Given a Banach space $\mathcal N$, as contraction semigroup is defined as a set of bounded operators $P^t$, $0\le t\le+\infty$ defined everywhere in $\mathcal N$, such that \begin{equation*} P^0=1, \hspace{5mm} P^tP^s=P^{t+s}, \hspace{5mm} t\ge 0, \;\; s\in\mathbb R \end{equation*} We can associate to these operators a norm defined by \begin{equation*} ||P^t||=\inf_{\beta\in\mathbb R^+}\left\{\beta:||P^t\phi||\le \beta||\phi||\; \forall \phi\in \mathcal V, ||\phi||\le 1\right\} \end{equation*} and we can define generators of such contractions as \begin{align}\label{gen_Pt} A\psi&=\lim_{t\to 0}\frac{P^t\psi-\psi}{t} \end{align} Now, I am trying to see whether given these definitions it also holds $P^t=e^{At}$. I think this should be valid, as it seems to me that a contraction groups equipped with such norm is also a strongly continuous semigroup. For this last class of semigroups for the identity with the exponential of the generator is known to hold. Strong continuity should in particular be valid as \begin{align} \lim_{t\to 0+}||{P^t-1}||&=\lim_{t\to 0+}\inf_{\beta\in\mathbb R^+}\{\beta:||{(P^t-1)\phi}||\le \beta||{\phi}||\; \forall \phi\in\mathcal N,\||{\phi}||\le 1\}=\\ &= \inf_{\beta\in\mathbb R^+}\{\beta:\lim_{t\to 0+}||{(P^t-1)\phi||}\le \beta||{\phi}||\; \forall \phi\in\mathcal N,||{\phi}||\le 1\}= \label{inflim}\\ &= \inf_{\beta\in\mathbb R^+}\{\beta: 0\le \beta||{\phi}||\; \forall \phi\in\mathcal N,||{\phi}||\le 1\}=0 \end{align} I assume the exchage between the $\lim$ and $\inf$ should hold as their argument is continuous in both parameters.

Are these arguments correct?



  • A semigroup on a given Banach space $\mathcal N$ is a family $(P^t)_{0\leq t<\infty}$ of bounded linear operators defined everywhere in $\mathcal N$ such that $$P^0=1,\qquad P^tP^s=P^{t+s}, \qquad t,s\ge 0$$ where $1$ is the identity operator on $\mathcal{N}$.

  • A contraction semigroup is a semigroup (in the sense defined above) which satisfies the extra condition $$\|P^t\|_{\mathcal{L}}\leq 1,\quad\forall\ t\geq 0$$ where $\|P^t\|_{\mathcal{L}}$ is the norm defined in your post (which is the usual operator norm).

  • The generators of a semigroup is defined by $$ A\psi=\lim_{t\to 0^+}\frac{P^t\psi-\psi}{t} $$ with domain $D(A)=\{\psi\mid \text{the above limit exists (in the sense of $\mathcal{N}$)}\}$.

  • A semigroup is strongly continuous if it satisfies $$\lim_{t\to 0^+} \|P^t\phi-\phi\|_{\mathcal{N}}=0,\quad\forall \ \phi\in\mathcal{N}.$$

  • A semigroup is uniformly continuous if it satisfies $$\lim_{t\to 0^+} \|P^t-1\|_{\mathcal{L}}=0.$$

Now, I am trying to see whether given these definitions it also holds $P^t=e^{At}$.

  • If the right-hand side means the exponential of a bounded operator, it holds if and only if the semigroup is uniformly continuous (see Theorem 1.2, Theorem 1.3 and Corollary 1.4 in Section 1.1 of Pazy's book).

  • Often the right-hand side is used only as a notation.

  • If you are using definitions which are different from the definitions in my post, please clarify.

  • $\begingroup$ Thanks for the clarification. Looking to Theorem 1.2 in Pazy and Corollary 1.4 b), I deduce that the boundness of the generator $A$ of a semigroup $P^t$ suffices to prove strong continuity and that $P^t=e^{-At}$ is well defined as expansion of the $A$, correct?I am in particular interested in using these notions in qm; the generators are the pot. and kin. energy, the semigroups are the kinetic and potential part of the Gibbs density $e^{-\beta\hat V( \hat q)}$ and $e^{-\beta\hat T(\hat p)}$. Shall I check if $\hat V$ and $\hat T$ are bounded in order to expand the exponentials? $\endgroup$ – Graz Feb 15 at 8:00
  • 2
    $\begingroup$ @Graz Suppose we have a semigroup $P^t$ with generator $A$. If we want an exponential representation $\displaystyle P_t=e^{tA}=\sum_{n=0}^\infty \frac{(t A)^n}{n!}$, we have to (i) show that $P^t$ is uniformly continuous or (ii) show that $A$ is bounded. Corollary 1.4(b) proves that (i) is enough. Theorem 1.3 proves that (ii) is enough. $\endgroup$ – Pedro Feb 15 at 14:45

In general it is not true that a contraction semigroup is uniformly continuous (meaning continuous in the operator-norm topology). For example:

Consider the shift semigroup $S_t:L^2(\mathbb{R})\rightarrow L^2(\mathbb{R})$ which acts on square integrable functions as: $S_t(f)(x)=f(x+t)$, where $x\in\mathbb{R}\;,t\geq 0$ . The members of this semigroup are isometries on $L^2(\mathbb{R})$ due to translation invariance of Lebesgue measure. Hence for all $t\geq 0$: $\left\Vert S_t\right\Vert=1$, which makes it a contraction $C_0$-semigroup. However, $S_t$ does not converge to the identity operator in the operator-norm topology. A hint for the proof can be found here: The lack of uniform continuity of shift operator on $L^2$

The right condition which guarantees that a semigroup is given by the exponential of its generator is uniform continuity, which implies boundedness of the generator. The latter guarantees that the operator $e^{tA}$ is well-defined by a convergent series.

Note: $C_0$ here means continuous in the strong operator topology, i.e. for all $x\in\mathcal{N}$: $\left\Vert (S_t-Id)x \right\Vert_{\mathcal{N}}\rightarrow 0$ as $t\rightarrow 0^+$. This is what is usually called strongly continuous semigroup.


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