The Compactness of Right Inverse of Imaginary Part of Reggeon Field Theory Hamiltonian on Bargmann Space

Abstract

The Hamiltonian of Reggeon field theory is defined by ${\displaystyle H_{\mu ,\lambda }=\mu A^{*}A + }$${\displaystyle i\lambda A^{*}(A+A^{*})A}$, where A and $A^{*}$ are the annihilation and creation operators satisfying $[A,A^{*}]=I$ and $\mu$, $\lambda$ are real parameters, and $i^2=-1$. This operator acts on Bargmann space $\mathbb{B}$ where $\mathbb{B}$ is a Hilbert space of holomorphic square integrable functions with respect to the Gaussian-weighted Lebesgue measure. In this work, we consider the operator ${\displaystyle H_{\lambda }=i\lambda A^{*}(A+A^{*})A}$ with maximum domain ${\displaystyle D\left(H_{\lambda }\right)=\{\phi \in \mathbb{B};H_{\lambda }\phi \in \mathbb{B}\}}$. If we limit the domain to polynomials and take the closure of the obtained operator, we denote it by ${\displaystyle H_{\lambda }^{min},}$ of which $H_{\lambda }$ is obviously an extension. Contrary to what happens for $\mu \ne 0$, it is well known that these two operators are different. The main purpose of the present work is to show that ${\displaystyle H_{\lambda }}$ admits a right-inverse ${\displaystyle K_{\lambda }}$, i.e., ${\displaystyle H_{\lambda }{K}_{\lambda }=I}$ on negative imaginary axis and that ${\displaystyle K_{\lambda }}$ is compact.

1. Introduction

We recall that the Hamiltonian of Reggeon Field Theory is defined by

$$H_{\mu ,\lambda }=\mu A^{*}A+i\lambda A^{*}(A+A^{*})A$$

(1)

where A and $A^{*}$ are the annihilation and creation operators, $\mu$ and $\lambda$ are real parameters, and $i^2=-1$.

This operator is considered to act on Bargmann space [1]:

$$\mathbb{B}=\{\phi :\mathbb{C}⟶\mathbb{C}\mathrm{entire};{\int }_{\mathbb{C}}|\phi \left(z\right){|}^2{e}^{-{\left|z\right|}^2}dxdy<\infty \}$$

(2)

where its usual basis is given by

$$e_n\left(z\right)={\displaystyle \frac{z^n}{\sqrt{n!}}};n\in \mathbb{N}$$

(3)

or

$${\tilde{e}}_n\left(z\right)={\displaystyle \frac{{\left(iz\right)}^n}{\sqrt{n!}}};n\in \mathbb{N}$$

(4)

and the annihilation operator A and creation operator $A^{*}$ are defined by

$$A\phi :={\displaystyle \frac{d\phi }{dz}}\phi \in \mathbb{B}$$

(5)

$$A^{*}\phi =z\phi \phi \in \mathbb{B}$$

(6)

Action of A and $A^{*}$ on usual basis ${\left(e_n\left(z\right)\right)}_{n\in \mathbb{N}}$ are given, respectively, by

$$A{e}_n=\sqrt{n}{e}_{n-1},n\ge 1\mathrm{and}A{e}_0=0$$

(7)

$$A^{*}{e}_n=\sqrt{n+1}{e}_{n+1},n\ge 0$$

(8)

Remark 1

(Some Specific Properties on Bargmann’s Space).

-

In [1] it is well known that $L^2\left(\mathbb{R}\right)$ is related to $\mathbb{B}$ by an unitary transform (called Bargmann transform) of $L^2\left(\mathbb{R}\right)$ onto $\mathbb{B}$, given by the following transform

$$\mathcal{B}\left(f\right):=\phi \left(z\right)=c{\int }_{\mathbb{R}}{e}^{-{\displaystyle \frac{1}{2}}{z}^2-{\displaystyle \frac{1}{2}}{q}^2+\sqrt{2}zq}f\left(q\right);f\in L^2\left(\mathbb{R}\right)\mathrm{and}c>0$$

-

In [2], I established several properties on Bargmann space:

(a) 

$\mathbb{B}$ is related to ${\mathbb{B}}_s$ by an unitary transform of $\mathbb{B}$ onto ${\mathbb{B}}_s$, given by the following transform:

$$\mathcal{I}:\mathbb{B}⟶{\mathbb{B}}_s;\mathcal{I}\left(\phi \right)=({\displaystyle \frac{1}{n!}}{\phi }^{\left(n\right)}\left(0\right){)}_0^{\infty }:={\left(a_n\right)}_0^{\infty }$$

where

$${\mathbb{B}}_s=\{{\left(a_n\right)}_0^{\infty }\in \mathbb{C}\mathit{such}\mathit{that}\sum _{n=0}^{\infty }n!{\left|a_n\right|}^2<\infty \}.$$

(b) 

Lemmas 1 and 2 and in particular Lemma 3 link a function of the Bargmann space with its derivative. This lemma plays a fundamental role in the study of the Hamiltonian of reggeon field theory:

Lemma 1

(Lemma 0.3 [2]).

In Bargmann space, we have ${\displaystyle \phi \in \mathbb{B}\iff z⟶\frac{{\phi }^{\prime }\left(z\right)-{\phi }^{\prime }\left(0\right)}{z}\in \mathbb{B}.}$

The expressions of $H_{\mu ,\lambda }$ and $H_{\lambda }$ are given, respectively, by

$$H_{\mu ,\lambda }\phi =\mu z{\displaystyle \frac{d\phi }{dz}}+i\lambda (z{\displaystyle \frac{d^2\phi }{d{z}^2}}+z^2{\displaystyle \frac{d\phi }{dz}})\phi \in \mathbb{B}$$

(9)

$$H_{\lambda }\phi =i\lambda (z{\displaystyle \frac{d^2\phi }{d{z}^2}}+z^2{\displaystyle \frac{d\phi }{dz}})\phi \in \mathbb{B}$$

(10)

• In [2], we have given a complete spectral analysis of the following $\mathbb{C}$-symmetric matrices which play an important role in Reggeon field theory

  $${\mathbb{H}}_n^{\mu ,\lambda }=\left(\begin{array}{cccccccc}\mu & i\lambda \sqrt{2}& 0& .& .& .& .& \\ i\lambda \sqrt{2}& 2\mu & i\lambda 2\sqrt{3}& \ddots & .& .& .\\ 0& i\lambda 2\sqrt{3}& 3\mu & i\lambda 3\sqrt{4}& \ddots & .& .\\ .& \ddots & \ddots & \ddots & \ddots & \ddots & .& \\ .& .& \ddots & \ddots & \ddots & \ddots & 0& \\ .& .& .& \ddots & \ddots & \ddots & i\lambda (n-1)\sqrt{n}& \\ .& .& .& .& 0& i\lambda (n-1)\sqrt{n}& n\mu & \end{array}\right).$$

  (11)

  where $\mu$, $\lambda$ are real parameters and $i^2=-1$.

In his consideration of the classical Carathéodory–Féjer problem in function theory, Takagi [3] observed the relevance of the anti-linear eigenvalue problem $TX=\lambda \overline{X}$, where T is an $n\times n$ symmetric complex matrix and X denotes complex conjugation of a vector X in ${\mathbb{C}}^n$. He noted that this equation implies that ${\displaystyle T^{*}TX={\left|\lambda \right|}^{2}X}$ and hence that $\left|\lambda \right|$ is an eigenvalue of ${\displaystyle \left|T\right|=\sqrt{T^{*}T}}$.

According this observation, Garcia and Putnar give in [4] the following definition:

Definition 1

(Complexe-Symmetric Operators). A bounded linear operator T on a complex Hilbert space $\mathcal{H}$ is called complex symmetric if $T=C{T}^{*}C$, where C is a conjugation (an isometric, anti-linear involution) of $\mathcal{H}$.

Remark 2.

(i) 

${\displaystyle {\mathbb{H}}_n^{\mu ,\lambda }\ne \overline{{}^t\mathbb{H}_n^{\mu ,\lambda }}}$ (i.e., ${\displaystyle {\mathbb{H}}_n^{\mu ,\lambda }}$ is non-self adjoint).

(ii) 

These $\mathbb{C}$-symmetric matrices ${\mathbb{H}}_n^{\mu ,\lambda }$ approximate our unbounded operator $H_{\mu ,\lambda }$ in the sense that $[{\mathbb{H}}_n^{\mu ,\lambda }{]}^{-1}$ approximate $H_{\mu ,\lambda }^{-1}$ (see [2], Theorem 2.1 page 345).

(iii) 

It is well known that for $\mu \ne 0$, $H_{\mu ,\lambda }$ is invertible and has compact resolvent because its maximal domain $D\left(H_{\mu ,\lambda }\right)$ is compactly embedded in Bargmann space $\mathbb{B}$ (see [2]). Now for $\mu =0$, $H_{\lambda }:=H_{0,\lambda }$ is not inversible because its spectrum $\sigma \left(H_{\lambda }\right)=\mathbb{C}$ [5]. Hence it is natural to ask whether $H_{\lambda }$ admits a left inverse or a right inverse, and if either of them exists, to see whether it is unbounded or bounded, and even better, whether it is compact.

• It is well known for several years that the eigenvalues of this operator $H_{\mu ,\lambda };\mu \ne 0$ are real and that, recently in 2023, we have shown the completeness of its generalized eigenvectors (for this topic see [6]).

Let

$${\displaystyle {\mathbb{B}}_0=\{\phi \in \mathbb{B};\phi \left(0\right)=0\}}$$

(12)

Then, it is well known that on ${\displaystyle {\mathbb{B}}_0}$, for $\psi \in {\mathbb{B}}_0$, an explicit inverse of $H_{\mu ,\lambda }$ restricted on imaginary axis; $y\in [0,+\infty [$ is given (see [7], Proposition 9) by

$${\displaystyle K_{\mu ,\lambda }:=H_{\mu ,\lambda }^{-1}\psi (-iy)={\int }_0^{\infty }{\mathcal{N}}_{\mu ,\lambda }(y,s)\psi (-is)ds}$$

(13)

where

$${\displaystyle {\mathcal{N}}_{\mu ,\lambda }(y,s)=\frac{1}{\lambda s}{e}^{-\frac{s^2}{2}-\frac{\mu }{\lambda }s}{\int }_0^{min(y,s)}{e}^{\frac{u^2}{2}+\frac{\mu }{\lambda }u}du}$$

(14)

By taking $\mu =0$ and $\lambda \ne 0$, we obtain the following:

$${\displaystyle K_{\lambda }:=H_{\lambda }^{-1}\psi (-iy)={\int }_0^{\infty }{\mathcal{N}}_{\lambda }(y,s)\psi (-is)ds}$$

(15)

where

$${\displaystyle {\mathcal{N}}_{\lambda }(y,s)=\frac{1}{\lambda s}{e}^{-\frac{s^2}{2}}{\int }_0^{min(y,s)}{e}^{\frac{u^2}{2}}du}$$

(16)

or by taking $\lambda =1$ and $\mu \ne 0$ (see Lemme 4 in [8])

$${\displaystyle K_{\mu }:=H_{\mu ,1}^{-1}\psi (-iy)={\int }_0^{\infty }{\mathcal{N}}_{\mu }(y,s)\psi (-is)ds}$$

(17)

where

$${\displaystyle {\mathcal{N}}_{\mu }(y,s)=\frac{1}{s}{e}^{-\frac{s^2}{2}-\mu s}{\int }_0^{min(y,s)}{e}^{\frac{u^2}{2}+\mu u}du}$$

(18)

The kernel of operator ${\displaystyle H_{\mu ,1}^{-1}}$ is analytic with respect to $\mu$. The integral operator defined by this kernel extends into a compact operator on a space $L_2$ with weight, including for negative values of $\mu$.

In particular, we have the following.

Proposition 1

(See Proposition 9 in [7,8] or Lemmas 4 and 8 in [8]).

Let ${\displaystyle L_2\left(\right[0,\infty [,e^{-x^2-2\frac{\mu }{\lambda }x}dx)}$ with $\lambda \ne 0$ be a space of square integrable functions with respect the measure ${\displaystyle e^{-x^2-2\frac{\mu }{\lambda }x}dx)}$, then we have:

(i) 

For all $\mu >0$, ${\displaystyle H_{\mu ,\lambda }^{-1}}$ can be extended to a Hilbert–Shmidt operator of ${\displaystyle L_2\left(\right[0,\infty [,e^{-x^2-2\frac{\mu }{\lambda }x}dx)}$ to itself.

(ii) 

The map $\mu ⟶K_{\mu ,\lambda }$ is analytic on $[0,+\infty [$ in Hilbert–Schmidt norm operators on $L_2\left(\right[0,\infty [,e^{-x^2-2{\displaystyle \frac{\mu }{\lambda }}x}dx)$.

(iii) 

For $\mu >0$, the smallest eigenvalue ${\sigma }_0\left(\mu \right)$ of $H_{\mu ,\lambda }$ which is simple can be extended to a real positive analytical function and is increasing with respect to μ on entire real axis.

Remark 3.

$H_{\mu ,\lambda }$ is non-normal operator; nevertheless, it has several properties analogous to those of the self-adjoint operators.

${•}_1$

In 1987, we have given in [7] (see also [2]) many non-trivial spectral properties of $H_{\mu ,\lambda }$ for $\mu >0$:

(i)

$\rho \left(H_{\mu ,\lambda }\right)\ne \varnothing$ where $\rho \left(H_{\mu ,\lambda }\right)$ is the resolvent set of $H_{\mu ,\lambda }$.

(ii)

$D_{min}\left(H_{\mu ,\lambda }\right)=D_{max}\left(H_{\mu ,\lambda }\right)$ where $D_{max}\left(H_{\mu ,\lambda }\right)=\{\phi \in \mathbb{B};H_{\mu ,\lambda }\phi \in \mathbb{B}\}$.

(iii)

$D_{min}\left(H_{\mu ,\lambda }\right)=\{\phi \in \mathbb{B};\exists p_n\in \mathcal{P},\exists \psi \in \mathbb{B};\underset{n⟶+\infty }{lim}{p}_n=\phi$ and $\underset{n⟶+\infty }{lim}{H}_{\mu ,\lambda }{p}_n=\psi \}$.

(iv)

For $\mu >0$ ${\displaystyle \left|\right|H_{\mu ,\lambda }\phi \left|\right|\ge \mu \left|\right|A\phi \left|\right|\forall \phi \in D_{max}\left(H_{\mu ,\lambda }\right)}$.

(v)

The injection of ${\displaystyle D_{max}\left(H_{\mu ,\lambda }\right)}$ in ${\displaystyle D\left(A\right)=\{\phi \in \mathbb{B};\frac{d}{dz}\phi \in \mathbb{B}\}}$ is continuous.

(vi)

The injections of $D\left(A\right)$ and of ${\displaystyle D_{max}\left(H_{\mu ,\lambda }\right)}$ in $\mathbb{B}$ are compact.

(vii)

The existence of the smallest eigenvalue ${\sigma }_0\ne 0$ and asymptotic expansion of semigroup ${\displaystyle e^{-t{H}_{\mu ,\lambda }}}$ and the elements of the diffusion matrix as $t⟶+\infty$.

More precisely, for $\mu >0$, ${\sigma }_0$ and ${\sigma }_1$, denote the smallest and the second eigenvalue of $H_{\mu ,\lambda }$, respectively, then we have the following:

$$\mid \mid e^{-t{H}_{\mu ,\lambda }}\mid \mid =e^{-t{\sigma }_0}+e^{-ϵt},\forall ϵ<{\sigma }_1$$

(19)

$$<e^{-t{H}_{\mu ,\lambda }}\varphi ,\psi >=<\varphi ,{\varphi }_0^{*}><{\varphi }_0,\psi >e^{-{\sigma }_{0}t}+O\left(e^{-{\sigma }_{1}t}\right),\forall \varphi ,\psi \in \mathbb{B}$$

(20)

where $H_{\mu ,\lambda }{\varphi }_0={\sigma }_0{\varphi }_0$ and ${\varphi }_0^{*}$ is the eigenfunction of $H_{\mu ,\lambda }^{*}$ associated to ${\sigma }_0$ such that $<{\varphi }_0,{\varphi }_0^{*}>=1$.

Remark 4.

(i) 

The operator ${\displaystyle H_{\lambda }=i\lambda A^{*}(A+A^{*})A}$ with maximal domain ${\displaystyle D\left(H_{\lambda }\right)=\{\phi \in ;H_{\lambda }\phi \in \mathbb{B}\}}$ is formally anti-adjoint, i.e., it is equal to the opposite of its adjoint.

(ii) 

The maximal domain of $H_{\lambda }$ is not equal to its minimal domain.

(iii) 

For $\mu =0$ in $H_{\mu ,\lambda }$, the $\mathbb{C}$-symmetric matrices:

$${\mathbb{H}}_n^{\lambda }=\left(\begin{array}{cccccccc}0& i\lambda \sqrt{2}& 0& .& .& .& .& \\ i\lambda \sqrt{2}& 0& i\lambda 2\sqrt{3}& \ddots & .& .& .\\ 0& i\lambda 2\sqrt{3}& 0& i\lambda 3\sqrt{4}& \ddots & .& .\\ .& \ddots & \ddots & \ddots & \ddots & \ddots & .& \\ .& .& \ddots & \ddots & \ddots & \ddots & 0& \\ .& .& .& \ddots & \ddots & \ddots & i\lambda (n-1)\sqrt{n}& \\ .& .& .& .& 0& i\lambda (n-1)\sqrt{n}& 0& \end{array}\right).$$

(21)

approximate $H_{\lambda }$:

$${\displaystyle H_{\lambda }=i\lambda A^{*}(A+A^{*})A=i\lambda [z\frac{d^2}{d{z}^2}+z^2\frac{d}{dz}]}$$

(22)

where λ is real parameter and $i^2=-1$.

(iv) 

${\mathbb{H}}_n^{\lambda }$ is equivalent to

$${\mathbb{H}}_n^{\lambda }=\left(\begin{array}{cccccccc}0& -2{\lambda }^2& 0& .& .& .& .& \\ 1& 0& -2^2.3{\lambda }^2& \ddots & .& .& .\\ 0& 1& 0& -3^2.4{\lambda }^2& \ddots & .& .\\ .& \ddots & \ddots & \ddots & \ddots & \ddots & .& \\ .& .& \ddots & \ddots & \ddots & \ddots & 0& \\ .& .& .& \ddots & \ddots & \ddots & -n{(n-1)}^2{\lambda }^2& \\ .& .& .& .& 0& 1& 0& \end{array}\right).$$

(23)

then ${\mathbb{H}}_n^{\lambda }$ is not symmetric but is $\mathbb{C}$-symmetric.

Remark 5.

Let ${\displaystyle {\alpha }_n=1,n=1,2,\dots }$, ${\displaystyle {\beta }_n=0; n=1,2,\dots \mathit{and} }$${\displaystyle |{\gamma }_n{|=n(n-1)}^2{\lambda }^2;n=2,3,\dots }$, then we have

(i) 

${\displaystyle \underset{n⟶\infty }{lim}\left|{\gamma }_n\right|=\infty }$ and ${\displaystyle |{\gamma }_n|\sim n^3}$

(ii) 

${\displaystyle \sum _{n=1}^{\infty }\frac{1}{|{\gamma }_n|}<\infty }$

In the sequel, we will consider the following integral operator:

$${\displaystyle K_{\lambda }\psi (-iy)={\int }_0^{\infty }{\mathcal{N}}_{\lambda }(y,s)\psi (-is)ds}$$

(24)

where

$${\displaystyle {\mathcal{N}}_{\lambda }(y,s)=\frac{1}{\lambda s}{e}^{-\frac{s^2}{2}}{\int }_0^{min(y,s)}{e}^{\frac{u^2}{2}}du}$$

(25)

The cubic term $H_{\lambda }$ parameterized by $\lambda$ is analogous to the Lindblad operators for describing non-unitarity in open quantum systems.

${•}_2$

In 1998, we have given in [9] the boundary conditions at infinity for a description of all maximal dissipative extensions in Bargmann space of the minimal Heun’s operator $H_I=z{\displaystyle \frac{d^2}{d{z}^2}}+z^2{\displaystyle \frac{d}{dz}}$. The characteristic functions of the dissipative extensions were computed and some completeness theorems were obtained for the system of generalized eigenvectors. It is well known that the restriction $H_I^{min}$ of the closure of $H_I$ on the polynomials set $\mathcal{P}$ is symmetric. But the minimal domain $D\left(H_I^{min}\right)=\{\phi \in \mathbb{B},\exists p_n\in \mathcal{P},p_n⟶\phi ,\exists \psi \in \mathbb{B};H_I{p}_n⟶\psi \}$ of $H_I$ is different of its maximal domain $D\left(H_I\right)=\{\phi \in \mathbb{B};H_I\phi \in \mathbb{B}\}$.

${•}_3$

It is also well known that ${\displaystyle H_I}$ is chaotic operator in Devaney’s sense [10] (see reference [5]. In particular its spectrum is $\sigma \left(H_I\right)=\mathbb{C}$. It follows that its action on the standard orthonormal basis ${\displaystyle {\left\{e_n\right\}}_{n\ge 0}}$ in the Bargmann $\mathbb{B}$ is given by

$${\displaystyle H_I^{min}{e}_n=b_n{e}_{n+1}+a_n{e}_n+b_{n-1}{e}_{n-1};a_n=0,b_n=n\sqrt{n+1}>0,n\ge 1}$$

(26)

Then $H_I^{min}$ can be represented in ${\displaystyle {\mathbb{B}}_0=\{\phi \in \mathbb{B};\phi \left(0\right)=0\}}$ by an infinite tridiagonal matrix

$${\mathcal{H}}_I=\left(\begin{array}{ccccccc}0& \sqrt{2}& 0& 0& \dots & \dots & \\ \sqrt{2}& 0& 2\sqrt{3}& 0& \dots & \dots & \\ 0& 2\sqrt{3}& 0& 3\sqrt{4}& \dots & \dots & \\ 0& 0& 3\sqrt{4}& 0& \ddots & \dots & \\ ⋮& ⋮& ⋮& \ddots & \ddots & \ddots & \\ ⋮& ⋮& ⋮& ⋮& \ddots & \ddots & \end{array}\right).$$

(27)

known as the Jacobi–Gribov matrix.

${•}_4$

In 2025, we studied in [11] the deficiency numbers of the generalized operator ${\displaystyle H^{p,m}=A^{{*}^p}(A^m+A^{{*}^m})A^p;p,m=1,2,\dots acting}$ on Bargmann space where ($H^{1,1}=H_I$) and we gave some conditions on the parameters p and m such that $H^{p,m}$ must be completely indeterminate. It follows from these conditions that $H^{p,m}$ is entirely of minimal type and is connected to chaotic operator.

We now describe briefly the contents of this paper, section by section.

In Section 2, according to Askey and Wilson in [12] on some hypergeometric orthogonal polynomials, we give some spectral properties associated with $H_I$; in Section 3, we give some properties of action of $H_{\lambda }$ on the Bargmann basis ${\tilde{e}}_n\left(z\right)={\displaystyle \frac{{\left(iz\right)}^n}{\sqrt{n!}}}$ where $z=-iy$ with $y>0$ and in Section 4, we show that $H_{\lambda }$ has a right-inverse $K_{\lambda }$ on negative imaginary axis and that $K_{\lambda }$ is compact.

2. Some Spectral Properties Associated to ${\mathit{H}}_{\mathit{I}}$ in the Askey–Wilson Sense

Let $H_I=A^{*}(A+A^{*})A$ with domain $D\left(H_I\right)=\{\phi \in {\mathbb{B}}_0;H_I\phi \in {\mathbb{B}}_0\}$.

Now, we denote $H_{I_{{|}_{\mathcal{P}}}}$ the above operator if we limit its domain to polynomials ${\mathcal{P}}_0$ where ${\mathcal{P}}_0=\{p\in \mathcal{P};p\left(0\right)=0\}$ and we denote the closure of this restriction by $H_I^{min}$.

Hence $H_I$ is obviously an extension to this closure.

It is well known that the minimal domain of $H_I$ is different of its maximal domain contrary to minimal and maximal domains of $H_{\mu ,\lambda }$ which coincide for $\mu \ne 0$.

According to Askey and Wilson [12] on some hypergeometric orthogonal polynomials, we deduce that the above Jacobi matrix associated with $H_I$ is related to the set of polynomials $P_n\left(x\right)$ of degree n satisfying the recurrent relation,

$$b_n{P}_{n+1}\left(x\right)+a_n{P}_n\left(x\right)+b_{n-1}{P}_{n-1}\left(x\right)=x{P}_n\left(x\right),n\ge 2$$

(28)

with the following initial conditions:

$$P_0\left(x\right)=0P_1\left(x\right)=1$$

(29)

where $a_n=0,\mathrm{and}{b}_n=n\sqrt{n+1}$.

Definition 2.

The polynomials $P_n\left(x\right)$, which solve to above recurrence relation and subject to the initial conditions $P_1\left(x\right)=1\mathrm{and}{P}_0\left(x\right)=0$, are called polynomials of the first kind.

The polynomials $Q_n\left(x\right)$, which satisfy the above recurrence relation and are subject to the initial conditions $Q_1\left(x\right)=1Q_2\left(x\right)={\displaystyle \frac{1}{\sqrt{2}}}$, are called polynomials of the second kind.

Remark 6

(discrete Wronskian).

Let $b_n{u}_{n+1}+b_{n-1}{u}_{n-1}=\sigma u_n,n\ge 2$ the spectral equation for $H_I$. Then for every two solutions $u=\left(u_n\right)$ and $v=\left(v_n\right)$ of above equation with the same parameter σ, the discrete Wronskian is defined by

$$W(u,v)=b_n(u_n{v}_{n+1}-u_{n+1}{v}_n)$$

(30)

which is independent of n.

For the polynomials $\left(P_n\left(x\right)\right)$ and $\left(Q_n\left(x\right)\right)$, the wronskian is equal to one for every x. It follows that

$$P_n{Q}_{n+1}-P_{n+1}{Q}_n={\displaystyle \frac{1}{b_n}}$$

(31)

Definition 3

(Borzov [13]).

A polynomial set ${\left\{{\psi }_n\left(x\right)\right\}}_{n=1}^{\infty }$ is called a canonical polynomial system if it is defined by the following recurrence relations:

$$c_{n-1}{\psi }_{n-1}\left(x\right)+c_n{\psi }_{n+1}=x{\psi }_n\left(x\right)n\ge 1,c_0=0$$

(32)

$${\psi }_1\left(x\right)=1$$

(33)

where the positive sequence ${\displaystyle {\left\{c_n\right\}}_{n=1}^{\infty }}$ is given.

Lemma 2.

(i) 

The polynomial set ${\displaystyle {\left\{P_n\left(x\right)\right\}}_{n=1}^{\infty }}$ is a canonical polynomial system.

(ii) 

The polynomials $P_n\left(x\right)$ have real coefficients and fulfill the following parity conditions

$$P_n(-x)={(-1)}^{n-1}{P}_n\left(x\right)$$

(34)

(iii) 

$${\displaystyle b_{n-1}{b}_{n+1}\le b_n^2}$$

(35)

(iv) 

$${\displaystyle \sum _{n=1}^{\infty }\frac{1}{b_n}<\infty }$$

(36)

(v) 

The operator $H_I$ has the deficiency indices $(1,1)$.

Proof. 

(i)

As the sequence $b_n$ is positive, then the proof of this property is trivial by taking $c_n=b_n$.

(ii)

As the coefficients of $P_1\left(x\right)=1$ and the coefficients of $P_2\left(x\right)={\displaystyle \frac{x}{b_1}}$ are real, then we deduce by recurrence that the polynomials $P_n\left(x\right)$ have real coefficients. Now as $P_2(-x)=-{\displaystyle \frac{x}{b_1}}=(-1)P_2\left(x\right)$, then if we suppose that $P_n(-x)={(-1)}^{n-1}{P}_n\left(x\right)$ and $P_{n-1}(-x)={(-1)}^{n-2}{P}_{n-1}\left(x\right)$, then from the recurrence relation

$${\displaystyle b_n{P}_{n+1}(-x)=-x{P}_n(-x)-b_{n-1}{P}_{n-1}(-x)}$$

we deduce that

$${\displaystyle b_n{P}_{n+1}(-x)=-x{(-1)}^{n-1}{P}_n\left(x\right)-b_{n-1}{(-1)}^{n-2}{P}_{n-1}\left(x\right)}$$

and

$${\displaystyle b_n{P}_{n+1}(-x)={(-1)}^n[x{P}_n\left(x\right)-b_{n-1}{P}_{n-1}\left(x\right)]={(-1)}^n{b}_n{P}_{n+1}\left(x\right)}$$

It follows that

$${\displaystyle P_{n+1}(-x)={(-1)}^n{P}_{n+1}\left(x\right).}$$

(iii)

Let $b_n=n\sqrt{n+1}$ then $b_{n-1}=(n-1)\sqrt{n}$ and $b_{n+1}=(n+1)\sqrt{n+2}$. It follows that $b_{n-1}{b}_{n+1}=(n-1)(n+1)\sqrt{n(n+2)}$ and $b_n^2=n^2(n+1)$. This implies that

$${\displaystyle b_{n-1}{b}_{n+1}\le b_n^2\iff {(n-1)}^2\le n^3\iff 3n\ge 2.}$$

As $3n\ge 2$ holds for all $n\ge 1$, we deduce that ${\displaystyle b_{n-1}{b}_{n+1}\le b_n^2}$.

(iv)

It is well known that the following series ${\displaystyle \zeta \left(\alpha \right)=\sum _{n=1}^{\infty }\frac{1}{n^{\alpha }}}$ converges for $\alpha >1$, where $\alpha$ is real, then as ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{b_n}=\sum _{n=1}^{\infty }\frac{1}{n\sqrt{n+1}}\equiv \sum _{n=1}^{\infty }\frac{1}{n^{\frac{3}{2}}}}$, it follows that ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{b_n}<\infty }$.

(v)

By using theorem 1.5, Ch.VII [14].

For any operator T, we denote by $\sigma \left(T\right)$, ${\sigma }_{ess}\left(T\right)$ and ${\sigma }_p\left(T\right)$ the spectrum, the essential spectrum and the point spectrum of T, respectively. □

Proposition 2.

Let $H_I=z{\displaystyle \frac{d^2}{d{z}^2}}+z^2{\displaystyle \frac{d}{dz}}$ acting on Bargmann space ${\mathbb{B}}_0$. For all ${\displaystyle \phi \left(z\right)=\sum _{n=1}^n{\phi }_n\frac{z^n}{\sqrt{n!}}}$, $H_I$ can be defined by

$$\left\{\begin{array}{c}{\displaystyle \left[H_I\phi \right]\left(z\right)=\sum _{n=1}^{\infty }{\left(H_I\right)}_n\frac{z^n}{\sqrt{n!}}}\\ \\ \mathrm{where}\\ \\ {\left(H_I\phi \right)}_1=\sqrt{2}{\phi }_2\\ \\ {\left(H_I\phi \right)}_n=(n-1)\sqrt{n}{\phi }_{n-1}+n\sqrt{n+1}{\phi }_{n+1}\end{array}\right.$$

(37)

Then the point spectrum of $H_I$ is $\mathbb{C}$.

Direct Proof. 

This direct proof is based on the following classical proposition: □

Proposition 3.

The Bertrand series ${\displaystyle \sum _{n>1}{u}_n}$ with general term $u_n={\displaystyle \frac{1}{nln{\left(n\right)}^{\beta }}}$ converges if and only if $\beta >1$.

Now, let us consider the sequence ${\displaystyle \left(u_n\left(\xi \right)\right),n=1,2,\dots ;\xi \in \mathbb{C};}$ defined by the recurrence relation:

$$\left\{\begin{array}{c}{\displaystyle u_1\left(\xi \right)=1}\\ \\ {\displaystyle u_2\left(\xi \right)=\frac{\xi }{\sqrt{2}}}\\ \\ {\displaystyle (n-1)\sqrt{n}{u}_{n-1}\left(\xi \right)+n\sqrt{n+1}{u}_{n+1}\left(\xi \right)=\xi u_n\left(\xi \right)n\ge 2}\end{array}\right.$$

(38)

and let ${\displaystyle {\phi }_{\xi }\left(z\right)=\sum _{n=1}^{\infty }{u}_n\left(\xi \right)\frac{z^n}{\sqrt{n!}}}$, where ${\displaystyle u_n\left(\xi \right)}$ is defined by (38).

It is clear that ${\displaystyle H_I{\phi }_{\xi }=\xi {\phi }_{\xi }}$.

${\phi }_{\xi }$ belongs to ${\mathbb{B}}_0$ we must prove that for all $\xi \in \mathbb{C}$ we have $\left(u_n\left(\xi \right)\right)\in {\mathsf{\ell }}_2\left(\mathbb{N}\right)$. For this, we will show that

$${\displaystyle \exists n_0\mathrm{such}\mathrm{that}\forall n\ge n_0\mathrm{we}\mathrm{have}|u_n\left(\xi \right)|\le \frac{M}{\sqrt{n}ln\left(n\right)}}$$

(39)

where M is a constant independent of n.

To determine $n_0$, we use the below relation when $n⟶\infty$

$${\displaystyle \frac{1}{2n^{\frac{3}{2}}ln\left(n\right)}}\sim {\displaystyle \frac{1}{\sqrt{n+1}ln(n+1)}}-{\displaystyle \frac{\sqrt{n-1}}{\sqrt{n(n-1)}ln(n-1)}}-{\displaystyle \frac{\left|\xi \right|}{n\sqrt{n(n+1)}ln\left(n\right)}}$$

(40)

which proves that the second part of the above equality is positive for $n\ge n_0$.

Now, we set ${\displaystyle M=max\left\{\right|u_{n_0}|ln\left(n_0\right),|u_{n_0-1}\left|ln(n_0-1)\right\}}$ and we deduce (39) by recurrence. We use the following inequalities

$$|u_{n+1}|\le \left|\xi \right|{\displaystyle \frac{|u_n|}{n\sqrt{n+1}}}+{\displaystyle \frac{(n-1)\sqrt{n}\left|u_{n-1}\right|}{n\sqrt{n+1}}}$$

and

$${\displaystyle \frac{(n-1)\sqrt{n}}{n\sqrt{(n+1)(n-1)}ln(n-1)}}+{\displaystyle \frac{\left|\xi \right|}{n\sqrt{n(n+1)}ln\left(n\right)}}\le {\displaystyle \frac{1}{\sqrt{(n+1)}ln(n+1)}}$$

It follows that

$$|u_n{\left(\xi \right)|}^2\le {\displaystyle \frac{M^2}{nl{n}^2\left(n\right)}}\mathrm{and}\sum _{n=1}^{\infty }{\left|u_n\left(\xi \right)\right|}^2\le M^2\sum _{n=1}^{\infty }{\displaystyle \frac{1}{nl{n}^2\left(n\right)}}<\infty$$

This implies that $\left(u_n\left(\xi \right)\right)\in {\mathsf{\ell }}_2\left(\mathbb{N}\right)$ for all $\xi \in \mathbb{C}$.

Second Proof. 

The second proof is based on the following classical lemma: □

Lemma 3

(Raabe-Duhamel test).

We suppose ${\displaystyle \forall n>0a_n>0}$.

If

$$\exists \alpha \in \mathbb{R},{\displaystyle \frac{a_{n+1}}{a_n}}=1-{\displaystyle \frac{\alpha }{n}}+o({\displaystyle \frac{1}{n}}),$$

then

(i) 

${\displaystyle \alpha >1⟹\sum a_n}$ converges

(ii) 

${\displaystyle \alpha <1⟹\sum a_n}$ diverges

Same conclusions if

$$\exists \alpha \in \mathbb{R},{\displaystyle \frac{a_{n+1}}{a_n}}=1-{\displaystyle \frac{1}{n}}-{\displaystyle \frac{\alpha }{nln\left(n\right)}}+o({\displaystyle \frac{1}{nln\left(n\right)}}),$$

and we apply Theorem 2.3 of the reference [11].

3. Some Properties of Integral Operator ${\mathit{K}}_{\mathit{\lambda }}$ Action on the Bargmann Basis ${\tilde{\mathit{e}}}_{\mathit{n}}\mathbf{\left(}\mathit{z}\mathbf{\right)}\mathbf{=}{\displaystyle \frac{{\mathbf{\left(}\mathit{iz}\mathbf{\right)}}^{\mathit{n}}}{\sqrt{\mathit{n}\mathbf{!}}}}$ Where $\mathit{z}\mathbf{=}\mathbf{-}\mathit{iy}$ with $\mathit{y}\mathbf{>}\mathit{0}$

We recall that on ${\mathbb{B}}_0=\{\phi \in \mathbb{B};\phi \left(0\right)=0\}$, it was well known that an explicit inverse of

$${\displaystyle H_{\mu ,\lambda }=i\lambda z\frac{d^2}{d{z}^2}+(i\lambda z^2+\mu z)\frac{d}{dz}}$$

(41)

restricted on imaginary axis, $y\in [0,+\infty [$ is given by

$$K_{\mu ,\lambda }\psi (-iy):=H_{\mu ,\lambda }^{-1}\psi (-iy)={\int }_0^{\infty }{\mathcal{N}}_{\mu ,\lambda }(y,s)\psi (-is)ds$$

where

$${\displaystyle {\mathcal{N}}_{\mu ,\lambda }(y,s)=\frac{1}{\lambda s}{e}^{-\frac{s^2}{2}-\frac{\mu }{\lambda }s}{\int }_0^{min(y,s)}{e}^{\frac{u^2}{2}+\frac{\mu }{\lambda }u}du.}$$

It follows that

$${\displaystyle K_{\mu ,\lambda }\psi (-iy):=H_{\mu ,\lambda }^{-1}\psi (-iy)={\int }_0^y{e}^{\frac{u^2}{2}+\frac{\mu }{\lambda }u}du{\int }_u^{\infty }\frac{1}{\lambda s}{e}^{-\frac{s^2}{2}-\frac{\mu }{\lambda }s}ds}$$

(42)

and for $\mu >0$, let $\sigma \left(\mu \right)$ be the smallest eigenvalue of the operator $H_{\mu ,\lambda }$, then it is well known in [8] that $\sigma \left(\mu \right)$ extends to a positive, increasing, analytic function on the whole real line and that the limit value $\sigma \left(0\right)$ is an eigenvalue of ${\displaystyle H_{\lambda }=i\lambda A^{*}(A+A^{*})A}$ in particular, $\sigma \left(0\right)\ne 0$.

Despite the difficulty of the absence of any relation between the domains of the self-adjoint and anti-adjoint parts of $H_{\mu ,\lambda }$, this operator has a fine spectral property in the Bargmann representation: if we restrict to an imaginary semi-axis, its inverse is an integral operator with a positive kernel, which allows us to apply the Krein–Rutman theorem [15] and the Jentzsch theorem [16].

For $\mu =0$, let

$${\displaystyle H_{\lambda }\phi \left(z\right)=i\lambda z(\frac{d}{dz}+z)\frac{d}{dz}\phi \left(z\right)}$$

(43)

then for ${\displaystyle z=-iy;y\in [0,\infty [\mathrm{and}\phi (-iy)=u\left(y\right)}$, we have the following:

$${\displaystyle H_{\lambda }u\left(y\right)=\lambda [-y{u}^{″}\left(y\right)+y^2{u}^{\prime }\left(y\right)]}$$

(44)

where $u^{\prime }\left(y\right)$ denotes the first derivative of $u\left(y\right)$, and $u^{″}\left(y\right)$ is its second derivative. And

$$K_{\lambda }\psi (-iy)={\displaystyle \frac{1}{\lambda }}{\int }_0^y{e}^{{\displaystyle \frac{u^2}{2}}}du{\int }_u^{\infty }{\displaystyle \frac{1}{s}}{e}^{-{\displaystyle \frac{s^2}{2}}}\psi (-is)ds={\displaystyle \frac{1}{\lambda }}{\int }_0^{y}du{\int }_u^{\infty }{\displaystyle \frac{1}{s}}{e}^{-{\displaystyle \frac{1}{2}}(s^2-u^2)}\psi (-is)ds$$

$$={\displaystyle \frac{1}{\lambda }}[-{\int }_0^{y}du{\int }_0^u{\displaystyle \frac{1}{s}}{e}^{-{\displaystyle \frac{1}{2}}(s^2-u^2)}\psi (-is)ds+{\int }_0^{y}du{\int }_0^{\infty }{\displaystyle \frac{1}{s}}{e}^{-{\displaystyle \frac{1}{2}}(s^2-u^2)}\psi (-is)ds$$

Remark 7.

${•}_1$

It is well known that $K_{\mu ,\lambda }$ is compact if $\mu >0$ see [8] or [7]. It belongs same to the Carleman-class ${\mathsf{\ell }}_{1+ϵ}\forall ϵ>0$, for this topic see the reference [17]. A compact operator K in the Hilbert space $\mathcal{H}$ is said to belong to the Carleman-class ${\mathsf{\ell }}_p,p>0$ if ${\displaystyle \sum _{n=1}^{\infty }{\left[s_n(\sqrt{K^{*}K})\right]}^p}$ converges where $s_n(\sqrt{K^{*}K})$ are the eigenvalues of compact operator $\sqrt{K^{*}K}$.

${•}_2$

It is well known that for $\mu =0$, $H_{\lambda }$ satisfies some properties which are to far form the properties of $H_{\mu ,\lambda };\mu \ne 0$:

(i) 

$\rho \left(H_{\lambda }\right)=\varnothing$ where $\rho \left(H_{\lambda }\right)$ is the resolvent set of $H_{\lambda }$.

(ii) 

$D_{min}\left(H_{\lambda }\right)\ne D_{max}\left(H_{\lambda }\right)$.

(iii) 

The spectrum of $H_{\lambda }$ is $\mathbb{C}$.

${•}_3$

If $e_n\left(z\right)={\displaystyle \frac{z^n}{\sqrt{n!}}};z=x+iy$ is the usual basis of Bargmann space then ${\tilde{e}}_n\left(z\right)={\displaystyle \frac{{\left(iz\right)}^n}{\sqrt{n!}}};z=x+iy$ is also an orthonormal basis of Bargmann space.

Now, if we restrict to an imaginary semi-axis, for example, $z=-iy,y\in [0,+\infty [$, we deduce that $e_n(-iy)={\displaystyle \frac{{(-iy)}^n}{\sqrt{n!}}}$ and ${\tilde{e}}_n(-iy)={\displaystyle \frac{{\left(i(-iy)\right)}^n}{\sqrt{n!}}}={\displaystyle \frac{y^n}{\sqrt{n!}}}:=u_n\left(y\right)$.

Lemma 4

(Action of $K_{\lambda }$ on the basis ${\tilde{e}}_n(-iy)$). Let ${\tilde{e}}_n(-iy)=u_n\left(y\right)={\displaystyle \frac{y^n}{\sqrt{n!}}}$ then

$$K_{\lambda }{u}_1:=v_1\left(y\right)={\displaystyle \frac{1}{\lambda }}{\int }_0^y{e}^{{\displaystyle \frac{u^2}{2}}}du{\int }_u^{\infty }{e}^{-{\displaystyle \frac{s^2}{2}}}ds$$

(45)

and

$$K_{\lambda }{u}_{n+1}={\displaystyle \frac{n-1}{\sqrt{n(n+1)}}}{K}_{\lambda }{u}_{n-1}+{\displaystyle \frac{u_n}{\lambda n\sqrt{n+1}}},n\ge 1$$

(46)

i.e.,

$$v_{n+1}={\displaystyle \frac{n-1}{\sqrt{n(n+1)}}}{v}_{n-1}+{\displaystyle \frac{u_n}{\lambda n\sqrt{n+1}}}$$

(47)

Proof. 

Let $v_{n+1}\left(y\right)=K_{\lambda }{u}_{n+1}={\displaystyle \frac{1}{\lambda }}{\int }_0^y{e}^{{\displaystyle \frac{u^2}{2}}}du{\int }_u^{\infty }{\displaystyle \frac{1}{s}}{e}^{-{\displaystyle \frac{s^2}{2}}}{\displaystyle \frac{s^{n+1}}{\sqrt{(n+1)!}}}ds$ $={\displaystyle \frac{1}{\lambda }}{\int }_0^y{e}^{{\displaystyle \frac{u^2}{2}}}du{\int }_u^{\infty }{e}^{-{\displaystyle \frac{s^2}{2}}}$${\displaystyle \frac{s^n}{\sqrt{(n+1)!}}}ds$$={\displaystyle \frac{1}{\lambda }}{\int }_0^y{e}^{{\displaystyle \frac{u^2}{2}}}du{\int }_u^{\infty }s{e}^{-{\displaystyle \frac{s^2}{2}}}{\displaystyle \frac{s^{n-1}}{\sqrt{(n+1)!}}}.$

Setting $v^{\prime }\left(s\right)=s{e}^{-{\displaystyle \frac{s^2}{2}}}$ and ${\displaystyle u\left(s\right)=s^{n-1}}$, this implies that $v\left(s\right)=-e^{-{\displaystyle \frac{s^2}{2}}}$ and $u^{\prime }\left(s\right)=(n-1)s^{n-2}$.

Now we use an integration by part to obtain:

$$\begin{array}{c}{\int }_u^{\infty }{e}^{-{\displaystyle \frac{s^2}{2}}}{s}^{n}ds={\left[-,e^{-{\displaystyle \frac{s^2}{2}}},s^{n-1}\right]}_u^{\infty }+(n-1){\int }_u^{\infty }{e}^{-{\displaystyle \frac{s^2}{2}}}{s}^{n-2}ds.\hfill \\ {\displaystyle =e^{-\frac{u^2}{2}}{u}^{n-1}+(n-1){\int }_u^{\infty }{e}^{-\frac{s^2}{2}}{s}^{n-2}ds.}\hfill \end{array}$$

It follows that $v_{n+1}\left(y\right)={\displaystyle \frac{1}{\lambda \sqrt{(n+1)!}}}[{\int }_0^y{e}^{{\displaystyle \frac{u^2}{2}}}{e}^{-{\displaystyle \frac{u^2}{2}}}{u}^{n-1}du+(n-1){\int }_0^y{e}^{{\displaystyle \frac{u^2}{2}}}{\int }_u^{\infty }{e}^{-{\displaystyle \frac{s^2}{2}}}{s}^{n-2}ds]$.

$={\displaystyle \frac{1}{\lambda \sqrt{(n+1)!}}}{\displaystyle \frac{y^n}{n}}+{\displaystyle \frac{n-1}{\lambda \sqrt{(n+1)!}}}{\int }_u^{\infty }{e}^{-{\displaystyle \frac{s^2}{2}}}{s}^{n-2}ds$.

$={\displaystyle \frac{1}{\lambda \sqrt{(n+1)!}}}{\displaystyle \frac{\sqrt{n!}{u}_n}{n}}+{\displaystyle \frac{n-1}{\sqrt{n(n+1)}}}{\displaystyle \frac{1}{\lambda \sqrt{(n-1)!}}}{\int }_u^{\infty }{e}^{-{\displaystyle \frac{s^2}{2}}}{s}^{n-2}ds$.

$={\displaystyle \frac{u_n}{\lambda n\sqrt{n+1}}}+{\displaystyle \frac{n-1}{\sqrt{n(n+1)}}}{v}_{n-1}$, i.e.,

$$K_{\lambda }{u}_{n+1}\left(y\right)={\displaystyle \frac{u_n}{\lambda n\sqrt{n+1}}}+{\displaystyle \frac{n-1}{\sqrt{n(n+1)}}}{K}_{\lambda }{u}_{n-1}.$$

(48)

□

Definition 4 

(Right (or Left) invertibility of unbounded operators).

Let $\mathcal{H}$ be a Hilbert space and let T be a linear operator with domain $D\left(T\right)\subset \mathcal{H}$.

We say that T is right invertible if there exists an everywhere defined $R\in \mathcal{B}\left(\mathcal{H}\right)$ (the space of bounded operators on $\mathcal{H}$) such that $TR=I$; and we say that T is left invertible if there is an everywhere defined $L\in \mathcal{B}\left(\mathcal{H}\right)$ such that $LT\subset I$.

Remark 8

(Examples).

(i) 

Let $\mathcal{P}\left(\mathbb{C}\right)$ be the set of polynomials, and $A={\displaystyle \frac{d}{dz}}$. We would like to undo differentiation, so we integrate the following:

$$\mathbb{J}\left(p\right)={\int }_0^{z}p\left(\xi \right)d\xi .$$

The fundamental theorem of calculus says that the derivative of this integral is p; that is, $A\mathbb{J}=I_{\mathcal{P}\left(\mathbb{C}\right)}$. So $\mathbb{J}$ is a right inverse of A; it provides a solution, not the only one of the differential equation:

$${\displaystyle \frac{d}{dz}}q\left(z\right)=p\left(z\right)$$

If we try things in the other direction, there is a problem:

$$\mathbb{J}A\left(p\right)={\int }_0^z{p}^{\prime }\left(\xi \right)d\xi =p\left(z\right)-p\left(0\right).$$

That is, $\mathbb{J}A$ sends p to $p-p\left(0\right)$, which is not the same as p. So $\mathbb{J}$ is not a left inverse to A; since A has a nonzero null space, we will see that no left inverse can exist.

(ii) 

Consider the space E of real sequences, the linear mapping T that maps a sequence $(a_0,a_1,\dots )$ to the sequence $(0,a_0,a_1,\dots )$ and the linear mapping S that maps a sequence $(a_0,a_1,a_2,\dots )$ to the sequence $(a_1,a_2,\dots )$. It is clear that $ST=I$. Now consider the sequence $a=(1,0,0,0,\dots )$. We have $S\left(a\right)=0$ where 0 is the sequence that vanishes identically and also $TS\left(a\right)=0$ hence $TS\ne I$.

(iii) 

Consider the usual (unilateral) shift S on ${\mathsf{\ell }}_2$. Then $S^{*}S=I,S{S}^{*}\ne$ I and S is hyponormal.

Hence

(1) 

S is left invertible without being invertible.

(2) 

Also, $\sigma \left(S^{*}S\right)=\left\{1\right\}\ne \sigma \left(S{S}^{*}\right)=0,1.$ where $\sigma \left(T\right)$ denotes the spectrum of an operator T.

Remark 9.

$v_1\left(y\right)$ is a primitive of function defined by

$$\varphi \left(u\right)=e^{{\displaystyle \frac{u^2}{2}}}{\int }_u^{\infty }{e}^{-{\displaystyle \frac{s^2}{2}}}ds$$

(49)

The function $\varphi \left(u\right)$ has interesting properties (see [18]).

${•}_1$

For every non negative integer n, there exists an unique couple $(P_n,Q_n)$ of polynomials that satisfy

$$\forall u\in \mathbb{R},{\varphi }^{\left(n\right)}\left(u\right)=P_n\left(u\right)\varphi \left(u\right)-Q_n\left(u\right)$$

(50)

Moreover, these polynomials are defined, starting from $(P_0,Q_0)=(1,0)$, by the recurrence relations

$${\displaystyle \forall n\in \mathbb{N},P_{n+1}=x{P}_n+{P^{\prime }}_n}$$

(51)

$${\displaystyle \forall n\in \mathbb{N},Q_{n+1}=P_n+{Q^{\prime }}_n}$$

(52)

and by Lemma 1 and Proposition 3 of Kouba in [19], the sequence ${\displaystyle {\left(P_n\left(x\right)\right)}_{n\in \mathbb{N}}}$ satisfies:

$$\forall n\in \mathbb{N},P_n\left(x\right)=e^{-{\displaystyle \frac{x^2}{2}}}{\displaystyle \frac{d^n}{d{x}^n}}{e}^{{\displaystyle \frac{x^2}{2}}}$$

(53)

The polynomials ${\displaystyle P_n\left(x\right)}$ are linked to the well-known physicist’s Hermite polynomials by the following relation

$${\displaystyle P_n\left(x\right)=(-\frac{i}{\sqrt{2}})H_n\left(\frac{i}{\sqrt{2}}x\right)}$$

(54)

In [18], some similar properties of the generating function ${\displaystyle \phi \left(u\right)=e^{u^2}{\int }_u^{\infty }{e}^{-s^2}ds}$ are given to study the “Plasma dispersion function widely used in the field of plasma physics”.

In particular

${•}_2$

The n-th derivative of ${\displaystyle \phi \left(u\right)=e^{u^2}{\int }_u^{\infty }{e}^{-s^2}ds}$ is given by:

$${\displaystyle {\phi }^{\left(n\right)}\left(u\right)=P_n\left(u\right)\phi \left(u\right)-Q_n\left(u\right)}$$

(55)

where $P_n$ and $Q_n$ are given by the following recurrence relations:

$$\left\{\begin{array}{c}{P}_0\left(u\right)=1,P_1\left(u\right)=2x\\ Q_0\left(u\right)=0,Q_1\left(u\right)=1\\ P_{n+1}\left(u\right)=2x{P}_n\left(u\right)+2n{P}_{n-1}\left(u\right)\\ Q_{n+1}\left(u\right)=2x{Q}_n\left(u\right)+2n{Q}_{n-1}\left(u\right)\end{array}\right.$$

(56)

Furthermore we have the following identity

$$Q_{n+1}\left(u\right)P_n\left(u\right)-P_{n+1}\left(u\right)P_n\left(u\right)={(-2)}^{n}n!$$

(57)

and the following expression for $P_n\left(u\right)$:

$$P_n\left(u\right)=e^{-u^2}{\displaystyle \frac{d^n}{d{u}^n}}{e}^{u^2}$$

(58)

In following lemma, we give an explicit expression of $v_1\left(y\right)$ and we show that it belongs to Bargmann space.

Lemma 5

(An explicit expression of $v_1\left(y\right)$).

$$v_1\left(y\right)=\sum _{n=0}^{\infty }{(-1)}^n{a}_n{u}_{2n+1}-\sum _{n=0}^{\infty }{b}_n{u}_{2(n+1)}\left(y\right)$$

(59)

where ${\displaystyle a_n=\frac{\sqrt{\left(2n\right)!\frac{\pi }{2}}}{2^{n}n!\sqrt{2n+1}}\sim \frac{1}{2}{n}^{-\frac{3}{4}}{\pi }^{\frac{1}{4}}}$ and ${\displaystyle b_n=\frac{2^{n}n!}{\sqrt{2(n+1)(2n+1)!}}\sim \frac{1}{2}{n}^{-\frac{3}{4}}{\pi }^{\frac{1}{4}}}$

Proof. 

We begin by writing the following:

$${\displaystyle \begin{array}{l}{\displaystyle v_1\left(y\right):=K_{\lambda }{u}_1\left(y\right)=K_{\lambda }y=\frac{1}{\lambda }{\int }_0^y{e}^{\frac{u^2}{2}}du{\int }_u^{\infty }{e}^{-\frac{s^2}{2}}ds}\hfill \\ {\displaystyle =\frac{1}{\lambda }\{{\int }_0^y{e}^{\frac{u^2}{2}}du[{\int }_0^{\infty }{e}^{-\frac{s^2}{2}}ds-{\int }_0^u{e}^{-\frac{s^2}{2}}ds]\}}\\ {\displaystyle =-\frac{1}{\lambda }{\int }_0^{y}du{\int }_0^u{e}^{-\frac{1}{2}(s^2-u^2)}ds+\frac{1}{\lambda }{\int }_0^y{e}^{\frac{u^2}{2}}du{\int }_0^{\infty }{e}^{-\frac{s^2}{2}}ds}\\ {\displaystyle =f\left(y\right)+g\left(y\right)}\end{array}}$$

where ${\displaystyle f\left(y\right)=-\frac{1}{\lambda }{\int }_0^{y}du{\int }_0^u{e}^{-\frac{1}{2}(s^2-u^2)}ds}$ and ${\displaystyle g\left(y\right)=\frac{1}{\lambda }{\int }_0^y{e}^{\frac{u^2}{2}}du{\int }_0^{\infty }{e}^{-\frac{s^2}{2}}ds}$.

Let ${\displaystyle f\left(y\right)=--\frac{1}{\lambda }{\int }_0^{y}du{\int }_0^u{e}^{-\frac{1}{2}(s^2-u^2)}ds}$.

Setting ${\displaystyle t=\frac{s}{u}}$ then we have

$${\displaystyle \begin{array}{c}{\displaystyle {\int }_0^u{e}^{-\frac{1}{2}(s^2-u^2)}ds={\int }_0^1{e}^{-\frac{1}{2}(t^2-1)u^2}udt}\hfill \\ {\displaystyle ={\int }_0^1\sum _{n=0}^{\infty }\frac{{(-\frac{1}{2})}^n{(t^2-1)}^n{u}^{2n+1}}{n!}}\hfill \\ {\displaystyle =\sum _{n=0}^{\infty }\frac{u^{2n+1}}{2^{n}n!}{\int }_0^1{(1-t^2)}^{n}dt}\hfill \\ {\displaystyle =\sum _{n=0}^{\infty }\frac{u^{2n+1}}{2^{n}n!}{W}_{n+1}}\hfill \end{array}}$$

where ${\displaystyle W_{n+1}=\frac{2^{2n}{(n!)}^2}{(2n+1)!}}$ is Wallis formula).

It follows that

$${\displaystyle \begin{array}{c}{\displaystyle {\int }_0^u{e}^{-\frac{1}{2}(s^2-u^2)}ds=\sum _{n=0}^{\infty }\frac{u^{2n+1}}{2^{n}n!}\frac{2^{2n}{(n!)}^2}{(2n+1)!}}\hfill \\ {\displaystyle =\sum _{n=0}^{\infty }\frac{2^{n}n!\sqrt{(2n+1)!}{u}_{2n+1}\left(u\right)}{(2n+1)!}}\hfill \\ {\displaystyle =\sum _{n=0}^{\infty }\frac{2^{n}n!}{\sqrt{(2n+1)!}}{u}_{2n+1}(u)}\hfill \end{array}}$$

Now as ${\displaystyle {\int }_0^y{u}_{2n+1}\left(u\right)du=\frac{1}{\sqrt{2(n+1)}}{u}_{2(n+1)}\left(y\right)}$ then we deduce that

$${\displaystyle f\left(y\right)=-\frac{1}{\lambda }{\int }_0^{y}du{\int }_0^u{e}^{-\frac{1}{2}(s^2-u^2)}ds=-\frac{1}{\lambda }\sum _{n=0}^{\infty }\frac{2^{n}n!}{\sqrt{2(n+1)(2n+1)!}}{u}_{2(n+1)}\left(y\right)}$$

i.e.,

$${\displaystyle f\left(y\right)=-\frac{1}{\lambda }\sum _{n=0}^{\infty }{a}_n{u}_{2(n+1)}\mathrm{where}{a}_n=\frac{2^{n}n!}{\sqrt{2(n+1)(2n+1)!}}}$$

(60)

Now, we consider ${\displaystyle g\left(y\right)=\frac{1}{\lambda }{\int }_0^y{e}^{\frac{u^2}{2}}du{\int }_0^{\infty }{e}^{-\frac{s^2}{2}}ds}$, as ${\displaystyle {\int }_0^{\infty }{e}^{-\frac{s^2}{2}}ds=\sqrt{2\pi }}$, then we deduce that

$${\displaystyle g\left(y\right)=\frac{\sqrt{2\pi }}{\lambda }{\int }_0^y{e}^{\frac{u^2}{2}}du=\frac{\sqrt{2\pi }}{\lambda }{\int }_0^y\sum _{n=0}^{\infty }\frac{u^{2n}}{2^{n}n!}=\frac{\sqrt{2\pi }}{\lambda }\sum _{n=0}^{\infty }\frac{y^{2n+1}}{2^n(2n+1)n!}}$$

As ${\displaystyle y^n=\sqrt{n!}{u}_n\left(y\right)}$, then it follows that

$${\displaystyle \begin{array}{c}{\displaystyle g\left(y\right)=\frac{\sqrt{2\pi }}{\lambda }\sum _{n=0}^{\infty }\frac{\sqrt{(2n+1)!}{u}_{2n+1}\left(y\right)}{2^n(2n+1)n!}}\hfill \\ {\displaystyle =\frac{\sqrt{2\pi }}{\lambda }\sum _{n=0}^{\infty }\frac{\sqrt{\left(2n\right)!}{u}_{2n+1}\left(y\right)}{2^{n}n!\sqrt{2n+1}}}\hfill \end{array}}$$

i.e.,

$${\displaystyle g\left(y\right)=\frac{\sqrt{2\pi }}{\lambda }\sum _{n=0}^{\infty }{b}_n{u}_{2n+1}\left(y\right)\mathrm{where}{b}_n=\frac{\sqrt{\left(2n\right)!}}{2^{n}n!\sqrt{2n+1}}}$$

(61)

□

Corollary 1.

(α) 

The function $v_1\left(y\right)$ belongs to Bargmann space, i.e., ${\displaystyle \sum _{n=0}^{\infty }{\left|a_n\right|}^2<\infty }$ and ${\displaystyle \sum _{n=0}^{\infty }{\left|b_n\right|}^2<\infty }$

(β) 

${\displaystyle f\left(y\right)=g\left(y\right)-v_1}$

Remark 10. 

Let ${\displaystyle a_n^2=\frac{2^{2n}{(n!)}^2}{2(n+1)(2n+1)!}}$, ${\displaystyle b_n^2=\frac{\left(2n\right)!}{2^n{(n!)}^2(2n+1)}}$.

Then by observing that ${\displaystyle (2n+1)!=\left(2n\right)!(2n+1)}$ and by setting ${\displaystyle c_n=\frac{\left(2n\right)!}{2^n{(n!)}^2}}$, we deduce that ${\displaystyle b_n^2=\frac{c_n}{2n+1}}$, ${\displaystyle a_n^2=\frac{1}{2(n+1)(2n+1)c_n}}$ and

$${\displaystyle a_n^2{b}_n^2=\frac{1}{2(n+1){(2n+1)}^2}\sim \frac{1}{n^3}}$$

(62)

In particular,

$${\displaystyle a_n{b}_n\sim \frac{1}{n^{\frac{3}{2}}}}$$

(63)

Below we show that ${\displaystyle b_n\sim \frac{1}{n^{\frac{3}{2}}}}$ to deduce the convergence of ${\displaystyle \sum _{n=0}^{\infty }{\left|b_n\right|}^2}$ and of ${\displaystyle \sum _{n=0}^{\infty }{\left|a_n\right|}^2}$.

The convergence of the series ${\displaystyle \sum _{n=1}^{\infty }{\left|b_n\right|}^2}$ requires Stirling’s approximation, $\underset{n⟶+\infty }{lim}{\displaystyle \frac{1}{n!}}{\left({\displaystyle \frac{n}{e}}\right)}^n$$\sqrt{2\pi n}=1$ and $\underset{n⟶+\infty }{lim}{\displaystyle \frac{1}{{(n!)}^2}}{\left({\displaystyle \frac{n}{e}}\right)}^{2n}2\pi n=1^2.$

We give two methods to prove this convergence: the first method uses Stirling’s approximation by establishing the convergence with limits, and for the second method, we use an upper bound of $n!$ using the trapezoidal method and a lower bound of $n!$ by the median point method.

(i)

First method: Establishing convergence with limits

Let ${\displaystyle a_n^2=\frac{\left(2n\right)!}{2^{2n}{(n!)}^2(2n+1)}}$ and ${\displaystyle c_n^2=\frac{1}{n^{\frac{3}{2}}}}$.

Then,

$${\displaystyle \begin{array}{c}{\displaystyle \underset{n⟶+\infty }{lim}\frac{a_n^2}{c_n^2}=\underset{n⟶+\infty }{lim}=\frac{\frac{\left(2n\right)!}{2^{2n}{(n!)}^2(2n+1)}}{\frac{1}{n^{3/2}}}}\hfill \\ {\displaystyle =\underset{n⟶+\infty }{lim}\frac{\left(2n\right)!n^{3/2}}{2^{2n}{(n!)}^{2}n}(byusing2n+1\sim 2n)}\hfill \\ {\displaystyle =\underset{n⟶+\infty }{lim}\frac{\left(2n\right)!\sqrt{n}}{2^{2n}{(n!)}^2}}\hfill \\ {\displaystyle =\underset{n⟶+\infty }{lim}\frac{\left(2n\right)!\sqrt{n}}{2^{2n}{(n!)}^2}\frac{1}{1^2}}\hfill \\ {\displaystyle =\underset{n⟶+\infty }{lim}\frac{\left(2n\right)!\sqrt{n}}{2^{2n}{(n!)}^2}\frac{\frac{1}{(2n)!}{(\frac{2n}{e})}^{2n}\sqrt{4\pi n}}{{(\frac{1}{n!}{(\frac{n}{e})}^n\sqrt{2\pi n})}^2}}\hfill \\ {\displaystyle =\underset{n⟶+\infty }{lim}\frac{\sqrt{n}}{2^{2n}}\frac{{(\frac{2n}{e})}^{2n}\sqrt{4\pi n}}{{(\frac{n}{e})}^{2n}(2\pi n)}}\hfill \\ {\displaystyle =\underset{n⟶+\infty }{lim}\frac{\sqrt{n}}{2^{2n}}\frac{2^{2n}\sqrt{4\pi n}}{(2\pi n)}}\hfill \\ {\displaystyle =\underset{n⟶+\infty }{lim}\frac{\sqrt{n}\sqrt{4\pi n}}{2\pi n}}\hfill \\ {\displaystyle =\frac{1}{\sqrt{\pi }}}.\hfill \end{array}}$$

It follows that ${\displaystyle a_n^2\sim \frac{1}{n^{\frac{3}{2}}}}$.

(ii)

Second method: lower bound and upper bound of $n!$

With the upper bound of $n!$, by using the trapezoidal method, we get

$$n!\le e{n}^n{e}^{-n}\sqrt{n}$$

(64)

This implies that

$$\left(2n\right)!\le \alpha {\left(2n\right)}^{2n}{e}^{-2n}\sqrt{2n}$$

(65)

where $\alpha$ is a positive constant.

It follows that

$$\sqrt{\left(2n\right)!}\le \sqrt{\alpha }{\left(2n\right)}^n{e}^{-n}{\left(2n\right)}^{{\displaystyle \frac{1}{4}}}$$

(66)

With the lower bound of $n!$, by using the median point method, we get

$$n!\ge c{n}^n{e}^{-n}\sqrt{n}$$

(67)

where c is a positive constant.

This implies that

$$\left(2n\right)!\ge \beta {\left(2n\right)}^{2n}{e}^{-2n}\sqrt{2n}$$

(68)

where $\beta$ is a positive constant. And

$$\sqrt{\left(2n\right)!}\ge \sqrt{\beta }{\left(2n\right)}^n{e}^{-n}{\left(2n\right)}^{{\displaystyle \frac{1}{4}}}$$

(69)

From (64) and (67) we deduce the following:

$$c{n}^n{e}^{-n}\sqrt{n}\le n!\le e{n}^n{e}^{-n}\sqrt{n}$$

(70)

From (66) and (69) we deduce the following:

$$\sqrt{\beta }{\left(2n\right)}^n{e}^{-n}{\left(2n\right)}^{{\displaystyle \frac{1}{4}}}\le \sqrt{\left(2n\right)!}\le \sqrt{\alpha }{\left(2n\right)}^n{e}^{-n}{\left(2n\right)}^{{\displaystyle \frac{1}{4}}}$$

(71)

Now from (70) and (71) we deduce the following:

$${\displaystyle \frac{\sqrt{\beta }{\left(2n\right)}^n{e}^{-n}{\left(2n\right)}^{\frac{1}{4}}}{e{n}^n{e}^{-n}\sqrt{n}}}\le {\displaystyle \frac{\sqrt{\left(2n\right)!}}{n!}}\le {\displaystyle \frac{\sqrt{\alpha }{\left(2n\right)}^n{e}^{-n}{\left(2n\right)}^{\frac{1}{4}}}{c{n}^n{e}^{-n}\sqrt{n}}}$$

(72)

(Note in above denominators that ${\displaystyle e{n}^n{e}^{-n}\sqrt{n}}$ is upper bound of $n!$ and ${\displaystyle c{n}^n{e}^{-n}\sqrt{n}}$ is lower bound of $n!$). It follows that

$${\displaystyle \frac{\sqrt{\beta }{2}^{n+\frac{1}{4}}}{e{n}^{\frac{1}{4}}}}\le {\displaystyle \frac{\sqrt{\left(2n\right)!}}{n!}}\le {\displaystyle \frac{\sqrt{\alpha }{2}^{n+\frac{1}{4}}}{c{n}^{\frac{1}{4}}}}$$

(73)

Now as ${\displaystyle b_n=\frac{\sqrt{\left(2n\right)!}}{2^{n}n!\sqrt{2n+1}}}$, then we deduce that

(i)

$${\displaystyle \frac{\sqrt{\beta }{2}^{\frac{1}{4}}}{e{n}^{\frac{1}{4}}\sqrt{2n+1}}}\le b_n\le {\displaystyle \frac{\sqrt{\alpha }{2}^{\frac{1}{4}}}{c{n}^{\frac{1}{4}}\sqrt{2n+1}}}$$

(74)

(ii)

$$\sum _{n=1}^{\infty }{b}_n^2\le {\displaystyle \frac{\alpha \sqrt{2}}{c^2}}\sum _{n=1}^{\infty }{\displaystyle \frac{1}{\sqrt{n}(2n+1)}}<\infty$$

(75)

As ${\displaystyle b_n\sim \frac{1}{n^{\frac{3}{4}}}\in {\mathsf{\ell }}_2}$ and ${\displaystyle a_n{b}_n\sim \frac{1}{n^{\frac{3}{2}}}}$ then ${\displaystyle a_n\sim \frac{1}{n^{\frac{3}{4}}}\in {\mathsf{\ell }}_2}$ and $v_1\in {\mathbb{B}}_0$.

Remark 11.

(α) 

$b_n=O({\displaystyle \frac{1}{n^{\frac{3}{4}}}})$, $b_n^2=O({\displaystyle \frac{1}{n^{\frac{3}{2}}}})$, $a_n=O({\displaystyle \frac{1}{n^{\frac{3}{4}}}})$ and $a_n^2=O({\displaystyle \frac{1}{n^{\frac{3}{2}}}}).$

(β) 

As ${\displaystyle v_1=f+g=\frac{1}{\lambda }{\int }_0^y{e}^{\frac{u^2}{2}}du{\int }_u^{\infty }{e}^{-\frac{s^2}{2}}ds}$ then $f=g-v_1$ where ${\displaystyle g(y)=\frac{\sqrt{2\pi }}{\lambda }\sum _{n=0}^{\infty }{b}_n}$$u_{2n+1}(y)\mathit{such}\mathit{that}{b}_n={\displaystyle \frac{\sqrt{(2n)!}}{2^{n}n!\sqrt{2n+1}}}$$=O({\displaystyle \frac{1}{n^{\frac{3}{4}}}}).$

Corollary 2.

$v_n$ belongs to Bargmann space for all $n=1,2,\dots$

Proof. 

As $v_n$ satisfies the recurrence relation:

$${\displaystyle v_{n+1}=\frac{n-1}{\sqrt{n(n+1)}}{v}_{n-1}+\frac{u_n}{\lambda n\sqrt{n+1}}\mathrm{for}\mathrm{all}n=1,2,\dots }$$

where $u_n$ belongs to Bargmann space.

Then we deduce that $v_2={\displaystyle \frac{u_1}{\lambda \sqrt{2}}}$, this implies that $v_2\in {\mathbb{B}}_0$, and by recurrence, we deduce that $v_n\in {\mathbb{B}}_0$ for all $n=1,2,\dots$ □

Proposition 4

(Determination explicit of $v_n$ with respect to $u_n$ and $v_1$).

Let

$$\left\{\begin{array}{c}{\displaystyle v_{n+1}=A_n{u}_n+B_n{v}_{n-1}}\\ \mathit{where}\\ {\displaystyle A_n=\frac{1}{\lambda n\sqrt{n+1}}}\\ \mathit{and}\\ {\displaystyle B_n=\frac{n-1}{\sqrt{n(n+1)}}}\end{array}\right.$$

Then, we have

$$v_n=P_{n-1}+{\alpha }_n{v}_1$$

(76)

with

$$\left\{\begin{array}{c}{\displaystyle {\alpha }_n=0\mathrm{if}n=2p;p\in \mathbb{N}}\\ \\ {\displaystyle {\alpha }_{2n+1}=\prod _{j=1}^p{B}_{2j}\mathrm{if}n=2p+1;p\in \mathbb{N}}\\ \\ P_0=0\\ \\ {\displaystyle P_1=\frac{u_1}{\lambda \sqrt{2}}}\\ {\displaystyle P_n=\frac{n-1}{\sqrt{n(n+1)}}{P}_{n-2}+\frac{u_n}{\lambda n\sqrt{n+1}}}\\ or\\ {\displaystyle P_n=B_n{P}_{n-2}+A_n{u}_n}\end{array}\right.$$

Proof.  

$v_2=A_1{u}_1=P_1+{\alpha }_2{v}_1;P_1=A_1{u}_1$ and ${\alpha }_2=0$

$v_3=A_2{u}_2+B_2{v}_1=P_2+{\alpha }_3{v}_1$; $P_2=A_2{u}_2$ and ${\alpha }_3=B_2$

$v_4=A_3{u}_3+B_3{v}_2$$=A_3{u}_3+B_3{A}_1{u}_1=P_3+{\alpha }_4{v}_1$; $P_3=A_3{u}_3+B_3{A}_1{u}_1=B_3{P}_1+A_3{u}_3$ and ${\alpha }_4=0$

$v_5=A_4{u}_4+B_4{v}_3$$=A_4{u}_4+B_4[A_2{u}_2+B_2{v}_1]$$=A_4{u}_4+B_4{A}_2{u}_2+B_4{B}_2{v}_1$$=P_4+{\alpha }_5{v}_1$: $P_4=A_4{u}_4+B_4{A}_2{u}_2=B_4{P}_2+A_4{u}_4$ and ${\displaystyle {\alpha }_5=B_4{B}_2=\prod _{j=1}^2{B}_{2j}}$

$v_6=A_5{u}_5+B_5{v}_4$$=A_5{u}_5+B_5[A_3{u}_3+B_3{A}_1{u}_1]$$=A_5{u}_5+B_5{A}_3{u}_3+B_5{B}_3{A}_1{u}_1$$=P_5+{\alpha }_6{v}_1$; $P_5=A_5{u}_5+B_5{A}_3{u}_3+B_5{B}_3{A}_1{u}_1$$=B_5{P}_3+A_5{u}_5$ and ${\alpha }_6=0$

$v_7=A_6{u}_6+B_6{v}_5$$=A_6{u}_6+B_6[A_4{u}_4+B_4{A}_2{u}_2+B_4{B}_2{v}_1]$$=A_6{u}_6+B_6{A}_4{u}_4+B_6{B}_4{A}_2{u}_2+B_6{B}_4{B}_2{v}_1$$=P_6+{\alpha }_7{v}_1$; $P_6=B_6{P}_4+A_6{u}_6$ and ${\displaystyle {\alpha }_7=\prod _{j=1}^3{B}_{2j}}$.

By recurrence with respect to p, we suppose that

$v_{2p}=P_{2p-1}+{\alpha }_{2p}{v}_1$ such that $P_{2p}=B_{2p}{P}_{2p-2}+A_{2p}{u}_{2p}$ and ${\alpha }_{2p}=0$.

$v_{2p+1}=P_{2p}+{\alpha }_{2p+1}{v}_1$ such that $P_{2p+1}=B_{2p+1}{P}_{2p-1}+A_{2p+1}{u}_{2p+1}$ and ${\displaystyle {\alpha }_{2p+1}=\prod _j^p{B}_{2j}}$.

Then we deduce that

${•}_{2(p+1)}$

$${\displaystyle \begin{array}{c}{\displaystyle v_{2(p+1)}=A_{2p+1}{u}_{2p+1}+B_{2p+1}{v}_{2p}}\hfill \\ {\displaystyle =A_{2p+1}{u}_{2p+1}+B_{2p+1}[P_{2p-1}+{\alpha }_{2p}{v}_1]}\hfill \\ {\displaystyle =A_{2p+1}{u}_{2p+1}+B_{2p+1}{P}_{2p-1}+{\alpha }_{2p}{B}_{2p+1}{v}_1}\hfill \\ {\displaystyle =B_{2p+1}{P}_{2p-1}+A_{2p+1}{u}_{2p+1}(because{\alpha }_{2p}=0)}\hfill \\ {\displaystyle =B_{2p+1}{P}_{2p-1}+A_{2p+1}{u}_{2p+1}+{\alpha }_{2(p+1)}{v}_1}.\hfill \end{array}}$$

This implies

$${\alpha }_{2(p+1)}=0$$

and

$${\displaystyle P_{2p+1}=B_{2p+1}{P}_{(2p+1)-2}+A_{2p+1}{u}_{2p+1}.}$$

i.e.,

$${\displaystyle v_{2(p+1)}=P_{2p+1}+{\alpha }_{2(p+1)}{v}_1}$$

${•}_{2(p+1)+1}$

$${\displaystyle \begin{array}{c}{\displaystyle v_{2(p+1)+1}=A_{2(p+1)}{u}_{2(p+1)}+B_{2(p+1)}{v}_{2p+1}}\hfill \\ {\displaystyle =A_{2(p+1)}{u}_{2(p+1)}+B_{2(p+1)}[P_{2p}+{\alpha }_{2p+1}{v}_1]}\hfill \\ {\displaystyle =A_{2(p+1)}{u}_{2(p+1)}+B_{2(p+1)}{P}_{2p}+B_{2(p+1)}{\alpha }_{2p+1}{v}_1}\hfill \\ {\displaystyle =A_{2(p+1)}{u}_{2(p+1)}+B_{2(p+1)}{P}_{2p}+B_{2(p+1)}\prod _{j=1}^p{B}_{2j}{v}_1}\hfill \\ {\displaystyle =B_{2(p+1)}{P}_{2p}+A_{2(p+1)}{u}_{2(p+1)}+\prod _{j=1}^{p+1}{B}_{2j}{v}_1}.\hfill \end{array}}$$

Then we deduce that

$${\displaystyle {\alpha }_{2(p+1)+1}=\prod _{j=1}^{p+1}{B}_{2j}{v}_1}$$

and

$${\displaystyle P_{2(p+1)}=B_{2(p+1)}{P}_{2p}+A_{2(p+1)}{u}_{2(p+1)}}$$

i.e.,

$${\displaystyle v_{2(p+1)+1}=P_{2(p+1)}+{\alpha }_{2(p+1)+1}{v}_1.}$$

□

Lemma 6.

Let ${\displaystyle u_n\left(y\right)=\frac{y^n}{\sqrt{n!}}}$ with norm $\left|\right|u_n\left|\right|=1$ and $P_n\left(y\right)$ is a polynomial which satisfies:

$${\displaystyle P_n\left(y\right)=B_n{P}_{n-2}\left(y\right)+A_n{u}_n\left(y\right)}$$

where ${\displaystyle B_n=\frac{n-1}{\sqrt{n(n+1)}}}$ and ${\displaystyle A_n=\frac{1}{\lambda n\sqrt{n+1}}}$

Then

(i) 

$P_{n-2}$ is orthogonal to $u_n$.

(ii) 

Let ${\displaystyle p_n=\left|\right|P_n{\left|\right|}^2}$ then ${\displaystyle p_n=B_n^2{p}_{n-2}+A_n^2}$.

(iii) 

${\displaystyle p_n=O(\frac{1}{n^{\frac{3}{2}}})}$.

Proof. 

(i)

As the degree of $P_{n-2}$ is $n-2$ then ${\displaystyle <P_{n-2},u_n>=0}$.

(ii)

We have

$${\displaystyle \begin{array}{c}{\displaystyle p_n=\left|\right|P_n{\left|\right|}^2=<B_n{P}_{n-2}\left(y\right)+A_n{u}_n\left(y\right),B_n{P}_{n-2}\left(y\right)+A_n{u}_n\left(y\right)>}\hfill \\ {\displaystyle =<B_n{P}_{n-2}\left(y\right),B_n{P}_{n-2}\left(y\right)>+<B_n{P}_{n-2}\left(y\right),A_n{u}_n\left(y\right)>}\hfill \\ {\displaystyle +<A_n{u}_n\left(y\right),B_n{P}_{n-2}\left(y\right)>+<A_n{u}_n,A_n{u}_n>}\hfill \end{array}}$$

By using (i) and $\left|\right|u_n\left|\right|=1$, we deduce that ${\displaystyle p_n=B_n^2{p}_{n-2}+A_n^2}$.

(iii)

is deduced from (ii) and Lemma 5.

□

4. On Right-Inverse ${\mathit{K}}_{\mathit{\lambda }}$ of ${\mathit{H}}_{\mathit{\lambda }}$ and Its Compactness

Proposition 5.

Let $\mathcal{H}$ be a Hilbert space and let T be a linear operator with domain $D\left(T\right)\subset \mathcal{H}$.

Let $N\left(T\right)$ be the kernel of T and $R\left(T\right)$ its range.

(i) 

T is right invertible ⇔ $R\left(T\right)=\mathcal{H}.$

(ii) 

The inverse of an unbounded operator could be compact, or bounded, or unbounded, as one can easily see by considering multiplication operators on L2 spaces.

(iii) 

The left inverse of a compact operator on an infinite dimensional Hilbert space, if it exists, must be unbounded.

Proof. 

(i)

• For $\begin{array}{|c|}\hline \Leftarrow \\ \hline\end{array}$: assume T is surjective. Then, for all $\psi \in \mathcal{H}$, there exists ${\phi }_{\psi }\in \mathcal{H}$ such that $T\left({\phi }_{\psi }\right)=\psi$. Define $R:\mathcal{H}\to \mathcal{H}$ to be the function which maps each $\psi \in \mathcal{H}$ to such $\phi \in \mathcal{H}$ (if there is more than one $\phi$, then the function R maps $\psi$ to one of them chosen in an arbitrary way). This excludes the possibility that R maps $\psi$ to two distinct values, in which case it would not be a function). It follows that

  $${\displaystyle \forall \psi \in \mathcal{H},(T\circ R)\left(\psi \right)=T\left(R\left(\psi \right)\right)=T\left({\psi }_{\phi }\right)=\psi }$$

  and $T\circ R=i{d}_{\mathcal{H}}$.
• For $\begin{array}{|c|}\hline \Rightarrow \\ \hline\end{array}$, assume $T:D\left(T\right)\subset \mathcal{H}\to \mathcal{H},R:\mathcal{H}\to \mathcal{H}$ are such that $T\circ R=i{d}_D\left(T\right)$. Then, for each $\psi \in \mathcal{H}$, ${\phi }_{\psi }\stackrel{def}{=}R\left(\psi \right)\in D\left(T\right)$ is a pre-image of $\psi$ by T, as $T\left({\phi }_{\psi }\right)=f\circ R\left(\psi \right)=i{d}_{\mathcal{H}}\left(\psi \right)=\psi$. Hence, T is surjective.
• $K_{\lambda }$ and $H_{\lambda }$ are the examples satisfying (ii) and (iii) (see the following propositions).

□

Proposition 6.

(i) 

$K_{\lambda }$ is a right inverse of $H_{\lambda }$, i.e., $H_{\lambda }{K}_{\lambda }=I$.

(ii) 

$H_{\lambda }$ is surjective.

(iii) 

$H_{\lambda }$ is non-injective.

Proof. 

(i)

As ${\displaystyle u_n=\frac{y^n}{\sqrt{n!}}}$ then we deduce that:

$${\displaystyle H_{\lambda }{u}_n=\lambda [-\frac{n(n-1)y^{n-1}}{\sqrt{n!}}+\frac{n{y}^{n+1}}{\sqrt{n!}}]}$$

and

$${\displaystyle \frac{H_{\lambda }{u}_n}{n\sqrt{n+1}}=\lambda [-\frac{(n-1)u_{n-1}}{\sqrt{n(n+1)}}+u_{n+1}]}$$

Now, from the following recurrence relation:

$${\displaystyle v_{n+1}=\frac{u_n}{\lambda n\sqrt{n+1}}+\frac{n-1}{\sqrt{n(n+1)}}{v}_{n-1}.}$$

(77)

we deduce that

$${\displaystyle K_{\lambda }{u}_{n+1}=\frac{u_n}{\lambda n\sqrt{n+1}}+\frac{n-1}{\sqrt{n(n+1)}}{K}_{\lambda }{u}_{n-1}}$$

(78)

and

$${\displaystyle \begin{array}{c}{\displaystyle H_{\lambda }{K}_{\lambda }{u}_{n+1}=\frac{H_{\lambda }{u}_n}{\lambda n\sqrt{(n+1)}}+\frac{n-1}{\sqrt{n(n+1)}}{H}_{\lambda }{K}_{\lambda }{u}_{n-1}.}\hfill \\ {\displaystyle =-\frac{(n-1)}{\sqrt{n(n+1)}}{u}_{n-1}+u_{n+1}+\frac{n-1}{\sqrt{n(n+1)}}{H}_{\lambda }{K}_{\lambda }{u}_{n-1}.}\hfill \end{array}}$$

Now as $K_{\lambda }{u}_2\left(y\right)={\displaystyle \frac{u_1\left(y\right)}{\lambda \sqrt{2}}}$ then ${\displaystyle H_{\lambda }{K}_{\lambda }{u}_2\left(y\right)=\frac{y^2}{\sqrt{2}}=u_2\left(y\right)}$.

And by recurrence we deduce that:

${\displaystyle H_{\lambda }{K}_{\lambda }{u}_{n+1}=-\frac{(n-1)}{\sqrt{n(n+1)}}{u}_{n-1}+u_{n+1}+\frac{n-1}{\sqrt{n(n+1)}}{u}_{n-1}=u_{n+1}\forall n}$.

By taking the range of $K_{\lambda }$ as domain of $H_{\lambda }$, It follows that:

$${\displaystyle H_{\lambda }{K}_{\lambda }=I}$$

(79)

(ii)

It is well known that if an operator T has a right inverse, then T is surjective (see above proposition). Then $H_{\lambda }$ is surjective. Conversely, if T is surjective, then T has a right inverse (see above proposition or [20]).

(iii)

$H_{\lambda }$ is non-injective since it cancels out the function ${\displaystyle g\left(y\right)=\frac{\sqrt{2\pi }}{\lambda }{\int }_0^y{e}^{\frac{u^2}{2}}du}$. In fact, as ${\displaystyle H_{\lambda }u\left(y\right)=\lambda [-y{u}^{″}\left(y\right)+y^2{u}^{\prime }\left(y\right)]}$ and as ${\displaystyle g^{\prime }\left(y\right)=\frac{\sqrt{2\pi }}{\lambda }{e}^{\frac{y^2}{2}}}$ and ${\displaystyle g^{″}\left(y\right)=\frac{\sqrt{2\pi }}{\lambda }y{e}^{\frac{y^2}{2}}}$ then ${\displaystyle H_{\lambda }(g\left(y\right)=\sqrt{2\pi }[-y^2{e}^{\frac{y^2}{2}}+y^2{e}^{\frac{y^2}{2}}]=0}$.

□

Proposition 7

(Compactness of $K_{\lambda }$).

Let ${\displaystyle u=\sum _{n=1}^{\infty }{c}_n{u}_n}$, ${\displaystyle K_{\lambda }u=\sum _{n=1}^{\infty }{c}_n{v}_n=\sum _{n=1}^{\infty }{c}_n{P}_{n-1}+c_1{v}_1}$ where $c_1=\sqrt{2}⟨\phi ,g-v_1⟩$ and let ${\displaystyle K_{\lambda ,m}u=\sum _{n=1}^m{c}_n{P}_{n-1}+\sqrt{2}⟨\phi ,g-v_1⟩v_1.}$ where $g\left(y\right)={\displaystyle \frac{\sqrt{2\pi }}{\lambda }}{\sum }_{n=0}^{\infty }{b}_n{u}_{2n+1}\left(y\right)\mathit{such}\mathit{that}{b}_n={\displaystyle \frac{\sqrt{\left(2n\right)!}}{2^{n}n!\sqrt{2n+1}}}$ (see Formula (59)).

Then,

(i) 

${\displaystyle K_{\lambda ,m}⟶K_{\lambda }}$ as $m⟶\infty$ (in operator norm).

(ii) 

$K_{\lambda }$ is compact.

Proof. 

(i)

Let ${\displaystyle u=\sum _{n=1}^{\infty }{c}_n{u}_n}$ then ${\displaystyle K_{\lambda }u=\sum _{n=1}^{\infty }{c}_n{P}_{n-1}+\sqrt{2}<\phi ,g-v_1>v_1}$ and ${\displaystyle K_{\lambda ,m}u=\sum _{n=1}^m{c}_n}$$P_{n-1}+\sqrt{2}<\phi ,g-v_1>v_1$.

Let us estimate the remainder of the series using Cauchy–Schwarz: ${\displaystyle ||\sum _{n=m+1}^{\infty }{c}_n{P}_{n-1}||\le }$${\displaystyle \sum _{n=m+1}^{\infty }|c_n|.||P_{n-1}||}$${\displaystyle \le (\sum _{n=m+1}^{\infty }|c_n{|}^2{)}^{\frac{1}{2}}.(\sum _{n=m+1}^{\infty }||P_{n-1}{||}^2{)}^{\frac{1}{2}}}$

${\displaystyle \le (\sum _{n=m+1}^{\infty }|<u,u_n{>|}^2{)}^{\frac{1}{2}}.(\sum _{n=m}^{\infty }||p_n{||)}^{\frac{1}{2}}}$${\displaystyle \le ||u||.(\sum _{n=m}^{\infty }||p_n{||)}^{\frac{1}{2}}}$

Then, ${\displaystyle K_{\lambda ,m}⟶K_{\lambda }}$ as $m⟶\infty$ (in operator norm).

(ii)

As $K_{\lambda }$ is an operator norm limit of finite rank operators, $K_{\lambda }$ is a compact operator.

□

Remark 12.

(i) 

We can take the range of $K_{\lambda }$ as domain of $H_{\lambda }$. This is possible because $R\left(K_{\lambda }\right)$ is dense in ${\mathbb{B}}_0$. To do this, let us take a function $\phi \in {\mathbb{B}}_0$ orthogonal to $R\left(K_{\lambda }\right)$; in particular, it is orthogonal to all $v_n$. Starting with $v_1$ from ${\displaystyle v_n=P_{k-1}+{\alpha }_n{v}_1}$ where ${\alpha }_{2n}=0$ and ${\alpha }_{2n+1}=\sqrt{2}{b}_n$ we deduce the following:

$${\displaystyle <P_k,\phi >=(v_{n+1},\phi )=0.}$$

(80)

Since $P_n$ is of degree n exactly, the set of $P_n$ is an algebraic basis for the space of polynomials that are zero at the origin, so it is the total in ${\mathbb{B}}_0$ then $\phi =0$.

(ii) 

$H_{\lambda }$ is an imaginary part of $H_{\mu ,\lambda }$, we observe that it is also the imaginary part of $H_{{\lambda }^{\prime },\mu ,\lambda }={\lambda }^{\prime }{A}^{{*}^2}{A}^2+H_{\mu ,\lambda }$. In [21], we have presented an interesting “non-linear” factorization of this operator ${\displaystyle H_{{\lambda }^{\prime },\mu ,\lambda };{\lambda }^{\prime }\ne 0.}$

5. Conclusions

Reggeon field theory (RFT) is an attempt to predict the high-energy behavior 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.

Originally, RFT was formulated 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:

$${\displaystyle S=\int dy\{{\psi }^{+}\partial y\psi -\mu {\psi }^{+}\psi +i\lambda {\psi }^{+}({\psi }^{+}+\psi )\psi \}}$$

(81)

The Hamiltonian of this theory is given by ${\displaystyle H_{\mu ,\lambda }=\mu A^{*}A+i\lambda A^{*}(A+A^{*})A}$ acting on Bargmann space, where $\mu \ne 0$ is intercept coupling, $\lambda$ is the triple Pomeron coupling, and A and $A^{*}$ are the annihilation and creation operators satisfying $[A,A^{*}]=I$. Many spectral properties of $H_{\mu ,\lambda }$ have been studied by the author and recalled in this work. The mathematical difficulties of this problem come of course from the non-self-adjointness of $H_{\mu ,\lambda }$. 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.

The main purpose of the present work (Section 4) is to give some new spectral properties of anti-adjoint part ${\displaystyle H_{\lambda }=i\lambda A^{*}(A+A^{*})A;i^2=-1}$ of $H_{\mu ,\lambda }$, in particular, to show that $H_{\lambda }$ admits a right-inverse $K_{\lambda }$, i.e., $H_{\lambda }{K}_{\lambda }=I$ on negative imaginary axis and that $K_{\lambda }$ is compact. This gives us an original example of an operator that admits a right inverse but does not admit a left inverse.