# Szegö Limit Theorem on the Heisenberg Group

Shyam Swarup Mondal  and  Jitendriya Swain  Department of Mathematics, Indian
Institute of Technology Guwahati, Guwahati 781039, India.
<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

Let $\mathcal{H}_{0}=V,\mathcal{H}_{1}=B+V$ and
$\mathcal{H}_{2}=\mathcal{L}+V$ be the operators on the Heisenberg group
$\mathbb{H}^{n}$, where $V$ is the operator of multiplication growing like
$|g|^{\kappa},0<\kappa<1$, $B$ is a bounded linear operator and $\mathcal{L}$
is the sublaplacian on $\mathbb{H}^{n}$. In this paper we prove Szegö limit
theorem for the operators $\mathcal{H}_{0},\mathcal{H}_{1}$ and
$\mathcal{H}_{2}$ on $L^{2}(\mathbb{H}^{n}).$

###### Key words and phrases:

Balian-Low Theorem; time-frequency analysis; Weyl transform; Zak transform

###### 2010 Mathematics Subject Classification:

Primary 42C15; Secondary 47B38.

##  1\. Introduction

The classical Szegö limit theorem describes the asymptotic distribution of
eigenvalues of the operator $P_{n}T_{f}P_{n}$, where $T_{f}$ is the
multiplication operator on $L^{2}((0,2\pi))$, associated with a positive
function $f\in\mathcal{C}^{1+\alpha}[0,2\pi],\alpha>0$ and the orthogonal
projections $\\{P_{n}\\}$ of $L^{2}[0,2\pi]$ onto a linear subspace spanned by
the functions $\\{e^{im\theta}:0\leq m\leq n;0\leq\theta<2\pi\\}$. For such a
triple $\left(f,T_{f},\\{P_{n}\\}\right),$ Szegö proved that

(1.1) $\lim_{n\to\infty}\frac{1}{n+1}\log\det
P_{n}T_{f}P_{n}=\frac{1}{2\pi}\int_{0}^{2\pi}\log f(\theta)d\theta.$

The orthogonal projections $P_{n}$ coincide with the spectral projections
$\pi_{\lambda}$ of the self-adjoint operator $-\frac{d^{2}}{dx^{2}}$ on
$L^{2}[0,2\pi]$, with a periodic boundary condition, corresponding to the
interval $[0,\lambda)$ with $n<\lambda<n+1.$ The result in equation (1.1) is
the well known as Szegö limit theorem. We refer to [4, 16] for details and
related results. More specifically, for a bounded real-valued integrable
function $f$, Szegö limit theorem is generalized to continuous functions $F$
(instead of the logarithm in (1.1)) defined on $[\inf f,\sup f]$, containing
the eigenvalues $\\{\lambda_{i}^{n}\\}_{i=1}^{n}$ of $P_{n}T_{f}P_{n}$ in
Sect. 5.3 of [4]. The result for such $F$ is,

$\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n}F(\lambda_{i}^{n})=\frac{1}{2\pi}\int_{0}^{2\pi}F(f(\theta))d\theta.$

Notice that the left hand side here can be seen to be the limit of

$tr(F(P_{n}T_{f}P_{n}))/tr(P_{n})$

where $tr(X)$ denotes the trace of the operator $X$, $e^{im\theta}$ is an
eigenfunction of $\Delta=-\frac{d^{2}}{dx^{2}}$, with a periodic boundary
condition and the asymptotic of the functional

$\rho_{\lambda}(F)=tr(\pi_{\lambda}F(\pi_{\lambda}T_{f}\pi_{\lambda})\pi_{\lambda})=\sum_{k}F(\mu_{k}(\lambda))$

is precisely the sum of Dirac measures located at the eigenvalues
$\mu_{k}(\lambda)$ of the operator $\pi_{\lambda}T_{f}\pi_{\lambda}$. Similar
results were obtained for various classes of differential and pseudo-
differential operators in [5], [6], [7], [15], [19] and [20].

In [20], Zelditch considered a Schrödinger operator on $\mathbb{R}^{n}$ of the
form $H=-\frac{1}{2}\Delta+V$, where $V$ is a smooth positive function that
grows like $V_{0}|x|^{\kappa},~{}\kappa>0$ at infinity. He took a $0$-th order
self-adjoint pseudo-differential operator $A$ associated with a symbol
$a(x,\xi)$ relative to Beals-Fefferman weights
$\varphi_{1}(x,\xi)=1,\varphi_{2}(x,\xi)=(1+|\xi|^{2}+V(x))^{1/2}$ and proved
the following Szegö type theorem: For any continuous function $f$,

$\lim_{\lambda\rightarrow\infty}\frac{{\rm
tr}f(\pi_{\lambda}A\pi_{\lambda})}{\mbox{rank}~{}(\pi_{\lambda})}=\lim_{\lambda\rightarrow\infty}\frac{\int_{H(x,\xi)\leq\lambda}f(a(x,\xi))~{}dxd\xi}{\displaystyle\mbox{Vol}{(H(x,\xi)\leq\lambda)}}$

where $H(x,\xi)=\frac{1}{2}|\xi|^{2}+V(x)$, assuming one of the limits exists.
Such asymptotic spectral formulae expressing the relation between functions of
pseudo-differential operators and their symbols is an important and
interesting problem in mathematical analysis.

In [15], the authors consider the operator of the form $H=\Delta+V$ on the
lattice, where the discrete Laplacian operator $(\Delta
u)(k)=\sum_{|k-j|=1}u(j)+2nu(k)$ and $V$ is multiplication by a positive
sequence

$V(n)=\left\\{\begin{array}[]{ll}1,&n=0\\\ |n|^{\kappa},&\kappa>0\\\
\end{array}\right.$

and used the Tauberian theorem of Grishin-Poedintseva [3] to compare such
limits. In this paper we aim to develop similar formulae for some pseudo-
differential operators on $\mathbb{H}^{n}$. Unlike the symbols on
$\mathbb{R}^{n}$ or $\mathbb{Z}^{n}$ the symbols on the Heisenberg group are
operator valued. Establishing Szegö limit theorem is therefore bit more
complicated.

To establish a Szegö type theorem, we need to consider the ratios of
distribution functions associated to different measures and their asymptotic
behavior. The asymptotic limit of such ratios is computed using a Tauberian
theorem where some transform of these measures is considered and the limit
taken for such transforms. For example, Zelditch [20] used the Laplace
transform (via Karamata’s Tauberian theorem ([18], p-192)) whereas Robert [11]
suggested the use of Stieltjes transform (via Keldysh Tauberian theorem) in
[8]. We use the Tauberian theorem of Grishin-Poedintseva (see theorem 8 of
[3]) and a theorem of Laptev-Safarov (see [9] and [10]), for estimating the
errors, to prove our main theorems (Theorem 1.1, 1.2 and 1.3).

There is extensive work on the Szegö’s theorem associated with orthogonal
polynomials in $L^{2}(\mathbb{T},d\mu)$ with $\mu$ some probability measure on
$\mathbb{T}$, we refer to the monumental work of Barry Simon [1] for the
details.

In this article we prove Szegö limit theorem for the operators
$\mathcal{H}_{0}=V,\mathcal{H}_{1}=B+V$ and $\mathcal{H}_{2}=\mathcal{L}+V$ on
the Heisenberg group $\mathbb{H}^{n}$, where $V$ is the operator of
multiplication growing like $|g|^{\kappa},\kappa>0$, $B$ is a bounded linear
operator and $\mathcal{L}$ is the sublaplacian on $\mathbb{H}^{n}$. Throughout
this article we assume $V(g)=|g|^{\kappa}$, where

(1.2) $\displaystyle|g|=(|x|^{4}+|t|^{2})^{\frac{1}{4}},\quad
g=(x,t)\in\mathbb{H}^{n},$

defining the homogenous norm on $\mathbb{H}^{n}$. Since the resolvent
operators $(\mathcal{H}_{j}-iI)^{-1}$ are compact for $j=0,1$ and $2,$ so the
operators $\mathcal{H}_{0},\mathcal{H}_{1}$ and $\mathcal{H}_{2}$ have
discrete spectrum. The eigenfunctions of $H_{j}$ form a complete orthogonal
basis for $L^{2}(\mathbb{H}^{n}).$

Then our main theorems are the following. We consider the operators
$\mathcal{H}_{0}=V$ and $M_{b}$, the operators of multiplication by $V$ and
$b$ respectably on $L^{2}(\mathbb{H}^{n})$, where $b$ is a bounded real valued
integrable function on the Heisenberg group and obtain the following theorem:

######  Theorem 1.1.

Let $\mathcal{H}_{0}=V$ and $V(g)=|g|^{\kappa},\kappa\geq 0.$ Let
$\pi_{r_{0}}$ be the orthogonal projection of $L^{2}(\mathbb{H}^{n})$ onto the
space of eigenfunctions of $\mathcal{H}_{0}$ with eigenvalue $\leq r_{0}.$ Let
$b$ be a bounded real valued integrable function on $\mathbb{H}^{n}$ and
$M_{b}$ be the operator of multiplication by $b$ on $L^{2}(\mathbb{H}^{n})$.
Then for any $f\in C(\mathbb{R})$ we have

$\displaystyle\lim_{r_{0}\to\infty}\frac{{tr}~{}{f(\pi_{r_{0}}M_{b}\pi_{r_{0}})}}{tr~{}(\pi_{r_{0}})}$
$\displaystyle=\int_{\mathbb{H}^{n}}f(b(g))\,dg.$

Further, we consider the operator $\mathcal{H}_{1}=B+V$, where $B$ is a
bounded linear operator on $L^{2}(\mathbb{H}^{n})$ and replace the
multiplication operator $M_{b}$ by a 0-th order self-adjoint pseudo-
differential operator on $L^{2}(\mathbb{H}^{n})$ and obtain the following
variation of Theorem 1.1:

######  Theorem 1.2.

Consider the operator $\mathcal{H}_{1}=B+V$ on the Heisenberg group
$\mathbb{H}^{n}$, where $B$ is a bounded operator on $\mathbb{H}^{n}$ and
$V(g)=|g|^{\kappa},0<\kappa<1.$ Let $\pi_{r_{1}}$ be the orthogonal projection
of $L^{2}(\mathbb{H}^{n})$ onto the space of eigenfunctions of
$\mathcal{H}_{1}$ with eigenvalue $\leq r_{1}$; let $A$ be a 0-th order self-
adjoint pseudo-differential operator on $L^{2}(\mathbb{H}^{n})$ with symbol
$a(g,{\lambda})$, where $g\in\mathbb{H}^{n},\lambda\in\mathbb{R}^{*}$ and let
$f\in C(\mathbb{R})$. Then

$\displaystyle\lim_{r_{1}\to\infty}\frac{{tr}~{}{f(\pi_{r_{1}}A\pi_{r_{1}})}}{tr~{}(\pi_{r_{1}})}$
$\displaystyle=\lim_{r_{1}\to\infty}\frac{\int_{\mathbb{R}^{*}}\int_{V(g)\leq
r_{1}}{tr}~{}f(a(g,{\lambda}))\,dg\,d\mu(\lambda)}{{r_{1}\int_{-r_{1}}^{r_{1}}\,d\mu(\lambda)}\int_{V(g)\leq
r_{1}}\,dg}.$

(Assuming one limit exists)

Finally we consider the Schrödinger operator $\mathcal{H}_{2}=\mathcal{L}+V$
on the Heisenberg group with the assumptions of Theorem 1.2 and obtain the
following theorem on the Heisenberg group:

######  Theorem 1.3.

Consider the Schrödinger operator of the form $\mathcal{H}_{2}=\mathcal{L}+V$
on the Heisenberg group $\mathbb{H}^{n}$, where $\mathcal{L}$ is the
sublaplacian associated with the Heisenberg group and
$V(g)=|g|^{\kappa},0<\kappa<1.$ Let $\pi_{r_{2}}$ be the orthogonal projection
of $L^{2}(\mathbb{H}^{n})$ onto the space of eigenfunctions of
$\mathcal{H}_{2}$ with eigenvalue $\leq r_{2}$; let $A$ be a 0-th order self-
adjoint pseudo-differential operator on $L^{2}(\mathbb{H}^{n})$ with symbol
$a(g,{\lambda})$ where $g\in\mathbb{H}^{n},\lambda\in\mathbb{R}^{*}$ and let
$f\in C(\mathbb{R})$. Then

$\displaystyle\lim_{r_{2}\to\infty}\frac{tr~{}{f(\pi_{r_{2}}A\pi_{r_{2}})}}{tr~{}(\pi_{r_{2}})}$
$\displaystyle=\lim_{r_{2}\to\infty}\frac{\int_{G^{r_{2}}}f(a_{g,{\lambda}}(\xi,x))\,d\xi\,dx\,dg\,d\lambda}{\int_{G^{r_{2}}}\,d\xi\,dx\,dg\,d\lambda}$

(Assuming one limit exists)
where
$G^{r_{2}}=\\{(g,\lambda,\xi,x)\in\mathbb{H}^{n}\times\mathbb{R}^{*}\times\mathbb{R}^{n}\times\mathbb{R}^{n}:|\lambda|(|\xi|^{2}+|x|^{2})+V(g)\leq
r_{2},V(g)\leq r_{2}^{\frac{1}{2}},|\lambda|>{{r_{2}}^{-\frac{1}{2}}}\\}$ and
$a(g,{\lambda})=Op^{W}(a_{g,{\lambda}}).$

Also we show that the above theorems are valid under a compact perturbation of
the pseudo-differential operator $A$ in Corollary 5.6.

We organize the paper as follows. In section 2, we provide necessary
background for the Szegö limit theorem on the Heisenberg group, estimate the
difference of the symbol of the composition and composition of the
corresponding symbols of two pseudo-differential operator on the Heisenberg
group. Finally we prove the Szegö limit theorem for the operators
$\mathcal{H}_{0},\mathcal{H}_{1}$ and $\mathcal{H}_{2}$ in section 3, 4 and 5
respectively.

##  2\. Notations and Background

The main aim of this section is to discuss the symbol classes on the
Heisenberg group via the left invariant vector fields and their correspondence
with the symbol classes on $\mathbb{R}^{n}.$

###  2.1. Hermite Operator

Let $H_{k}$ denote the Hermite polynomial on $\mathbb{R}$, defined by

$H_{k}(x)=(-1)^{k}\frac{d^{k}}{dx^{k}}(e^{-x^{2}})e^{x^{2}},\quad
k=0,1,2,\cdots,$

and $h_{k}$ denote the normalized Hermite functions on $\mathbb{R}$ defined by

$h_{k}(x)=(2^{k}\sqrt{\pi}k!)^{-\frac{1}{2}}H_{k}(x)e^{-\frac{1}{2}x^{2}},\quad
k=0,1,2,\cdots,$

The Hermite functions $\\{h_{k}\\}$ are the eigenfunctions of the Hermite
operator $H=-\frac{d^{2}}{dx^{2}}+x^{2}$ with eigenvalues
$2k+1,k=0,1,2,\cdots$. These functions form an orthonormal basis for
$L^{2}(\mathbb{R})$. The higher dimensional Hermite functions denoted by
$\Phi_{\alpha}$ are then obtained by taking tensor product of one dimensional
Hermite functions. Thus for any multi-index $\alpha\in\mathbb{N}_{0}^{n}$ and
$x\in\mathbb{R}^{n}$, we define
$\Phi_{\alpha}(x)=\prod_{j=1}^{n}h_{\alpha_{j}}(x_{j}).$ The family
$\\{\Phi_{\alpha}\\}$ is then an orthonormal basis for
$L^{2}(\mathbb{R}^{n})$. They are eigenfunctions of the Hermite operator
$H=-\Delta+|x|^{2}$ corresponding to eigenvalues $(2|\alpha|+n)$, where
$|\alpha|=\sum_{j=1}^{n}\alpha_{j}$.

###  2.2. Pseudo-Differential Operator

A linear differential operator $p(x,D)=\displaystyle\sum_{|\alpha|\leq
m}a_{\alpha}(x)\partial^{\alpha}$ with variable coefficient $a_{\alpha}$ can
be expressed as

$\displaystyle p(x,D)f(x)$
$\displaystyle=(2\pi)^{-\frac{n}{2}}\int_{\mathbb{R}^{n}}e^{ix\cdot\xi}p(x,\xi)\hat{f}(\xi)\,d\xi,\quad\forall
x\in\mathbb{R}^{n}$

where the Fourier transform of $f$ is defined by

$\displaystyle\hat{f}(\xi)$
$\displaystyle=(2\pi)^{-\frac{n}{2}}\int_{\mathbb{R}^{n}}{f}(x)e^{-ix\cdot\xi}\,dx,\quad\forall\xi\in\mathbb{R}^{n}.$

When $p(x,\xi)=a(x,\xi)$ is a measurable function on
$\mathbb{R}^{n}\times\mathbb{R}^{n}$, (not necessarily a polynomial) then the
corresponding operator $T_{a}$ associated with the function $a(x,\xi)$ is
given by

$\displaystyle T_{a}f(x)$
$\displaystyle=a(x,D)f(x)=(2\pi)^{-\frac{n}{2}}\int_{\mathbb{R}^{n}}e^{ix\cdot\xi}a(x,\xi)\hat{f}(\xi)\,d\xi,\quad\forall
x\in\mathbb{R}^{n}.$

The operator $T_{a}$ is called pseudo-differential operator and the function
$a(x,\xi)$ is called the symbol of the pseudo-differential operator $T_{a}$.
Let $m\in\mathbb{R},0\leq\delta<\rho\leq 1.$ Then the symbol class
$S_{\rho,\delta}^{m}$ consists of those functions $a(x,\xi)\in
C^{\infty}(\mathbb{R}^{n}\times\mathbb{R}^{n})$ satisfying

(2.1) $\displaystyle|\partial_{x}^{\alpha}\partial_{\xi}^{\beta}a(x,\xi)|\leq
C_{\alpha,\beta}\,(1+|\xi|^{2})^{\frac{{m-\delta|\alpha|+\rho|\beta|}}{2}}$

for all multi-indices $\alpha,\beta$. We take $\rho=1$ and $\delta=0$ through
out the paper and denote the symbol class by $S^{m}$.

The Weyl quantization $Op^{W}$ for a “reasonable” symbol $a$ in
$\mathbb{R}^{n}\times\mathbb{R}^{n}$ is given by

$Op^{W}(a)f(u)=(2\pi)^{-n}\int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}e^{i(u-v)\cdot\xi}a\left(\xi,\frac{u+v}{2}\right){f}(v)\,dv\,d\xi,\quad\forall
u\in\mathbb{R}^{n},$

for all Schwartz class functions $f$ on $\mathbb{R}^{n}$.

###  2.3. Heisenberg Group

One of the simple and natural example of non-abelian, non-compact group is the
famous Heisenberg group $\mathbb{H}^{n}$, which plays an important role in
several branches of mathematics. The Heisenberg group $\mathbb{H}^{n}$ is a
Nilpotent Lie group whose underlying manifold is $\mathbb{R}^{2n+1}$ and the
group operation is defined by

$(x,y,t)(x^{\prime},y^{\prime},t^{\prime})=(x+x^{\prime},y+y^{\prime},t+t^{\prime}+\frac{1}{2}(xy^{\prime}-x^{\prime}y)),$

where $(x,y,t)$ and $(x^{\prime},y^{\prime},t^{\prime})$ are in
$\mathbb{R}^{n}\times\mathbb{R}^{n}\times\mathbb{R}$. Moreover,
$\mathbb{H}^{n}$ is a unimodular Lie group on which the Haar measure is the
usual Lebesgue measure $\,dx\,dy\,dt.$ The canonical basis for the Lie algebra
$\mathfrak{h}_{n}$ of $\mathbb{H}^{n}$ is given by left-invariant vector
fields:

(2.2) $\displaystyle X_{j}$
$\displaystyle=\partial_{x_{j}}-\frac{y_{j}}{2}\partial_{t},$ $\displaystyle
Y_{j}=\partial_{y_{j}}+\frac{x_{j}}{2}\partial_{t},\quad j=1,2,\dotsc
n,~{}\mathrm{and}~{}T=\partial_{t},$

satisfying the commutator relation $[X_{j},Y_{j}]=T,\quad j=1,2,\dotsc n.$

The sublaplacian on the Heisenberg group is given by

$\mathcal{L}=\sum_{j=1}^{n}(X_{j}^{2}+Y_{j}^{2})=\sum_{j=1}^{n}\bigg{(}\bigg{(}\partial_{x_{j}}-\frac{y_{j}}{2}\partial_{t}\bigg{)}^{2}+\bigg{(}\partial_{y_{j}}+\frac{x_{j}}{2}\partial_{t}\bigg{)}^{2}\bigg{)}.$

By Stone-von Neumann theorem, the only infinite dimensional unitary
irreducible representations (up to unitary equivalence) are given by
$\pi_{\lambda}$, $\lambda$ in $\mathbb{R}^{*}$, where $\pi_{\lambda}$ is
defined by

$\pi_{\lambda}(x,y,t)f(u)=e^{i\lambda(t+\frac{1}{2}xy)}e^{i\sqrt{\lambda}yu}f(u+\sqrt{|\lambda|x}),\quad\forall
u\in\mathbb{R}^{n},\forall f\in
L^{2}(\mathbb{R}^{n})\mbox{~{}and~{}}(x,y,t)\in\mathbb{H}^{n}.$

We use the convention

$\sqrt{\lambda}:={\rm
sgn}(\lambda)\sqrt{|\lambda|}=\begin{cases}\sqrt{\lambda},~{}~{}~{}\lambda>0\\\
-\sqrt{|\lambda|},~{}~{}\lambda<0.\end{cases}$

For each ${\lambda}\in\mathbb{R}^{*}$, the group Fourier transform of $f\in
L^{1}(\mathbb{H}^{n})$ is a bounded linear map on $L^{2}(\mathbb{R}^{n})$
defined by

$\displaystyle\hat{f}(\lambda)\equiv\pi_{\lambda}(f)=\int_{\mathbb{H}^{n}}f(x,y,t)\pi_{\lambda}^{*}(x,y,t)\,dx\,dy\,dt.$

We denote $B(L^{2}(\mathbb{R}^{n}))$ to be the set of all bounded operators on
$L^{2}(\mathbb{R}^{n})$. If $f\in L^{2}(\mathbb{H}^{n})$, then
$\hat{f}(\lambda)$ is a Hilbert-Schmidt operator on $L^{2}({\mathbb{R}}^{n})$
and satisfies the Plancherel formula

$\int_{\mathbb{R}^{*}}\|\hat{f}(\lambda)\|_{B_{2}}^{2}\,d\mu(\lambda)=\|{f}\|_{L^{2}(\mathbb{H}^{n})},$

where $\|.\|_{B_{2}}$ stands for the norm in the Hilbert space $B_{2}$ of all
Hilbert-Schmidt operators on $L^{2}(\mathbb{R}^{n})$ and
$d\mu(\lambda)=c_{n}{|\lambda|}^{n}\,d\lambda$ where $c_{n}$ is a constant.

######  Theorem 2.1.

For all Schwartz class functions on $\mathbb{H}^{n}$, the following inversion
formula holds:

$\displaystyle f(g)$
$\displaystyle=\int_{\mathbb{R}^{*}}tr~{}(\pi_{\lambda}(g)\hat{f}(\lambda))\,d\mu(\lambda),\quad\forall
g\in\mathbb{H}^{n}.$

For a detailed study on the Heisenberg group we refer to Thangavelu [17].

###  2.4. Quantization and Pseudo-differential operator on the Heisenberg
group

Since the Fourier transform on $\mathbb{H}^{n}$ is operator valued, the symbol
of a pseudo-differential operator on the Heisenberg group is also operator
valued. We define the symbol class $S_{\rho,\delta}^{m}({\mathbb{H}^{n})}$ for
the Heisenberg group as in [12]. The symbol class
$S_{\rho,\delta}^{m}(\mathbb{H}^{n})$ is defined by the set of all symbols
$\sigma$ for which the following quantities are finite:

$\|\sigma\|_{{S_{\rho,\delta}^{m}},a,b,c}:=\sup_{{g\in{\mathbb{H}^{n}}},{\lambda\in\mathbb{R}^{*}}}\|\sigma(g,\lambda)\|_{S_{\rho,\delta}^{m}a,b,c},\quad
a,b,c\in\mathbb{N}_{0}$

where

(2.3)
$\|\sigma(g,\lambda)\|_{S_{\rho,\delta}^{m}a,b,c}:=\sup_{\begin{subarray}{c}[\alpha]\leq
a\\\ [\beta]\leq b\\\ [\gamma]\leq
c\end{subarray}}\|\pi_{\lambda}(I-\mathcal{L})^{\frac{\rho[\alpha]-m-\delta[\beta]+\gamma}{2}}X_{g}^{\beta}\Delta^{\prime\alpha}\sigma(g,\lambda)\pi_{\lambda}(I-\mathcal{L})^{-\frac{\gamma}{2}}\|_{op}$

with

$\alpha=(\alpha_{1},\alpha_{2},\alpha_{3})=(\alpha_{11},\alpha_{12}\dotsc\alpha_{1n},\alpha_{21},\alpha_{22}\dotsc\alpha_{2n},\alpha_{3})\in\mathbb{N}_{0}^{n}\times\mathbb{N}_{0}^{n}\times\mathbb{N}_{0}$

and $[\alpha]=|\alpha_{1}|+|\alpha_{2}|+2\alpha_{3}$. For each $\alpha$ we
write

$X^{\alpha}=X^{\alpha_{1}}Y^{\alpha_{2}}T^{\alpha_{3}},~{}\mathrm{where}~{}X^{\alpha_{1}}=X_{1}^{\alpha_{11}}X_{2}^{\alpha_{12}}\dotsc
X_{n}^{\alpha_{1n}}~{}\mathrm{and}~{}Y^{\alpha_{1}}=Y_{1}^{\alpha_{11}}Y_{2}^{\alpha_{12}}\cdots
Y_{n}^{\alpha_{1n}}$

and

$\Delta^{\prime\alpha}:=\Delta_{x}^{\alpha_{1}}\Delta_{y}^{\alpha_{2}}\Delta_{t}^{\alpha_{3}},~{}\mbox{where}~{}\Delta_{x}^{\alpha_{1}}=\Delta_{x_{1}}^{\alpha_{11}}\Delta_{x_{2}}^{\alpha_{12}}\cdots\Delta_{x_{n}}^{\alpha_{1n}},~{}\Delta_{y}^{\alpha_{2}}=\Delta_{y_{1}}^{\alpha_{21}}\Delta_{y_{2}}^{\alpha_{22}}\dotsc\Delta_{y_{n}}^{\alpha_{2n}},$

where the difference operators are defined (see [12] or [14]) as

$\Delta_{x_{j}}\hat{f}(\pi_{\lambda})=\pi_{\lambda}(x_{j}f),\Delta_{y_{j}}\hat{f}(\pi_{\lambda})=\pi_{\lambda}(y_{j}f)~{}~{}\mbox{and}~{}~{}\Delta_{t}\hat{f}(\pi_{\lambda})=\pi_{\lambda}(tf).$

The operator corresponding to the symbol $a$ is denoted by $Op(a)$ and is
defined by

(2.4) $\displaystyle Op(a)f(g)=(T_{a}f)(g)$
$\displaystyle=\int_{\mathbb{R}^{*}}{tr}~{}(\pi_{\lambda}^{*}(g)a(g,\lambda)\hat{f}(\lambda))\,d\mu(\lambda),$

for all Schwartz class functions $f$ on $\mathbb{H}^{n}$. The relation between
the Weyl quantization and the representations of the Heisenberg group enables
to work with the symbols in $S_{\rho,\delta}^{m}$ by
$\sigma(g,\lambda)=Op^{W}(a_{g,\lambda})$ which can be seen in the following
theorem.

######  Theorem 2.2.

[14] Let $\rho$, $\delta$ be real numbers such that $1\geq\rho\geq\delta\geq
0$ and $(\rho,\delta)\neq(0,0).$

  1. (1)

if $\sigma={\sigma(g,\lambda)}$ is in $S_{\rho,\delta}^{m}(\mathbb{H}^{n})$,
then there exist a smooth function $a=a(g,\lambda,\xi,u)=a_{g,\lambda}(\xi,u)$
on
$\mathbb{H}^{n}\times\mathbb{R}^{*}\times\mathbb{R}^{n}\times\mathbb{R}^{n}$
such that :

(2.5) $\displaystyle\sigma(g,\lambda)=Op^{W}(a_{g,\lambda}).$

It satisfies for any
$\alpha,\beta\in\mathbb{N}_{0}^{n},\bar{\beta}\in\mathbb{N}_{0}^{2n+1}$ and
$\bar{\alpha}\in\mathbb{N}_{0},$

(2.6)
$\displaystyle|\partial_{\xi}^{\alpha}\partial_{u}^{\beta}\tilde{\partial}_{\lambda,\xi,u}^{\bar{\alpha}}X_{g}^{\bar{\beta}}a_{g,\lambda}(\xi,u)|\leq
C_{\alpha,\beta,\bar{\alpha},\bar{\beta}}|\lambda|^{\rho{\frac{|\alpha|+|\beta|}{2}}}(1+|\lambda|(1+|\xi|^{2}+|u|^{2}))^{\frac{m-2\rho\bar{\alpha}+\delta[\bar{\beta}]-\rho(|\alpha|+|\beta|)}{2}},$

where the operator $\tilde{\partial}_{\lambda,\xi,u}$ was defined in [14].

  2. (2)

Conversely, if $a=\\{a{(g,\lambda,\xi,u)}=a_{g,\lambda}(\xi,u)\\}$ is a smooth
function on
$\mathbb{H}^{n}\times\mathbb{R}^{*}\times\mathbb{R}^{n}\times\mathbb{R}^{n}$
satisfying (2.6) for every $\alpha,\beta\in\mathbb{N}_{0}^{n}$ and
$\bar{\alpha}\in\mathbb{N}_{0}$, then there exist a unique symbol $\sigma\in
S_{\rho,\delta}^{m}(\mathbb{H}^{n})$ such that (2.5) holds.

In order to prove the Szegö limit theorem we restrict ourselves to a special
symbol class on the Heisenberg group
$S_{\rho,\delta,\Phi}^{m}(\mathbb{H}^{n})$, where
$\Phi(g,\lambda)=|\lambda|+V(g)$,

$\displaystyle\quad\quad S_{\rho,\delta,\Phi}^{m}(\mathbb{H}^{n}):$
$\displaystyle=$
$\displaystyle\left\\{\sigma(g,\lambda):|\partial_{\xi}^{\alpha}\partial_{u}^{\beta}\tilde{\partial}_{\lambda,\xi,u}^{\bar{\alpha}}X_{g}^{\bar{\beta}}a_{g,\lambda}(\xi,u)|\right.$
$\displaystyle\leq$
$\displaystyle\left.C_{\alpha,\beta,\bar{\alpha},\bar{\beta}}|\lambda|^{\rho{\frac{|\alpha|+|\beta|}{2}}}(1+\Phi(g,\lambda)(1+|\xi|^{2}+|u|^{2}))^{\frac{m-2\rho\bar{\alpha}+\delta[\bar{\beta}]-\rho(|\alpha|+|\beta|)}{2}}\right\\}$

and $\sigma(g,\lambda),a_{g,\lambda}$ are related by (2.5). Thus symbol class
$S_{\rho,\delta,\Phi}^{m}(\mathbb{H}^{n})$ is defined by the set of all
symbols $\sigma$ for which the following quantities are finite:

$\|\sigma\|_{{S_{\rho,\delta,\Phi}^{m}},a,b,c}:=\sup_{g\in{\mathbb{H}^{n}},\lambda\in\mathbb{R}^{*}}\|\sigma(g,\lambda)\|_{S_{\rho,\delta,\Phi}^{m}a,b,c},\quad
a,b,c\in\mathbb{N}_{0}$

where

(2.8)
$\|\sigma(g,\lambda)\|_{S_{\rho,\delta,\Phi}^{m}a,b,c}:=\sup_{\begin{subarray}{c}[\alpha]\leq
a\\\ [\beta]\leq b\\\ [\gamma]\leq
c\end{subarray}}\|(\pi_{\lambda}(I-\mathcal{L})+V(g))^{\frac{\rho[\alpha]-m-\delta[\beta]+\gamma}{2}}X_{g}^{\beta}\Delta^{\prime\alpha}\sigma(g,\lambda)(\pi_{\lambda}(I-\mathcal{L})+V(g))^{-\frac{\gamma}{2}}\|_{op}.$

We take $\rho=1$ and $\delta=0$ throughout the article and denote the symbol
classes $S_{1,0}^{m}(\mathbb{H}^{n})$ and $S_{1,0,\Phi}^{m}(\mathbb{H}^{n})$
by $S^{m}(\mathbb{H}^{n})$ and $S_{\Phi}^{m}(\mathbb{H}^{n})$ respectively.
Note that if $\sigma(g,\lambda)\in S_{\Phi}^{m}(\mathbb{H})$ then
$\|\sigma\|_{{S_{\rho,\delta}^{m}},a,b,c}<\infty$ for all
$a,b,c\in\mathbb{N}_{0}$. For example, for each $\lambda\in\mathbb{R}^{*},$
the operators $V(g)I,\lambda I,|\lambda|H+V(g)I\in
S_{\Phi}^{2}(\mathbb{H}^{n}).$

###  2.5. Composition of symbols

In this section we discuss the composition formula for two pseudo-differential
operators in different symbol classes in our setup and estimate the difference
of the symbol of the composition with the composition of the corresponding
symbols of the pseudo-differential operators on $\mathbb{H}^{n}.$

######  Theorem 2.3.

Let $a\in S_{\Phi}^{\mu}(\mathbb{H}^{n})$ and $b\in
S_{\Phi}^{\nu}(\mathbb{H}^{n})$. Then the composition $Op(a)\circ Op(b)$ is a
pseudo-differential operator with symbol $a\\#_{\mathbb{H}^{n}}b\in
S_{\Phi}^{\mu+\nu}(\mathbb{H}^{n})$. Moreover $a\\#_{\mathbb{H}^{n}}b$ has the
following asymptotic formula:

$\displaystyle
a\\#_{\mathbb{H}^{n}}b=b\\#a+\frac{1}{2\sqrt{\lambda}}\sum_{1\leq j\leq
n}X_{j}b\\#T_{j}a+Y_{j}b\\#T_{j}^{{}^{\prime}}a$ $\displaystyle+$
$\displaystyle\frac{1}{8|\lambda|}\sum_{1\leq j,k\leq
n}X_{j}X_{k}b\\#T_{j}T_{k}a+Y_{j}Y_{k}b\\#T_{j}^{{}^{\prime}}T_{k}^{{}^{\prime}}a+{X_{j}}Y_{k}b\\#T_{j}T_{k}^{{}^{\prime}}a+Y_{j}{X_{k}}b\\#T_{j}^{{}^{\prime}}T_{k}a$
$\displaystyle+$ $\displaystyle R_{2},$

where $R_{2}$ depends only on $X^{\alpha}$ for $\alpha\geq 2$ and

(2.10) $\displaystyle\quad\quad
a\\#b=\sum_{s=0}^{N}\frac{1}{s!}\Big{(}{\frac{i}{2}}\Big{)}^{s}a\Big{(}{\overleftarrow{\frac{\partial}{\partial\xi}}\overrightarrow{\frac{\partial}{\partial
u}}-\overrightarrow{\frac{\partial}{\partial\xi}}\overleftarrow{\frac{\partial}{\partial
u}}\Big{)}^{s}b}+S_{N}$

where $S_{N}\in S^{|\mu|+|\nu|-N}$ with $T_{j}a=-\partial_{u_{j}}a$ and
$T_{j}^{{}^{\prime}}a=\partial_{\xi_{j}}a.$

###### Proof.

The proof follows along the similar lines as in Proposition 3.5, page 57 of
[2], with appropriate modifications. ∎

######  Corollary 2.4.

Let $m\leq 0$, $a(g,\lambda)\in S_{\Phi}^{m}(\mathbb{H}^{n})$ and
$b(g,\lambda)=(|\lambda|H+V(g)+r_{2}I)^{-1}\in S_{\Phi}^{-2}(\mathbb{H}^{n})$.
Then $a\\#_{\mathbb{H}^{n}}b(g,\lambda)\approx b(g,\lambda)\circ
a(g,\lambda)+R(g,\lambda)$, where
$\|R(g,\lambda)\|_{B(L^{2}(\mathbb{R}^{n}))}\rightarrow 0$ as
$r_{2}\rightarrow\infty$.

###### Proof.

Using (2.4) and (2.10) we have

$\displaystyle\quad\quad|a\\#b(\xi,u)-b(\xi,u)a(\xi,u)|$ $\displaystyle\leq$
$\displaystyle\sum_{j=1}^{\infty}\frac{1}{2^{s}s!}\left|a(\xi,u)\Big{(}{\overleftarrow{\frac{\partial}{\partial\xi}}\overrightarrow{\frac{\partial}{\partial
u}}-\overrightarrow{\frac{\partial}{\partial\xi}}\overleftarrow{\frac{\partial}{\partial
u}}\Big{)}^{s}b(\xi,u)}\right|$ $\displaystyle\leq$
$\displaystyle\frac{1}{r_{2}}\sum_{s=1}^{\infty}\frac{1}{2^{s}s!}\frac{1}{r_{2}^{s/2}}.$

Therefore $\|a\\#b-ab\|=O(r_{2}^{-1})$. Again using the above estimate
repeatedly in (2.3) and using the fact that $n<<r_{2},$ we have

$\displaystyle\quad\quad\|a\\#_{\mathbb{H}^{n}}b-ba\|_{B(L^{2}(\mathbb{R}^{n}))}$
$\displaystyle\leq$
$\displaystyle\frac{1}{r_{2}}\sum_{s=0}^{\infty}\left(\frac{2n}{r_{2}}\right)^{s}\to
0\quad\mbox{as}\quad r_{2}\to\infty.$

∎

######  Lemma 2.5.

Let $m\leq 0$ and $a\in S_{\Phi}^{m}(\mathbb{H}^{n})$. Then for sufficiently
large $N\in\mathbb{N}$, $g,g_{1}\in\mathbb{H}^{n},h\in L^{2}(\mathbb{H}^{n})$
and for $B_{1}\in B(L^{2}(\mathbb{R}^{n}))$ we have

  1. (1)

$\left|tr(\pi_{\lambda}^{*}(g)a(g,\lambda)B_{1})\right|\leq
M|{tr}~{}\big{(}\pi_{\lambda}(I-\mathcal{L})^{-2N}\big{)}|\|a(g,\lambda)\|\|B_{1}\|$.

  2. (2)

$\left|\displaystyle\int_{\mathbb{R}^{*}}~{}tr(\pi_{\lambda}^{*}(g)a(g,\lambda)\hat{h}(\lambda))d\mu(\lambda)\right|\leq
C\|h\|_{2}.$

Proof. For any $N\in\mathbb{N},$ we make use of the identity

(2.11) $\displaystyle\pi_{\lambda}(g)=T^{4N}(\lambda^{-4N}I)\pi_{\lambda}(g)$

to obtain the required bound, where the vector field
$T=\frac{\partial}{\partial t}$. For $h\in L^{2}(\mathbb{H}^{n})$, consider

$\displaystyle|{tr}~{}\big{(}\pi_{\lambda}^{*}(g)~{}a(g,\lambda)B_{1}\big{)}|$
$\displaystyle=$
$\displaystyle|{tr}~{}\big{(}T^{4N}(\lambda^{-4N}I)\pi_{\lambda}^{*}(g)~{}a(g,\lambda)B_{1}\big{)}|$
$\displaystyle=$
$\displaystyle|{tr}~{}\big{(}\pi_{\lambda}(I-\mathcal{L})^{-4N}\pi_{\lambda}(I-\mathcal{L})^{4N}T^{4N}(\lambda^{-4N}I)\pi_{\lambda}^{*}(g)~{}a(g,\lambda)B_{1}\big{)}|$
$\displaystyle\leq$
$\displaystyle|{tr}~{}\big{(}\pi_{\lambda}(I-\mathcal{L})^{-2N}\big{)}|\|\pi_{\lambda}(I-\mathcal{L})^{4N}T^{4N}(\lambda^{-4N}I)\pi_{\lambda}^{*}(g)\|a(g,\lambda)\|\|B_{1}\|.$

Since $(\lambda^{-4N}I)\pi_{\lambda}^{*}(g)$ as a pseudo-differential operator
of order $-8N$ on $\mathbb{H}^{n}$, the operator norm of
$\pi_{\lambda}(I-\mathcal{L})^{4N}T^{4N}(\lambda^{-4N}I)\pi_{\lambda}^{*}(g)$
is uniformly bounded by $M$(say) (see (2.8)). Using the above fact we have
$|{tr}~{}\big{(}\pi_{\lambda}^{*}(g)a(g,\lambda)\pi_{\lambda}(g_{1})\big{)}|\leq
M|{tr}~{}\big{(}\pi_{\lambda}(I-\mathcal{L})^{-2N}\big{)}|\|a(g,\lambda)\|\|B_{1}\|.$
This proves part (a).

For part (b), choose sufficiently large $N$ such that the sum
$\displaystyle\sum_{\alpha}\frac{1}{(2|\alpha|+n)^{N}}$ and the integral
$\displaystyle\int_{\mathbb{R}^{*}}\frac{\,d\mu(\lambda)}{(1+|\lambda|)^{4N}}$
converges. Since $\pi_{\lambda}(I-\mathcal{L})=1+|\lambda|H$, we have

$\displaystyle
tr~{}(1+|\lambda|H)^{-2N}=\sum_{\alpha}\langle(1+|\lambda|H)^{-2N}\Phi_{\alpha},\Phi_{\alpha}\rangle$
$\displaystyle=\sum_{\alpha}\frac{1}{(1+|\lambda|(2|\alpha|+n))^{2N}}$
$\displaystyle\leq\sum_{\alpha}\frac{1}{(1+|\lambda|)^{N}(2|\alpha|+n))^{N}}$
$\displaystyle=\frac{1}{(1+|\lambda|)^{N}}\sum_{\alpha}\frac{1}{(2|\alpha|+n))^{N}}.$

Thus by part $(a)$ we have

$\displaystyle\int_{\mathbb{R}^{*}}\left|{tr}~{}\left(\pi_{\lambda}^{*}(g)a(g,\lambda)\widehat{h}(\lambda)\right)\right|\,d\mu(\lambda)$
$\displaystyle\leq
C\int_{\mathbb{R}^{*}}\|\widehat{h}(\lambda)\|~{}tr~{}\left(\pi_{\lambda}(I-\mathcal{L})^{-4N}\right)\,d\mu(\lambda)$
$\displaystyle\leq
C\sum_{\alpha}\frac{1}{(2|\alpha|+n)^{2N}}\int_{\mathbb{R}^{*}}\|\widehat{h}(\lambda)\|_{B_{2}}~{}(1+|\lambda|)^{-2N}\,d\mu(\lambda)$
$\displaystyle\leq
C\|{h}\|_{2}~{}\left(\int_{\mathbb{R}^{*}}(1+|\lambda|)^{-4N}\,d\mu(\lambda)\right)^{\frac{1}{2}}<\infty.$

∎

######  Corollary 2.6.

Under the assumptions of Corollary 2.4 one has

  1. (1)

$\|Op(a\\#_{\mathbb{H}^{n}}b)-Op(b\circ a)\|_{B(L^{2}(\mathbb{H}^{n}))}\to 0$
as $r_{2}\to\infty.$

  2. (2)

For $\ell\in\mathbb{N}$,
$\|[Op(b)]^{\ell}-Op(b^{\ell})\|_{B(L^{2}(\mathbb{H}^{n}))}\to 0$ as
$r_{2}\to\infty.$

###### Proof.

Take $h\in L^{1}\cap L^{2}(\mathbb{H}^{n})$. Since
$a\\#_{\mathbb{H}^{n}}b-ba\in S_{\Phi}^{0}(\mathbb{H}^{n}),$ applying part
$(1)$ of Lemma 2.5 and Corollary 2.4 we get

$\displaystyle|Op(a\\#_{\mathbb{H}^{n}}b)-Op(b\circ a)h(g)|$ $\displaystyle=$
$\displaystyle\left|\int_{\mathbb{R}^{*}}{tr}~{}\left(\pi_{\lambda}^{*}(g)[a\\#_{\mathbb{H}^{n}}b(g,\lambda)-(b\circ
a)(g,\lambda)]\hat{h}(\lambda)\right)\,d\mu(\lambda)\right|$
$\displaystyle\leq$
$\displaystyle\frac{C}{r_{2}}\int_{\mathbb{R}^{*}}\|\hat{h}(\lambda)\|_{B_{2}}\frac{1}{(1+|\lambda|)^{N-1}}\,d\mu(\lambda).$

An application of Cauchy-Schwarz inequality proves part (1) by choosing
sufficiently large $N$. To prove $(2)$ we use mathematical induction on
$\ell$. For $\ell=1$, the hypothesis is trivially true. Assume that $(2)$
holds for $[Op(b)]^{\ell-1}$. Using the composition formula (2.3) and
Corollary 2.4 we have

$(b^{\ell-1}\\#_{\mathbb{H}^{n}}b)(g,\lambda)\approx
b^{\ell-1}(g,\lambda)\circ b(g,\lambda)+R_{\ell}(g,\lambda),$

where $\|R_{\ell}(g,\lambda)\|_{B(L^{2}(\mathbb{R}^{n}))}\to 0$ as
$r_{2}\to\infty.$ This proves $(2)$. ∎

Further, using the asymptotic formula (2.3), we prove the that the $\ell$-fold
composition of $Op(a)\in S_{\phi}^{0}(\mathbb{H}^{n})$ with itself is
asymptotically equal to $Op(a^{\ell})$ in the following lemma.

######  Lemma 2.7.

Let $A$ be the pseudo-differential operator on $L^{2}(\mathbb{H}^{n})$ with
symbol $a(g,\lambda)\in S_{\Phi}^{0}(\mathbb{H}^{n})$. Then for any
$\ell\in\mathbb{N}$, the symbol $a_{\ell}(g,\lambda)$ of the operator
$A^{\ell}$ can be written as

$a_{\ell}(g,\lambda)=\left(a(g,\lambda)\right)^{\ell}+E_{l}(g,\lambda)$

with $E_{\ell}(g,\lambda)\in B(L^{2}(\mathbb{R}^{n})).$

Proof. We prove the lemma using mathematical induction on $\ell.$ For $\ell=1$
our hypothesis holds trivially. Assume that our hypothesis is true for the
symbol $a_{\ell-1}$. That means
$a_{\ell-1}(g,\lambda)=\left(a(g,\lambda)\right)^{\ell-1}+E_{\ell-1}(g,\lambda)$
with $E_{\ell-1}(g,\lambda)\in B(L^{2}(\mathbb{R}^{n})).$ But by (2.3) and
(2.10) we have $a_{\ell}\approx(a_{\ell-1}\\#a)+T_{1}\approx
a^{\ell}+S_{1}+S_{2},$ where $S_{1}=\frac{1}{2\sqrt{\lambda}}\sum_{1\leq j\leq
n}X_{j}b\\#T_{j}a+Y_{j}b\\#T_{j}^{{}^{\prime}}a+\frac{1}{8|\lambda|}\sum_{1\leq
j,k\leq
n}X_{j}X_{k}b\\#T_{j}T_{k}a+Y_{j}Y_{k}b\\#T_{j}^{{}^{\prime}}T_{k}^{{}^{\prime}}a+{X_{j}}Y_{k}a\\#T_{j}T_{k}^{{}^{\prime}}a+Y_{j}{X_{k}}b\\#T_{j}^{{}^{\prime}}T_{k}a+R_{2}$
and
$S_{2}(\xi,u)=\displaystyle\sum_{s=1}^{\infty}\frac{1}{2^{s}s!}a_{\ell-1}(\xi,u)\Big{(}{\overleftarrow{\frac{\partial}{\partial\xi}}\overrightarrow{\frac{\partial}{\partial
u}}-\overrightarrow{\frac{\partial}{\partial\xi}}\overleftarrow{\frac{\partial}{\partial
u}}\Big{)}^{s}a(\xi,u)}.$

Since

$\displaystyle\left|\frac{\partial^{s_{1}}}{\partial\xi^{s_{1}}}\frac{\partial^{s_{2}}}{\partial
u^{s_{2}}}X_{g}^{\alpha}a_{\ell-1}(\xi,u)\right|\leq
C_{s_{1},s_{2}}|\lambda|^{\frac{s_{1}+s_{2}}{2}}((1+\Phi(g,\lambda))(1+|\xi|^{2}+|u|^{2}))^{-{\frac{s_{1}+s_{2}}{2}}},$

and

$\displaystyle\left\|\frac{\partial^{s_{1}}}{\partial\xi^{s_{1}}}\frac{\partial^{s_{2}}}{\partial
u^{s_{2}}}{T_{j}^{{}^{\prime}}}^{\beta}T_{k}^{\gamma}a\right\|\leq
C_{s_{1},s_{2}}{|\lambda|^{\frac{|\beta|+|\gamma|+s_{1}+s_{2}}{2}}}{(1+\Phi(g,\lambda)(1+|\xi|^{2}+|u|^{2}))^{-\frac{|\beta|+|\gamma|+s_{1}+s_{2}}{2}}},$

we have

(2.12) $\displaystyle\|S_{2}\|\leq
C\sum_{s=0}^{\infty}\dfrac{|\lambda|^{s}}{{\|(1+|\lambda|H)}\|^{s}}<\infty$

and

$\displaystyle\|X_{j}a_{\ell-1}\\#{T_{j}}a\|$ $\displaystyle\leq
C\dfrac{n|\lambda|^{\frac{1}{2}}}{{\|(1+|\lambda|H)}\|^{\frac{1}{2}}}\displaystyle\sum_{s=0}^{\infty}\left(\dfrac{\lambda}{{\|(1+|\lambda|H)}\|}\right)^{s}.$

Again

$\displaystyle\|X_{j}X_{k}a_{\ell-1}\\#{T_{j}}T_{k}a\|$
$\displaystyle\leq\dfrac{nC_{j}|\lambda|}{{\|(1+|\lambda|H)}\|}\displaystyle\sum_{s=0}^{\infty}\left(\dfrac{\lambda}{{\|(1+|\lambda|H)}\|}\right)^{s}.$

Therefore

$\displaystyle\|S_{1}\|$ $\displaystyle\leq
C\left[\sum_{s=0}^{\infty}\left(\dfrac{\lambda}{{\|(1+|\lambda|H)}\|}\right)^{s}\right]\left(\dfrac{n^{2}}{{\|(1+|\lambda|H)}\|^{\frac{1}{2}}}+\dfrac{n^{3}}{{\|(1+|\lambda|H)}\|}+\dfrac{n^{4}}{{\|(1+|\lambda|H)}\|^{\frac{3}{2}}}+\cdots\right)$
$\displaystyle\leq
C\dfrac{n^{2}}{{\|(1+|\lambda|H)}\|^{\frac{1}{2}}}\sum_{s=0}^{\infty}\left(\dfrac{\lambda}{{\|(1+|\lambda|H)}\|}\right)^{s}\displaystyle\sum_{l=0}^{\infty}\left(\dfrac{1}{{\|(1+|\lambda|H)}\|^{\frac{1}{4}}}\right)^{l}.$

Consequently $T_{2}$ is a bounded operator and
$\|a_{l}-a^{l}\|_{B(L^{2}(\mathbb{R}^{n}))}<\infty.$ ∎

##  3\. Szegö limit theorem for $\mathcal{H}_{0}$

We need to observe the following facts before proving Theorem 1.1.

######  Lemma 3.1.

Let $M_{b}$ be the multiplication operator defined in Theorem 1.1, then
${tr}{f(\pi_{r_{0}}M_{b}\pi_{r_{0}})}={tr}{(\pi_{r_{0}}f(M_{b})\pi_{r_{0}})}$
for any $f\in C(\mathbb{R})$.

Proof. Notice that
$\|(I-\pi_{r_{0}})M_{b}\pi_{r_{0}}\|_{B_{2}}^{2}={tr}~{}{(\pi_{r_{0}}M_{b}(I-\pi_{r_{0}})M_{b}\pi_{r_{0}})}={tr}~{}{(\pi_{r_{0}}M_{b}^{2}\pi_{r_{0}})}-{tr}~{}{(\pi_{r_{0}}M_{b}\pi_{r_{0}})^{2}}.$
The operator $\pi_{r_{0}}M_{b}^{2}\pi_{r_{0}}$ is an operators on
$L^{2}(\mathbb{H}^{n})$ with kernel
$K_{1}(g,g_{1})=\displaystyle\sum_{k_{1},k_{2}\leq r_{0}}\langle
b^{2}e_{k_{1}},e_{k_{2}}\rangle e_{k_{2}}(g)\overline{e_{k_{1}}(g_{1})}$, for
any orthonormal basis $\\{e_{k}\\}$ of $L^{2}(\mathbb{H}^{n})$.

Therefore
${tr}~{}{(\pi_{r_{0}}M_{b}^{2}\pi_{r_{0}})}=\int_{\mathbb{H}^{n}}K_{1}(g,g)dg=\displaystyle\sum_{k\leq
r_{0}}\langle b^{2}e_{k},e_{k}\rangle$. Further,
${tr}~{}{(\pi_{r_{0}}M_{b}\pi_{r_{0}})^{2}}=\int_{\mathbb{H}^{n}}K_{2}(g,g)dg=\displaystyle\sum_{k\leq
r_{0}}\langle b^{2}e_{k},e_{k}\rangle,$ where the operator
$\pi_{r_{0}}M_{b}\pi_{r_{0}}M_{b}\pi_{r_{0}}$ is an integral operator with
kernel $K_{2}(g,g_{1})=\displaystyle\sum_{k_{1},k_{2},k_{3}\leq r_{0}}\langle
be_{k_{1}},e_{k_{2}}\rangle\langle be_{k_{2}},e_{k_{3}}\rangle
e_{k_{3}}(g)\overline{e_{k_{1}}(g_{1})}$. So
${tr}~{}{(\pi_{r_{0}}M_{b}^{2}\pi_{r_{0}})}={tr}~{}{(\pi_{r_{0}}M_{b}\pi_{r_{0}})^{2}}.$
So $\|(I-\pi_{r_{0}})M_{b}\pi_{r_{0}}\|_{B_{2}}^{2}=0$.

Observe that for $n\in\mathbb{N}$,
$\pi_{r_{0}}M_{b}^{n}\pi_{r_{0}}=\pi_{r_{0}}M_{b}(\pi_{r_{0}}+(I-\pi_{r_{0}}))M_{b}\cdots
M_{b}\pi_{r_{0}}=(\pi_{r_{0}}M_{b}\pi_{r_{0}})^{n}$\+ terms with a factor of
$(I-\pi_{r_{0}})M_{b}\pi_{r_{0}}$. By Cauchy-Schwarz inequality,
$tr(\mbox{terms with a factor of}$ $(I-\pi_{r_{0}})M_{b})$ is dominated by
some constant (depending on $b$) times
$\|(I-\pi_{r_{0}})M_{b}\pi_{r_{0}}\|_{B_{2}}$. Therefore
$|{tr}~{}{(\pi_{r_{0}}M_{b}^{n}\pi_{r_{0}})}-{tr}~{}{(\pi_{r_{0}}M_{b}\pi_{r_{0}})^{n}}|=0$.
Thus
${tr}~{}{f(\pi_{r_{0}}M_{b}\pi_{r_{0}})}={tr}~{}{(\pi_{r_{0}}f(M_{b})\pi_{r_{0}})}$
for $f(x)=x^{n}$, $\forall n\in\mathbb{N}$ and this result can be extended to
continuous functions as an application of the Weierstrass approximation
theorem and spectral theorem. ∎

######  Lemma 3.2.

For $r_{0}>0$ define $I_{r_{0}}:L^{2}(\mathbb{H}^{n})\rightarrow
L^{2}(\mathbb{H}^{n})$ by
$I_{r_{0}}(\phi)(g)=\displaystyle\int_{-r_{0}}^{r_{0}}tr(\pi_{\lambda}^{*}(g)\hat{\phi}(\lambda))d\mu(\lambda)$.
Then

(3.1)
$\displaystyle\lim_{r_{0}\to\infty}\frac{{tr}~{}{(\pi_{r_{0}}f(M_{b})\pi_{r_{0}})}}{tr~{}(\pi_{r_{0}})}$
$\displaystyle=\lim_{r_{0}\to\infty}\frac{{tr}~{}{(\pi_{r_{0}}I_{r_{0}}f(M_{b})\pi_{r_{0}}I_{r_{0}})}}{tr~{}(\pi_{r_{0}}I_{r_{0}})}.$

Proof. We know that if $X$ is a positive trace class operator and $Y$ is a
bounded operator on $L^{2}(\mathbb{H}^{n})$ then
$|tr~{}(XYX)|\leq\|Y\||tr~{}(X^{2}).$ Using this inequality we get
$\big{|}tr~{}(\pi_{r_{0}})-tr~{}(\pi_{r_{0}}I_{r_{0}})\big{|}=\big{|}tr~{}(\pi_{r_{0}}(I-I_{r_{0}}))\big{|}\leq\|I-I_{r_{0}}\||tr~{}(\pi_{r_{0}})|.$
But for $\psi\in L^{2}(\mathbb{H}^{n})$, an application of Plancherel formula
gives
$\|(I-I_{r_{0}})\psi\|_{2}^{2}=\displaystyle\int_{|\lambda|>r_{0}}\|\hat{\psi}(\lambda)\|_{B_{2}}^{2}\,d\mu(\lambda)\rightarrow
0~{}\mathrm{as}~{}r_{0}\rightarrow\infty.$ Therefore,

(3.2)
$\displaystyle\bigg{|}\frac{tr~{}(\pi_{r_{0}}I_{r_{0}})}{tr~{}(\pi_{r_{0}})}-1\bigg{|}\leq\|I-I_{r_{0}}\|\rightarrow
0~{}\mathrm{as}~{}r_{0}\rightarrow\infty.$

We add a suitable constant to make the operator $M_{b}$ positive and any $f\in
C(\mathbb{R})$ can be written as the difference of two positive functions
namely the positive and the negative part of $f$. So without loss of
generality we take $f(M_{b})$ as a positive operator. Further,

$\displaystyle\big{|}{tr}~{}{(\pi_{r_{0}}f(M_{b})\pi_{r_{0}})-{tr}~{}(\pi_{r_{0}}I_{r_{0}}f(M_{b})\pi_{r_{0}}I_{r_{0}})}\big{|}$
$\displaystyle=\big{|}{tr}~{}(\pi_{r_{0}}f(M_{b})\pi_{r_{0}}(I-I_{r_{0}}))-{tr}~{}{(\pi_{r_{0}}(I-I_{r_{0}})f(M_{b})\pi_{r_{0}}I_{r_{0}})}\big{|}$
$\displaystyle\leq\big{|}{tr}~{}(\pi_{r_{0}}f(M_{b})\pi_{r_{0}})\big{|}\|I-I_{r_{0}}\|+\big{|}{tr}~{}{(\pi_{r_{0}}f(M_{b})\pi_{r_{0}}I_{r_{0}})}\big{|}\|I-I_{r_{0}}\|.$

Therefore,

(3.3)
$\displaystyle\bigg{|}\frac{{tr}~{}(\pi_{r_{0}}I_{r_{0}}f(M_{b})\pi_{r_{0}}I_{r_{0}})}{{tr}~{}(\pi_{r_{0}}f(M_{b})\pi_{r_{0}})}-1\bigg{|}\leq\left(1+\bigg{|}\frac{{tr}~{}{(\pi_{r_{0}}f(M_{b})\pi_{r_{0}}I_{r_{0}})}}{{tr}~{}(\pi_{r_{0}}f(M_{b})\pi_{r_{0}})}\bigg{|}\right)\|I-I_{r_{0}}\|\rightarrow
0,$

as $r_{0}\rightarrow\infty$. Combining (3.2) and (3.3) we get (3.1). ∎

Proof of theorem 1.1:
The operator $\pi_{r_{0}}I_{r_{0}}f(M_{b})\pi_{r_{0}}I_{r_{0}}$ is an integral
operator with kernel

$K_{r_{0}}(g,g_{1})=\int_{-r_{0}}^{r_{0}}{tr}~{}(\pi_{\lambda}^{*}(g)I_{r_{0}\times
r_{0}}f(b(g_{1}))\pi_{\lambda}(g_{1}))\,d\mu(\lambda).$

Therefore

$\displaystyle{tr}~{}{(\pi_{r_{0}}I_{r_{0}}f(M_{b})\pi_{r_{0}}I_{r_{0}})}=\int_{\mathbb{H}^{n}}K_{r_{0}}(g,g)\,dg$
$\displaystyle=r_{0}\int_{-r_{0}}^{r_{0}}d\mu(\lambda)\int_{\mathbb{H}^{n}}f(b(g))\,dg$

and ${tr}~{}{(\pi_{r_{0}}I_{r_{0}})}=\displaystyle\sum_{i\leq
r_{0}}\langle\widehat{I_{r_{0}}\phi_{i}},\hat{\phi_{i}}\rangle=\sum_{i\leq
r_{0}}\langle\hat{\phi_{i}}\mathcal{X}_{[-r_{0},r_{0}]},\hat{\phi_{i}}\rangle=\sum_{i\leq
r_{0}}\int_{-r_{0}}^{r_{0}}{tr}~{}(\hat{\phi_{i}}^{*}(\lambda)\hat{\phi_{i}}(\lambda))\,d\mu(\lambda)=r_{0}\int_{-r_{0}}^{r_{0}}\,d\mu(\lambda),$
for any orthonormal basis $\\{\phi_{i}\\}_{i=1}^{\infty}$ of
$L^{2}(\mathbb{H}^{n}).$ Thus

$\displaystyle\lim_{r_{0}\to\infty}\frac{{tr}~{}{(\pi_{r_{0}}f(M_{b})\pi_{r_{0}})}}{tr~{}(\pi_{r_{0}})}$
$\displaystyle=\lim_{r_{0}\to\infty}\frac{{tr}~{}{(\pi_{r_{0}}I_{r_{0}}f(M_{b})\pi_{r_{0}}I_{r_{0}})}}{tr~{}(\pi_{r_{0}}I_{r_{0}})}=\int_{\mathbb{H}^{n}}f(b(g))\,dg.$

##  4\. Szegö limit theorem for $\mathcal{H}_{1}$

Consider the operators $\mathcal{H}_{1}$ and $V$ as defined in Theorem 1.2.
Since the operators $(\mathcal{H}_{1}+r_{1}I)^{-1}$ and $(V+r_{1}I)^{-1}$ are
compact for $r_{1}>0$, we choose a suitable $m\in\mathbb{N}$ such that
$(\mathcal{H}_{1}+r_{1}I)^{-m}$ and $(V+r_{1}I)^{-m}$ are trace class
operators on $L^{2}(\mathbb{H}^{n})$. We observe the following facts before
proving Theorem 1.2.

######  Lemma 4.1.

Consider the self-adjoint operators $V$ and $\mathcal{H}_{1}$ as defined in
Theorem 1.2. Then
(a)

$\displaystyle\bigg{|}\frac{tr~{}((\mathcal{H}_{1}+r_{1}I)^{-m})}{tr~{}((V+r_{1}I)^{-m})}-1\bigg{|}\rightarrow
0~{}\mathrm{as}~{}r_{1}\rightarrow\infty.$

(b) If $B$ is any bounded operator on $L^{2}(\mathbb{H}^{n})$, then

$\displaystyle\bigg{|}\frac{tr~{}(B(\mathcal{H}_{1}+r_{1}I)^{-m})}{tr~{}(B(V+r_{1}I)^{-m})}-1\bigg{|}\rightarrow
0~{}\mathrm{as}~{}r_{1}\rightarrow\infty.$

Proof. Without loss of generality we prove the result the positive operator
$B$ by adding a suitable constant $c>0$ which makes the operator $B+cI$
positive.

(a) Since $B$ and $(V+r_{1}I)^{-1}$ are bounded and positive operators, we
have

(4.1)
$\displaystyle(\mathcal{H}_{1}+r_{1}I)=(V+r_{1}I)^{\frac{1}{2}}((V+r_{1}I)^{-\frac{1}{2}}(B)(V+r_{1}I)^{-\frac{1}{2}}+1)(V+r_{1}I)^{\frac{1}{2}}.$

Therefore

(4.2)
$\displaystyle(\mathcal{H}_{1}+r_{1}I)^{-m}=(V+r_{1}I)^{-m}+(V+r_{1}I)^{-\frac{m}{2}}((1+K_{r_{1}})^{-m}-1)(V+r_{1}I)^{-\frac{m}{2}},$

where $K_{r_{1}}=(V+r_{1}I)^{-\frac{1}{2}}(B)(V+r_{1}I)^{-\frac{1}{2}}$. Here
$K_{r_{1}}$ is a positive operator and $\|(I+K_{r_{1}})^{-1}\|\leq 1,$ for any
$r_{1}>0.$ Thus

$\displaystyle~{}\big{|}tr~{}\big{(}(\mathcal{H}_{1}+r_{1}I)^{-m})-{tr~{}((V+r_{1}I)^{-m}\big{)}}\big{|}=$
$\displaystyle~{}\big{|}tr~{}\big{(}(V+r_{1}I)^{-\frac{m}{2}}((1+K_{r_{1}})^{-m}-1)(V+r_{1}I)^{-\frac{m}{2}}\big{)}\big{|}$
$\displaystyle\leq$
$\displaystyle~{}tr~{}\big{(}(V+r_{1}I)^{-m}\big{)}\big{\|}((1+K_{r_{1}})^{-m}-1)\big{\|}$
$\displaystyle\leq$
$\displaystyle~{}m\big{\|}K_{r_{1}}\big{\|}tr~{}\big{(}(V+r_{1}I)^{-m}\big{)}$
$\displaystyle\leq$
$\displaystyle~{}m\big{\|}B\big{\|}\big{\|}(V+r_{1}I)^{-1}\big{\|}tr~{}\big{(}(V+r_{1}I)^{-m}\big{)}.$

Therefore,

$\displaystyle\bigg{|}\frac{tr~{}((\mathcal{H}_{1}+r_{1}I)^{-m})}{tr~{}((V+r_{1}I)^{-m})}-1\bigg{|}\leq\big{\|}B\big{\|}\big{\|}(V+r_{1}I)^{-1}\big{\|}\rightarrow
0~{}\mathrm{as}~{}r_{1}\rightarrow\infty.$

(b) Using (4.2) we have

$\displaystyle\big{|}tr~{}\big{(}B(\mathcal{H}_{1}+r_{1}I)^{-m})-{tr~{}(B(V+r_{1}I)^{-m}\big{)}}\big{|}$
$\displaystyle=\big{|}tr~{}\big{(}B(V+r_{1}I)^{-\frac{m}{2}}((1+K_{r_{1}})^{-m}-1)(V+r_{1}I)^{-\frac{m}{2}}\big{)}\big{|}$
$\displaystyle=\big{|}tr~{}\big{(}(V+r_{1}I)^{-\frac{m}{2}}B(V+r_{1}I)^{-\frac{m}{2}}((1+K_{r_{1}})^{-m}-1)\big{)}\big{|}$
$\displaystyle=\big{|}tr~{}\big{(}W_{r_{1}}((1+K_{r_{1}})^{-m}-1)\big{)}\big{|}$
$\displaystyle=\big{|}tr~{}\big{(}W_{r_{1}}^{\frac{1}{2}}((1+K_{r_{1}})^{-m}-1)W_{r_{1}}^{\frac{1}{2}}\big{)}\big{|}$
$\displaystyle\leq
m\big{\|}B\big{\|}\big{\|}(V+r_{1}I)^{-1}\big{\|}tr\big{(}B(V+r_{1}I)^{-m}\big{)},$

where $W_{r_{1}}=(V+r_{1}I)^{-\frac{m}{2}}B(V+r_{1}I)^{-\frac{m}{2}}$ is a
positive and trace class operator on $L^{2}(\mathbb{H}^{n})$. ∎

Therefore,

$\displaystyle\bigg{|}\frac{tr~{}(B(\mathcal{H}_{1}+r_{1}I)^{-m})}{tr~{}(B(V+r_{1}I)^{-m})}-1\bigg{|}\leq\|B\|\big{\|}(V+r_{1}I)^{-1}\big{\|}\rightarrow
0~{}\mathrm{as}~{}r_{1}\rightarrow\infty.$

######  Corollary 4.2.

Consider the self-adjoint operators $V$ and $\mathcal{H}_{1}$ as given in
Theorem 1.2. Let $\phi_{\mathcal{H}_{1}}(r_{1})=tr(\pi_{r_{1}})$ and
$\phi_{V}(r_{1})=Vol(\\{g\in\mathbb{H}^{n}:V(g)\leq r_{1}\\})$, then we have
the following asymptotic :

  1. (1)

$\phi_{V}$ is multiplicatively continuous.

  2. (2)

$\phi_{\mathcal{H}_{1}}(r_{1})\approx Cr_{1}{{}^{\frac{2(n+1)}{\kappa}}}$ as
$r_{1}\to\infty.$

  3. (3)

$\sup_{\mu\leq r_{1}}[tr(\mu+r)-tr(\mu)]\leq
tr(\pi_{r_{1}})\left(\frac{2(n+1)}{\kappa}{{\frac{r}{r_{1}}+\mathcal{O}\left(\frac{1}{r_{1}}\right)^{2}}}\right)$,
as $r_{1}\to\infty.$

Proof. (1) The function $\phi_{V}$ is given by

$\phi_{V}(r_{1})=Vol(\\{g\in\mathbb{H}^{n}:V(g)\leq
r_{1}\\})=Vol(\\{g\in\mathbb{H}^{n}:|g|\leq r_{1}^{\frac{1}{k}}\\})\approx
r_{1}^{\frac{2(n+1)}{k}}$
$\displaystyle\displaystyle\lim_{\begin{subarray}{c}r_{1}\to\infty\\\ \tau\to
1\end{subarray}}\frac{\phi_{V}(\tau
r_{1})}{\phi_{V}(r_{1})}=\displaystyle\lim_{\begin{subarray}{c}r_{1}\to\infty\\\
\tau\to 1\end{subarray}}\frac{(\tau
r_{1}){{}^{\frac{2(n+1)}{\kappa}}}}{r_{1}{{}^{\frac{2(n+1)}{\kappa}}}}=\displaystyle\lim_{\tau\to
1}\tau{{}^{\frac{2(n+1)}{\kappa}}}=1$

Therefore $\phi_{V}$ is multiplicatively continuous function. Further from
part (a) of Lemma 4.1

$\int_{0}^{\infty}\frac{r_{1}^{m}}{(u+r_{1})^{m}}d\phi_{\mathcal{H}_{1}}(u)/\int_{0}^{\infty}\frac{r_{1}^{m}}{(u+r_{1})^{m}}d\phi_{V}(u)\to
1,~{}\mbox{as}~{}r_{1}\to\infty.$

Now, applying Theorem 8 of Grishin-Poedintseva [3], part (3) follows directly
from part (2):

$\displaystyle\sup_{\mu\leq r_{1}}[tr(\pi_{\mu+r})-tr(\pi_{\mu})]\leq
tr(\pi_{r_{1}})~{}\Bigg{(}\frac{2(n+1)}{\kappa}{{\frac{r}{r_{1}}+\mathcal{O}\left(\frac{1}{r_{1}}\right)^{2}}}\Bigg{)}.$

∎

######  Lemma 4.3.

Let $V$ and $\mathcal{H}$ defined as in Theorem 1.2, then

$\displaystyle\lim_{r_{1}\to\infty}\frac{tr~{}(B(\mathcal{H}_{1}+r_{1}I)^{-m})}{tr~{}((\mathcal{H}_{1}+r_{1}I)^{-m})}=\lim_{r_{1}\to\infty}\frac{tr~{}(B(V+r_{1}I)^{-m})}{tr~{}((V+r_{1}I)^{-m})}.$

the above equality valid in the sense that if one of limits exist then the
other also does and the limits are same.

Proof. For each $r_{1}>0$, we have

(4.3)
$\displaystyle\frac{\bigg{(}\frac{tr~{}(B(\mathcal{H}_{1}+r_{1}I)^{-m})}{tr~{}(B(V+r_{1}I)^{-m})}\bigg{)}}{\bigg{(}\frac{tr~{}((\mathcal{H}_{1}+r_{1}I)^{-m})}{tr~{}((V+r_{1}I)^{-m})}\bigg{)}}=\frac{\bigg{(}\frac{tr~{}(B(\mathcal{H}_{1}+r_{1}I)^{-m})}{tr~{}((\mathcal{H}_{1}+r_{1}I)^{-m})}\bigg{)}}{\bigg{(}\frac{tr~{}((B(V+r_{1}I)^{-m})}{tr~{}((V+r_{1}I)^{-m})}\bigg{)}}.$

Since the left hand side has limit 1 (by part (b) of Lemma 4.1), the right
hand side limit in (4.3) exists and equal to 1. Therefore if the numerator or
the denominator in the fraction in the right hand side has a limit in (4.3),
then the other also has a limit and they both agree. Therefore,
$\displaystyle\lim_{r_{1}\to\infty}\frac{tr~{}(B(\mathcal{H}_{1}+r_{1}I)^{-m})}{tr~{}((\mathcal{H}_{1}+r_{1}I)^{-m})}=\lim_{r_{1}\to\infty}\frac{tr~{}(B(V+r_{1}I)^{-m})}{tr~{}((V+r_{1}I)^{-m})}.$
∎

######  Lemma 4.4.

Define $\pi_{V,r_{1}}:L^{2}(\mathbb{H}^{n})\rightarrow L^{2}(\mathbb{H}^{n})$
by $\pi_{V,r_{1}}(h)(g)=h(g)\mathcal{X}_{\\{g\in\mathbb{H}^{n}:V(g)\leq
r_{1}}(g)$ for all $h\in L^{2}(\mathbb{H}^{n})$. Then
$\bigg{|}\frac{{tr}~{}{(\pi_{r_{1}}f(A)\pi_{r_{1}})}}{tr~{}(\pi_{r_{1}})}-\frac{{tr}~{}{f(\pi_{V,r_{1}}A\pi_{V,r_{1}})}}{tr~{}(\pi_{V,r_{1}})}\bigg{|}\rightarrow
0~{}~{}\mathrm{as}~{}~{}r_{1}\rightarrow\infty.$

Proof. Without loss of generality add a suitable constant to make the function
$f$ is positive. Then $f(A)$ is a positive operator on
$L^{2}(\mathbb{H}^{n})$.

Let $\phi_{\mathcal{H}_{1}}(r_{1})=tr~{}{(\pi_{r_{1}})}$ and
$\phi_{V}(r_{1})=tr~{}{(\pi_{\pi_{V,r_{1}}})}$. Set
$\phi_{\mathcal{H}_{1},f}(r_{1})=tr~{}(\pi_{r_{1}}f(A)\pi_{r_{1}})=tr~{}(f(A)^{\frac{1}{2}}\pi_{r_{1}}f(A)^{\frac{1}{2}})$
and
$\phi_{V,f}(r_{1})=tr~{}(\pi_{V,r_{1}}f(A)\pi_{V,r_{1}})=tr~{}(f(A)^{\frac{1}{2}}\pi_{V,r_{1}}f(A)^{\frac{1}{2}}).$
Using the spectral theorem for both $\mathcal{H}_{1}$, $V$ and Lemma 4.3 we
get

$\displaystyle\lim_{r\to\infty}\frac{\int_{0}^{\infty}\frac{\phi_{\mathcal{H}_{1},f}(u)}{(1+\frac{u}{r})^{m+1}}\,du}{\int_{0}^{\infty}\frac{\phi_{\mathcal{H}_{1}}(u)}{(1+\frac{u}{r})^{m+1}}\,du}=\lim_{r_{1}\to\infty}\frac{\phi_{\mathcal{H}_{1},f}(r_{1})}{\phi_{\mathcal{H}_{1}}(r_{1})}=\lim_{r_{1}\to\infty}\frac{\phi_{V,f}(r_{1})}{\phi_{V}(r_{1})}=\lim_{r\to\infty}\frac{\int_{0}^{\infty}\frac{\phi_{V,f}(u)}{(1+\frac{u}{r})^{m+1}}\,du}{\int_{0}^{\infty}\frac{\phi_{V}(u)}{(1+\frac{u}{r})^{m+1}}\,du}.$

Therefore
$\bigg{|}\frac{{tr}~{}{(\pi_{r_{1}}f(A)\pi_{r_{1}})}}{tr~{}(\pi_{r_{1}})}-\frac{{tr}~{}{f(\pi_{V,r_{1}}A\pi_{V,r_{1}})}}{tr~{}(\pi_{V,r_{1}})}\bigg{|}\rightarrow
0~{}~{}\mathrm{as}~{}~{}r_{1}\rightarrow\infty.$ ∎

######  Lemma 4.5.

Let $I_{r_{1}}$ be defined as in Lemma 3.2, then

  1. (a)

$\bigg{|}\frac{tr~{}(\pi_{V,r_{1}}I_{r_{1}})}{tr~{}(\pi_{V,r_{1}})}-1\bigg{|}\rightarrow
0~{}\mathrm{as}~{}r_{1}\rightarrow\infty.$

  2. (b)

$\bigg{|}\frac{tr~{}(\pi_{r_{1}}\pi_{V,r_{1}}I_{r_{1}}\pi_{r_{1}})}{tr~{}(\pi_{V,r_{1}}I_{r_{1}})}-1\bigg{|}\rightarrow
0~{}\mathrm{as}~{}r_{1}\rightarrow\infty.$

Proof. The proof of the lemma follows similarly as in Lemma 3.2. ∎

######  Lemma 4.6.

(a) Let $\mathcal{H}_{1},A$ be the operators defined in Theorem 1.2 and
$k\in(0,1)$, then the operator $[\mathcal{H}_{1},A]$ is bounded on
$L^{2}(\mathbb{H}^{n}).$

(b) Under the assumption of Theorem 1.2, we have

$\bigg{|}\frac{{tr}~{}{f(\pi_{r_{1}}A\pi_{r_{1}})}}{tr~{}(\pi_{r_{1}})}-\frac{{tr}~{}{(\pi_{r_{1}}f(A)\pi_{r_{1}})}}{tr~{}(\pi_{r_{1}})}\bigg{|}\rightarrow
0~{}\text{as}~{}r_{1}\rightarrow\infty.$

Proof. Since $[B,A]$ is bounded on $L^{2}(\mathbb{H}^{n})$, the boundedness of
the operator $[V,A]$ will imply boundedness of the operator
$[\mathcal{H}_{1},A]$ on $L^{2}(\mathbb{H}^{n})$. For $h\in
L^{2}(\mathbb{H}^{n})$,

$[A,V]h(g)=(AV-VA)h(g)=\int_{\mathbb{H}^{n}}K_{3}(g,g_{1})h(g_{1})\,dg_{1},$

where
$K_{3}(g,g_{1})=\big{(}|g|^{k}-|g_{1}|^{k}\big{)}\int_{\mathbb{R}^{*}}{tr}~{}\big{(}\pi_{\lambda}^{*}(g)~{}a(g,\lambda)\pi_{\lambda}(g_{1})\big{)}\,d\mu(\lambda).$
By Lemma 2.5 we have
$|{tr}~{}\big{(}\pi_{\lambda}^{*}(g)~{}a(g,\lambda)\pi_{\lambda}(g_{1})\big{)}|\leq
M|{tr}~{}\big{(}\pi_{\lambda}(I-\mathcal{L})^{-2N}\big{)}|\|(1+|\lambda|H)^{-2N}\pi_{\lambda}(g_{1}g^{-1})\pi_{\lambda}^{*}(g_{1}g^{-1})\|\|a(g,\lambda)\|$.
Since $(1+|\lambda|H)^{-2N}\pi_{\lambda}(g_{1}g^{-1})$ has order $-4N$, (2.4)
gives

(4.4)
$\displaystyle\|(1+|\lambda|H)^{-2N}\pi_{\lambda}(g_{1}g^{-1})\|\leq\frac{c}{(1+V(g_{1}g^{-1}))^{2N}}.$

Using the above inequality, the identity
$\pi_{\lambda}(I-\mathcal{L})=1+|\lambda|H$ and the fact that $\kappa\in(0,1)$
we get

$|K_{3}(g,g_{1})|\leq
c_{N}\frac{\big{|}gg_{1}^{-1}\big{|}^{\kappa}}{(1+V(g_{1}g^{-1}))^{2N}},$

and

$\displaystyle\|[A,V]h\|_{2}^{2}=$
$\displaystyle\int_{\mathbb{H}^{n}}\bigg{|}\int_{\mathbb{H}^{n}}K_{3}(g,g_{1})h(g_{1})\,dg_{1}\bigg{|}^{2}\,dg$
$\displaystyle\leq
c_{N}\int_{\mathbb{H}^{n}}\bigg{(}\int_{\mathbb{H}^{n}}\bigg{|}\frac{\big{|}gg_{1}^{-1}\big{|}^{k}h(g_{1})}{(1+V(g_{1}g^{-1}))^{2N}}\bigg{|}\,dg_{1}\bigg{)}^{2}\,d\
g=\||h|\ast K\|_{2}^{2}$

where $K(g)=\dfrac{|g|^{k}}{(1+|g|)^{2N}}$. Since for a sufficiently large
$N$, $K\in L^{1}({\mathbb{H}^{n})}$, an application of Minkowski’s inequality
gives $\|[A,V]h\|_{2}\leq C\|K\|_{1}\|h\|_{2}.$

(b) Since $A$ is bounded self-adjoint, the spectrum of $A$, $\sigma(A)$ is a
compact subset of $\mathbb{R}.$ Since any continuous function can be
approximated in the supremum norm by smooth functions, it is enough to if we
assume that $f\in{C^{2}(\sigma(A))}.$ By Theorem 1.6 of Laptev-Safarov [9], we
get
(setting $A=\mathcal{H}_{1},B=A,\chi=0,\psi=f,P_{\lambda}=\pi_{r_{1}}$)

$|{tr}~{}(\pi_{r_{1}}f(A)\pi_{r_{1}}-\pi_{r_{1}}f(\pi_{r_{1}}A\pi_{r_{1}})\pi_{r_{1}})|$
$\leq\frac{1}{2}\|f^{\prime\prime}\|_{\infty}N_{r}(r_{1})\Big{(}{\|\pi_{r_{1}}A\|^{2}+\frac{\pi^{2}}{6r^{2}}\|\pi_{r_{1}-r}[A,\mathcal{H}_{1}]\|^{2}}\Big{)}.$

Dividing both sides by $tr(\pi_{r_{1}})$ and setting
$r=r_{1}^{\alpha},\alpha\in(0,1)$ and using the boundness of
$A,[A,\mathcal{H}_{1}]$ we have

$\frac{|{tr}~{}(\pi_{r_{1}}f(A)\pi_{r_{1}}-\pi_{r_{1}}f(\pi_{r_{1}}A\pi_{r_{1}})\pi_{r_{1}})|}{tr(\pi_{r_{1}})}\leq
C\frac{N_{r_{1}^{\alpha}}(r_{1})}{tr(\pi_{r_{1}})}\to
0~{}\mbox{as}~{}r_{1}\to\infty$

by (3) of Corollary 4.2, where $N_{r}(r_{1})=\displaystyle\sup_{\mu\leq
r_{1}}\left(tr(\pi_{\mu}-\pi_{\mu-r})\right)$. ∎

Proof of theorem 1.2: Using Lemma 4.4, Lemma 4.5 and Lemma 4.6 we have

$\displaystyle\lim_{r_{1}\to\infty}\frac{{tr}~{}{f(\pi_{r_{1}}A\pi_{r_{1}})}}{tr~{}(\pi_{r_{1}})}$
$\displaystyle=\lim_{r_{1}\to\infty}\frac{{tr}~{}{(\pi_{r_{1}}f(A)\pi_{r_{1}})}}{tr~{}(\pi_{r_{1}})}$
$\displaystyle=\lim_{r_{1}\to\infty}\frac{{tr}~{}{(\pi_{V,r_{1}}f(A)\pi_{V,r_{1}})}}{tr~{}(\pi_{V,r_{1}})}$
$\displaystyle=\lim_{r_{1}\to\infty}\frac{{tr}~{}{(\pi_{V,r_{1}}f(A)\pi_{V,r_{1}})}}{tr~{}(\pi_{V,r_{1}}I_{r_{1}})}$
$\displaystyle=\lim_{r_{1}\to\infty}\frac{{tr}~{}{(\pi_{V,r_{1}}f(A)\pi_{V,r_{1}})}}{tr~{}(\pi_{r_{1}}\pi_{V,r_{1}}I_{r}\pi_{r_{1}})}.$

Let $\ell\in\mathbb{N}$ and $\psi\in L^{2}(\mathbb{H}^{n})$. Now applying
Lemma 2.7 we have

$\displaystyle\pi_{V,r_{1}}A^{\ell}\pi_{V,r_{1}}\psi(g)$
$\displaystyle=\pi_{V,r_{1}}A^{\ell}\psi(g)\mathcal{X}_{V\leq r_{1}}(g)$
$\displaystyle=\pi_{V,r_{1}}\int_{\mathbb{R}^{*}}{tr}~{}(\pi_{\lambda}^{*}(g)a_{\ell}(g,{\lambda})(\psi\mathcal{X}_{V\leq
r_{1}}(g)\hat{)}(\lambda))\,d\mu(\lambda)$ $\displaystyle=\int_{V(g)\leq
r_{1}}\int_{\mathbb{R}^{*}}{tr}~{}(\pi_{\lambda}^{*}(g)a_{\ell}(g,{\lambda})\psi(g_{1})\pi_{\lambda}(g_{1}))\,d\mu(\lambda)\,dg_{1}$
$\displaystyle=\int_{\mathbb{H}^{n}}\bigg{(}\mathcal{X}_{V\leq
r_{1}}\int_{\mathbb{R}^{*}}{tr}~{}(\pi_{\lambda}^{*}(g)a_{\ell}(g,{\lambda})\pi_{\lambda}(g_{1}))\,d\mu(\lambda)\bigg{)}\psi(g_{1})\,dg_{1}$
$\displaystyle=\int_{\mathbb{H}^{n}}K_{4}(g,g_{1})\psi(g_{1})\,dg_{1},$

where the kernel of the operator $\pi_{V,r_{1}}A^{\ell}\pi_{V,r_{1}}$ is given
by

$K_{4}(g,g_{1})=\mathcal{X}_{V\leq
r_{1}}\int_{\mathbb{R}^{*}}{tr}~{}(\pi_{\lambda}^{*}(g)a_{\ell}(g,{\lambda})\pi_{\lambda}(g_{1}))\,d\mu(\lambda).$

Thus,

(4.5) $\displaystyle{tr}~{}{(\pi_{V,r_{1}}A^{\ell}\pi_{V,r_{1}})}$
$\displaystyle=\int_{\mathbb{H}^{n}}K_{4}(g,g))\,dg$
$\displaystyle=\int_{\mathbb{H}^{n}}\mathcal{X}_{V\leq
r_{1}}\int_{\mathbb{R}^{*}}{tr}~{}(\pi_{\lambda}^{*}(g)a_{\ell}(g,{\lambda})\pi_{\lambda}(g))\,d\mu(\lambda)\,dg$
$\displaystyle=\int_{V\leq
r_{1}}\int_{\mathbb{R}^{*}}{tr}~{}(a_{\ell}(g,{\lambda}))\,d\mu(\lambda)\,dg$
$\displaystyle=\int_{V\leq
r_{1}}\int_{\mathbb{R}^{*}}{tr}~{}(a(g,{\lambda})^{\ell}+E_{\ell}(g,\lambda))\,d\mu(\lambda)\,dg$
$\displaystyle=\int_{V\leq
r_{1}}\int_{\mathbb{R}^{*}}{tr}~{}(a(g,{\lambda})^{\ell})\,d\mu(\lambda)\,dg+\int_{V\leq
r_{1}}\int_{\mathbb{R}^{*}}{tr}~{}(E_{\ell}(g,\lambda))\,d\mu(\lambda)\,dg.$

Since $\int_{V\leq
r_{1}}\int_{\mathbb{R}^{*}}{tr}~{}\left(E_{\ell}(g,{\lambda}\right)\,d\mu(\lambda)\,dg=\int_{V\leq
r_{1}}\int_{\mathbb{R}^{*}}{tr}~{}\left(\pi_{\lambda}^{*}(g)\pi_{\lambda}(g)E_{\ell}(g,{\lambda})\right)\,d\mu(\lambda)\,dg$
and $E_{\ell}(g,{\lambda})$ is a bounded operator (see Lemma 2.7), part $(a)$
of Lemma 2.5 and (4.4) gives

(4.6) $\displaystyle\int_{V\leq
r_{1}}\int_{\mathbb{R}^{*}}{tr}~{}(E_{\ell}(g,\lambda))\,d\mu(\lambda)\,dg<\infty.$

Since $\pi_{r_{1}}I_{r_{1}}=I_{r_{1}}\pi_{r_{1}}$,

$\displaystyle{\pi_{r_{1}}\pi_{V,r_{1}}I_{r}\pi_{r_{1}}\psi(g)}=\int_{\mathbb{H}^{n}}K_{5}(g,g_{1})\psi(g_{1})\,dg_{1},$

where

$K_{5}(g,g_{1})=\mathcal{X}_{V\leq
r_{1}}(g)\int_{-r_{1}}^{r}{tr}~{}(\pi_{\lambda}^{*}(g)I_{r_{1}\times
r_{1}}f(b(g_{1}))\pi_{\lambda}(g_{1}))\,d\mu(\lambda)$

Therefore

$\displaystyle{tr}~{}{({\pi_{r_{1}}\pi_{V,r_{1}}I_{r_{1}}\pi_{r_{1}}})}$
$\displaystyle=\int_{\mathbb{H}^{n}}K_{5}(g,g)\,dg$ $\displaystyle=\int_{V\leq
r_{1}}\int_{-r_{1}}^{r_{1}}{tr}~{}(\pi_{\lambda}^{*}(g)I_{r_{1}\times
r_{1}}\pi_{\lambda}(g))\,d\mu(\lambda)\,dg$ $\displaystyle=\int_{V\leq
r_{1}}\int_{-r_{1}}^{r_{1}}{tr}~{}(I_{r_{1}\times r_{1}})\,d\mu(\lambda)\,dg$
$\displaystyle=r_{1}\int_{-r_{1}}^{r_{1}}d\mu(\lambda)\int_{V\leq r_{1}}\,dg.$

Therefore

$\displaystyle\frac{{tr}~{}{(\pi_{V,r_{1}}A^{\ell}\pi_{V,r_{1}})}}{tr~{}(\pi_{r_{1}}\pi_{V,r_{1}}I_{r_{1}}\pi_{r_{1}})}$
$\displaystyle=\frac{\int_{V\leq
r_{1}}\int_{\mathbb{R}^{*}}{tr}~{}(a(g,{\lambda})^{\ell})\,d\mu(\lambda)\,dg}{{r_{1}\int_{-r_{1}}^{r_{1}}\,d\mu(\lambda)}\int_{V\leq
r_{1}}\,dg}+E(r_{1}),$

where

$\displaystyle|E(r_{1})|$ $\displaystyle\leq\frac{\int_{V\leq
r_{1}}\int_{\mathbb{R}^{*}}{tr}~{}(E_{\ell}(g,\lambda))\,d\mu(\lambda)\,dg}{{r_{1}\int_{-r_{1}}^{r_{1}}\,d\mu(\lambda)}}\to
0$

as $r_{1}\to\infty.$ This result can be extended to continuous functions as an
application of the Weierstrass approximation theorem and spectral theorem.
This proves our Theorem 1.2.

##  5\. Szegö limit theorem for $\mathcal{H}_{2}$

In this section we prove Szegö limit theorem for $\mathcal{H}_{2}$ under
certain assumptions on the symbol $a(g,\lambda)$ as in [20]. We assume that

(5.1) $\displaystyle\displaystyle\lim_{E\to\infty}\bar{a}(E)=a,$

where
$\bar{a}(E)=\dfrac{1}{S(E)}\displaystyle\int_{G_{E}}a_{g,\lambda}(\xi,x)\,d\xi\,dx\,dg\,d\lambda$
and $S(E)=\displaystyle\displaystyle\int_{G_{E}}\,d\xi\,dx\,dg\,d\lambda,$
with
$G_{E}=\\{(g,\lambda,\xi,x)\in\mathbb{H}^{n}\times\mathbb{R}^{*}\times\mathbb{R}^{n}\times\mathbb{R}^{n}:|\lambda|(|\xi|^{2}+|x|^{2})+V(g)=E,V(g)\leq
E^{\frac{1}{2}},|\lambda|\geq{E^{-\frac{1}{2}}}\\}.$

###  5.1. Proposition

Let
$G^{E}=\\{(g,\lambda,\xi,x)\in\mathbb{H}^{n}\times\mathbb{R}^{*}\times\mathbb{R}^{n}\times\mathbb{R}^{n}:|\lambda|(|\xi|^{2}+|x|^{2})+V(g)\leq
E,V(g)\leq E^{\frac{1}{2}},|\lambda|\geq{E^{-\frac{1}{2}}}\\}$. Then volume of
$G^{r_{2}}=Vol(G^{r_{2}})\approx
Cr_{2}{{}^{\frac{n+1}{\kappa}+\frac{5n-4}{2}}}$ as $r_{2}\rightarrow\infty$.

###### Proof.

: Evaluating the volume of $G^{E}$ using the homogeneous norm on
$\mathbb{H}^{n}$ by changing to spherical coordinate we have
$Vol(G^{r_{2}})=\displaystyle\int_{G^{r_{2}}}\,d\xi\,dx\,dg\,d\mu(\lambda)$
grows asymptotically similar to that of
$Cr_{2}{{}^{\frac{n+1}{\kappa}+\frac{5n-4}{2}}}.$ ∎

######  Lemma 5.1.

Let $\phi(r_{2})={tr}({\pi_{r_{2}}})$ and
$\psi(r_{2})=tr({\pi_{r_{2}}A\pi_{r_{2}}})$. Then under the assumption (5.1)
we have

$\Phi(u)=\int_{0}^{\infty}\frac{\phi(r_{2})}{(r_{2}+u)^{m+1}}\,dr_{2}=\int_{0}^{\infty}\frac{\displaystyle\int_{G^{E}}\,d\xi\,dx\,dg\,d\mu(\lambda)}{(E+u)^{m+1}}\,dE+E_{1}(u)\\\
$

and

$\Psi(u)=\int_{0}^{\infty}\frac{\psi(r_{2})}{(r_{2}+u)^{m+1}}\,dr_{2}=\int_{0}^{\infty}\frac{\displaystyle\int_{G^{E}}a_{g,\lambda}(\xi,x)\,d\xi\,dx\,dg\,d\mu(\lambda)}{(E+u)^{m+1}}\,dE+E_{2}(u),$

where
$E_{1}(u)=m\displaystyle\int_{\mathbb{H}^{n}\times\mathbb{R}^{*}}tr~{}(E_{1}(g,\lambda,u))~{}dg\,d\mu(\lambda)$
and
$E_{2}(u)=m\displaystyle\int_{\mathbb{H}^{n}\times\mathbb{R}^{*}}tr~{}(E_{2}(g,\lambda,u))~{}dg\,d\mu(\lambda)$
with $|E_{i}(u)|\to 0$ as $u\to\infty,~{}i=1,2.$

###### Proof.

The operator $\mathcal{H}_{2}$ has discrete spectrum of eigenvalues $0\leq
c_{1}\leq c_{2}\cdots\infty$. Let $\\{\psi_{j}\\}_{j=1}^{\infty}$ be the
complete set of eigenfunctions on corresponding to the eigenvalues
$\\{c_{j}\\}$ on $L^{2}(\mathbb{H}^{n})$. Then
$\phi(r_{2})=tr({\pi_{r_{2}}})=\displaystyle\sum_{c_{j}\leq
r_{2}}\langle\psi_{j},\psi_{j}\rangle$ and
$\phi^{\prime}(r_{2})=\displaystyle\sum_{j=1}^{\infty}\langle\psi_{j},\psi_{j}\rangle\delta(r_{2}-c_{j})$.
Now

$\Phi(u)=\int_{0}^{\infty}\frac{\phi(r_{2})}{(r_{2}+u)^{m+1}}\,dr_{2}=m\sum_{j=1}^{\infty}\langle\psi_{j},\psi_{j}\rangle~{}\int_{0}^{\infty}\frac{\delta(r_{2}-c_{j})}{(r_{2}+u)^{m}}\,dr_{2}=m\sum_{j=1}^{\infty}\langle\psi_{j},\psi_{j}\rangle~{}\frac{1}{(c_{j}+u)^{m}}.$

Now applying Corollary 2.4 on the symbols of $\mathcal{H}_{2}$ we get

$\displaystyle\Phi(u)$ $\displaystyle=$ $\displaystyle
m~{}{tr}~{}\Bigg{(}{{(\mathcal{H}_{2}+u)^{-m}}}\Bigg{)}$ $\displaystyle=$
$\displaystyle
m\int_{\mathbb{H}^{n}\times\mathbb{R}^{*}}{tr}~{}\Bigg{(}{{\big{(}{|\lambda|H+V(g)}+u\big{)}^{-m}}+E_{1}(g,\lambda,u)}\Bigg{)}\,dg\,d\mu(\lambda)$
$\displaystyle=$ $\displaystyle
m\int_{\mathbb{H}^{n}\times\mathbb{R}^{*}}\int_{\mathbb{R}^{n}\times\mathbb{R}^{n}}\left({\big{(}|\lambda|(|\xi|^{2}+|x|^{2})+V(g)+u\big{)}^{-m}}\right)\,dg\,d\mu(\lambda)+E_{1}(u)$
$\displaystyle=$
$\displaystyle\int_{0}^{\infty}\frac{\displaystyle\int_{G^{E}}\,d\xi\,dx\,dg\,d\mu(\lambda)}{(E+u)^{m+1}}\,dE+E_{1}(u),$

where
$E_{1}(u)=m\displaystyle\int_{\mathbb{H}^{n}\times\mathbb{R}^{*}}{tr}~{}(E_{1}(g,\lambda,u))~{}dgd\mu(\lambda)$.
Similarly taking $\psi(r_{2})={tr}~{}{(\pi_{r_{2}}A\pi_{r_{2}}})$, we have

$\displaystyle\Psi(u)$
$\displaystyle=\int_{0}^{\infty}\frac{\displaystyle\int_{G^{E}}a_{g,\lambda}(\xi,x)\,d\xi\,dx\,dg\,d\mu(\lambda)}{(E+u)^{m+1}}\,dE+E_{2}(u)$

where
$E_{2}(u)=m\displaystyle\int_{\mathbb{H}^{n}\times\mathbb{R}^{*}}{tr}~{}(E_{2}(g,\lambda,u))~{}dgd\mu(\lambda)$.
Observe that $E_{1}(g,\lambda,u)$ is a symbol on $\mathbb{H}^{n}.$ Therefore
by Lemma 2.5 and (4.4) we have $E_{1}(u)\to 0$ as $u\to\infty.$ Similarly
$E_{2}(u)\to 0$ as $u\to\infty.$ ∎

In order to prove the Szegö limit theorem for the Schrödinger operator
$\mathcal{H}_{2}$, we need to estimate the asymptotic growth of the measures
$tr(\pi_{r_{2}}A\pi_{r_{2}})$ and $tr(\pi_{r_{2}})$. Therefore we apply
Keldysh Tauberian Theorem to compare the measures and obtain the following
Corollary.

######  Corollary 5.2.

Consider the self-adjoint operator $\pi_{r_{2}}$ and $v(r_{2})$ as given in
Theorem 1.3 and Proposition 5.1 respectively. Let
$\phi(r_{2})=tr(\pi_{r_{2}}),$ $\psi(r_{2})=tr(\pi_{r_{2}}A\pi_{r_{2}})$ then
we have the following asymptotic :

  1. (1)

$v(r_{2})\approx Cr_{2}{{}^{\frac{n+1}{\kappa}+\frac{5n-4}{2}}}$ as
$r_{2}\to\infty.$

  2. (2)

$v$ is multiplicatively continuous.

  3. (3)

$tr(\pi_{r_{2}})\approx Cr_{2}{{}^{\frac{n+1}{\kappa}+\frac{5n-4}{2}}}$ as
$r_{2}\to\infty.$

  4. (4)

$\displaystyle\sup_{\mu\leq r_{2}}[tr(\mu+r)-tr(\mu)]\leq
tr(\pi_{r_{2}})\left(\frac{(n+1)}{k}{{\frac{r}{r_{2}}+\mathcal{O}\left(\frac{1}{r_{2}}\right)^{2}}}\right)$,
as $r_{2}\to\infty.$

  5. (5)

$\psi$ is multiplicatively continuous.

Proof. Clearly (1) directly follows from Proposition 5.1. Now

$\displaystyle\displaystyle\lim_{\begin{subarray}{c}r_{2}\to\infty\\\ \tau\to
1\end{subarray}}\frac{v(\tau
r_{2})}{v(r_{2})}=\displaystyle\lim_{\begin{subarray}{c}r_{2}\to\infty\\\
\tau\to 1\end{subarray}}\frac{(\tau
r_{2}){{}^{\frac{n+1}{\kappa}+\frac{5n-4}{2}}}}{r_{2}^{\frac{n+1}{\kappa}+\frac{5n-4}{2}}}=\displaystyle\lim_{\tau\to
1}\tau{{}^{\frac{n+1}{\kappa}+\frac{5n-4}{2}}}=1$

Therefore $v$ is multiplicatively continuous function. We choose sufficiently
large $m$ such that the operator $(\mathcal{H}_{2}+uI)^{-m}$ is in trace
class. Therefore by Lemma 5.1 and Theorem 8 of Grishin-Poedintseva [3], we get
$\phi(r_{2})/v(r_{2})\to 1~{}\mbox{as}~{}r_{2}\to\infty.$ This proves (3).
Using the asymptotic in (3), it is easy to check that

$\displaystyle\sup_{\mu\leq r_{2}}[tr(\mu+r)-tr(\mu)]\leq
tr(r_{2})~{}\Bigg{(}\frac{(n+1)}{k}{{\frac{r}{r_{2}}+\mathcal{O}\left(\frac{1}{r_{2}}\right)^{2}}}\Bigg{)}.$

To prove (5), notice that if $\varphi$ and $\chi$ are two distribution
functions satisfying
$\displaystyle\lim_{r\to\infty}\frac{\varphi(r)}{\chi(r)}=1$ then $\varphi$ is
multiplicatively continuous whenever $\chi$ is. Therefore $\psi$ is also a
multiplicatively continuous function. ∎

######  Theorem 5.3.

Under the assumption (5.1) we have

$\displaystyle\lim_{r_{2}\to\infty}\frac{{tr}~{}{(\pi_{r_{2}}A\pi_{r_{2}})}}{tr~{}(\pi_{r_{2}})}$
$\displaystyle=\lim_{r_{2}\to\infty}\frac{\int_{G^{r_{2}}}a_{g,\lambda}(\xi,x)\,d\xi\,dx\,dg\,d\mu(\lambda)}{\int_{G^{r_{2}}}\,d\xi\,dx\,dg\,d\mu(\lambda)}.$

###### Proof.

The proof follows directly by Lemma 5.1, as all the requirements (by our the
assumption (5.1) on the symbol $a(g,\lambda)$) of Theorem 8 of Grishin-
Poedintseva [3] are satisfied. ∎

######  Corollary 5.4.

Let $P(r)$ be a polynomial in $\mathbb{R}.$ Then

$\displaystyle\lim_{r_{2}\to\infty}\frac{{tr}~{}({\pi_{r_{2}}P(A)\pi_{r_{2}}})}{tr~{}(\pi_{r_{2}})}$
$\displaystyle=\lim_{r_{2}\to\infty}\frac{\int_{G^{r_{2}}}P(a_{g,\lambda}(\xi,x))\,d\xi\,dx\,dg\,d\mu(\lambda)}{\int_{G^{r_{2}}}\,d\xi\,dx\,dg\,d\mu(\lambda)}$

Proof. The operator $P(A)$ has symbol $P(a(g,\lambda))+E_{-1}(g,\lambda)$
where $E_{-1}(g,\lambda)\in B(L^{2}(\mathbb{R}^{n})).$ Proceeding exactly as
in (4.6) we get
$\displaystyle\int_{\mathbb{H}^{n}\times\mathbb{R}^{*}}{tr}~{}(E_{-1}(g,{\lambda}))\,dg\,d\mu(\lambda)<\infty.$
Therefore we get the corollary as $r_{2}\to\infty.$ ∎

######  Lemma 5.5.

Let $\mathcal{H}_{2},A$ be the operators defined in Theorem 1.3 and
$\kappa\in(0,1)$, then

(a) the operator $[\mathcal{H}_{2},A]$ is bounded on $L^{2}(\mathbb{H}^{n}).$

(b) Under the assumptions of Theorem 1.3, we have

$\bigg{|}\frac{{tr}~{}{f(\pi_{r_{2}}A\pi_{r_{2}})}}{tr~{}(\pi_{r_{2}})}-\frac{{tr}~{}{(\pi_{r_{2}}f(A)\pi_{r_{2}})}}{tr~{}(\pi_{r_{2}})}\bigg{|}\rightarrow
0~{}\text{as}~{}r_{2}\rightarrow\infty.$

Proof. For $\kappa\in(0,1)$ the operator $[V,A]$ is bounded (see part $(a)$ of
Lemma 4.6). The boundedness of the operator $[\mathcal{L},A]$ will imply
boundedness of the operator $[\mathcal{H}_{2},A]$ on $L^{2}(\mathbb{H}^{n})$.
Using the identity (2.11) we get

$\displaystyle A\mathcal{L}h(g)$
$\displaystyle=\int_{\mathbb{R}^{*}}{tr}~{}\left(\pi_{\lambda}^{*}(g)a(g,\lambda)\widehat{\mathcal{L}h}(\lambda)\right)\,d\mu(\lambda)$
$\displaystyle=\int_{\mathbb{R}^{*}}{tr}~{}\bigg{(}\pi_{\lambda}(I-\mathcal{L})^{4N}T^{4N}\lambda^{-4N}\pi_{\lambda}^{*}(g)a(g,\lambda)\hat{h}(\lambda)|\lambda|{H}\pi_{\lambda}(I-\mathcal{L})^{-4N}\bigg{)}\,d\mu(\lambda)$
$\displaystyle\leq
C(1+V(g))^{-2N+2}\int_{\mathbb{R}^{*}}\|\hat{h}(\lambda)\|_{B_{2}}\frac{1}{(1+|\lambda|)^{N-1}}\,d\mu(\lambda).$

For sufficiently large $N$, an application of Cauchy-Schwarz inequality gives
$\|A\mathcal{L}h\|_{2}\leq M_{1}\|h\|_{2}.$ Similarly,

$\displaystyle\mathcal{L}Ah(g)$
$\displaystyle=\int_{\mathbb{R}^{*}}{tr}~{}\left(\pi_{\lambda}^{*}(g)\widehat{\mathcal{L}Ah}(\lambda)\right)\,d\mu(\lambda)$
$\displaystyle=\int_{\mathbb{R}^{*}}{tr}~{}\left(\pi_{\lambda}^{*}(g)|\lambda|\widehat{Ah}(\lambda){H}\right)\,d\mu(\lambda)$
$\displaystyle=\int_{\mathbb{R}^{*}}\int_{\mathbb{H}^{n}}Ah(g_{1})~{}{tr}~{}\bigg{(}\pi_{\lambda}^{*}(g)\pi_{\lambda}(g_{1})|\lambda|{H}\bigg{)}\,d\mu(\lambda)\,dg_{1}$
$\displaystyle=\int_{\mathbb{R}^{*}}\int_{\mathbb{H}^{n}}\left(\int_{\mathbb{R}^{*}}{tr}~{}\left(\pi_{\lambda_{1}}^{*}(g)a(g_{1},\lambda_{1})\widehat{h}(\lambda_{1})\right)\,d\mu(\lambda_{1})\right)~{}{tr}~{}\bigg{(}\pi_{\lambda}^{*}(g)\pi_{\lambda}(g_{1})|\lambda|{H}\bigg{)}\,d\mu(\lambda)\,dg_{1}$

By Lemma 2.5 we get,
$\displaystyle\int_{\mathbb{R}^{*}}\left|{tr}~{}\left(\pi_{\lambda_{1}}^{*}(g)a(g_{1},\lambda_{1})\widehat{h}(\lambda_{1})\right)\right|\,d\mu(\lambda_{1})\leq
C\|{h}\|_{2}.$ Further, choose sufficiently large $N$ we get

$\int_{\mathbb{R}^{*}}\int_{\mathbb{H}^{n}}~{}\left|{tr}~{}\bigg{(}\pi_{\lambda}^{*}(g)\pi_{\lambda}(g_{1})|\lambda|{H}\bigg{)}\right|\,d\mu(\lambda)\,dg_{1}\leq
C(1+V(g))^{-4N+4}\int_{\mathbb{R}^{*}}\int_{\mathbb{H}^{n}}\frac{(1+V(g_{1}))^{-2N+2}}{(1+|\lambda|)^{N-1}}d\mu(\lambda)dg_{1}.$

Therefore $\|[A,\mathcal{H}_{2}]h\|_{2}\leq M\|h\|_{2},$ which completes part
(a). Part (b) follows similarly as in the proof of part (b) of Lemma 4.6. ∎

######  Corollary 5.6.

Theorem 1.1, 1.2 and 1.3 also holds under the compact perturbation of the
pseudo-differential operator $A$.

###### Proof.

To prove the above result, enough to show
$\displaystyle\lim_{r\rightarrow\infty}\frac{tr(\pi_{r}A^{n}\pi_{r})}{tr~{}(\pi_{r})}=\displaystyle\lim_{r\rightarrow\infty}\frac{tr(\pi_{r}(A+B)^{n}\pi_{r})}{tr~{}(\pi_{r})}$
for any compact operator $B$ on $L^{2}(\mathbb{H}^{n})$. Notice that
$(A+B)^{n}=A^{n}$\+ terms with factor $A^{p}B^{n-p}$ or $B^{p}A^{n-p}$ where
$p\in\\{1,2,\cdots,n\\}$. Since the class of compact operators form a two
sided ideal of the class of bounded operators, $(A+B)^{n}=A^{n}+$ a compact
operator. We are done if we can prove that for a compact operator $T$,
$\displaystyle\lim_{r\rightarrow\infty}\frac{tr(\pi_{r}T\pi_{r})}{tr~{}(\pi_{r})}=0.$
Since T is a compact operator, for given $\epsilon>0$ there exist a finite
rank operator $T_{k}$ such that $\|T_{k}-T\|\rightarrow 0~{}\text{as
}k\rightarrow\infty$. Then
$\left|\dfrac{tr(\pi_{r}T\pi_{r})-tr(\pi_{r}T_{k}\pi_{r})}{tr(\pi_{r})}\right|\leq\|T-T_{k}\|~{}\rightarrow
0~{}\text{as }k\rightarrow\infty.$ Therefore for given $\epsilon>0$ there
exist $N_{0}\in\mathbb{N}$ such that
$\left|\dfrac{tr(\pi_{r}T\pi_{r})-tr(\pi_{r}T_{k}\pi_{r})}{tr(\pi_{r})}\right|<\dfrac{\epsilon}{2}$
for $k\geq N_{0}$. Further,
$\bigg{|}\frac{tr(\pi_{r}T_{N_{0}}\pi_{r})}{tr~{}(\pi_{r})}\bigg{|}\rightarrow
0~{}\text{as }r\rightarrow\infty$ i.e, for given
$\epsilon>0,~{}~{}\exists~{}N_{1}\in\mathbb{N}$ such that
$\bigg{|}\frac{tr(\pi_{r}T_{N_{0}}\pi_{r})}{tr~{}(\pi_{r})}\bigg{|}<\dfrac{\epsilon}{2},~{}~{}\forall~{}r>N_{1}.$
Thus
$\bigg{|}\frac{tr(\pi_{r}T\pi_{r})}{tr~{}(\pi_{r})}\bigg{|}\leq\bigg{|}\frac{tr(\pi_{r}T_{N_{0}}\pi_{r})}{tr~{}(\pi_{r})}\bigg{|}+\|T-T_{N_{0}}\|<\epsilon\quad\forall~{}r\geq
N_{1}.$ ∎

######  Remark 5.7.

  1. (1)

The result in Theorem 1.1 is valid if the operator $\mathcal{H}_{0}$ is
replaced by $\mathcal{H}_{0}^{{}^{\prime}}:B+V$ where $B$ is a bounded
operator on $L^{2}(\mathbb{H}^{n})$ for all $\kappa>0$.

  2. (2)

The range of $\kappa$ in (0,1) is required to show that the boundedness of the
commutator $[V,A]$ (see the proof of $(a)$ in Lemma 4.6). However if we
proceed along the similar lines of [20] then the boundedness of commutator
$[V,A]$ can be proved for $\kappa>0.$

## Acknowledgments

The first author wishes to thank the Ministry of Human Resource Development,
India for the research fellowship and Indian Institute of Technology Guwahati,
India for the support provided during the period of this work.

##  6\. Appendix

We collect few definitions and theorems of Grishin-Poedintseva [3], that we
use in our paper for the reader’s convenience.

######  Definition 6.1.

Let $\phi$ be a positive function on the half line $[0,\infty)$. Let

$S=\\{\alpha:\exists M,R~{}\mathrm{with}~{}\phi(tr)\leq
Mt^{\alpha}\phi(r),~{}\mathrm{for~{}all~{}}t\geq 1,r\geq R\\}$

and

$G=\\{\beta:\exists M,R~{}\mathrm{with}~{}\phi(tr)\geq
Mt^{\beta}\phi(r),~{}\mathrm{for~{}all~{}}t\geq 1,r\geq R\\}$

Then the numbers $\alpha(\phi):=\inf S$ and $\beta(\phi):=\sup G$ are called
the upper and lower Matushevskaya index of $\phi$ respectively.

######  Theorem 6.2.

([3],Theorem 2)

Let $m>-1$. Assume that $\varphi$ is positive measurable function on
$[0,\infty)$ that does not vanish identically in any neighborhood of infinity.
Let $\Phi(r)=\displaystyle\int_{0}^{\infty}\dfrac{\varphi(rt)}{(1+t)^{m+1}}dt$
be finite. Then the functions $\varphi$ and $\Phi$ have same growth at
infinity if and only if $\beta(\varphi)>-1$ and $\alpha(\varphi)<m$.

######  Definition 6.3.

A function $\varphi$ is said to be multiplicatively continuous at infinity if
it satisfies $\displaystyle\lim_{\begin{subarray}{c}r\rightarrow\infty\\\
\tau\rightarrow 1\end{subarray}}\dfrac{\varphi(\tau r)}{\varphi(r)}=1.$

######  Theorem 6.4.

([3],Theorem 8) Let $\varphi$ and $\psi$ be positive functions on $[0,\infty)$
satisfying the following conditions:

  1. (1)

the functions $\varphi$ and $\psi$ do not vanish identically in any
neighborhood of infinity;

  2. (2)

the function $\varphi$ is multiplicatively continuous at infinity and
$\beta(\varphi)>-1$;

  3. (3)

the function $\psi$ is increasing;

  4. (4)

at least one of the inequalities $\alpha(\varphi)<m$ and $\alpha(\psi)<m$
holds, where $m>-1$;

  5. (5)

the functions

$\Phi(r)=\int_{0}^{\infty}\dfrac{\varphi(ru)}{(1+u)^{m+1}}du~{}~{}and~{}~{}\Psi(r)=\int_{0}^{\infty}\dfrac{\psi(ru)}{(1+u)^{m+1}}du$

are finite and $\displaystyle\lim_{r\to\infty}\dfrac{\Psi(r)}{\Phi(r)}=1$ then
$\displaystyle\lim_{r\to\infty}\dfrac{\psi(r)}{\varphi(r)}=1$.

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