq 0$  $H_{\lambda',\mu,\lambda}^{-1}$ can be extended to Hilbert-Schmidt operator of $L_{2}([0, \rho'], r(y)dy)$ to $L_{2}([0, \rho'], r(y)dy)$.

Proof (see  follow)

The integral operator  $H_{\lambda',\mu, \lambda}^{-1}$ on $\displaystyle{[0, -i\frac{\lambda}{\lambda'}]}$  is extended to Hilbert-Scmidt operator on  $\displaystyle{L_{2}([0, -i\frac{\lambda}{\lambda'}], r(y)dy)}$

We wiil give an explicit inversion of $H_{\lambda',\mu,\lambda}$ on $[0, -i\frac{\lambda}{\lambda'}]$.

Setting $\rho' = \frac{\lambda}{\lambda'}$, $\rho = \frac{\mu}{\lambda}$ and $\delta = \rho'(\rho + \rho') -1$ for $\lambda' \neq 0$ and $\lambda \neq 0$.

We begin by to explicit the inverse of  $H_{\lambda',\mu,\lambda}$ on $[0, -i\frac{\lambda}{\lambda'}]$.

Let $\psi \in \mathbb{B}_{0}$ and $\phi \in D(H_{\lambda',\mu,\lambda}$, we consider the equation  $H_{\lambda',\mu,\lambda}\phi = \psi$ which can be written under following form :

$\displaystyle{(\lambda'z^{2} + i\lambda z)\phi''(z) = (i\lambda z^{2}  + \mu z)\phi'(z) = \psi(z)}$   (B.1)

Let $\psi \in \mathbb{B}_{0}$ and choosing the straight line connecting $-i\rho' , z \in \mathbb{C}$ parametrized  by

$\displaystyle{\gamma : [0, 1] \rightarrow \mathbb{C}, \gamma(t) = -i\rho' + t(z + i\rho') \quad \quad (\gamma(0) = -i\rho' , \gamma(1) = z)}$

If we define $\displaystyle{\int_{-i\rho'}^{z}\psi(\xi)d\xi:=\int_{\gamma}\psi(\xi)d\xi:= \int_{0}^{1}}\psi(\gamma(t))\gamma'(t)dt$ then the equation (B.1) can be transformed to following integral equation :

$\displaystyle{\phi(z) = \frac{1}{\lambda'}\int_{-i\rho'}^{z}e^{-i\rho'\eta}(\eta + i\rho')^{-(\delta +1)} [\int_{-i\rho'}^{\eta}e^{i\rho'\xi}(\xi + i\rho')^{\delta}\frac{\psi(\xi)}{\xi}d\xi]d\eta}$   (B.2)

However, the integral representation of $\phi(z)$ in this last equation (B.2) is hard to study in $\mathbb{C}$ for some existing results on eigenvalues and eigenfunctions of our operators. To overcome this difficulty, the study of (B.2) is restricted on negative imaginary axis by setting :

$ z = -iy$, $u(y) = \phi(-iy)$ and $ f(y) = \psi(-iy)$ with $y \in [0, \rho']$ then (B.1) can be written in the following form:

$\displaystyle{(\lambda'y^{2} - \lambda y)u''(y) + (\lambda y^{2} + \mu y) u'(y) = f(y)}$   (B.3)

As $\psi \in \mathbb{B}_{0}$ then $f(0) = 0$ and if we put $\eta = -is$ and $\xi = y_{1}$ with , $s \in [0, y]$ and $y_{1} \in [s, \rho']$ then (B.3) can be transformed to  following integral equation

$\displaystyle{\phi(-iy) = \frac{1}{\lambda'}\int_{0}^{y}e^{-\rho's}(\rho' - s)^{-(\delta +1)}\int_{s}^{\rho'}e^{\rho'y_{1}}(\rho' - y_{1})^{\delta}\frac{\psi(-iy_{1})}{y_{1}}dy_{1}ds}$   (B.4)

then we get

$\displaystyle{K_{\lambda',\mu,\lambda}\psi(-iy) = \int_{0}^{\rho'}N_{\lambda',\mu,\lambda}(y, y_{1})\psi(-iy_{1})dy_{1}}$   (B.5)

$\displaystyle{N_{\lambda',\mu,\lambda}(y, y_{1}) = \frac{1}{\lambda'y_{1}} e^{\rho'y_{1}}(\rho' - y_{1})^{\delta}\int_{0}^{min(y,y_{1})}e^{-\rho's}( \rho' - s)^{-(\delta +1)}ds}$   (B.6)

$\displaystyle{N_{\lambda',\mu,\lambda}(y, y_{1}) = \frac{1}{\lambda y_{1}} e^{\rho'y_{1}}(1 - \frac{y_{1}}{\rho'})^{\delta}\int_{0}^{min(y,y_{1})}e^{-\rho's}(1 - \frac{s}{\rho'})^{-(\delta +1)}ds}$   (B.7)

Let $ \Upsilon = min(y,y_{1})$ and $ \displaystyle{\Theta(\Upsilon) = \int_{0}^{\Upsilon}e^{-\rho's}(1 - \frac{s}{\rho'})^{-(\delta +1)}ds}$

then we get the following elementary properties on the function  $\Theta(\Upsilon)$  near to points zero and $\rho'$ respectively:

Let $ \Upsilon = min(y,y_{1})$ et $ \displaystyle{\Theta(\Upsilon) = \int_{0}^{\Upsilon}e^{-\rho's}(1 - \frac{s}{\rho'})^{-(\delta +1)}ds}$ then

(i) $\displaystyle{\Theta(\Upsilon) \equiv \Upsilon}$ near zero.

(ii) $\displaystyle{\Theta(\Upsilon) \sim (\rho' -\Upsilon)^{-\delta}}$ near $\rho'$.

To give an extension of $K_{\lambda',\mu,\lambda}$ to $L_{2}([0, \rho'], r(y)dy)$, we consider

Let $L_{2}([0, \rho'], r(y)dy)$ be  the space of square integrable  functions with respect to the weight  $r(y) = e^{2\rho'y}(1 - \frac{y}{\rho'})^{2\delta}; \rho' = \frac{\lambda}{\lambda'}$ and $\delta = \rho'(\rho + \rho') - 1$.

By using the above lemma, we deduce the following theorem:

For $\lambda' > 0$, $\mu > 0$ and $\delta \geq 0$  $K_{\lambda',\mu,\lambda}$ can be extended to Hilbert-Schmidt operator of  $L_{2}([0, \rho'], r(y)dy)$ to $L_{2}([0, \rho'], r(y)dy)$.

Let $\displaystyle{r(y) = e^{2\rho'y}(1 - \frac{y}{\rho'})^{2\delta}; \delta = \rho'(\rho + \rho') - 1}$ then

$\displaystyle{K_{\lambda',\mu,\lambda}\psi(-iy) = \int_{0}^{\rho'}N_{\lambda',\mu,\lambda}(y, y_{1})\psi(-iy_{1})dy_{1}}$

$\displaystyle{N_{\lambda',\mu,\lambda}(y, y_{1}) = \frac{1}{\lambda y_{1}} e^{\rho'y_{1}}(1 - \frac{y_{1}}{\rho'})^{\delta}\int_{0}^{min(y,y_{1})}e^{-\rho's}(1 - \frac{s}{\rho'})^{-(\delta +1)}ds}$

can be written under the following form :

$\displaystyle{K_{\lambda',\mu,\lambda}\psi(-iy) = \int_{0}^{\rho'}\tilde{N}_{\lambda',\mu,\lambda}(y, y_{1})\psi(-iy_{1})r(y_{1})dy_{1}}$   (B.8)

$\displaystyle{\tilde{N}_{\lambda',\mu,\lambda}(y, y_{1}) = \frac{1}{\lambda y_{1}\sqrt{r(y_{1})}} \int_{0}^{min(y,y_{1})}e^{-\rho's}(1 - \frac{s}{\rho'})^{-(\delta +1)}ds}$   (B.9)

$\displaystyle{\tilde{N}_{\lambda',\mu,\lambda}(y, y_{1}) = \frac{1}{\lambda y_{1}\sqrt{r(y_{1})}} \Theta(min(y,y_{1}))}$   (B.10)

Now we consider the following integral:

$\displaystyle{\mathbb{I} = \int_{0}^{\rho'}\int_{0}^{\rho'}\tilde{N}_{\lambda',\mu,\lambda}^{2}(y, y_{1})r(y)r(y_{1})dydy_{1}}$   (B.11)

$\displaystyle{\mathbb{I} = \int_{0}^{\rho'}\int_{0}^{\rho'}\frac{1}{\lambda^{2} y_{1}^{2}} \Theta^{2}(min(y,y_{1}))r(y)dydy_{1}}$   (B.12)

By applying Fubini theoren then (B.10) can be written as :

$\displaystyle{\mathbb{I} = \int_{0}^{\rho'}\int_{0}^{y_{1}}\frac{1}{\lambda^{2} y_{1}^{2}} \Theta^{2}(min(y,y_{1}))r(y)dydy_{1} + \int_{0}^{\rho'}\int_{y_{1}}^{\rho'}\frac{1}{\lambda^{2} y_{1}^{2}} \Theta^{2}(min(y,y_{1}))r(y)dydy_{1}}$

and we have:

$\displaystyle{\mathbb{I} = \int_{0}^{\rho'}\int_{0}^{y_{1}}\frac{1}{\lambda^{2} y_{1}^{2}} \Theta^{2}(y)r(y)dydy_{1} + \int_{0}^{\rho'}\int_{y_{1}}^{\rho'}\frac{1}{\lambda^{2} y_{1}^{2}} \Theta^{2}(y_{1})r(y)dydy_{1} = \mathbb{I}_{1} + \mathbb{I}_{2}}$

$\displaystyle{\mathbb{I}_{1} = \int_{0}^{\rho'}\frac{1}{\lambda^{2} y_{1}^{2}}\int_{0}^{y_{1}} \Theta^{2}(y)r(y)dydy_{1}}$    (B.13)

$\displaystyle{\mathbb{I}_{2} = \int_{0}^{\rho'}\frac{\Theta^{2}(y_{1})}{\lambda^{2} y_{1}^{2}}\int_{y_{1}}^{\rho'}r(y)dydy_{1}}$   (B.14)

Now we remark that

(i) For $y_{1}$ near zero we have $\displaystyle{\int_{0}^{y_{1}} \Theta^{2}(y)r(y)dy \equiv y_{1}^{3}}$

(ii) For $y_{1}$ near $\rho'$, the function $ y_{1} \rightarrow \displaystyle{\int_{0}^{y_{1}} \Theta^{2}(y)r(y)dy}$ is continuous.

then from (i) and (ii), we deduce that the first integral (B.11) converges. In similar way, we verify that the second integral (B.12) converges and consequently the operator $K_{\lambda',\mu,\lambda}$ is Hilbert-Schmidt on $L_{2}([0, \rho'], r(y)dy)$.   $\square$

We summary some above results in the following theorem :

i) Let $\displaystyle{N_{\rho'}(y, y_{1}) = \frac{1}{\lambda y_{1}\sqrt{r(y_{1})}} \int_{0}^{min(y,y_{1})}e^{-\rho's}(1 - \frac{s}{\rho'})^{-(\delta +1)}ds}$ then

the function $\rho' \rightarrow N_{\rho'}$ is decreasing.

ii) Let $\displaystyle{N_{\rho'}(y, y_{1}) = \frac{1}{\lambda  y_{1}}e^{\frac{- y_{1}^{2}}{2} - \frac{\mu}{\lambda} y_{1}}\int_{0}^{min(y,  y_{1})}e^{\frac{s^{2}}{2} + \frac{\mu}{\lambda}s}ds}$ then

$\displaystyle{Lim \quad \int_{0}^{\rho'}N_{\rho'}(y, y_{1})\psi(-iy_{1})dy_{1} = \int_{0}^{+\infty}N_{\mu,\lambda}(y, y_{1})\psi(-iy_{1})dy_{1}}$ as $\rho' \longrightarrow +\infty$

iii) The integral operator $K_{\rho'}$ can be extended to Hilbert-Schmidt operator of
$\displaystyle{L_{2}([0, +\infty[, r_{\infty}(y)dy)}$ to $\displaystyle{L_{2}([0, +\infty[, r_{\infty}(y)dy)}$.

iv) On $\displaystyle{L_{2}([0, +\infty[, r_{\infty}(y)dy)}$, the integral operator $K_{\rho'}$ converges to  integral operator $K_{0,\mu,\lambda}$ as $\rho'$ goes to infinity with respect to Hilbert-Schmidt norm.

v) Let $\Omega(\lambda',\mu,\lambda)$ be the spectral radius of $K_{\lambda',\mu,\lambda}$  of kernel $N_{\lambda',\mu,\lambda}$ and let $\Omega(0,\mu,\lambda)$ be the spectral radius of $K_{0,\mu,\lambda}$ of kernel $N_{0,\mu,\lambda}$ then

$Lim \quad \Omega(\lambda',\mu,\lambda) = \Omega(0,\mu,\lambda)$ as $\lambda' \rightarrow 0$ or $\rho' \rightarrow +\infty$.

vi) The functions $\mu \rightarrow \tilde{N}_{\lambda',\mu,\lambda}$ and $\mu \rightarrow \Omega(\lambda',\mu,\lambda)$ are creasing with respect to $\mu$.

vii) The function $\mu \rightarrow \Omega(\lambda',\mu,\lambda)$ is analytic on the whole real line with respect to $\mu$ and the function $\mu \rightarrow \Omega(0,\mu,\lambda)$ is also analytic on the whole real line with respect to $\mu$.

i) By observing that the map $\displaystyle{\rho' \rightarrow e^{\frac{a}{\rho'}}}$ is decreasing where $a$ is an positive constant, we deduce that the map $\rho' \rightarrow  N_{\rho'}$ is also decreasing.

ii) We obtain this property from an above lemma.

iii) If we consider the integral

then we have:

$\displaystyle{\int_{0}^{+\infty}\int_{0}^{+\infty}N_{\rho'}^{2}\frac{r_{\infty}(y)}{r_{\infty}(y_{1})}dydy_{1}}$ = $\displaystyle{\int_{0}^{\rho'}\int_{0}^{\rho'}\tilde{N}_{\lambda', \mu, ,\lambda}^{2}e^{-(y^{2} - y_{1}^{2}) - 2\rho(y - y_{1})}dydy_{1}}$

$\displaystyle{\leq C_{\rho'}\int_{0}^{\rho'}\int_{0}^{\rho'}\tilde{N}_{\lambda', \mu, ,\lambda}^{2}dydy_{1}}$ where $C_{\rho'}$ is a constant because the function $\displaystyle{e^{-(y^{2} - y_{1}^{2}) - 2\rho(y - y_{1})}}$ is bounded.

As $\displaystyle{\int_{0}^{\rho'}\int_{0}^{\rho'}\tilde{N}_{\lambda', \mu, ,\lambda}^{2}dydy_{1} < +\infty} $ then we deduce the property iii) of this theorem.

iv) Let $(\mathbb{X},\Sigma, \tau) $ be a measure space with positive measure $\tau$. An version of the classical monotone convergence theorem of Beppo Levi formulated in terms of functions of the Lebesgue space $L_{2}(X,\Sigma, \tau)$ reads as follows: If $(f_{n }: n \geq 1)$ is an decreasing sequence (that is, $f_{n}(x) \geq f_{n+1}(x)$ for every $n$ and almost every $x$ in $\mathbb{X}$) of nonnegative,square integrable functions on $\mathbb{X}$ such that $\displaystyle{\int_{X}\mid f_{1}(x)\mid^{2}d\tau < +\infty}$

$f_{n}$ converges to some square integrable function $f$ both almost everywhere and in $\displaystyle{L_{2}}$ norm as $n \rightarrow+\infty$.

By applying this theorem, (ii) the above lemma , (i) and (ii) of theorem 3.5 we deduce that:

$\displaystyle{lim\int_{0}^{+\infty}\int_{0}^{+\infty}G_{\rho'}^{2}\frac{r_{\infty}(y)}{r_{\infty}(y_{1})}dydy_{1}}$ $= \displaystyle{\int_{0}^{+\infty}\int_{0}^{+\infty}N_{\mu,\lambda}^{2}\frac{r_{\infty}(y)}{r_{\infty}(y_{1})}dydy_{1}}$ as $\rho' \rightarrow+\infty$

$K_{\rho'}$ converges to $K(\mu, \lambda)$ as $\rho'$ goes to infinity in Hilbert-Schmidt norm on $\displaystyle{L_{2}([0, +\infty[, r_{\infty}(y)dy)}$.

v) Let $\Omega(\lambda',\mu,\lambda)$ be the spectral radius of $K_{\lambda',\mu,\lambda}$  of kernel $N_{\lambda',\mu,\lambda}$ and let $\Omega(0,\mu,\lambda)$ be the spectral radius of $K_{0,\mu,\lambda}$ of kernel $N_{0,\mu,\lambda}$ then

As $N_{\lambda',\mu,\lambda}$ is positive decreasing sequence of kernels which converges to the positive kernel $N_{0,\mu,\lambda}$ as goes to infinity we deduce that :

$\Omega(\mu,\lambda) \leq Lim \quad \Omega(\lambda',\mu,\lambda) $ as $\rho' \rightarrow + \infty$.

Thus it suffices to show that