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This article is cited in 1 scientific paper (total in 1 paper) Operator-norm Trotter product formula on Banach spaces V. A. Zagrebnovabc a Institut de Mathématiques de Marseille b Université d'Aix-Marseille – CNRS c CMI – Technopôle Château-Gombert, Marseille, France
Bibliographic databases:
     In memory of Academician Vasilii Sergeevich Vladimirov on the 100th anniversary of his birthday § 1. IntroductionIn this paper we collect results concerning the operator-norm convergent Trotter product formula for two semigroups $\{\mathrm{e}^{- t A}\}_{t\geqslant 0}$, $\{\mathrm{e}^{- t B}\}_{t\geqslant 0}$, with densely defined generators $A$ and $B$ in a Banach space. Although the strong convergence in Banach space for contraction semigroups is known since the seminal paper by Trotter (1959), which after more than three decades was extended to convergence in the operator-norm topology in Hilbert spaces by Rogava (1993), the operator-norm convergence in a Banach space was established only in 2001. For the first time this result was established under hypothesis that one of the involved into the product formula contraction semigroups, for example, $\{\mathrm{e}^{- t A}\}_{t\geqslant 0}$, is holomorphic together with certain conditions of smallness on generators $B$ and $B^*$ with respect to generators $A$ and $A^*$. Note that in spite of a quite strong assumptions on operators $A$ and $B$ the proof of the operator-norm convergent Trotter product formula on a Banach space is (unexpectedly) involved and technical. To elucidate the question of how far these conditions are from optimal ones, we show an example of the operator-norm convergent Trotter product formula for two semigroups $\{\mathrm{e}^{- t A}\}_{t\geqslant 0}$ and $\{\mathrm{e}^{- t B}\}_{t\geqslant 0}$ on a Banach space, where hypothesis on operator $A$ is relaxed to condition that $A$ is generator of a contraction semigroup. § 2. Preliminaries2.1. Bounded semigroups on $\mathfrak{X}$For what follows, the properties of holomorphic (contraction) semigroups on a Banach space $\mathfrak{X}$ are essential. Therefore, we start with a suitable for our aim recall of details concerning the bounded, holomorphic semigroups, and fractional powers of their generators. We begin with definitions and properties to introduce certain notations adapted in this section for semigroups on $\mathfrak{X}$. Definition 2.1. Recall that a family $\{U(t)\}_{t\geqslant 0}$ of bounded linear operators on a Banach space $\mathfrak{X}$ is called a one-parameter strongly continuous semigroup if it satisfies the conditions: (i) $U(0)=\mathbb{I}$; (ii) $U(s+t) = U(s)U(t)$ for all $s,t\geqslant 0$; (iii) $\lim_{t\to +0} U(t)x = x$ for all $x\in{\mathfrak{X}}$. We recall some immediate consequences of this definition. $\bullet$ There are constants $C_A\geqslant 1$ and $\gamma_A\in\mathbb{R}$, depending on the generator of the semigroup, such that $\|U(t)\| \leqslant C_A \mathrm{e}^{\gamma_A t}$ for all $t\geqslant 0$. $\bullet$ $t\mapsto U(t)$ is a strongly continuous function from $[0,+\infty)$ onto the algebra $\mathcal{L}(\mathfrak{X})$ of bounded linear operators on $\mathfrak{X}$. $\bullet$ There exists a closed densely defined linear operator $A$ on $\mathfrak{X}$ with domain $\operatorname{dom}(A)$, called the generator of the semigroup, such that $\lim_{t\to +0}(U(t)x-x)/t = -Ax$, for any $x\in\operatorname{dom}(A)$, that is, by convention $U(t):=\mathrm{e}^{-tA}$. $\bullet$ The resolvent of the generator satisfies the estimate
$$
\begin{equation*}
\|R_{A}(-\lambda)\|\,{=}\,\|(A\,{+}\,\lambda \mathbb{I})^{-1}\|\leqslant C_A / (\operatorname{Re}(\lambda) - \gamma_A)
\end{equation*}
\notag
$$
for all $\lambda$ with $\operatorname{Re}(\lambda)>\gamma_A$, thus the open half plane with $\operatorname{Re}(z) < -\gamma_A$ is contained in the resolvent set of the operator $A$, which is defined as $\rho(A) = \{z\in\mathbb{C}\colon\|R_{A}(z)\| < +\infty \}$. $\bullet$ If $\gamma_A\leqslant 0$, $U(t)$ is called a bounded semigroup (otherwise, $U(t)$ is called a quasi-bounded semigroup of type $\gamma_A>0$). For any strongly continuous semigroup, we can construct a bounded semigroup by adding some constant $\eta\geqslant\gamma_A$ to its generator. Let $\widetilde{A} = A+\eta \mathbb{I}$, then for the semigroup $\widetilde{Q}(t)$ generated by $\widetilde{A}$, one has $\|\widetilde{Q}(t)\|\leqslant C_A$, $t\geqslant 0$, and the open half-plane $\operatorname{Re}(\lambda)<\eta-\gamma_A$ is included into the resolvent set $\rho(\widetilde{A})$ of $\widetilde{A}$. So it is not restrictive to suppose that the considered semigroup $U(t)$ is bounded and that the set $\{z\in\mathbb{C}\colon \operatorname{Re}(z)\leqslant 0\} \subseteq\rho(A)$. $\bullet$ If $\|U(t)\|\leqslant 1$, $t\geqslant 0$, the semigroup is called a contraction semigroup. We comment that the method of the preceding remark does not permit to construct a contraction semigroup from a bounded semigroup in general, since it can not change the value of the constant $C_A$. Below we need a characterization of generators of these contraction semigroups. First we recall that the space of linear bounded functionals $\mathfrak{X}^* = \mathcal{L}(\mathfrak{X}, \mathbb{C})$ is a dual of the Banach space $\mathfrak{X}$ and that $\mathfrak{X}^*$ is itself a Banach space. Recall that a linear operator $A$ in $\mathfrak{X}$ is accretive if for all pairs $(u, \phi)\in\operatorname{dom}(A)\times\mathfrak{X}^*$ with $\|u\|_{\mathfrak{X}}=1$, $\|\phi\|_{\mathfrak{X}^*}=1$, $(u, \phi) =1$, one has $\operatorname{Re}(Au,\phi) \geqslant 0$. We also add that a densely defined in $\mathfrak{X}$ accretive operator $A$ is generator of contraction semigroup if the range of $\lambda \mathbb{I} + A$ is $\mathfrak{X}$ for some $\lambda>0$. Now we prove a series of estimates indispensable throughout this paper. Lemma 2.2. Let $U(t)$ be a bounded semigroup with boundedly invertible generator $A$. Then, for all $t\geqslant 0$ and for any $n\in\mathbb{N}$, Proof. We proceed by induction, and we first prove that
$$
\begin{equation}
(U(t)-\mathbb{I})x = -\int_0^t d\tau \, U(\tau)Ax, \qquad x\in\operatorname{dom}(A).
\end{equation}
\tag{2.3}
$$
Note that for any $\epsilon>0$ the semigroup properties yields the representation
$$
\begin{equation*}
\begin{aligned} \, \int_0^t ds \, U(s){U(\epsilon)-\frac{\mathbb{I}}{\epsilon}} &= \int_0^t ds \, {U(s+\epsilon) - \frac{U(s)}{\epsilon}} \\ &= \int_t^{t+\epsilon}ds \, \frac{U(s)}{\epsilon} - \int_0^\epsilon ds \, \frac{U(s)}{\epsilon} = (U(t)-\mathbb{I}) \, \frac{1}{\epsilon} \int_0^\epsilon ds \, U(s). \end{aligned}
\end{equation*}
\notag
$$
Moreover, one also gets
$$
\begin{equation*}
\begin{gathered} \, \lim_{\epsilon\to 0} \frac{1}{\epsilon} \int_0^\epsilon ds \, U(s)x = x, \qquad x\in{\mathfrak{X}}, \\ \lim_{\epsilon\to 0} \frac{U(\epsilon)-\mathbb{I}}{\epsilon} \, x = - Ax, \qquad x\in\operatorname{dom}(A). \end{gathered}
\end{equation*}
\notag
$$
This proves (2.3), and since $A$ is boundedly invertible, we obtain (2.1) for $n=0$. Furthermore, since $U(t)$ is bounded by $C_A$, we prove (2.2) for $n=0$.
Suppose that (2.1) and (2.2) are true for some $n$, then a straightforward calculation leads to (2.1) for $n+1$. Hence, using the representation (2.1) and (2.2) for $n$ to estimate the integrand, we obtain (2.2) for $n+1$, which completes the proof by induction. $\Box$ Similarly, we obtain a representation for a restricted development of $(\mathbb{I} + A)^{-1}$. Lemma 2.3. Let $A$ be as in Lemma 2.2. Then, for any $n\geqslant 0$, Proof. For $n=0$, the representation (2.4) follows from the resolvent formula
$$
\begin{equation}
(\mathbb{I} + A)^{-1} - A^{-1} = -(\mathbb{I} + A)^{-1}A^{-1}.
\end{equation}
\tag{2.5}
$$
Suppose that (2.4) holds for an integer $n > 1$, then
$$
\begin{equation}
(\mathbb{I} + A)^{-1}A^{-n-2} = \biggl(\sum_{k=0}^n (-A)^k \biggr)A^{-n-2} + (-1)^{n+1}(\mathbb{I} + A)^{-1}A^{-1}.
\end{equation}
\tag{2.6}
$$
Applying (2.5) to the last term of (2.6), we get the representation (2.4) for $n+1$, and thus for any $n$ by induction. $\Box$ Proof. By Lemma 2.2,
$$
\begin{equation}
\biggl\|(U(t)-\mathbb{I}+ tA)\frac{1}{t^2} A^{-2}\biggr\| \leqslant\frac{C_A}{2}.
\end{equation}
\tag{2.8}
$$
On the other hand, by Lemma 2.3, we have
$$
\begin{equation}
\biggl\|\bigl((\mathbb{I} + tA)^{-1} - \mathbb{I} + tA \bigr) \, \frac{1}{t^2} \, A^{-2}\biggr\| = \|(\mathbb{I} + tA)^{-1}\| \leqslant C_A.
\end{equation}
\tag{2.9}
$$
Here, the last estimate follows from $(\mathbb{I} + tA)^{-1} = (1/t)R_{A}(-1/t)$ and $\|R_{A}(-\lambda)\|\leqslant C_A /(\lambda+\delta)$, $\delta\geqslant 0$, which is valid for bounded semigroups with boundedly invertible generators. Hence (2.7) follows from (2.8) and (2.9). $\Box$ 2.2. Holomorphic contraction semigroups on $\mathfrak{X}$Now let $U\colon z \mapsto U(z)$ be a family of operators with $z$ taking their values in the sector of the complex plane
$$
\begin{equation}
S_\theta = \{ z\in\mathbb{C}\colon \ z\neq 0 \text{ and } |{\arg(z)}|<\theta\},
\end{equation}
\tag{2.10}
$$
where $0<\theta\leqslant\pi/2$. Definition 2.5. Recall that the family of operators $\{U(z)\}_{z\in S_\theta}$ is a bounded holomorphic semigroup of semi-angle $\theta\in (0, \pi/2]$ on a Banach space $\mathfrak{X}$ if it satisfies the following conditions: (i) if $0<\epsilon<\theta$, then $\|U(z)\|\leqslant M_\epsilon$ for all $z\in S_{\theta-\epsilon}$ and some $M_\epsilon<\infty$; (ii) $U(z_1)U(z_2) = U(z_1+z_2)$ for all $z_1,z_2\in S_\theta$; (iii) $U\colon z \mapsto U(z)$ is analytic function of $z\in S_\theta$; (iv) if $x\in{\mathfrak{X}}$ and $0<\epsilon<\theta$, then $\lim_{z\to 0}U(z)x = x$ provided $z \in S_{\theta-\epsilon}$. Let $\sigma(A)=\mathbb{C}\setminus\rho(A)$ denote the spectrum of $A$. It can be used for the following characterisation of the holomorphic semigroup generators (see Chap. IX in [1]). Proposition 2.6. An operator $A$ in a Banach space $\mathfrak{X}$ is the generator of a bounded holomorphic semigroup of semi-angle $\theta \in (0, \pi/2]$ if and only if $A$ is a closed operator with a dense domain $\operatorname{dom}(A)$ such that and where $0< \epsilon < \theta$. For applications the following property is also useful, which is an alternative characterisation of the semigroups of this kind. Proposition 2.7. If $U(z)$ is a bounded holomorphic semigroup of semi-angle $\theta$ with generator $A$, then $U(z){\mathfrak{X}}\subseteq\operatorname{dom}(A^n)$ for all $z\in S_{\theta}$ and $n\in\mathbb{N}$. Moreover, there are positive constants $C_A'$, $C_A^{(n)}$ such that, for $t>0$, Remark 2.8. Similarly to strongly continuous semigroups, a family $U(z)$, $z\,{\in}\,S_\theta$, is called a quasi-bounded holomorphic semigroup of semi-angle $\theta$ if there exists a constant $\beta >0$ such that restriction of $\{\mathrm{e}^{-\beta z} U(z)\}_{z\in S_\theta}$ to $\mathbb{R}^+_0$ is a bounded $C_0$-semigroup. The class of semigroups that we consider here is restricted to holomorphic contraction semigroups. Definition 2.9. We say that $\{U_A(z)\}_{z\in S_\theta}$ is a holomorphic contraction semigroup with generator $A \in \mathscr{H}_{\mathrm{c}}(\theta,0)$ if its restriction $\{U_A(t)\}_{t \geqslant 0}$ to $\mathbb{R}^+_0$ is a contraction $C_0$-semigroup, that is $\mathscr{H}_{\mathrm{c}}(\theta, 0):= \mathscr{H}(\theta, 0) \cap \mathscr{G}(1,0)$. Note that this class of semigroups is not empty and the corresponding generators have the following properties. (i) Let $\{U_A(t)\}_{t \geqslant 0}$ be a contraction semigroup with generator $A\in \mathscr{G}(1,0)$ in a Banach space $\mathfrak{X}$ such that $U_A(t)\mathfrak{X}\subseteq\operatorname{dom}(A)$ for $t>0$. If $\|AU(t)\|\leqslant M_1 t^{-1}$ for some $M_1 > 0$ and all $t>0$, then there exists $\theta = \arcsin (e M_1)^{-1}$ ($< \pi/2)$ such that $U_A(t)$ may be analytically continued to contraction holomorphic semigroup of semi-angle $\theta$. (ii) Let $A$ be a sectorial operator in a Hilbert space $\mathfrak{H}$, that is, its numerical range $W= \{(Au,u)\colon u\in\operatorname{dom}(A) \text{ and } \|u\|=1\}\subset S_{\pi/2-\theta}$ for $0<\theta\leqslant \pi/2$. If the operator $A$ is closed, then it is generator of the holomorphic contraction semigroup of semi-angle $\theta$. (iii) Let $A$ be a generator of holomorphic semigroup on a Banach space $\mathfrak{X}$. If $A$ is accretive, then $A$ generates a holomorphic contraction semigroup. (iv) If $A$ is the generator of a strongly continuous group $\{U_A(t)\}_{t \in \mathbb{R}}$ of contractions $\|U_A(t)\| \leqslant 1$, then $\pm A\in \mathscr{G}(1,0)$, and $A^2 \in \mathscr{H}_{\mathrm{c}}(\pi/2,0)$, see Corollary II.4.9 in [2]. 2.3. Fractional powers of generatorsWe scrutinise in this section some properties of fractional powers of the generators for bounded semigroups in a Banach space, see, for example, Chap. IX in [3]. Recall that fractional power $A^{\alpha}$, $0<\alpha<1$, of generator of a bounded $C_0$-semigroup $U(t)$ ($\|U(t)\| \leqslant C_A$) can be expressed by the integral (when it is well-defined)
$$
\begin{equation}
A^{\alpha}x = \frac1{\Gamma(-\alpha)}\int_0^\infty d\lambda \, \lambda^{-\alpha-1} (U(\lambda) -\mathbb{I})x, \qquad x\in \operatorname{dom}(A),
\end{equation}
\tag{2.12}
$$
where $\Gamma(\,{\cdot}\,)$ is the Gamma-function and $\lambda^{\alpha}$ is chosen to be positive for $\lambda>0$. Since for any $x\in\operatorname{dom}(A)$ and $0 < \alpha \leqslant 1$, the integral (2.12) is convergent, $\operatorname{dom}(A)\subseteq\operatorname{dom}(A^\alpha)$. We set $A^0:=\mathbb{I}$ and define $A^\alpha=A^{\alpha-[\alpha]}A^{[\alpha]}$ for any $\alpha>0$, where $[\alpha]$ denotes the integer part of $\alpha$. Proposition 2.10. For each $\alpha\in [0,1]$, there exists a constant $C_{A,\alpha}$, depending only on $C_A$ and $\alpha$, such that for all $\mu>0$, Proof. For $\alpha=0$ or $\alpha=1$, the result follows directly from the estimate of the resolvent. Let $0<\alpha<1$ and $x\in{\mathfrak{X}}$. Note that $\operatorname{ran}(A+\mu \mathbb{I})^{-1} = \operatorname{dom}(A) \subseteq \operatorname{dom}(A^\alpha)$. Then
$$
\begin{equation}
A^{\alpha}(A+\mu \mathbb{I})^{-1} x = \frac{1}{\Gamma(-\alpha)}\int_0^\infty d\lambda \, \lambda^{-\alpha-1} (U(\lambda) - \mathbb{I}))(A + \mu \mathbb{I})^{-1} x.
\end{equation}
\tag{2.14}
$$
We divide the integral (2.14) into two parts: $0<\lambda\leqslant\mu^{-1}$ and $\lambda>\mu^{-1}$, and we use the representation (2.3):
$$
\begin{equation*}
\begin{aligned} \, A^{\alpha}(A+\mu \mathbb{I})^{-1}x &= \frac1{\Gamma(-\alpha)}\int_0^{\mu^{-1}} d\lambda \, \lambda^{-\alpha-1} \int_0^\lambda dt \, (-U(t)) (\mathbb{I}-\mu(A+\mu \mathbb{I})^{-1})x \\ &\qquad + \frac1{\Gamma(-\alpha)}\int_{\mu^{-1}}^\infty d\lambda \, \lambda^{-\alpha-1}(U(\lambda) - \mathbb{I})(A + \mu \mathbb{I})^{-1}x. \end{aligned}
\end{equation*}
\notag
$$
Now by the estimate of the resolvent $\|(A+\mu )^{-1}\|\leqslant C_A/\mu$ for all $\mu>0$, we have
$$
\begin{equation*}
\begin{aligned} \, \|A^{\alpha}(A+\mu \mathbb{I})^{-1}x\| & \leqslant \frac{C_A(1+C_A)\ \|x\|}{\Gamma(-\alpha)} \biggl(\int_0^{\mu^{-1}} d\lambda \, \lambda^{-\alpha} + \frac1{\mu}\int_{\mu^{-1}}^\infty d\lambda \, \lambda^{-\alpha-1} \biggr) \\ &\leqslant \frac{C_A(1+C_A)\mu^{\alpha-1}}{\alpha(1-\alpha)\Gamma(-\alpha)} \|x\|. \end{aligned}
\end{equation*}
\notag
$$
Setting $C_{A,\alpha}:= C_A(1+C_A)/(\alpha(1-\alpha)\Gamma(-\alpha))$, we obtain estimate (2.13). $\Box$ Next we recall the following well-known property of the semigroup generator $A$. Lemma 2.11. $\operatorname{dom}((A+\delta \mathbb{I})^\alpha) = \operatorname{dom}(A^\alpha)$ for all $\delta>0$ and $0<\alpha<1$. Theorem 2.12. Let $U_A(t)$ be a bounded holomorphic semigroup with generator $A$. Then, for any real $\alpha>0$, Proof. Let $0<\alpha<1$. Since $\operatorname{dom}(A)\subseteq \operatorname{dom}(A^\alpha)$, we have $\operatorname{dom}(A^\alpha U_A(t)) = {\mathfrak{X}}$. Hence by (2.12)
Now we split the integral (2.16) in two parts: $0< \lambda \leqslant t$ and $\lambda>t$, and we use the estimate of the derivative of the holomorphic semigroup (see Proposition 2.7) to obtain that
$$
\begin{equation}
\|U_A(t+\lambda) - U_A(t)\| \leqslant \lambda\sup_{t\leqslant\tau\leqslant t+\lambda} \| \partial_{\tau}U_A(\tau)\| \leqslant \lambda \frac{C_A'}{t}.
\end{equation}
\tag{2.17}
$$
This leads to the estimate
$$
\begin{equation*}
\begin{aligned} \, \|A^\alpha U_A(t)\| &\leqslant \frac1{\Gamma(-\alpha)}\biggl(\int_0^t d\lambda \, \lambda^{-\alpha}\,\frac{C_A'}{t} + \int_t^\infty d\lambda \, 2C_A\lambda^{-\alpha-1}\biggr) \\ &\leqslant \frac{t^{-\alpha}}{\Gamma(-\alpha)}\biggl(\frac{C_A'}{1-\alpha} + \frac{2C_A}{\alpha}\biggr). \end{aligned}
\end{equation*}
\notag
$$
Now we get (2.15) for $0\,{<}\,\alpha\,{<}\,1$ by setting $M_\alpha:= \Gamma(-\alpha)^{-1}({C_A'/(1-\alpha)}+2C_A/\alpha)$.
For integer powers $\alpha$, (2.15) follows directly from Proposition 2.7. Notice that by Proposition 2.7, $\operatorname{ran}(U_A(t))\subseteq \operatorname{dom}(A^n)$ for $t>0$. Now (2.15) follows for any non integer $\alpha>1$ from the observation that $\operatorname{dom}(A^\alpha = A^{\alpha-[\alpha]}A^{[\alpha]}) \supset \operatorname{dom}(A^{[\alpha]+1})$, representation (2.16), and estimate (2.11) of derivatives of order $[\alpha]+1$. $\Box$ § 3. Operator-norm Trotter product formula3.1. Perturbation of holomorphic contraction semigroupsIn this section we prove the operator-norm convergence of the Trotter product formula in a Banach space $\mathfrak{X}$. Note that convergence of the abstract version of this formula in the strong operator topology is known since [4]. The proof goes via estimate of the rate of convergence in the case of small perturbations of the holomorphic contraction semigroups, cf. [5]. We require that operator $A$ generates a holomorphic contraction semigroup, and that perturbation $B$ satisfies the following hypothesis: (H1) $B$ is generator of a contraction semigroup on $\mathfrak{X}$; (H2) there is a real $\alpha\in[0,1)$ such that $\operatorname{dom}(A^{\alpha})\subseteq\operatorname{dom}(B)$ and that $\operatorname{dom}(A^*)\subseteq\operatorname{dom}(B^*)$ adjoint operator in the dual space $\mathfrak{X}^*$. Notice that we can suppose the operator $A$ boundedly invertible. If it is not the case, one considers $A+\eta$ for some $\eta>0$. Then, by Lemma 2.11, we have $\operatorname{dom}((A+\eta)^\alpha) = \operatorname{dom}(A^\alpha) \subseteq \operatorname{dom}(B)$. Remark 3.1. We note that assumption (H2) implies that $B$ is relatively bounded with respect to $A$ with the relative bound equals to zero. Indeed, for $\eta>0$, since $\operatorname{dom}(A+\eta \mathbb{I})\subseteq\operatorname{dom}(A^{\alpha})\subseteq\operatorname{dom}(B)$ and by Proposition 2.10, (2.13), it follows that (here, the operator $A$ is supposed to be boundedly invertible) For this class of (small) perturbations of holomorphic contraction semigroup with generator $A \in \mathscr{H}_{\mathrm{c}}(\theta, 0)$ (Definition 2.9) we have the following result. Lemma 3.2. Let $\{\mathrm{e}^{-zA}\}_{z\in S_\theta}$ be a holomorphic contraction semigroup of semi-angle $\theta$ on $\mathfrak{X}$ and perturbation $B$ satisfy the hypothesis (H1) and (H2). Then the algebraic sum $A+B$ of operators defined on $\operatorname{dom}(A+B) = \operatorname{dom}(A)$ is also a generator of holomorphic contraction semigroup with the same semi-angle, that is, $(A + B) \in \mathscr{H}_{\mathrm{c}}(\theta, 0)$. Proof. To this end we verify conditions of Proposition 2.6. Let $\epsilon \in (0,\theta)$. Then, by (3.2), we obtain inequality
$$
\begin{equation*}
\|B(A+z \mathbb{I})^{-1}\| \leqslant \frac{C_\alpha \|BA^{-\alpha}\|}{\eta^{1-\alpha}} \|A(A+z \mathbb{I})^{-1}\| + \eta^\alpha C_\alpha \|BA^{-\alpha}\| \, \|(A+z \mathbb{I})^{-1}\|,
\end{equation*}
\notag
$$
for $|{\arg(z)}|<\theta + \pi/2 -\epsilon$. Seeing that $\|(z\mathbb{I} +A)^{-1}\|\leqslant N_\epsilon |z|^{-1}$ for $N_\epsilon > 0$ and $z \in S_{\theta + \pi/2 -\epsilon}$, this inequality leads to
$$
\begin{equation}
\|B(A+z \mathbb{I})^{-1}\| \leqslant \frac{C_\alpha\|BA^{-\alpha}\|}{\eta^{1-\alpha}} (1+N_\epsilon) + \eta^\alpha C_\alpha \|BA^{-\alpha}\| \frac{N_\epsilon}{|z|},
\end{equation}
\tag{3.3}
$$
for $z \in S_{\theta + \pi/2 -\epsilon}$. Therefore, the Neumann series for $(A+B+z \mathbb{I})^{-1}$ converges if the right-hand side of (3.3) is smaller than $1$. Since we can choose $\eta$ and $z \in S_{\theta + \pi/2 -\epsilon}$ such that the right-hand side of the estimate (3.3) becomes smaller than $1$, we have
$$
\begin{equation*}
\|(A+B+z \mathbb{I})^{-1}\| \leqslant \frac{M}{|z-\gamma|}.
\end{equation*}
\notag
$$
Here, $M$ and $\gamma$ are some positive constants. Then, by Proposition 2.6, we conclude that the operator $(A+B) \in \mathscr{H}(\theta, \gamma)$, that is, it generates a quasi-bounded holomorphic semigroup $\{U_{A+B}(z)\}_{z\in S_{\theta}}$.
On the other hand, the conditions of lemma imply that $A$ and $B$ are accretive, thus operator $A+B$ is also accretive. Since, for $\lambda <0$, and $|\lambda|$ sufficiently large ($|\lambda|>\gamma$), the point $\lambda$ is in the resolvent set $\rho(A+B)$, we conclude that $(A+B)\in \mathscr{G}(1,0)$ generates a contraction semigroup. Since by $(A+B) \in \mathscr{H}(\theta, \gamma)$ this semigroup is also holomorphic, one finally obtains the assertion. $\Box$ The proof of the main theorem of this section involves three technical lemmata. For the two of them we need only that $B$ ($B^*$) are $A$($A^*$)-bounded in the Kato sense, that is, there are positive constants $a$ and $b$ such that
$$
\begin{equation}
x \in\operatorname{dom}(A)\subseteq\operatorname{dom}(B), \qquad \|Bx\| \leqslant a\|Ax\| + b\|x\|,
\end{equation}
\tag{3.4}
$$
$$
\begin{equation}
\phi \in\operatorname{dom}(A^*)\subseteq\operatorname{dom}(B^*), \qquad \|B^*\phi\| \leqslant a\|A^*\phi\| + b\|\phi\|.
\end{equation}
\tag{3.5}
$$
If $A$ is boundedly invertible, then we can put $b=0$ with the relative bound $a + b\|A^{-1}\|$ instead of $a$. Lemma 3.3. Let boundedly invertible $A$ and operator $B$ be generators of bounded semigroups. Let $B$ and $B^*$ verify (3.4), (3.5) and suppose that operator $H=(A\,{+}\,B)$ with $\operatorname{dom}(H)=\operatorname{dom}(A)$ is the boundedly invertible generator of a bounded semigroup. Then there exists a constant $L_1$ such that, for all $\tau\geqslant 0$, Proof. By virtue of the identity
$$
\begin{equation*}
\begin{aligned} \, A^{-1}\bigl(\mathrm{e}^{-\tau B}\mathrm{e}^{-\tau A} - \mathrm{e}^{-\tau(A+B)}\bigr) &= A^{-1}(\mathrm{e}^{-\tau B} - \mathbb{I})\mathrm{e}^{-\tau A} \\ &\qquad +A^{-1}(\mathrm{e}^{-\tau A}-\mathbb{I}) + A^{-1}HH^{-1}(\mathbb{I} - \mathrm{e}^{-\tau H}) \end{aligned}
\end{equation*}
\notag
$$
and by Lemma 2.2, we obtain (3.6):
$$
\begin{equation*}
\begin{aligned} \, \|A^{-1}(\mathrm{e}^{-\tau B}\mathrm{e}^{-\tau A}-\mathrm{e}^{-\tau H})\| &\leqslant \biggl\|\int_0^\tau ds \, A^{-1}B \mathrm{e}^{-sB}\biggr\| + \|A^{-1}(\mathrm{e}^{-\tau A}-\mathbb{I})\| \\ &\qquad + \|A^{-1}H\|\, \|H^{-1}(\mathbb{I} - \mathrm{e}^{-\tau H})\| \\ &\leqslant \|A^{-1}B\|C_B\tau + C_A\tau + \|A^{-1}H\|C_H\tau, \end{aligned}
\end{equation*}
\notag
$$
where the coefficients $C_B$ and $C_H$ are defined similarly to $C_A$ in § 2.1.
Finally, we remark that (3.5) implies the boundedness of the closed operator $A^{-1}B$, and that $\|A^{-1}H\|\leqslant \|\mathbb{I}+ A^{-1}B\|\leqslant 1+a + b\|A^{-1}\|$. To prove (3.7), one has to use (3.4), and the same line of reasoning as above to put finally $L_1 =C_B a' + C_A + C_H (1+a')$, where $a'=a+b\|A^{-1}\|$. $\Box$ Lemma 3.4. Let $A$, $B$, and $H=A+B$ be the same as in Lemma 3.3. Then there exists a constant $L_2$ such that, for all $\tau\geqslant 0$, Proof. By virtue of
$$
\begin{equation*}
\begin{aligned} \, \mathrm{e}^{-\tau B}\mathrm{e}^{-\tau A}-\mathrm{e}^{-\tau H} &= (\mathbb{I}-\mathrm{e}^{-\tau B})(\mathbb{I}-\mathrm{e}^{-\tau A}) + \bigl(\mathrm{e}^{-\tau A} - (\mathbb{I} + \tau A)^{-1}\bigr) \\ &\qquad + \bigl(\mathrm{e}^{-\tau B} - (\mathbb{I} + \tau B)^{-1}\bigr) + \bigl((\mathbb{I} + \tau H)^{-1} - \mathrm{e}^{-\tau H}\bigr) \\ &\qquad + \tau H(\mathbb{I} + \tau H)^{-1} - \tau A(\mathbb{I} + \tau A)^{-1} - \tau B(\mathbb{I} + \tau B)^{-1} \end{aligned}
\end{equation*}
\notag
$$
and by the identity
$$
\begin{equation*}
\begin{aligned} \, &A^{-1}\bigl(\tau H(\mathbb{I} + \tau H)^{-1} - \tau A(\mathbb{I} + \tau A)^{-1} - \tau B(\mathbb{I} + \tau B)^{-1}\bigr)A^{-1} \\ &\qquad =\tau^2\bigl((\mathbb{I} +\tau A)^{-1} + A^{-1}B(\mathbb{I} +\tau B)^{-1}BA^{-1} - A^{-1}H(\mathbb{I} + \tau H)^{-1}HA^{-1}\bigr), \end{aligned}
\end{equation*}
\notag
$$
we obtain the representation
$$
\begin{equation*}
\begin{aligned} \, &A^{-1}(\mathrm{e}^{-\tau B}\mathrm{e}^{-\tau A}-\mathrm{e}^{-\tau H})A^{-1} = A^{-1}(\mathbb{I} - \mathrm{e}^{-\tau B})(\mathbb{I} - \mathrm{e}^{-\tau A})A^{-1} \\ &\qquad + \bigl(\mathrm{e}^{-\tau A} - (\mathbb{I} + \tau A)^{-1}\bigr)A^{-2} + A^{-1}\bigl(\mathrm{e}^{-\tau B} - (\mathbb{I} +\tau B)^{-1}\bigr)A^{-1} \\ &\qquad+A^{-1}H\bigl((\mathbb{I} + \tau H)^{-1} - \mathrm{e}^{-\tau H}\bigr)H^{-2}HA^{-1} +\tau^2(\mathbb{I} + \tau A)^{-1} \\ &\qquad+ \tau^2A^{-1}B(\mathbb{I} + \tau B)^{-1}BA^{-1} - \tau^2A^{-1}H(\mathbb{I} + \tau H)^{-1}HA^{-1}. \end{aligned}
\end{equation*}
\notag
$$
This presentation yields the estimate
$$
\begin{equation*}
\begin{aligned} \, &\frac{1}{\tau^2}\|A^{-1}(\mathrm{e}^{-\tau B}\mathrm{e}^{-\tau A}-\mathrm{e}^{-\tau H})A^{-1}\| \leqslant \frac{1}{\tau} \|A^{-1}B\|\biggl\|\int_0^\tau ds \, \mathrm{e}^{-sB}\biggr\| \, \frac{1}{\tau}\, \|(\mathbb{I} - \mathrm{e}^{-\tau A})A^{-1}\| \\ &\qquad + \frac{1}{\tau^2} \bigl\|\bigl(\mathrm{e}^{-\tau A} - (\mathbb{I} + \tau A)^{-1}\bigr)A^{-2}\bigr\| \\ &\qquad + \frac{1}{\tau^2} \|A^{-1}B\|\biggl\|\int_0^\tau ds_1 \int_0^{s_1} ds_2 \, \mathrm{e}^{-s_2 B} - \tau^2(\mathbb{I}+\tau B)^{-1}\biggr\| \|BA^{-1}\| \\ &\qquad + \frac{1}{\tau^2} \|A^{-1}H\|\bigl\|\bigl((\mathbb{I} + \tau H)^{-1} - \mathrm{e}^{-\tau H}\bigr)H^{-2}\bigr\| \|HA^{-1}\| \\ &\qquad + 1 + \|A^{-1}B\|\, \|BA^{-1}\| + \|A^{-1}H\|\, \|HA^{-1}\|. \end{aligned}
\end{equation*}
\notag
$$
Now Lemmas 2.2 and 2.4 together with (3.4), (3.5) imply (3.8), where we can take $L_2 = a'C_A C_B + {3C_A/ 2} + 3C_B a'^2/2 + 3C_H (1+a')^2/2 + 1 + a'^2 + (1+a')^2$ with $a'=a+b\|A^{-1}\|$. Similarly one obtains (3.9). $\Box$ Note that for the proof of the third lemma (Lemma 3.4) we do need conditions (H2), as well as requirement that semigroups are contractive. Lemma 3.5. Let $A$ be a boundedly invertible generator of holomorphic contraction semigroup. If $B$ is a generator of a contraction semigroup and there exists $\alpha\in [0,1)$ such that $\operatorname{dom}(A^{\alpha})\subseteq\operatorname{dom}(B)$, then, for any $k\geqslant 1$ and $\tau>0$, Proof. We start with the following chain of estimates:
$$
\begin{equation*}
\begin{aligned} \, \bigl\|(\mathrm{e}^{-\tau B}\mathrm{e}^{-\tau A})^k A\bigr\| &\leqslant \bigl\|\bigl((\mathrm{e}^{-\tau B}\mathrm{e}^{-\tau A})^k - \mathrm{e}^{-k\tau A}\bigr)A\bigr\| + \|\mathrm{e}^{-k\tau A}A\| \\ &\leqslant \biggl\|\sum_{j=0}^{k-1}(\mathrm{e}^{-\tau B}\mathrm{e}^{-\tau A})^{k-1-j} (\mathrm{e}^{-\tau B}-I)\mathrm{e}^{-\tau A} \mathrm{e}^{-j\tau A}A\biggr\| + \|\mathrm{e}^{-k\tau A}A\| \\ &\leqslant \sum_{j=0}^{k-1}\biggl\|\int_0^\tau ds \, \mathrm{e}^{-sB}BA^{-\alpha}\biggr\| \bigl\|A^{\alpha}\mathrm{e}^{-(j+1) \tau A}A\bigr\| + \|\mathrm{e}^{-k\tau A}A\|. \end{aligned}
\end{equation*}
\notag
$$
Notice that the second inequality holds in particular due to contractions of $\mathrm{e}^{-tA}$ and $\mathrm{e}^{-tB}$, and to equation (2.3) of Lemma 2.2. From the hypothesis $\operatorname{dom}(A^{\alpha}) \subseteq\operatorname{dom}(B)$ we deduce that $\|BA^{-\alpha}\|\leqslant d$, see Remark 3.1. By Propositions 2.7 and Theorem 2.12, we obtain, respectively,
$$
\begin{equation*}
\|\mathrm{e}^{-k\,\tau A}A\| \leqslant \frac{C_A'}{k\,\tau}\quad \text{and} \quad \bigl\|A^{1+\alpha}\mathrm{e}^{-(j+1)\tau A}\bigr\|\leqslant \frac{M_\alpha}{((j+1)\tau)^{1+\alpha}}.
\end{equation*}
\notag
$$
Therefore, we conclude that
$$
\begin{equation*}
\bigl\|(\mathrm{e}^{-\tau B}\mathrm{e}^{-\tau A})^k A\bigr\|\leqslant \frac{M_\alpha d}{\tau^{\alpha}} \sum_{j=0}^{k-1} \frac{1}{(j+1)^{1+\alpha}} + \frac{C_A'}{k\,\tau}.
\end{equation*}
\notag
$$
Since $\alpha>0$, this gives the announced result (3.10) with
$$
\begin{equation*}
L_3 = dM_\alpha \sum_{j=1}^{\infty}\biggl(\frac1{j}\biggr)^{1+\alpha},
\end{equation*}
\notag
$$
and (3.11) for $\alpha=0$ with $\widetilde{L}_3 = \|B\| C_A'$. $\Box$ Note that since $\operatorname{dom}(A^\alpha)\subseteq\operatorname{dom}(B)$ implies $\operatorname{dom}(A^{\alpha'}) \subseteq \operatorname{dom}(B)$ for $\alpha'\geqslant\alpha$, estimate (3.10) is valid in fact for any $\alpha'\geqslant \alpha$. 3.2. Convergence rateTheorem 3.6. Let $\{\mathrm{e}^{-zA}\}_{z \in S_\theta}$ be a holomorphic contraction semigroup, that is, $A \in \mathscr{H}_{\mathrm{c}}(\theta, 0)$. Let $B$ be a generator of a contraction semigroup. If there exists $\alpha\in [0,1)$ such that $\operatorname{dom}(A^{\alpha})\subseteq\operatorname{dom}(B)$ and $\operatorname{dom}(A^*)\subseteq\operatorname{dom}(B^*)$, then there are constants $M_1$, $M_2$, $\widetilde{M}_2$, $\eta>0$, such that, for any $t\geqslant 0$ and $n>2$, Proof. Since $B$ satisfies hypothesis (H1) and (H2), by Lemma 3.2 the operator $H = (A+B)$ is the generator of a holomorphic contraction semigroup. If the operator $A$ has no bounded inverse, we set $\widetilde{A}:=A + \eta$ and $\widetilde{H}:=\widetilde{A} + B$ for some $\eta >0$ (see Remark 3.1). Hence both operators are boundedly invertible. As indicated above, these changes of generators do not modify the domain inclusions. If we want to obtain $\|B\widetilde{A}^{-1}\|\,{<}\,1$, then, by the estimate (3.1), we have to choose a sufficiently large shift parameter $\eta > 0$. This gives us the estimate $\|\widetilde{A}\widetilde{H}^{-1}\| = \|(\mathbb{I} + B\widetilde{A}^{-1})^{-1}\| \leqslant {1/ (1-a)}$, where we set $a=\|B\widetilde{A}^{-1}\|$.
Now we put $\tau:= t/n$, $\widetilde{U}(t):=\mathrm{e}^{-t\widetilde{H}}$, and $\widetilde{T}(\tau):= \mathrm{e}^{-\tau B}\mathrm{e}^{-\tau\widetilde{A}}$. To estimate the left-hand side of (3.12), we use
$$
\begin{equation}
\bigl(\mathrm{e}^{-tB/n}\mathrm{e}^{-tA/n}\bigr)^n - \mathrm{e}^{-t(A + B)} = (\widetilde{T}^n(\tau) - \widetilde{U}^n(\tau))\mathrm{e}^{t\eta},
\end{equation}
\tag{3.14}
$$
and the telescopic identity
$$
\begin{equation*}
\begin{aligned} \, &\widetilde{T}(\tau)^n-\widetilde{U}(\tau)^n = \sum_{m=0}^{n-1}\widetilde{T}(\tau)^{n-m-1}(\widetilde{T}(\tau)-\widetilde{U}(\tau)) \widetilde{U}(\tau)^m \\ &\qquad= \widetilde{T}(\tau)^{n-1}\widetilde{A}\widetilde{A}^{-1}\bigl(\widetilde{T}(\tau) -\widetilde{U}(\tau)\bigr)+ \bigl(\widetilde{T}(\tau) -\widetilde{U}(\tau)\bigr)\widetilde{A}^{-1}\widetilde{A}\widetilde{H}^{-1}\widetilde{H} \widetilde{U}(\tau)^{n-1} \\ &\qquad\qquad +\sum_{m=1}^{n-2}\widetilde{T}(\tau)^{n-m-1}\widetilde{A}\widetilde{A}^{-1} \bigl(\widetilde{T}(\tau)-\widetilde{U}(\tau)\bigr)\widetilde{A}^{-1} \widetilde{A}\widetilde{H}^{-1}\widetilde{H}\widetilde{U}(\tau)^m, \end{aligned}
\end{equation*}
\notag
$$
which implies
$$
\begin{equation*}
\begin{aligned} \, &\bigl\|\widetilde{T}(\tau)^n-\widetilde{U}(\tau)^n\bigr\| \leqslant \bigl\|\widetilde{T}(\tau)^{n-1}\widetilde{A}\bigr\| \, \bigl\|\widetilde{A}^{-1}(\widetilde{T}(\tau)-\widetilde{U}(\tau))\bigr\| \\ &\qquad + \bigl\|\bigl(\widetilde{T}(\tau)-\widetilde{U}(\tau)\bigr)\widetilde{A}^{-1}\bigr\| \, \bigl\|\widetilde{A}\widetilde{H}^{-1}\bigr\| \, \bigl\|\widetilde{H}\widetilde{U}(\tau)^{n-1}\bigr\| \\ &\qquad + \sum_{m=1}^{n-2}\bigl\|\widetilde{T}(\tau)^{n-m-1}\widetilde{A}\bigr\| \, \bigl\|\widetilde{A}^{-1}\bigl(\widetilde{T}(\tau)-\widetilde{U}(\tau)\bigr) \widetilde{A}^{-1}\bigr\| \, \bigl\|\widetilde{A}\widetilde{H}^{-1}\bigr\| \, \bigl\|\widetilde{H}\widetilde{U}(\tau)^m\bigr\|. \end{aligned}
\end{equation*}
\notag
$$
Hence, by Lemmas 3.3–3.5 (it is at this point that we use the hypothesis of contraction), and by Proposition 2.7, we obtain the estimate
$$
\begin{equation}
\begin{aligned} \, &\bigl\|\widetilde{T}(\tau)^n-\widetilde{U}(\tau)^n\bigr\| \leqslant \biggl(\frac{L_3}{\tau^\alpha} + \frac{C_A'}{(n-1)\tau}\biggr)L_1\tau + \frac{L_1}{1-a}\, \frac{C_H'}{n-1} \nonumber \\ &\qquad\qquad +\sum_{m=1}^{n-2}\biggl(L_3 \tau^{1-\alpha} + \frac{C_A'}{n-m-1}\biggr)\frac{L_2}{1-a}\, \frac{C_H'}{m} \nonumber \\ &\qquad\leqslant L_3 L_1 \frac{t^{1-\alpha}}{n^{1-\alpha}} + \frac{L_1}{n-1}\biggl(C_A'+\frac{C_H'}{1-a}\biggr) + \frac{L_3 L_2 C_H'}{1-a}\, \frac{t^{1-\alpha}}{n^{1-\alpha}} \sum_{m=1}^{n-2}\frac{1}{m} \nonumber \\ &\qquad\qquad+ \frac{L_2 C_H' C_A'}{1-a} \sum_{m=1}^{n-2}\frac{1}{n-m-1}\,\frac{1}{m} \nonumber \\ &\qquad \leqslant L_3 L_1 \frac{t^{1-\alpha}}{n^{1-\alpha}} + \frac{L_1}{n-1}\biggl(C_A'+\frac{C_H'}{1-a}\biggr) \nonumber \\ &\qquad\qquad + 2\frac{L_3 L_2 C_H'}{1-a}\, t^{1-\alpha} \frac{\ln n}{n^{1-\alpha}} + 4\frac{L_2 C_H' C_A'}{1-a}\, \frac{\ln n}{n}. \end{aligned}
\end{equation}
\tag{3.15}
$$
Here, we used that
$$
\begin{equation*}
\sum_{m=1}^{n-1} \frac{1}{(n-m)m} = \frac{2}{n}\sum_{m=1}^{n-1} \frac{1}{m} \leqslant \frac{2}{n}\bigl(1 + \ln(n-1)\bigr) \leqslant 4\frac{\ln n}{n}.
\end{equation*}
\notag
$$
The estimate (3.15) and (3.14) imply the announced result (3.12) for $\alpha>0$, with
$$
\begin{equation*}
M_1=4L_1\biggl(C_A'+ \frac{C_H'}{1-a}\biggr) + 4\frac{L_2 C_H' C_A'}{1-a}\quad\text{and} \quad M_2 = 2L_3 L_1+2\frac{L_3 L_2 C_H'}{1-a}.
\end{equation*}
\notag
$$
In a similar way one obtains also an estimate for $\alpha=0$:
$$
\begin{equation*}
\begin{aligned} \, &\bigl\|\widetilde{T}(\tau)^n-\widetilde{U}(\tau)^n\bigr\| \leqslant \widetilde{L}_3\bigl(1+\ln(n-1)\bigr)L_1\frac{t}{n} + \frac{L_1 C_A'}{n-1} + \frac{L_1{C}_H'}{1-a}\, \frac{1}{n-1} \\ &\qquad + \sum_{m=1}^{n-2}\biggl(\widetilde{L}_3\frac{t}{n} \bigl(1+\ln(n-m-1)\bigr) + \frac{C_A'}{n-m-1}\biggr)\frac{L_2}{1-a}\, \frac{C_H'}{m}. \end{aligned}
\end{equation*}
\notag
$$
This estimate together with (3.14) yield (3.13) for $\Box$ Theorem 3.7. Let $\{\mathrm{e}^{-zA}\}_{z\in S_\theta}$ be a holomorphic contraction semigroup, that is, $A \in \mathscr{H}_{\mathrm{c}}(\theta, 0)$. Let $B$ be a generator of a contraction semigroup, and there exists $\alpha\in [0,1)$ such that $\operatorname{dom}((A^{\alpha})^*)\subseteq\operatorname{dom}(B^*)$ and $\operatorname{dom}(A) \subseteq\operatorname{dom}(B)$. If in addition $\operatorname{dom}(A^*)\subseteq\operatorname{dom}(B^*)$ (for the case, when the space $\mathfrak{X}$ is not reflexive), then there are constants $M_3$, $M_4$, and $\widetilde{M}_4$, $\eta>0$, such that, for any $t\geqslant 0$ and $n>2$, for any $t\geqslant 0$ and $n>2$. Proof. Let $\widetilde{F}(\tau):=\mathrm{e}^{-\tau \widetilde{A}}\mathrm{e}^{-\tau B}$. Then, by the same arguments as in the proof of Theorem 3.6, we have
$$
\begin{equation*}
\begin{aligned} \, &\widetilde{U}(\tau)^n-\widetilde{F}(\tau)^n = \sum_{m=0}^{n-1}\widetilde{U}(\tau)^{n-m-1}\bigl(\widetilde{U}(\tau)-\widetilde{F}(\tau)\bigr) \widetilde{F}(\tau)^m \\ &\qquad= \widetilde{U}(\tau)^{n-1}\widetilde{H}\widetilde{H}^{-1}\widetilde{A} \widetilde{A}^{-1}\bigl(\widetilde{U}(\tau)-\widetilde{F}(\tau)\bigr) + \bigl(\widetilde{U}(\tau)-\widetilde{F}(\tau)\bigr)\widetilde{A}^{-1}\widetilde{A} \widetilde{F}(\tau)^{n-1} \\ &\qquad\qquad + \sum_{m=1}^{n-2}\widetilde{U}(\tau)^{n-m-1}\widetilde{H}\widetilde{H}^{-1}\widetilde{A} \widetilde{A}^{-1}\bigl(\widetilde{U}(\tau)-\widetilde{F}(\tau)\bigr) \widetilde{A}^{-1}\widetilde{A}\widetilde{F}(\tau)^m. \end{aligned}
\end{equation*}
\notag
$$
Notice that the Lemmas 3.3 and 3.4 hold for $\widetilde{F}(\tau)$. By a simple modification of Lemma 3.5, where one uses $\|\widetilde{A}^{-\alpha}B\| = \|B^*(\widetilde{A}^{-\alpha})^*\|<\infty$, we find that
$$
\begin{equation*}
\begin{alignedat}{2} \bigl\|\widetilde{A}\bigl(\mathrm{e}^{-\tau \widetilde{A}}\mathrm{e}^{-\tau B}\bigr)^k\bigr\| &\leqslant \frac{L_4}{\tau^\alpha} + \frac{C_A'}{k\tau}, &\qquad \alpha&>0, \\ \bigl\|\widetilde{A}\bigl(\mathrm{e}^{-\tau \widetilde{A}}\mathrm{e}^{-\tau B}\bigr)^k\bigr\| &\leqslant \widetilde{L}_4(1+\ln k) + \frac{C_A'}{k\tau}, &\qquad \alpha&=0. \end{alignedat}
\end{equation*}
\notag
$$
These ingredients ensure the estimates (3.16) and (3.17). $\Box$ Corollary 3.8. Under the same conditions as in Theorem 3.6, we have the operator-norm convergence of the symmetrised Trotter formula, that is, there exists $M_5$, $M_6$, and $\widetilde{M}_6$, $\eta>0$, such that, for any $t\geqslant 0$ and $n>2$, Proof. Since Lemmas 3.3–3.5 can be easily extended to the symmetrized product $\mathrm{e}^{-\tau A/2}\mathrm{e}^{-\tau B}\mathrm{e}^{-\tau A/2}$, the proof of the Theorem 3.6 carries through verbatim to obtain (3.18) and (3.19). $\Box$ Remark 3.9. Seeing that in Theorems 3.6, 3.7 and in Corollary 3.8 the perturbation $B$ of dominating operator $A$ is either infinitesimally $A$-small, or simply bounded, the corresponding results in Banach space $\mathfrak{X}$ are weaker than those in Hilbert space $\mathfrak{H}$, see [6]–[9]. Recall that in [6], [7] the perturbation $B$ in $\mathfrak{H}$ is Kato-small with respect to operator $A$ for relative bound $b < 1$. The fractional condition (H2) in a Hilbert space $\mathfrak{H}$ was introduced in [10]. Note that in the both cases: $\mathfrak{X}$ and $\mathfrak{H}$, the dominating operator $A$ is supposed to be generator of a holomorphic semigroup. § 4. ExampleResuming Remark 3.9, the question arises: whether the Trotter product formula converges in the operator-norm topology if the condition on dominating generator $A \in \mathscr{H}_{\mathrm{c}}(\theta, 0)$ is relaxed to hypothesis that $A \in \mathscr{G}(1,0)$, that is, it is generator of a contraction (but not holomorphic!) semigroup and $B$ is a bounded generator? The aim of this section is to give an answer to this question using example of a certain class of generators and semigroups. It turns out that appropriate for this purpose is the class of generators of evolution semigroups. 4.1. Evolution semigroups for nACP and Trotter product formulaTo proceed further, we need some key notions from the evolution semigroups theory and in particular the notion of solution operator for non-autonomous Cauchy problem (nACP), see [11]. A strongly continuous mapping $U(\,{\cdot}\,,{\cdot}\,)\colon \Delta \to \mathcal{L}(\mathfrak{X})$, where the domain $\Delta:= \{(t,s)\colon 0 < s \leqslant t \leqslant T\}$ and $\mathcal{L}(\mathfrak{X})$ is the set of bounded operators on a separable Banach space $\mathfrak{X}$, is called a solution operator if the conditions
$$
\begin{equation*}
\begin{aligned} \, &\textrm{(i)}\ \sup_{(t,s)\in\Delta}\|U(t,s)\|_{\mathcal{L}(\mathfrak{X})} < \infty, \\ &\textrm{(ii)}\ U(t,s) = U(t,r)U(r,s),\qquad 0 < s \leqslant r \leqslant t \leqslant T, \end{aligned}
\end{equation*}
\notag
$$
are satisfied. Let us consider the Banach space $L^p(\mathcal{I},\mathfrak{X})$ for $\mathcal{I}:= [0,T]$ and $p \in [1,\infty)$. The operator $\mathcal{K}$ is an evolution generator of the evolution semigroup $\{\mathcal{U}(\tau):= \mathrm{e}^{-\tau \mathcal{K}}\}_{\tau\geqslant0}$ if there is a solution operator such that the Howland–Evans–Neidhardt representation (see [12]–[15])
$$
\begin{equation}
(\mathrm{e}^{-\tau \mathcal{K}}f)(t) = U(t,t-\tau)\chi_{\mathcal{I}}(t-\tau)f(t-\tau), \qquad f \in L^p(\mathcal{I},\mathfrak{X}),
\end{equation}
\tag{4.1}
$$
holds for almost all $t \in \mathcal{I}$ and $\tau \geqslant 0$. Seeing that on account of (4.1) the semigroup $\{\mathrm{e}^{-\tau \mathcal{K}}\}_{\tau\geqslant0}$ is nilpotent: $\mathrm{e}^{-\tau \mathcal{K}} f = 0$ for $\tau \geqslant T$, the evolution generator $\mathcal{K}$ can never be generator of a holomorphic semigroup. A simple example of an evolution generator is the differentiation operator (cf. [11])
$$
\begin{equation}
\begin{aligned} \, (D_0f)(t) &:= \partial_{t} f(t), \\ f \in \operatorname{dom}(D_0) &:= \{f \in W^{1,p}(\mathcal{I},\mathfrak{X})\colon f(0) = 0\}, \end{aligned}
\end{equation}
\tag{4.2}
$$
where $W^{1,p}(\mathcal{I},\mathfrak{X})$ is the Sobolev space of order $(1,p)$ of Bochner $p$-integrable functions. Now, by (4.2), one obviously gets the contraction shift semigroup
$$
\begin{equation}
(\mathrm{e}^{-\tau D_0}f)(t) = \chi_{\mathcal{I}}(t-\tau)f(t-\tau), \qquad f \in L^p(\mathcal{I},\mathfrak{X}),
\end{equation}
\tag{4.3}
$$
for almost all $t \in \mathcal{I}$ and $\tau \geqslant 0$. Hence (4.1) implies that the corresponding solution operator of the non-holomorphic evolution semigroup $\{\mathrm{e}^{-\tau D_0}\}_{\tau\geqslant0}$ is given by $U_{D_0}(t,s) = \mathbb{I}$ for all $(t,s) \in \Delta$. Below we consider the operator $\mathcal{K}_0:= \overline{D_0 + \mathcal{A}}$, where $\mathcal{A}$ is the multiplication operator induced by generator $A$ of a holomorphic contraction semigroup on $\mathfrak{X}$. More precisely,
$$
\begin{equation*}
\begin{gathered} \, (\mathcal{A} f)(t) := Af(t) \quad \text{and} \quad (\mathrm{e}^{-\tau \mathcal{A}} f)(t) = \mathrm{e}^{-\tau A} f(t), \\ f \in \operatorname{dom}(\mathcal{A}) := \{f \in L^p(\mathcal{I},\mathfrak{X})\colon Af \in L^p(\mathcal{I},\mathfrak{X})\}. \end{gathered}
\end{equation*}
\notag
$$
Now the perturbation of the shift semigroup (4.3) by $\mathcal{A}$ corresponds to the semigroup with generator $\mathcal{K}_0$. One easily checks that $\mathcal{K}_0$ is an evolution generator of a contraction semigroup on $L^p(\mathcal{I},\mathfrak{X})$, that is never holomorphic [11]. Indeed, since the generators $D_0$ and $\mathcal{A}$ commute, the representation (4.1) for evolution semigroup $\{\mathrm{e}^{-\tau \mathcal{K}_0}\}_{\tau\geqslant0}$ takes the form
$$
\begin{equation*}
(\mathrm{e}^{-\tau \mathcal K_0}f)(t) = \mathrm{e}^{-\tau A}\chi_{\mathcal{I}}(t-\tau)f(t-\tau), \qquad f \in L^p(\mathcal{I},\mathfrak{X}),
\end{equation*}
\notag
$$
for almost all $t \in \mathcal{I} = [0,T]$ and $\tau \geqslant 0$. Hence by (4.1), the solution operator $U_{0}(t,s) = \mathrm{e}^{-(t-s) A}$. Therefore, $\mathrm{e}^{-\tau \mathcal{K}_0} f = 0$ for $\tau \geqslant T$, that is, the semigroup $\{\mathrm{e}^{-\tau \mathcal{K}_0}\}_{\tau\geqslant0}$ is nilpotent. Furthermore, if $\{B(t)\}_{t \in \mathcal{I}}$ is a strongly measurable family of generators of contraction semigroups on $\mathfrak{X}$, that is, $B(\,{\cdot}\,)\colon \mathcal{I} \to \mathcal{G}(1,0)$, then the induced multiplication operator $\mathcal{B}$
$$
\begin{equation}
\begin{gathered} \, (\mathcal{B} f)(t) := B(t) f(t), \\ \begin{aligned} \, f \in \operatorname{dom}(\mathcal{B}) &:= \{f \in L^p(\mathcal{I},\mathfrak{X})\colon f(t) \in \operatorname{dom}(B(t)) \text{ for almost all }t \in \mathcal{I}, \\ &\qquad\quad B(t)f(t) \in L^p(\mathcal{I},\mathfrak{X})\}, \end{aligned} \nonumber \end{gathered}
\end{equation}
\tag{4.4}
$$
is a generator of a contraction semigroup on $L^p(\mathcal{I},\mathfrak{X})$. In the next § 4.2, we consider perturbation of the generator $\mathcal{K}_0$ by the multiplication operator $\mathcal{B}$ (see (4.4)). Thereupon we construct by means of the Trotter product formula approach a corresponding perturbed semigroup. Remark 4.1. We conclude by remarks concerning some notations and definitions that we use below throughout § 4. 1. For characterising the rate of convergence, we use, so-called, the Landau symbols:
$$
\begin{equation*}
\begin{alignedat}{3} g(n) &= O(f(n)) \quad &\Longleftrightarrow\quad &\limsup_{n\to\infty} \biggl|\frac{g(n)}{f(n)}\biggr| < \infty, \\ g(n) &= o(f(n)) \quad &\Longleftrightarrow\quad &\limsup_{n\to\infty} \biggl|\frac{g(n)}{f(n)}\biggr| = 0, \\ g(n) &= \Theta(f(n)) \quad &\Longleftrightarrow\quad &0 < \liminf_{n\to\infty} \biggl|\frac{g(n)}{f(n)}\biggr| \leqslant \limsup_{n\to\infty} \biggl|\frac{g(n)}{f(n)}\biggr| < \infty, \\ g(n) &= \omega(f(n)) \quad &\Longleftrightarrow\quad &\limsup_{n\to\infty} \biggl|\frac{g(n)}{f(n)}\biggr| = \infty. \end{alignedat}
\end{equation*}
\notag
$$
2. We also use the notation $C^{0,\beta}(\mathcal I):=\{f\colon \mathcal I\to \mathbb{C} \colon \text{there exists some } K>0$ such that $|f(x)-f(y)|\leqslant K |x-y|^\beta \text{ for any } x,y \in \mathcal I \text{ and } \beta \in (0,1]\}$. 3. Below we consider the Banach space $L^p(\mathcal{I},\mathfrak{X})$ for $\mathcal{I}:= [0, T]$ and $p \in [1,\infty)$. 4.2. Trotter product formulaWe reminisce (cf. § 4.1) that the semigroup $\{\mathcal{U}(\tau)\}_{\tau \geqslant 0}$, on the Banach space $L^p(\mathcal{I},\mathfrak{X})$ is called the evolution semigroup if there is a solution operator $\{U(t,s)\}_{(t,s)\in\Delta}$ such that representation (4.1) holds. Let $\mathcal{K}_0$ be the generator of an evolution semigroup $\{\mathcal{U}_0(\tau)\}_{\tau \geqslant 0}$ and let $\mathcal{B}$ be a multiplication operator induced by a measurable family $\{B(t)\}_{t\in\mathcal{I}}$ of generators of contraction semigroups. Note that in this case the multiplication operator $\mathcal{B}$ (4.4) is a generator of a contraction semigroup $(\mathrm{e}^{- \tau\mathcal{B}} f)(t) = \mathrm{e}^{- \tau B(t)} f(t)$, on the Banach space $L^p(\mathcal{I},\mathfrak{X})$. Since $\{\mathcal{U}_0(\tau)\}_{\tau \geqslant 0}$ is an evolution semigroup, then, by definition (4.1), there is a propagator $\{U_0(t,s)\}_{(t,s) \in \Delta}$ such that the representation
$$
\begin{equation*}
(\mathcal{U}_0(\tau)f)(t) = U_0(t,t-\tau) \chi_\mathcal{I}(t-\tau)f(t-\tau), \qquad f \in L^p(\mathcal{I},\mathfrak{X}),
\end{equation*}
\notag
$$
holds for almost all $t \in \mathcal{I}$ and $\tau \geqslant 0$. Next, we define $\tau_{n}:= (t-s)/n$, for $n \in \mathbb{N}$, and
$$
\begin{equation*}
G_j(t,s;n):= U_0\bigl(s + j\tau_{n}, \, s + (j-1)\tau_{n}\bigr) \mathrm{e}^{-\tau_{n}B(s+(j-1)\tau_{n})}, \qquad (t,s) \in \Delta,
\end{equation*}
\notag
$$
where $j \in \{1,2,\dots,n\}$, $n \in \mathbb{N}$, $(t,s) \in \Delta$, and we set
$$
\begin{equation*}
V_n(t,s):= \prod^{1}_{j=n}G_j(t,s;n), \qquad n \in \mathbb{N}, \quad (t,s) \in \Delta.
\end{equation*}
\notag
$$
That is, the product is increasingly ordered in $j$ from the right to the left. Now a straightforward computation shows that the representation
$$
\begin{equation}
\bigl(\bigl(\mathrm{e}^{-\tau \mathcal{K}_0/n}\mathrm{e}^{-\tau \mathcal{B}/n}\bigr)^n f\bigr)(t) = V_n(t,t-\tau)\chi_\mathcal{I}(t-\tau)f(t-\tau),
\end{equation}
\tag{4.5}
$$
$f \in L^p(\mathcal{I},\mathfrak{X})$, holds for each $\tau \geqslant 0$ and almost all $t \in \mathcal{I}$. Theorem 4.2. Let $\mathcal{K}$ and $\mathcal{K}_0$ be generators of evolution semigroups on the Banach space $L^p(\mathcal{I},\mathfrak{X})$ for some $p \in [1,\infty)$. Further, let $\{B(t)\in \mathcal{G}(1,0)\}_{t\in \mathcal{I}}$ be a strongly measurable family of generators of contraction semigroups on $\mathfrak{X}$. Then Proof. Let $\{L(\tau)\}_{\tau \geqslant 0}$ be the left-shift semigroup on the Banach space $L^p(\mathcal{I},\mathfrak{X})$: Using that, we obtain for $\tau \geqslant 0$ and almost all $t \in \mathcal{I}$.
It turns out that, for each $n \in \mathbb{N}$, the operator $L(\tau)(\mathrm{e}^{-\tau \mathcal{K}} - (\mathrm{e}^{-\tau/n \mathcal{K}_0}\mathrm{e}^{-\tau \mathcal{B}/n})^n)$ is a multiplication operator induced by $\{(U(t+\tau,t) - V_n(t+\tau,t)) \chi_\mathcal{I}(t+\tau)\}_{t\in\mathcal I}$. As a consequence,
$$
\begin{equation*}
\begin{aligned} \, &\bigl\|L(\tau)\bigl(\mathrm{e}^{-\tau \mathcal{K}} - \bigl(\mathrm{e}^{-\tau\mathcal{K}_0/n}\mathrm{e}^{-\tau \mathcal{B}/n}\bigr)^n\bigr)\bigr\|_{\mathcal{L}(L^p(\mathcal{I},\mathfrak{X}))} \\ &\qquad =\operatorname*{ess\,sup}_{t\in \mathcal{I}}\|U(t+\tau,t) - V_n(t+\tau,t)\|_{\mathcal{L}(\mathfrak{X})}\chi_\mathcal{I}(t+\tau) \end{aligned}
\end{equation*}
\notag
$$
for each $\tau \geqslant 0$. Note that
$$
\begin{equation*}
\begin{aligned} \, &\sup_{\tau \geqslant 0}\bigl\|L(\tau)\bigl(\mathrm{e}^{-\tau \mathcal{K}} - \bigl(\mathrm{e}^{-\tau\mathcal{K}_0/n}\mathrm{e}^{-\tau \mathcal{B}/n}\bigr)^n\bigr)\bigr\|_{\mathcal{L}(L^p(\mathcal{I},\mathfrak{X}))} \\ &\qquad= \operatorname*{ess\,sup}_{\tau \geqslant 0}\bigl\|L(\tau)\bigl(\mathrm{e}^{-\tau \mathcal{K}} - \bigl(\mathrm{e}^{-\tau\mathcal{K}_0/n}\mathrm{e}^{-\tau \mathcal{B}/n}\bigr)^n\bigr)\bigr\|_{\mathcal{L}(L^p(\mathcal{I},\mathfrak{X}))}. \end{aligned}
\end{equation*}
\notag
$$
This is based on the fact that if $F(\,{\cdot}\,)\colon \mathbb{R}^+ \to \mathcal{L}(\mathfrak{X})$ is strongly continuous, then $\sup_{\tau \geqslant 0}\|F(\tau)\|_{\mathcal{L}(\mathfrak{X})} = \operatorname*{ess\,sup}_{\tau \geqslant 0}\|F(\tau)\|_{\mathcal{L}(\mathfrak{X})}$. Hence, we find
$$
\begin{equation*}
\begin{aligned} \, &\sup_{\tau \geqslant 0}\bigl\|L(\tau)\bigl(\mathrm{e}^{-\tau \mathcal{K}} - \bigl(\mathrm{e}^{-\tau\mathcal{K}_0/n}\mathrm{e}^{-\tau \mathcal{B}/n}\bigr)^n\bigr)\bigr\|_{\mathcal{L}(L^p(\mathcal{I},\mathfrak{X}))} \\ &\qquad =\operatorname*{ess\,sup}_{\tau \geqslant 0}\, \operatorname*{ess\,sup}_{t\in \mathcal{I}}\|U(t+\tau,t) - V_n(t+\tau,t)\|_{\mathcal{L}(\mathfrak{X})} \chi_\mathcal{I}(t+\tau). \end{aligned}
\end{equation*}
\notag
$$
Further, if $\Phi(\,{\cdot}\,,{\cdot}\,)\colon \mathbb{R}^+ \times \mathcal{I} \to \mathcal{L}(\mathfrak{X})$ is a strongly measurable function, then
$$
\begin{equation*}
\operatorname*{ess\,sup}_{(\tau,t)\in\mathbb{R}^+\times \mathcal{I}}\|\Phi(\tau,t)\|_{\mathcal{L}(\mathfrak{X})} = \operatorname*{ess\,sup}_{\tau \geqslant 0} \, \operatorname*{ess\,sup}_{t\in\mathcal{I}} \|\Phi(\tau,t)\|_{\mathcal{L}(\mathfrak{X})}.
\end{equation*}
\notag
$$
Now, taking into account two last equalities, we find that
$$
\begin{equation*}
\begin{aligned} \, &\sup_{\tau \geqslant 0}\bigl\|L(\tau)\bigl(\mathrm{e}^{-\tau \mathcal{K}} - \bigl(\mathrm{e}^{-\tau\mathcal{K}_0/n}\mathrm{e}^{-\tau \mathcal{B}/n}\bigr)^n\bigr)\bigr\|_{\mathcal{L}(\mathfrak{X})} \\ &\qquad =\operatorname*{ess\,sup}_{(\tau,t)\in\mathbb{R}^+\times\mathcal{I}}\|U(t+\tau,t) - V_n(t+\tau,t)\|_{\mathcal{L}(\mathfrak{X})} \chi_\mathcal{I}(t+\tau) \\ &\qquad = \operatorname*{ess\,sup}_{(t,s)\in \Delta}\|U(t,s) - V_n(t,s)\|_{\mathcal{L}(\mathfrak{X})}, \end{aligned}
\end{equation*}
\notag
$$
which proves (4.6). $\Box$ We study bounded perturbations of the evolution generator $\mathcal{K}_0 = D_0$ (4.2). To this aim, we consider $\mathcal I =[0,1]$, $\mathfrak{X}= \mathbb{C}$ and denote by $L^p(\mathcal I)$ the Banach space $L^p(\mathcal I, \mathbb{C})$. For $t \in \mathcal I$, let $q\colon t \mapsto q(t) \in L^\infty(\mathcal I)$. Then $q$ induces a bounded multiplication operator $\mathcal{B} = Q$ on the Banach space $L^p(\mathcal I)$, For simplicity, we assume that $q\geqslant 0$. Hence $Q$ generates on $L^p(\mathcal I)$ a contraction semigroup $\{\mathrm{e}^{- \tau Q}\}_{\tau \geqslant 0}$. Since the generator $Q$ is bounded, the closed operator $\mathcal A:= D_0 + Q$, with domain $\operatorname{dom}(\mathcal A) = \operatorname{dom}(D_0)$ is a generator of a $C_0$-semigroup on $L^p(\mathcal I)$. By the Trotter product formula in the strong operator topology, it follows immediately that
$$
\begin{equation}
\lim_{n \to \infty}\bigl(\mathrm{e}^{-\tau D_0/n}\mathrm{e}^{-\tau Q/n}\bigr)^n f = \mathrm{e}^{-\tau (D_0+Q)}f,\qquad f\in L^p(\mathcal{I}),
\end{equation}
\tag{4.7}
$$
uniformly in $\tau\in[0,T]$ on bounded time intervals. Next, we define on the Banach space $\mathfrak{X} = \mathbb{C}$ a family of bounded operators $\{V(t)\}_{t\in\mathcal I}$ by
$$
\begin{equation*}
V(t): = \exp\biggl\{- \int_0^t ds \, q(s) \biggr\}, \qquad t\in\mathcal I.
\end{equation*}
\notag
$$
Note that, for almost every $t\in \mathcal I$, these operators are positive. Hence $V^{-1}(t)$ exists and it has the form
$$
\begin{equation*}
V^{-1}(t) = \exp\biggl\{ \int_0^t ds \, q(s) \biggr\}, \qquad t\in\mathcal I.
\end{equation*}
\notag
$$
The operator families $\{V(t)\}_{t\in\mathcal I}$ and $\{V^{-1}(t)\}_{t\in\mathcal I}$ induce two bounded multiplication operators $\mathcal V $ and $\mathcal V ^{-1}$ on $L^p(\mathcal I)$, respectively. Now the invertibility implies that $\mathcal V\mathcal V^{-1} = \mathcal V^{-1}\mathcal V = \mathbb{I} \big|_{L^p(\mathcal I)}$. Using the operator $\mathcal V$, one easily verifies that $D_0+Q$ is similar to $D_0$, that is,
$$
\begin{equation*}
\mathcal V^{-1}(D_0 + Q)\mathcal V= D_0\quad \text{or}\quad D_0 + Q= \mathcal VD_0 \mathcal V^{-1}.
\end{equation*}
\notag
$$
Hence the semigroup generated on $L^p(\mathcal I)$ by $D_0 + Q$ gets the explicit form
$$
\begin{equation}
\bigl(\mathrm{e}^{-\tau(D_0 + Q)}f\bigr)(t) = \bigl(\mathcal V\mathrm{e}^{-\tau D_0} \mathcal V^{-1} f\bigr)(t) = \exp\biggl\{-\int_{t-\tau}^t dy \, q(y)\biggr\} f(t-\tau) \chi_{\mathcal I}(t-\tau).
\end{equation}
\tag{4.8}
$$
Since by (4.1) the solution operator $U(t,s)$ corresponding to the evolution semigroup (4.8) is defined by the equation
$$
\begin{equation*}
\bigl(\mathrm{e}^{-\tau(D_0 + Q)}\bigr)f(t) = U(t, t-\tau) f(t-\tau)\chi_{\mathcal I}(t-\tau),
\end{equation*}
\notag
$$
we deduce that it is equal to Now we study the corresponding Trotter product formula. For a fixed $\tau \geqslant 0$ and $n\in \mathbb{N}$, we define approximates $\{V_n\}_{n\geqslant 1}$ by
$$
\begin{equation*}
\bigl(\bigl(\mathrm{e}^{- \tau D_0/n}\mathrm{e}^{- \tau Q/n} \bigr)^n f\bigr)(t) =: V_n(t,t-\tau)\chi_\mathcal{I}(t-\tau)f(t-\tau).
\end{equation*}
\notag
$$
Hence by straightforward calculations, which are similar to (4.5), we find that the approximants have the following explicit form
$$
\begin{equation*}
V_n(t,s)= \exp\biggl\{-\tau_n \sum_{k=0}^{n-1} q(s + k\tau_n)\biggr\}, \qquad (t,s) \in \Delta, \quad \tau_n = \frac{t-s}{n}, \quad n\in \mathbb{N}.
\end{equation*}
\notag
$$
Theorem 4.3. Let $q \in L^\infty(\mathcal{I})$ be non-negative. Then Proof. First, by Theorem 4.2 and by we obtain
$$
\begin{equation}
\begin{aligned} \, &\sup_{\tau \geqslant 0}\bigl\|\mathrm{e}^{-\tau(D_0 + Q)} - \bigl(\mathrm{e}^{-\tau D_0/n}\mathrm{e}^{-\tau Q/n}\bigr)^n\bigr\|_{\mathcal{L} (L^p(\mathcal I))} \nonumber \\ &\qquad =\operatorname*{ess\,sup}_{(t,s)\in \Delta}\biggl|\exp\biggl\{-\int^t_s dy \, q(y)\biggr\} - \exp\biggl\{- \tau_n \sum_{k=0}^{n-1} q(s + k\tau_n)\biggr\}\biggr|. \end{aligned}
\end{equation}
\tag{4.9}
$$
Next, using the inequality
$$
\begin{equation*}
\mathrm{e}^{-\max\{x,y\}} |x-y| \leqslant |\mathrm{e}^{-x} - \mathrm{e}^{-y}| \leqslant |x-y|, \qquad 0 \leqslant x, y,
\end{equation*}
\notag
$$
for $0 \leqslant s < t \leqslant 1$, we get the estimates
$$
\begin{equation*}
\mathrm{e}^{-\|q\|_{L^\infty}} R_n(t,s;q) \leqslant \biggl|\exp\biggl\{-\int^t_s dy \, q(y)\biggr\} - \exp\biggl\{- \tau_n \sum_{k=0}^{n-1} q(s + k\tau_n)\biggr\}\biggr| \leqslant R_n(t,s;q),
\end{equation*}
\notag
$$
where
$$
\begin{equation}
R_n(t,s,q):= \biggl|\int^t_s dy \, q(y) - \tau_n \sum_{k=0}^{n-1} q(s + k\tau_n)\biggr|, \qquad (t,s) \in \Delta, \quad n\in \mathbb{N}.
\end{equation}
\tag{4.10}
$$
Hence for the left-hand side of (4.9), we obtain the estimate where $R_n(q):= \operatorname*{ess\,sup}_{(t,s)\in \Delta}R_n(t,s;q)$, $n \in \mathbb{N}$. These estimates together with the definition of $\Theta$ prove the assertion. $\Box$ Note that by virtue of (4.10) and Theorem 4.3, the operator-norm convergence rate of the Trotter product formula for the pair $\{D_0, Q\}$ coincides with the convergence rate of the integral Darboux–Riemann sum approximation of the Lebesgue integral. 4.3. Rate of convergenceFirst, we consider the case of a real Hölder-continuous function $q\in C^{0,\beta}(\mathcal{I})$. Proof. For $\tau_n = (t-s)/n$ and $n\in \mathbb{N}$, we have
$$
\begin{equation*}
\int^t_s dy \, q(y) - \tau_n \sum^{n-1}_k q(s+k\tau_n)= \sum^{n-1}_{k=0}\int^{(k+1)\tau_n}_{k\tau_n} dy \, \bigl(q(s+y) - q(s+k\tau_n)\bigr),
\end{equation*}
\notag
$$
which yields the estimate
$$
\begin{equation*}
\biggl|\int^t_s dy \, q(y) - \tau_n \sum^{n-1}_k q(s + k\tau_n)\biggr| \leqslant\sum^{n-1}_{k=0}\int^{(k+1)\tau_n}_{{k}\tau_n}dy \, |q(s+y)-q(s+k\tau_n)|.
\end{equation*}
\notag
$$
Since $q \in C^{0,\beta}(\mathcal{I})$, there is a constant $L_\beta > 0$ such that, for $y\in[k \tau_n, (k+1) \tau_n]$,
$$
\begin{equation*}
|q(s+y) - q(s+ k\tau_n)| \leqslant L_\beta |y- k\tau_n|^\beta \leqslant L_\beta \frac{(t-s)^\beta}{n^\beta}.
\end{equation*}
\notag
$$
Hence
$$
\begin{equation*}
\biggl|\int^t_s dy \, q(y) - \tau_n \sum^{n-1}_k q(s + k\tau_n)\biggr| \leqslant L_\beta \frac{(t-s)^{1+\beta}}{n^\beta} \leqslant L_\beta \frac{1}{n^\beta},
\end{equation*}
\notag
$$
which proves (cf. Remark 4.1)
$$
\begin{equation*}
\operatorname*{ess\,sup}_{(t,s)\in \Delta}\biggl|\int^t_s dy \, q(y) - \tau_n \sum^{n-1}_k q(s + k\tau_n)\biggr| = O\biggl(\frac{1}{n^\beta}\biggr).
\end{equation*}
\notag
$$
Applying now Theorem 4.3, one completes the proof. $\Box$ The next natural question is as follows: what happens, when function $q$ is only continuous? Theorem 4.5. If $q\colon \mathcal I \to \mathbb{C}$ is continuous and non-negative, then as $n\to \infty$. Proof. Seeing that $q$ is continuous, for any $\varepsilon > 0$, there is $\delta > 0$ such that for $y,x \in \mathcal{I}$ and $|y-x| < \delta$ we have $|q(y) - q(x)| < \varepsilon$. Therefore, if $1/n < \delta$, then, for $y \in (k\tau_n, (k+1)\tau_n)$,
$$
\begin{equation*}
|q(s+y) - q(s+ k\tau_n)| < \varepsilon, \qquad (t,s) \in \Delta.
\end{equation*}
\notag
$$
Hence
$$
\begin{equation*}
\biggl|\int^t_s dy \, q(y) - \tau_n \sum^{n-1}_k q(s + k\tau_n)\biggr| \leqslant \varepsilon (t-s) \leqslant \varepsilon,
\end{equation*}
\notag
$$
which yields (cf. Remark 4.1)
$$
\begin{equation*}
\operatorname*{ess\,sup}_{(t,s)\in\Delta}\biggl|\int^t_s dy \, q(y) - \tau_n \sum^{n-1}_k q(s + k\tau_n)\biggr| = o(1).
\end{equation*}
\notag
$$
Now it remains only to apply Theorem 4.3. $\Box$ Here, it is worth to note that for general continuous function $q$ one can say nothing about the convergence rate. Indeed, it can be shown that in (4.11) the convergence to zero can be arbitrary slow (4.12) for a bounded perturbation $Q$. This is drastically different to the case, when dominating generator corresponds to a holomorphic semigroup and perturbation operator is bounded, cf. (3.17), (3.19) for $\alpha = 0$, or to the case of unbounded perturbation, when $0 < \alpha < 1$, see (3.16), (3.18). Theorem 4.6. Let $\delta_n>0$ be a sequence with $\delta_n \to 0$ as $n \to \infty$. Then there exists a continuous function $q\colon \mathcal{I} = [0,1] \to \mathbb{R}$ such that Proof. Taking into account the Walsh–Sewell theorem (see [16], Theorem 6), we find that, for any sequence $\{\delta_n\}_{n\in\mathbb{N}}$, $\delta_n > 0$, satisfying $\lim_{n\to\infty}\delta_n = 0$, there exists a continuous function $f\colon [0,2\pi] \to \mathbb{R}$ such that
$$
\begin{equation*}
\biggl|\int^{2\pi}_0 dx \, f(x) - \frac{2\pi}{n}\sum^n_{k=1}f\biggl(\frac{2k\pi}{n}\biggr)\biggr| = \omega(\delta_n),
\end{equation*}
\notag
$$
as $n \to \infty$. Setting $q(y):= f(2\pi(1-y))$ for $y \in [0,1]$, we get a continuous function $q\colon [0,1] \to \mathbb{R}$ such that
$$
\begin{equation*}
\biggl|\int^{1}_0 dy \, q(y)- \frac{1}{n}\sum^{n-1}_{k=0}q\biggl(\frac{k}{n}\biggr)\biggr| = \omega(\delta_n).
\end{equation*}
\notag
$$
Given that function $q$ is continuous, we find
$$
\begin{equation*}
\operatorname*{ess\,sup}_{(t,s)\in \Delta}\biggl|\int^t_s dy \, q(y) - \tau_n \, \sum^{n-1}_{n=0} q(s +k\tau_n)\biggr| \geqslant \biggl|\int^{1}_0 dy \, q(y) - \frac{1}{n}\sum^{n-1}_{k=0}q\biggl(\frac{k}{n}\biggr)\biggr|,
\end{equation*}
\notag
$$
which yields
$$
\begin{equation*}
\operatorname*{ess\,sup}_{(t,s)\in \Delta}\biggl|\int^t_s dy \, q(y) - \tau_n \sum^{n-1}_{n=0}q(s +k \tau_n)\biggr| = \omega(\delta_n).
\end{equation*}
\notag
$$
Applying now Theorem 4.3, we prove (4.12). $\Box$ Our final comment concerns the case, where the function $q\colon [0,1] \to \mathbb{R}$ is only measurable. In this case it can happen that the Trotter product formula for that pair $\{D_0, Q\}$ does not converge in the operator-norm topology. Proof. Let us introduce the open intervals
$$
\begin{equation*}
\begin{aligned} \, \Delta_{0,n} &:= \biggl(0,\frac{1}{2^{2n+2}}\biggr), \\ \Delta_{k,n} &:= \biggl(t_{k,n} - \frac{1}{2^{2n +2}}, \, t_{k,n} + \frac{1}{2^{2n +2}}\biggr), \qquad k = 1,2,\dots,2^n-1, \\ \Delta_{2^n,n} &:= \biggl(1-\frac{1}{2^{2n+2}}, \, 1\biggr), \end{aligned}
\end{equation*}
\notag
$$
$n \in \mathbb{N}$, where
$$
\begin{equation*}
t_{k,n} = \frac{k}{2^n}, \qquad k = 0,\dots, n,\quad n \in \mathbb{N}.
\end{equation*}
\notag
$$
Notice that $t_{0,n} = 0$ and $t_{2^n,n} = 1$. One easily checks that the intervals $\Delta_{k,n}$, $k = 0,\dots,2^n$, are mutually disjoint. We introduce the open sets
$$
\begin{equation*}
\mathcal{O}_n = \bigcup^{2^n}_{k = 0} \Delta_{k,n} \subseteq \mathcal{I}, \qquad n \in \mathbb{N},
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\mathcal{O} = \bigcup_{n\in\mathbb{N}}\mathcal{O}_n \subseteq \mathcal{I}.
\end{equation*}
\notag
$$
Now it is clear that Therefore, the Lebesgue measure of the closed set $\mathcal{C}:= \mathcal{I}\setminus\mathcal{O} \subseteq \mathcal{I}$ is be estimated by Using the characteristic function $\chi_{\mathcal{C}}(\,{\cdot}\,)$ of the set $\mathcal{C}$, we define The function $q$ is measurable and satisfies $0 \leqslant q(t) \leqslant 1$, $t \in \mathcal{I}$.
Let $\varepsilon \in (0,1)$. We choose $s \in (0,\varepsilon)$, $t \in (1-\varepsilon,1)$, and define
$$
\begin{equation*}
\xi_{k,n}(t,s):= s + k \frac{t-s}{2^n}, \qquad k = 0,\dots,2^n-1, \quad n \in \mathbb{N}, \quad (t,s) \in \Delta.
\end{equation*}
\notag
$$
Note that $\xi_{k,n}(t,s) \in (0,1)$, $k = 0,\dots,2^n-1$, $n \in \mathbb{N}$. Moreover, we have
$$
\begin{equation*}
t_{k,n} - \xi_{k,n}(t,s) = k \frac{1}{2^n} -s - k \frac{t-s}{2^n} = k \frac{1-t+s}{2^n} -s,
\end{equation*}
\notag
$$
which leads to the estimate
$$
\begin{equation*}
|t_{k,n} - \xi_{k,n}(t,s)| \leqslant \varepsilon\biggl(1 + \frac{k}{2^{n-1}}\biggr), \qquad k =0,\dots,2^n-1, \quad n \in \mathbb{N}.
\end{equation*}
\notag
$$
Hence,
$$
\begin{equation*}
|t_{k,n} - \xi_{k,n}(t,s)| \leqslant 3\varepsilon, \qquad k =0,\dots,2^n-1, \quad n \in \mathbb{N}.
\end{equation*}
\notag
$$
Let $\varepsilon_n:= {1}/{(3\cdot 2^{2n+2})}$ for $n \in \mathbb{N}$. Then we get that $\xi_{k,n}(t,s) \in \Delta_{k,n}$ for $k = 0,\dots,2^n-1$, $n \in \mathbb{N}$, $s \in (0,\varepsilon_n)$ and for $t \in (1-\varepsilon_n,1)$.
Now let
$$
\begin{equation*}
S_n(t,s;q):= \tau_n \sum^{n-1}_{k=0}q(s + k\tau_n), \qquad (t,s) \in \Delta, \quad \tau_n = \frac{t-s}{n}, \quad n \in \mathbb{N}.
\end{equation*}
\notag
$$
We consider
$$
\begin{equation*}
S_{2^n}(t,s;q) = \frac{t-s}{2^n} \sum^{2^n-1}_{k=0}q\biggl(s+k\frac{t-s}{2^n}\biggr) = \frac{t-s}{2^n} \sum^{2^n-1}_{k=0}q\bigl(\xi_{k,n}(t,s)\bigr),
\end{equation*}
\notag
$$
$n \in \mathbb{N}$, $(t,s) \in \Delta$. If $s \in (0,\varepsilon_n)$ and $t \in (1-\varepsilon_n,1)$, then $S_{2^n}(t,s;q) = 0$, $n \in \mathbb{N}$ and
$$
\begin{equation*}
\biggl|\int^t_s dy \, q(y) - S_{2^n}(t,s;q)\biggr| = \int^t_s dy \, q(y), \qquad n \in \mathbb{N},
\end{equation*}
\notag
$$
for $s \in (0,\varepsilon_n)$ and $t \in (1-\varepsilon_n,1)$. In particular, this yields
$$
\begin{equation*}
\operatorname*{ess\,sup}_{(t,s)\in \Delta}\biggl|\int^t_s dy \, q(y) - S_{2^n}(t,s;q)\biggr| \geqslant \operatorname*{ess\,sup}_{(t,s)\in\Delta}\int^t_s dy \, q(y) \geqslant \int_{\mathcal{I}} dy \, \chi_\mathcal{C}(y) \geqslant \frac{1}{2}.
\end{equation*}
\notag
$$
Hence, we obtain
$$
\begin{equation*}
\limsup_{n\to\infty}\,\operatorname*{ess\,sup}_{(t,s)\in \Delta}\biggl|\int^t_s dy \, q(y) - S_{2^n}(t,s;q)\biggr| \geqslant \frac{1}{2},
\end{equation*}
\notag
$$
and applying Theorem 4.3, we finish the prove of (4.13). $\Box$ Remark 4.8. We note that Theorem 4.7 does not exclude the convergence of the Trotter product formula for the pair $\{D_0, Q\}$ in the strong operator topology. We would remind that the same kind of dichotomy is known for the Trotter product formula on a Hilbert space, see the Hiroshi Tamura example in [17], Theorem B. By virtue of (4.7) and (4.13), Theorem 4.7 yields an example of this dichotomy on a Banach space. Again, there is a drastic difference between the origin of these conclusions in a Hilbert space (see [17], Theorem B) for unbounded perturbation of the holomorphic semigroup and in a Banach space (Theorem 4.7) for bounded perturbation of a (non-holomorphic) contractive semigroup. § 5. NotesNotes to § 2. Characterisation of holomorphic contraction semigroups at the end of § 2.2 (iv), is due to [2], Corollary II.4.9. For the proof of Lemma 2.11, see, for example Lemma 2.3.5 in [18]. Notes to § 3. Here, we extend to the operator-norm convergence of the product formula on a Banach space for perturbation $B$ with a relative zero $A$-bound for holomorphic semigroup $\{\mathrm{e}^{-tA}\}_{t\geqslant 0}$ some of the Trotter–Chernoff results, cf. [4], [19], [9]. This shows that hypothesis of self-adjointness in the case of a Hilbert space [10] has only a technical importance. On the other hand, the operator-norm topology is “natural” for holomorphic $C_0$-semigroups, which may lead one to think that it is an essential hypothesis for the operator-norm convergence of the Trotter product formula. In § 4, we showed that this hypothesis is also technical, but we have to assume contraction of semigroup $\{\mathrm{e}^{-tA}\}_{t\geqslant 0}$. We would like to remark that demand of contraction is not as superfluous as one could suppose. For demonstration, we address the reader to instructive example by Trotter [4], where it is called “the norm condition”. This section contains a revision of the result in § 3 of [5], where the operator-norm convergence of the Trotter product formula on a Banach space $\mathfrak{X}$ has been proved (up to our knowledge) for the first time. For a survey of similar results in this direction see [20]. Notes to § 4. In contrast to § 3, where operator-norm convergence holds true if the dominating operator $A \in \mathscr{H}_{\mathrm{c}}(\theta, 0)$ generates a holomorphic contraction semigroup and operator $B$ is an $A$-infinitesimally small generator of a contraction semigroup (in particular, if $B$ is a bounded operator), we present Example that this is also possible if condition on the generator $A$ is relaxed. The conditions (cf. [21]) are the following: (1) the operator $A = \mathcal{K}_0$ generates a contractive (not holomorphic!) semigroup; (2) $B = \mathcal{B}$ is a bounded operator. There it is also demonstrated that the operator-norm convergence generally fails (even for bounded operators $\mathcal{B}$) if unbounded $\mathcal{K}_0$ is not a holomorphic generator (Theorem 4.7), or that operator-norm convergence of the Trotter product formula can be arbitrary slow, cf. Theorem 4.6. This is again very different to the holomorphic case: $A \in \mathscr{H}_{\mathrm{c}}(\theta, 0)$ (cf. § 3.2), where the rate of the operator-norm convergence is of the order $O({(\ln n)^2}/{n})$ for any bounded perturbation $B$ ($\alpha =0$), see Theorems 3.6, 3.7, and Corollary 3.8. 
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