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To appear in Integral Equations Operator Theory. 
AMS-Code: 47D08, 47D06, 35J10
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Exponetial product formula, Trotter product formula,  
Trotter--Kato product formula, pointwise integral kernel convergence, 
harmonic oscillator.
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\begin{document}

\title{Note on Norm and Pointwise Convergence of Exponential Products
and their Integral Kernels for the Harmonic Oscillator}

\author{Yoshiki Azuma and Takashi Ichinose}


\date{}

\maketitle

\medskip\noindent
{\small {\bf Abstract.} 
The aim of this paper is to study the exponential symmetric product formula
for the semigroup of the one-dimensional harmonic oscillator
to discuss its convergence pointwise of the integral kernels as well as
in norm with sharp optimal error bound.}


\medskip\noindent
{\small {\bf Mathematics Subject Classification (2000).}
 Primary 47D08; Secondary 47D06, 35J10}

\medskip\noindent
{\small {\bf Keywords.}
Exponetial product formula, Trotter product formula, 
Trotter--Kato product formula, pointwise integral kernel convergence,
harmonic oscillator.}




%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Introduction and Results}

The Trotter or Trotter--Kato product formula or the
exponential product formula is a useful tool to approximate the
semigroup for a generator being a sum of two operators (\cite{Tr}, \cite{Ka}). 
One of the typical
cases is: 
if $A$ and $B$ are nonnegative selfadjoint operators
in a Hilbert space ${\mathcal H}$ and if the {\it operator sum}
$H :=A+B$ with domain $D[H] := D[A] \cap D[B]$ is selfadjoint 
in ${\mathcal H}$, it holds in {\it strong operator topology} that
$$
 \lim_{n\rightarrow \infty}
  [e^{-\frac{t}{n}A}e^{-\frac{t}{n}B}]^n
 = \lim_{n\rightarrow \infty}
  [e^{-\frac{t}{2n}B}e^{-\frac{t}{n}A}e^{-\frac{t}{2n}B}]^n
 = e^{-tH},
$$
uniformly on each bounded $t$-interval in the closed half-line 
$[0, \infty)$.

It has been shown in \cite{IT}, \cite{ITTZ} that it holds even
{\it in operator norm}, namely,  that for $n = 1, 2, \dots$,
\begin{eqnarray}
 \|[e^{-\frac{t}{n}A}e^{-\frac{t}{n}B}]^n -e^{-tH}\|
     &\!=\!& O(n^{-1}), \label{nonsymprod}\\
 \|[e^{-\frac{t}{2n}B}e^{-\frac{t}{n}A}e^{-\frac{t}{2n}B}]^n
    - e^{-tH}\| &\!=\!& O(n^{-1}), \label{symprod}
\end{eqnarray}
uniformly on each bounded $t$-interval in the closed half-line 
$[0, \infty)$.
Further, if $H$ is strictly positive, the convergence is uniform
in the whole closed half-line $[0, \infty)$.
Here the error bound $O(n^{-1})$ is optimal, in the sense
that there exists a pair of unbounded selfadjoint operators
$A$ and $B$ with their sum $H$ being selfadjoint such that
the left-hand side of the non-symmetric/symmetric
product case (\ref{nonsymprod})/(\ref{symprod})
is bounded from below by $c(t)n^{-1}$ with 
a nonnegative continuous function $c(t)$ in $t\geq 0$, positive in
$t>0$ with $c(0)=0$. Note that, for both $A$ and $B$ bounded 
operators, the error bound in the non-symmetric product case is 
$O(n^{-1})$, while in the symmetric product case $O(n^{-2})$.

This result applies, needless to say,  to the Schr\"odinger 
operator $H =A+B$ in $L^{2}({\bf R}^d)$ with 
$A :=H_0 \equiv -\frac12 \Delta$ the Laplacian and 
$B :=V(x)$ a potential function growing
polynomially  as $|x|\rightarrow \infty$,
i.e. satisfying $V(x)\geq C (1+|x|^2)^{\rho/2} ,\ \ 
|\partial_x^{\alpha}V(x)| 
\leq C_{\alpha} (1+|x|^2)^{(\rho-\delta|\alpha|)_+/2}$
for some constants $C,\, C_{\alpha} \geq 0$ and 
$\rho \geq 0,\,0<\delta \leq 1$. In this case, 
Takanobu \cite{Ta} showed the integral kernels
also converge pointwise uniformly with error bound 
$O(n^{-\rho/2})$, using the Feynman--Kac formula.
Recently we proved in \cite{IT2} (cf. \cite{IT3}) that 
the symmetric product formula (\ref{symprod})
holds even with  the error bound $O(n^{-2})$, sharper than the general
optimal $O(n^{-1})$,  both pointwise for the 
integral kernels and in operator norm, the convergence taking place
uniformly on each {\it compact} $t$-interval in the {\it open} 
half-line $(0, \infty)$. As it had been anticipated that there be also
a pair of unbounded selfadjoint operators $A$ and $B$ for which
the error bound is $O(n^{-2})$, this problem was thus
settled.

The aim of this paper is to specifically consider 
the one-dimensional harmonic oscillator
 $H := -\frac{1}{2}\Delta+ \frac12 x^2$ with
 $H_0 := -\frac{1}{2}\Delta, \,\, V := \frac12 x^2$, 
in $L^{2}({\bf R})$ to improve slightly these results in \cite{IT2}.
In fact, we show in the {\it symmetric} product case, apart from the
{\it non-symmetric} case, that 
the integral kernels converge pointwise {\it uniformly 
on each bounded interval} $(0, T]$ in the {\it open} half-line 
$(0,\infty)$ {\it just} with the error bound $O(n^{-2})$.
As a by-product, we also show that the error bound $O(n^{-2})$
for the symmetric product formula in norm is optimal not only from above
but also from below.
It is our main aim to settle this problem on optimal error bounds.
We put the {\it symmetric} product as
\begin{equation} \label{symprodkernel}
K^{(n)}(t) :=  \bigl[e^{-\frac{t}{2n}V} e^{-\frac{t}{n}H_0}
      e^{-\frac{t}{2n}V}\bigr]^n,
\end{equation}
and its integral kernel as $K^{(n)}(t,x,y)$. 
The integral kernel $K(t,x,y)$ of the semigroup $K(t) := e^{-tH}$
is known and {\it explicitly} given (e.g. \cite{GJ}) by
\begin{equation}  \label{Moellerkernel} 
K(t,x,y)=\frac{1}{\sqrt{\pi(e^t-e^{-t})}}
\exp\Big[\frac{4xy- (e^{t}+e^{-t})(x^2+y^2)}
{2(e^{t}-e^{-t})}\Big];
\end{equation}
this fact is crucial in the process of this paper.
The proof is made in an elementary way with Taylor's theorem,
first obtaining an explicit expression of $K^{(n)}(t,x,y)$ 
and then estimating its difference from $K(t,x,y)$.

We give two theorems.
The first theorem concerns the pointwise convergence
of the integral kernels of the exponential symmetric product formula.

\begin{theorem}%[Main Theorem]
There exist uniformly bounded sequences 
$\{R^{(n)}(t,x,y)\}_{n=1}^{\infty}$ ${}$ and 
$\{Q^{(n)}(t,x,y)\}_{n=1}^{\infty}$ of 
continuous functions in $(0,\infty)\times {\bf R}\times {\bf R}$
satisfing the following conditions:
it holds for every fixed $T>0$ and for sufficiently large $n$
that 
\begin{eqnarray}
 &\!K^{(n)}(t,x,y)- K(t,x,y) =  R^{(n)}(t,x,y)n^{-2},\!& 
                                          \label{eq1-thm1}\\
 &\!R^{(n)}(t,x,y) = R(t,x,y) + Q^{(n)}(t,x,y)n^{-1},\!&
                                             \label{eq2-thm1}
\end{eqnarray}
in $(t,x,y) \in (0,T] \times {\bf R}\times {\bf R}$,
where $R(t,x,y)$ is a bounded continuous function
in $(0,\infty)\times {\bf R}\times {\bf R}$ given by
\begin{eqnarray}
R(t,x,y)  %\nonumber\\
&\!=\!& K(t,x,y)
    \Big[\frac{t^3}{12}\Big(\frac1{4}\frac{e^t+e^{-t}}{e^t-e^{-t}}
        +\frac{(e^t+e^{-t})xy-(x^2+y^2)}{(e^t-e^{-t})^2}\Big)
                                     \nonumber\\
&&\qquad\qquad\quad\quad   + \frac{t^2}{16}
    \Big(1+\frac{4xy-(e^t+e^{-t})(x^2+y^2)}{e^t-e^{-t}}\Big)\Big]. 
                                  \label{eq3-thm1}
\end{eqnarray}
If $t>0$, $R(t,x,y)$ can become positive and negative. $R(t,x,y)$,
$Q^{(n)}(t,x,y)$ and so $R^{(n)}(t,x,y)$ satisfy
\begin{eqnarray} 
 &&\lim_{t\rightarrow 0} \sup_{x,y} |R(t,x,y)| 
    = \lim_{t\rightarrow \infty} \sup_{x,y} |R(t,x,y)| = 0,
                                 \label{lim0/infty-of-R}   \\
 &&\lim_{t\rightarrow 0} \sup_n \sup_{x,y} |Q^{(n)}(t,x,y)|= 0, 
                                 \label{lim0/infty-of-Q^{(n)}}\\
 &&\lim_{t\rightarrow 0} \sup_n \sup_{x,y} |R^{(n)}(t,x,y)|  = 0.
                                 \label{lim0/infty-of-R^{(n)}}
\end{eqnarray}

\end{theorem} 

\bigskip
In this theorem it is interesting that $R(t,x,y)$ 
appearing in the main part of the error term can be
identified as (\ref{eq3-thm1}). 
Note that it determines $Q^{(n)}(t,x,y)$ 
as  well as $R^{(n)}(t,x,y)$ through the relations
(\ref{eq1-thm1}) and (\ref{eq2-thm1}).

\bigskip\noindent
{\it Remark}.
 As $t \rightarrow 0+$, the integral kernel 
$K(t,x,y)$ converges to $\delta(x-y)$ in the sense of 
distributions in ${\bf R}^2$.  The approximation $K^{(n)}(t,x,y)$ can 
cancel out well the singularity of $K(t,x,y)$ in the limit
when $t$ approaches zero, since the right-hand side of (\ref{eq1-thm1})
is uniformly bounded in 
$(0,T] \times {\bf R}\times {\bf R}$ and for sufficiently large $n$,
i.e.
$R(t,x,y)$ is uniformly bounded in $(0, \infty) \times {\bf R}\times {\bf R}$,
and $Q^{(n)}(t,x,y)$ uniformly bounded in 
$(0,T] \times {\bf R}\times {\bf R}$ and for sufficiently large $n$,
and satisfy 
(\ref{lim0/infty-of-R}) and (\ref{lim0/infty-of-Q^{(n)}}), respectively,
although both $K^{(n)}(t,x,y)$ and $K(t,x,y)$ are unbounded as functions
in ${\bf R}^2$ and $t>0$. 
We should like to emphasize  that this point is clarified by the theorem. 
In this respect, Theorem 1.1 turns out to be an improvement of our 
previous result in \cite{IT2}.

\medskip
From Theorem 1.1 we can also show the second theorem on the
convergence in $L^2$- operator norm with sharp error bounds.

\begin{theorem}%Theorem 1.2
There exist nonnegative, bounded 
continuous functions $C(t)$ and  $c(t)$ in $t\geq 0$, 
positive except $t=0$ with $C(0)=c(0)=0$,
independent of $n$, such that, for every fixed $T>0$ and
for sufficiently large $n$,
\begin{equation} \label{eq-thm2}
 c(t)n^{-2} \leq 
\|[e^{-\frac{t}{2n}V}e^{-\frac{t}{n}H_0}e^{-\frac{t}{2n}V}]^n
    - e^{-tH}\| \leq C(t) n^{-2}, 
\end{equation}
in $t \in [0,T]$. 
\end{theorem}


Of the two bounds in (\ref{eq-thm2}), the lower one  and 
the upper with such $t$-dependent $C(t)$ are new.
In \cite{IT2} the upper bound was established, for more general
potentials growing at infinity,  
as a sharp upper error bound $C n^{-2}$ uniformly on each compact 
$t$-interval inside the open half line $(0,\infty)$ 
with  $C$ being a positive constant possibly depending on this interval.
We shall show Theorem 1.2 by exploiting the integral kernels 
to be obtained in the proof of Theorem 1.1.

\medskip
We note that the first results on the norm convergence
were obtained by Rogava \cite{R} for a more restricted pair
of selfadjoint operators $A$ and $B$ with error bound 
$O(n^{-1/2}\log n)$, and by B.~Helffer \cite{He} for 
the Schr\"odinger operator $H=H_0+V(x)$ as above
with  $\rho =2, \delta=0$ with error bound $O(n^{-1})$.
The latter result was extended in \cite{ITak} and \cite{DIT}
 to the general $\rho$ and $\delta$. For some result on the
{\it form sum} of two selfadjoint operators we refer to \cite{INZ},
and for further extensive references on related subjects 
to \cite{Ich}, \cite{IT} and \cite{Za}.

%\medskip
The present paper is a slightly enlarged version of the results 
obtained by the first author \cite{A}.

Section 2 is devoted to derivation of the expression of 
the integral kernel of the product $K^{(n)}(t)$ through 
an recursion relation. In Section 3 we find $R(t,x,y)$,
and prove the results described in Section 1. 
Section 4 briefly remarks another derivation of $R(t,x,y)$
by a sophisticated method of commutators as used in [9].
 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\setcounter{equation}{0}

\section{The integral kernel of $K^{(n)}(t)$}

Let $t>0$. Since the semigroup $e^{-tH_0}$ in (\ref{symprodkernel}) has the 
integral kernel
\begin{equation}
e^{-tH_0}(x,y) = e^{\frac{t}{2}\Delta}(x,y)
 =(2\pi t)^{-\frac{1}{2}}e^{-(x-y)^2/2t},
\end{equation}
the integral kernel $K^{(n)}(t,x,y)$ of 
$K^{(n)}$ can be written as the following integral:
\begin{eqnarray}
&& K^{(n)}(t,x,y) \label{1stexpresprodkern}\\
&\!=\!& \Big(\frac{2\pi t}{n}\Bigr)^{-\frac{n}{2}}
        \overbrace{\int_{\bf{R}}\cdots\int_{\bf{R}}}^{\mbox{$(n-1)$ times}}
        \prod_{j=1}^n\Big[e^{-\frac{t}{4n}x_j^2} 
        e^{-\frac{(x_j-x_{j-1})^2}{2t/n}}
        e^{-\frac{t}{4n}x_{j-1}^2}\Big]
         dx_1\cdots dx_{n-1} \nonumber
\end{eqnarray}
with $x_0=y$�C$x_n=x$.
The  $(n-1)$-ple integral on the right of (\ref{1stexpresprodkern}) 
is calculated in the following lemma.


\bigskip
\begin{lemma}
The integral kernel of $K^{(n)}$ is given by
\begin{eqnarray}
&&K^{(n)}(t,x,y)   \nonumber\\
&\!=\!& \frac{1}{\sqrt{\pi}}\Biggl(\frac{\sqrt{1+\frac{t^2}{4n^2}}}
  {\Big(1+\frac{t}{n}\sqrt{1+\frac{t^2}{4n^2}}+\frac{t^2}{2n^2}\Big)^n
-\Big(1-\frac{t}{n}\sqrt{1+\frac{t^2}{4n^2}}+\frac{t^2}{2n^2}\Big)^n}
\Biggr)^{1/2}      \nonumber\\
&&\times 
\exp{\Biggl[\frac{2\sqrt{1+\frac{t^2}{4n^2}}}
{\Big(1+\frac{t}{n}\sqrt{1+\frac{t^2}{4n^2}}+\frac{t^2}{2n^2}\Big)^n
-\Big(1-\frac{t}{n}\sqrt{1+\frac{t^2}{4n^2}}+\frac{t^2}{2n^2}\Big)^n}}xy
\Biggr]             \label{3rdexpresprodkern-n}\\
&&\times 
\exp\Biggl\{\Biggl[-\frac{t}{4n}-\frac{n}{2t} \Biggl(1- \nonumber\\
&& 
\frac{\Big(1+\frac{t}{n}\sqrt{1+\frac{t^2}{4n^2}}+\frac{t^2}{2n^2}\Big)^{n-1}
-\Big(1-\frac{t}{n}\sqrt{1+\frac{t^2}{4n^2}}+\frac{t^2}{2n^2}\Big)^{n-1}}
{\Big(1+\frac{t}{n}\sqrt{1+\frac{t^2}{4n^2}}+\frac{t^2}{2n^2}\Big)^n
-\Big(1-\frac{t}{n}\sqrt{1+\frac{t^2}{4n^2}}+\frac{t^2}{2n^2}\Big)^n}
  \Biggr)\Biggr](x^2+y^2)\Biggr\}. \nonumber
\end{eqnarray}

\end{lemma}

\smallskip
\noindent {\em Proof}.
 For any integer $k$ with $2\leq k\leq n$, put
\begin{eqnarray}
&&K_k^{(n)}(t,x,y)   \label{1stexpresprodkern-k}\\
&\!=\!& \Big(\frac{2\pi t}{n}\Bigr)^{-\frac{k}{2}}
        \overbrace{\int_{\bf{R}}\cdots\int_{\bf{R}}}^{\mbox{$(k-1)$ times}}
        \prod_{j=1}^k \Big[e^{-\frac{t}{4n}x_j^2}
        e^{-\frac{(x_j-x_{j-1})^2}{2t/n}}
        e^{-\frac{t}{4n}x_{j-1}^2}\Big]
        \,dx_1\cdots dx_{k-1},   \nonumber
\end{eqnarray}
with $x_0=y$�C$x_k=x$.

We are going to derive an explicit expression of $K_k^{(n)}(t,x,y)$ 
by induction on $k$. 
Put $a=\frac{t}{4n},\,b=\frac{n}{2t}$. Though they satisfy 
$b=\frac{1}{8a}$, we shall not use this relation for a while.
We use the formula for the Gaussian integral 
$ \int_{\bf R} e^{-Ax^2+Bx}dx 
=\sqrt{\frac{\pi}{A}}e^{{B^2}/{4A}}$
to perform calculation.

For $k=2$, we have with $x_0=y$ and $x_2=x$�C
\begin{eqnarray*}
&&K_2^{(n)}(t,x,y)\\
&\!=\!& \Bigl(\frac{2\pi t}{n}\Bigr)^{-1} \int \Big(
e^{-\frac{t}{4n}x_2^2}e^{-\frac{(x_2-x_1)^2}{2t/n}} 
e^{-\frac{t}{4n}x_1^2}\Big)\Big(
e^{-\frac{t}{4n}x_1^2}e^{-\frac{(x_1-x_0)^2}{2t/n}} 
e^{-\frac{t}{4n}x_0^2}\Big)\, dx_1 \\
&\!=\!&  \Bigl(\frac{2\pi t}{n}\Bigr)^{-1}\sqrt{\frac{\pi}{2(a+b)}}
\exp\Big[-\Big(a+b-\frac{b^2}{2(a+b)}\Big)(x_0^2+x_2^2)\Big]
\exp{\Big[\frac{b^2}{a+b}x_0x_2\Big]}.
\end{eqnarray*}

If $k=3$,  we have with $x_0=y$ and $x_3=x$�C�@
\begin{eqnarray*}
K_3^{(n)}(t,x,y)
&\!=\!& \Bigl(\frac{2\pi t}{n}\Bigr)^{-{3}/{2}}
\int\int \prod_{j=1}^3\Big[e^{-\frac{t}{4n}x_j^2}
e^{-\frac{(x_j-x_{j-1})^2}{2t/n}} 
e^{-\frac{t}{4n}x_{j-1}^2}\Big]dx_1 dx_2\\
&\!=\!& \Big(\frac{2\pi t}{n}\Big)^{-{3}/{2}}
\sqrt{\frac{\pi}{2(a+b)}}
\sqrt{\frac{\pi}{2(a+b)-\frac{b^2}{2(a+b)}}} \\
&& \times 
 \exp\biggl[-\Big(a+b-\frac{b^2}{2(a+b)-\frac{b^2}{2(a+b)}}\Big)x_3^2\\
&& -\Big(a+b-\frac{b^2}{2(a+b)}-\frac{(\frac{b^2}{a+b})^2}
{4\bigl(2(a+b)-\frac{b^2}{2(a+b)}\bigr)}\Big)x_0^2\biggr]\\
&&\times 
\exp\biggl[\frac{\frac{b^3}{a+b}}{2(a+b)-\frac{b^2}{2(a+b)}}x_0x_3\biggr].
\end{eqnarray*}

We note, in the expression for $k=3$ above, that
the coefficient of $x_3^2$ coincides with that of $x_0^2$, because 
$$%\begin{eqnarray*}
\frac{b^2}{2(a+b)}+\frac{(\frac{b^2}{a+b})^2}
{4\big(2(a+b)-\frac{b^2}{2(a+b)}\big)}
    =\frac{b^2}{2(a+b)-\frac{b^2}{2(a+b)}}.
$$%\end{eqnarray*}

Now define the finite continued fractions $A_k$ 
by the recursion relation
\begin{equation}  \label{recursion}
 A_k=A_2-{\frac{b^2}{A_{k-1}}}, \quad k\geq 3,
 \quad   A_2=2(a+b).
\end{equation}
Note that $A_3= A_2-\frac{b^2}{A_2}=2(a+b)-\frac{b^2}{2(a+b)}$
and $A_4= A_2-\frac{b^2}{A_3}$.
As a result, we turn to have 
\begin{eqnarray*}
K_2^{(n)}(t,x,y) 
&\!\!\!\!=\!\!\!\!& \Big(\frac{2\pi t}{n}\Big)^{-1}
\Big(\frac{\pi}{A_2}\Big)^{{1}/{2}}
\exp\Big[-\bigl(\frac{A_2}{2}-\frac{b^2}{A_{2}}\bigr)
  (x^2+y^2)\Big]
\exp\Big[\frac{2b^2}{A_2}\,xy \Big],\\
K_3^{(n)}(t,x,y) 
&\!\!\!\!=\!\!\!\!& \Big(\frac{2\pi t}{n}\Big)^{-3/2}
\Big(\frac{\pi^2}{A_2A_3}\Big)^{{1}/{2}}
 \exp\Big[-\bigl(\frac{A_2}{2}-\frac{b^2}{A_{3}}\bigr)
                    (x^2+y^2)\Big]           
 \exp\Big[\frac{2b^3xy}{A_2A_3}\, \Big].
\end{eqnarray*}
Then we want to prove that 
for general $k$ with $2\leq k\leq n$ with $x_0=y$ and $x_k=x$�C
\begin{eqnarray}
&& \!\!\!\! K_{k}^{(n)}(t,x,y) \nonumber\\
&\!\!\!\!=\!\!\!\!& \Big(\frac{2\pi t}{n}\Big)^{-{k}/{2}}
{\int\cdots\int} \prod_{j=1}^{k}
\Big[e^{-\frac{t}{4n}x_j^2}e^{-\frac{(x_j-x_{j-1})^2}{2t/n}} 
e^{-\frac{t}{4n}x_{j-1}^2}\Big] dx_1\cdots dx_{k-1}
                                          \nonumber\\ 
&\!\!\!\!=\!\!\!\!& \Big(\frac{2\pi t}{n}\Big)^{-{k}/{2}}
\Big(\frac{\pi^{k-1}}{A_2 A_3\cdots A_k}\Big)^{{1}/{2}} \nonumber\\ 
 &&\qquad\quad \times 
  \exp\Big[\bigl(-\frac{A_2}{2}+\frac{b^2}{A_{k}}\bigr)
     (x_{k}^2+x_0^2)\Big]
\exp\Big[\frac{2b^{k}}{A_2A_3\cdots A_{k}}x_0x_{k}\Big].
                                     \label{2nsexpesprod-k}
\end{eqnarray}
We have seen above that this is valid for $k=2$ and also
for $k=3$.

Now assuming the validity for $k$, we shall show the case $k+1$. 
Then  we have by induction hypothesis for $k$
with $x_0=y$ and $x_{k+1}=x$,
\begin{eqnarray*}
&&K_{k+1}^{(n)}(t,x,y)\\
&\!=\!& \Big(\frac{2\pi t}{n}\Bigr)^{-\frac{k}{2}}
        \int_{\bf{R}}\cdots\int_{\bf{R}}
        \prod_{j=1}^{k}\Big[e^{-\frac{t}{4n}x_j^2}
        e^{-\frac{(x_j-x_{j-1})^2}{2{t}/{n}}}
        e^{-\frac{t}{4n}x_{j-1}^2}\Big] dx_1\cdots dx_{k-1}\\
&\!=\!& \Bigl(\frac{2\pi t}{n}\Bigr)^{-{(k+1)}/{2}}\int
         (e^{-ax_{k+1}^2}e^{-b(x_{k+1}-x_{k})^2}e^{-ax_k^2})
         \Bigl(\frac{\pi^{k-1}}
         {A_2 A_3\cdots A_k}\Bigr)^{{1}/{2}}\\
       &&\quad \times  
         \exp\Big[-(\frac{A_2}{2}-\frac{b^2}{A_{k}})
         (x_{k}^2+x_0^2)\Big]\exp\Big[\frac{2b^{k}}
         {A_2A_3\cdots A_{k}}x_0x_{k}\Big] dx_k\\
&\!=\!& \Bigl(\frac{2\pi t}{n}\Bigr)^{-{(k+1)}/{2}}
         \Bigl(\frac{\pi^{k-1}}
         {A_2 A_3\cdots A_k}\Bigr)^{{1}/{2}}\\
     &&\times
        \int \exp\Big[-A_{k+1}x_{k}^2
         +(2bx_{k+1}+\frac{2b^{k}}{A_1\cdots A_{k}}x_0)x_{k}\Bigr]dx_{k}\\
      &&\quad \times    
         \exp\Big[-\frac{A_2}{2}x_{k+1}^2
         -(\frac{A_2}{2}-\frac{b^2}{A_{k}})x_0^2\Big]\\
&\!=\!& \Bigl(\frac{2\pi t}{n}\Bigr)^{-{(k+1)}/{2}}
         \Bigl(\frac{\pi^{k}}{A_2 A_3\cdots A_{k+1}}\Bigr)^{{1}/{2}}
\exp\Big[\frac{2b^{k+1}}{A_2A_3\cdots A_{k+1}}x_0x_{k^+1}\Big]\\
&& \quad \times 
   \exp\Big[-(\frac{A_2}{2}-\frac{b^2}{A_{k+1}})x_{k+1}^2
     -(\frac{A_2}{2}-\frac{b^2}{A_k}
     -\frac{b^{2k}}{(A_2A_3 \cdots A_{k})^2A_{k+1}})x_0^2\Big].
\end{eqnarray*}
Here we can see that the coefficients of $x_0^2$ 
and $x_{k+1}^2$ coincide. In fact, it holds that
\begin{equation} \label{coeff-agree}
\frac{b^2}{A_{k}}+\frac{b^{2k}}{(A_2A_3 \cdots A_{k})^2A_{k+1}}
=\frac{b^2}{A_{k+1}}, \quad  n\geq 3,
\end{equation}
where we recall to have put
$
A_k=A_2-\frac{b^2}{A_{k-1}},\ \ A_2=2(a+b)
$
in (\ref{recursion}). 
To show it, since the left-hand side of (\ref{coeff-agree})
is equal to
$$%\begin{eqnarray*}
\frac{(A_2 \cdots A_{k-1})^2A_kA_{k+1}b^2+b^{2k}}
  {(A_2A_3 \cdots A_k)^2A_{k+1}}
= \frac{b^2}{A_{k+1}}\frac{(A_2 \cdots A_{k-1})^2A_kA_{k+1}+b^{2(k-1)}}
      {(A_2A_3 \cdots A_k)^2}.
$$%\end{eqnarray*}
we need to confirm that
\begin{equation} \label{identity}
\frac{(A_2 \cdots A_{k-1})^2A_kA_{k+1}+b^{2(k-1)}}
      {(A_2A_3 \cdots A_k)^2}=1,
\end{equation}
or
$$
(A_2 \cdots A_{k-1})^2A_kA_{k+1}+b^{2(k-1)} = (A_2A_3 \cdots A_k)^2.
$$
In view of the recursion relation, it is equivalent to
\begin{eqnarray*}%$$
&&A_2(A_2 \cdots A_{k-1})^2A_k - (A_2 \cdots A_{k-1})^2b^2 + b^{2(k-1)}\\
 &\!=\!&  A_2(A_2 \cdots A_{k-1})^2 A_k 
    - (A_2\cdots A_{k-2})^2A_{k-1}A_k b^2.
\end{eqnarray*}%$$
In the same way, it is equivalent to
\begin{eqnarray*}%$$
&&A_2(A_2 \cdots A_{k-2})^2A_{k-1}b^2 
  - (A_2 \cdots A_{k-3})^2A_{k-2}A_{k-1}b^4 +b^{2(k-1)}\\
&\!=\!& A_2(A_2 \cdots A_{k-2})^2 A_{k-1}b^2 
    - (A_2 \cdots A_{k-2})^2 b^4,
\end{eqnarray*}%$$
and  again to
$$
(A_2\cdots A_{k-3})^2b^6 +b^{2(k-1)}
 = (A_2\cdots A_{k-4})^2A_{k-3}A_{k-2}b^6,
$$
and further again to
$$
(A_2\cdots A_{k-5})^2A_{k-4}A_{k-3}b^8 +b^{2(k-1)}
 = (A_2\cdots A_{k-4})^2b^8.
$$
Thus at last we reach its equivalent
$$
(A_2)^2A_3A_4 b^{2(k-2)}+ b^{2(k-1)}
 = (A_2A_3)^2b^{2(k-3)},
$$
whence 
$$
(A_2)^2 b^{2(k-2)}- b^{2(k-1)}
 = A_2A_3b^{2(k-2)}.
$$
The last equation holds, because $A_3 =A_2 - \frac{b^2}{A_2}$.
Therefore we have proved (\ref{identity}) to yield
the  expression of $K^{(n)}_{k+1}(t,x,y)$, thus establishing
the desired expression (\ref{2nsexpesprod-k}) for any $k\leq n$.

\bigskip
Finally, we determine $A_n$ from their recursion relation
(\ref{recursion}).
Let $p$ and $q$ be the two solutions of 
$$
 z^2 -2(a+b)z+ b^2 =0.
$$
We have
$$
p=(a+b)+\sqrt{a^2+2ab}, \quad q=(a+b)-\sqrt{a^2+2ab},
$$
so that $p+q=2(a+b), \,pq=b^2$.
It follows that
\begin{eqnarray*}
  A_{n+1}-p
&\!=\!& \frac{2(a+b)A_n-b^2-pA_n}{A_n}  =\frac{qA_n-b^2}{A_n}\\
&\!=\!& \frac{q(A_n-{b^2}/{q})}{A_n}=\frac{q(A_n-p)}{A_n}, 
\end{eqnarray*}
and similarly,
$$A_{n+1}-q=\frac{p(A_n-q)}{A_n}.
$$
Therefore
$$\frac{A_n-p}{A_n-q}=\frac{q}{p}\frac{A_n-p}{A_n-q}
      =\Big(\frac{q}{p}\Big)^{n-2}\frac{A_2-p}{A_2-q}.
$$
Hence we obtain
\begin{eqnarray*}
A_n&\!=\!& \frac{p^n-q^n}{p^{n-1}-q^{n-1}}\\
   &\!=\!& \frac{1}{2}
\frac{\Bigl(\frac{t}{2n}+\frac{n}{t}+\sqrt{\frac{t^2}{4n^2}+1}\Bigr)^n
-\Bigl(\frac{t}{2n}+\frac{n}{t}-\sqrt{\frac{t^2}{4n^2}+1}\Bigr)^n}
{\Bigl(\frac{t}{2n}+\frac{n}{t}+\sqrt{\frac{t^2}{4n^2}+1}\Bigr)^{n-1}
-\Bigl(\frac{t}{2n}+\frac{n}{t}-\sqrt{\frac{t^2}{4n^2}+1}\Bigr)^{n-1}}.
\end{eqnarray*}
Substituting this $A_n$ into $K_{k}^{(n)}(t,x,y)$,
we have
\begin{eqnarray*}
&&\Bigl(\frac{2\pi t}{n}\Bigr)^{-{k}/{2}}
 \Bigl(\frac{\pi^{k-1}}{A_2 A_3\cdots A_k}\Bigr)^{{1}/{2}}\\
&\!=\!&  \Biggl[\frac{n^k}{2^k\pi^kt^k}
 \frac{\pi^{k-1}2^k\sqrt{1+\frac{t^2}{4n^2}}}
   {\Big(\frac{t}{2n}+\frac{n}{t} +\sqrt{\frac{t^2}{n^2}+1}\Big)^k
 -\Big(\frac{t}{2n}+\frac{n}{t}-\sqrt{\frac{t^2}{4n^2}+1}\Big)^k}
 \Biggr]^{{1}/{2}}\\
&\!=\!& \Biggl[
 \frac{1}{\pi}\frac{\sqrt{1+\frac{t^2}{4n^2}}}{\Big(1+\frac{t^2}{2n^2}+
 \frac{t}{n}\sqrt{1+\frac{t^2}{4n^2}}\Big)^k
 -\Big(1+\frac{t^2}{2n^2}-\frac{t}{n}\sqrt{1+\frac{t^2}{4n^2}}\Big)^k}
                         \Biggr]^{{1}/{2}}.
\end{eqnarray*}
We have also
\begin{eqnarray*}
\frac{2b^k}{A_2A_3\cdots A_k}
&\!\!=\!\!& 2\Big(\frac{n}{2t}\Big)^k
\frac{2^k\sqrt{1+\frac{t^2}{4n^2}}}{\Big(\frac{t}{2n}+\frac{n}{t}
+\sqrt{\frac{t^2}{n^2}+1}\Big)^k
-\Big(\frac{t}{2n}+\frac{n}{t}-\sqrt{\frac{t^2}{4n^2}+1}\Big)^k}\\
&\!\!=\!\!& \frac{2\sqrt{1+\frac{t^2}{4n^2}}}
{\Big(1+\frac{t^2}{2n^2}+\frac{t}{n}\sqrt{1+\frac{t^2}{4n^2}}\Big)^k
-\Big(1+\frac{t^2}{2n^2}-\frac{t}{n}\sqrt{1+\frac{t^2}{4n^2}}\Big)^k}\,,\\
\end{eqnarray*}
and
\begin{eqnarray*}
&&\!\!\!-\frac{A_2}{2}+\frac{b^2}{A_k}\\
&\!\!\!\!=\!\!\!\!& -(a+b)+\frac{b^2}{A_k}\\
&\!=\!& -\frac{t}{4n}-\frac{n}{2t}+(\frac{n}{2t})^2
2\frac{\Big(\frac{t}{2n}+\frac{n}{t}+\sqrt{\frac{t^2}{4n^2}+1}\Big)^{k-1}
-\Big(\frac{t}{2n}+\frac{n}{t}-\sqrt{\frac{t^2}{4n^2}+1}\Big)^{k-1}}
{\Big(\frac{t}{2n}+\frac{n}{t}+\sqrt{\frac{t^2}{4n^2}+1}\Big)^k
-\Big(\frac{t}{2n}+\frac{n}{t}-\sqrt{\frac{t^2}{4n^2}+1}\Big)^k}\\
&\!\!\!\!=\!\!\!\!& -\frac{t}{4n}-\frac{n}{2t}+\frac{n}{2t}
\frac{\Big(1+\frac{t^2}{2n^2}+\frac{t}{n}\sqrt{1+\frac{t^2}{4n^2}}\Big)^{k-1}
-\Big(1+\frac{t^2}{2n^2}-\frac{t}{n}\sqrt{1+\frac{t^2}{4n^2}}\Big)^{k-1}}
{\Big(1+\frac{t^2}{2n^2}+\frac{t}{n}\sqrt{1+\frac{t^2}{4n^2}}\Big)^k
-\Big(1+\frac{t^2}{2n^2}-\frac{t}{n}\sqrt{1+\frac{t^2}{4n^2}}\Big)^k}\\
&\!\!\!\!=\!\!\!\!& -\frac{t}{4n}  \\
  &&\quad -  
 \frac{n}{2t}\Biggl(1-
\frac{\Big(1+\frac{t^2}{2n^2}+\frac{t}{n}\sqrt{1+\frac{t^2}{4n^2}}\Big)^{k-1}
-\Big(1+\frac{t^2}{2n^2}-\frac{t}{n}\sqrt{1+\frac{t^2}{4n^2}}\Big)^{k-1}}
{\Big(1+\frac{t^2}{2n^2}+\frac{t}{n}\sqrt{1+\frac{t^2}{4n^2}}\Big)^k
-\Big(1+\frac{t^2}{2n^2}-\frac{t}{n}\sqrt{1+\frac{t^2}{4n^2}}\Big)^k}\Biggr).
\end{eqnarray*}
Hence
\begin{eqnarray*}
&&K_k^{(n)}(t,x,y)  \\
&\!=\!& \frac{1}{\sqrt{\pi}}\Biggl(\frac{\sqrt{1+\frac{t^2}{4n^2}}}
  {\Big(1+\frac{t}{n}\sqrt{1+\frac{t^2}{4n^2}}+\frac{t^2}{2n^2}\Big)^k
-\Big(1-\frac{t}{n}\sqrt{1+\frac{t^2}{4n^2}}+\frac{t^2}{2n^2}\Big)^k}
\Biggr)^{1/2}      \nonumber\\
&&\times 
\exp{\Biggl[\frac{2\sqrt{1+\frac{t^2}{4n^2}}}
{\Big(1+\frac{t}{n}\sqrt{1+\frac{t^2}{4n^2}}+\frac{t^2}{2n^2}\Big)^k
-\Big(1-\frac{t}{n}\sqrt{1+\frac{t^2}{4n^2}}+\frac{t^2}{2n^2}\Big)^k}}xy
\Biggr]             \\
&&\times 
\exp\Biggl\{\Biggl[-\frac{t}{4n}-\frac{n}{2t} \Biggl(1- \\
&& 
\frac{\Big(1+\frac{t}{n}\sqrt{1+\frac{t^2}{4n^2}}+\frac{t^2}{2n^2}\Big)^{k-1}
-\Big(1-\frac{t}{n}\sqrt{1+\frac{t^2}{4n^2}}+\frac{t^2}{2n^2}\Big)^{k-1}}
{\Big(1+\frac{t}{n}\sqrt{1+\frac{t^2}{4n^2}}+\frac{t^2}{2n^2}\Big)^k
-\Big(1-\frac{t}{n}\sqrt{1+\frac{t^2}{4n^2}}+\frac{t^2}{2n^2}\Big)^k}
  \Biggr)\Biggr](x^2+y^2)\Biggr\}. 
\end{eqnarray*}
Putting $k=n$�Cso that $x_k=x_n=x$ and $x_0=y$, we have obtained
$K^{(n)}(t,x,y)$ in (\ref{3rdexpresprodkern-n}) in Lemma 2.1. 
$\Box$%\end{proof}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\setcounter{equation}{0}

\section{Proof of the theorems}

We shall prove first Theorem 1.1 and next Theorem 1.2.

\subsection{Proof of Theorem 1.1}

Introduce
\begin{eqnarray}
E_n(t) &\!\!\!\!:=\!\!\!\!\!&
  \Big(1+\frac{t}{n}\sqrt{1+\frac{t^2}{4n^2}}+\frac{t^2}{2n^2}\Big)^n
 -\Big(1-\frac{t}{n}\sqrt{1+\frac{t^2}{4n^2}}+\frac{t^2}{2n^2}\Big)^n,
        \label{expression-of-En}\\
F_n(t) &\!\!\!\!:=\!\!\!\!\!&
  \Big(1+\frac{t}{n}\sqrt{1+\frac{t^2}{4n^2}}+\frac{t^2}{2n^2}\Big)^{n-1}
 -\Big(1-\frac{t}{n}\sqrt{1+\frac{t^2}{4n^2}}+\frac{t^2}{2n^2}\Big)^{n-1}.
             \label{expression-of-Fn}
\end{eqnarray}
Then we may write $K^{(n)}(t,x,y)$  
in (\ref{3rdexpresprodkern-n}) in Lemma 2.1 as the product of the 
following three factors $a(t,x,y)$, $b(t,x,y)$ and $c(t,x,y)$:
\begin{eqnarray}
a(t,x,y)
&\!=\!& \frac{1}{\sqrt{\pi}}\biggl[\frac{\bigl(1+\frac{t^2}{4n^2}\bigr)^{1/2}}
  {E_n(t)}\biggr]^{1/2}, \label{eq-of-a}\\
b(t,x,y)
&\!=\!& \exp\biggl[\frac{2 \bigl(1+\frac{t^2}{4n^2}\bigr)^{1/2}}
    {E_n(t)}\, xy\biggr], \label{eq-of-b}\\
c(t,x,y)
&\!=\!& \exp\biggl[-\Big(\frac{t}{4n}+\frac{n}{2t}
\bigl(1- \frac{F_n(t)}{E_n(t)}\bigr)\Big)(x^2+y^2)\biggr]. \label{eq-of-c}
\end{eqnarray}

We want to estimate the difference between $K^{(n)}(t,x,y)$ and
$K(t,x,y)$. To this end, put with $R(t,x,y)$ in (\ref{eq3-thm1})
\begin{eqnarray}
R^{(n)}(t,x,y)
 &\!\!=\!\!& n^2\big(K^{(n)}(t,x,y) - K(t,x,y)\big), \label{def-of-R^{(n)}}\\
Q^{(n)}(t,x,y)
 &\!\!=\!\!& n(R^{(n)}\big(t,x,y) - R(t,x,y)\big).   \label{def-of-Q^{(n)}}
\end{eqnarray}

\medskip
First we shall study the property of $R(t,x,y)$, accepting its explicit 
expression as in (\ref{eq3-thm1}) in Proposition 3.1,  and next derive 
this expression in the proof of Proposition 3.2. 
It will turn out to be given through
$$
  \frac1{2 !} R(t,x,y) = \frac{\partial^2}{\partial \varepsilon^2}
               K^{\varepsilon}(t,x,y)\Big|_{\varepsilon =0}\,,
$$ 
where $K^{\varepsilon}(t,x,y)$ is defined through 
$$%\begin{equation} \label{def-of-K^{varepsilon}}
K^{\varepsilon}(t,x,y) := K^{(n)}(t,x,y)
$$%\end{equation}
first by putting $\varepsilon = 1/n$ and then by letting $\varepsilon$
be an arbitrary positive number. We can also see that
$K^{\varepsilon}(t,x,y)$ makes sense for $\varepsilon <0$.
Defining $K^{\varepsilon}(t,x,y) =K(t,x,y)$ for
$\varepsilon =0$, because 
$\lim_{\varepsilon \rightarrow 0} K^{\varepsilon}(t,x,y)
= K(t,x,y)$, we see further that $K^{\varepsilon}(t,x,y)$
is real-analytic in $\varepsilon \in {\bf R}$. Here note that the identity
\begin{equation} \label{inverse-rel}
 1-\tau\sqrt{1+\frac{\tau^2}{4}} +\frac{\tau^2}{2}
 = \Big(1+\tau\sqrt{1+\frac{\tau^2}{4}} +\frac{\tau^2}{2}\Big)^{-1}
\end{equation}
holds for $\tau$ real, and 
$$
 \lim_{\varepsilon \rightarrow 0}
\Big[\Big(1+t\varepsilon\sqrt{1+\frac{t^2\varepsilon^2}{4}}
  +\frac{t^2\varepsilon^2}{2}\Big)^{1/\varepsilon}
 - \Big(1-t\varepsilon\sqrt{1+\frac{t^2\varepsilon^2}{4}}
  +\frac{t^2\varepsilon^2}{2}\Big)^{1/\varepsilon}\Big] =e^t-e^{-t},
$$
so that 
$\lim_{n \rightarrow \infty} E_n(t) 
=\lim_{n \rightarrow \infty} F_n(t) = e^t -e^{-t}$.
Another derivation of $R(t,x,y)$ through the
commutator method is given in Section 4.

\bigskip

\begin{proposition}%{\bf Proposition 3.1.} 
If $t>0$, $R(t,x,y)$ takes positive and negative
values, and satisfies $|R(t,x,y)| \leq C_1(t)$ in 
$(0,\infty)\times {\bf R}\times {\bf R}$, where
$C_1(t)$  is a bounded continuous function in $t\geq 0$ such that 
$\lim_{t\rightarrow 0+} C_1(t)= \lim_{t\rightarrow \infty} C_1(t) =0$.
In particular, $R(t,x,y)$ satisfies (\ref{lim0/infty-of-R}).

\end{proposition}

\smallskip
\noindent {\em Proof}.
We may rewrite the expression of $R(t,x,y)$ in (\ref{eq3-thm1}) as
\begin{equation} \label{eq-of-R}
 R(t,x,y)= K(t,x,y)s(t,x,y),
\end{equation}
with
\begin{eqnarray}
s(t,x,y)  
&\!\!\!:=\!\!\!& 
  \frac{t^3}{12}\Big(\frac1{4}\frac{e^t+e^{-t}}{e^t-e^{-t}}
        +\frac{(e^t+e^{-t})xy-(x^2+y^2)}{(e^t-e^{-t})^2}\Big)
                                     \nonumber\\
&&\qquad\qquad\qquad\quad   + \frac{t^2}{16}
    \Big(1+\frac{4xy-(e^t+e^{-t})(x^2+y^2)}{e^t-e^{-t}}\Big)
                                \label{expression1-of-s} \\
&\!\!\!=\!\!\!& 
\frac{t^2}{2}\Big(\frac{1}{8} 
     +\frac{t}{24}\frac{e^t+e^{-t}}{e^t-e^{-t}}\Big) 
                                 \nonumber\\
&\!\!+\!\!\!&
  \frac{4xy-(e^t+e^{-t})(x^2+y^2)}{2(e^t-e^{-t})}
   t^2\Big(\frac{1}{8}+ \frac{t}{24}
           \frac{e^t+e^{-t}}{e^t-e^{-t}}\Big)
           + \frac{t^3}{48}(x^2+y^2)\,. \nonumber\\
&&                           \label{expression2-of-s} 
\end{eqnarray}
In this section we use the second expression 
(\ref{expression2-of-s}).  Put
\begin{equation} \label{expression-of-r}
 r(t,x,y) := -\frac{4xy-(e^t+e^{-t})(x^2+y^2)}{2(e^t-e^{-t})},
\end{equation}
which is nonnegative for all $x,\, y$ and every $t>0$, so that 
$K(t,x,y)= \frac{e^{-r(t,x,y)}}{\sqrt{\pi(e^t-e^{-t})}}$.
Put also 
\begin{eqnarray}
q(t,x,y) 
&\!\!:=\!\!& -\frac{4xy-(e^t+e^{-t})(x^2+y^2)}{2(e^t-e^{-t})}
      t^2\Big(\frac{1}{8}+ \frac{t}{24} \frac{e^t+e^{-t}}{e^t-e^{-t}}\Big)
 - \frac{t^3}{48}(x^2+y^2) \nonumber\\
&\!\!=\!\!& t^2\Big[r(t,x,y)
 \Big(\frac{1}{8}+ \frac{t}{24} \frac{e^t+e^{-t}}{e^t-e^{-t}}\Big)
 - \frac{t}{48}(x^2+y^2)\Big], \label{expression-of-q}
\end{eqnarray}
so that from (\ref{expression2-of-s})
\begin{equation}
 s(t,x,y)
 = \frac{t^2}{2}\Big(\frac{1}{8}
 +\frac{t}{24}\frac{e^t+e^{-t}}{e^t-e^{-t}}\Big) -q(t,x,y)\,.            
                                    \label{expression22-of-s} 
\end{equation}

The coefficients of $x^2$ and $y^2$ in $q(t,x,y)$ 
in (\ref{expression-of-q})
are nonnegative, because they are equal and given by  
$$%\begin{eqnarray*}
\frac{(e^t+e^{-t})t^2\Big(\frac{1}{8}
  + \frac{t}{24}\frac{e^t+e^{-t}}{e^t-e^{-t}}\Big)
     -\frac{t^3}{24}(e^t-e^{-t})}{2(e^t-e^{-t})}
 = \frac{t^2\big(3(e^{2t}-e^{-2t})+4t\big)}{48(e^t-e^{-t})^2}.
$$%\end{eqnarray*}
Further, $q(t,x,y)$ is nonnegative.
To see it, note that for $t>0$ and  $(x,y) \not= (0,0)$,
$$%\begin{eqnarray*}
\frac{q(t,x,y)}{t^2r(t,x,y)} 
%&\!=\!& 
%\frac{-\frac{4xy-(e^t+e^{-t})(x^2+y^2)}{2(e^t-e^{-t})}
%t^2\Big(\frac{1}{8}+ \frac{t}{24} \frac{e^t+e^{-t}}{e^t-e^{-t}}\Big)
%      - \frac{t^3}{48}(x^2+y^2)}
% {-\frac{4xy-(e^t+e^{-t})(x^2+y^2)}{2(e^t-e^{-t})}t^2}         \\
= \frac{1}{8}+\frac{t}{24}\frac{e^t+e^{-t}}{e^t-e^{-t}}
    + \frac{t}{24}\frac{e^t-e^{-t}}{\frac{4xy}{x^2+y^2}-(e^t+e^{-t})}.
$$%\end{eqnarray*}
Since
$-2 \leq \frac{4xy}{x^2+y^2} \leq 2$ for $(x,y) \not= (0,0)$,
it follows that
\begin{eqnarray*}
\frac{1}{8}+\frac{t}{24}\frac{e^t+e^{-t}}{e^t-e^{-t}}
    - \frac{t}{24}\frac{e^t-e^{-t}}{(e^t+e^{-t})-2}
&\! \leq \!& 
\frac{q(t,x,y)}{t^2r(t,x,y)}\\
&\! \leq \!&  
\frac{1}{8}+\frac{t}{24}\frac{e^t+e^{-t}}{e^t-e^{-t}}
    - \frac{t}{24}\frac{e^t-e^{-t}}{(e^t+e^{-t})+2}\,,
\end{eqnarray*}
which is equivalent to that
$$%\begin{eqnarray*}
\frac{1}{8}-\frac{t}{12}\frac{1}{e^t-e^{-t}}
\leq 
\frac{q(t,x,y)}{t^2r(t,x,y)}
\leq  
\frac{1}{8}+\frac{t}{12}\frac{1}{e^t-e^{-t}}.
$$%\end{eqnarray*}
Then, since the first member of the above inequality is equal to
$
\frac{1}{12}+ \frac{(e^t-e^{-t})-2t}{24(e^t-e^{-t})} 
  \geq \frac{1}{12},
$
and the last member bounded by $\frac{1}{6}$, 
we obtain
\begin{equation} \label{ratio-of-q/r}
\frac{1}{12} \leq \frac{q(t,x,y)}{t^2r(t,x,y)} 
\leq \frac{1}{6}.
\end{equation}
In particular, we have shown  that $q(t,x,y)$ is nonnegative,
since $r(t,x,y)$ is nonnegative.
Thus, in view of (\ref{expression22-of-s}),
we see that if $t>0$, $s(t,x,y)$ takes positive and negative
values, and so does $R(t,x,y)$.

Next, to see that $R(t,x,y)$ is bounded in
$(0,\infty)\times {\bf R}\times {\bf R}$ and satisfies
(\ref{lim0/infty-of-R}),  
we note that for fixed $m \geq 1 >\alpha >0$,
\begin{eqnarray} \label{wedge}
&& \frac{t^m}{(e^t-e^{-t})^{\alpha}} 
\leq c(\alpha,m) [t^{m-\alpha} \wedge 1], 
                      \qquad\qquad\qquad\qquad\qquad t>0,\\
&&t^m \frac{e^t+e^{-t}}{e^t-e^{-t}} 
\leq  \frac{e+e^{-1}}{e-e^{-1}}\,[1 \vee t]t^{m-1}
\leq  \frac{e+e^{-1}}{e-e^{-1}}\,[1 \vee t^m], 
                      \quad\, t>0,  \label{vee}
\end{eqnarray}
with a postive constant $c(\alpha,m)$ depending on 
$\alpha$ and $m$,
where we write $a \wedge b := \min\{a,b\}$ and 
$a\vee b := \max\{a,b\}$
for positive numbers $a$ and $b$. Indeed, for (\ref{wedge})
%we have $\frac{t^m}{(e^t-e^{-t})^{\alpha}} 
%\leq  2^{-\alpha}t^{m-\alpha}$ for $0<t\leq 1$, and 
%$\frac{t^m}{(e^t-e^{-t})^{\alpha}} \leq
%2^{\alpha}(m/\alpha e)^m$ for $t\geq 1$, where 
we have used the fact that $\tau^a e^{-\tau} \leq (a/e)^a$ for 
all $\tau > 0$ and $a >0$.
%, while for (\ref{vee}) we have
%$t^m \frac{e^t+e^{-t}}{e^t-e^{-t}}
% \leq \frac{e+e^{-1}}{e-e^{-1}}t^{m-1}$
%for $0<t\leq 1$, and $t^m\frac{e^t+e^{-t}}{e^t-e^{-t}} \leq 
%\frac{e+e^{-1}}{e-e^{-1}}t^m$ for $t\geq 1$.

Then we have by (\ref{wedge}) and (\ref{vee})
\begin{eqnarray}
 |R(t,x,y)| 
&\!\!\leq \!\!& 
 \frac{e^{-r(t,x,y)}}{\sqrt{\pi(e^t-e^{-t})}}
 \frac{t^2}{2}\Big(\frac{1}{8} 
     +\frac{t}{24}\frac{e^t+e^{-t}}{e^t-e^{-t}}\Big)
                                          \nonumber\\
&&\!\!   + \frac{1}{\sqrt{\pi(e^t-e^{-t})}}
                   \Big(e^{-r(t,x,y)}r(t,x,y)\Big)\,
     \frac{q(t,x,y)}{t^2r(t,x,y)}\, t^2  \nonumber\\
&\!\!\leq\!\!& \frac{1}{\sqrt{\pi(e^t-e^{-t})}}
\Big(\frac{1}{16}+\frac{1}{48}\,\frac{e+e^{-1}}{e-e^{-1}}\,
[1 \vee t] +\frac{1}{6e}\Big)t^2 \nonumber\\
&\!\!=:\!\!& C_1(t).  \label{bound-of-R}
\end{eqnarray}
Then $C_1(t)$ is bounded in $t>0$, because 
by (\ref{wedge}), we have
$C_1(t) \leq C_1 \frac{t^2[1 \vee t]}{(e^t-e^{-t})^{1/2}}$,
with a positive constant $C_1$ independent $t$.
It also satisfies $\lim_{t\rightarrow 0+}C_1(t) =0$, and so
we may define $C_1(0)=0$.
Here we have used (\ref{ratio-of-q/r}).
We can also see $\lim_{t\rightarrow \infty} C_1(t) =0$.
This shows that $R(t,x,y)$ is bounded in 
$(0,\infty)\times {\bf R}\times {\bf R}$, also
satisfying (\ref{lim0/infty-of-R}).
This proves Proposition 3.1. 
$\Box$%\end{proof}

%\bigskip
Now we are going to estimate $R^{(n)}(t,x,y)$
when  $0<t \leq n^{1/2}$. After proving it, 
we shall show it yields Theorem 1.1.
We note  that $0<t \leq n^{1/2}$ if and only it
$t(t/n) \leq 1$, which also implies 
$t(t/n)^j \leq n^{-(j-1)/2}\leq 1$ for $j \geq 2$. 


\bigskip%\noindent
\begin{proposition}%{\bf Proposition 3.2.} 
Let $0< t\leq n^{1/2}$. Then
$Q^{(n)}(t,x,y)$ and $R^{(n)}(t,x,y)$ in
(\ref{def-of-R^{(n)}})/(\ref{def-of-Q^{(n)}}) or
(\ref{eq1-thm1})/(\ref{eq2-thm1}) satisfy
\begin{eqnarray}
 &&|Q^{(n)}(t,x,y)| \leq C_2(t), \label{bd-of-kerQ^{(n)}}\\ 
 &&|R^{(n)}(t,x,y)| \leq C_1(t) + C_2(t),
                                       \label{bd-of-kerR^{(n)}}
\end{eqnarray}
in $(x,y) \in {\bf R}\times {\bf R}$
for  large $n$, where $C_1(t)$ is the same bounded 
continuous function in Proposition 3.1, i.e. (\ref{bound-of-R}),
and $C_2(t)$ another bounded continuous function in $t\geq 0$
such that
$\lim_{t\rightarrow 0+} C_2(t) =0$.

\end{proposition}


\bigskip
Before proving this proposition, 
we provide two lemmas to get expansions of
$E_n(t),\,\, F_n(t)$ in (\ref{expression-of-En}), 
(\ref{expression-of-Fn}) with respect to the powers of $t/n$
with $0< t\leq n^{1/2}$.


\bigskip

\begin{lemma}% Lemma 3.3 
Let $\tau$ be real. Then
\begin{eqnarray*}
\hbox{\rm (i)}
 &&\sqrt{1+\frac{\tau^2}{4}}= 1+ \frac{\tau^2}{8} +  O(|\tau|^4)\,;\\
\hbox{\rm (ii)}
 &&1+\tau\sqrt{1+\frac{\tau^2}{4}}+ \frac{\tau^2}{2}
= 1+\tau +\frac{\tau^2}{2}+ \frac{\tau^3}{8} + O(|\tau|^5),\\
 &&\Big(1+\tau\sqrt{1+\frac{\tau^2}{4}}+ \frac{\tau^2}{2}\Big)^{-1}
     = 1- \tau+ \frac{\tau^2}{2} -\frac{\tau^3}{8} +  O(|\tau|^5).
\end{eqnarray*}
\end{lemma}

\smallskip
\noindent {\em Proof}.
The assertion (i) and the first of (ii) follow from Taylor's theorem 
$$
(1+x)^{1/2}
  = 1+\frac{x}{2}-\frac{x^2}{8}(1+\theta x)^{-3/2}, \quad x >-1,
$$
for some $0<\theta <1$, with $x = \frac 14 \tau^{2}$, 
and the second from (\ref{inverse-rel}). 
$\Box$%\end{proof}

\bigskip%\noindent
\begin{lemma}% Lemma 3.4 
Let $0<|t| \leq n^{1/2}$. Then
\begin{eqnarray*}
\hbox{\rm (i)}
&&\!\!
 \Big(1+ \frac{t}{n}\sqrt{1+\frac{t^2}{4n^2}}+\frac{t^2}{2n^2}\Big)^{n}
= e^t\Big[1- \frac{t}{24} (t/n)^2
      + (t+t^2)\, O \big(|t/n|^4\big)\Big]\,;\\
\hbox{\rm (ii)}
&&\!\!
 \Big(1+\frac{t}{n}\sqrt{1+\frac{t^2}{4n^2}}+\frac{t^2}{2n^2}\Big)^{n-1}\\
&\!=\!& e^{t}\Big[1- (t/n) +\Big(\frac12 -\frac{t}{24}\Big)(t/n)^2
  +\Big(-\frac18 + \frac{t}{24}\Big)(t/n)^3 
   + (t+t^2)\,O\big(|t/n|^4\big)\Big]\,.
\end{eqnarray*}
\end{lemma}


\smallskip
\noindent {\em Proof}.
 (i) First note that by Taylor's theorem we have for $x >-1$ 
$$
 \log(1+x)= x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}
            + \frac{x^5}{5}(1+\theta x)^{-5}
$$
for some $0<\theta <1$, and next
that if $\tau \geq -1$, then
$
 \tau\sqrt{1+\frac{\tau^2}{4}}+ \frac{\tau^2}{2}
          \geq -\frac{2}{1+\sqrt{5}} > -\frac23,
$
so that
$
 \Big(1+ \theta 
   \bigl(\tau\sqrt{1+\frac{\tau^2}{4}}+\frac{\tau^2}{2}\bigr)\Big)^{-1}
   \leq \Big(1-\frac{2}{1+\sqrt{5}}\Big)^{-1}
 = \frac{3+\sqrt{5}}{2}. 
$
Then, for $\tau \geq -1$ or, in particular, for $|\tau|\leq 1$, we obtain 
\begin{eqnarray*}
&&\log \Big(1+\tau\sqrt{1+\frac{\tau^2}{4}}+\frac{\tau^2}{2}\Big)\\
%&\!=\!& \sum _{j=1}^4 \frac{(-1)^{j-1}}{j}
%   \Big(\tau\sqrt{1+\frac{\tau^2}{4}}+\frac{\tau^2}{2}\Big)^{j}\\
% &&  + \frac{1}{5}\Big(\tau\sqrt{1+\frac{\tau^2}{4}}
%     +\frac{\tau^2}{2}\Big)^{5}
%     \Big(1+ \theta 
%     \bigl(\tau\sqrt{1+\frac{\tau^2}{4}}+\frac{\tau^2}{2}\bigr)
%     \Big)^{-5}\\ 
&\!=\!& \sum _{j=1}^4 \frac{(-1)^{j-1}}{j}
   \Big(\tau+\frac{\tau^2}{2}+ \frac{\tau^3}{8}+O(|\tau|^5)\Big)^{j}\\
 &&  + \frac{1}{5}
     \Big(\tau+\frac{\tau^2}{2}+ \frac{\tau^3}{8}+O(|\tau|^5)\Big)^{5}
     \Big(1+ \theta 
   \bigl(\tau\sqrt{1+\frac{\tau^2}{4}}+\frac{\tau^2}{2}\bigr)\Big)^{-5}, 
\end{eqnarray*}
so that
$
  \log \Big(1+\tau\sqrt{1+\frac{\tau^2}{4}}+\frac{\tau^2}{2}\Big)
= \tau- \frac{\tau^3}{24} + O(|\tau|^5).
$
Hence
$$%\begin{eqnarray*}
\frac{1}{\tau} 
 \log \Big(1+\tau\sqrt{1+\frac{\tau^2}{4}}+\frac{\tau^2}{2}\Big)
= 1- \frac{\tau^2}{24} + O(|\tau|^4).
$$%\end{eqnarray*}
Then, by taking $\tau= t/n$, we obtain for 
$|t| \leq n^{1/2}$ or $|t||t/n| \leq 1$
\begin{eqnarray*}
\biggl(1+\frac{t}{n}\sqrt{1+\frac{t^2}{4n^2}}+\frac{t^2}{2n^2}\biggr)^{n}
&\!=\!& \Big(1+\tau\sqrt{1+\frac{\tau^2}{4}}+\frac{\tau^2}{2}\Big)^{n}\\
&\!=\!& \exp\Big[\frac{t}{\tau}
 \log (1+\tau\sqrt{1+\frac{\tau^2}{4}}+\frac{\tau^2}{2})\Big]\\
&\!=\!& \exp\Big[t\bigl(1- \frac{\tau^2}{24} +O(|\tau|^4)\bigr)\Big]\\
&\!=\!& e^t\exp\Big[t\big(- \frac{1}{24}(t/n)^2+O((t/n)^4)\big)\Big]\\
&\!=\!& e^t\Big[1- \frac{t}{24} (t/n)^2+(|t|+|t|^2)O((t/n)^4)\Big].
\end{eqnarray*}
In the last equality above we note that since $|t| \leq n^{1/2}$,
we have 
$|t|\big|\frac{1}{24}(t/n)^2- O(|\frac{t}{n}|^4)\big| 
 = n^{-1/2}O(1)$, which becomes less than $1$ for $n$ large.

\medskip
(ii) In the same way as in (i) above with Lemma 3.3 (ii), we have 
\begin{eqnarray*}
&&\biggl(1+\frac{t}{n}\sqrt{1+\frac{t^2}{4n^2}}+\frac{t^2}{2n^2}\biggr)^{n-1}\\
&\!\!=\!\!& 
\biggl(1+\frac{t}{n}\sqrt{1+\frac{t^2}{4n^2}}+\frac{t^2}{2n^2}\biggr)^{n}
\biggl(1+\frac{t}{n}\sqrt{1+\frac{t^2}{4n^2}}+\frac{t^2}{2n^2}\biggr)^{-1}\\
&\!\!=\!\!& e^t\Big[1- \frac{t}{24} (t/n)^2 + (|t|+|t|^2) O((t/n)^4)\Big]\\
&&\quad \times
 \Big(1-\frac{t}{n}+ \frac12 (t/n)^2
        -\frac18 (t/n)^3 + O(|t/n|^5)\Big)\\
&\!\!=\!\!& e^t\Big[1- (t/n) +\Big(\frac12-\frac{t}{24}\Big) (t/n)^2
   +\Big(-\frac18 + \frac{t}{24}\Big)(t/n)^3 %\\ 
%&&\qquad\qquad\qquad\qquad   
 +(|t|+|t|^2)O((t/n)^4)\Big],
\end{eqnarray*}
because for the coefficient of $O((t/n)^4)$ we have
$|t|+|t|^2+|t/n|+|t||t/n| \leq 2(|t|+|t|^2)$ when $|t|\leq  n^{1/2}$.
This proves the desired expansions. 
$\Box$%\end{proof}

Now we start the proof of Proposition 3.2.

\medskip\noindent
{\it Proof of Proposition 3.2.}
We need to seek, when $0<t\leq n^{1/2}$, expansions of 
the factors $a(t,x,y)$, $b(t,x,y)$and $c(t,x,y)$ in (\ref{eq-of-a}),
(\ref{eq-of-b}) and (\ref{eq-of-c}) of $K^{(n)}(t,x,y)$. 
Since that $0<t\leq n^{1/2}$ implies that 
$t(t/n)^j \leq n^{-(j-1)/2} \leq 1,\, j\geq 2$,
we see by use of Lemma 3.4 (i) and (ii) that
$E_n(t)$ and $F_n(t)$ in (\ref{expression-of-En}) and 
(\ref{expression-of-Fn}) have the following  expansions 
\begin{eqnarray}
E_n(t) 
&\!\!=\!\!& e^t-e^{-t}-\frac{t}{24}(e^t+e^{-t})(t/n)^2
  +(t+t^2)(e^t+e^{-t}) O((t/n)^4))    \nonumber\\
&\!\!=\!\!& (e^t-e^{-t})\Big[1-\frac{t}{24}
    \frac{e^t+e^{-t}}{e^t-e^{-t}}(t/n)^2
  +[1 \vee t^2] O((t/n)^4)\Big], 
                                       \label{expression2-of-En}\\
F_n(t) 
&\!\!=\!\!& e^t-e^{-t}-(e^t+e^{-t})(t/n)
  +\Big(\frac{e^t-e^{-t}}{2}-{t}\frac{e^t+e^{-t}}{24}\Big)(t/n)^2
                                       \nonumber\\
&&\!\!
-t\Big(\frac{e^t+e^{-t}}{8}
    -t\frac{e^t-e^{-t}}{24}\Big)(t/n)^3 
     + (t+t^2)(e^{t}+e^{-t})O((t/n)^4) \nonumber\\  
&\!\!=\!\!& 
(e^t-e^{-t})\Big[1-\frac{e^t+e^{-t}}{e^t-e^{-t}}(t/n)
  +\Big(\frac{1}{2}-\frac{t}{24}\frac{e^t+e^{-t}}{e^t-e^{-t}}\Big)
         (t/n)^2                              \nonumber\\
&&\!\!
-t\Big(\frac{1}{8}\frac{e^t+e^{-t}}{e^t-e^{-t}}
    -\frac{t}{24}\Big)(t/n)^3 
+ [1 \vee t^2] O((t/n)^4)\Big].
                                 \label{expression2-of-Fn}                  
\end{eqnarray}
Here note by (\ref{vee}) that 
$
 (t+t^2)\frac{e^t+e^{-t}}{e^t-e^{-t}}\leq [1 \vee t]+ [1 \vee t^2]
  \leq 2 [1 \vee t^2].
$

%%%%%%%
\medskip
By using these new expressions (\ref{expression2-of-En}) and 
(\ref{expression2-of-Fn}), we 
find expansions for $a(t,x,y)$, $b(t,x,y)$ and $c(t,x,y)$.
As to $a(t,x,y)$, we obtain with Lemma 3.3 (i) 
\begin{eqnarray*}
a(t,x,y)
&\!\!=\!\!& \frac{1}{\sqrt{\pi}}
  \biggl[\frac{\bigl(1+\frac{t^2}{4n^2}\bigr)^{1/2}}{E_n(t)}\biggr]^{1/2}\\
&\!\!=\!\!&
 \frac{1}{\sqrt{\pi}}\biggl[\frac{1+\frac{1}{8}\big(\frac{t}{n}\big)^2
          +O\big(\big(\frac{t}{n}\big)^4\big)}
    {(e^t-e^{-t})\big[1
    -\frac{t}{24}\frac{e^t+e^{-t}}{e^t-e^{-t}}\big(\frac{t}{n}\big)^2
    +[1 \vee t^2] O\big(\big(\frac{t}{n}\big)^4\big)}
                                            \biggr]^{1/2}\\
&\!\!=\!\!& \frac{1}{\sqrt{\pi(e^t-e^{-t})}}
 \biggl[1+\frac{1}{2} \Big(\frac{1}{8} 
     +\frac{t}{24}\frac{e^t+e^{-t}}{e^t-e^{-t}}\Big)(t/n)^2 
   +[1 \vee t^2] O((t/n)^4)\biggr], 
\end{eqnarray*}
because the function $\sqrt{\frac{1+\tau}{1+\sigma}}$ is analytic in
$|\tau|<1$ and $|\sigma|<1$, and has an expansion
\begin{eqnarray*}
 \sqrt{\frac{1+\tau}{1+\sigma}}
&\!=\!& \Big(1+\frac{\tau}{2}-\frac{\tau^2}{8}
  +\frac{\tau^3}{16}- \cdots\Big)
  \Big(1-\frac{\sigma}{2}+\frac{3\sigma^2}{8}
   -\frac{5\sigma^3}{13}+ \cdots\Big)\\
&\!=\!& 1+\frac{\tau}{2}-\frac{\sigma}{2}-\frac18(\tau^2-\sigma^2)
     -\frac{\tau\sigma}{4}+ \cdots.
\end{eqnarray*}
The last equality in the equation for $a(t,x,y)$ above is due to that
one can think of the inverse function of $E_n(t)$ in $t>0$
for large $n$, because,  on the 
right of $E_n(t)$ in (\ref{expression2-of-En}),
the function inside the blacket $[\cdots]$ is invertible 
for large $n$,
since 
$|-\frac{t}{24}
    \frac{e^t+e^{-t}}{e^t-e^{-t}}(t/n)^2 +[1 \vee t^2] O((t/n)^4)|
$
is less than $1$  by (\ref{wedge}) and (\ref{vee}) for 
$0<t\leq n^{1/2}$, if $n$ is large.
The same note will apply to equations for $b(t,x,y)$ and
$c(t,x,y)$ which come below.

\medskip
As to $b(t,x,y)$, we have
\begin{eqnarray*}
b(t,x,y)
&\!\!=\!\!& 
\exp\biggl[
 \frac{2\bigl(1+\frac{t^2}{4n^2}\bigr)^{1/2}}{E_n(t)}\, xy\biggr]\\
&\!\!=\!\!&
\exp\Biggl[\frac{2xy}{e^t-e^{-t}}\Biggr(\frac
{1+\frac{1}{8}\big(\frac{t}{n}\big)^2+O(\big(\frac{t}{n}\big)^4)}
{1-\frac{t}{24}\frac{e^t+e^{-t}}{e^t-e^{-t}}\big(\frac{t}{n}\big)^2
 +[1 \vee t^2] O\big(\big(\frac{t}{n}\big)^4\big)}\Biggr)\Biggr]\\
&\!\!=\!\!&
\exp\Big(\frac{2xy}{e^t-e^{-t}}\Big)
 \exp\biggl[\frac{2xy}{e^t-e^{-t}} \biggl(\Big(\frac{1}{8} 
  +\frac{t}{24}\frac{e^t+e^{-t}}{e^t-e^{-t}}\Big)(t/n)^2\\
 &&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad
    +[1 \vee t^2] O\big(t/n)^4\big)\biggr].
\end{eqnarray*}

\medskip
As to $c(t,x,y)$, first we observe
with (\ref{expression2-of-En}) and (\ref{expression2-of-Fn})
\begin{eqnarray*}
&&\!\!\frac{n}{2t}\Big(1-\frac{F_n(t)}{E_n(t)}\Big)\\
&\!\!=\!\!&
\frac{e^t+e^{-t}}{2(e^t-e^{-t})}
\frac{1-\frac{1}{2}\frac{e^t-e^{-t}}{e^t+e^{-t}}\frac{t}{n}
  +\big(\frac18-\frac{t}{24}\frac{e^t-e^{-t}}{e^t+e^{-t}}\big)
    \big(\frac{t}{n}\big)^2
    + [1 \vee t^2] O\big(\big(\frac{t}{n}\big)^3\big)}
  {1-\frac{t}{24} \frac{e^t+e^{-t}}{e^t-e^{-t}}\big(\frac{t}{n}\big)^2
     +[1 \vee t^2] O\big(\big(\frac{t}{n}\big)^4\big)}\\
&\!\!=\!\!&
\frac{e^t+e^{-t}}{2(e^t-e^{-t})}
 \Big[1-\frac{1}{2}\frac{e^t-e^{-t}}{e^t+e^{-t}}(t/n) \\
&&\quad\qquad
   +\Big(\frac{1}{8}+ \frac{t}{24}\frac{e^t+e^{-t}}{e^t-e^{-t}}
   -\frac{t}{24} \frac{e^t-e^{-t}}{e^t+e^{-t}}\Big)(t/n)^2
   + [1 \vee t^2] O(({t}/{n})^3)\Big].
\end{eqnarray*}
Hence
\begin{eqnarray*}
&&\!\!\frac{t}{4n}+ \frac{n}{2t}\Big(1-\frac{F_n(t)}{E_n(t)}\Big)\\
&\!\!=\!\!&
\frac{e^t+e^{-t}}{2(e^t-e^{-t})}\\
&&\quad \times
   \biggl[1+\Big(\frac{1}{8}
  + \frac{t}{24}\frac{e^t+e^{-t}}{e^t-e^{-t}}
    -\frac{t}{24} \frac{e^t-e^{-t}}{e^t+e^{-t}}\Big)
                               ({t}/{n})^2
  + [1 \vee t^2] O(({t}/{n})^3)\biggr]\,.
\end{eqnarray*}
so that
\begin{eqnarray*}
&&\!\!c(t,x,y) \\
&\!\!=\!\!&
\exp\Bigl(-\frac{e^t+e^{-t}}{2(e^t-e^{-t})}(x^2+y^2)\Bigr)\\
&&\!\! \times
\exp\Biggl[-\frac{e^t+e^{-t}}{2(e^t-e^{-t})}
\Big(\frac{1}{8}+ \frac{t}{24}\frac{e^t+e^{-t}}{e^t-e^{-t}}\Big)
   (t/n)^2(x^2+y^2) \\
&&\qquad\qquad\qquad\qquad
  +\frac{t}{48}(t/n)^2(x^2+y^2)
 -[1 \vee t^2]O\big((t/n)^3\big)(x^2+y^2)\Biggr].
\end{eqnarray*}


\medskip
Therefore, from the expansions of $a(t,x,y)$, $b(t,x,y)$ 
and $c(t,x,y)$ obtained above with $K(t,x,y)$ in (\ref{Moellerkernel}),
we get to the following expansion for $K^{(n)}(t,x,y)$:
\begin{eqnarray}
K^{(n)}(t,x,y)
&\!\!\!=\!\!\!& a(t,x,y)b(t,x,y)c(t,x,y) \nonumber\\
&\!\!\!=\!\!\!& \frac{e^{-r(t,x,y)}}{\sqrt{\pi(e^t-e^{-t}})} 
   \Big[1+\frac{1}{2} \Big(\frac{1}{8} 
     +\frac{t}{24}\frac{e^t+e^{-t}}{e^t-e^{-t}}\Big)(t/n)^2 %\nonumber\\
 % &&\qquad\qquad\qquad\qquad\qquad\qquad\qquad
     +[1 \vee t^2]O((t/n)^4)\Big]            \nonumber\\
&&\times
\exp\Biggl[\frac{4xy-(e^t+e^{-t})(x^2+y^2)}{2(e^t-e^{-t})}
 \biggl(\Big(\frac{1}{8}+ \frac{t}{24} \frac{e^t+e^{-t}}{e^t-e^{-t}}
 \Big) (t/n)^2         \nonumber\\
&&\qquad\qquad\qquad\qquad   
 +[1 \vee t^2]O(({t}/{n})^3)\biggr)  
 +\frac{t}{48}({t}/{n})^2(x^2+y^2)\Biggr]\,
                                            \nonumber\\
&\!\!\!=\!\!\!& 
 K(t,x,y)e^{-q(t,x,y)/n^2}
 \exp\Big[-r(t,x,y)[1 \vee t^2]
      O(({t}/{n})^3)\Big]  \nonumber\\
 &&\quad \times
  \Big[1+\frac{1}{2} \Big(\frac{1}{8} 
          +\frac{t}{24}\frac{e^t+e^{-t}}{e^t-e^{-t}}\Big)(t/n)^2
          +[1 \vee t^2] O(({t}/{n})^4)\Big]\,.
                                     \label{expansion2-of-K^{(n)}}
\end{eqnarray}
Here in the last equality we have used (\ref{expression-of-q}).
Then
\begin{eqnarray}
R^{(n)}(t,x,y)                         
&\!\!\!=\!\!\!& n^2\big(K^{(n)}(t,x,y) -K(t,x,y)\big) \nonumber\\
&\!\!\!=\!\!\!& n^2 e^{-q(t,x,y)/n^2}K(t,x,y)
  \Big[e^{-r(t,x,y)[1 \vee t^2]
      O(({t}/{n})^3)} -1\Big] \nonumber\\
&&\quad\times 
  \Big[1+\frac{1}{2} \Big(\frac{1}{8} 
      +\frac{t}{24}\frac{e^t+e^{-t}}{e^t-e^{-t}}\Big)(t/n)^2
      +[1 \vee t^2] O(({t}/{n})^4)\Big]\nonumber\\
&& + n^2K(t,x,y) \big(e^{-q(t,x,y)/n^2} -1\big)           \nonumber\\
&&\quad \times 
  \Big[1+\frac{1}{2} \Big(\frac{1}{8} 
          +\frac{t}{24}\frac{e^t+e^{-t}}{e^t-e^{-t}}\Big)(t/n)^2
          +[1 \vee t^2] O(({t}/{n})^4)\Big]\nonumber\\
&& + n^2K(t,x,y)
   \Big[\frac{1}{2} \Big(\frac{1}{8} 
          +\frac{t}{24}\frac{e^t+e^{-t}}{e^t-e^{-t}}\Big)(t/n)^2
          +[1 \vee t^2] O(({t}/{n})^4)\Big]\nonumber\\
&\!\!\!=:\!\!\!& R^{(n)}_1(t,x,y) +R^{(n)}_2(t,x,y) +R^{(n)}_3(t,x,y).
                                       \label{eq-of-R^{(n)}}          
\end{eqnarray}

Recall we are under the condition $0<t \leq n^{1/2}$.
For $R^{(n)}_1(t,x,y)$ we have
\begin{eqnarray}
&\!\!\!& R^{(n)}_1(t,x,y)           \nonumber\\
&\!\!\!=\!\!\!& 
  \Big[1+\frac{1}{2} \Big(\frac{1}{8} 
      +\frac{t}{24}\frac{e^t+e^{-t}}{e^t-e^{-t}}\Big)(t/n)^2
      +[1 \vee t^2] O(({t}/{n})^4)\Big] \nonumber\\
&&\quad \times  \frac{n^2e^{-q(t,x,y)/n^2}}{[\pi(e^t-e^{-t})]^{1/2}}
 \int_0^1 \frac{d}{d\theta}
    e^{-r(t,x,y)(1+\theta [1 \vee t^2]O((t/n)^3)))}d\theta \nonumber\\
%&\!\!\!=\!\!\!& 
% -\frac{(1+O(n^{-1/2}))n^2e^{-q(t,x,y)/n^2}}{[\pi(e^t-e^{-t})]^{1/2}}\\
% && \qquad \times
%    \int_0^1 [1 \vee t^2]O((t/n)^3)) r(t,x,y)
%   e^{-r(t,x,y)(1+\theta [1 \vee t^2]O((t/n)^3)))}d\theta \nonumber\\
&\!\!\! =\!\!\!& O(n^{-1})e^{-q(t,x,y)/n^2}
    \frac{[1 \vee t^2]t^3}{[\pi(e^t-e^{-t})]^{1/2}} \int_0^1     
    r(t,x,y)e^{-r(t,x,y)(1+ \theta [1 \vee t^2]O((t/n)^3)))} d\theta\,.
                                                            \nonumber\\
&&                                               \label{eq-of-R^{(n)}_1}
\end{eqnarray}
Here we have used (\ref{wedge}) with  $t(t/n)^2 \leq n^{-1/2}$
in the second equality, and note that
$
[1 \vee t^2] (t/n)^4 \leq [1 \vee t^2] (t/n)^3 \leq n^{-1/2},
$
so that we have
$|[1 \vee t^2]O((t/n)^3)| < \frac12
$ for large $n$, 
and hence
$$e^{-r(t,x,y)(1+ \theta [1 \vee t^2]O((t/n)^3))}
\leq e^{-r(t,x,y)/2}.
$$
It follows that for large $n$,
\begin{eqnarray} 
|R^{(n)}_1(t,x,y)|
&\!\! \leq \!\!& O(n^{-1})\frac{[1 \vee t^2]t^3}{[\pi(e^t-e^{-t})]^{1/2}}
                 2[\frac{r(t,x,y)}{2}e^{-r(t,x,y)/2}]  \nonumber\\
&\!\! \leq \!\!& \frac{2}{e}\frac{[1 \vee t^2]t^3}{[\pi(e^t-e^{-t})]^{1/2}}
                 O(n^{-1})\,,   \label{bd-of-R^{(n)}_1} 
\end{eqnarray}
where we have used that $\tau e^{-\tau} \leq 1$ for
$\tau \geq 0$, and 
note that the coefficient of $O(n^{-1})$ is  bounded in $t>0$, by (\ref{wedge}). 
%%%%%%%%%%%

For one of the factors in $R^{(n)}_2(t,x,y)$, we have
$$
e^{-q(t,x,y)/n^2} -1
= -q(t,x,y)n^{-2}
            + \frac{q(t,x,y)^2}{2}O(n^{-4}),
$$
by Taylor's theorem that
$
 e^{-x}= 1-x +\frac{x^2}{2} e^{-\theta x} \,\, 0< \theta <1.
$
Then 
\begin{eqnarray}
&&\!\!R^{(n)}_2(t,x,y)+R^{(n)}_3(t,x,y)  \nonumber\\
&\!\!=\!\!&
 K(t,x,y)[-q(t,x,y)+ \frac{q(t,x,y)^2}{2t^2}O((t/n)^2)] \nonumber\\
&&\qquad\times 
  \Big[1+\frac{1}{2} \Big(\frac{1}{8} 
    +\frac{t}{24}\frac{e^t+e^{-t}}{e^t-e^{-t}}\Big)({t}/{n})^2
     +[1 \vee t^2] O(({t}/{n})^4)\Big]\nonumber\\
&& + K(t,x,y)
    \Big[\frac{1}{2} \Big(\frac{1}{8} 
          +\frac{t}{24}\frac{e^t+e^{-t}}{e^t-e^{-t}}\Big)t^2
   +[1 \vee t^2]t^2 O(({t}/{n})^2)\Big]\nonumber\\
&\!\!=\!\!&
 K(t,x,y)\Big[\frac{t^2}{2} \Big(\frac{1}{8} 
          +\frac{t}{24}\frac{e^t+e^{-t}}{e^t-e^{-t}}\Big)-q(t,x,y)\Big]
                                                        \nonumber\\
&&\quad + K(t,x,y)\Big[[1 \vee t^2]t^2 O((t/n)^2)
    -\frac{1}{2} \Big(\frac{1}{8} 
    +\frac{t}{24}\frac{e^t+e^{-t}}{e^t-e^{-t}}\Big)q(t,x,y)(t/n)^2                                                                      \nonumber\\
&& +\frac{q(t,x,y)^2}{2t^2}O((t/n)^2)
   -[1 \vee t^2]q(t,x,y) O((t/n)^4)         \nonumber\\
&& +\frac{1}{2} \Big(\frac{1}{8} 
    +\frac{t}{24}\frac{e^t+e^{-t}}{e^t-e^{-t}}\Big)
          \frac{q(t,x,y)^2}{2t^2}O((t/n)^4)
    +[1 \vee t^2]\frac{q(t,x,y)^2}{2t^2}O((t/n)^6)\Big] \nonumber\\
&\!\!=:\!\!& R^{(n)}_{23,1}(t,x,y)+R^{(n)}_{23,2}(t,x,y)\,.
                                  \label{def-of-R^{(n)}_{23,12}}
\end{eqnarray}

First we note that $R^{(n)}_{23,1}(t,x,y)$, in fact 
is independent of $n$,  
is nothing but $R(t,x,y)$ in (\ref{eq3-thm1}) or (\ref{eq-of-R}).
Next for $R^{(n)}_{23,2}(t,x,y)$, we have
\begin{eqnarray}
R^{(n)}_{23,2}(t,x,y)      
&\!\!\!\!=\!\!\!\!& \frac{1}{[\pi(e^t-e^{-t})]^{1/2}}
  \Big\{e^{-r(t,x,y)}[1 \vee t^2]t^2 O((t/n)^2) \nonumber\\
&&\!\!\!\!  -[e^{-r(t,x,y)}r(t,x,y)]\frac{q(t,x,y)}{t^2r(t,x,y)} 
   \frac12 \Big(\frac{1}{8} 
  +\frac{t}{24}\frac{e^t+e^{-t}}{e^t-e^{-t}}\Big)t^2(t/n)^2\nonumber\\
&&\!\!\!\! +[e^{-r(t,x,y)}r(t,x,y)^2]
   \Big(\frac{q(t,x,y)}{t^2r(t,x,y)}\Big)^2\frac{t^2}{2}O((t/n)^2)
                                                        \nonumber\\
&&\!\!\!\! -[e^{-r(t,x,y)}r(t,x,y)]\frac{q(t,x,y)}{t^2r(t,x,y)}
    [1 \vee t^2]t^2 O((t/n)^4)             \nonumber\\
&&\!\!\!\! +[e^{-r(t,x,y)}r(t,x,y)^2]
   \Big(\frac{q(t,x,y)}{t^2r(t,x,y)}\Big)^2
    \frac{t^2}{2}\frac{1}{2} \Big(\frac{1}{8} 
  +\frac{t}{24}\frac{e^t+e^{-t}}{e^t-e^{-t}}\Big)O((t/n)^4)
                                           \nonumber\\
&&\!\!\!\! +[e^{-r(t,x,y)}r(t,x,y)^2]
   \Big(\frac{q(t,x,y)}{t^2r(t,x,y)}\Big)^2
    \frac{t^2}{2}[1 \vee t^2]O((t/n)^6) \Big\}\,.                                                                             \label{eq-of-R^{(n)}_{23,2}}
\end{eqnarray}
Then, by (\ref{ratio-of-q/r}), (\ref{wedge}) and by the fact that
$\tau^ae^{-\tau} \leq (a/e)^a$ for all $\tau >0$ and $a>0$, we have
\begin{eqnarray}
&&\!\!\!\!|R^{(n)}_{23,2}(t,x,y)|        \nonumber\\
&\!\!\!\! \leq \!\!\!\!& \frac{1}{[\pi(e^t-e^{-t})]^{1/2}} 
 \Big\{[1 \vee t^2]t^2 O((t/n)^2)
 +\frac{1}{6e}\frac{1}{2} \Big(\frac{1}{8} 
  +\frac{t}{24}\frac{e^t+e^{-t}}{e^t-e^{-t}}\Big)t^2(t/n)^2 \nonumber\\
&&+ \Big(\frac{2}{6e}\Big)^2\frac{t^2}{2}  O((t/n)^2)
  +\frac{1}{6e}[1 \vee t^2]t^2  O((t/n)^4)    \nonumber\\
&&+ \Big(\frac{2}{6e}\Big)^2\frac{t^2}{2}\frac{1}{2} \Big(\frac{1}{8} 
  +\frac{t}{24}\frac{e^t+e^{-t}}{e^t-e^{-t}}\Big) O((t/n)^4)
  + \Big(\frac{2}{6e}\Big)^2 \frac{[1 \vee t^2]t^2}{2}  O((t/n)^6)
                                        \Big\}   \nonumber\\ 
&\!\!\!\! \leq \!\!\!\!& \frac{1}{[\pi(e^t-e^{^t})]^{1/2}}
 \Big\{\Big[[1 \vee t^2]t^4 
+ \frac{1}{6e} \frac{1}{2} \Big(\frac{1}{8} 
  +\frac{t}{24}\frac{e^t+e^{-t}}{e^t-e^{-t}}\Big)t^4
  +\Big(\frac{1}{3e}\Big)^2 \frac{t^2}{2}\Big]O(n^{-2}) \nonumber\\ 
&&\qquad\qquad\qquad
 + \Big[\frac{[1 \vee t^2]t^6}{6e} + \Big(\frac{1}{3e}\Big)^2
    \frac{t^6}{2} \frac{1}{2} \Big(\frac{1}{8} 
  +\frac{t}{24}\frac{e^t+e^{-t}}{e^t-e^{-t}}\Big)\Big]O(n^{-4})
                                                       \nonumber\\
&&\qquad\qquad\qquad\qquad\qquad\qquad\qquad
  +  \Big(\frac{1}{3e}\Big)^2 \frac{[1 \vee t^2]t^8}{2} 
    O(n^{-6}) \Big\}\,.  \label{eq2-of-R^{(n)}_{23,2}} 
\end{eqnarray}
Hence, taking into account the relation 
$R^{(n)}(t,x,y) = R(t,x,y)+ Q^{(n)}(t,x,y)n^{-1}$, which
we get from 
(\ref{def-of-R^{(n)}}) and (\ref{def-of-Q^{(n)}}), we obtain
$$
 Q^{(n)}(t,x,y) = n\big(R^{(n)}_{1}(t,x,y)+R^{(n)}_{23,2}(t,x,y)\big),
$$
and
\begin{eqnarray}
|Q^{(n)}(t,x,y)|    
&\!\!\!\! \leq \!\!\!\!& \frac{1}{[\pi(e^t-e^{-t})]^{1/2}}
 \Big\{\frac2{e}[1 \vee t^2]t^3 O(1) \nonumber\\
&&\!\!\!\!+\Big[[1 \vee t^2]t^4 
  +\frac{1}{6e} \frac{1}{2} \Big(\frac{1}{8} 
  +\frac{t}{24}\frac{e^t+e^{-t}}{e^t-e^{-t}}\Big)t^4
  +\Big(\frac{1}{3e}\Big)^2\frac{t^4}{2}\Big]O(n^{-1}) \nonumber\\
&&\!\!\!\!
 + \Big[\frac{[1 \vee t^2]t^6}{6e}
   + \Big(\frac{1}{3e}\Big)^2\frac{t^6}{2} 
     \frac{1}{2} \Big(\frac{1}{8}
     +\frac{t}{24}\frac{e^t+e^{-t}}{e^t-e^{-t}}\Big)\Big]O(n^{-3}) 
                                                     \nonumber\\
&&\qquad\qquad\qquad\qquad\qquad\quad  
   +\Big(\frac{1}{3e}\Big)^2
     \frac{[1 \vee t^2]t^8}{2}O(n^{-5}) \Big\}\,.  
                                    \label{bound-of-Q^{(n)}}
\end{eqnarray} 
The right-hand side of (\ref{bound-of-Q^{(n)}}) is independent of
$x,\, y$, and bounded
in $t>0$ and $n\geq 1$, by (\ref{wedge}), and tends
to zero as $t\rightarrow 0+$. 
Therefore we see there exists a continuous function 
$C_2(t)$ bounded in $t \geq 0$, independent of $n$, 
such that 
\begin{equation}   \label{bound2-of-Q^{(n)}}
|Q^{(n)}(t,x,y)| \leq C_2(t),  
       \qquad (x,y) \in {\bf R}\times {\bf R}, 
\end{equation}
with $0<t \leq n^{1/2}$.
We may assume with (\ref{wedge}) that this $C_2(t)$ 
is so taken as to satisfy
$\lim_{t\rightarrow 0+} C_2(t) =0$.
Hence and from (\ref{bound-of-R}) we can also get the bound 
for $R^{(n)}(t,x,y)$, as
$|R^{(n)}(t,x,y)| \leq |R(t,x,y)|+|Q^{(n)}(t,x,y)|n^{-1}
\leq C_1(t)+C_2(t)$, namely, (\ref{bd-of-kerR^{(n)}}).
Thus we have shown the desired assertion of Proposition 3.2,
ending the proof of Proposition 3.2.

\bigskip
We are now in a position to prove Theorem 1.1.

\bigskip\noindent 
{\it Completion of Proof of Theorem 1.1.}
Let $T>0$. Then take a sufficiently large positive integer $N$ 
such that $T \leq N^{1/2}$. Then Proposition 3.2 holds
for all $n \geq N$ and for all $t \in (0,T]$.
This is nothing but the assertion of Theorem 1.1,
completing the proof of Theorem 1.1.

\bigskip
%%%%%%%%%%%%%%%%%%%
\subsection{Proof of Theorem 1.2}

We use the result in Theorem 1.1. Let $T>0$.
Denote by $R(t)$ and $Q^{(n)}(t)$ the bounded selfadjoint 
operators on 
$L^{2}({\bf R})$ with integral kernel $R(t,x,y)$ and 
$Q^{(n)}(t,x,y)$, respectively, so that
 $R^{(n)}(t) =R(t) + Q^{(n)}(t)n^{-1}$ is 
a selfadjoint operator with integral kernel $R^{(n)}(t,x,y)$.
We need to find a lower and upper bound of the norm 
of $R^{(n)}(t)$.

First, for a upper bound of the norm of 
$R^{(n)}(t)$, note 
from (\ref{bound-of-R}) and (\ref{bd-of-kerQ^{(n)}})
\begin{eqnarray}
\|R(t)\| &\!=\!& \sup_{\|f\|=1} |(f,R(t)f)|
         \leq \sup_{x,y}|R(t,x,y)| \leq C_1(t), \\ 
                                \label{upperbound-of-R}
 \|Q^{(n)}(t)\| 
   &\!=\!&  \sup_{\|f\|=1} |(f,Q^{(n)}(t)f)|
   \leq \sup_{x,y}|Q^{(n)}(t,x,y)| \leq C_2(t),
                                \label{bound2-of-Q^{(n)}}
\end{eqnarray}
where $C_1(t)$ and $C_2(t)$ are the bounded functions in
Propositions 3.1 and 3.2. It follows that for sufficiently 
large $n$ and for all $t \in (0,T]$,
$$
\|R^{(n)}(t)\|\leq \|R(t)\|+\|Q^{(n)}(t)\|n^{-1}
\leq C_1(t)+ C_2(t)=: C(t).
$$
This  $C(t)$ is a continuous function 
in $t \geq 0$, uniformly bounded, with $C(0)=0$ and 
positive in $t>0$. Thus we have shown the upper bound 
of the relation (\ref{eq-thm2}).

%%%%
\medskip
Next, we seek a lower bound of the norm of $R^{(n)}(t)$. 
Since the quadatic expression $q(t,x,y)$ in
(\ref{expression-of-q}) is nonnegative,
$s(t,x,y)$ in (\ref{expression2-of-s})/(\ref{expression22-of-s}) 
vanishes for $(t,x,y)$ such that
$\frac{t^2}{2}\Big(\frac{1}{8}
   +\frac{t}{24}\frac{e^t+e^{-t}}{e^t-e^{-t}}\Big)
  = q(t,x,y)$. For $t \geq 0$, let
$S_t := \{(x,y) \in {\bf R}^2\,;\, s(t,x,y) >0\}$.
If $t=0$, it reduces to one point: $S_0 =\{(0,0)\}$. But if
$t>0$, it is a bounded open set in ${\bf R}^2$ around the
origin. It contains, for given  $\delta(t)>0$, 
which is to determined later, a disc
$B_t := \{(x,y) \in {\bf R}^2\,;\, x^2+y^2 \leq \delta(t)^2 \}$
as well as a square
$Q_t := \{(x,y) \in {\bf R}^2\,;\, 
              |x| \leq \delta(t), |y|\leq \delta(t)\}$,
so that $B_t \subseteq Q_t$.
Let 
$f_0(x): = 1/\sqrt{2\delta(t)},\, |x|\leq \delta(t)\,;\,
 f_0(x)=0, \, |x|> \delta(t)$.
Then $f_0$ belongs to $L^{2}({\bf R})$ and has norm 
$\|f_0\|= (\int |f_0(x)|^2 dx)^{1/2}= 1$. 
Note that $R(t)f_0$ is nonnegative. Therefore
we have
$$
  \|R(t)\| = \sup_{\|f\|=1}|(f,R(t)f)| \geq (f_0,R(t)f_0).
$$
where
$$
(f_0,R(t)f_0) 
= \int\int_{Q_t}R(t,x,y)dxdy 
   = \int\int_{Q_t} s(t,x,y) K(t,x,y)dxdy.
$$
If $(x,y) \in Q_t$, then $x^2+y^2 \leq 2\delta (t)^2$, so that
by (\ref{ratio-of-q/r}) with (\ref{expression-of-r})
$$
q(t,x,y) \leq \frac{t^2}{6} r(t,x,y)
 = \frac{t^2}{6}\frac{(e^t+e^{-t})(x^2+y^2)-4xy}{e^t-e^{-t}}
\leq
\delta(t)^2 \frac{t^2}{3} \frac{e^t+e^{-t} +2}{e^t-e^{-t}}.
$$
Therefore, for each fixed $t>0$, we choose $\delta(t) >0$ 
such that
$$
 \delta(t)^2 \frac{t^2}{3} \frac{e^t+e^{-t} +2}{e^t-e^{-t}}
 = \frac12 \frac{t^2}{2}
  \Big(\frac{1}{8}+\frac{t}{24}\frac{e^t+e^{-t}}{e^t-e^{-t}}\Big),
$$
namely, 
\begin{equation}  \label{determine-delta(t)}
 \delta(t)^2 = \frac34 
\frac{\frac18(e^t-e^{-t})+\frac{t}{24}(e^t+e^{-t})}{e^t+e^{-t}+2}.
\end{equation}
It follows that $\delta(t)$ is of order $O(t^{1/2})$.
We have then for $(x,y) \in Q_t$
$$
 s(t,x,y) = 
\frac{t^2}{2}
    \Big(\frac{1}{8}+\frac{t}{24}\frac{e^t+e^{-t}}{e^t-e^{-t}}\Big)
    -q(t,x,y) %\\
\geq
\frac12 \frac{t^2}{2}
  \Big(\frac{1}{8}+\frac{t}{24}\frac{e^t+e^{-t}}{e^t-e^{-t}}\Big).
$$
We use the polar coordinates to bound from below 
\begin{eqnarray*}
(f_0,R(t)f_0) 
&\!\!\geq\!\!& \frac12 \frac{t^2}{2}
       \Big(\frac{1}{8}+\frac{t}{24}\frac{e^t+e^{-t}}{e^t-e^{-t}}\Big)
       \int\int_{B_t}K(t,x,y)dxdy \\
%&\!\!=\!\!& \frac{\frac{t^2}{4}
%       \Big(\frac{1}{8}+\frac{t}{24}\frac{e^t+e^{-t}}{e^t-e^{-t}}\Big)}
%        {\sqrt{\pi(e^t-e^{-t})}}  %\\
% %&&\qquad\quad  \times
%        \int\int_{B_t}\exp\Big[-\frac{(e^t+e^{-t})(x^2+y^2)-4xy}
%        {2(e^t-e^{-t})}\Big]dxdy \\
&\!\!=\!\!& \frac{\frac{t^2}{4}
        \Big(\frac{1}{8}+\frac{t}{24}\frac{e^t+e^{-t}}{e^t-e^{-t}}\Big)}
         {\sqrt{\pi(e^t-e^{-t})}}\\
 &&\qquad\quad   \times
  \int_0^{\delta(t)}\int_0^{2\pi}\exp
  \Big[-\frac{(e^t+e^{-t}-4\sin\theta\cos\theta)\rho^2}
   {2(e^t-e^{-t})}\Big]\rho d\rho d\theta \\
&\!\!\geq\!\!& \frac{\frac{t^2}{4}
         \Big(\frac{1}{8}+\frac{t}{24}\frac{e^t+e^{-t}}{e^t-e^{-t}}\Big)}
         {\sqrt{\pi(e^t-e^{-t})}} \int_0^{\delta(t)}\int_0^{2\pi}\exp
          \Big[-\frac{(e^t+e^{-t}+ 2)\rho^2}
          {2(e^t-e^{-t})}\Big]\rho d\rho d\theta.
\end{eqnarray*}
Calculating the integral on the right, we get with 
(\ref{determine-delta(t)})
\begin{eqnarray}
&&\!\!
(f_0,R(t)f_0) \nonumber \\
&\!\!\geq\!\!&
 \frac{\sqrt{\pi(e^t-e^{-t})}}{2}\,t^2\Big(\frac{1}{8}
   +\frac{t}{24}\frac{e^t+e^{-t}}{e^t-e^{-t}}\Big)
   \frac{1-\exp[-\frac38
   \frac{\frac18(e^t-e^{-t})+\frac{t}{24}(e^t+e^{-t})}{e^t-e^{-t}}]}
         {e^t+e^{-t}+2}  \nonumber \\
&\!\!=:\!\!& c_1(t). \label{lowerbound-of-R} 
\end{eqnarray}
Then $c_1(t)$ is a positive, continous function  in $t>0$, 
which we see with  (\ref{wedge}) is uniformly bounded.
Since $\lim_{t\rightarrow 0+}c_1(t) =0$, we may define $c_1(0)=0$.
So we have obtained a lower bound $\|R(t)\| \geq c_1(t)$.
Next, we get from 
(\ref{bound-of-Q^{(n)}})/(\ref{bound2-of-Q^{(n)}})
$$%\begin{eqnarray*}
(f_0, R^{(n)}(t)f_0)
 = (f_0,R(t)f_0) +(f_0,Q^{(n)}(t)f_0)n^{-1}
 \geq \bigl(c_1(t)  -C_2(t)n^{-1}\bigr).
$$%\end{eqnarray*}
We can see that $C_2(t)/c_1(t)$ is bounded in $t>0$, 
because all the seven functions on the right of 
(\ref{bound-of-Q^{(n)}})

$\frac{[1 \vee t^2]t^3}{(e^t-e^{-t})^{1/2}},\,\
 \frac{[1 \vee t^2]t^4}{(e^t-e^{-t})^{1/2}}, \,\
 \frac{\big(\frac{1}{8}
   +\frac{t}{24}\frac{e^t+e^{-t}}{e^t-e^{-t}}\big)t^4}
  {(e^t-e^{-t})^{1/2}},\,\
 \frac{t^4}{(e^t-e^{-t})^{1/2}},
$

$
 \frac{[1 \vee t^2]t^6}{(e^t-e^{-t})^{1/2}},\,\
 \frac{\big(\frac{1}{8}
   +\frac{t}{24}\frac{e^t+e^{-t}}{e^t-e^{-t}}\big)t^6}
  {(e^t-e^{-t})^{1/2}},\,\
  \frac{[1 \vee t^2]t^8}{(e^t-e^{-t})^{1/2}}
$

\noindent
divided by $c_1(t)$ are bounded there, 
by (\ref{wedge}) and (\ref{vee}).
It follows that $c_1(t)- C_2(t)n^{-1}$ is positive in $t>0$
for $n$ large. 
In fact, there exists a positive integer $N$, independent
of $t$,  such that $c_1(t)- C_2(t)n^{-1}\geq \frac12 c_1(t) >0$ for
 all $t>0$ and all $n\geq N$.
Take $c(t) := \frac12 c_1(t)$ so that
$(f_0, R^{(n)}(t)f_0) \geq c(t)$. The function
$c(t)$ is nonnegative, continuous and uniformly bounded
in $t \geq 0$ with $c(0) := \lim_{t\rightarrow 0+} c(t) = 0$.
We can also see $\lim_{t\rightarrow \infty} c(t) =0$.
This yields that $\|R^{(n)}(t)\| \geq c(t)$  in $t>0$
for $n \geq N$, and hence, as we may think, for
all $n \geq 1$, showing the lower bound of the desired 
relation (\ref{eq-thm2}).
This completes the proof of Theorem 1.2.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\setcounter{equation}{0}

\section{Concluding Remark}

The results of the present paper can be in fact obtained
in a more sophisticated way as in [9], by the commutator
method. Among others, we shall briefly mention only how by this method 
the bounded operator $R(t)$ with integral kernel $K(t,x,y)$
in (\ref{eq3-thm1}), Theorem 1.1, comes out.

By the Baker--Campbell--Hausdorff formula (e.g. \cite{ITTZ}),
one has, when $A$ and $B$ are {\it bounded} operators, 
\begin{eqnarray}
&&\!\!\!\! [e^{-tB/2n}e^{-tA/n}e^{-tB/2n}]^n -e^{-t(A+B)} 
                                      \nonumber\\
&\!\!=\!\!& \exp \big(-t(A+B) 
      - n^{-2} \frac{t^2}{24}[2A+B,[A,B]] - O_p(n^{-3})\big)
                                      \nonumber\\
&\!\!=\!\!& e^{-t(A+B)}-n^{-2} \frac{t^2}{24}
        \int_0^t e^{-(t-s)(A+B)}[2A+B,[A,B]]e^{-s(A+B)} ds
            + O_p(n^{-3}),            \nonumber\\
&&                                              \label{BCH}
\end{eqnarray}
where $[A,B] := AB-BA$, and
$O_p(n^{-3})$ stands for an operator with norm of order $O(n^{-3})$.
We cannot in general take unbounded operators as $A$ and $B$.
However, in our case with $A :=H_0=-\frac12 \Delta$ and $B:=V=\frac12 x^2$, 
though they are unbounded operators in $L^2({\bf R})$, we can 
show that (4.1) is still valid. We omit the proof. Instead,
we content ourselves to see that  
the second term in the last member of (\ref{BCH}) makes 
sense to yield $R(t)n^{-2}$. Namely, putting
\begin{equation} \label{op-expression-of-R}
\hat{R}(t) := -\frac{t^2}{24}\int_0^te^{-(t-s)H}
           [2H_0+V, [H_0,V]]e^{-sH} ds,                  
\end{equation}
we have

\begin{proposition}%Proposition 4.1
For the operators $H_0=-\frac12 \Delta$ and $V=\frac12 x^2$ in $L^2({\bf R})$,
$\hat{R}(t)$
has the same integral kernel $R(t,x,y)$ in (\ref{eq3-thm1})
%\begin{eqnarray}
%K(t,x,y)t^2
%    \Big[\frac{t}{12}\Big(\frac1{4}\frac{e^t+e^{-t}}{e^t-e^{-t}}
%        +\frac{(e^t+e^{-t})xy-(x^2+y^2)}{(e^t-e^{-t})^2}\Big)
%                                     \nonumber\\
%&&\qquad\qquad\qquad\quad   + \frac1{16}
%    \Big(1+\frac{4xy-(e^t+e^{-t})(x^2+y^2)}{e^t-e^{-t}}\Big)\Big].
%                    \label{op2-expression-of-R} 
%\end{eqnarray}
as $R(t)$, so that both the operators $\hat{R}(t)$ and $R(t)$ 
coincide with each other.

\end{proposition}

\medskip
In the following proof, we shall refer to (\ref{eq3-thm1}) 
related to the first one of the two expressions of $s(t,x,y)$
in (\ref{expression1-of-s})/(\ref{expression2-of-s}) 
with which $R(t,x,y)$ is rewritten as (\ref{eq-of-R}).

\smallskip
\noindent {\em Proof}. 
We have
\begin{eqnarray*}
 && [H_0,V] = -\frac12(1+2x \partial_x) \\
 && [H_0, [H_0,V]]= \partial_x^2 = -2H_0,
   \quad  [V, [H_0,V]]= x^2 =2V,
\end{eqnarray*}
so that
$$ 
  [2H_0 +V, [H_0,V]]= -4H_0 +2V = -4H+6V.
$$
Put $C := \frac{1}{2}(p\cdot x + x\cdot p)$ 
with $p=-i\partial_x$.
Then since $H= H_0+V = (p^2+x^2)/2$, we have
$i[H,C]=[(p^2+x^2)/2,iC] =(p^2-x^2)$, 
so that
$$
  \frac12 x^2 =\frac14\big((p^2+x^2)-(p^2-x^2)\big)
      =\frac14 (2H-i[H,C]).
$$
Then
\begin{eqnarray*}
  \int_0^te^{-(t-s)H}Ve^{-sH} ds
&\!=\!& \int_0^te^{-(t-s)H}(x^2/2)e^{-sH} ds\\
&\!=\!& \frac14 \int_0^te^{-(t-s)H} (2H- i[H,C])e^{-sH} ds\\
&\!=\!& \frac{t}{2}He^{-tH}- 
   \frac{i}{4}\int_0^t \frac{d}{ds}[e^{-(t-s)H}Ce^{-sH}]ds\\
&\!=\!& \frac{t}{2}He^{-tH}- 
   \frac{1}{4} (e^{-tH}(iC) -(iC)e^{-tH}).
\end{eqnarray*}
It follows that
\begin{eqnarray}  
 \hat{R}(t) 
&\!=\!& \frac{t^2}{24}\int_0^te^{-(t-s)H}
           (4H-6V)e^{-sH} ds        \nonumber\\
&\!=\!& -t^2\Big[(-\frac{4}{24}+\frac{3}{24})tHe^{-tH}
           +\frac1{16}(e^{-tH}(iC)-(iC)e^{-tH})\Big] \nonumber\\
    &\!=\!& t^2\Big[\frac{1}{24}tHe^{-tH}
           -\frac1{16}(e^{-tH}(iC)-(iC)e^{-tH})\Big].
                               \label{op-expression2-of-R}
\end{eqnarray}
Then, calculating both 
$-He^{-tH} = \frac{d}{dt}e^{-tH} = \frac{d}{dt}K(t)$ and 
\begin{eqnarray*}
e^{-tH}(iC)-(iC)e^{-tH}
&\!=\!& \frac12 [-y(\partial_y e^{-tH})-\partial_y(y e^{-tH})
    -x(\partial_xe^{-tH})-\partial_x(xe^{-tH})]\\
&\!=\!& -e^{-tH}- y(\partial_y e^{-tH}) -x(\partial_x e^{-tH})\\
&\!=\!&
   -\Big[1+\frac{4xy-(e^t+e^{-t})(x^2+y^2)}{e^t-e^{-t}}\Big]e^{-tH}
\end{eqnarray*}
on the right of (\ref{op-expression2-of-R}), 
we get the first and the second term on the right-hand side 
of (\ref{eq3-thm1}). 
$\Box$%\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{Acknowledgments}
The authors would like to thank Professor Hideo Tamura
for valuable discussion on the subject, in particular, 
the content of Section 4.
The research of Y.A. was supported  in part 
by the Teaching Assistant Fellowship,
and that of T.I. by the Grant-in-Aid for 
Scientific Research (B) No.~16340038, 
Japan Society for Promotion of Science.


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%%%%%%%%%%%%%%

\bigskip\noindent
Yoshiki Azuma

\noindent
Division of Mathematics and Physics, Graduate School
of Natural Science and Technology, Kanazawa University

\noindent 
Kanazawa, 920--1192, Japan;\\
Current Address: 149-4, Obara, Shimotsu-cho,
Kainan-city, 649--0162, Japan

\bigskip\noindent
Takashi Ichinose

\noindent
Division of Mathematics and Physics, Graduate School
of Natural Science and Technology, Kanazawa University

\noindent 
Kanazawa, 920--1192, Japan

\noindent
E-mail: ichinose@kenroku.kanazawa-u.ac.jp



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