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\begin{document}

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\begin{center}

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{\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 Trotter-Kato product formula and operator-norm convergence}


\vspace{2 cm}

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{\bf H. NEIDHARDT\footnote{Fachbereich Mathematik Universit\"{a}t Potsdam - Email : neidhard@rz.uni-potsdam.de}  
and
V.A. ZAGREBNOV\footnote{Universit\'e de la M\'editerran\'ee
(Aix-Marseille II) - Email : zagrebnov@cpt.univ-mrs.fr} }

\vspace{2,3 cm%
}

{\bf Abstract}
\end{center}

We find necessary and sufficient conditions for the {\em operator-norm}
convergence of the Trotter-Kato product formula. Using them we
prove that this convergence takes place: {\em (i)} if the resolvent of one 
of the involved operators is compact, either {\em (ii)} if one operator is 
relatively compact with respect to another one, or {\em (iii)} if the product 
of resolvents of the involved operators is compact.

\vspace{2 cm%
}


\bigskip

\noindent 
Mathematics Subject Classification: 47D03, 47B25, 47B07

\bigskip

\noindent Keywords : Trotter-Kato product formula, operator-norm convergence


\bigskip

\noindent
June 1998\\
CPT-98/P.3670

\bigskip

\noindent anonymous ftp :
ftp.cpt.univ-mrs.fr\\
web : www.cpt.univ-mrs.fr\\ 



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\section{Introduction}
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Let $A$ and $B$ be two non-negative self-adjoint operators on the separable 
Hilbert space $\kH$, i.e.
%
%
\be\la{1.1}
A \ge 0 \qquad \mbox{and} \qquad B \ge 0.
\ee
%
%
By $\kH_0$ we denote the subspace
%
%
\be\la{1.2}
\kH_0 := \overline{\dom(A^{1/2}) \cap \dom(B^{1/2})}.
\ee
%
%
It may happen that $\kH_0 = \{0\}$. The form sum $H = A \stackrel{.}{+} B$ is
well-defined on the subspace $\kH_0$. Kato has shown in \cite{Ka1} that under
these conditions the Trotter product formula \cite{Tr1} strongly converges to 
$e^{-tH}$ uniformly in $t \in [0,T]$, $0 < T < +\infty$, i.e., one has
%
%
\be\la{1.2-1}
s-\lim_{n\to\infty}\left(e^{-tA/n}e^{-tB/n}\right)^n = e^{-tH}P_0, 
\qquad t \ge 0,
\ee
%
%
where $P_0$ denotes the orthogonal projection from $\kH$ onto $\kH_0$. 
Moreover, it was found by Kato in \cite{Ka2,Ka1} that the Trotter product
formula is true not only for the exponential function $e^{-x}$, $x \ge 0$, but
for a whole class of Borel measurable functions $f(\cdot)$ and $g(\cdot)$ which
are defined on $\bR_+ := [0,\infty)$ and characterized by
%
%
\bea
& & 0 \le f(x) \le 1, \qquad f(0) = 1, \qquad f'(+0) = -1 \la{1.3}\\
& & 0 \le g(x) \le 1, \qquad g(0) = 1, \qquad g'(+0) = -1. \la{1.4}
\eea
%
%
and which are called {\em Kato-functions} in the following. In \cite{Ka1} it 
was proven that for Kato-functions $f(\cdot)$ and $g(\cdot)$ besides 
(\re{1.2-1}) one always has
%
%
\be\la{1.6}
s-\lim_{n\to\infty}\left(f(tA/n)g(tB/n)\right)^n = e^{-tH}P_0, \qquad t \ge 0,
\ee
%
%
uniformly in $t \in [0,T]$, $0 < T < +\infty$. In the following we call 
product formulae of this type {\em Trotter-Kato product formulae}. 

Naturally, the problem arises to find conditions such that the strong 
convergence in (\re{1.2-1}) and (\re{1.6}) can be replaced by the 
operator-norm convergence. A partial solution of this problem was found 
in \cite{NZ1}. Defining
%
%
\be\la{1.7}
0 \le \varphi_0(x) := \inf_{0 < s \le x}s^{-1}\left(\frac{1}{f(s)} - 1\right),
\ee
%
%
where $0^{-1} := +\infty$, and setting
%
%
\be\la{1.8}
f_0(x) := \left\{\begin{array}{ll} 1                        & x = 0 \\
                                   (1 + x\varphi_0(x))^{-1} & x > 0, \end{array} \right.
\ee
%
%
which again defines a Kato-function, it was shown in \cite{NZ1} that under  
condition
%
%
\be\la{1.9}
f_0(tA) \in \kL_p(\kH), \qquad t > 0, \qquad 1 \le p < +\infty,
\ee
%
%
one gets for Kato-functions $f(\cdot)$ and $g(\cdot)$
%
%
\be\la{1.10}
\lim_{n\to\infty}\left\|\left(f(tA/n)g(tB/n)\right)^n - e^{-tH}P_0\right\|_1 = 0
\ee
%
%
uniformly in $t \in [a,b]$ for any bounded interval 
$[a,b] \subseteq (0,\infty)$. Here 
$\kL_p(\kH)$ and $\|\cdot\|_p$,  $1 \le  p < +\infty$, denote the so-called 
von Neumann-Schatten ideals of compact operators and their norms ($p \ge 1$),
in particular, $\kL_\infty(\kH)$ denotes the set of compact operators.
Since trace-norm convergence (\re{1.10}) implies operator-norm convergence, the 
condition (\re{1.9}) implies the desired convergence. 
In particular, if $f(x) = e^{-x}$, $x \ge 0$, then $f_0(x) = (1 + x)^{-1}$ which
yields that condition
%
%
\be\la{1.11}
(I + tA)^{-1} \in \kL_p(\kH), \qquad t > 0, \quad 1 \le p < +\infty,
\ee
%
%
implies the operator-norm convergence. 

However, by the recent papers \cite{IT1}, \cite{NZ2}, \cite{NZ3} and \cite{Ro1} 
the assumption (\re{1.9}) is not necessary for the operator-norm 
convergence. For instance, in \cite{NZ2} it is proven that under 
the conditions $A \ge I$, $B \ge I$, 
%
%
\be\la{1.13}
\dom(A) \subseteq \dom(B)
\ee
%
%
and
%
%
\be\la{1.14}
\|Bf\| \le a\|Af\|, \qquad 0 < a < 1, \quad f \in \dom(A),
\ee
%
%
for a certain subclass of Kato-functions one has
%
%
\be\la{1.15}
\|(f(tA/n)g(tB/n))^n - e^{-tH}\| = {\Large O}(\ln(n)/n)
\ee
%
%
uniformly in $t \ge 0$. Moreover, replacing conditions (\re{1.13}) and 
(\re{1.14}) by fractional-power conditions 
%
%
\be\la{1.16}
\dom(A^\ga) \subseteq \dom(B^\ga), \qquad \frac{1}{2} < \ga < 1,
\ee
%
%
and
%
%
\be\la{1.17}
\|B^\ga f\| \le a\|A^\ga f\|, \qquad 0 < a < 1, \quad f \in \dom(A^\ga),
\ee
%
%
for a certain subclass of Kato-functions it was in particular found in 
\cite{NZ3} that 
%
%
\be\la{1.18}
\|(f(tA/n)g(tB/n))^n - e^{-tH}\| = {\Large O}(\ln(n)/n^{2\ga-1})
\ee
%
%
uniformly in $t \in [\epsilon,\infty)$ , where $\epsilon > 0$. Hence very 
different conditions can imply operator-norm convergence of the Trotter-Kato 
product formula. 

So, our first aim (see Section 2) will be to look for {\em necessary} and {\em sufficient} 
conditions which guarantee the {\em operator-norm} convergence of the Trotter-Kato 
product formula for given non-negative self-adjoint operators $A$ and $B$ 
and Kato-functions $f(\cdot)$ and $g(\cdot)$. However, in contrast to the
papers \cite{IT1}, \cite{IT2}, \cite{NZ2}, \cite{NZ3} and \cite{Ro1} we are 
{\em not} interested in error bounds. 

Using this approach we prove in Section 3 that under the condition  
%
%
\be\la{1.19}
f_0(t_0A) \in \kL_\infty(\kH)
\ee
%
%
for some $t_0 > 0$ one gets operator-norm convergence of the Trotter-Kato product 
formula for any non-negative self-adjoint operator $B$ and any Kato-function 
$g(\cdot)$. Notice, that condition (\re{1.19}) is weaker than (\re{1.9}).
Furthermore, it turns out that the simpler condition (cf. (\re{1.11})):
%
%
\be\la{1.19-1}
(I + t_0A)^{-1} \in \kL_\infty(\kH), \qquad t_0 > 0,
\ee
%
%
implies the operator-norm convergence for {\em any} $B \ge 0$, {\em any 
regular} Kato-function $f(\cdot)$ (for definitions see Section 3) and 
{\em any} Kato-function 
$g(\cdot)$. Moreover, the converse holds: If $A$ is a non-negative 
self-adjoint operator such that the operator-norm convergence of the 
Trotter-Kato product formula takes place for {\em any} $B \ge 0$, {\em any 
regular} 
Kato-function $f(\cdot)$ and {\em any} Kato-function $g(\cdot)$, 
then condition  (\re{1.19-1}) has to be satisfied for some $t = t_0 > 0$.

Notice that methods used in \cite{NZ2} and \cite{NZ3} do not cover the case
$\ga = 1/2$ (cf. (\re{1.16})):
%
%
\be\la{1.20}
\dom(A^{1/2}) \subseteq \dom(B^{1/2}).
\ee
%
%
Using the results of Section 2  we prove in Section 4 that operator-norm 
convergence takes place under the condition (\re{1.20}) provided that 
{\em in addition} one has
%
%
\be\la{1.21}
B^{1/2}A^{-1/2} \in \kL_\infty(\kH), \qquad A \ge I.
\ee
%
%
We note that under condition (\re{1.21}) the operator $B^{1/2}$ is relatively
$A^{1/2}$-bounded with {\em relative bound} $a = 0$ which clarifies the difference 
between conditions (\re{1.21}) and (\re{1.17}) for $\ga = 1/2$. 

Finally, neither conditions (1.14) or (1.16),  nor conditions 
(\re{1.19}) or (\re{1.21}) are 
applicable to the operator-norm convergence of the Trotter product formula for 
Schr\"odinger semigroups studied in \cite{DIT1}, \cite{ITa1}, \cite{ITa2}. It 
turns out that this situation can be treated on the abstract 
level and for the Trotter-Kato product case assuming 
%
%
\be\la{1.23}
(I + A)^{-1}(I + B)^{-1} \in \kL_\infty(\kH).
\ee
%
%
Using results of Section 2 we show in Section 5 that condition (\re{1.23})
guarantees the operator-norm convergence of the Trotter-Kato product formula,
however, without error bounds.


%\newpage

\section{Operator-norm convergence}
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We set
%
%
\bea
T(t) & := & f(tA)^{1/2}g(tB)f(tA)^{1/2}, \qquad t \ge 0, \la{2.0-1}\\
R(t) & := & \frac{1}{t}(I - T(t)), \qquad t > 0, \la{2.0-2}\\
F(t) & := & g(tB)^{1/2}f(tA)g(tB)^{1/2}, \qquad t \ge 0, \la{2.0-3}\\
S(t) & := & \frac{1}{t}(I - F(t)), \qquad t > 0, \la{2.0-4}.
\eea
%
%
Further, we set $\kH(t) := \overline{\ran(f(tA))} \subseteq \kH$, $t \ge 0$. 
By $Q(t)$, $t \ge 0$, we denote the orthogonal projection from $\kH$ onto 
$\kH(t)$. By (\re{1.3}) one obviously has that
%
%
\be\la{2.1}
s-\lim_{t\to+0} Q(t) = I.
\ee
%
%
We note that for each $t \ge 0$ the projection $Q(t)$ commutes with $A$.
We set $\tilde{f}(tA) := f(tA)|\kH(t): \kH(t) \longrightarrow \kH(t)$, 
$t \ge 0$. By definition we have $\ker(\tilde{f}(tA)) = \{0\}$ and 
%
%
\be\la{2.2}
0 \le \tilde{f}(tA) \le I(t), \qquad t \ge 0,
\ee
%
%
where $I(t) := I|\kH(t)$. Therefore, it makes sense to introduce the family 
of operators $\{M(t)\}_{t>0}$,
%
%
\be\la{2.5}
M(t) := \frac{1}{t}[\tilde{f}(tA)^{-1} - I(t)] + \frac{1}{t}Q(t)[I - g(tB)]Q(t): \kH(t) \longrightarrow \kH(t),
\ee
%
%
with domain 
%
%
\be\la{2.6}
\dom(M(t)) = \ran(f(tA)) \subseteq \kH(t).
\ee
%
%
Finally, we put $I_0 := I|\kH_0$, see (\re{1.2}). Note that in case 
$f(x) > 0$, $x \in \bR$, one readily gets 
%
%
\be\la{2.6-1}
\kH(t) = \kH \qquad \mbox{and} \qquad Q(t) = I.
\ee
%
%
\begin{theo}\la{II.1}
%
%
Let $A$ and $B$ be non-negative self-adjoint operators on the separable
Hilbert space $\kH$ and let $f(\cdot)$ and $g(\cdot)$ be Kato-functions. 
Then the following conditions are equivalent:
%
%
\begin{enumerate}

\item[(i)] $\lim_{r\to+\infty}\sup_{t \in [a,b]}\|T(t/r)^r - P_0e^{-tH}P_0\| = 0$, \qquad $[a,b] \subseteq (0,\infty)$, 

\item[(ii)] $\lim_{t\to+0}\|(\gl I + R(t))^{-1} - P_0(\gl I_0 + H)^{-1}P_0\| = 0$, \qquad $\gl > 0$,  

\item[(iii)] $\lim_{t\to+0}\|Q(t)(\gl I(t) + M(t))^{-1}Q(t) - P_0(\gl I_0 + H)^{-1}P_0\| = 0$, \qquad $\gl > 0$,

\item[(iv)] $\lim_{t\to+0}\|(\gl I + S(t))^{-1} - P_0(\gl I_0 + H)^{-1}P_0\| = 0$, \qquad $\gl > 0$,

\item[(v)] $\lim_{r\to+\infty}\sup_{t \in [a,b]}\|F(t/r)^r - P_0e^{-tH}P_0\| = 0$, \qquad $[a,b] \subseteq (0,\infty)$.

\end{enumerate}
%
%
\end{theo}
%
%
{Proof:} We prove $(i) \Rightarrow (ii) \Rightarrow (iii) \Rightarrow (iv) 
\Rightarrow (v) \Rightarrow (i)$.

$(i) \Rightarrow (ii)$. Since 
%
%
\be\la{2.7}
\sup_{x \in [0,1]}|x^r - e^{-r(1-x)}| \le \frac{1}{r}, \qquad r \ge 1,
\ee
%
%
by the spectral integral one gets that
%
%
\be\la{2.8}
\|T(t/r)^r - e^{-tR(t/r)}\| \le \frac{1}{r}, \qquad r \ge 1, \quad t \ge 0.
\ee
%
%
By
%
%
\be\la{2.9}
e^{-tR(t/r)} - P_0e^{-tH}P_0 = T(t/r)^r - P_0e^{-tH}P_0 + e^{-tR(t/r)} - T(t/r)^r
\ee
%
%
we find the estimate
%
%
\be\la{2.10}
\|e^{-tR(t/r)} - P_0e^{-tH}P_0\| \le \|T(t/r)^r - P_0e^{-tH}P_0\| + \|e^{-tR(t/r)} - T(t/r)^r\|. 
\ee
%
%
Applying $(i)$ and (\re{2.8}) we immediately find
%
%
\be\la{2.11}
\lim_{r\to+\infty}\|e^{-tR(t/r)} - P_0e^{-tH}P_0\| = 0
\ee
%
%
for each $t \ge 0$. Let  $\kA$  denote the algebra of all real-valued 
continuous functions (which are defined on $\bR_+$ and which tend to zero as 
$x \longrightarrow +\infty$) endowed with the supremum norm. Then the 
subalgebra 
generated by the function $e^{-x}$, $x \in \bR_+$, is denoted by $\kA_e$. The 
Stone-Weierstrass theorem implies that $\kA_e$ is dense in $\kA$. In 
particular, this yields that for each $\epsilon > 0$ there is a polynomial 
$\kP(\cdot)$ such that
%
%
\be\la{2.12}
\sup_{x \ge 0}|(\gl + x)^{-1} - \kP(e^{-x})| \le \epsilon
\ee
%
%
and some fixed $\gl > 0$. In particular, we have
%
%
\be\la{2.13}
\sup_{x \ge 0}|(\gl + tx)^{-1} - \kP(e^{-tx})| < \epsilon
\ee
%
%
uniformly for $t \ge 0$. Hence
%
%
\be\la{2.14}
\|(\gl I + tR(t/r))^{-1} - \kP(e^{-tR(t/r)})\| < \epsilon
\ee
%
%
and
%
%
\be\la{2.15}
\|(\gl I_0 + tH)^{-1} - \kP(e^{-tH})\| < \epsilon
\ee
%
%
uniformly for $t \ge 0$. By
%
%
\bea\la{2.16}
\lefteqn{\hspace{-2.0cm} (\gl I + tR(t/r))^{-1} - P_0(\gl I_0 + tH)^{-1}P_0 = 
(\gl I + tR(t/r))^{-1} - \kP(e^{-tR(t/r)}) + }\\
& & \kP(e^{-tR(t/r)}) - P_0\kP(e^{-tH})P_0 + P_0(\kP(e^{-tH}) - (\gl I_0 + tH)^{-1})P_0 \nonumber
\eea
%
%
we get the estimate
%
%
\bea\la{2.17}
\lefteqn{\hspace{-1.7cm} \|(\gl I + tR(t/r))^{-1} - P_0(\gl I_0 + tH)^{-1}P_0\| \le \|(\gl I + 
tR(t/r))^{-1} - \kP(e^{-tR(t/r)})\| + }\\
& & \hspace{0.5cm}\|\kP(e^{-tR(t/r)}) - P_0\kP(e^{-tH})P_0\| + \|P_0(\kP(e^{-tH}) - (\gl I_0 + tH)^{-1})P_0\|. \nonumber
\eea
%
%
Taking into account (\re{2.14}) and (\re{2.15}) we obtain
%
%
\be\la{2.18}
\|(\gl I + tR(t/r))^{-1} - P_0(\gl I_0 + tH)^{-1}P_0\| \le 2\epsilon + \|\kP(e^{-tR(t/r)}) - P_0\kP(e^{-tH})P_0\|.\nonumber
\ee
%
%
Since $\|e^{-tR(t/r)}\| \le 1$ and $\|e^{-tH}\| \le 1$ we find a constant
$C_\kP > 0$, which depends on the polynomial $\kP(\cdot)$, such that
%
%
\be\la{2.19}
\|\kP(e^{-tR(t/r)}) - P_0\kP(e^{-tH})P_0\| \le C_\kP\|e^{-tR(t/r)} - P_0e^{-tH}P_0\|.
\ee
%
%
for $t \ge  0$. Hence
%
%
\be\la{2.20}
\|(\gl I + tR(t/r))^{-1} - P_0(\gl I_0 + tH)^{-1}P_0\| \le 2\epsilon + C_{\kP}\|e^{-tR(t/r)} - P_0e^{-tH}P_0\| 
\ee
%
%
for $t \ge 0$. Thus, using (\re{2.11}) we get
%
%
\be\la{2.21}
\lim_{r\to+\infty}\|(\gl I + tR(t/r))^{-1} - P_0(\gl I_0 + tH)^{-1}P_0\| = 0.
\ee
%
%
for $t > 0$. Setting $t = 1$ and replacing $1/r$ by $t$ we obtain $(ii)$.

$(ii) \Rightarrow (iii)$. Since
%
%
\be\la{2.22}
R(t)  =  \frac{1}{t}\{I - f(tA) + f(tA)^{1/2}[I - g(tB)]f(tA)^{1/2}\},
\ee
%
%
we find
%
%
\be\la{2.23}
R(t) = \frac{1}{t}(I - Q(t)) + f(tA)^{1/2}M(t)f(tA)^{1/2} ,
\ee
%
%
which yields
%
%
\bea\la{2.24}
\lefteqn{\hspace{-1.0cm}(I + R(t))^{-1} =}\\
& & \frac{t}{1 + t}(I - Q(t)) + Q(t)\tilde{f}(tA)^{-1/2}\{\tilde{f}(tA)^{-1} + M(t)\}^{-1}\tilde{f}(tA)^{-1/2}Q(t). \nonumber
\eea
%
%
Hence
%
%
\be\la{2.25}
f(tA)^{1/2}(I + R(t))^{-1}f(tA)^{1/2} = Q(t)\{\tilde{f}(tA)^{-1} + M(t)\}^{-1}Q(t).
\ee
%
%
Since
%
%
\be\la{2.26}
\tilde{f}(tA)^{-1} + M(t) = (1 + t)\{M(t) + \frac{1}{1 + t}Q(t)g(tB)Q(t)\} \ge I(t) ,
\ee
%
%
we finally get
%
%
\bea\la{2.27}
\lefteqn{\hspace{-1.5cm} f(tA)^{1/2}(I + R(t))^{-1}f(tA)^{1/2} = }\\
& & \frac{1}{1 + t}Q(t)\{M(t) + \frac{1}{1 + t}Q(t)g(tB)Q(t)\}^{-1}Q(t).\nonumber
\eea
%
%
Now we prove that
%
%
\be\la{2.28}
\lim_{t\to+0}\|f(tA)^{1/2}(I + R(t))^{-1}f(tA)^{1/2} - P_0(I_0 + H)^{-1}P_0\| = 0.
\ee
%
%
To this end we use the representation
%
%
\bea\la{2.29}
\lefteqn{\hspace{-0.5cm} f(tA)^{1/2}(I + R(t))^{-1}f(tA)^{1/2} - P_0(I_0 + H)^{-1}P_0 =}\\
& & f(tA)^{1/2}\{(I + R(t))^{-1} - P_0(I_0 + H)^{-1}P_0\}f(tA)^{1/2} + \nonumber\\
& & [f(tA)^{1/2} - I]P_0(I_0 + H)^{-1}P_0f(tA)^{1/2} + P_0(I_0 + H)^{-1}P_0[f(tA)^{1/2} - I] ,\nonumber
\eea
%
%
which gives the estimate
%
%
\bea\la{2.30}
\lefteqn{\hspace{-0.5cm} \|f(tA)^{1/2}(I + R(t))^{-1}f(tA)^{1/2} - P_0(I_0 + H)^{-1}P_0\| \le}\\
& & \|(I + R(t))^{-1} - P_0(I_0 + H)^{-1}P_0\| + 2\|[I - f(tA)^{1/2}]P_0(I_0 + H)^{-1}P_0\|.\nonumber
\eea
%
%
Since by (\re{1.2}) $\dom(H^{1/2}) \subseteq \dom(A^{1/2})$, we have 
the representation
%
%
\bea\la{2.31}
\lefteqn{[I - f(tA)^{1/2}](I_0 + H)^{-1}h = }\\
& & t^{1/2}[I - f(tA)^{1/2}](tA)^{-1/2}A^{1/2}(I_0 + H)^{-1/2}(I_0 + H)^{1/2}(I_0 + H)^{-1}h,\qquad h \in \kH_0, \nonumber
\eea
%
%
which yields the estimate
%
%
\be\la{2.32}
\|[I - f(tA)^{1/2}](I_0 + H)^{-1}h\| \le t^{1/2}C_{1/2}\|h\|, \qquad h \in \kH,
\ee
%
%
where we have used that the Kato-function $f(\cdot)$ satisfies (cf.(1.4)):
%
%
\be\la{2.33}
C_{1/2} := \sup_{x > 0}\frac{1 - f(x)^{1/2}}{x^{1/2}} < +\infty.
\ee
%
%
Hence
%
%
\be\la{2.33-1}
\|[I - f(tA)^{1/2}](I_0 + H)^{-1}\| \le C_{1/2}t^{1/2}.
\ee
%
%
Then by $(ii)$, ({\re{2.30}) and (\re{2.33-1}) we obtain (\re{2.28}). 
Taking into account (\re{2.27}) and (\re{2.28}) one immediately finds that
%
%
\be\la{2.34}
\lim_{t\to+0}\left\|Q(t)\left\{M(t) + \frac{1}{1+t}Q(t)g(tB)Q(t)\right\}^{-1}Q(t) - P_0(I_0 + H)^{-1}P_0\right\| = 0.
\ee
%
%
To show that 
%
%
\be\la{2.35}
\lim_{t\to+0}\left\|\left\{M(t) + \frac{1}{1+t}Q(t)g(tB)Q(t)\right\}^{-1} - \{M(t) + I(t)\}^{-1}\right\| = 0
\ee
%
%
we use the identity
%
%
\bea\la{2.36}
\lefteqn{\hspace{-0.5cm} \{M(t) + \frac{1}{1+t}Q(t)g(tB)Q(t)\}^{-1} - \{M(t) + I(t)\}^{-1} =}\\
& & \{M(t) + I(t)\}^{-1}\left\{I(t) - \frac{1}{1+t}Q(t)g(tB)Q(t)\right\}\left\{M(t) + \frac{1}{1+t}Q(t)g(tB)Q(t)\right\}^{-1} \nonumber
\eea
%
%
to find the representation
%
%
\bea\la{2.37}
\lefteqn{\hspace{-0.8cm} \{M(t) + \frac{1}{1+t}Q(t)g(tB)Q(t)\}^{-1} - \{M(t) + I(t)\}^{-1} =}\\
& & \frac{t}{1+t}\{M(t) + I(t)\}^{-1}\left\{M(t) + \frac{1}{1+t}Q(t)g(tB)Q(t)\right\}^{-1} + \nonumber\\
& & \frac{1}{1+t}\{M(t) + I(t)\}^{-1}Q(t)[I - g(tB)]Q(t)\left\{M(t) + \frac{1}{1+t}Q(t)g(tB)Q(t)\right\}^{-1}.\nonumber
\eea
%
%
Therefore, the limit (\re{2.35}) takes place if 
%
%
\be\la{2.38}
\lim_{t\to+0}\left\|[I - g(tB)]Q(t)\left\{M(t) + \frac{1}{1+t}Q(t)g(tB)Q(t)\right\}^{-1}\right\| = 0.
\ee
%
% 
Since by (\re{1.2}) $\dom(H^{1/2}) \subseteq \dom(B^{1/2})$, one 
gets the representation
%
%
\bea\la{2.39}
\lefteqn{\hspace{-0.5cm} [I - g(tB)]Q(t)(M(t) + \frac{1}{1+t}Q(t)g(tB)Q(t))^{-1} = }\\
& & [I - g(tB)]\left\{Q(t)\left(M(t) + \frac{1}{1+t}Q(t)g(tB)Q(t)\right)^{-1}Q(t) - P_0(I_0 + H)^{-1}P_0\right\} + \nonumber\\
& & t^{1/2}[I + g(tB)^{1/2}][I - g(tB)^{1/2}](tB)^{-1/2}B^{1/2}(I_0 + H)^{-1/2}(I_0 + H)^{-1/2}P_0 \nonumber
\eea
%
%
and the estimate
%
%
\be\la{2.39-1}
\|[I - g(tB)^{1/2}]P_0(I_0 + H)^{-1}P_0\| \le S_{1/2}t^{1/2},
\ee
%
%
because for the Kato-function $g(\cdot)$  we have (cf.(1.5)):
%
%
\be\la{2.41}
S_{1/2} := \sup_{x > 0}\frac{1 - g(x)^{1/2}}{x^{1/2}} < +\infty.
\ee
%
%
Therefore, from (\re{2.39}) and (\re{2.39-1}) we obtain the estimate
%
%
\bea\la{2.40}
\lefteqn{\hspace{-0.5cm} \|[I - g(tB)]Q(t)\{M(t) + \frac{1}{1+t}Q(t)g(tB)Q(t)\}^{-1}\| \le }\\
& & \left\|Q(t)\left(M(t) + \frac{1}{1+t}Q(t)g(tB)Q(t)\right)^{-1}Q(t) - P_0(I_0 + H)^{-1}P_0\right\| + 2S_{1/2}t^{1/2}. \nonumber
\eea
%
%
Applying (\re{2.34}) to (\re{2.40}) we obtain (\re{2.38}) which yields 
(\re{2.35}). Then  from (\re{2.34}) and (\re{2.35}) we obtain $(iii)$ for
$\gl = 1$ and obviously for any $\gl > 0$.

$(iii) \Rightarrow (iv)$. The identity
%
%
\bea\la{2.42}
\lefteqn{\hspace{-0.5cm} g(tB)^{1/2}Q(t)(\gl I(t) + M(t))^{-1}Q(t)g(tB)^{1/2} = (\gl I + S(t))^{-1}F(t) + }\\
& & \gl (\gl I + S(t))^{-1}g(tB)^{1/2}Q(t)[I - f(tA)]Q(t)(\gl I(t) + M(t))^{-1}Q(t)g(tB)^{1/2} \nonumber 
\eea
%
%
yields the representation 
%
%
\bea\la{2.43}
\lefteqn{\hspace{-0.5cm} (\gl I + S(t))^{-1} = tS(t)(\gl I + S(t))^{-1} +}\\
& & g(tB)^{1/2}Q(t)(\gl I(t) + M(t))^{-1}Q(t)g(tB)^{1/2} + \nonumber\\
& & \gl (\gl I  + S(t))^{-1}g(tB)^{1/2}[f(tA) - I]Q(t)(\gl I(t) + M(t))^{-1}Q(t)g(tB)^{1/2}. \nonumber
\eea
%
%
Hence
%
%
\bea\la{2.44}
\lefteqn{(\gl I + S(t))^{-1} - P_0(\gl I_0 + H)^{-1}P_0 = tS(t)(\gl I + S(t))^{-1} + } \\
& & g(tB)^{1/2}\left\{Q(t)(\gl I(t) + M(t))^{-1}Q(t) - P_0(\gl I_0 + H)^{-1}P_0\right\}g(tB)^{1/2} + \nonumber\\
& & [g(tB)^{1/2} - I]P_0(\gl I_0 + H)^{-1}P_0g(tB)^{1/2} + P_0(\gl I_0 + H)^{-1}P_0[g(tB)^{1/2} - I] + \nonumber\\
& & \gl(\gl I + S(t))^{-1}g(tB)^{1/2}[f(tA) - I]\times \nonumber \\
& & \left\{Q(t)(\gl I(t) + M(t))^{-1}Q(t) - P_0(\gl I_0 + H)^{-1}P_0\right\}g(tB)^{1/2} + \nonumber\\
& & \gl(\gl I + S(t))^{-1}g(tB)^{1/2}[f(tA) - I]P_0(\gl I_0 + H)^{-1}P_0g(tB)^{1/2}. \nonumber 
\eea
%
%
Therefore, we get the estimate
%
%
\bea\la{2.45}
\lefteqn{\hspace{-0.8cm} \|(\gl I + S(t))^{-1} - P_0(\gl I_0 + H)^{-1}P_0\| \le }\\
& & \frac{t}{\gl} + 2\left\|Q(t)(\gl I(t) + M(t))^{-1}Q(t) - P_0(\gl I_0 + H)^{-1}P_0\right\| + \nonumber \\
& & 2\|[I - g(tB)^{1/2}]P_0(\gl I_0 + H)^{-1}P_0\| + \|[I - f(tA)]P_0(\gl I_0 + H)^{-1}P_0\|. \nonumber
\eea
%
%
Taking into account $(iii)$, (\re{2.33-1}) and (\re{2.39-1}) we obtain
$(iv)$.

$(iv) \Rightarrow (v)$. By $\kA_r$ we denote the subalgebra of $\kA$ 
generated by the function $(\gm + x)^{-1}$, $x \in \bR_+$, for a fixed 
$\gm > 0$. Again by the Stone-Weierstrass theorem the algebra $\kA_r$ is dense 
in $\kA$ in the supremum norm. Hence for any $\epsilon_0 > 0$ there is a 
polynomial $\kP(\cdot)$ such that
%
%
\be\la{2.46}
\sup_{x \ge 0}|e^{-x} - \kP((\gm + x)^{-1})| \le \epsilon_0.
\ee
%
%
Hence we find
%
%
\be\la{2.47}
\sup_{t \in [a,b]}\sup_{x \ge 0}|e^{-tx} - \kP((\gm + tx)^{-1}| \le \epsilon_0
\ee
%
%
for each interval $[a,b] \subseteq (0,\infty)$. Therefore,
%
%
\be\la{2.48}
\sup_{t \in [a,b]}\|e^{-tS(t/r)} - \kP((\gm I + tS(t/r))^{-1}\| \le \epsilon_0, \quad r \ge 1,
\ee
%
%
and
%
%
\be\la{2.49}
\sup_{t \in [a,b]}\|e^{-tH} - \kP((\gm I_0 + tH)^{-1}\| \le \epsilon_0
\ee
%
%
for the interval $[a,b] \subseteq (0,\infty)$. Since the representation
%
%
\bea\la{2.50}
\lefteqn{\hspace{-0.5cm} e^{-tS(t/r)} - P_0e^{-tH}P_0 = e^{-tS(t/r)} - \kP((\gm I + tS(t/r))^{-1}) + }\\
& & \kP((\gm I + tS(t/r))^{-1}) - P_0\kP((\gm I_0 + tH)^{-1})P_0 + P_0\{\kP((\gm I_0 + tH)^{-1}) - e^{-tH}\}P_0 \nonumber
\eea
%
%
yields the estimate
%
%
\bea\la{2.51}
\lefteqn{\hspace{-0.2cm} \|e^{-tS(t/r)} - P_0e^{-tH}P_0\| \le \|e^{-tS(t/r)} - \kP((\gm I + tS(t/r))^{-1})\| + }\\
& & \|\kP((\gm I + tS(t/r))^{-1}) - P_0\kP((\gm I_0 + tH)^{-1})P_0\| + \|P_0\{\kP((\gm I_0 + tH)^{-1}) - e^{-tH}\}P_0\|, \nonumber
\eea
%
%
by (\re{2.48}) and (\re{2.49}) we find
%
%
\bea\la{2.52}
\lefteqn{\hspace{-2.0cm}\|e^{-tS(t/r)} - P_0e^{-tH}P_0\| \le }\\
& & 2\epsilon_0 + \sup_{t \in [a,b]}\|\kP((\gm I + tS(t/r))^{-1}) - P_0\kP((\gm I_0 + tH)^{-1})P_0\|. \nonumber
\eea
%
%
Since $\|(\gm I + tS(t/r))^{-1}\| \le \frac{1}{\gm}$ and 
$\|(\gm I_0 + tH)^{-1}\| \le \frac{1}{\gm}$, $t \ge 0$, there is a constant
$C_\kP > 0$ , which depends only on $\kP(\cdot)$ and $\gm$, such that
%
%
\bea\la{2.53}
\lefteqn{\hspace{-0.5cm} \sup_{t \in [a,b]}\|e^{-tS(t/r)} - P_0e^{-tH}P_0\| \le }\\
& & 2\epsilon_0 + C_\kP\sup_{t \in [a,b]}\|(\gm I + tS(t/r))^{-1}) - P_0(\gm I_0 + tH)^{-1}P_0\|\nonumber
\eea
%
%
for each interval $[a,b] \subseteq (0,\infty)$. Using resolvent series
%
%
\be\la{2.55}
(\gm I + tS(t/r))^{-1} = \frac{1}{t}\sum^\infty_{k=0}(\gl - \frac{\gm}{t})^k(\gl I + S(t/r))^{-(k+1)}
\ee
%
%
and
%
%
\be\la{2.56}
(\gm I_0 + tH)^{-1} = \frac{1}{t}\sum^\infty_{k=0}(\gl - \frac{\gm}{t})^k(\gl I_0 + H)^{-(k+1)}
\ee
%
%
for $0 < \gm \le 2\gl a$,  one gets 
%
%
\bea\la{2.57}
\lefteqn{\hspace{-2.0cm} (\gm I + tS(t/r))^{-1} - P_0(\gm I_0 + tH)^{-1}P_0 =}\\
& & \frac{1}{t}\sum^\infty_{k=0}\left(\gl - \frac{\gm}{t}\right)^k\left\{(\gl I + S(t/r))^{-(k+1)} - P_0(\gl I_0 + H)^{-(k+1)}P_0\right\}.\nonumber
\eea
%
%
Therefore,
%
%
\bea\la{2.58}
\lefteqn{\hspace{-2.0cm}\|(\gm I + tS(t/r))^{-1} - P_0(\gm I_0 + tH)^{-1}P_0\| \le}\\
& & \frac{1}{t}\sum^\infty_{k=0}(k+1)\left|1 - \frac{\gm}{t\gl}\right|^k\|(\gl I + S(t/r))^{-1} - P_0(\gl I_0 + H)^{-1}P_0\| .\nonumber
\eea
%
%
If $0 < \gm < 2\gl a$, then
%
%
\be\la{2.59}
\eta := \sup_{t \in [a,b]}|1 - \frac{\gm}{t\gl}| < 1.
\ee
%
%
Since $\sum^\infty_{k=0}(k+1)\eta^k < +\infty$, one gets from (\re{2.58}) the 
estimate
%
%
\bea\la{2.60}
\lefteqn{\hspace{-1.5cm}\|(\gm I + tS(t/r))^{-1} - P_0(\gm I_0 + tH)^{-1}P_0\| \le}\\
& & \frac{1}{a}\|(\gl I + S(t/r))^{-1} - P_0(\gl I_0 + H)^{-1}P_0\|\sum^\infty_{k=0}(k+1)\eta^k. \nonumber
\eea
%
%
By $(iv)$ for any $\epsilon_1 > 0$ there exists a $\gd > 0$ such that for 
$0 < t/r \le \gd$ one has
%
%
\be\la{2.61}
\|(\gl I + S(t/r))^{-1} - P_0(\gl I_0 + H)^{-1}P_0\| \le \epsilon_1.
\ee
%
%
Therefore,
%
%
\be\la{2.62}
\sup_{t \in [a,b]} \|(\gl I + S(t/r))^{-1} - P_0(\gl I_0 + H)^{-1}P_0\| \le \epsilon_1
\ee
%
%
for $b/\gd \le r$. Hence, by (\re{2.60}) we get
%
%
\be\la{2.63}
\|(\gm I + tS(t/r))^{-1} - P_0(\gm I_0 + tH)^{-1}P_0\| \le \frac{\epsilon_1}{a}\sum^\infty_{k=0}(k+1)\eta^k
\ee
%
%
for sufficiently large $r$. The estimates (\re{2.53}) and (\re{2.63}) give
%
%
\be\la{2.64}
\sup_{t \in [a,b]}\|e^{-tS(t/r)} - P_0e^{-tH}P_0\| \le 2\epsilon_0 + \epsilon_1\frac{C_\kP}{a}\sum^\infty_{k=0}(k+1)\eta^k
\ee
%
%
for sufficiently large $r$. 

To prove $(v)$ one has only to remark that the 
estimate similar to (\re{2.7}) is valid for $F(t/r)^{r}$ 
and $\exp(-tS(t/r))$.

$(v) \Rightarrow (i)$. We start (see (\re{2.0-1}), (\re{2.0-3})) with the 
identity
%
%
\be\la{2.65}
F(t)g(tB)^{1/2}f(tA)^{1/2} = g(tB)^{1/2}f(tA)^{1/2}T(t)
\ee
%
%
for $t \ge 0$. Hence
%
%
\be\la{2.66}
F(t)^rg(tB)^{1/2}f(tA)^{1/2} = g(tB)^{1/2}f(tA)^{1/2}T(t)^r
\ee
%
%
for $t \ge 0$ and $r \ge 1$, or
%
%
\be\la{2.67}
f(tA/r)^{1/2}g(tB/r)^{1/2}F(t/r)^rg(tB/r)^{1/2}f(tA/r)^{1/2} = T(t)^{r+1}.
\ee
%
%
Then one gets 
%
%
\bea\la{2.68}
\lefteqn{\hspace{-2.0cm} T(t/r)^r = T(t/r)^r(I - T(t/r)) + }\\
& & f(tA/r)^{1/2}g(tB/r)^{1/2}F(t/r)^rg(tB/r)^{1/2}f(tA/r)^{1/2}, \nonumber
\eea
%
%
which leads to the representation
%
%
\bea\la{2.69}
\lefteqn{\hspace{-1.5cm} T(t/r)^r - P_0e^{-tH}P_0 = T(t/r)^r(I - T(t/r)) + }\\
& & f(tA/r)^{1/2}g(tB/r)^{1/2}\{F(t/r)^r - P_0e^{-tH}P_0\}g(tB/r)^{1/2}f(tA/r)^{1/2} + \nonumber\\
& & f(tA/r)^{1/2}[g(tB/r)^{1/2} - I]P_0e^{-tH}P_0g(tB/r)^{1/2}f(tA/r)^{1/2} + \nonumber\\
& & [f(tA/r)^{1/2} - I]P_0e^{-tH}P_0g(tB/r)^{1/2}f(tA/r)^{1/2} +\nonumber \\
& & P_0e^{-tH}P_0[g(tB/r)^{1/2} - I]f(tA/r)^{1/2} + P_0e^{-tH}P_0[f(tA/r)^{1/2} - I].\nonumber
\eea
%
%
This representation readily implies the estimate
%
%
\bea\la{2.70}
\lefteqn{\|T(t/r)^r - P_0e^{-tH}P_0\| \le \|T(t/r)^r(I - T(t/r))\| + }\\ 
& & \|F(t/r)^r - P_0e^{-tH}P_0\| + 2\|[I - g(tB/r)^{1/2}]P_0e^{-tH}P_0\| + 2\|[I - f(tA/r)^{1/2}]P_0e^{-tH}P_0\|.\nonumber
\eea
%
%
We note that by Lemma 2.1 of \cite{NZ2} one has
%
%
\be\la{2.71}
\|T(t/r)^{r}(I - T(t/r))\| \le \frac{1}{1 + r}, \qquad r \ge 1.
\ee
%
%
Using the representation
%
%
\bea\la{2.72}
\lefteqn{\hspace{-0.5cm} [I - f(tA/r)^{1/2}]e^{-tH}h = }\\
& & r^{-1/2}[I - f(tA/r)^{1/2}](tA/r)^{-1/2}(tA)^{1/2}(I_0 + tH)^{-1/2}(I_0 + tH)^{1/2}e^{-tH}h \nonumber
\eea
%
%
for $h \in \kH_0$, by (\re{2.33}) one gets (see Lemma 4.2 of \cite{NZ3}) that
%
%
\be\la{2.73}
\|[I - f(tA/r)^{1/2}]e^{-tH}h\| \le r^{-1/2}C_{1/2}\|(I_0 + tH)^{1/2}e^{-tH}\|\;\|h\| \le r^{-1/2}C_{1/2}\|h\| e
\ee
%
%
for $h \in \kH_0$. Hence
%
%
\be\la{2.74}
\|[I - f(tA/r)^{1/2}]P_0e^{-tH}P_0\| \le r^{-1/2}C_{1/2} e
\ee
%
%
for $t \ge 0$ and $r \ge 1$. Similarly (cf.(2.45)), we prove that
%
%
\be\la{2.75}
\|[I - g(tB/r)^{1/2}]P_0e^{-tH}P_0\| \le r^{-1/2}S_{1/2} e
\ee
%
%
for $t \ge 0$ and $r \ge 1$. Therefore, (\re{2.70}), (\re{2.71}) and (\re{2.74}), (\re{2.75}) imply
%
%
\bea\la{2.76}
\lefteqn{\hspace{-1.5cm} \sup_{t \in [a,b]}\|T(t/r)^r - P_0e^{-tH}P_0\| \le }\\
& & \sup_{t \in [a,b]}\|F(t/r)^r - P_0e^{-tH}P_0\| + \frac{1}{1 + r} + \frac{2e}{r^{1/2}}(C_{1/2} + S_{1/2}). \nonumber
\eea
%
% 
Now by $(v)$ we obtain $(i)$. \hfill $\Box$
%
%
\begin{cor}\la{II.2}
%
%
Under assumptions of Theorem \re{II.1} the conditions $(i) - (v)$ are
equivalent to 
%
%
\begin{enumerate}

\item[(vi)] $\lim_{n\to\infty}\sup_{t \in [a,b]}\|(f(tA/n)g(tB/n))^n - P_0e^{-tH}P_0\| = 0$, \quad $[a,b] \subseteq (0,\infty)$,

\item[(vii)] $\lim_{n\to\infty}\sup_{t \in [a,b]}\|(g(tB/n)f(tA/n))^n - P_0e^{-tH}P_0\| = 0$, \quad $[a,b] \subseteq (0,\infty)$,

\end{enumerate}
%
%
\end{cor}
%
%
{\em Proof:} By substitutions $A \leftrightarrow B$ and 
$f(\cdot) \leftrightarrow g(\cdot)$ it is enough to prove only 
$(i) \Leftrightarrow (vi)$. 

$(i) \Rightarrow (vi)$. Since
%
%
\be\la{2.77} 
(f(t/n)g(tB/n))^n = f(tA/n)^{1/2}T(t/n)^{n-1}f(tA/n)^{1/2}g(tB/n), \quad n = 1,2\ldots,
\ee
%
%
we get the representation
%
%
\bea\la{2.78}
\lefteqn{\hspace{-0.5cm} (f(tA/n)g(tB/n))^n - P_0e^{-tH}P_0 =}\\
& & f(tA/n)^{1/2}T(t/n)^{n-1}(I - T(t/n))f(tA/n)^{1/2}g(tB/n) + \nonumber\\ 
& & f(tA/n)^{1/2}(T(t/n)^n - P_0e^{-tH}P_0)f(tA/n)^{1/2}g(tB/n) + \nonumber\\ 
& & [f(tA/n)^{1/2} - I]P_0e^{-tH}P_0f(tA/n)^{1/2}g(tB/n) + \nonumber\\
& & P_0e^{-tH}P_0[f(tA/n)^{1/2} - I]g(tB/n) + P_0e^{-tH}P_0[g(tB/n)^{1/2} - I][g(tB/n)^{1/2} + I], \nonumber
\eea
%
%
which gives the estimate
%
%
\bea\la{2.79}
\lefteqn{\hspace{-1.5cm} \|(f(tA/n)g(tB/n))^n - P_0e^{-tH}P_0\| \le }\\
& & \|T(t/n)^{n-1}(I - T(t/n))\| + \|T(t/n)^n - P_0e^{-tH}P_0\| + \nonumber\\ 
& & 2\|[I - f(tA/n)^{1/2}]P_0e^{-tH}P_0\| + 
2\|[I - g(tB/n)^{1/2}] P_0e^{-tH}P_0\|.\nonumber
\eea
%
%
Using (\re{2.71}), (\re{2.74}) and (\re{2.75}) we get
%
%
\bea\la{2.80}
\lefteqn{\hspace{-1.5cm} \|(f(tA/n)g(tB/n))^n - P_0e^{-tH}P_0\| \le }\\
& & \|T(t/n)^n - P_0e^{-tH}P_0\| + \frac{1}{n}
+ \frac{2e}{n^{1/2}}(C_{1/2} +  S_{1/2}) \nonumber
\eea
%
%
which yields
%
%
\bea\la{2.82}
\lefteqn{\hspace{-1.5cm} \sup_{t \in [a,b]}\|(f(tA/n)g(tB/n))^n - P_0e^{-tH}P_0\| \le }\\
& & \sup_{t \in [a,b]} \|T(t/n)^n - P_0e^{-tH}P_0\| + 
\frac{1}{n} +\frac{2e}{n^{1/2}}(C_{1/2} + S_{1/2}). \nonumber
\eea
%
%
Therefore, $(i)$ implies $(vi)$.

$(vi) \Rightarrow (i)$. From
%
%
\be\la{2.83}
T(t/n)^{n+1} = f(tA/n)^{1/2}g(tB/n)(f(tA/n)g(tB/n))^nf(tA/n)^{1/2}, \quad n =1,2,\ldots,
\ee
%
%
we obtain the representation
%
%
\bea\la{2.84}
\lefteqn{\hspace{-1.5cm} T(t/n)^n - P_0e^{-tH}P_0 = T(t/n)^n(I - T(t/n)) + }\\
& & f(tA/n)^{1/2}g(tB/n)\left((f(tA/n)g(tB/n))^n - P_0e^{-tH}P_0\right)f(tA/n)^{1/2} +\nonumber\\
& & f(tA/n)^{1/2}[g(tB/n)^{1/2} + I][g(tB/n)^{1/2} - I]P_0e^{-tH}P_0f(tA/n)^{1/2} + \nonumber\\
& & [f(tA)^{1/2} - I]P_0e^{-tH}P_0f(tA/n)^{1/2} +
P_0e^{-tH}P_0[f(tA/n)^{1/2} - I].\nonumber
\eea
%
%
Hence
%
%
\bea\la{2.85}
\lefteqn{\hspace{-1.5cm} \|T(t/n)^n - P_0e^{-tH}P_0\| \le}\\
& & \|T(t/n)^n(I - T(t/n))\| + \|(f(tA/n)g(tB/n))^n - P_0e^{-tH}P_0\| +\nonumber\\
& & 2\|[I - g(tB/n)^{1/2}]P_0e^{-tH}P_0\| + 2\|[I -
f(tA)^{1/2}]P_0e^{-tH}P_0\|. \nonumber
\eea
%
% 
>From (\re{2.71}), (\re{2.74}) and (\re{2.75})  we obtain
%
%
\bea\la{2.86}
\lefteqn{\hspace{-1.0cm} \sup_{t \in [a/2,b]}\|T(t/n)^n - P_0e^{-tH}P_0\| \le}\\
& & \sup_{t \in [a/2,b]}\|(f(tA/n)g(tB/n))^n - P_0e^{-tH}P_0\| + \frac{1}{n+1} + \frac{2e}{n^{1/2}}(C_{1/2} + S_{1/2}) \nonumber
\eea
%
%
for {\em any} interval $[a/2,b] \subseteq (0,\infty)$. Hence, by 
$(vi)$ for each $\epsilon > 0$ and interval $[a/2,b] \subseteq (0,\infty)$ 
there is a $n_0 \ge 1$ such that
%
%
\be\la{2.87}
\sup_{t \in [a/2,b]}\|T(t/n)^n - P_0e^{-tH}P_0\| \le \epsilon
\ee
%
%
for $n \ge n_0 \ge 1$. To extend (\re{2.87}) to $(i)$ for $r \in \bR_+$ we proceed
as follows. Let $[r] := \sup\{n \in \bN: n \le r\}$, $\bN := \{1,2,\ldots\}$. 
If $t \in [a,b] \subseteq (0,\infty)$, then 
$[r]t/r \in [a/2,b]$, $r \ge 1$. Hence, by (\re{2.87}) one gets
%
%
\be\la{2.88}
\sup_{t \in [a,b]}\|T([r]t/r[r])^{[r]} - P_0e^{-[r]tH/r}P_0\| \le \epsilon
\ee
%
%
for $n \ge n_0 \ge 1$ and $t \in [a,b]$, which yields
%
%
\be\la{2.89}
\sup_{t \in [a,b]}\|T(t/r)^{[r]} - P_0e^{-[r]tH/r}P_0\| \le \epsilon
\ee
%
%
for $n \ge n_0 \ge 1$. Since
%
%
\be\la{2.90}
e^{-[r]tH/r}h - e^{-tH}h = \int^t_{[r]t/r}He^{-sH}h ds, \quad h \in \dom(H),
\ee
%
%
by Lemma 4.2 of \cite{NZ3} we get the estimate
%
%
\be\la{2.91}
\|e^{-[r]tH/r} - e^{-tH}\| \le \int^t_{[r]t/r}\frac{1}{s} ds = \ln(r/[r]), \qquad t > 0,
\ee
%
%
which implies
%
%
\be\la{2.92}
\lim_{r\to+\infty}\sup_{t \in [a,b]}\|e^{-[r]tH/r} - e^{-tH}\| = 0.
\ee
%
%
Let $X$ be a non-negative self-adjoint operator obeying $0 \le X \le I$. By
its spectral representation one gets
%
%
\be\la{2.93}
\|X^n(I - X^{\ga})\| \le \frac{1}{n}, \qquad n = 1,2,\ldots, \quad 0 \le \ga < 1.
\ee
%
%
Using that
%
%
\be\la{2.94}
T(t/r)^{[r]} - T(t/r)^r = T(t/r)^{[r]}(I - T(t/r)^{r - [r]})
\ee
%
% 
and applying (\re{2.93}) we get the estimate
%
%
\be\la{2.95}
\|T(t/r)^{[r]} - T(t/r)^r\| \le \frac{1}{[r]}, \qquad r \ge 1, \quad t \ge 0.
\ee
%
%
Since
%
%
\be\la{2.96}
T(t/r)^r - P_0e^{-tH}P_0 = T(t/r)^r - T(t/r)^{[r]} + T(t/r)^{[r]} - P_0e^{-tH}P_0, 
\ee
%
%
due to (\re{2.95}) we have the estimate
%
%
\be\la{2.97}
\sup_{t \in [a,b]}\|T(t/r)^r - P_0e^{-tH}P_0\| \le \sup_{t \in [a,b]}\|T(t/r)^{[r]} - P_0e^{-tH}P_0\| + \frac{1}{[r]}.
\ee
%
% 
Taking into account (\re{2.89}) and (\re{2.92}) we prove $(i)$. \hfill $\Box$

%\newpage

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

In this section we prove the operator-norm convergence of the Trotter-Kato
product formula under resolvent compactness assumptions. Below we say that the 
Trotter-Kato product formula {\em converges locally uniformly away from zero 
in the operator-norm for the family} $\{F(t)\}_{t \ge 0}$  if the limit 
$(i)$ 
 
(see Theorem 2.1) exists for any interval $[a,b] \subseteq (0,\infty)$. By Theorem \re{II.1} and 
Corollary \re{II.2} this is equivalent to the convergence in the 
for other families generated 
by $f(\cdot)$ and $g(\cdot)$, see $(v)$ and $(vi) - (vii)$. For short we simply
say that the Trotter-Kato product formula {\em converges locally 
uniformly away from zero in the operator-norm} if one of the limits 
$(i) - (vii)$ is established.

%
%
\begin{lem} \la{III.1}
%
%
(cf. \cite{Za1}) Let $\{Y(t)\}_{t>0}$ be a sequence of non-negative 
self-adjoint operators on $\kH$ and let $X$ be a non-negative compact operator 
on $\kH$ such that 
%
%
\be\la{3.9}
0 \le Y(t) \le X, \qquad t > 0.
\ee
%
%
If
%
%
\be\la{3.10}
w-\lim_{t\to+0} Y(t) = Y,
\ee
%
%
then $Y \in \kL_\infty(\kH)$ and, moreover,
%
%
\be\la{3.11}
\lim_{t\to+0}\|Y(t) - Y\| = 0.
\ee
%
%
\end{lem}
%
%
{\em Proof:} By (\re{3.9}) there is a sequence of contractions 
$\{\gG(t)\}_{t>0}$, $\gG(t): \overline{\ran(\sqrt{X})} \longrightarrow \kH$,
such that
%
%
\be\la{3.12}
\sqrt{Y(t)} = \gG(t)\sqrt{X}, \qquad t > 0.
\ee
%
%
We set
%
%
\be\la{3.13}
Z(t) := \gG(t)^*\gG(t) : \overline{\ran(\sqrt{X})} \longrightarrow \overline{\ran(\sqrt{X})}.
\ee
%
%
By (\re{3.9}) and (\re{3.10}) one gets $0 \le Y \le X$ which yields the 
existence of a contraction 
$\gG: \overline{\ran(\sqrt{X})} \longrightarrow \kH$ such that
%
%
\be\la{3.14}
\sqrt{Y} = \gG\sqrt{X}.
\ee
%
%
Hence $Y \in \kL_\infty(\kH)$. Let 
$Z := \gG^*\gG: \overline{\ran(\sqrt{X})} \longrightarrow \overline{\ran(\sqrt{X})}$. 
Then 
%
%
\be\la{3.15}
Y(t) - Y = \sqrt{X}(Z(t) - Z)\sqrt{X}, \qquad t > 0,
\ee
%
%
which yields
%
%
\be\la{3.16}
w-\lim_{t\to+0}(Z(t) - Z) = 0.
\ee
%
%
Since $\sqrt{X}$ is compact, by (3.7) and (3.8) one immediately gets (\re{3.11}). \hfill$\Box$

The next theorem can be regarded as a generalization of Theorem 3.4 of
\cite{NZ1} to compact operators, i.e., to $p = \infty$.
%
%
\begin{theo}\la{III.2}
%
%
Let $A$ and $B$ be non-negative self-adjoint operators on $\kH$ and let 
$f(\cdot)$ and $g(\cdot)$ be Kato-functions. If condition 
$f_0(t_0A) \in \kL_\infty(\kH)$ is satisfied for some $t_0 > 0$, then the 
Trotter-Kato formula converges locally uniformly away from zero in the 
operator-norm. Moreover, one has
%
%
\be\la{3.17}
(\gl I_0 + H)^{-1} \in \kL_\infty(\kH)
\ee
%
%
for $\gl > 0$.
%
%
\end{theo}
%
%
{\em Proof:} By \cite{Ch1}, \cite{Ch2} and \cite{Ka1} the above conditions
on $A$, $B$ and  $f(\cdot)$, $g(\cdot)$ (see (\re{1.1}), (\re{1.2}) (\re{1.3}),
(\re{1.4})) imply that
%
%
\be\la{3.18}
s-\lim_{t\to+0}(\gl I + S(t))^{-1} = P_0(\gl I_0 + H)^{-1}P_0
\ee
%
%
for $\gl > 0$. From the identity (\re{2.42}) we obtain 
%
%
\bea\la{3.18-1}
\lefteqn{\hspace{-0.5cm} g(tB)^{1/2}Q(t)(\gl I(t) + M(t))^{-1}Q(t)g(tB)^{1/2} = (\gl + S(t))^{-1}F(t) +} \nonumber\\
& & \gl g(tB)^{1/2}Q(t)(\gl I(t) + M(t))^{-1}Q(t)[I - f(tA)]g(tB)^{1/2}(\gl I + S(t))^{-1} \nonumber
\eea 
%
%
for $t > 0$ and $\gl > 0$. Using $s-\lim_{t\to+0}g(tB)^{1/2} = 
s-\lim_{t\to+0}f(tA) = s-\lim_{t\to+0}F(t) = s-\lim_{t\to+0}Q(t) = I$ and 
(\re{3.18}) we find
%
%
\be\la{3.19}
w-\lim_{t\to+0} Q(t)(\gl I(t) + M(t))^{-1}Q(t) = P_0(\gl I_0 + H)^{-1}P_0
\ee
%
%
for each $\gl > 0$. By definitions (\re{1.7}) and (\re{1.8}) one has
%
%
\be\la{3.25}
\varphi_0(x) \ge \varphi_0(y), \qquad 0 < x \le y,
\ee
%
%
and
%
%
\be\la{3.26}
0 \le f(x) \le f_0(x) \le 1, \qquad x > 0.
\ee
%
%
Therefore, we get
%
%
\be\la{3.27}
f(tA) \le f_0(tA), \qquad t \ge 0,
\ee
%
%
which yields (cf. (\re{2.1}), (\re{2.2}))
%
%
\be\la{3.28}
\tilde{f}(tA) \le f_0(tA)Q(t), \qquad t > 0.
\ee
%
%
Since $f_0(x) > 0$ for $x > 0$, we find
%
%
\be\la{3.29}
f_0(tA)^{-1}Q(t) \le \tilde{f}(tA)^{-1}, \qquad t > 0,
\ee
%
%
which yields
%
%
\be\la{3.30}
\frac{1}{t}[f_0(tA)^{-1} - I]Q(t) \le \frac{1}{t}[\tilde{f}(t)^{-1} - I(t)],
\qquad t > 0.
\ee
%
%
By (\re{2.5}) we finally get
%
%
\be\la{3.31}
\frac{1}{t}[f_0(tA)^{-1} - I]Q(t) \le M(t), \qquad t > 0,
\ee
%
%
or, equivalently (cf. (\re{1.8})), 
%
%
\be\la{3.31-1}
A\varphi_0(tA)Q(t) \le M(t), \qquad t > 0.
\ee
%
%
Hence
%
%
\bea\la{3.32}
\lefteqn{\hspace{-1.5cm} Q(t)(\gl I(t) + M(t))^{-1}Q(t) \le }\\
& & (\gl I + A\varphi_0(tA))^{-1}Q(t) \le (\gl I + A\varphi_0(tA))^{-1}, \quad t > 0, \nonumber
\eea
%
%
where we used that $Q(t)$ commutes with $A$. Since $\varphi_0(x)$ is a 
non-increasing function (3.12), we have
%
%
\be\la{3.33}
\varphi_0(tA) \ge \varphi_0(t_0A), \qquad 0 < t \le t_0,
\ee
%
%
which yields
%
%
\be\la{3.34}
(\gl I + A\varphi_0(t_0A))^{-1} \ge (\gl I + A\varphi_0(tA))^{-1}, \qquad 0 < t \le t_0.
\ee
%
%
Hence
%
%
\be\la{3.35}
Q(t)(\gl I(t) + M(t))^{-1}Q(t) \le (\gl I + A\varphi_0(t_0A))^{-1}, \qquad 
0 < t \le t_0.
\ee
%
%
Since
%
%
\be\la{3.36}
(\gl I + A\varphi_0(t_0A))^{-1} = t_0(I + (\gl t_0 - 1)f_0(t_0A))^{-1}f_0(t_0A),
\ee
%
%
we obtain
%
%
\be\la{3.37}
(\gl I + A\varphi_0(t_0A))^{-1} \in \kL_\infty(\kH)
\ee
%
%
for $\gl t_0 > 1$. Taking into account (\re{3.35})  we obtain that assumption
(\re{3.9}) is satisfied. Therefore, (\re{3.19}) and Lemma \re{III.1} give the 
proof of the 
condition $(iii)$ of Theorem \re{II.1} as well as (\re{3.17}). \hfill $\Box$

Notice that $0 \le \varphi_0(x) \le 1$, $x > 0$, implies
%
%
\be\la{3.38}
1 + x\varphi_0(x) \le 1 + x, \qquad x > 0,
\ee
%
%
which together with (\re{1.8}) yields
%
%
\be\la{3.39}
\frac{1}{1 + x} \le f_0(x), \qquad x > 0.
\ee
%
%
Therefore, $f_0(t_0A) \in \kL_\infty(\kH)$  implies that 
%
%
\be\la{3.40}
(I + t_0A)^{-1} \in \kL_\infty(\kH).
\ee
%
%
The converse is certainly {\em not} true as it is demonstrated by the following 
example.
%
%
\begin{exam}\la{III.3}
%
%
{\em
Let $\{\gD_n\}^\infty_{n=0}$ be a partition of $\bR_+$ given by 
%
%
\be\la{3.41}
\gD_0 = [0,1), \qquad \gD_n := [n^n,(n+1)^{n+1}), \qquad n = 1,2,\ldots.
\ee
%
%
We define a Borel function $f(\cdot)$ setting
%
%
\be\la{3.42}
\varphi(x) := \left\{\begin{array}{r@{\qquad:\qquad}l} 1             & x \in \gD_0 \\
                                                       \frac{1}{n^n} & x \in \gD_n, \qquad n = 1,2,\ldots \end{array} \right.
\ee
%
%
and
%
%
\be\la{3.43}
f(x) := (1 + x\varphi(x))^{-1}, \qquad x \in \bR_+.
\ee
%
%
A straightforward computation shows (see (\re{1.7}), (\re{1.8})) that 
$\varphi_0(x) = \varphi(x)$ and $f_0(x) = f(x)$, $x \in \bR_+$. Let $A$ be a 
non-negative self-adjoint operator with simple discrete spectrum given by 
$\{\gl_n\}^\infty_{n=0}$, $\gl_n = n^n$. Since 
$\lim_{n\to\infty}\gl_n = +\infty$ the resolvent of $A$ is compact. Further, 
we have
%
%
\be\la{3.44}
f_0(t_0\gl_n) = f(t_0\gl_n) = \left(1 + n^nn^{-n}\right)^{-1} = \frac{1}{2}, \qquad n = 1,2,\ldots, \quad t_0 = 1,
\ee
%
%
which yields $\lim_{n\to\infty}f_0(\gl_n) = 1/2$. Therefore, the essential 
spectrum of $f_0(A)$ contains the point $\{1/2\}$ which is impossible if 
$f_0(A)$ is compact. By the way, if the spectrum of $A$ is given by 
$\gl_n := (n+1)^{n+1}-1$, then $\lim_{n\to\infty}f_0(\gl_n) = 0$. Hence 
$f_0(A) \in \kL_\infty(\kH)$.} \hfill $\Box$
%
%
\end{exam}
%
%
Let $f(\cdot)$ be a Borel function obeying (\re{1.3}). We set
%
%
\be\la{3.45}
b(x) := \sup_{0 \le s \le x} sf(s), \qquad x \in \bR_+,
\ee
%
%
and
%
%
\be\la{3.46}
r(x) := \sup_{s \in [x,\infty)} f(s), \qquad x \in \bR_+.
\ee
%
%
We note that $b(\cdot)$ is non-decreasing and $0 \le b(x) \le x$,
$x \in \bR_+$, while $r(x)$ is non-increasing such that $0 \le r(x) \le
1$, $x \in \bR_+$.
%
%
\begin{lem}\la{III.4}
%
%
Let $f(\cdot)$ be a Kato-function. For any non-negative operator $A$ obeying 
$(I + A)^{-1} \in \kL_\infty(\kH)$ one has $f_0(A) \in \kL_\infty(\kH)$
if and only if
%
%
\be\la{3.47}
\lim_{x\to+\infty} \frac{b(x)}{x} = 0
\ee
%
%
and
%
%
\be\la{3.48}
0 \le r(x) < 1
\ee
%
%
for $x > 0$.
%
%
\end{lem}
%
%
{\em Proof:} If $(I + A)^{-1} \in \kL_\infty(\kH)$, then the spectrum of
$A$ consists of isolated non-negative eigenvalues of finite multiplicity
with the only accumulation point at infinity. Denoting by 
$\{\gl_n\}^\infty_{n=1}$ the spectrum of $A$ counting multiplicities we deduce 
that the operator $f_0(A)$ is compact if and only if a unique accumulation 
point of $\{f_0(\gl_n)\}^\infty_{n=1}$ is at zero. This yields
$\lim_{n\to\infty}f_0(\gl_n) = 0$. Hence $(I +A)^{-1} \in \kL_\infty(\kH)$
implies $f_0(A) \in \kL_\infty(\kH)$ if and only if for any
sequence of non-negative numbers $\{\gl_n\}^\infty_{n=1}$ tending to
infinity as $n \longrightarrow \infty$ one has that 
$\lim_{n\to\infty}f_0(\gl_n) = 0$. This yields 
%
%
\be\la{3.49}
\lim_{x\to\infty}f_0(x) = 0.
\ee
%
%
Using the representation (\re{1.8}) one gets that $f_0(t_0A) \in \kL_\infty(\kH)$
is equivalent to
%
%
\be\la{3.50}
\lim_{x\to\infty}x\varphi_0(x) = +\infty.
\ee
%
%
Since $\varphi_0(x)$ is non-increasing one has that $\varphi_0(x_0) = 0$
for some $x_0 \ge 0$ yields $\varphi_0(x) = 0$ for all $x \ge x_0$. Therefore,
if $f_0(t_0A) \in \kL_\infty(\kH)$, then $\varphi_0(x) > 0$ for all $x \in
\bR_+$. By (\re{3.26}) one has
%
%
\be\la{3.51}
f(x) \le f_0(x) = \frac{1}{1 + x\varphi_0(x)}, \qquad x \in \bR_+,
\ee
%
%
and consequently 
%
%
\be\la{3.52}
xf(x) \le \frac{x}{1 + x\varphi_0(x)} \le \frac{1}{\varphi_0(x)}
\ee
%
%
for $x \in \bR_+$. Hence
%
%
\be\la{3.53}
b(x) = \sup_{s \in [0,x]}sf(s) \le \frac{1}{\varphi_0(x)}, \qquad x \in
\bR_+. 
\ee
%
%
Therefore,
%
%
\be\la{3.54}
\frac{b(x)}{x} \le \frac{1}{x\varphi_0(x)}, \qquad x \in \bR_+.
\ee
%
%
Applying (\re{3.50}) we obtain (\re{3.47}). If there is a $x_0 > 0$ such
that $r(x_0) = 1$, then there is a non-increasing sequence
$\{x_n\}^\infty_{n=1}$, $x_0 \le x_n$, tending to $x_0$ such that
$\lim_{n\to\infty}f(x_n) = 1$. Since by (\re{1.7}) 
%
%
\be\la{3.55}
\varphi_0(x_1) \le \frac{1}{x_n}\left(\frac{1}{f(x_n)} - 1\right), \qquad n =
1,2,\ldots,
\ee
%
%
one gets that $\varphi_0(x_1) = 0$. Thus  $\varphi_0(x) = 0$ for all $x \ge
x_1$ which contradicts to (\re{3.50}). Hence (\re{3.48}) is true.

Conversely, by (\re{1.3}) there is a $\gd > 0$ such that for $x \in [0,\gd]$ 
we have 
%
%
\be\la{3.57}
\frac{1}{x}\left(\frac{1}{f(x)} - 1\right) \ge \frac{1}{2}.
\ee
%
%
Since by (\re{3.45}) 
%
%
\be\la{3.58}
\frac{b(x)}{x} \ge f(x), \qquad x \in \bR_+,
\ee
%
%
we get
%
%
\be\la{3.59}
\frac{1}{f(x)} \ge \frac{x}{b(x)}, \qquad x \in \bR_+.
\ee
%
%
Then, using $1 - f(x) \ge 1 - r(\gd)$ for $x \ge \gd$, we obtain
%
%
\be\la{3.60}
\frac{1 - f(x)}{xf(x)} \ge \frac{1 -r(\gd)}{b(x)}, \qquad x \ge \gd.
\ee
%
%
Therefore, by (\re{3.57}) and (\re{3.60}) one gets
%
%
\be\la{3.61}
\varphi_0(x) \ge \min\left\{\frac{1}{2},\frac{1 - r(\gd)}{b(x)}\right\},
\qquad x \in \bR_+,
\ee
%
%
where we have used that $1/b(x)$ is non-increasing. If
$\lim_{x\to\infty}b(x) = b_0 < ~\infty$, then
%
%
\be\la{3.62}
\varphi_0(x) \ge c_0 := \min\left\{\frac{1}{2},\frac{1 -
r(\gd)}{b_0}\right\}, \qquad x \in \bR_+.
\ee
%
%
Hence (\re{3.50}) is proven. If $\lim_{x\to+\infty}b(x) = +\infty$, then
%
%
\be\la{3.63}
\varphi_0(x) \ge \frac{1 - r(\gd)}{b(x)}
\ee
%
%
for sufficiently large $x$. Hence
%
%
\be\la{3.64}
x\varphi_0(x) \ge (1 -r(\gd))\frac{x}{b(x)}
\ee
%
%
for sufficiently large $x$. Then applying (\re{3.47}) and (\re{3.48}) to 
(\re{3.64}) we again get (\re{3.50}). \hfill $\Box$
%
%
\begin{rem}\la{III.4-1}
%
%
{\em In the following a Kato-function is called {\em regular} if it satisfies
(\re{3.47}) and (\re{3.48}). For instance, $f(x) = \frac{1}{1 + x}$ and 
$f(x) = e^{-x}$ are regular Kato-functions while $f(x) = 1 - |\sin(x)|$ is 
{\em not} a regular one.}
%
%
\end{rem}
%
%
An obvious consequence of Theorem \re{III.2} and Lemma \re{III.4} is that the 
Trotter-Kato product formula converges locally uniformly away from zero in the 
operator-norm  for {\em any} regular Kato-function $f(\cdot)$, {\em any} 
Kato-function $g(\cdot)$ and {\em any} non-negative self-adjoint operator $B$ 
if the resolvent of operator $A$ is {\em compact}. It turns out that these 
assumptions are not only {\em sufficient} but also 
{\em necessary}. To prove this we need the following lemma.
%
%
\begin{lem}\la{III.5}
%
%
Let $f(x) = \frac{1}{1 + x}$ and let $g(\cdot)$ be another Kato-function.
Further, let $D$ be a non-negative self-adjoint operator such that $\|D\| < 1$.
If the Trotter-Kato product formula converges locally uniformly away from zero 
in the operator-norm for a couple of non-negative self-adjoint operators $A$ and $B$, 
then the same holds for the couple of operators $A + D$ and 
$B$.
%
%
\end{lem}
%
%
{\em Proof:} We are going to verify the condition $(iii)$ of Theorem \re{II.1}. 
Since the Trotter-Kato product formula converges in the operator-norm for 
self-adjoint operators $A$ and $B$, by Theorem \re{II.1}$(iii)$ for 
the Kato-function $f(x) = (1 + x)^{-1} > 0$ on $\bR_+$, i.e. for $Q(t) = I$ 
(see (\re{2.6-1})), one has
%
%
\be\la{3.65}
\lim_{t\to+0}\|(\gl I + M(t))^{-1} - P_0(\gl I_0 + H)^{-1}P_0\| = 0, \qquad \gl > 0,
\ee
%
%
where by (\re{2.5}) and (\re{2.6-1}) we have
%
%
\be\la{3.66}
M(t) := A + \frac{1}{t}[I - g(tB)], \qquad t > 0.
\ee
%
%
The lemma will be proven if we show that (\re{3.65}) implies
%
%
\be\la{3.67}
\lim_{t\to+0}\|(\gl I + M(t) + D)^{-1} - P_0(\gl I_0 + H + D_0)^{-1}P_0\| = 0
\ee
%
%
for $\gl > 0$, where $D_0 := P_0DP_0$. Let $\gl \ge 1$. Since $\|D\| < 1$, one
has the representations
%
%
\be\la{3.68}
(\gl I + M(t) + D)^{-1} = (\gl I + M(t))^{-1}\sum^\infty_{k=0}(-1)^k(D(\gl I + M(t))^{-1})^k
\ee
%
%
and
%
%
\be\la{3.69}
(\gl I_0 + H + D_0)^{-1} = (\gl I_0 + H)^{-1}\sum^\infty_{k=0}(-1)^k(D_0(\gl I_0 + H)^{-1})^k.
\ee
%
%
Since $\gl \ge 1$ and $\|D\| < 1$, for any $\epsilon > 0$ there is a $k_0$ such
that for $k \ge k_0$ one has
%
%
\be\la{3.70}
\|(\gl I + M(t))^{-1}\sum^\infty_{k=k_0}(-1)^k(D(\gl I + M(t))^{-1})^k\| \le \epsilon
\ee
%
%
and
%
%
\be\la{3.71}
\|(\gl I_0 + H)^{-1}\sum^\infty_{k=k_0}(-1)^k(D_0(\gl I_0 + H)^{-1})^k\| \le \epsilon
\ee
%
%
uniformly in $t > 0$. Moreover, by (\re{3.65}) there is $\gd > 0$ such that 
one has
%
%
\be\la{3.72}
\|(\gl I + M(t))^{-1}(D(\gl I + M(t))^{-1})^k - P_0(\gl I_0 + H)^{-1}(D_0(\gl I_0 + H)^{-1})^kP_0\|  \le \frac{\epsilon}{k_0}
\ee
%
%
for $k = 0,1,\ldots,k_0 - 1$ and $t \in (0,\gd]$. Summing up 
(\re{3.68}) - (\re{3.72}) we get
%
%
\be\la{3.73}
\|(\gl I + M(t) + D)^{-1} - P_0(\gl I_0 + H + D_0)^{-1}P_0\| \le 3\epsilon
\ee
%
%
for $t \in (0,\gd]$. Therefore, the limit (\re{3.67}) is proven. \hfill $\Box$
%
%
\begin{theo}\la{III.6}
%
%
Let $A$ be an non-negative self-adjoint operator. The Trotter-Kato product 
formula converges locally uniformly away from zero in the operator-norm for any
regular Kato-function $f(\cdot)$, any Kato-function $g(\cdot)$ as well as 
for any non-negative self-adjoint operator $B$ if and only if one has
$(I + A)^{-1} \in \kL_\infty(\kH)$.
%
%
\end{theo}
%
%
{\em Proof:} If $(I + A)^{-1} \in \kL_\infty(\kH)$, then by 
Lemma \re{III.4} one has $f_0(A) \in \kL_\infty(\kH)$. Applying 
Theorem \re{III.2} we verify the sufficient part of the theorem.

Conversely, assume that the Trotter-Kato product formula converges
in the operator-norm for {\em any} regular Kato-function $f(\cdot)$, for 
{\em any}
Kato-function $g(\cdot)$ and for {\em any} $B \ge 0$. In particular, this holds for
$f(x) = (1 + x)^{-1}$. Therefore, one gets that (\re{3.65}) is valid for
{\em each} Kato-function $g(\cdot)$ and {\em each} $B \ge 0$, where $M(t)$ is given by
(\re{3.66}). By the von Neumann theorem (see e.g. \cite{Ka3}) for any $\epsilon > 0$ 
there is a Hilbert-Schmidt operator $V$ with the Hilbert-Schmidt norm 
$\|V\|_2 \le \epsilon$ such that the spectrum of $A + V$ is purely pointwise. 
Let $\epsilon < 1/2$. Setting $D = \frac{1}{2}I + V$ we construct a 
non-negative self-adjoint operator such that $\|D\| < 1$ and the spectrum of 
$A + D$ is purely pointwise. Note that 
$\gs_{ess}(A + D) = \gs_{ess}(A + \frac{1}{2}I) = \frac{1}{2} + \gs_{ess}(A)$. 
If $(I + A)^{-1} \in \kL_\infty(\kH)$ is not satisfied, then 
$\gs_{ess}(A) \not= \emptyset$. 
Hence $\gs_{ess}(A + D) \not= \emptyset$. Let $\gm \in \gs_{ess}(A + D)$.
Of course $\gm \ge \frac{1}{2}$. There is an orthonormal sequence of eigenfunctions
$\{\varphi_n\}^\infty_{n=1}$ and a sequence of eigenvalues 
$\{\gm_n\}^\infty_{n=1}$ such that
%
%
\be\la{3.75}
(A + D)\varphi_n = \gm_n\varphi_n, \qquad n =1,2,\ldots,
\ee
%
%
and
%
%
\be\la{3.76}
\lim_{n\to\infty}\gm_n = \gm.
\ee
%
%
Notice that the function $g(x) = 1 - |\sin(x)|$ is from the Kato class. By 
$\kK$ we denote the closed linear span of $\{\varphi_n\}^\infty_{n=1}$. Setting
%
%
\be\la{3.77}
B\varphi_n = 2^{n+1}\pi\varphi_n, \qquad n = 1,2,\ldots,
\ee
%
%
and
%
%
\be\la{3.78}
Bf = 0, \qquad f \in \kH \ominus \kK,
\ee
%
% 
we define a non-negative self-adjoint operator $B$. Obviously, we have
%
%
\be\la{3.79}
(I - g(t_nB))\varphi_n = |\sin(t_nB)|\varphi_n = |\sin(t_n2^{n+1}\pi)|\varphi_n = 0
\ee
%
%
for $t_n := 2^{-n}$, $n = 1,2,\ldots$. Therefore, by (\re{3.66}), (\re{3.75})
and (\re{3.79}) one gets
%
%
\be\la{3.80}
(\gl I + M(t_n) + D)\varphi_n = (\gl + \gm_n)\varphi_n, \qquad n = 1,2,\ldots\,.
\ee
%
%
Since $P_0 = I$, due to (\re{3.75}) and (\re{3.77}), we have
%
%
\be\la{3.81}
(\gl I + H + D)\varphi_n = (\gl + \gm_n + 2^{n+1}\pi)\varphi_n, \qquad n = 1,2,\ldots.
\ee
%
%
Therefore,
%
%
\bea\la{3.84}
\lefteqn{\hspace{-3.0cm}\|(\gl I + M(t_n) + D)^{-1}\varphi_n - (\gl I + H + D)^{-1}\varphi_n\| \le}\\
& & |(\gl + \gm_n)^{-1} - (\gl + \gm_n + 2^{n+1}\pi)^{-1}|.\nonumber
\eea
%
%
By (\re{3.76}) we find
%
%
\be\la{3.85}
\lim_{n\to\infty}\|(\gl I + M(t_n) + D)^{-1}\varphi_n - (\gl + H + D)^{-1}\varphi_n\| = (\gl + \gm)^{-1}.
\ee
%
%
Now taking into account (\re{3.67}) we obtain
%
%
\be\la{3.86}
\lim_{n\to\infty}\|(\gl I + M(t_n) + D)^{-1}\varphi_n - (\gl + H + D)^{-1}\varphi_n\| = 0,
\ee
%
%
which contradicts (\re{3.85}). Therefore, the assumption 
$\gs_{ess}(A) \not= \emptyset$ is false which implies immediately that
$(I + A)^{-1} \in \kL_\infty(\kH)$. \hfill $\Box$

%\newpage

\section{Relative Compactness}
\setcounter{equation}{0}
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\setcounter{theo}{0}
\renewcommand{\thetheo}{\arabic{section}.\arabic{theo}}

In this section we treat the case $\ga = 1/2$ which was not covered by our 
paper
\cite{NZ3}.

\begin{theo}\la{IV.1}
%
%
Let $A \ge I$ and $B \ge 0$ be two self-adjoint operators such that
%
%
\be\la{4.1}
\dom(A^{1/2}) \subseteq \dom(B^{1/2}).
\ee
%
%
Further, let $f(\cdot)$ and $g(\cdot)$ be Kato-functions. If
%
%
\be\la{4.2}
C_0 := \sup_{x > 0}\frac{xf(x)}{1 - f(x)} < +\infty
\ee
%
%
and
%
%
\be\la{4.3}
B^{1/2}A^{-1/2} \in \kL_\infty(\kH),
\ee
%
%
then the Trotter-Kato product formula converges locally uniformly away from 
zero in the opera\-tor-norm.
%
%
\end{theo}
%
%
{\em Proof:} From (\re{4.2}) it follows that
%
%
\be\la{4.4}
0 \le f(x) \le \frac{C_0}{C_0 + x}, \qquad x > 0.
\ee
%
%
First we show that
%
%
\be\la{4.5}
\lim_{t\to+0}\left\|\left(I + \frac{1}{t}[I - f(tA)]\right)^{-1} - (I + A)^{-1}\right\| = 0.
\ee
%
%
Notice that (\re{4.4}) implies
%
%
\be\la{4.6}
1 - f(x) \ge 1 - \frac{C_0}{C_0 + x}, \qquad x > 0,
\ee
%
%
which yields
%
%
\be\la{4.7}
\frac{1}{t}[1 - f(tx)] \ge \frac{x}{C_0 + tx}, \qquad x > 0, \quad t \ge 0,
\ee
%
%
or, due to $C_0 \ge 1$ (see (1.4), (\re{4.2})),
%
%
\be\la{4.8}
\left(1 + \frac{1}{t}[1 - f(tx)]\right)^{-1} \le \frac{C_0}{C_0 + x} + t, \qquad x > 0, \quad t \ge 0.
\ee
%
%
Therefore,
%
%
\be\la{4.9}
\sup_{x \ge N}\left(1 + \frac{1}{t}[1 - f(tx)]\right)^{-1} \le \frac{C_0}{C_0 + N} + t, \qquad t \ge 0.
\ee
%
%
Since
%
%
\bea\la{4.11}
\lefteqn{\hspace{-2.0cm} \left(1 + \frac{1}{t}[1 - f(tx)]\right)^{-1} - (1 + x)^{-1} =}\\
& & \left(1 + \frac{1}{t}[1 - f(tx)]\right)^{-1}\left(1 - \frac{1}{xt}[1 - f(tx)]\right)\frac{x}{1 + x}, \nonumber
\eea
%
%
we get the estimate
%
%
\be\la{4.12}
\left|\left(1 + \frac{1}{t}[1 - f(tx)]\right)^{-1} - (1 + x)^{-1}\right| \le \left|1 - \frac{1}{xt}[1 - f(tx)]\right|.
\ee
%
%
By (\re{1.3}) for each $\epsilon > 0$ there is a $\gd > 0$ such that
%
%
\be\la{4.13}
\left|1 - \frac{1}{y}[1 - f(y)]\right| \le \epsilon
\ee
%
%
for $y \in (0,\gd]$. Hence for $t \in (0,\gd/N]$ and $x \in (0,N]$ one gets
%
%
\be\la{4.14}
\left|1 - \frac{1}{xt}[1 - f(xt)]\right| \le \epsilon.
\ee
%
%
Therefore, for any $\epsilon > 0$ and $N > 0$, there is a
$\gd' > 0$ such that
%
%
\be\la{4.15}
\sup_{x \in (0,N]}\left|1 - \frac{1}{xt}[1 - f(tx)]\right| \le \epsilon
\ee
%
%
for $t \in (0,\gd']$. Hence, (\re{4.12}) yields
%
%
\be\la{4.16}
\lim_{t\to+0}\sup_{x \in (0,N]}\left|\left(1 + \frac{1}{t}[1 - f(xt)]\right)^{-1} - \frac{1}{1 + x}\right| = 0
\ee
%
%
for each $N$, $0 < N < +\infty$. Since by (\re{4.9})
%
%
\bea\la{4.17}
\lefteqn{\hspace{-1.0cm}\sup_{x > 0}\left|\left(1 + \frac{1}{t}[1 - f(xt)]\right)^{-1} - \frac{1}{1 + x}\right| \le} \\
& & \max\left\{\sup_{x \in (0,N]}\left|\left(1 + \frac{1}{t}[1 - f(xt)]\right)^{-1} - \frac{1}{1 + x}\right|,\,\frac{C_0}{C_0 + N} + t,\, \frac{1}{1 + N}\right\}, \nonumber
\eea
%
%
one gets that
%
%
\be\la{4.18}
\lim_{t\to+0}\sup_{x > 0}\left|\left(1 + \frac{1}{t}[1 - f(xt)]\right)^{-1} - \frac{1}{1 + x}\right| = 0
\ee
%
%
which proves (\re{4.5}).

Next we verify the condition $(ii)$ of Theorem \re{II.1}. Let $\gl =1$.
>From (\re{2.5}) and (\re{2.24}) we find the representation
%
%
\be\la{4.19}
I + R(t) = \left(I + \frac{1}{t}[I - f(tA)]\right)^{1/2}\{I + X(t)\}\left(I + \frac{1}{t}[I - f(tA)]\right)^{1/2}
\ee
%
%
where
%
%
\bea\la{4.20}
\lefteqn{\hspace{-1.5cm} X(t) := f(tA)^{1/2}A^{1/2}\left(I + \frac{1}{t}[I - f(tA)]\right)^{-1/2}A^{-1/2}B^{1/2} \times}\\
& & [I - g(tB)](t\tilde{B})^{-1}B^{1/2}A^{-1/2}f(tA)^{1/2}A^{1/2}\left(I + \frac{1}{t}[I - f(tA)]\right)^{-1/2}, \nonumber
\eea
%
%
where $\tilde{B} = B|\ran(B): \overline{\ran(B)} \longrightarrow 
\overline{\ran(B)}$. Hence one obtains
%
%
\be\la{4.21}
(I + R(t))^{-1} = \left(I + \frac{1}{t}[I - f(tA)]\right)^{-1/2}\{I + X(t)\}^{-1}\left(I + \frac{1}{t}[I - f(tA)]\right)^{-1/2}.
\ee
%
%
Since by (\re{4.8})
%
%
\be\la{4.22}
\sup_{t > 0}\left\|f(tA)^{1/2}A^{1/2}\left(I + \frac{1}{t}[I - f(tA)]\right)^{-1/2}\right\| \le C^{1/2}_0,
\ee
%
%
in virtue of (\re{4.5}) one easily gets that
%
%
\be\la{4.23}
s-\lim_{t\to+0}f(tA)^{1/2}A^{1/2}\left(I + \frac{1}{t}[I - f(tA)]\right)^{-1/2} = A^{1/2}(I + A)^{-1/2}.
\ee
%
%
Since (\re{1.4}) implies
%
%
\be\la{4.24}
s-\lim_{t\to+0}[I - g(tB)](t\tilde{B})^{-1}h = h, \qquad h \in \ran(B),
\ee
%
%
by (\re{4.20}), (\re{4.23}) and (\re{4.24}) we get
%
%
\be\la{4.25}
s-\lim_{t\to+0}X(t) = (I + A)^{-1/2}B(I + A)^{-1/2}.
\ee
%
%
Moreover, by condition (\re{4.3}) the operator $B^{1/2}A^{-1/2}$ is compact. 
Since it is involved into (\re{4.20}), by standard arguments 
(see e.g. \cite{Si1}, Chapter 2) and by  (\re{4.23}), (\re{4.24}) we get, 
instead of (4.24), the limit 
%
%
\be\la{4.26}
\lim_{t\to+0}\|X(t) - (I + A)^{-1/2}B(I + A)^{-1/2}\| = 0.
\ee
%
%
Using (\re{4.26}) we find that
%
%
\be\la{4.27}
\hspace{-0.5cm} \lim_{t\to+0}\|(I + X(t))^{-1} - (I + (I + A)^{-1/2}B(I + A)^{-1/2})^{-1}\| = 0.
\ee
%
%
Taking into account (\re{4.5}), (\re{4.21}) and (\re{4.27}) we obtain
%
%
\be\la{4.28}
\lim_{t\to+0}\|(I + R(t))^{-1} - (I + A)^{-1/2}(I + (I + A)^{-1/2}B(I + A)^{-1/2})^{-1}(I + A)^{-1/2}\| = 0
\ee
%
%
which yields
%
%
\be\la{4.29}
\lim_{t\to+0}\|(I + R(t))^{-1} - (I + H)^{-1}\| = 0.
\ee
%
%
Then a reference to Theorem \re{II.1} $(ii)$  completes the proof. \hfill $\Box$

%\newpage

\section{Resolvent Product Compactness}
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In this section we consider another kind of compactness condition.
Since the product of resolvents is involved in this condition, 
we call it the {\em resolvent product compactness}. We start with two technical
lemmata.
%
%
\bl\la{V.0}
%
%
Let $X$ be a non-negative self-adjoint operator on the separable Hilbert space
$\kH$ and let $Q$ be an orthogonal projection on $\kH$. Then one has
%
%
\be\la{5.1}
Q(I_Q + QXQ)^{-1}Q \le (I + X)^{-1},
\ee
%
%
where $I_Q := I|Q\kH$.
%
%
\el
%
%
{\em Proof:} Obviously, one has
%
%
\be\la{5.1-1}
I + X \le I + n(I-Q) + X, \qquad n = 0,1,2,\ldots,
\ee
%
%
which implies
%
%
\be\la{5.1-2}
(I + n(I-Q) + X)^{-1} \le (I + X)^{-1}, \qquad n = 0,1,2,\ldots.
\ee
%
%
A straightforward computation shows that
%
%
\be\la{5.1-3}
s-\lim_{n\to\infty}(I + n(I-Q) + X)^{-1} = Q(I_Q + QXQ)^{-1}Q,
\ee
%
%
where $I_Q := I|Q\kH$. Hence from (\re{5.1-2}) and (5.4) we obtain 
(\re{5.1}). \hfill $\Box$
%
%
\bl\la{V.0-1}
%
%
Let $\{X(t)\}_{t \ge 0}$ be a sequence of non-negative self-adjoint operators
on the subspace $\{\kH(t)\}_{t\ge 0}$ of the separable Hilbert space $\kH$ 
and let 
$\{Q(t)\}_{t \ge 0}$ be the corresponding sequence of orthogonal projections 
from 
$\kH$ onto $\kH(t)$. If for each $\gl > 0$ the weak resolvent convergence takes
place, i.e.,
%
%
\be\la{5.1-4}
w-\lim_{t\to+0}Q(t)(\gl I(t) + X(t))^{-1}Q(t) = Q(0)(I(0) + X(0))^{-1}Q(0),
\ee
%
%
then for each $\gl > 0$ the strong resolvent convergence holds, i.e.,
%
%
\be\la{5.1-5}
s-\lim_{t\to+0}Q(t)(\gl I(t) + X(t))^{-1}Q(t) = Q(0)(\gl I(0) + X(0))^{-1}Q(0),
\ee
%
%
where $I(t) = I|\kH(t)$, $t \ge 0$. In particular, one has
%
%
\be\la{5.1-6}
s-\lim_{t\to+0}Q(t)(\gl I(t) + X(t))^{-1/2}Q(t) = Q(0)(\gl I(0) + X(0))^{-1/2}Q(0),
\ee
%
%
for $\gl > 0$.
%
%
\el
%
%
{\em Proof:} The proof is based on two observations. Firstly, we note 
that (\re{5.1-4}) implies 
%
%
\be\la{5.1-7}
\lim_{t\to+0}\|Q(t)(\gl I(t) + X(t))^{-1/2}Q(t)h\|^2 = \|Q(0)(I(0) + X(0))^{-1/2}Q(0)h\|^2
\ee
%
%
for $h \in \kH$ and $\gl > 0$. Secondly, by the representation
%
%
\be\la{5.1-8}
Q(t)(\gl I(t) + X(t))^{-1/2}Q(t) = \frac{1}{\pi}\int^{+\infty}_0 \frac{dx}{\sqrt{x}}\;Q(t)(I(t) + x(\gl I(t) + X(t)))^{-1}Q(t), \quad t \ge 0,
\ee
%
%
we find from (\re{5.1-4}) that
%
%
\be\la{5.1-9}
w-\lim_{t\to+0}Q(t)(\gl I(t) + X(t))^{-1/2}Q(t) = Q(0)(\gl I(0) + X(0))^{-1/2}Q(0)
\ee
%
%
for each $\gl > 0$. By a standard argument (see e.g. [12]) we obtain that 
(\re{5.1-7})
and (\re{5.1-9}) imply the strong convergence of the resolvents in 
(\re{5.1-8}), i.e., that (\re{5.1-6}) is valid. However, the relation
(\re{5.1-6}) immediately implies (\re{5.1-5}). \hfill$\Box$

The following theorem is  motivated by the fact that neither {\em resolvent 
compactness} or {\em relative compactness} (see (1.19), (1.21)), nor 
{\em smallness conditions} (see (1.13), (1.16))
allow to prove the operator-norm convergence of the Trotter-Kato product 
formula, for example (see \cite{DIT1}, \cite{ITa1}, \cite{ITa2}),
in the case of increasing potentials of the Schr{\"o}dinger semigroup generators. See also \cite{DSch1} and \cite{He1} - \cite{He3}. The following theorem treats 
this 
case on an {\em abstract} level.
% 
%
\begin{theo}\la{V.1}
%
%
Let $A$ and $B$ be non-negative self-adjoint operators and  let $f(\cdot)$ and $g(\cdot)$
be Kato-functions such that conditions (\re{4.2}) and 
%
%
\be\la{5.2}
S_0 := \sup_{x > 0}\frac{xg(x)}{1 - g(x)} < +\infty
\ee
%
%
are satisfied. If
%
%
\be\la{5.3}
(I + A)^{-1}(I + B)^{-1} \in \kL_\infty(\kH),
\ee
%
%
then the Trotter-Kato product formula converges locally uniformly away from 
zero in the opera\-tor-norm . Moreover, one has 
$(I_0 + H)^{-1} \in \kL_\infty(\kH_0)$.
%
%
\end{theo}
%
%
{\em Proof:} Notice that (\re{2.5}) implies
%
%
\be\la{5.4}
(I(t) + M(t))^{-1} \le \left(I(t) + \frac{1}{t}[\tilde{f}(tA)^{-1} - I(t)]\right)^{-1}, \qquad t > 0.
\ee
%
%
Hence, there is a family of contractions $\{\gG(t)\}_{t>0}$, $\gG(t): \kH(t) \longrightarrow \kH(t)$,
such that
%
%
\be\la{5.5}
(I(t) + M(t))^{-1/2} = \gG(t)\left(I(t) + \frac{1}{t}[\tilde{f}(tA)^{-1} - I(t)]\right)^{-1/2}, \qquad t > 0.
\ee
%
%
Again by (\re{2.5}) we get
%
%
\be\la{5.6}
(I(t) + M(t))^{-1} \le \left(I(t) + \frac{1}{t}Q(t)[I - g(tB)]Q(t)\right)^{-1}, \qquad t > 0.
\ee
%
%
Applying Lemma \re{V.0} we obtain
%
%
\be
Q(t)\left(I(t) + \frac{1}{t}Q(t)[I - g(tB)]Q(t)\right)^{-1}Q(t) \le \left(I + \frac{1}{t}[I - g(tB)]\right)^{-1}, \qquad t > 0.
\ee
%
%
Therefore, there exists a family of contractions 
$\{\gP(t)\}_{t>0}$, $\gP(t): \kH \longrightarrow \kH(t)$, such that
%
%
\be\la{5.8}
(I(t) + M(t))^{-1/2}Q(t) = \gP(t)\left(I + \frac{1}{t}[I - g(tB)]\right)^{-1/2}, \qquad t > 0.
\ee
%
%
Setting 
%
%
\be\la{5.9}
\gG h := (I_0 + H)^{-1/2}P_0(I + A)^{1/2}h, \qquad h \in \dom(A^{1/2}),
\ee
%
%
and
%
%
\be\la{5.10}
\gP h := (I_0 + H)^{-1/2}P_0(I + B)^{1/2}h, \qquad h \in \dom(B^{1/2}),
\ee
%
%
we define two contractions acting from $\kH$ into $\kH_0$. Therefore, we find
the representations
%
%
\be\la{5.11}
(I_0 + H)^{-1/2}P_0f = \gG(I + A)^{-1/2}f, \qquad f \in \kH,
\ee
%
%
and
%
%
\be\la{5.12}
(I_0 + H)^{-1/2}P_0f = \gP(I + B)^{-1/2}f, \qquad f \in \kH.
\ee
%
%
Hence, one gets the representation
%
%
\be\la{5.13}
P_0(I_0 + H)^{-1}P_0 = \gG(I + A)^{-1/2}(I + B)^{-1/2}\gP^*.
\ee
%
%
By assumption (\re{5.3}) we immediately obtain 
$(I_0 + H)^{-1} \in \kL_\infty(\kH_0)$. Notice that by definition of $Q(t)$ we 
have
%
%
\be\la{5.15}
Q(t)\left(I(t) + \frac{1}{t}[\tilde{f}(tA)^{-1} - I(t)]\right)^{-1}Q(t) = \left(I + \frac{1-t}{t}[I - f(tA)]\right)^{-1}f(tA)
\ee
%
%
for $t > 0$. Since
%
%
\bea\la{5.16}
\lefteqn{\left(I + \frac{1}{t}[I - f(tA)]\right)^{-1} - \left(I + \frac{1-t}{t}[I - f(tA)]\right)^{-1}f(tA) = }\\
& & \frac{1}{1-t}\frac{1-t}{t}[I - f(tA)]\left(I + \frac{1-t}{t}[I - f(tA)]\right)^{-1}[I - f(tA)]\left(I + \frac{1}{t}[I - f(tA)]\right)^{-1},
\nonumber
\eea
%
%
one gets the representation
%
%
\bea\la{5.16-1}
\lefteqn{\hspace{-0.5cm} \left(I + \frac{1}{t}[I - f(tA)]\right)^{-1} - Q(t)\left(I(t) + \frac{1}{t}[\tilde{f}(tA)^{-1} - I(t)]\right)^{-1}Q(t) =}\\ 
& & \frac{1}{1-t}\frac{1-t}{t}[I - f(tA)]\left(I + \frac{1-t}{t}[I - f(tA)]\right)^{-1}[I - f(tA)]\times \nonumber \\ 
& & \left\{(I + \frac{1}{t}[I - f(tA)])^{-1} -  (I + A)^{-1}\right\} + \nonumber\\
& & \frac{t}{1-t}\frac{1-t}{t}[I - f(tA)]\left(I + \frac{1-t}{t}[I - f(tA)]\right)^{-1}[I - f(tA)](tA)^{-1}A(I + A)^{-1} \nonumber
\eea
%
%
for $t > 0$. Hence, we obtain the estimate
%
%
\bea\la{5.17}
\lefteqn{\hspace{-1.0cm} \left\|\left(I + \frac{1}{t}[I - f(tA)]\right)^{-1} - \left(I(t) + \frac{1}{t}[\tilde{f}(tA)^{-1} - I(t)]\right)^{-1}Q(t)\right\| \le}\\ 
& & \frac{1}{1-t}\left\|\left(I + \frac{1}{t}[I - f(tA)]\right)^{-1} - (I + A)^{-1}\right\| + C_1\frac{t}{1-t} \nonumber
\eea
%
%
for $t > 0$, where $C_1 := \sup_{x > 0}(1 - f(x))/x < +\infty$. Then taking 
into account (\re{4.5}) we get from (5.26) that
%
%
\be\la{5.18}
\lim_{t\to+0}\left\|\left(I + \frac{1}{t}[I - f(tA)]\right)^{-1} - \left(I(t) + \frac{1}{t}[\tilde{f}(tA)^{-1} - I(t)]\right)^{-1}Q(t)\right\| = 0.
\ee 
%
%
By (\re{5.18}) and again by (\re{4.5}) we finally obtain
%
%
\be\la{5.19}
\lim_{t\to+0}\left\|\left(I(t) + \frac{1}{t}[\tilde{f}(tA)^{-1} - I(t)]\right)^{-1}Q(t) - (I + A)^{-1}\right\| = 0.
\ee
%
%
Furthermore, applying the same argument as in (\re{4.6}) - (\re{4.18}), but now 
for 
the operator $B$, by (\re{5.2}) we obtain the analogue of (\re{4.5}):
%
%
\be\la{5.20}
\lim_{t\to+0}\left\|\left(I + \frac{1}{t}[I - g(tB)]\right)^{-1} - (I + B)^{-1}\right\| = 0.
\ee
%
%
Both relations (\re{5.19}) and (\re{5.20}) imply
%
%
\be\la{5.21}
\lim_{t\to+0}\left\|Q(t)\left(I(t) + \frac{1}{t}[\tilde{f}(tA)^{-1} - I(t)]\right)^{-1/2}Q(t) - (I + A)^{-1/2}\right\| = 0.
\ee
%
%
and 
%
%
\be\la{5.22}
\lim_{t\to+0}\left\|\left(I + \frac{1}{t}[I - g(tB)]\right)^{-1/2} - (I + B)^{-1/2}\right\| = 0.
\ee
%
%
Since under conditions of the Theorem the
results of \cite{Ch1}, \cite{Ch2} and \cite{Ka1} imply (\re{3.19})(see the 
Proof of Theorem 3.2), 
applying Lemma \re{V.0-1} to (\re{3.19}) we find that
%
%
\be\la{5.24}
s-\lim_{t\to+0} Q(t)(I(t) + M(t))^{-1/2}Q(t) = P_0(I_0 + H)^{-1/2}P_0.
\ee
%
%
Taking into account (\re{5.5}), (\re{5.21}) and (\re{5.24}) we obtain
%
%
\be\la{5.25}
s-\lim_{t\to+0}\gG(t)Q(t)(I + A)^{-1/2} = P_0(I_0 + H)^{-1/2}P_0.
\ee
%
%
Using (\re{5.11}) one gets
%
%
\be\la{5.26}
s-\lim_{t\to+0} \gG(t)Q(t)(I + A)^{-1/2} = \gG(I + A)^{-1/2},
\ee
%
%
which yields
%
%
\be\la{5.27}
s-\lim_{t\to+0}\gG(t)Q(t) = \gG.
\ee
%
%
Similarly, from (\re{5.8}), (\re{5.12}), (\re{5.22}) and (\re{5.24})  we find
%
%
\be\la{5.28}
s-\lim_{t\to+0} \gP(t) = \gP.
\ee
%
%
>From (\re{5.5}) and (\re{5.8}) we obtain the representation
%
%
\bea\la{5.29}
\lefteqn{\hspace{-1.0cm} Q(t)(I(t) + M(t))^{-1}Q(t) =}\\
& & \gG(t)\left(I(t) + \frac{1}{t}[\tilde{f}(tA)^{-1} - I(t)]\right)^{-1/2}Q(t)\left(I + \frac{1}{t}[I - g(tB)]\right)^{-1/2}\gP(t)^*\nonumber
\eea
%
%
for $t > 0$. Using the identity
%
%
\bea\la{5.30}
\lefteqn{\hspace{-1.0cm} Q(t)(I(t) + M(t))^{-1}Q(t) - P_0(I_0 + H)^{-1}P_0 =}\\
& & \gG(t)Q(t)\left\{\left(I(t) + \frac{1}{t}[\tilde{f}(tA)^{-1} - I(t)]\right)^{-1/2}Q(t) - (I + A)^{-1/2}\right\}\times \nonumber \\
& & \left(I + \frac{1}{t}[I - g(tB)]\right)^{-1/2}\gP(t)^* + \nonumber \\
& & \gG(t)Q(t)(I + A)^{-1/2}\left\{\left(I + \frac{1}{t}[I - g(tB)]\right)^{-1/2} - (I + B)^{-1/2}\right\}\gP(t)^* + \nonumber \\
& & (\gG(t)Q(t) - \gG)(I + A)^{-1/2}(I + B)^{-1/2}\gP(t)^* + \nonumber\\
& & \gG(I + A)^{-1/2}(I + B)^{-1/2}(\gP(t)^* - \gP^*) \nonumber
\eea
%
% 
one gets the estimate
%
%
\bea\la{5.31}
\lefteqn{\|Q(t)(I(t) + M(t))^{-1}Q(t) - P_0(I_0 + H)^{-1}P_0\| \le}\\
& & \left\|\gG(t)Q(t)\left(I(t) + \frac{1}{t}[\tilde{f}(tA)^{-1} - I(t)]\right)^{-1/2}Q(t) - (I + A)^{-1/2}\right\| + \nonumber \\
& & \left\|\left(I + \frac{1}{t}[I - g(tB)]\right)^{-1/2} - (I + B)^{-1/2}\right\| + \nonumber \\
& & \|(\gG(t)Q(t) - \gG)(I + A)^{-1/2}(I + B)^{-1/2}\| + \|(I + A)^{-1/2}(I + B)^{-1/2}(\gP(t)^* - \gP^*)\| \nonumber 
\eea
%
%
for $t > 0$. Since $(I + A)^{-1/2}(I + B)^{-1/2} \in \kL_\infty(\kH)$, by 
(\re{5.27}), (\re{5.28}) and standard arguments (\cite{Si1}, Chapter 2) one  
obtains that
%
%
\be\la{5.35}
\lim_{t\to+0}\|(\gG(t)Q(t) - \gG)(I + A)^{-1/2}(I + B)^{-1/2}\| = 0
\ee
%
%
and
%
%
\be\la{5.36}
\lim_{t\to+0}\|(I + A)^{-1/2}(I + B)^{-1/2}(\gP(t)^* - \gP^*)\| = 0.
\ee
%
%
Taking into account (\re{5.21}) and (\re{5.22}) we finish the proof because  
(5.39)-(5.41)  guarantee the condition $(iii)$ of Theorem 2.1.\hfill $\Box$

%\newpage

\section{Remarks}
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We conclude this paper by some remarks.

\begin{enumerate}

\item In the present paper we study {\em only} the {\em operator-norm  
convergence}
of the Trotter-Kato product formula (for $t$ away from zero and infinity) and
{\em not} the estimate of {\em error bounds} for the convergence as it is done 
in \cite{IT1}, \cite{IT2}, \cite{NZ2} - \cite{Ro1}. 

\item Notice that in fact the necessary and sufficient conditions $(ii)$ and $(iv)$ 
of 
Theorem \re{II.1} are a {\em norm-operator} analogous of conditions for the
{\em strong} convergence of the exponential Trotter formula formulated by 
Chernoff 
in \cite{Ch1} and \cite{Ch2}.

\item The condition of type $(iii)$ of Theorem \re{II.1} was used at the first
time by Kato in \cite{Ka2} in order to prove the {\em strong} convergence of the 
Trotter-Kato product formula.

\item A necessity of condition $(I + A)^{-1} \in \kL_\infty(\kH)$
in the  Theorem \re{III.6} is due to the fact that the Kato-function $g(\cdot)$
is {\em not regular}. It would be interested to prove the similar result under 
the assumption that the {\em both} Kato-functions $f(\cdot)$ and $g(\cdot)$ are 
{\em regular}.

\item Theorem \re{IV.1} extends results of \cite{NZ3} to the case
$\ga = 1/2$ assuming in {\em addition} that $B^{1/2}$ is relatively compact with 
respect to $A^{1/2}$ (see also remark 3 in Section 7 of \cite{NZ3}), however, 
without 
error bounds. It seems that Theorem \re{IV.1} remains true under a
weaker assumption that $B^{1/2}$ is relatively $A^{1/2}$-bounded with the 
{\em relative zero-bound}.


\item The resolvent product compactness condition of Section 5 has never been 
used before. It applies very well to the case of increasing (at infinity) 
potentials for Schr\"{o}dinger semigroup generators. However, the resolvent 
product 
compactness does not give error bounds for convergence of the Trotter product 
formula similar to \cite{DIT1}, \cite{ITa1} and \cite{ITa2}. It would be 
interested to reformulate this our observation in such 
a way that besides the fact of the norm-convergence one also obtains the 
corresponding error bounds.


\end{enumerate}

\section*{Acknowledgements}
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The first author (H.N.) thanks the Center de Physique Th\'{e}orique - CNRS - 
Luminy
and Universit\'{e} de Provence (Marseille) as well as Deutsche 
Forschungsgemeinschaft (DFG) and Deutsche Akademische Austauschdienst (DAAD)
for financial support and for hospitality during his visits to CPT - Luminy. 
We thank Professor Ichinose for discussions of the Trotter product formula
for Schr\"{o}dinger semigroups.

%\newpage

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\end{document}






