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\newtheorem{theo}{Theorem}
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\begin{document}
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\font\fifteen=cmbx10 at 15pt \font\twelve=cmbx10 at 12pt 

\begin{titlepage} 

\begin{center}

\renewcommand{\thefootnote}{\fnsymbol{footnote}}

{\twelve Centre de Physique Th\'eorique\footnote{Unit\'e Propre de
Recherche 7061 }, CNRS Luminy, Case 907}

{\twelve F-13288 Marseille -- Cedex 9}

\vspace{2 cm}

{\fifteen ACCRETIVE PERTURBATIONS AND ERROR ESTIMATES
FOR THE TROTTER PRODUCT FORMULA\footnote{\bf To the memory of Tosio Kato}}

\vspace{1.5 cm}

\setcounter{footnote}{0}
\renewcommand{\thefootnote}{\arabic{footnote}}

{\bf V. CACHIA\footnote{Allocataire-moniteur at Universit\'e de la 
M\'editerran\'ee (Aix-Marseille II)\\ Email : cachia@cpt.univ-mrs.fr},
H. NEIDHARDT\footnote{Weierstra\ss\  Institut f\"ur angewandte Analysis und 
Stochastik, Mohrenstr. 39, D-10117 Berlin, Germany - Email : 
neidhard@wias-berlin.de} and
V.A. ZAGREBNOV\footnote{Universit\'e de la M\'editerran\'ee
(Aix-Marseille II) - Email : zagrebnov@cpt.univ-mrs.fr} }

\vspace{1.5 cm}

{\bf Abstract}

\end{center}

We study the error bound estimate in the operator-norm topology for the
exponential Trotter product formula in the case of accretive
perturbations. Let $A$ be a semibounded from below self-adjoint operator on a 
separable Hilbert space. Let $B$ be a closed maximal accretive operator which 
is, together with $B^*$, Kato-small with respect to $A$ with relative bounds 
less than one. We show that in this case the operator-norm error bound estimate 
for the exponential Trotter product formula is the same as for the self-adjoint 
$B$ \cite{NZ1}:
$$\left\|\left(e^{-tA/n}e^{-tB/n}\right)^n - e^{-t(A+B)}\right\| \leq L {\ln 
n\over n},\ n = 2,3,\ldots\,.$$
We verify that the operator $-(A+B)$ generates a holomorphic contraction 
semigroup. One gets a similar result when $B$ is substituted by $B^*$.


\vspace{1 cm}


\bigskip

\noindent Keywords :  Semigroups, Trotter product formula, accretive operators, 
operator-norm estimates.

\noindent Mathematics Subject Classification : 47D03, 47B25, 35K22, 41A80



\bigskip
\bigskip


\noindent anonymous ftp :
ftp.cpt.univ-mrs.fr\\
web : www.cpt.univ-mrs.fr\\
February 2000\\
CPT-2000/P.3961

\renewcommand{\thefootnote}{\fnsymbol{footnote}}

\end{titlepage}

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\newpage


\section{Introduction and Setup of the Problem}
\setcounter{equation}{0}
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Let $-A$ and $-B$ be generators of strong continuous semigroups on some 
separable Hilbert space ${\frak H}$. For those operators the convergence of the 
Trotter product formula
% 
\begin{equation}\Label{1}
\tau -\lim_{n\rightarrow +\infty}(e^{-tA/n}e^{-tB/n})^{n} = e^{-tH},
\end{equation}
%
where $H$ is the sum of the operators $A$ and $B$ defined in a suitable sense, 
was of interest for many years only with respect to the strong topology $\tau 
=s$, see \cite{Ch1}, \cite{Ch2}, \cite{Tr1}. A final result in this direction 
was obtained by Kato \cite{Ka2}, \cite{Ka1} who has shown that the convergence 
in the strong topology always takes place if the operators $A$ and $B$ are 
self-adjoint and non-negative, or even more generally they may be m-sectorial. 
Here and below we follow the terminology of the Kato's book \cite{Kato}. 
Moreover, he identified $H$ as the form-sum of $A$ and $B$ and generalized the 
product formula (\ref{1}) to the form in which the exponential functions are 
replaced by functions of a certain class which is often called the Kato-class. 
However, certain applications of the Trotter  or the more general Trotter-Kato 
product formul\ae\  require that
%
%
\begin{enumerate}
%
%
\item[(a)] the  convergence in (\Ref{1}) is in the operator-norm topology or
in stronger norms, for instance, in the trace- or the Hilbert-Schmidt-norm,
%
%
\item[(b)] error estimates in these norms, in particular, in operator-norm, are 
available,
%
%
\item[(c)] the convergence and the error estimates are valid not only for 
self-adjoint operators $A$ and $B$.
%
%
\end{enumerate}


Until now the problems (a) and (b) were solved in a satisfactory manner only for 
non-negative self-adjoint operators $A$ and $B$. In papers \cite{NZ0}-\cite{NZ4} 
the conditions (partially necessary and sufficient) are found for convergence of 
the Trotter-Kato product formula in the operator-norm or in norms of 
symmetrically-normed ideals of compact operators, see also \cite{Hi1}. Error 
bound estimates in the operator-norm for non-negative self-adjoint operators 
were given in \cite{IT1}, \cite{NZ1}, \cite{NZ2} and \cite{Ro1} and for the 
trace-norm in \cite{IT2} and \cite{NZ4}. On the other hand there are only few 
results concerning the problem (c). The first result in this direction was that 
of Lie who has shown the convergence of the exponential product formula 
(\ref{1}) for arbitrary matrices $A$ and $B$. In fact, his proof provides the 
operator-norm error bound estimate for a couple of bounded operators $A$ and $B$ 
in a complete normed space:
%
%
\begin{equation}\Label{1.4-1}
\left\|(e^{-tA/n}e^{-tB/n})^{n} - e^{-tH}\right\| \le \frac{C}{n}, \quad n = 
1,2,\ldots,
\end{equation}
%
%
where $H = A + B$, see \cite{RS1}. So the Lie's result solves problems (a) - (c) 
for bounded operators in a Banach space. In \cite{CZ2} the problem (a) in a 
Hilbert space is solved in the operator-norm topology $\tau = \|\cdot\|$ for a 
class of unbounded m-sectorial operators $A$ and $B$ with semi-angles $\theta_A, 
\theta_B\in (0,\pi/2)$. A solution of the problem (b) in a Banach space was 
presented in \cite{CZ1} for particular domain conditions. Namely, let $-A$ be 
generator of a holomorphic contraction semigroup and let $-B$ be generator of a 
contraction semigroup satisfying the conditions: $\dom(B)\supseteq 
\dom(A^\alpha)$ for some $\alpha\in [0,1)$ and $\dom(B^*)\supseteq \dom(A^*)$. 
then there is a constant $C_\alpha$ such that
%
%
\begin{equation}\Label{1.5}
\left\|\left(e^{-tB/n}e^{-tA/n}\right)^{n}-e^{-tH}\right\| \leq 
C_\alpha\frac{(\ln(n))^{s(\alpha)}}{n^{1-\alpha}}, \quad n = 2,3,\ldots, 
\end{equation}
%
%
uniformly in $t \in [0,T]$, $0 < T < +\infty$, where $H = A +B$ on $\dom(H) = 
\dom(A)$, and $s(\alpha=0)=2$, $s(\alpha\neq 0)=1$. The condition 
$\dom(B)\supseteq \dom(A^\alpha)$ implies that $B$ is relatively bounded with 
respect to $A$ with the relative bound $a=0$. Thus, comparing with the 
corresponding results in the self-adjoint case on a Hilbert space, see 
\cite{NZ1} and \cite{NZ2}, the result (\ref{1.5}) is not optimal. For instance 
in \cite{NZ1} it is proven that if $A\geq I$, $B\geq 0$ and $B$ is relatively 
bounded with respect to $A$ with $a<1$, then there is a constant $C$ such that

\begin{equation}\Label{1.5-2}
\left\|\left(e^{-tA/n}e^{-tB/n}\right)^{n} - e^{-tH}\right\| \leq 
C\frac{\ln(n)}{n}, \quad n = 2,3,\ldots,
\end{equation}
%
%
uniformly in $t \ge 0$, where $H=A+B$ on $\dom(H) = \dom(A)$.


In what follows we suppose that $A=A^{\ast }$ be a semibounded from below
densely defined self-adjoint operator on the Hilbert space ${\kH}$ which
generates a semigroup $\{U_{A}(t)= e^{-tA}\}_{t\geq 0}$. Without loss of
generality one can suppose that $A\geq I$. Then it is known (see, e.g. the 
Kato's book \cite{Kato}) that $U_{A}(t)$ is
a contraction holomorphic semigroup of angle $\omega =\pi /2$, i.e. $\Vert
U_{A}(t\geq 0)\Vert \leq 1$ and $U_{A}(t)$ has a holomorphic extension
from $t\geq 0$ into the open sector of the complex plane: $S_{\omega =\pi
/2}=\{z\in {\bC}:z\neq 0$ and $|{\arg }(z)|<\pi /2\}.$

About the generator $B$ we suppose that :

\begin{enumerate}
\item[$(i)$] domain $\dom(B)\supseteq \dom(A)$ and there exists $a>0$ such that
\begin{equation}
\|Bu\| \leq a\|Au\|,\ u\in\dom(A);\label{1.1}
\end{equation}
\item[$(ii)$] domain $\dom(B^*)\supseteq \dom(A)$ and there exists $a_*> 0$ such 
that
\begin{equation}
\|B^*u\| \leq a_*\|Au\|,\ u\in\dom(A);\label{1.2}
\end{equation}
\item[$(iii)$] $B$ is a m-accretive (i.e. closed and maximal accretive) operator 
on ${\kH}$, i.e. for $\lambda>0$ one has
\begin{equation}
\Vert (B+\lambda)^{-1}\Vert \leq \lambda^{-1}.
\label{1.3}
\end{equation}
\end{enumerate}
The latter implies that $\{U_{B}(t)= e^{-tB}\}_{t\geq 0}$ is a
(non self-adjoint) contraction semigroup, but $B$ is not necessarily sectorial: 
$\theta_B$ may be $\pi/2$.

Notice that under conditions $(i)$-$(iii)$ the convergence of the Trotter
product formula (\ref{1}) is known, due to \cite{Ch1}, \cite{Ch2}, \cite{Tr1}, 
only in the strong topology $\tau =s$. In our main theorem (see Section 3) we 
prove that under conditions $(i)$-$(iii)$ and $a,a_*<1$ there is a constant 
$L>0$ such that uniformly in $t\geq 0$ one gets (cf (\ref{1.5-2})
\begin{equation}
\left\| \left(e^{-tA/n}e^{-tB/n}\right)^{n}-e^{-tH}\right\| \leq L{\ln (n)\over 
n}\ ,\ n=1,2,3,\dots\quad .
\label{1.6}
\end{equation}
Here $H=A+B$ is defined on $\dom(H)=\dom(A)$. Similar result holds for $B^{\ast
}$ and $H^{\ast }$.

The paper is organized as follows. In Section 2 we prove some technical results, 
which have an interest in itself, and which are crucial for the proof of the 
main theorem, which is presented in Section 3. The proof of this main theorem 
(while inspired by) needs a considerable deviation from the line of reasoning 
known for the self-adjoint case. Finally, in Section 4 we make some remarks and 
comments.



\section{Preliminaries}
\setcounter{equation}{0}
\renewcommand{\theequation}{\arabic{section}.\arabic{equation}}
\setcounter{theo}{0}
\renewcommand{\thetheo}{\arabic{section}.\arabic{theo}}

Let $-H$ be the generator of a semigroup $\{U_H(t)=e^{-tH}\}_{t \ge 0}$. The 
semigroup $U_H(t)$ is called a bounded holomorphic semigroup with angle $\omega 
\in (0,\pi/2]$ if
%
%
\begin{enumerate}
%
\item[($\alpha$)] $\{U(t)\}_{t \ge 0}$ is the restriction to the positive real 
axis of an analytic family of bounded operators $\{U(z)\}_{z \in S_\omega}$, 
$S_\omega = \{z \in {\mathbb C}: z \not= 0, \; |\arg z| < \omega\}$, obeying 
$U(z + z') = U(z)U(z')$, $z,z' \in S_\omega$, and

\item[($\beta$)] for every $\theta \in (0,\omega)$ the family $\{U(z)\}_{z \in 
S_\omega}$ is uniformly bounded in the sector $S_\theta$ and $U(z)f 
\longrightarrow f$ as $z \longrightarrow 0$ in $z \in S_\theta$ for each $f \in 
{\frak H}$.
\end{enumerate}

It turns out that this is the case if and only if the spectrum $\sigma(H)$ of 
$H$ is contained in the closed sector $\overline{S_{\pi/2-\omega}}$, i.e. 
$\sigma(H) \subseteq \overline{S_{\pi/2-\omega}}$, and for each $\theta\in 
[0,\omega)$ there is a constant $M_\theta > 0$ such that 
%
%
\begin{equation}\Label{2.1}
\|(H + z)^{-1}\| \le \frac{M_\theta}{|z|}
\end{equation}
%
%
for $z \in S_{\pi/2+\theta}$, see e.g. \cite[Ch.IX]{Kato}.
%
%
\begin{theo}\Label{II.1}
%
%
Let $A \ge I$ and let $B$ be a closed accretive operator satisfying condition 
(i) with $a < 1$. Then the operator $-H=-(A+B)$ is the boundedly invertible 
generator of a contraction holomorphic semigroup of angle $\omega = \arccos a$, 
in the sense that $\|e^{-tH}\|\leq 1$ for $t\geq 0$.
%
%
\end{theo}
%
%
\begin{preuve} Our goal is to verify (\Ref{2.1}) for $z\in S_{\pi/2+\theta}$, 
$\theta\in (0,\omega)$, where $\omega = \arccos a < \pi/2$. To this end we use 
the representation
%
%
\begin{equation}\Label{2.2}
(H + z)f = \{I + BA^{-1}A(A + z)^{-1}\}(A + z)f, \qquad f \in \dom(H) = \dom(A).
\end{equation}
%
%
Since
%
%
\begin{equation}\Label{2.3}
\|A(A + z)^{-1}\| \le \sup_{\lambda \ge 1}\left|\frac{\lambda}{\lambda + 
z}\right| \leq \sup_{\lambda \ge 0}\left|\frac{\lambda}{\lambda + z}\right|
\end{equation}
%
%
a straightforward computation shows that 
%
%
\begin{equation}\Label{2.4}
\|A(A + z)^{-1}\| \le \left\{\begin{array}{l} 
1 \mbox{ for } \real z\geq 0 \\
|z|/|\imag z| \mbox{ for } \real z<0,\ \imag z\neq 0                             
                                   \end{array}\right.
\end{equation}
%
%
For $z \in S_{\pi/2+\theta}$ and $\real z < 0$, then
%
%
\begin{equation}\Label{2.5}
\frac{|z|}{|\imag z|} \leq {1\over\cos\theta}.
\end{equation}
%
%
Hence (\ref{2.4}) leads to the estimate
%
%
\begin{equation}\Label{2.6}
\|A(A + z)^{-1}\| \le \frac{1}{\cos\theta}
\end{equation}
%
%
for $z \in S_{\pi/2+\theta}$ and $\real z<0$. If $\real z\geq 0$ then 
(\ref{2.4}) yields the estimate
%
\begin{equation}\Label{2.6-1}
\|A(A + z)^{-1}\| \le 1 \le \frac{1}{\cos\theta}.
\end{equation}
Therefore the estimate (\ref{2.6}) holds for any $z\in S_{\pi/2+ \theta}$.

Using the condition $(i)$ we find the estimate
%
%
\begin{equation}\Label{2.8}
\|BA^{-1}A(A + z)^{-1}\| \le \|BA^{-1}\|\|A(A + z)^{-1}\|  \le 
\frac{a}{\cos\theta} < 1
\end{equation}
%
%
for $z \in S_{\pi/2+\theta}$. Hence the operator $I + B(A + z)^{-1}$ is 
invertible for $z \in S_{\pi/2+\theta}$ and we have
the estimate
%
%
\begin{equation}\Label{2.9}
\|(I + B(A + z)^{-1})^{-1}\| \le \frac{\cos\theta}{\cos\theta - a}
\end{equation}
%
%
for $z \in S_{\pi/2+\theta}$. Then from (\Ref{2.2}) we obtain that $z \in 
S_{\pi/2+\theta} \subseteq \rho(-H)$ the resolvent set of the operator $-H$ and 
the representation
%
%
\begin{equation}\Label{2.10}
(H + z)^{-1} = (A + z)^{-1}(I + B(A + z)^{-1})^{-1},
\end{equation}
%
and the estimate
%
\begin{equation}\Label{2.11}
\|(H + z)^{-1}\| \le \frac{\cos\theta}{\cos\theta - a}\|(A + z)^{-1}\|
\end{equation}
for $z \in S_{\pi/2+\theta}$.
%
Since $A$ is self-adjoint, we have

\begin{equation}
\|(z+A)^{-1}\| \leq {1\over\cos\theta}{1\over|z|}
\end{equation}
for any $z\in S_{\pi/2+\theta}$. Thus we conclude

\begin{equation}\Label{2.20}
\|(H + z)^{-1}\| \le \frac{M_\theta}{|z|}
\end{equation}
for all $z\in S_{\pi/2+\theta}$ with
\begin{equation}\Label{2.19}
M_\theta := \frac{1}{\cos\theta - a}
\end{equation}
which shows that $-H$ is the generator of a holomorphic semigroup with angle 
$\omega = \arccos a$.

Since $B$ is accretive and $A$ is self-adjoint positive, $H=A+B$ is accretive 
and since $\lambda+H$ is boundedly invertible for $\lambda>0$, $H$ is 
m-accretive, i.e. $-H$ generates a contraction semigroup. Finally by the 
representation $H=(I+BA^{-1})A$ with $\|BA^{-1}\|\leq a<1$, the operator $H$ is 
boundedly invertible.

\end{preuve}

{\noindent\bf Remark:} Since $\sigma(H)\subseteq S_{\pi/2-\omega}$ implies 
$\sigma(H^*)\subseteq S_{\pi/2 - \omega}$, the characterisation (\ref{2.1}) 
entails that $-H^*$ is also the generator of a holomorphic semigroup.

Below we denote by $ \bD = \{z \in {\mathbb C}: |z| \le 1\}$ and $\mbox{D}_{1/2} 
= \{z \in {\mathbb C}: \left|z - 1/2\right| \le 1/2\}$, two discs in the complex 
plane $\bC$.
%
\begin{lem}\Label{II.2}
%
%
Let $A \ge I$. Then for any $\tau > 0$
%
%
\begin{equation}\Label{2.21}
\|(I - e^{-\tau A})(z - e^{-\tau A})^{-1}\| \le \left\{\begin{array}{l} |z|^{-1} 
\qquad z \in \mathbb{D} \setminus \mathrm{D}_{1/2} \\
|1 - z|/|\Im{\mathrm m}\, z| \qquad z \in \mathrm{D}_{1/2} \setminus [0,1]
\end{array}\right.
\end{equation}
%
%
%\begin{equation}\Label{2.22}                          
%\mbox{D}_{1/2} := \left\{z \in {\mathbb C}: \left|z - \frac{1}{2}\right| \le 
%\frac{1}{2}\right\}.
%\end{equation}
%
%
\end{lem}
%
%
{\em Proof:} We use the functional calculus for self-adjoint operators. 
Obviously, we have
%
%
\begin{equation}\Label{2.23}
\|(I - e^{-\tau A})(z - e^{-\tau A})^{-1}\| \le \sup_{\lambda \ge 1} \left|{1 - 
e^{-\tau\lambda}\over z - e^{-\tau\lambda}}\right| \le \sup_{\lambda \ge 0} 
\left|{1 - e^{-\lambda}\over z - e^{-\lambda}}\right|.
\end{equation}
%
%
Let $z = x + iy$ and
%
%
\begin{equation}\Label{2.24}
m(\lambda) := \frac{1 - e^{-\lambda}}{|z - e^{-\lambda}|} = \frac{1 - 
e^{-\lambda}}{\sqrt{(x - e^{-\lambda})^2 + y^2}}, \quad \lambda \ge 0.
\end{equation}
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%
If $z \in \bD \setminus \mbox{D}_{1/2}$, then $m'(\lambda) \ge 0$ for $\lambda 
\ge 0$. Therefore $m(\lambda)$ is a non-decreasing function such that
%
%
\begin{equation}\Label{2.26}
\sup_{\lambda\geq 0}m(\lambda) = \lim_{\lambda \to +\infty} m(\lambda) = 
\frac{1}{|z|},
\end{equation}
%
%
which proves the first estimate (\Ref{2.21}). If $z \in \mbox{D}_{1/2} \setminus 
[0,1]$, then $m(\lambda)$ attains its maximum at $\lambda_0$ such that
%
%
\begin{equation}\Label{2.27}
e^{-\lambda_0} = \frac{\frac{1}{4} - (x - \frac{1}{2})^2 - y^2}{1 - x}.
\end{equation}
%
%
Inserting (\Ref{2.27}) into (\Ref{2.24}) we obtain
%
%
\begin{equation}\Label{2.28}
m(\lambda_0) = \frac{|1 - z|}{|\imag z|},
\end{equation}
%
%
which yields
%
%
\begin{equation}\Label{2.29}
0 \le m(\lambda) \le \frac{|1 - z|}{|\imag z|},
\end{equation}
%
%
The inequality (\ref{2.29}) verifies the second estimate in (\Ref{2.21}). 
\hfill$\Box$

Since for $A\geq I$ and an m-accretive $B$ the operator
%
%
\begin{equation}\Label{2.30}
T(\tau) = e^{-\tau A}e^{-\tau B}, \qquad \tau > 0,
\end{equation}
%
%
is a contraction, its spectrum $\sigma(T(\tau)) \subseteq \bD$, $\tau > 0$. 
However, for $B$ small relative to $A$ we can show more. To this end we 
introduce the family of closed convex sets in $\bC$, for $r\in [0,1]$
\begin{equation}\Label{2.33}
\Sigma_{r} := \Sigma^0_{r} \cup \Sigma^1_{r},
\end{equation}
%
where
%
\begin{equation}\Label{2.31}
\Sigma^0_{r} := \{z \in \mbox{D}_{1/2}:\ |\imag z| \le r|1 - z|\}
\end{equation}
%
and
%
\begin{equation}\Label{2.32}
\Sigma^1_r := \{z\in\bD\setminus\mbox{D}_{1/2}:\ |z|\leq r\},
\end{equation}
%
see Figure 1. We remark that the family $\Sigma_r$ is increasing with $r$: for 
$r<r'$ one gets $\Sigma_r\subset\Sigma_{r'}$.

On the other hand, we denote by $R_r$ the decreasing family $\bD \setminus 
\Sigma_r$.
%
%
%
\begin{theo}\Label{II.3}
%
%
Let $A \ge I$ and let $B$ be an m-accretive operator satisfying the condition 
(ii) with $a_*<1$. Then $\sigma(T(\tau)) \subseteq \Sigma_{a_*}$ for each $\tau 
> 0$. Moreover, for any $r\in(a_*,1)$, one obtains the estimate
\begin{equation}\Label{2.61}
\|(z - T(\tau))^{-1}\| \le \frac{r}{r - a_*}\|(z - e^{-\tau A})^{-1}\|,\ z\in 
R_r=\bD\setminus\Sigma_r.
\end{equation}
%
%
\end{theo}
%
%
{\em Proof:} One has to prove that each point $z \in \bD \setminus 
\Sigma_{a_*}=R_{a_*}$ belongs to the resolvent set $\rho(T(\tau))$ of $T(\tau)$ 
for any $\tau > 0$.
To this end we use the representation
%
%
\begin{equation}\Label{2.34}
z - T(\tau) = z - e^{-\tau A} + e^{-\tau A}(I - e^{-\tau B}).
\end{equation}
%
%
If $\tau > 0$, then the operator $(I - e^{-\tau A})$ is invertible. Let
%
%
\begin{equation}\Label{2.35}
X(\tau,z) := (I - e^{-\tau A})(z - e^{-\tau A})^{-1}
\end{equation}
%
%
and 
%
%
\begin{equation}\Label{2.36}
Y(\tau) := (I - e^{-\tau A})^{-1}e^{-\tau A}(I - e^{-\tau B}).
\end{equation}
%
%
Then one obviously gets the representation
%
%
\begin{equation}\Label{2.37}
z - T(\tau) = (z - e^{-\tau A})\{I + X(\tau,z)Y(\tau)\}.
\end{equation}
%
%
For each $f \in \dom(B)$ one has
%
%
\begin{equation}\Label{2.38}
(I - e^{-\tau B})f = \int^\tau_0 ds Be^{-s B}f.
\end{equation}
%
%
Hence we get
%
%
\begin{equation}\Label{2.39}
A^{-1}(I - e^{-\tau B})f = \int^\tau_0 ds A^{-1}Be^{-s B}f.
\end{equation}
%
%
This leads to the estimate
%
%
\begin{equation}\Label{2.40}
\|A^{-1}(I - e^{-\tau B})\| \le \tau a_*
\end{equation}
%
%
Using for (\ref{2.36}) the representation
%
%
\begin{equation}\Label{2.42}
Y(\tau) = (I - e^{-\tau A})^{-1}Ae^{-\tau A}A^{-1}(I - e^{-\tau B})
\end{equation}
%
%
we obtain
%
%
\begin{equation}\Label{2.43}
\|Y(\tau)\| \le \tau a_* \|(I - e^{-\tau A})^{-1}Ae^{-\tau A}\|.
\end{equation}
%
%
By the functional calculus for self-adjoint operators one verifies that 
%
%
\begin{equation}\Label{2.44}
\tau\|(I - e^{-\tau A})^{-1}Ae^{-\tau A}\| \le 1
\end{equation}
%
%
for each $\tau > 0$ and by consequence
%
%
\begin{equation}\Label{2.45}
\|Y(\tau)\| \le a_*\ .
\end{equation}
%
%
Let $r>a_*$. To estimate $X(\tau,z)$ we have to consider two cases:

1. Let $z \in \mbox{D}_{1/2}\cap R_{r}$. Then $|\imag z| > r|1 - z|$.
%
%
Since $z \in \mbox{D}_{1/2}$ by Lemma \Ref{II.2} we find
%
%
\begin{equation}\Label{2.47}
\|X(\tau,z)\| \le \frac{|1 - z|}{|\imag z|}.
\end{equation}
%
%
Hence we have
%
\begin{equation}\Label{2.50}
\|X(\tau,z)\| <  \frac{1}{r}
\end{equation}
%
%
for $z \in R_{r}\cap\mbox{D}_{1/2}$ and $\tau > 0$.

2. Let $z \in R_{r} \setminus \mbox{D}_{1/2}$. Since $|z|>r$ by Lemma \Ref{II.2} 
we find
%
%
\begin{equation}\Label{2.52}
\|X(\tau,z)\| \le \frac{1}{|z|} < \frac{1}{r}.
\end{equation}
%
Finally we see that the estimate
%
%
\begin{equation}\Label{2.55}
\|X(\tau,z)\| < \frac{1}{r}
\end{equation}
%
%
is valid for any $z \in R_{r}$ and $\tau > 0$. Moreover, by (\Ref{2.45}) and 
(\Ref{2.55}) we have
%
%
\begin{equation}\Label{2.56}
\| X(\tau,z)Y(\tau)\| \le \|X(\tau,z)\|\|Y(\tau)\| < {a_*\over r} \leq  1
\end{equation}
%
%
for $z \in R_{r}$, $\tau > 0$. This yields that the operator $\{I + Y(\tau) 
X(\tau,z) \}$ is invertible for 
$z \in R_{a_*}=\bigcup_{r>a_*}R_r$ and $\tau > 0$, with the following estimate 
for $z\in R_r$, $r>a_*$: 
\begin{equation}
\|(I+X(\tau,z)Y(\tau,z))^{-1}\| \leq {1\over 1-a_*/r},\ \tau>0.
\end{equation}

Since the operator $(z - e^{-\tau A})$ is invertible for $z \in R_{a_*}$ and 
$\tau > 0$, we obtain from the representation (\Ref{2.37}) that the operator 
$T(\tau)$ is invertible for $z \in R_{a_*}$, $\tau > 0$, and the announced 
estimate for $z\in R_r$, $r\in(a_*,1)$. 
\hfill $\Box$

Let us define the closed curve $\Gamma_r=\partial\Sigma_r$. Outside the disc 
$\mbox{D}_{1/2}$, it coincides with the arc of centre $0$ and of radius $r$, 
whereas inside $\mbox{D}_{1/2}$, this curve consists of two segments of tangents 
to the preceding arc passing the point $1$ (see Figure 1).


\begin{figure}[H]
\epsfysize=10cm

$$\epsfbox{CNZ1.eps}$$
\caption{Illustration of the set $\Sigma_{a_*}$ (shaded domain) with boundary  
$\Gamma_{a_*}=\partial\Sigma_{a_*}$, where $a_* = \sin \alpha$, as well as of 
our choice of the contour $\Gamma_r$ in the resolvent set $\rho(F(\tau))$, where 
$r=\sin \beta >a_*$. The contour $\Gamma_r$ consists of two segments of tangent 
straight lines $(1,A)$ and $(1,B)$ and the arc $(A,B)$ of radius $r$. The dotted 
circle $\partial \mbox{D}_{1/2}$ corresponds to the set of tangent points for 
the different values of $r\in[0,1]$.}

\end{figure}




\begin{theo}\Label{II.5}
%
%
Let $A \ge I$ and let $B$ be an m-accretive operator obeying condition (ii) with 
$a_*<1$. Then there is a constant $C$ such that
%
%
\begin{equation}\Label{2.68}
\|T(\tau)^k(I - T(\tau))\| \le \frac{C}{k + 1}, \qquad k = 0,1,2,\ldots,
\end{equation}
%
%
holds for any $\tau > 0$.
%
%
\end{theo}
%
%
{\em Proof:} By Theorem \ref{II.3}, one gets that for any $r>a_*$ and $\tau > 
0$, $\Gamma_r\in\rho(T(\tau))$, the resolvent set of $T(\tau)$. Using the 
Dunford-Taylor calculus we get the representation
%
%
\begin{equation}\Label{2.69}
T(\tau)^k(I - T(\tau)) = \frac{1}{2\pi i}\int_{\Gamma_r} z^k(1 - z)(z - 
T(\tau))^{-1} dz.
\end{equation}
%
%
We estimate the integral (\Ref{2.69}) with help of the estimate (\ref{2.61}):

\begin{equation}\label{esinteg1}
\left\|\frac{1}{2\pi i}\int_{\Gamma_r} z^k(1 - z)(z - T(\tau))^{-1} dz\right\| 
\leq {1\over 2\pi}\int_{\Gamma_r} |z^k(1-z)|{r\over r-a_*}\|(z-e^{-\tau 
A})^{-1}\||dz|.
\end{equation}

Since the integral in the right hand side of (\ref{esinteg1}) is invariant with 
respect to conjugation $z\rightarrow\overline{z}$, it is sufficient to estimate 
the integral on the path $\Gamma^+_r$: the branch of $\Gamma_r$ with $\imag 
z>0$. This branch consists of segment $(1,A)$ and arc $(\pi/2-\beta,\pi)$ of 
radius $r$, see Figure 1.

Let us parametrize $(1,A)$ by
\begin{equation}\label{para1}
z = 1 - s e^{-i\beta},\ 0\leq s\leq \cos\beta,\mbox{ with } r=\sin\beta.
\end{equation}
Since for (\ref{para1}) and the self-adjoint $e^{-\tau A}$ one has 
$\|(z-e^{-\tau A})^{-1}\| \leq (\imag z)^{-1} \leq (s \sin\beta)^{-1}$, we get
\begin{equation}\label{esinteg2}
\int_{(1,A)}|z^k(1-z)|\|(z-e^{-\tau A})^{-1}\||dz| \leq \int_0^{\cos\beta} ds 
|1-s e^{-i\beta}|^k{1\over\sin\beta}
\end{equation}
Notice that for $0\leq s\leq \cos\beta$ one gets by convexity that
\begin{equation}
|1-s e^{-i\beta}| \leq 1 - s{1-\sin\beta\over\cos\beta}
\end{equation}
Therefore, we obtain for (\ref{esinteg2}) the estimate
\begin{equation}\label{esinteg3}
\int_{(1,A)}|z^k(1-z)|\|(z-e^{-\tau A})^{-1}\||dz| \leq {\cos\beta\over\sin\beta 
 (1-\sin\beta)}{1-(\sin\beta)^k\over k+1}
\end{equation}
%
For the arc $(\pi/2-\beta,\pi)$ of radius $r$, one has the estimate 
$\|(z-e^{-\tau A})^{-1}\| \leq (r\cos\beta)^{-1}$. Then we find:
\begin{equation}\label{esinteg4}
\int_{\Gamma^+_r\setminus (1,A)}|z^k(1-z)|\|(z-e^{-\tau A})^{-1}\||dz| \leq 
(\pi/2 + \beta)r{2r^k\over r\cos\beta} \leq {2\pi\over\cos\beta}{1/r\over e \ln 
(1/r)} {1\over k+1}.
\end{equation}
In the last inequality we use that: $\sup_{k\geq 0} (k+1) e^{-k\ln x} = x/(e\ln 
x)$.

Finally, combining (\ref{esinteg1}), (\ref{esinteg3}) and (\ref{esinteg4}) we 
obtain:
\begin{eqnarray}
& & \hspace{-2cm} \left\|\frac{1}{2\pi i}\int_{\Gamma_r} z^k(1 - z)(z - 
T(\tau))^{-1} dz\right\| \leq \\
& & 2\left({\cos\beta\over 2\pi\sin\beta(1-\sin\beta)} + {1/r\over e\cos\beta\ln 
(1/r)}\right){r\over r-a_*}{1\over k+1} \nonumber
\end{eqnarray}
which gives the announced estimate (\ref{2.68}) with
\begin{equation}
C=\left({r\cos\beta\over 2\pi\sin\beta(1-\sin\beta)} + {1\over e\cos\beta\ln 
(1/r)}\right){2\over r-a_*}.
\end{equation} \hfill $\Box$

\noindent{\bf Remark:} If $T(\tau) = e^{-\tau A}e^{-\tau B}$ is replaced by 
$F(\tau) = e^{-\tau B}e^{-\tau A}$, then the above Theorems \ref{II.3} and 
\ref{II.5} remain valid if $B$ satisfies, instead of $(ii)$, the condition $(i)$ 
with $a<1$. For the proof one has to consider the adjoint family $F(\tau)^* = 
e^{-\tau A}e^{-\tau B^*}$. In this case we have simply to replace $a_*$ by $a$ 
in the course of proofs.




\section{Proof of the main Theorem}
\setcounter{equation}{0}
\renewcommand{\theequation}{\arabic{section}.\arabic{equation}}
\setcounter{theo}{0}
\renewcommand{\thetheo}{\arabic{section}.\arabic{theo}}


First we recall the following property of holomorphic semigroups (see e.g. 
\cite[Ch.IX]{Kato}):

\begin{prop}\Label{III.2}
%
%
Let $-H$ be the generator of a bounded holomorphic semigroup. Then $\ran 
(e^{-tH}) \subseteq \dom(H)$, and we have the estimate:
%
%
\begin{equation}\Label{CH}
\left\|He^{-tH}\right\| \le \frac{C_H}{t}, \qquad t > 0.
\end{equation}
%
%
\end{prop}


\begin{lem}\label{III.4}
%
%
Let $A$ be a self-adjoint operator, $A\geq I$. Let operator $B$ be m-accretive, 
satisfying (ii) with $a_*<1$. Then operator $I - T(\tau)$ is boundedly 
invertible for each $\tau > 0$ and
%
%
\begin{equation}\Label{3.16}
\left\|(I - T(\tau))^{-1}u\right\| \le \frac{1}{1 - a_*}\left(\|u\| + 
\frac{1}{\tau}\|A^{-1}u\|\right), \qquad u \in \kH,
\end{equation}
%
%
\end{lem}
%
%
{\em Proof:} We use representation 
%
%
\begin{equation}\Label{3.17}
I - T(\tau) = I - e^{-\tau A} + e^{-\tau A}(I - e^{-\tau B}).
\end{equation}
%
%
Since $\|e^{-\tau A}\|\leq e^{-\tau}<1$, the operator $I - e^{-\tau A}$ is 
boundedly invertible provided $\tau > 0$. Hence, one gets representation
%
%
\begin{equation}\Label{3.18}
I - T(\tau) = (I - e^{-\tau A})\left(I + Y(\tau)\right).
\end{equation}
%
where we have used the notation (\Ref{2.36}). By (\Ref{2.45}) and $a_* < 1$ we
obtain that $I - T(\tau)$ is invertible. Moreover, we have the estimate :

\begin{equation}\label{es3.5}
\|(I - T(\tau))^{-1}u\| \leq {1\over 1-a_*} \left\|\left(I - e^{-\tau 
A}\right)^{-1}u\right\|, \quad u\in \kH
\end{equation}
By Lemma 2.3 of \cite{NZ1}, we have for any $u\in{\kH}$ and $\tau>0$ that
\begin{equation}\label{es3.6}
\left\|\left(I - e^{-\tau A}\right)^{-1}u\right\| \leq \|u\| + {1\over 
\tau}\|A^{-1}u\|.
\end{equation}
The estimates (\ref{es3.5}), (\ref{es3.6}) lead to the desired inequality 
(\Ref{3.16}) \hfill$\Box$



\begin{lem}\label{III.5}
Let $-A$, boundedly invertible, and $-B$ be generators of bounded semigroups, 
with $B$ satisfying the conditions (i), (ii) for some (not necessarily bounded 
by 1) $a$ and $a_*$. Assume that operator $-H=-(A+B)$ is the boundedly 
invertible generator of a bounded semigroup. Then there exists a constant $L_1$ 
such that for all $\tau>0$:
\begin{eqnarray}
\left\|A^{-1}\left(e^{-\tau A}e^{-\tau B} - e^{-\tau(A+B)}\right)\right\| & \leq 
&  L_1\tau \label{L11}\\
\left\|\left(e^{-\tau A}e^{-\tau B} - e^{-\tau(A+B)}\right)A^{-1}\right\| & \leq 
&  L_1\tau \label{L12}
\end{eqnarray}
\end{lem}

\begin{preuve}
See Lemma 5 in \cite{CZ1}.
\end{preuve}

\begin{lem}\label{III.6}
Let $A$, $B$ and $H=A+B$ be the same as in Lemma \ref{III.5}. Then there exists 
a constant $L_2$ such that for all $\tau\geq 0$ :
\begin{equation}
\left\|A^{-1}\left(e^{-\tau A}e^{-\tau B} - e^{-\tau(A+B)}\right)A^{-1}\right\| 
\leq L_2\tau^2 \label{L21}
\end{equation}
\end{lem}

\begin{preuve}
See Lemma 6 in \cite{CZ1}.
\end{preuve}


Now we are in position to prove our main Theorem.

%
\begin{theo}\Label{III.7}
%
%
Let $A$ be a self-adjoint operator and $A\geq I$. Let $B$ satisfy conditions 
(i), (ii), (iii), and $a,a_*<1$. Then there is a constant $L > 0$ such that the 
estimate
%
%
\begin{equation}\Label{3.24}
\left\|\left(e^{-tA/n}e^{-tB/n}\right)^n - e^{-tH}\right\| \le 
L\;\frac{\ln(n)}{n},
\qquad n = 3,4,\ldots,
\end{equation}
%
%
holds uniformly in $t \ge 0$, where $H=A+B$ with $\dom(H) = \dom(A)$.
%
%
\end{theo}
%
%
{\em Proof:} Let $\tau = t/n$ and $U(\tau) = e^{-\tau H}$, which is a 
holomorphic contraction semigroup by Theorem \ref{II.1}. We start with the 
representation
%
%
\begin{eqnarray}\Label{3.25}
\lefteqn{T(\tau)^n - U(\tau)^n} \\
& = & T(\tau)^{n-1}(T(\tau) - U(\tau)) + \sum^{n-1}_{m=1}T(\tau)^{n-m-1}(T(\tau) 
- U(\tau))U(\tau)^m. \nonumber
\end{eqnarray}
%
%
Since by Lemma \Ref{III.4} the operator $(I - T(\tau))^{-1}$ exists for any 
$\tau>0$, we find
%
%
\begin{eqnarray}\Label{3.26}
\lefteqn{T(\tau)^n - U(\tau)^n = T(\tau)^{n-1}(T(\tau) - U(\tau))}\\
& &  + \sum^{n-1}_{m=1}T(\tau)^{n-m-1}(I - T(\tau))(I - T(\tau))^{-1}(T(\tau) - 
U(\tau))U(\tau)^m. \nonumber
\end{eqnarray}
%
%
This leads to the estimate
%
%
\begin{eqnarray}\Label{3.27}
\lefteqn{\|T(\tau)^n - U(\tau)^n\| \le  \|T(\tau)^{n-1}(T(\tau) - U(\tau))\|}\\
& &  + \sum^{n-1}_{m=1}\|T(\tau)^{n-m-1}(I - T(\tau))\|\|(I - 
T(\tau))^{-1}(T(\tau) - U(\tau))U(\tau)^m\|. \nonumber
\end{eqnarray}
%
%
Then by (\Ref{3.16}) we obtain
%
%
\begin{eqnarray}\Label{3.28}
\lefteqn{\|(I - T(\tau))^{-1}(T(\tau) - U(\tau))U(\tau)^m\|}\\
& \le & \frac{1}{1 - a_*}\left\{\|(T(\tau) - U(\tau))U(\tau)^m\| + 
        \frac{1}{\tau}\left\|A^{-1}(T(\tau) - U(\tau))U(\tau)^m\right\|\right\}. 
\nonumber
\end{eqnarray}
%
%
Since 
%
%
\begin{equation}\Label{3.29}
(T(\tau) - U(\tau))U(\tau)^m = (T(\tau) - U(\tau))A^{-1}AH^{-1}HU(\tau)^m,
\end{equation}
%
%
one gets 
%
%
\begin{equation}\Label{3.30}
\|(T(\tau) - U(\tau))U(\tau)^m\| \le  \left\|(T(\tau) - 
U(\tau))A^{-1}\right\|\|AH^{-1}\|\|HU(\tau)^m\|.
\end{equation}
%
%
Applying now Proposition \Ref{III.2} we obtain
%
%
\begin{equation}\Label{3.31}
\|(T(\tau) - U(\tau))U(\tau)^m\| \le  \left\|A^{-1}(T(\tau) - 
U(\tau))\right\|\|AH^{-1}\|\frac{C_H}{\tau}\frac{1}{m}.
\end{equation}
%
%
By $AH^{-1} = (I + BA^{-1})^{-1}$ and $\|BA^{-1}\|\leq a<1$ one gets
%
%
\begin{equation}\Label{3.32}
\|AH^{-1}\| \le \frac{1}{1 - a}.
\end{equation}
%
%
Hence 
%
%
\begin{equation}\Label{3.34}
\|(T(\tau) - U(\tau))U(\tau)^m\| \le  \left\|A^{-1}(T(\tau) - 
U(\tau))\right\|\frac{C_H}{(1 - a)\tau}\frac{1}{m}.
\end{equation}
%
%
Similarly we get
%
%
\begin{equation}\Label{3.35}
\|A^{-1}(T(\tau) - U(\tau))U(\tau)^m\| \le  \left\|A^{-1}(T(\tau) - 
U(\tau))A^{-1}\right\|\frac{C_H}{(1 - a)\tau}\frac{1}{m}.
\end{equation}
%
%
Inserting (\Ref{3.34}) and (\Ref{3.35}) into (\Ref{3.28}) we find
%
%
\begin{eqnarray}\Label{3.40}
\lefteqn{\|(I - T(\tau))^{-1}(T(\tau) - U(\tau))U(\tau)^m\|}\\
& \le & \left\{\frac{1}{\tau}\|(A^{-1}(T(\tau) - U(\tau))\| + 
        \frac{1}{\tau^2}\left\|A^{-1}(T(\tau) - 
U(\tau))A^{-1}\right\|\right\}\frac{C_H}{(1 - a_*)(1-a)}\frac{1}{m}. \nonumber
\end{eqnarray}
%
%
Applying to (\Ref{3.40}) Lemma \Ref{III.5} and Lemma \Ref{III.6} we find
%
%
\begin{equation}\Label{3.41}
\|(I - T(\tau))^{-1}(T(\tau) - U(\tau))U(\tau)^m\| \le C_H\frac{L_1 + L_2}{(1 - 
a_*)(1-a)}\frac{1}{m} = \frac{L_3}{m},
\end{equation}
%
%
where we put $L_3 := C_H(L_1 + L_2)/(1-a_*)(1-a)$. Further, by Theorem 
\Ref{II.5} we have
%
%
\begin{equation}\Label{3.42}
\|T(\tau)^{n-m-1}(I - T(\tau))\| \le \frac{C}{n - m}, \qquad n = 2,3,\ldots, 
\qquad m = 0,1,2,\ldots,n-1.
\end{equation}
%
%
Inserting (\Ref{3.41}) and (\Ref{3.42}) into (\Ref{3.27}) we get
%
%
\begin{equation}\Label{3.43}
\|T(\tau)^n - U(\tau)^n\| \le  \|T(\tau)^{n-1}(T(\tau) - U(\tau))\|  + C 
L_3\sum^{n-1}_{m=1}\frac{1}{(n-m)m}.
\end{equation}
%
%
Since
%
%
\begin{eqnarray}\Label{3.44}
\lefteqn{\sum^{n-1}_{m=1}\frac{1}{(n-m)m}}\\
& = & \frac{2}{n}\sum^{n-1}_{m=1}\frac{1}{m} \le \frac{2}{n}\left(1 + 
\ln(n-1)\right) \le 4\frac{\ln(n)}{n}, 
\qquad n = 2,3,\dots, \nonumber
\end{eqnarray}
%
%
one has
%
%
\begin{equation}\Label{3.45}
\|T(\tau)^n - U(\tau)^n\| \le  \|T(\tau)^{n-1}(F(\tau) - U(\tau))\|  + 4C 
L_3\frac{\ln(n)}{n}.
\end{equation}
%
%
It remains to estimate $\|T(\tau)^{n-1}(T(\tau) - U(\tau))\|$. Again by 
invertibility of $I - T(\tau)$ (see Lemma 3.2) we find
%
%
\begin{equation}\Label{3.46}
\|T(\tau)^{n-1}(T(\tau) - U(\tau))\| \le \|T(\tau)^{n-1}(I - T(\tau)\|\|(I - 
T(\tau))^{-1}(T(\tau) - U(\tau))\|.
\end{equation}
%
%
By (\Ref{3.16}) we get
%
%
\begin{eqnarray}\Label{3.47}
\lefteqn{\|(I - T(\tau))^{-1}(T(\tau) - U(\tau))\|}\\
& \le & \frac{1}{1 - a_*}\left\{\|T(\tau) - U(\tau)\| + 
\frac{1}{\tau}\left\|A^{-1}(T(\tau) - U(\tau))\right\|\right\} \nonumber \\
& \le &\frac{1}{1 - a_*}\left\{2 + \frac{1}{\tau}\left\|A^{-1}(T(\tau) - 
U(\tau))\right\|\right\}. \nonumber 
\end{eqnarray}
%
% 
Applying to the right-hand side of (\Ref{3.47}) Lemma \Ref{III.5} we find
%
%
\begin{equation}\Label{3.63}
\|(I - T(\tau))^{-1}(T(\tau) - U(\tau))\| \le \frac{2 + L_1}{1 - a_*}.
\end{equation}
%
%    
Inserting (\Ref{3.63}) into (\Ref{3.46}) one gets
%
%
\begin{equation}\Label{3.64}
\|T(\tau)^{n-1}(T(\tau) - U(\tau))\| \le \frac{2 + L_1}{1 - 
a_*}\|T(\tau)^{n-1}(I - F(\tau)\|.
\end{equation}
%
%
Then applying Theorem \Ref{II.5} to the right-hand side of (\Ref{3.64}) we 
obtain
%
%
\begin{equation}\Label{3.65}
\|T(\tau)^{n-1}(T(\tau) - U(\tau))\| \le \frac{2 + L_1}{1 - a_*}\;\frac{C}{n},\ 
n=1,2,\dots\ .
\end{equation}
%
%
Inserting (\Ref{3.65}) into (\Ref{3.45}) we find the estimate (\Ref{3.24}):
%
%
\begin{equation}\Label{3.66}
\|T(\tau)^n - U(\tau)^n\| \le L\;\frac{\ln(n)}{n}, \qquad n = 3,4,\dots\ .
\end{equation}
%
%
Here $L := C(2 + L_1)(1 - a_*)^{-1} + 4C L_3$ and we used $1/n<\ln(n)/n$ to 
estimate (\ref{3.65}), for $n=3,4,\dots$ \hfill$\Box$
%

\begin{cor}\label{cor1}
Let $A$ and $B$ satisfy the conditions of Theorem \ref{III.7}. Then there are 
constants  $L', L''$ such that
\begin{eqnarray}
\left\|\left(e^{-tB/n}e^{-tA/n}\right)^n - e^{-tH}\right\| & \leq & 
L'\;\frac{\ln(n)}{n}, \qquad n = 3,4,\ldots,\label{escor11}\\
\left\|\left(e^{-tA/2n}e^{-tB/n}e^{-tA/2n}\right)^n - e^{-tH}\right\| & \leq & 
L''\;\frac{\ln(n)}{n}, \qquad n = 3,4,\ldots\ .\label{escor12}
\end{eqnarray}
holds uniformly in $t\geq 0$.
\end{cor}

\begin{preuve}
Since $B^*$ satisfies conditions of Theorem \ref{III.7}, we have
\begin{equation}
\left\|\left(e^{-tA/n}e^{-tB^*/n}\right)^n - e^{-tH^*}\right\| \leq 
L'\,\frac{\ln (n)}{n}.
\end{equation}
Notice that $H^* = A+B^*$ is well defined and m-accretive. By taking the adjoint 
sequence of operators, we find (\ref{escor11}). In order to obtain 
(\ref{escor12}), one has to observe that the line of reasoning in the proof of 
Theorem \ref{III.7} follows thought verbatim for the symmetrized formula.
\end{preuve}



\section{Conclusion}
\setcounter{equation}{0}
\renewcommand{\theequation}{\arabic{section}.\arabic{equation}}
\setcounter{theo}{0}
\renewcommand{\thetheo}{\arabic{section}.\arabic{theo}}


The aim of the present paper was to show that the operator-norm error bound 
estimates for the exponential Trotter product formula persist if for a couple 
{$A = A^*, B$} we pass from a self-adjoint $A$-small perturbation $B$ to an 
m-accretive $A$-small operator $B$. This generalizes the previous result 
\cite{NZ1} to the case of a non-self-adjoint $B$ under conditions that $\|Bu\| 
\leq a\|Au\|$, $u\in\dom(A)$ with $a\in(0,1)$ and $\|B^*u\|\leq a_*\|Au\|$, 
$u\in\dom(A)$ with $a_*\in(0,1)$, see Theorem \ref{III.7}.


Let us consider the following special case: $B(z) = zV$ where $V$ is a 
self-adjoint operator which is relatively small with respect to $A \ge I$, i.e. 
$\dom(A) \subseteq \dom(V)$ and
%
%
\begin{equation}\Label{4.0}
\|Vf\| \le a\|Af\|, \quad f \in \dom(A), \quad \mbox{for some} \quad a \in 
(0,1).
\end{equation}
%
%
Obviously, the operators $B(z) = zV$ and $B(z)^* = \overline{z}V$ are 
m-accretive and satisfy the conditions $(i)$-$(ii)$ for $\real z\geq 0$. If, in 
addition, $|z|\leq 1$, then the constants $a,a_*$ for $B(z)$ remain smaller than 
$1$. Applying Theorem \Ref{III.7} we find, for instance, that the error estimate
%
%
\begin{equation}\Label{4.0-1}
\|(e^{-tA/n}e^{-tzV/n})^n - e^{-tH(z)}\| \le L\frac{\ln(n)}{n}, \quad n = 
2,3\ldots,
\end{equation}
%
%
holds uniformly in $t \ge 0$ and in $z$ such that $\real z\geq 0$, $|z|\leq 1$, 
where $H(z) = A+zV$ on $\dom(H(z)) = \dom(A)$. In the case $z=i$, we consider 
the perturbation of a self-adjoint contraction semigroup by a unitary group.
%
%

Since $a,a^*$ may be strictly positive, the present paper is also a 
generalisation of \cite{CZ1}, where the estimate (\ref{1.5}) was found for $B$ 
which is {\it infinitesimally} Kato-small with respect to $A$. Notice that in 
\cite{CZ1} we did not suppose that the operator $A$ is self-adjoint but only 
that it generates a contraction holomorphic semigroup. It is the case in 
particular when $A$ is an m-sectorial operator with vertex in $0$, see 
\cite[Ch.IX]{Kato}.\\

Therefore, it is relevant to ask: why do we need $A = A^*$ in the main Theorem 
of the present paper ? First, this condition is important for estimate 
$\|(z-e^{-\tau A})^{-1}\|$, see (\ref{esinteg2}) and (\ref{esinteg4}) in the
proof of Theorem 2.4. Second, it makes possible to use the spectral theorem to 
obtain the estimates (\ref{2.21}), see Lemma 2.2, and (\ref{3.16}), see Lemma 
3.2.\\

Suppose, for example, that instead to be  self-adjoint the operator $A$ is 
m-sectorial. Then operator-norm continuity of the holomorphic semigroup 
$U_{A}(t)= e^{-tA}$ for $ t > 0$, allows to localize the {\it spectrum} of this 
semigroup $\sigma(e^{-tA})$ but {\it not} the numerical range $\Theta(e^{-tA})$ 
! However to estimate the norm of the resolvent $(z-e^{-\tau A})^{-1}$, the 
spectrum is not sufficient in general, except for normal (in particular 
self-adjoint) $A$, for which $\overline{\Theta(e^{-tA})}$ is the convex hull of 
$\sigma(e^{-tA})$. Thus we solve the problem here only for the couple of 
self-adjoint $A$ and m-accretive $B$, i.e. the limiting case of angles $\theta_A 
= 0$ and $\theta_B = \pi/2$.\\

Although the above mentioned result of \cite{CZ1}  allows to pass to an 
m-sectorial operator $A$ with $0 < \theta_A < \pi/2$, the price is that $B$ and 
$B^*$ should be infinitesimally $A-$small if one insists on the {\it error bound 
estimate}. If {\it not}, then \cite{CZ2} for two m-sectorial operators $A$ and 
$B$ ( $\theta_A,\theta_B\in(0,\pi/2)$ ) such that:

-- either $(I+A)^{-1}$ or $(I+\real A)^{-1/2}(I+\real B)^{1/2}$ is compact,

-- or $A$ and $B$ satisfy conditions
%
\begin{enumerate}
\item[$(i)$] ${\Re \mbox{\rm e }}A , {\Re \mbox{\rm e }}B \geq \gamma > 0$ ;
%
\item[$(ii)$] $\dom({\Re \mbox{\rm e }}A) \subseteq \dom({\Im \mbox{\rm m }}A)$ 
;
%
\item[$(iii)$] $\dom({\Re \mbox{\rm e }}B) \subseteq \dom({\Im \mbox{\rm m }}B)$ 
;
%
\item[$(iv)$] $\dom(({\Re \mbox{\rm e }}A)^\alpha) \subseteq \dom(({\Im 
\mbox{\rm m }}B)^\alpha)$   for some $\alpha\in(1/2,1]$ ;
%
\item[$(v)$] $\|({\Re \mbox{\rm e }}B)^\alpha u\| \leq  c \|({\Re \mbox{\rm e 
}}A)^\alpha u\|$ , $u\in \dom (({\Re \mbox{\rm e }}A)^\alpha)$, $0 < c < 1$ and 
some $\alpha\in(1/2,1]$,
%
\end{enumerate}
one gets   
\begin{equation}
\|.\|-\lim_{n\rightarrow +\infty }(e^{-tA/n}e^{-tB/n})^{n}=e^{-tH}  .
\label{4.1}
\end{equation}   
The operator-norm convergence in (\ref{4.1}) is uniform in $t$ for any compact 
set of the sector $S_{\omega}=\{z\in {\bC}:z\neq 0$ and $|{\arg }(z)|<\omega\}$, 
where ${\omega}= {\pi/2}- \max\{\theta_A,\theta_B\}$ and $H$ is the form-sum of 
$A$ and $B$.\\

We conclude by remark that our main Theorem 3.5 is valid if, instead of 
$e^{-tA/n}$, one considers operator $ f(tA/n)$. Here $f : [0,+\infty) 
{\rightarrow} [0,1]$ is a Borel function such that, cf.\cite{NZ1} :\\
\begin{eqnarray}
C_0 & := &\sup_{x > 0} \left\{\frac{x\sqrt{f(x)}}{1 - f(x)}\right\} < + \infty   
                              \Label{4.2} \\[2mm]
C_1 & := & \sup_{x > 0} \left\{ \frac{1 - f(x)}{x}\right\} < + \infty            
                                \Label{4.3} \\[2mm]
C_2 & := & \sup_{x > 0} \left|\left(f(x) - \frac{1}{1 + 
x}\right)\frac{1}{x^2}\right| < + \infty \Label{4.4} \\[2mm] 
\end{eqnarray}
%
%
as well as
%
%
\begin{equation}
0 < C_0a < 1.
\Label{4.5}
\end{equation}
Then we get 
\begin{equation}
\left\|\left(f(tA/n)e^{-tB/n}\right)^n - e^{-t(A+B)}\right\| \leq L_0 {\ln 
n\over n},\ n\geq 3
\Label{4.6}
\end{equation}
uniformly in $t\geq 0$. The proof needs only minor modifications of Lemmata 
3.2-3.4 as it is outlined in \cite{NZ1}.



\section*{Acknowledgements}
H.N. would like to thank Deutsche Forschungsgemeinschaft for financial support 
and Centre de Physique Th\'{e}orique for hospitality.



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\end{thebibliography}


\end{document}
